PDA

View Full Version : Cones - Center of Gravity, Center of Mass, etc.



Pages : [1] 2

xylophobe
2012-Mar-14, 11:32 PM
In wikipedia and elsewhere it is stated that: The center of mass of a conic solid of uniform density lies one-quarter of the way from the center of the base to the vertex, on the straight line joining the two. (http://en.wikipedia.org/wiki/Cone_(geometry)#Center_of_mass)

But when I set up the problem as shown in this image, with "x" being the location of the center of mass then I get:

x = h*(1 - 1/(2)1/3) = 0.2063*h

I essentially set the volume (it is uniform density) of the small cone, V2 = V3, the mass of the frustum, then solve for "x"

Known: V3 = V1 - V2

I used centroids and areas to create a solid/volume of revolution: http://en.wikipedia.org/wiki/Pappus's_centroid_theorem

What am I missing or is wikipedia wrong?

16529

pzkpfw
2012-Mar-15, 12:03 AM
I'm way out of practice with this stuff, but for fun tried a different approach. I figured, apply a scale factor (f) to the height and radius of a cone, to make the volume half what it was.

V1 = (0.333...).Pi.r2.h

V2 = (0.333...).Pi.(f.r)2.(f.h) = 0.5 V1

f3.(0.333...).Pi.r2.h = 0.5 (0.333...).Pi.r2.h

f3 = 0.5

f = 0.793701

Which would place x, 0.206 from the base of the cone. That's "close" to the "one quarter" on that Wiki page, but not very close. (But does match your number).


(Googling, it seems the answer involves integration...)

antoniseb
2012-Mar-15, 12:12 AM
It's been a while since I solved for center of gravity in for something other than a bar or a sphere, but I think that the center of gravity is not just a question of finding the dividing line about which there are two equal masses, but that the length of the lever arm is also a factor. Hence the volume v0 at the tip has a greater influence than the same volume in the thin wafer at the bottom which is closer to the CG.

Jeff Root
2012-Mar-15, 12:36 AM
Isn't the center of gravity of an isoceles triangle at the same point
as the center of mass of a cone with the same cross-section?
If so, you can cut a triangle out of cardboard and balance it on
a rounded point to find the center by trial and error. That should
be more than accurate enough to distinguish between 0.206 h
and 0.25 h.

-- Jeff, in Minneapolis

antoniseb
2012-Mar-15, 01:03 AM
Isn't the center of gravity of an isoceles triangle at the same point
as the center of mass of a cone with the same cross-section?
No the Third dimension makes this comparison wrong.

swampyankee
2012-Mar-15, 01:10 AM
Here's a site with it all worked out: http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm

Jeff Root
2012-Mar-15, 01:34 AM
At least my first guess for the figure that I thought would apply to
the cone appears to apply to the triangle: I get 1/3 h with a paper
cutout. That would apply to prisms, too. Funny it doesn't work
for cones.

-- Jeff, in Min Minneapolis

Jeff Root
2012-Mar-15, 01:49 AM
Aha! The triangle has 1/2 the area of the enclosing rectangle,
but the cone has only 1/3 the volume of the enclosing cylinder!

-- Jeff, in Minneapolis

xylophobe
2012-Mar-15, 02:36 AM
It's been a while since I solved for center of gravity in for something other than a bar or a sphere, but I think that the center of gravity is not just a question of finding the dividing line about which there are two equal masses, but that the length of the lever arm is also a factor. Hence the volume v0 at the tip has a greater influence than the same volume in the thin wafer at the bottom which is closer to the CG.

For a cone I figured that the center of mass was different than the center of gravity due to the moment arm being greater at the point than at the base so I did my calculation to exactly divide the mass into two parts - an equal number of atoms on each side of that point on the x-axis (I already surmised that the CoM was on the x-axis).

When I use the centroid (http://en.wikipedia.org/wiki/List_of_centroids) of a triangle and revolve it then it seems that the center of gravity is at h/3

I can derive the formula for the volume of the cone using the second theorem of Pappus' centroid theorem and I can determine that the cone has only 1/3 the volume of the enclosing cylinder ... I don't see why this method does not give the center of gravity for the cone.

grapes
2012-Mar-15, 06:30 PM
For a cone I figured that the center of mass was different than the center of gravityIn a constant gravity field (approximately true for small objects near the surface of the earth), the center of gravity and the center of mass are the same, by definition. The center of mass is the point where, to first degree, you can treat the entire mass as a point mass.

Your approach is problematic anyway. Consider three points, A, B, and C, having the same mass. The center of mass, as you've taken it, would be somewhere on the line including the one in the middle, B. However, if the points form a triangle, you can move your point-of-view so that the two points A and C align so your center of mass must be along the line between them--and you can do this for any of the pairs of points.

ShinAce
2012-Mar-15, 07:06 PM
I have a test tomorrow in calculus regarding iterated integration. Once i've written the test, ill come back and show that method for solving this problem. I suspect you have the right answer with 0.206 from the base, height=1. Note that a multiple integral has density as a function of xyz, so we can solve for non uniform density cones as well.

frankuitaalst
2012-Mar-15, 07:06 PM
Here's a site with it all worked out: http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm
Correct .
The formula in this link simply says : X cgravity = integral ( x * dm ) / Total mass , where dm is an infetisimal mass increase over the section ( area) with a thickness dx .
dm here can be written as rho * Area * dx , where Area on its turn again is a function of x .
This formula applies for every centroid .
As different centroids have other functions of "area" as function of x the height of the center of mass is different for each centroid . In case of a cone we get indeed 0.75 of the height , if we measure from the point .

xylophobe
2012-Mar-16, 04:12 PM
Okay, I set up the problem as shown in the image, where x=% of h (% in decimal format) and I get the following polynomial:

2x3-7x2+9x-2=0

that results in x=0.2773

I also did an engineering analysis (using real values instead of variables), that gives the same value for x=0.2773 , by finding the composite centroid of the frustum area and the centroid of the Small Cone area. Further, as a check, I used these centroids to create volumes that sum to the same Frustum volume as achieved using the regular cone volume formula:

Frustum Volume = Large Cone Volume - Small Cone Volume

Cone Volume Formula = pi()r2h/3

To get the above polynomial I multiplied the volume of the Small Cone times the moment arm of the Small Cone the result of which should equal the volume of the Frustum times the moment arm of the Frustum. I then solved for "x".

M2=M3

where:
M2=V2*C2
M3=V3*C3
V2=pi()r2h(1-x)3
V3=pi()r2h(1-(1-x)3)
C2=h(1-x)/3 --- (location of the centroid of the Small Cone)
C3=xh*(3-x)/3(2-x) --- (location of the centroid of the Frustum)



16531

frankuitaalst
2012-Mar-16, 06:55 PM
I thnk the flaw in the approach above lies in the fact of assuming that the center of gravity of a centroid lies at 1/3 of the hieght , just as it does in a flat figure :
The assumption : CG = h(1-x)/3 for instance applies for a flat triangle , not essentially for a centroid .

grapes
2012-Mar-16, 07:23 PM
You're trying to find the centroid of the large cone, and assume it is at xh. That's fine, but then the centroid of the small cone should be at a distance from there of h(1-x)x, whereas you label it h(1-x)/3.

How you're going to determine the formula to use for the frustrum is beyond me. :)

However, the link provided by swampyankee has the steps all in one line, and it looks like it might be correct.

xylophobe
2012-Mar-16, 07:31 PM
I thnk the flaw in the approach above lies in the fact of assuming that the center of gravity of a centroid lies at 1/3 of the hieght , just as it does in a flat figure :
The assumption : CG = h(1-x)/3 for instance applies for a flat triangle , not essentially for a centroid .

I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998430#post1998430).

As far as your statement it seems to be at odds with this statement: If a physical object has uniform density, then its center of mass is the same as the centroid of its shape. (http://en.wikipedia.org/wiki/Centroid)

So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website (http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm) why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?

grapes
2012-Mar-16, 07:43 PM
I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998430#post1998430). Kudos for that, but I'm wondering about your certainty of your wrong answer in the face of the "right answer" :)


As far as your statement it seems to be at odds with this statement: If a physical object has uniform density, then its center of mass is the same as the centroid of its shape. (http://en.wikipedia.org/wiki/Centroid)He meant, that the centroid of a cone is different than the centroid of a triangle.


So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website (http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm) why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?That's the moment arm.

frankuitaalst
2012-Mar-16, 07:44 PM
I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998430#post1998430).

As far as your statement it seems to be at odds with this statement: If a physical object has uniform density, then its center of mass is the same as the centroid of its shape. (http://en.wikipedia.org/wiki/Centroid)

So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website (http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm) why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?
The "x" is there to account for the moment arm for the same reason as you do in your formulae .
Further I agree with Grapes when he states that when you assume the CG to be at x for the whole then this also applies for the upper part of the cone , being h(1-x) x instead of h(1-x)/3

xylophobe
2012-Mar-16, 08:17 PM
..., but then the centroid of the small cone should be at a distance from there of h(1-x)x, whereas you label it h(1-x)/3. ...

I know what you are getting at because it seems to be a circular logic - I guess I could substitue h(1-x)/4 to see if it all works out to be x=4.0

xylophobe
2012-Mar-16, 11:38 PM
Ok, my math skills are limited so I shall resort to a graphical reproof of the cone CoG=1/4h

In the picture below I show a progression from a cylinder, to half a cylinder, to the cone in question.

We can all agree that the CoG for the cylinder is in the middle.

If I cut the cylinder in half then the CoG for the half moves to 1/4h.

So if I cut off the outer portion shaded red and add the inner portion shaded green then, by area, I have added the same amount but since these areas are rotated around the centerline then I am subtracting much more volume with the red than is added by the green so therefore the CoG for the cone can not remain at the 1/4h position as shown when the cylinder is cut in half.

The CoG for a cone is at the 1/3h position.


16536

grapes
2012-Mar-17, 01:08 AM
The CoG for a cone is at the 1/3h position.But, it's actually at the 1/4h position??


In wikipedia and elsewhere it is stated that: The center of mass of a conic solid of uniform density lies one-quarter of the way from the center of the base to the vertex, on the straight line joining the two. (http://en.wikipedia.org/wiki/Cone_(geometry)#Center_of_mass)

Hornblower
2012-Mar-17, 01:37 AM
I just performed the integration, using the formula given here,
http://en.wikipedia.org/wiki/Center_of_mass#Definition
and got exactly 1/4 of the height.

I am completely confident that Wiki has it right, in accordance with the definition of center of mass. Can anyone who got a different result explain, in appropriate mathematical detail, why they think otherwise?

xylophobe
2012-Mar-17, 02:42 AM
According to this picture and using Pappus's 2nd centroid theorem (http://en.wikipedia.org/wiki/Pappus's_centroid_theorem) the volume of the red area is 5x the volume of the green area.

16537

Jeff Root
2012-Mar-17, 03:33 AM
xylophobe,

Your graphical analysis starts out by saying that you are looking
at a cylinder cut in half, and ends by saying that you locate the
centroid of a cone, but the intervening steps appear to be looking
only at plane figures, not solids. As was pointed out to me early
in the thread, the centroid of a cone is different from the centroid
of the triangle which is the cross-section of the cone. That became
apparent to me when I saw that the triangle has 1/2 the area of the
enclosing rectangle, but the cone has only 1/3 the volume of the
enclosing cylinder. The difference shifts the location of the cone's
centroid proportionally toward the base.

I just don't see that your diagrams show what you describe them
as showing.

Correction:

For some reason I said "and ends by saying that you locate the
centroid of a cone" But actually xylophobe's post #20 ends by
saying that he locates the "CoG" (center of gravity) of the cone.
Not that there's any difference between the centroid and the CoG --
there isn't -- but I said what I was thinking instead of what I read.

-- Jeff, in Minneapolis

xylophobe
2012-Mar-17, 03:57 AM
xylophobe,

..., and ends by saying that you locate the
centroid of a cone, ...

-- Jeff, in Minneapolis

Hi Jeff,

I did not say that I located the center of gravity of the cone ... but I do know where the centroid of the area is which is needed in order to use Pappus's 2nd theorem to find the volumes.

That is why I show the r/6 and the 5r/6 dimensions (http://www.bautforum.com/attachment.php?attachmentid=16537&d=1331952103) to the centroids because these y-axis locations are the "r" in the Pappus calculation.

Because 5/6 is 5 times larger than 1/6 then I know that the mass of the red area is 5 times more than the mass of the green area (after revolving around the x-axis).

Hornblower
2012-Mar-17, 01:39 PM
I stand behind my conclusion in post 22, and nothing you have said since then shakes my confidence.


Ok, my math skills are limited so I shall resort to a graphical reproof of the cone CoG=1/4h

In the picture below I show a progression from a cylinder, to half a cylinder, to the cone in question.

We can all agree that the CoG for the cylinder is in the middle.

If I cut the cylinder in half then the CoG for the half moves to 1/4h.

So if I cut off the outer portion shaded red and add the inner portion shaded green then, by area, I have added the same amount but since these areas are rotated around the centerline then I am subtracting much more volume with the red than is added by the green so therefore the CoG for the cone can not remain at the 1/4h position as shown when the cylinder is cut in half.

The CoG for a cone is at the 1/3h position.


16536

My bold. From that remark, along with everything else you have posted, I infer that you do not know how to do the calculus that yields an exact analytic solution. You appear to be attempting rough-and-dirty numerical integrations on much too coarse a scale. For problems that do not have analytic solutions, and there are such things, numerical integration is a good plan B, but the increments have to be very small in proportion to the whole body to get reasonably good approximations.

There is no shame in having trouble with calculus. I had some rough moments before progressing from a C after one quarter in my senior year in high school to an A in college.

frankuitaalst
2012-Mar-17, 03:35 PM
16540

One way to calculate the CM "graphically" or "numerically" is to divide the cone in several slices .
A division in two sections is shown in annex . We get a ratio of 0.7H .
As we divide the cone section in more slices the ratio increases ....
If we would do the same for 100 ,1000,... sections we get close to the 0.75 H .
In fact this is what numerical integration or calculus does...

profloater
2012-Mar-17, 05:01 PM
what a funny thread! the integral of x^3 divided by the integral of x^2 as given earlier in the posts is clearly the way to calculate the moment of each disc of the cone divided by the volume of each disc; so the 3/4 result is obviously right.

publiusr
2012-Mar-17, 08:01 PM
O/T but there was an interesting shape called a superegg that looked interesting...

Jeff Root
2012-Mar-17, 11:29 PM
Is a supergg based on the superellipse? The superellipse
was given its name while I was in junior high school.

I'm surprised and rather shocked that finding the center of
a cone apparently requres calculus. Finding the center of
a cylinder is as easy as falling off a cylindrical log. Why
is finding the center of a cone so much more involved?

-- Jeff, in Minneapolis

frankuitaalst
2012-Mar-18, 12:54 PM
what a funny thread! the integral of x^3 divided by the integral of x^2 as given earlier in the posts is clearly the way to calculate the moment of each disc of the cone divided by the volume of each disc; so the 3/4 result is obviously right.
right . If in another thread the question comes about the CG of a paraboloid it's obvious it is 5/6 (INT x5/ INT x4) .

swampyankee
2012-Mar-18, 01:04 PM
Is a supergg based on the superellipse? The superellipse
was given its name while I was in junior high school.

I'm surprised and rather shocked that finding the center of
a cone apparently requres calculus. Finding the center of
a cylinder is as easy as falling off a cylindrical log. Why
is finding the center of a cone so much more involved?

-- Jeff, in Minneapolis

Perhaps because it doesn't have a constant cross-section?

You could always find the center of mass empirically.

Perikles
2012-Mar-18, 01:28 PM
Is a supergg based on the superellipse? Yes. See here (http://en.wikipedia.org/wiki/Superegg). I have one on my desk, by the way. Small, but perfectly formed.

grapes
2012-Mar-18, 04:42 PM
I'm surprised and rather shocked that finding the center of
a cone apparently requres calculus. Finding the center of
a cylinder is as easy as falling off a cylindrical log. Why
is finding the center of a cone so much more involved?
A log has central symmetry, a cone doesn't. It also has curves, but we can get rid of them using Cavalieri's principle.

http://en.m.wikipedia.org/wiki/Cavalieri's_principle

So we can take a small block of length h and chop off pieces until it is a pyramid: two half-blocks sliced off on a diagonal, but then we have to add back in another pyramid whose base is the other side. The mass of the block 3P is three times that of the pyramid P, so to find the moment arm x of the pyramid, the torque of the pyramid is

P xh = 3P h/2 - 2 3P/2 2h/3 + P (1-x)h

2P xh = P h/2

x = 1/4

No calculus necessary! Unless you count Cavalieri's principle (which, btw, I do :) )

xylophobe
2012-Mar-21, 01:32 AM
I stand behind my conclusion in post 22, and nothing you have said since then shakes my confidence.


My bold. From that remark, along with everything else you have posted, I infer that you do not know how to do the calculus that yields an exact analytic solution. You appear to be attempting rough-and-dirty numerical integrations on much too coarse a scale. For problems that do not have analytic solutions, and there are such things, numerical integration is a good plan B, but the increments have to be very small in proportion to the whole body to get reasonably good approximations.

There is no shame in having trouble with calculus. I had some rough moments before progressing from a C after one quarter in my senior year in high school to an A in college.

I learned calculus (integration) and understand how to do it but I have not practiced the art in a long time so I am re-learning it. I usually shy away from calculus and just do a quick excel file dividing things into many small sections. After cutting the cone into numerous horizontal slices I come to the same answer as wikipedia and the site (http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm) posted by swampyankee. I am beginning re-realize the ease of using calculus to find solutions (I always relied on 3-D computer models to do math).

So, for the cone it is true that the

center of mass = 0.2063*h

;whereas, the

center of gravity = 0.25*h

Hornblower
2012-Mar-21, 02:03 AM
I learned calculus (integration) and understand how to do it but I have not practiced the art in a long time so I am re-learning it. I usually shy away from calculus and just do a quick excel file dividing things into many small sections. After cutting the cone into numerous horizontal slices I come to the same answer as wikipedia and the site (http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm) posted by swampyankee. I am beginning re-realize the ease of using calculus to find solutions (I always relied on 3-D computer models to do math).

So, for the cone it is true that the

center of mass = 0.2063*h

;whereas, the

center of gravity = 0.25*h

Please review for us your explicit definitions of center of mass and center of gravity, in appropriate mathematical detail. You must be using an unorthodox definition for the former. I stand by my finding, having reviewed the calculus myself on the basis of what I understand to be the standard definition.

grapes
2012-Mar-21, 03:16 AM
So, for the cone it is true that the

center of mass = 0.2063*h

;whereas, the

center of gravity = 0.25*hThe usual definition of center of mass is *not* where half the mass is on one side, and half is on the other. Besides, that point .2063h along the axis is not always at the halfway point--just tilt the cone a little and you can see that the halfway plane moves off that point.

a1call
2012-Mar-21, 04:05 PM
As a mental simulation (or if you have the time actual measurement) Consider:
*- When you suspend an object from any point, the center of mass lies along a vertical line
Passing through the point of suspension.
*- Visualize suspending a cone (or any object) from two points and getting the crossing point of the two vertical lines. This will be the center of mass.

BTW, for a cone this is 1/4 of the way from the base along the center line (as already pointed out).

profloater
2012-Mar-21, 04:13 PM
As a mental simulation (or if you have the time actual measurement) Consider:
*- When you suspend an object from any point, the center of mass lies along a vertical line
Passing through the point of suspension.
*- Visualize suspending a cone (or any object) from two points and getting the crossing point of the two vertical lines. This will be the center of mass.

BTW, for a cone this is 1/4 of the way from the base along the center line (as already pointed out).
a balanced argument

grapes
2012-Mar-21, 04:19 PM
Technically, that describes the center of gravity, but in a uniform gravity field that will also be the center of mass.

profloater
2012-Mar-21, 04:22 PM
Technically, that describes the center of gravity, but in a uniform gravity field that will also be the center of mass. I think that will also be the point, regardless of gravity, at which a force will cause acceleration without rotation, assuming infinite stiffness.

grapes
2012-Mar-21, 05:26 PM
I think that will also be the point, regardless of gravity, at which a force will cause acceleration without rotation, assuming infinite stiffness.That would be the center of mass, I think, not the center of gravity. :)

The OP has visualized the center of mass as having half on one side, half on the other, which is wrong.

profloater
2012-Mar-21, 05:39 PM
Now I am confused, the C of g is the point through which the force of gravity would act, accelerating without rotation, if it (the object) were replaced by a point mass which surely is the same as the c of M? Clearly the OP idea of mass is wrong, the moment of each element of mass is key to the definition. Aha I see you refer to a constant gravity field which I just assumed, no problem

a1call
2012-Mar-21, 06:46 PM
*- Both centers of gravity and mass lie along a vertical line through the suspension point at the movement of equilibrium

*- Suspension from a different point on the object may alter center-of-gravity but not center-of-mass

*- Crossing point of two such vertical lines through suspension points will be center-of-mass and not (necessarily) center-of-gravity, which is not a constant/single point if the object is moved.

grapes
2012-Mar-21, 09:35 PM
*- Both centers of gravity and mass lie along a vertical line through the suspension point at the movement of equilibrium

*- Suspension from a different point on the object may alter center-of-gravity but not center-of-mass

*- Crossing point of two such vertical lines through suspension points will be center-of-mass and not (necessarily) center-of-gravity, which is not a constant/single point if the object is moved.Those statements are self-contradictory. One of them has to be wrong.

If the center of mass is not coincident with the center of gravity, it doesn't have to be vertically below the point of suspension. The notion of "vertical" may even be ambiguous.

a1call
2012-Mar-21, 10:41 PM
Two incoincident points don't "have to" lie on a vertical line, but they may do so. In case of centers of gravity and mass they both lie on a line that passes through them and the center of the gravity field (Earth) i.e. vertical.
I fail to see any contradiction in the statements. Please be more specific.

John Mendenhall
2012-Mar-21, 11:49 PM
Gentle folk, you are spending much time discussing ideas which can be quickly referenced.

From Wiki, 'Center of Mass':

"In physics, the center of mass or barycenter of a body is a point in space where, for the purpose of various calculations, the entire mass of a body is concentrated. A center of gravity is a related point where the gravitational weight of a body acts as if it were concentrated. In a uniform gravitational field, the center of mass is a center of gravity, and in common usage, the two phrases are used as synonyms. In a non-uniform field, the center of mass no longer serves as an exact center of gravity, so one can distinguish the center of gravity as a separate concept." (my italics)

Bang. Done.

Regards, John M.

Jeff Root
2012-Mar-22, 02:59 AM
Are there actual, real situations where the center of gravity of
something is *measurably* different from its center of mass?

-- Jeff, in Minneapolis

caveman1917
2012-Mar-22, 04:04 AM
Are there actual, real situations where the center of gravity of
something is *measurably* different from its center of mass?

-- Jeff, in Minneapolis

Sure, any tidal effects will do.

Jeff Root
2012-Mar-22, 08:02 AM
I assumed that more than mere tidal effects were required.

Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?

-- Jeff, in Minneapolis

grapes
2012-Mar-22, 09:59 AM
Two incoincident points don't "have to" lie on a vertical line, but they may do so. In case of centers of gravity and mass they both lie on a line that passes through them and the center of the gravity field (Earth) i.e. vertical.
A straight line aligned with local vertical on earth might miss the center of mass of the earth by 20 km. Half of that is due to rotation though.

I suppose that answers Jeff's question too. :)

Jeff Root
2012-Mar-22, 11:26 AM
So, can somebody help me with the missing number near
the bottom of the first (main) section of this web page?

http://www.freemars.org/jeff/Earth/down.htm

It says "0.00 degree." I need to fill in the correct number.
Calculating it requires elliptical trigonometry, I think.
Should be easy for several people here. But I'd like to
see it worked out so I know what I'm talking about.

I made a diagram of the geometry, below. On a monitor
with close pixel spacing, it may be best viewed at 200%.
That's zoomed in 6 times in Firefox! Yuk. Intermediate
sizes make it messy.

I'm assuming a perfectly ellipsoidal Earth.

Earth's equatorial radius is 6,378 km.
Earth's polar radius is 6,357 km.

I seem to have forgotten since I made the diagram, but I think
S is the distance along the surface of a perfectly spherical
Earth with radius R, from the north pole to the 45-degree line.

theta is 45 degrees.

g is the gravitational force at the 45-degree line.

f is the centrifugal force at the 45-degree line..

phi is the angle I want to know.

-- Jeff, in Minneapolis

.

ShinAce
2012-Mar-22, 01:40 PM
I assumed that more than mere tidal effects were required.

Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?

-- Jeff, in Minneapolis

When i saw tidal effects, i went "ohhh of course".

The measurable difference seems obvious. The tides cause equal bulges on both sides, so the center of mas is the same. But far enough away from the earth, say from the moon, the center of gravity is closer and offset to the side. Hence the torque which causes the moon to migrate.

grapes
2012-Mar-22, 04:34 PM
Jeff, the gravitational effect of the equatorial bulge is almost as much as the centrifugal force at a latitude of 45, so the line you've labeled as g does not point to the center of the earth.

By definition, latitude is determined by the horizon angle with the axis of the earth (ETA: in theory, but international and national interests come into play), so it is the tangent of the ellipsoid.

The ellipsoid is \frac{x^2}{E^2} + \frac{y^2}{P^2} = 1 and its derivative is \frac{dy}{dx} = - \frac{P^2 x}{E^2 y} which is equal to one (or negative one) at latitudes of 45, and so \frac{y}{x} = \frac{P^2}{E^2} --but that is the slope of the angle to the center of the ellipse. Its arctangent will give you an angle that deviates slightly from 45, and the difference is the angle you seek.

ETA: I plugged "(arctan ( 6357^2/6378^2)) radians in degrees" into wolframalpha, and the result says your number is about 11 minutes. I multiplied it by 6368 and got 21.00 kilometers, which is about how far it misses the center of the earth. Of course, it intersects the plane of the equator much farther than that from the earth center.

EETA: I was almost sure we'd worked on this before.
http://www.bautforum.com/showthread.php/39047-The-shape-the-Earth-is-in?p=702186&highlight=semimajor#post702186

caveman1917
2012-Mar-22, 10:30 PM
I assumed that more than mere tidal effects were required.

No, in fact every instance of the center of gravity not being the same as the center of mass is going to be an instance of a tidal effect. Think about it, the center of gravity equals the center of mass in a uniform gravitational field. What we call a tidal effect is nothing more than saying that you can't consider the gravitational field as uniform over some extended body, for example the tides on earth because the gravitational pull from the moon is not uniform across the earth. So finding examples of measurable differences is the same as finding examples of measurable tidal effects.

xylophobe
2012-Mar-25, 02:32 PM
The usual definition of center of mass is *not* where half the mass is on one side, and half is on the other. Besides, that point .2063h along the axis is not always at the halfway point--just tilt the cone a little and you can see that the halfway plane moves off that point.

Maybe there is another term for what I describe as the center of mass - what I am describing is the point through which the 3 orthogonal planes pass whereby those planes evenly divide the subject mass into two equal masses. So if I melted down either half then I could pour it into a mold of the second and exactly fill it creating two exact replicas.

The center of gravity being the balance point of a uniform object in a uniform gravitational field.

profloater
2012-Mar-25, 03:21 PM
This point you are after is very important when sharing an ice cream cone and want exactly half, ditto all irregular shapes that might be divided equally, sorry I don't know the term for the half mass point.

Perikles
2012-Mar-25, 04:00 PM
This point you are after is very important when sharing an ice cream cone and want exactly half,.The problem there is that you have to make the decision before knowing how far down the cone the ice cream goes. Or, you bite the small end off and work upwards, but it can get very messy.

profloater
2012-Mar-25, 04:10 PM
That's true, cones have so many interesting properties and being part filled is difficult to include in the calculus but I propose a portable balance with attached nomogram, used to find the balance point of the actual ice cream cone and then giving the half mass point. The nomogram would have lines for various standard cone biscuit weights. Named the "Econometer" However I am stumped by the 99 (with chocolate flake) How could you ever find half of that? It's an irregular apparently random cyclinder.

swampyankee
2012-Mar-25, 04:24 PM
I assumed that more than mere tidal effects were required.

Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?

-- Jeff, in Minneapolis

Gravitationally stabilized satellites.

Jeff Root
2012-Mar-25, 05:26 PM
As is too often the case, I'm skipping the obviously involved
posts and going for the seemingly easy one...




Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?
Gravitationally stabilized satellites.
So, can you describe it?

Is a satellite's center of gravity different from its center
of mass because it is gravitationally stabilized? Or is the
center of gravity of the Earth changed because satellites
are gravitationally stabilized? Or what? :)

-- Jeff, in Minneapolis

Hornblower
2012-Mar-25, 06:55 PM
As is too often the case, I'm skipping the obviously involved
posts and going for the seemingly easy one...


So, can you describe it?

Is a satellite's center of gravity different from its center
of mass because it is gravitationally stabilized?No, that would have the causality backward. The satellite is stabilized because the center of mass and the center of gravity are separated, and there is a position at which this creates a stable equilibrium.
Or is the
center of gravity of the Earth changed because satellites
are gravitationally stabilized? Or what? :)

-- Jeff, in Minneapolis

The center of gravity of the Earth is unaffected by the presence of satellites, whether they are stabilized or not.

ShinAce
2012-Mar-25, 07:09 PM
I did the integral and got 1/4 of height from the base. I setup the integral so that the line x=0 is at the center of gravity. Then torque from negative x balances torque from positive x values. So i integrated from -a to 0 on lhs and then 0 to b on right hand side. Since i took disk cross sections, the force from torque is rF. But r=x and F=mg where m=area*dx. I dropped the linear density as it is constant. We can drop acceleration due to gravity as its constant and appears on both sides. The integral is trivial but the solution gives a polynomial in a and b of degree 4.

edit: As is always the case, make a nice post and it will be destroyed. Since I lost everything, we all get shafted.

Short answer is integral is actually evaluated from (tip)-a to 0 and also from 0 to b(base).

The resulting polynomial is:
0= (a^2*b^2)/2 + 2(a*b^3)/3 + (b^4)/4 - (a^4)/12

which has two roots:
a=3b
a= -b

but the limits of integration forbid a = -b. I doubt I'm allowed to start cutting the cone and flipping pieces around.

The height h=a+b so a=3h/4 . Which means the CG is 3/4 of the way down from the tip.

QED. density is uniform, gravity is uniform, and center of gravity is along central axis by symmetry.

profloater
2012-Mar-25, 09:43 PM
congratualtions let's move to second moment of inertia and the energy in a spinning cone??

ShinAce
2012-Mar-25, 10:23 PM
Easy enough. Assume that the central axis is the axis of rotation. Take slices of the cone as thin cylinders concentric with the axis of rotation. You go from cylinders with height h and radius 0 to ones with height 0 and radius r. You will get (3/10)*m*r^2 . Where m is the cylinder's total mass and r is the radius. For any other axis of rotation, find a parallel one which passes through the center of mass, the distance between both axes is d. The new moment of inertia, I = (3/10)*m*r^2 + m*d^2. This is the parallel axis theorem.

Kinetic energy follows from Ke = (1/2)*I*angular velocity^2, where the angular velocity is in radians/second.

Once you have the center of mass, the only other part that requires integration is the moment of inertia about the central axis. But the relation between height and radius is a linear one.

Again, the cone is solid and of uniform density.

profloater
2012-Mar-25, 10:34 PM
Ah but the while the M of I is always through the C of G it is different in the three axes for any non symmetrical body so there are three Is (or only two for a cone) but for a general shape, three. Then you can find a unique M of I For any axis, knowing these three.

Hornblower
2012-Mar-26, 01:32 AM
We can coin the expression "center of mass" and define it anyway we wish that might be useful for a task at hand. A physicist or engineer likely will be more interested in finding a balance point than in calculating the division of an ice cream cone into equal masses of ice cream.

As has been pointed out, there is a unique point about which an extended body will be balanced in any orientation in a uniform gravitation field when suspended by that point. There is an explicit vector integration formula for calculating the location of that point if we know the mass distribution. The expression "center of gravity" is a reasonable choice of words for this point.

In the absence of external gravitation, we can use a thruster of some sort to push the body at that same point. The same sort of calculation that was done with the gravitation shows that the body will be balanced in any orientation and will not turn. Thus, the expression "center of mass". This is an example of the principle of equivalence at work.

Some further thoughts:

In a non-uniform gravitational field, the center of gravity is not necessarily a fixed point on the body. It can change with changes in orientation relative to the gravitational gradient. Suppose we hang a long dumbbell by its center of mass in the Earth's field. If the dumbbell is horizontal it will be in equilibrium, but it will be unstable. If it is tilted the least bit, the lower ball experiences more downward force than the upper one and creates a net torque that results in a runaway increase in the tilt. At some particular tilt angle we could balance it by moving the suspension toward the lower ball, but it will still be unstable. A further increase in the tilt increases the difference between the forces on the two balls and requires a further movement of the suspension point to get a momentary equilibrium. Clearly there is not a fixed point we can call the center of gravity regardless of the amount of tilt.

Of course, the torque drops to zero as the dumbbell reaches the vertical position, and this will be a stable equilibrium if it is still hanging from the center of mass. The center of gravity is now at the maximum distance below the center of mass. As long as it hangs from any point above this center of gravity it will be stable. If hung from the center of mass it will be stable with either ball down, even if they are of different masses.

grapes
2012-Mar-26, 01:58 PM
Maybe there is another term for what I describe as the center of mass - what I am describing is the point through which the 3 orthogonal planes pass whereby those planes evenly divide the subject mass into two equal masses. So if I melted down either half then I could pour it into a mold of the second and exactly fill it creating two exact replicas.Possibly, there is no such term. The reason might be because it's not a unique point--that point changes position as you tilt the cone.

If you tilt it far enough, the ice cream falls out. Another complicating factor.

ShinAce
2012-Mar-26, 02:36 PM
Ah but the while the M of I is always through the C of G it is different in the three axes for any non symmetrical body so there are three Is (or only two for a cone) but for a general shape, three. Then you can find a unique M of I For any axis, knowing these three.

Crap, that's right. I will take a pass on calculating the moment of inertia for the axis of rotation perpendicular to the central radial axis. How about I integrate e^x^2 in polar coordinates instead? All the dazzle, half the hassle.

profloater
2012-Mar-26, 03:22 PM
Actually for a cone it's easy peasy using the perpendicular axis theorem Ix =Iy + Iz. By inspection the two orthogonal axes are the same so the I about them is half the I about the long axis. and as you said before you can then shift the centre of rotation using the parallel axis theorem.

xylophobe
2012-Mar-29, 06:29 PM
So if I were to drawn lines around the cone (depicted as a triangle) to represent equal gravitational potential/acceleration could I just scale up the cone outline using the CoG as the scaling origin?

For a sphere (depicted as a circle) these lines would be concentric circles so to double the size of the circle I can scale it to 2.0 using the center of the circle.

I would like to visualize the gravitational field around the cone (and other irregularly shaped objects).

This image is sort of what I am talking about.

16630

a1call
2012-Mar-29, 11:12 PM
So if I were to drawn lines around the cone (depicted as a triangle) to represent equal gravitational potential/acceleration could I just scale up the cone outline using the CoG as the scaling origin?

Except for the fact that center of mass (and in most cases center of gravity) is quarter of the way from the base and not where you have depicted, yes (CoM not CoG) as long as scaling is uniform. This is because scaling cone about a point 1/4 of the way from the base along the centerline will result in a cone which has the same point as the 1/4 of the way from its base along the centerline.
As for CoG it will gradually move away more and more from the previous CoG as you scale it up. As an extreme visualize a super low density cone, thousands of times bigger than Earth sitting on it. The CoG will be much closer than the CoM in this case because the far points will be much less attracted than the close points.

Jeff Root
2012-Mar-30, 12:41 AM
The gravitational field around a cone looks more like my
modification of your diagram below, but it still isn't very
accurate. I'm pretty sure the contours would not parallel
the sides of the cone. (They *would* parallel the base.)

(It looks really ugly in Firefox when the window is zoomed
to anything other than a multiple of 100%. Yechhh.)

-- Jeff, in Minneapolis

.

Jeff Root
2012-Mar-30, 11:24 AM
This is probably closer to it.

-- Jeff, in Minneapolis

profloater
2012-Mar-30, 01:26 PM
at a small distance from the cone like one length, the lines would start to look like a sphere centred on the c of g I would think; it seems very unlikely they would follow the cone especially at the point where there is very little mass.

xylophobe
2012-Mar-30, 02:42 PM
at a small distance from the cone like one length, the lines would start to look like a sphere centred on the c of g I would think; it seems very unlikely they would follow the cone especially at the point where there is very little mass.

I was thinking along those lines - that it converges to a spherical field at increasing distances from the CoG, sort of like Jeff Root drew - but then there is also more mass because of the points so I think the lines would be parallel to the original form. Also, considering the 1/r2 decrease of gravity then the surface area has to increase proportionally to the distance. Jeff Root's depiction does not meet the squaring requirement for area.

For a sphere the area increases by the square of the size which is also true for cones.

In the image below the red is the actual solid mass while the green is where the area is doubled which means the gravitational effect is 1/4.

16647

btw: I fixed the location of the CoG per a1call's observation.

grapes
2012-Mar-30, 04:34 PM
The equipotential surfaces of the gravity field of a cone are not even a cone at the very surface of the cone! It's close to spherical even before the surfaces stop intersecting the cone.

As an example, on the earth's surface, ocean water tends to conform to an equipotential surface (the geoid), which has little islands sticking out of it (and large continents).

xylophobe
2012-Mar-30, 06:42 PM
The equipotential surfaces of the gravity field of a cone are not even a cone at the very surface of the cone! It's close to spherical even before the surfaces stop intersecting the cone.

As an example, on the earth's surface, ocean water tends to conform to an equipotential surface (the geoid), which has little islands sticking out of it (and large continents).

So you're saying that if I poured water onto our cone-planet then it would accumulate as shown here:

16648

and leave "mountains" at points "A", "B", and "C"?

Jeff Root
2012-Mar-30, 10:50 PM
This is probably as close as I can get. Somebody needs to
use mathcad or some such to go for the kill.

16651

-- Jeff, in Minneapolis

grapes
2012-Mar-31, 05:00 PM
So you're saying that if I poured water onto our cone-planet then it would accumulate as shown here:

and leave "mountains" at points "A", "B", and "C"?That's the general idea, but the equipotential surfaces will never be exactly spherical, no matter how far you get from the cone.

Notice, too, in your illustration, that the figure is rotated about the axis--A and B are not mountain peaks, but are both points on the edge of a "mesa" (the bottom of the cone would not be "flat", it's center would be lower than its edge.

Jeff Root
2012-Mar-31, 05:21 PM
The bottom of the cone *would* be flat, but nevertheless it would
be uphill in every direction from the center of that flat mesa to the
circular edge.

-- Jeff, in Minneapolis

xylophobe
2012-Apr-03, 12:27 AM
That's the general idea, but the equipotential surfaces will never be exactly spherical, no matter how far you get from the cone.

Notice, too, in your illustration, that the figure is rotated about the axis--A and B are not mountain peaks, but are both points on the edge of a "mesa" (the bottom of the cone would not be "flat", it's center would be lower than its edge.

I'm gonna think on the "... never be exactly spherical ..." for a while.

Maybe Jeff Root's latest drawing is closer to reality.

Anyhow, here is another picture of the cone which is more representative of an ice cream cone:

16666

I'm thinking that if I drill a hole through then the water will accumulate at the CoG.

a1call
2012-Apr-03, 03:43 AM
If the cone is a shell with a thickness of zero, and without a base, then the centroid will be 1/3rd of the way from the base rather than 1/4 of the way from the base.

http://en.wikipedia.org/wiki/Centroid#Centroid_of_cone_or_pyramid

The CoM will converge towards this point as the sell gets thinner.

xylophobe
2012-Apr-03, 02:46 PM
If the cone is a shell with a thickness of zero, and without a base, then the centroid will be 1/3rd of the way from the base rather than 1/4 of the way from the base.

http://en.wikipedia.org/wiki/Centroid#Centroid_of_cone_or_pyramid

The CoM will converge towards this point as the shell gets thinner.

I found the CoG for my hollowed-out cone by multiplying the mass of the smaller subtracted cone by the CoG location for the small cone then multiplying the mass of the larger cone by its CoG location then dividing the result by the mass of the hollowed cone.

If the cone is zero thickness then it will have zero mass and no CoG - just to clarify.

Here is a corrected picture with a hole depicting the water draining to the CoG.

16667

Hornblower
2012-Apr-03, 04:12 PM
I found the CoG for my hollowed-out cone by multiplying the mass of the smaller subtracted cone by the CoG location for the small cone then multiplying the mass of the larger cone by its CoG location then dividing the result by the mass of the hollowed cone.Have you checked up on the accuracy of your simple method by doing the necessary calculus? Caution: For all we know some integrations might not be doable analytically. We may need to do a numerical integration in small increments with a powerful computer.


If the cone is zero thickness then it will have zero mass and no CoG - just to clarify.We are concerned with thin walls and the limit as the thickness of the walls approaches zero.


Here is a corrected picture with a hole depicting the water draining to the CoG.

16667Have you done any integration to determine the gravitational vectors in different parts of the hollow. For all we know they may not all line up with a single point.

grapes
2012-Apr-03, 05:26 PM
I have to repeat: the equipotential surface will never be a perfect sphere.

And water will not "flow" down to the COG like that. The COG doesn't gravitationally attract the water--just consider that there is *no* gravitational force inside a hollow sphere, but its COG is its center.

The formula for that "cone-hollow inside a cone" shape is .25 (1 + \frac{1}{1/h+1/h^2+1/h^3}) , where h is the ratio of the smaller cone to the larger cone. If h is zero, the COG is .25; if h is 1, the COG is 1/3.

xylophobe
2012-Apr-04, 07:37 PM
Have you checked up on the accuracy of your simple method by doing the necessary calculus? ...

Calculus is not necessary for such a simple CoG calculation - I used the method for finding a composite centroid (http://en.wikipedia.org/wiki/Centroid#By_geometric_decomposition) of an area only I found the composite centroid of a volume and because it is a uniform solid then volume = mass which means the centroid of the volume is the CoG for the solid.

I get the exact same answer for the location of the CoG using my method as I get using grapes's formula in his post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2004742#post2004742).

"If an object has uniform density then its center of mass is the same as the centroid of its shape. (http://en.wikipedia.org/wiki/Center_of_mass)"

xylophobe
2012-Apr-04, 08:19 PM
...And water will not "flow" down to the COG like that. The COG doesn't gravitationally attract ...

If there are two identical bodies then a spacecraft can orbit the CoG which is halfway between the two planets. The spacecraft can also "fall" towards the CoG if the spacecraft fires a retrograde burn (http://www.orbiterwiki.org/wiki/retrograde) (see picture below).

16671

Grey
2012-Apr-04, 09:24 PM
If there are two identical bodies then a spacecraft can orbit the CoG which is halfway between the two planets. The spacecraft can also "fall" towards the CoG if the spacecraft fires a retrograde burn (http://www.orbiterwiki.org/wiki/retrograde) (see picture below).Yes, but if the spacecraft is not exactly on the plane between the two spheres, it will not generally be attracted to the center of gravity (in particular, if it's fairly close to one of the two objects, it's acceleration will be toward that object, largely ignoring the effects of the other). It's not that an object is never attracted toward the center of gravity of some object or system, it's that you cannot safely assume that it will in all cases.

grapes
2012-Apr-04, 09:42 PM
Grey is right (surely, I've said that many times before). I didnt mean to imply that the attraction is never towards the COG--there are plenty of examples where it is. Just not in this case.

xylophobe
2012-Apr-04, 10:02 PM
Has there ever been experiments conducted to see if the CoG of an open cone is gravitationally special?

For instance using a set-up similar to the Cavendish experiment (http://en.wikipedia.org/wiki/File:Cavendish_Torsion_Balance_Diagram.svg) but using hollow cones as attractors instead of spheres?

It would seem that the test spheres would be pulled into the cone(s) until the CoG is reached and then the test spheres would be repelled outwardly from the cone(s) after passing the CoG.

Hornblower
2012-Apr-04, 10:29 PM
Has there ever been experiments conducted to see if the CoG of an open cone is gravitationally special?

For instance using a set-up similar to the Cavendish experiment (http://en.wikipedia.org/wiki/File:Cavendish_Torsion_Balance_Diagram.svg) but using hollow cones as attractors instead of spheres?

It would seem that the test spheres would be pulled into the cone(s) until the CoG is reached and then the test spheres would be repelled outwardly from the cone(s) after passing the CoG.

Science and math have shown over and over again that "it would seem" is not always reliable. It would take some serious calculus to determine the orientation and magnitude of the gravity vector at various points inside a hollow cone. The forces are just too weak for any practical test here. It is hard enough with simple solid lead balls.

xylophobe
2012-Apr-05, 03:08 AM
If two gravitating objects are to exert the same force upon a test mass then the mass per unit area (or mass per surface element) should be equal, correct?

Considering a gravitating mass to be a point mass then a spherical shell around that point with a radius equal to the distance between the point mass and the test mass will have a definite mass per unit area.

If there are two gravitating masses

m1=1.000

and

m2=4.000

and

r1=1.0

while

r2=2.0

then the

spherical area1 = 4*pi()*12

while

spherical area2 = 4*pi()*22

and the

mass per unit area1 = 1.000 / (4*pi()*12) = mass per unit area2 = 4.000 / (4*pi()*22)

Grey
2012-Apr-05, 02:02 PM
If two gravitating objects are to exert the same force upon a test mass then the mass per unit area (or mass per surface element) should be equal, correct?If I'm understanding what you're saying (and it's certainly possible that I'm misunderstanding you), then I believe that you're correct. It's pretty much a direct consequence of the fact that the gravitational force on a test mass is proportional to the mass of the source and inversely proportional to the square of the distance.

xylophobe
2012-Apr-09, 04:22 PM
Will the cone balance on the CoG if it is tilted in the uniform gravity field (shown by blue arrows) as shown by the green cone on the right in this image?

We already determined that the cone will balance on the CoG if it is positioned as shown by the red cone on the left.

16701

ShinAce
2012-Apr-09, 04:47 PM
Yes, of course.

When I found that the COG is 1/4 of the way up, the solution is actually a point, and not a disk in the cone. Imagine the cone is standing on its tip. For it to be balanced, it must be perfectly vertical. Then we can see from symmetry that the same distribution is on the left and right. Likewise, front to back. As a matter of fact, the cone could be spinning like a top and nothing changes. This rotational symmetry means that the COG has to be on the line that passes right down the center.

With the integration of the cone on its side complete, we have that the COG is somewhere on a line that is parallel to the base, at 1/4 of the height from the base, and passes through the center of any circular cross section.

The two lines intersect at a single point, the COG. Imagine that you could take a very thin circular slice out of the cone, at 1/4 of the height from the base. The center of gravity is the point at the origin of that circle.

In physics, you learn that as long as the force is acting through the cog, then only its linear momentum is affected.

In your diagram, the third blue arrow from the left(for the green 'cone') hits the point marked COG. If you extended this line straight down through the other side of the cone, that is the place where a needle must be held vertically to balance the cone.

edit: with a ruler to the screen, I see that you carefully put the COG at 1/4(h). Very nice!

grapes
2012-Apr-09, 04:49 PM
Yes, it will balance at the COM

xylophobe
2012-Apr-09, 06:38 PM
Below is a snapshot of my MS Excel spreadsheet that explores cone volumes, masses, CoGs, and the gravitational effect of cone segments.

It is interesting to note that the mass of a cone increases by the cube of the factor of the change in cone height (h); whereas, the CoG location ("r" squared) increases by the square of the factor of the change in the cube height.

For instance the difference between V1 and V2 is an increase in height by a factor of 2 while the mass increases by a factor of 8 but the r-squared of the CoG changes by a factor of 4.

Then for the frustums, which are labeled: "segments"; the masses do not increase by the same rate as the r-squared location of the CoG.

This indicates to me that the "segments" do not possess the same gravitational influence upon the vertex point "P" so if we double the height of the cone then the gravitational influence of the cone does not double.

I also looked at the mass per unit area for the vertex-side of the frustums.

16702

grapes
2012-Apr-09, 07:08 PM
It is interesting to note that the mass of a cone increases by the cube of the factor of the change in cone height (h); whereas, the CoG location ("r" squared) increases by the square of the factor of the change in the cube height.
Not sure where you're going with this.

Volume increases by the cube of one dimension, if all other increase is proportional--because there are three dimensions. That's true for any shape, not just cones.

The CoG location doesn't increase by the square, it is directly proportional. But it looks like you may have confused it with the effect of gravity, which does decrease as the square of distance. Which means, if you double the size of the object, its mass will increase 2x2x2, 8 times, but since the distance doubles also, the gravitational effect will only be 2x2x2/2x2, or twice.

This indicates to me that the "segments" do not possess the same gravitational influence upon the vertex point "P" so if we double the height of the cone then the gravitational influence of the cone does not double.Maybe we're talking about two different things, but the same holds true for a cone--the gravitational strength at a particular point will double.

ETA: I retract my last statement, after looking over your diagram. The force of gravity is not necessarily towards the center of mass, of course.

xylophobe
2012-Apr-10, 01:26 AM
...The CoG location doesn't increase by the square, it is directly proportional. But it looks like you may have confused it with the effect of gravity, which does decrease as the square of distance. ...

Yeah, it was kinda confusing what I was trying to point out - the effect of gravity decreases as the square of the distance, with the distance being from the vertex of the cone (or where the vertex of the frustrum would be) and the CoG of either the cone or the frustum.

I guess the point I am making is that the segments (frustums) do not have the same gravitational effect upon the point "P" which is at the vertex (unlabeled in my previous picture).

Below is a picture that makes it more clear, I think. From my calculations it would be segment 4 (frustum 4) that would exert the greatest gravitational force upon point "P" and in fact none of the segments on either side has the same gravitational effect as any of the others.

Part of this is due to the CoGs for the frustums not following a strict increment; whereas, for the cones it progresses in a manner directly proportional to the increase in the cone size.

16706

grapes
2012-Apr-10, 02:09 AM
The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}

if I've set it up right. Then you can divide by the mass of the cone times G

\frac{G \delta \pi r^2 h}{3}

and take the inverse square root of that to get the effective distance

ShinAce
2012-Apr-10, 02:47 AM
I'm studying iterated integrals right now as I have a final on wednesday.

That looks like two things at once. It looks like a surface area for a single integral since I see 2*pi*y in it. But that can't be right. It's a double integral in dydx, so I can't see how it's the gravitational potential for a 3d object.

I need formula for the region of integration to be able to visualize it. It's actually a straightforward evaluation of the integral. The x and y upstairs should disappear. Looks like the original function is 8*pi*sqrt(x^2 + y^2) derived twice.

grapes
2012-Apr-10, 08:30 AM
I'm studying iterated integrals right now as I have a final on wednesday.

That looks like two things at once. It looks like a surface area for a single integral since I see 2*pi*y in it. But that can't be right. It's a double integral in dydx, so I can't see how it's the gravitational potential for a 3d object. It's not potential, it's acceleration (I should have included G in there). The 2*pi*y is circumference of a ring whose cross section is dxdy--opposite sides of the ring cancel, except for the component of the acceleration parallel to the cone axis.

I need formula for the region of integration to be able to visualize it. It's actually a straightforward evaluation of the integral. The x and y upstairs should disappear. Looks like the original function is 8*pi*sqrt(x^2 + y^2) derived twice.Wolframalpha says it ( Integrate[ G delta * 2 * Pi * x * y /(x^2+y^2)^(3/2), {x,0,h},{y,0,r*x/h}] ) evaluates to

2 \pi G \delta h \left( 1 - \frac{h}{\sqrt{h^2+r^2}} \right)

Dividing that by

\frac{G \delta \pi r^2 h}{3}

leaves

\frac{6}{r^2} \left( 1 - \frac{h}{\sqrt{h^2+r^2}} \right)

So the effective distance depends upon the ratio between h and r, and not just h. If r = h/2, which would be a cone as wide as it is tall, and taking the inverse square root, we get .628h.

So, if I've done nothing wrong (what are the odds of that?), such a cone acts on its point as if its entire mass were concentrated at .628h, and not at the COM at .75h. I must've done something wrong. :)

ETA: The integral is easy, by hand. One interesting result is that when r = 1.0096h, the mass of the cone acts as if it were concentrated at .75h.

EETA: of course I made a mistake--ignore that last result! It would seem there is no ratio that would allow that.

frankuitaalst
2012-Apr-10, 05:16 PM
The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}

if I've set it up right. Then you can divide by the mass of the cone times G


\frac{G \delta \pi r^2 h}{3}

and take the inverse square root of that to get the effective distance
Shouldn't the nominator of the first integral not read as ...ydydx instead of ...yxdydx ?
EDIT : Ignore my remark . The integral is correct ...

grapes
2012-Apr-10, 07:41 PM
Thanks for checking!

If r=xh, then the effective distance seems to be sqrt( (1+x^2+sqrt(1+x^2))/6), which agrees with my answer for x=1/2, but also allows an effective distance of .75 when x=.985422

I dunno what I did wrong, yet. :) But the effective distance grows without bound as the ratio increases, essentially as the cone becomes more like a large disk next to the point. I guess that makes sense.

frankuitaalst
2012-Apr-10, 09:05 PM
I got the same result Grapes , after calculating the integral . So nothing is wrong with your approach or calculation :-).
The only thing I can think of is the assumption that the force exerted on a mass this close does not behave as 1/r .
Remember : only at "big distances" a body may be considered to have it's mass centered in the COG .
I think this was proven above :-)

Grey
2012-Apr-11, 02:57 AM
But the effective distance grows without bound as the ratio increases, essentially as the cone becomes more like a large disk next to the point. I guess that makes sense.Actually, I think that does make sense. The gravitational force of an infinite flat plane is constant, with no dependence on distance. As the cone flattens out, it approaches an infinite plane as a limit. (I think you have to let the mass go infinite, too, or else it approaches an infinite plane of zero density, which has no gravitational effect at all).

grapes
2012-Apr-11, 07:58 AM
I got the same result Grapes I appreciate the cross check!

Remember : only at "big distances" a body may be considered to have it's mass centered in the COG .
I think this was proven above :-)Proven, and I've mentioned several times this thread. :)

Actually, I think that does make sense. The gravitational force of an infinite flat plane is constant, with no dependence on distance. As the cone flattens out, it approaches an infinite plane as a limit. (I think you have to let the mass go infinite, too, or else it approaches an infinite plane of zero density, which has no gravitational effect at all).Yeah, that's what I was thinking. We let \delta stay constant, instead of the total mass.

A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same. The x integral is of just a constant--that's where the factor of h comes from.

Grey
2012-Apr-11, 02:21 PM
A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same.That's very cool. That only works for flat objects, though, right? So that the total mass is proportional to the area subtended. For nonflat objects, the mass increases faster than the area, so it's no longer exactly balanced by the distance squared factor. For example, the Sun and the Moon subtend almost precisely the same angle in the sky, and the Moon actually has a higher mean density, but the gravitational force from the Sun is much higher.

profloater
2012-Apr-11, 02:23 PM
surely the gravitational effect inside a mass is only inversely proportional to the distance rather than the square. That's what I was taught for calculating forces inside the Earth.

xylophobe
2012-Apr-11, 02:33 PM
...

A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same. ...

If I am reading this correctly then because, from point "P", the cone segments subtend the same area of the sky then they should have the same gravitational effect upon point "P" even though the masses of the cone segments are differing by factors not directly related to the distance-squared? Or does this area principle only apply to spherical objects?

grapes
2012-Apr-11, 02:38 PM
As Grey points out, they'd have to have the same depth.

grapes
2012-Apr-11, 02:41 PM
surely the gravitational effect inside a mass is only inversely proportional to the distance rather than the square. That's what I was taught for calculating forces inside the Earth.That would be inside a sphere of constant density, which the earth is not--the downward force of gravity stays pretty constant, almost to the core-mantle boundary, about halfway to the center of the earth.

ShinAce
2012-Apr-11, 03:43 PM
If I am reading this correctly then because, from point "P", the cone segments subtend the same area of the sky then they should have the same gravitational effect upon point "P" even though the masses of the cone segments are differing by factors not directly related to the distance-squared? Or does this area principle only apply to spherical objects?

Imagine staring down at a small pizza. Under it is a medium and under that, a large. If they're lined up eith matching edges, like an eclipse, then the force of gravity from any one pizza on you is a constant. Same thickness, different radius.

Radius goes linearly with distance, but area would be proportional to distance squared. Area is mass in this case so mass goes up by d^2 while gravity goes down by d^2. Unity.

xylophobe
2012-Apr-11, 05:32 PM
Imagine staring down at a small pizza. Under it is a medium and under that, a large. If they're lined up eith matching edges, like an eclipse, then the force of gravity from any one pizza on you is a constant. Same thickness, different radius.

Radius goes linearly with distance, but area would be proportional to distance squared. Area is mass in this case so mass goes up by d^2 while gravity goes down by d^2. Unity.

If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders.

There is a difference, I think, between the frustums and the cylinders because the frustums have a triangular cross-section on the periphery of the cylinder which changes the amount of mass per cone segment (or frustum). And as the diameters increase the mass-contribution of the triangular section increases more due to the larger radius of rotation. I tried to show that by showing the mass per unit area --- the area increases proportionally to the square of the distance but the mass increases even more than that.

As for the sun and moon analogy descibed by Grey the gravitational flux of both bodies goes through essentially the same-size subtended area but the mass per unit area is much greater for the sun than the moon (even though the sun is much farther away) so the sun is gravitationally stronger than the moon, from our Earth-based perspective.

frankuitaalst
2012-Apr-11, 05:43 PM
The equipotential surfaces of the gravity field of a cone are not even a cone at the very surface of the cone! It's close to spherical even before the surfaces stop intersecting the cone.

As an example, on the earth's surface, ocean water tends to conform to an equipotential surface (the geoid), which has little islands sticking out of it (and large continents).

Indeed , it is . Equipotential lines generally do not follow the surface . Exception is the well known spherical surface .

Maybe somewhat out of topic relative to the OP :
In annex a visualisation of the equipotential lines of a simple rod freely floating in space .
The red line represents the rod .
The green lines represent the gravitational potential of the rod . The animation starts near the rod . Each frame increases the potential
In the rod the gravitational potential is -infinity .
The surface of the equipotential field may represent the shape of a water volume on the rod ( assuming we neglect the self gravitation of the water ) .
It's nice to see how a small volume of water stays centered around the center of the rod .

xylophobe
2012-Apr-11, 05:56 PM
...

There is a difference, I think, between the frustums and the cylinders because the frustums have a triangular cross-section on the periphery of the cylinder which changes the amount of mass per cone segment (or frustum). And as the diameters increase the mass-contribution of the triangular section increases more due to the larger radius of rotation. I tried to show that by showing the mass per unit area --- the area increases proportionally to the square of the distance but the mass increases even more than that.

...

I hate to quote myself but ...... to illustrate the above here is a picture that shows the radius of revolution of the triangular portions, per Pappus's centroid theorem (http://en.wikipedia.org/wiki/Pappus's_centroid_theorem), which shows that the centroids (http://en.wikipedia.org/wiki/List_of_centroids) of the triangles do not line up (red lines) with respect to point "P" which means the mass of the cone segments is not directly proportional to the distance from point "P".

16714

ShinAce
2012-Apr-11, 07:36 PM
If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders..

No. Mass varies as distance^2.

Edit: see posts 108 and 114

xylophobe
2012-Apr-12, 02:46 AM
No. Mass varies as distance^2.

Edit: see posts 108 and 114

I did the math and the mass of the frustums does not vary as distance2

Refer back to the very first post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998394#post1998394) in this thread and it outlines a simple way to calculate the mass of a frustum (V3) = large cone (V1) minus small cone (V2)

grapes
2012-Apr-12, 03:39 AM
I did the math and the mass of the frustums does not vary as distance2
You must be talking about two different things

If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders.
If, as you say, the radius varies directly proportional with distance, then since mass is proportional to the square of the radius, mass must be proportional to the square of distance. No? Why would mass be directly proportional to diameter?

xylophobe
2012-Apr-12, 12:42 PM
... Area is mass in this case so mass goes up by d^2 while gravity goes down by d^2. Unity.


...If, as you say, the radius varies directly proportional with distance, then since mass is proportional to the square of the radius, mass must be proportional to the square of distance. No? Why would mass be directly proportional to diameter?

I hate this "area" argument!! The simple fact is that area does not equal mass nor does area equal volume nor does area equal gravity.

Check the units: when area (m2) is multiplied by density (kg/m3) we end up with kg/m which is non-sense.

I tried to show to ShinAce that the pizza argument is only applicable to cylinders with a fixed length but diameter varying with distance and now grapes is confused - we are talking about frustums not cylinders.

ShinAce
2012-Apr-12, 01:42 PM
Now this is comical. Grapes and i do these types of calculations often enough. I'm majoring in physics and 80% of what i do is setup integrals.

We have density, depth, and radius. We assume that each slice of pizza weighs the same. Then you take pi*radius^2 to get the area of the pizza. Multiply that area by depth and you have volume. Multiply that by density and you get mass. Density is mass per volume.

What we're doing is ignoring depth because we only consider pizzas with equal depth. Let's say they have a depth of one. Then all pizza areas have to be multiplied by one to get volume. Tada! That doesnt change the answer when u multiply by one.

This lets us compare areas as if they were mass equivalent. As long as density is constant everywhere, no fancy math is needed. Area is proportional to mass in this case.

grapes
2012-Apr-12, 02:00 PM
I hate this "area" argument!! The simple fact is that area does not equal mass nor does area equal volume nor does area equal gravity.

Check the units: when area (m2) is multiplied by density (kg/m3) we end up with kg/m which is non-sense.
I think we all agree with that. :)

When they said "area is mass", in context, they clearly meant that area is proportional to mass, by a constant in the case of a fixed depth of the "pizza".

I tried to show to ShinAce that the pizza argument is only applicable to cylinders with a fixed length but diameter varying with distance and now grapes is confused - we are talking about frustums not cylinders.The one thing that I am confused about is what I mentioned in the last post:

If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders.
I'm pretty sure you're talking about cylinders here. :)

Why would you say that the mass of a cylinder be directly proportional to diameter? It's directly proportional to the square of the diameter.

xylophobe
2012-Apr-12, 02:21 PM
Why would you say that the mass of a cylinder be directly proportional to diameter? It's directly proportional to the square of the diameter.

Yes, you are correct, that was a mistake on my part.

As for the area equals mass argument in the context of the previous discussions about cylinders with fixed lengths then the comparison is valid but if we compare the cross-sections it is quite obvious that a frustum does not equal a cylinder.

The problem with the area-argument is that it confuses people and prevents them from actually doing the volume calculations.

Again, if you do the math per my first post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998394#post1998394) using the familiar volume of a cone formula

V = pi()*r2*h/3

V3 = V1 - V2

and then compare with the distance you can see that the volume/mass of a frustum is not directly related to the distance-squared nor directly related to the distance.

ShinAce
2012-Apr-12, 04:10 PM
Isn't a frustrum just a slice of a cylinder?

If we're integrating infinitesimal slices, then a frustrum only changes the limits of integration. The cylinder is the full thing and a frustrum is a piece. I can take the hard work grapes put in, change the limits, and solve any frustrum i want.

His point stands regarding gravitational influence. He has every right to reduce it as he saw it.

grapes
2012-Apr-12, 04:45 PM
As for the area equals mass argument in the context of the previous discussions about cylinders with fixed lengths then the comparison is valid but if we compare the cross-sections it is quite obvious that a frustum does not equal a cylinder.
Did anyone really say otherwise?

The problem with the area-argument is that it confuses people and prevents them from actually doing the volume calculations.
I prefer to think that we confuse ourselves, and I don't really see how it *prevents* people from doing a calculation.

Again, if you do the math per my first post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=1998394#post1998394) using the familiar volume of a cone formula

V = pi()*r2*h/3

V3 = V1 - V2

and then compare with the distance you can see that the volume/mass of a frustum is not directly related to the distance-squared nor directly related to the distance.

Jeff Root
2012-Apr-12, 11:26 PM
We have density, depth, and radius. We assume
that each slice of pizza weighs the same.
Only for a rather tricky definition of "weigh".

You assume that each slice of pizza has the same
density and the same depth. That gives larger pizza
slices greater mass, or "weight". But if by "weighs"
you mean force (per unit mass) applied at point P,
then yes, each slice weighs the same at point P.




Isn't a frustrum just a slice of a cylinder?

A frustum is not a slice of a cylinder, it is a slice of a cone.
But my immediate intuitive impression is that frustums and
cylinders should work equally well. So should a variety of
other shapes in which the "depth" or thickness is the same
for the different slices.

-- Jeff, in Minneapolis

xylophobe
2012-Apr-13, 02:14 PM
The image below is mathematical proof that the volume/mass of frustums are not proportional to distance.

What this means is that the mass per unit area is not equal for the segments (see post #100 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2006659#post2006659)) which means that the gravitational influence upon point "P" differs for each segment of a cone.

The reason the volume/mass of frustums are not proportional to distance is because of the centroid of the triangular portion of each segment --- it is a constant r/3 and by definition constants are not variables so adding a constant to a variable eliminates the proportionality.

The math below is the application of Pappus's centroid theorem (http://en.wikipedia.org/wiki/Pappus's_centroid_theorem) for determining volumes.

16724

Hornblower
2012-Apr-13, 06:03 PM
The image below is mathematical proof that the volume/mass of frustums are not proportional to distance.

What this means is that the mass per unit area is not equal for the segments (see post #100 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2006659#post2006659)) which means that the gravitational influence upon point "P" differs for each segment of a cone.

The reason the volume/mass of frustums are not proportional to distance is because of the centroid of the triangular portion of each segment --- it is a constant r/3 and by definition constants are not variables so adding a constant to a variable eliminates the proportionality.

The math below is the application of Pappus's centroid theorem (http://en.wikipedia.org/wiki/Pappus's_centroid_theorem) for determining volumes.

16724

You are dealing with a few thick frustums. If we take a large number of thin slices, as an exercise in rough and dirty calculus, the ratio of the incremental gravitational contributions at the vertex should approach a constant as a limit as h approaches zero.

grapes
2012-Apr-13, 06:26 PM
What this means is that the mass per unit area is not equal for the segments (see post #100 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2006659#post2006659)) which means that the gravitational influence upon point "P" differs for each segment of a cone.
But the integral in post #101 seems to say that is wrong, that the gravitational effect at the apex is the same for each frustum having the same height.

It has nothing to do with mass bring proportional to any power of distance, it's just a calculation of the effect itself.

ETA: In other words, if you were put your segment one on the left, and any *one* of the other three on the right, at their appropriate distance, the gravitational acceleration at the apex due to the two segments would sum to zero.

xylophobe
2012-Apr-13, 06:46 PM
The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}

if I've set it up right. Then you can divide by the mass of the cone times G

\frac{G \delta \pi r^2 h}{3}

and take the inverse square root of that to get the effective distance

If I am reading your integrals correctly you are calculating for the whole cone because your limits start from zero for both integrals - to calculate the first frustum the limits should be between h and 2h and r and 2r

I am not an expert at integrals but I can solve for the volume of a frustum in only one line using an integral with limits between

0 and h for segment 1
h and 2h for segment 2
2h and 3h for segment 3
3h and 4h for segment 4

My results using integrals are the same as I receive using Pappus's centroid theorem - using two methods and getting the same answer is more persuasive than using one method alone.

ShinAce
2012-Apr-13, 06:49 PM
Thanks Jeff. I did mean cone instead of cylinder. Guess i overstudied for my calc exam. I just kept thinking how easy it is to solve this stuff in cylindrical coordinates.

+1 to concensus. When you have half of the cone on one side and the other half opposite, their gravity cancels. The division of half and half is measured by the height of the slices, not their mass. You can end up with more mass on one side, but most of that mass is further away.

Jeff Root
2012-Apr-13, 06:51 PM
Hornblower,

Dividing the cone into negligibly-thin frustums may work to find
the solution, but I doubt it answer's xylophobe's objection. I think
some step of his analysis is just plain wrong. Which step? Is it
the assumption that the centroids of the triangular sections are
relevant?

-- Jeff, in Minneapolis

xylophobe
2012-Apr-13, 07:00 PM
... I think
some step of his analysis is just plain wrong. Which step? Is it
the assumption that the centroids of the triangular sections are
relevant?

-- Jeff, in Minneapolis

You have to find the centroid of the component areas in order to use Pappus's centroid theorem - I used this method because it shows why the volume/mass of frustums are not proportional to the distance.

Hornblower
2012-Apr-13, 07:03 PM
You have to find the centroid of the component areas in order to use Pappus's centroid theorem - I used this method because it shows why the volume/mass of frustums are not proportional to the distance.

Has anyone said otherwise? If so, please direct me to it.

xylophobe
2012-Apr-13, 07:04 PM
By the way: I did make a mistake with my page of calculations because I forgot to carry pi() to the answers so here is the fixed version.

16725

xylophobe
2012-Apr-13, 07:09 PM
Has anyone said otherwise? If so, please direct me to it.

This link (http://en.wikipedia.org/wiki/Centroid#By_geometric_decomposition) will get you to how to find a composite centroid for an area.

grapes
2012-Apr-13, 07:12 PM
If I am reading your integrals correctly you are calculating for the whole cone because your limits start from zero for both integrals - to calculate the first frustum the limits should be between h and 2h and r and 2r
The x limits would go from h to 2h, but the y limits would stay the same--zero up to a max of 2r


I am not an expert at integrals but I can solve for the volume of a frustum in only one line using an integral with limits between

I wasn't solving for mass or volume, I was calculating gravitational effect

My results using integrals are the same as I receive using Pappus's centroid theorem - using two methods and getting the same answer is more persuasive than using one method alone.What is your answer about the gravitational effect?

Jeff Root
2012-Apr-13, 07:26 PM
Hornblower,

It seems apparent that I suggested otherwise, but I was writing
out of ignorance. If you agree that the centroids of the triangular
sections can be used to determine the total gravitational effect
of the slices, then I accept that you are both right. Though I'm
not going to study the Wikipedia article deeply enough to grok it.
Even though it is admirably brief and apparently concise.

Well... in order to write the above, I had to look at that section
of the article again. Although the math looks complex at first
glance, it is really very simple. Obviously it works for plane
figures. Maybe it works for solids, too.

-- Jeff, in Minneapolis

xylophobe
2012-Apr-13, 07:37 PM
I wasn't solving for mass or volume, I was calculating gravitational effect What is your answer about the gravitational effect?

I have an idea how to calculate the gravitational effect of the frustums but I do not have the integral skill to do so, yet.

My understanding is that the mass is supposed to scale up with increasing distance so that the increase in mass exactly balances the increase in distance - I do not see that with frustums but it is quite obvious for cones (segment 1 is a cone).

grapes
2012-Apr-13, 07:45 PM
If you agree that the centroids of the triangular
sections can be used to determine the total gravitational effect
of the slices,
They cannot. Objects with the same mass and centroid can have different gravitational effects.

then I accept that you are both right. Right about what? :)

ShinAce
2012-Apr-13, 08:07 PM
I have an idea how to calculate the gravitational effect of the frustums but I do not have the integral skill to do so, yet.

My understanding is that the mass is supposed to scale up with increasing distance so that the increase in mass exactly balances the increase in distance - I do not see that with frustums but it is quite obvious for cones (segment 1 is a cone).

The integration is very helpful. If you want volume, replace everything inside the integral with the number 1. If you want mass, replace 1 with the density function. So far, we've stuck to a constant. In othet words, the material of the solid isn't wood in one case and metal in another. If you want gravitational effect, which grapes started with, you add the force expression to the integration. What you then find is what we've been arguing for a while now. That it doesn't matter if its a round cone, an elliptical cone or a slice of any type of pyramid. As long as the walls of that solid are projecting directly away from the origin, you get that any particular slice of equal height contributes an equal gravitational force.

There's another implication as we let the frustrum have infinite length, but it's too early to cover that.

Jeff Root
2012-Apr-13, 08:35 PM
Grapes,

Should I reply to that? The smiley is telling me that I still
don't understand what's going on...



As long as the walls of that solid are projecting directly away
from the origin, you get that any particular slice of equal height
contributes an equal gravitational force.
As I said twice above, my impression is that it should work
for many other shapes as well, in which the sides are not
projecting directly away from the origin. For example, your
cylindrical pizzas.

-- Jeff, in Minneapolis

ShinAce
2012-Apr-13, 09:43 PM
If we're talking about a cylinder in the sense of a pipe, then no. Each slice dx has equal mass and volume, but not gravitation influence. Only symmetrical slices cancel.

If we talk about a "cylindrical cone". Let's call it a line passing through and pivoting about the origin. Take this line and move it around. Once you get back to where you started, you'll have a shape that is both 'cylindrical' and 'conical'. The reason I call it cylindrical is because you would integrate it in cylindrical coordinates. These shapes obey the law that slices of equal thickness have equal gravitational influence. But then they have varying masses and volumes. As long as it's a shape you can create by pivoting that line through the origin, it doesn't matter if you later remove sections.

Such shapes include(the shapes are actually doubled, tip to tip): the cone with a circular base, the cone with an elliptical base, a pyramid of any base{triangle, square, hexagon, etc...). In plain english, the shapes are 'conical' but not necessarily with a circular base.

Straight old cylinders don't work. As you approach infinity, the distace is so large as to make the gravitational influence zero. If there was such a thing as infinity/2 , integrating from 0 to infinity/2 would not be equivalent to integrating from infinity/2 to infinity. For the cones, those integrals are equivalent.

xylophobe
2012-Apr-14, 02:42 PM
Here is an image that shows the volume/mass calculations that I did before only using the familiar cone volume formula = 1/3pi()r2h

It also includes the volume/mass divided by area which shows that the volumes/masses are not proportional to the distance.

ShinAce, In regard to your post #142 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2007824#post2007824) I would appreciate a more detailed description of the gravitational effect double-integral (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2006674#post2006674) (by grapes) showing the parts and how it is calculated, I am trying to re-learn calculus.

I have been trying to give links and show my calculations so that others can follow and I appologize to Jeff Root for not explaining the importance of calculating the composite centroid for use with the Pappus calculation. The Pappus calculation is basically multiplying the total cross-sectional area by the distance that the composite centroid would travel and since it is a circle then it is simply the Y-value of the composite centroid, as the radius, used in the arc length formula = 2*pi()*r

The reason the mass of frustums do not scale proportionally with distance is that the "r" in the Pappus calculation has the constant r/3 added to it.

The reason frustums differ from cones is because cone cross-sections only have 3 lines and 3 defining parameters: h, r, r/3

h is the height of the cone
r is the radius of the base
r/3 is the y-location of the centroid for the area

whereas,

frustum cross-sections have 4 lines and 4 defining parameters: h, r1, r2, r-composite.

h is the height of the cone or frustum and for our discussion, to make is simple, is always equal for the cones and frustums (as shown in my pictures)
r1 is the front (small) face of the frustum
r2 is the back (large) face of the frustum
r-composite is the composite centroid y-location

I suspect that if the double-integral is calculated with the correct limits for x in regard to segment 2 (using the limits h and 2h) then the result will not equal the result for segment 3 (using the limits 2h and 3h) or segment 4 (using the limits 3h and 4h) or segment 1 (using the limits 0 and h).


16727

grapes
2012-Apr-14, 04:27 PM
ShinAce, In regard to your post #142 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2007824#post2007824) I would appreciate a more detailed description of the gravitational effect double-integral (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2006674#post2006674) (by grapes) showing the parts and how it is calculated, I am trying to re-learn calculus.
I'm happy to do that myself! :)

The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}
G is the gravitational constant, delta is the density, 2pi y is the circumference of a ring which when multiplied by its cross-section dx dy gives the volume of an infinitesimal ring (multiplied by density gives mass), which is then multiplied by G and divided by distance squared to get the gravitational effect at the apex. However, opposite sides of the ring oppose each other, and cancel some of that force, leaving only the force parallel to the axis, which is found by multiplying by x divided by the distance.


I suspect that if the double-integral is calculated with the correct limits for x in regard to segment 2 (using the limits h and 2h) then the result will not equal the result for segment 3 (using the limits 2h and 3h) or segment 4 (using the limits 3h and 4h) or segment 1 (using the limits 0 and h).

The nice thing about the evaluation of that integral is that the y-integral is a constant! The x variable disappears:

\int_{y=0}^{r x/h} \frac{ y x dy}{(x^2+y^2)^{3/2}} = 1- \frac{h}{\sqrt{h^2+r^2}}

Multiplying by the other four constants ( G \delta 2 \pi ) is still a constant, so the effect of the x-integral is just to multiply that resulting constant by the difference in the x limits. In other words, in your cases, h, each time. So, all four segments have the same answer.

HUb'
2012-Apr-14, 10:29 PM
ok? {maybe not {{ the {{{ i mean my QUestion is ? can U (Um}? map the above
integral form to a (Um2}? " Differential form "

http://www.principiadiscordia.com/forum/index.php/topic,27356.90/msg,1141630.html
riday 1/27 / 13 Chuen 2815 24/22.5 = .3sB
Integral form ...-> Differential form ...-> Matrix form
so the Question becomes ? wiLL i ever return?
i donno but i did find appendix I, so i may as
well try: starting from the integral form page 225
_/` E.dl = -dOb/dt & _/`B.dl=uo[i+eodoE/dt]
---------------------------------------------
E { electric field
B { magnetic field
O { flux
i { current

No, i have no idea
this was in January ? it is now April ?/?

BigDon
2012-Apr-15, 03:17 AM
My only contribution to the thread is to say that the reason Indian tepees are conical is because a cone is a shape that wouldn't blow over in the prairie winds.

a1call
2012-Apr-15, 05:28 AM
Hopefully the etc. in the title makes the following valid comments.
*- Any cone (- the base) will unravel to a section of a circle
*- IIRC, the area of any such subsection is equal (radius x arc-length)/2
*- The conical aspect of eggs prevents them from rolling off shallow slopes. There is a correlation between the conical-ness of bird's eggs and how high they nest on cliffs. If a cone doesn't slip it will simply roll around the apex and reach a stable equilibrium which will stop any further rolling.

ETA:From Archimedes:

Any sphere is (in respect of solid content) four times the cone with base equal to a great circle of the sphere and height equal to its radius
Source (http://www.cut-the-knot.org/pythagoras/Archimedes.shtml)

Jeff Root
2012-Apr-15, 08:23 AM
As long as the walls of that solid are projecting directly away
from the origin, you get that any particular slice of equal height
contributes an equal gravitational force.
As I said twice above, my impression is that it should work
for many other shapes as well, in which the sides are not
projecting directly away from the origin. For example, your
cylindrical pizzas.



If we're talking about a cylinder in the sense of a pipe,
then no. Each slice dx has equal mass and volume, but
not gravitation influence. Only symmetrical slices cancel.
I'm totally confused.

Everybody knows what a cylinder is. But "a cylinder in the
sense of a pipe" makes it sound as though a hole is a feature
of what we're talking about. While a hole appears to be a
feature of xylophobe's analysis of frustums, it certainly isn't a
feature I had in mind.

Your idealized pizzas, which you suggested in post #114

http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2007151#post2007151

are cylindrical, aren't they? They have parallel sides, not sloped.
They don't have holes in them, do they?

They do not have "equal mass and volume". The farther the
pizza is from the origin, the larger its diameter, volume, and
mass.



If we talk about a "cylindrical cone".
A what??? What the heck kind of terminology is that???



Let's call it a line passing through and pivoting about the origin.
That sounds like the surface of a cone.



Take this line and move it around.
A second time? Or are you re-stating the "pivoting" of the
preceeding sentence?



Once you get back to where you started, you'll have a shape
that is both 'cylindrical' and 'conical'.
No. I haven't a clue. Gimme a diagram.



The reason I call it cylindrical is because you would integrate
it in cylindrical coordinates.
I would? Maybe if I knew what that meant.



These shapes obey the law that slices of equal thickness
have equal gravitational influence.
Well, you appear to be talking about the right subject, so I'm
trying to follow along.

But then they have varying masses and volumes.[/quote]
Yes, naturally.



As long as it's a shape you can create by pivoting that line
through the origin, it doesn't matter if you later remove sections.
That seems to be a description of what xylophobe is doing in
his diagrams of two cones divided up into frustums, and makes
sense to me, so maybe I'm not completely lost.



Such shapes include (the shapes are actually doubled, tip to tip):
the cone with a circular base, the cone with an elliptical base, a
pyramid of any base{triangle, square, hexagon, etc...). In plain
english, the shapes are 'conical' but not necessarily with a circular
base.
I think I understand that. I vaguely recall trying unsuccessfully
to find a general term for such a family of shapes, myself, a few
years ago.



Straight old cylinders don't work.
I would think they would, although perhaps the distances would
be described slightly differently: Rather than measuring to the
near surface or the far surface or the center of mass, it might
be necessary to measure to the center of gravity. Yes???



As you approach infinity, the distace is so large as to make the
gravitational influence zero.
What? Are we talking about the same thing?? The farther
away the disk is, the bigger it is. The gravitational influence
on the origin point should be the same from all disks, nomatter
how far away. Yes???



If there was such a thing as infinity/2 , integrating from 0 to
infinity/2 would not be equivalent to integrating from infinity/2
to infinity.
Not that I understand what you are saying there, but are you
sure about that?



For the cones, those integrals are equivalent.
Okay. How about you try to describe these integrals for me
in simple terms? (Although both you and grapes probably
already did so more than once, above...) Then apply it to
your "cylindrical cone". By which I'll now guess you simply
meant a stack of cylinders whose envelope is a cone.
Is that what you meant??????? That would make so much
sense. Please tell me that's what you meant!

And / or draw a diagram! Diagrams rule!

-- Jeff, in Minneapolis

ShinAce
2012-Apr-15, 03:43 PM
Any shape can be hollow or solid. This thread is about solid objects.

Don't take the pizzas too literally. We start with a cone and slice it up. The walls of that pizza are not parallel. They are approximately parallel for thin pizzas. You can then slice the pizza into triangles. As long as each triangle has the same surface area, it will have the same weight. Weight is not mass. Weight is a force. We want to calculate forces, so the gravitational force is weight.

A cylinder is anything with parallel walls, which we don't have. The majority of pools, including kidney shaped pools are cylinders. The base can be any closed 2d surface. But cylinders don't reduce to a point at the origin.

A cone is any closed surface made by pivoting a line through the origin. Mathematically speaking, a pyramid is a type of cone. But you would integrate a straight sided pyramid in cartesian coordinates and elliptical based cones in cylindrical coordinates.

The wiki page for cones covers their definition.

xylophobe
2012-Apr-16, 05:14 PM
First, thanks for the explanation of the double-integral ... I'll probably have questions as is I think about it.


... 2pi y is the circumference of a ring which when multiplied by its cross-section dx dy gives the volume of an infinitesimal ring ...

This statement indicates that the formula is using the Pappus calculation to calculate the volume which if done correctly then the "y" should be y/2 because the centroid of the cross-sectional area, which is a rectangle (with y for height), is half the height value (the x-loc is also half the x-length but that is not used in the Pappus volume calculation).

If the integration takes equal horizontal slices then the centroid for the 5th slice is located at 4*dy + dy/2 which is the "r" for the area [dx*dy] times the distance the centroid travels [circumference = 2*pi()*r] around the axis which equals 2*pi()*(4*dy + dy/2)

Centroid of Rectangle (http://www.wisc-online.com/Objects/ViewObject.aspx?ID=ENG11504) (see slide #4 of the presentation)

grapes
2012-Apr-16, 05:51 PM
No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

HUb'
2012-Apr-16, 06:53 PM
No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

On that note, i will just say " i have no idea,,,, this was in January ? it is now April ?/? "
& travel elsewhere: what i would like to find, is an explanation of perigee.bas
which computes the days when the moon reaches PERIGEE & apogee also,
I do have a version of that sky & telescope program from years gone by
i could upload it {Maybe) it is rather long? & maybe i already did? it begins
REM CONSTANTS
DIM A(30), B(30), R(30), A1(30)
DIM C0(30), C1(30), C2(30), C3(30)
DIM C4(30):

THUS: there are 30 parameters ?{objects) of
:Line " 253 DATA 51818,0,1,0,0,0 " of six parameters each
My thought {at this time)i would like to add A 31'st {oh never mind | time is short
AUTO SAVE in effect | Link to where i went to later {Hasta Lego ToV (http://www.bautforum.com/showthread.php/130276-Transit-of-Venus-2012?p=2008595#post2008595)

xylophobe
2012-Apr-16, 08:35 PM
No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

Does the integral account for the exponential increase in mass due to increasing radius. The exponential increase in the below picture is only shown in the cylindrical portion of the frustum but it continues into the triangular section which shifts the CoG towards the larger end.


16745

Jeff Root
2012-Apr-16, 10:42 PM
xylophobe,

I'm glad if you are learning from this (it's your thread,
so that's the first priority), but I just keep getting loster
and loster. Maybe HUb' can explain it to me.

Your diagram showing "exponentially increasing mass"
makes absolutely no sense to me. The density of the
hatching seems to be an indicator of material density.
But you have given no reason for the density to vary.
And you show it increasing radially outward from the
centerline of the cone. What on Earth is that about???
Am I completely misinterpreting the diagram and what
you are saying about increasing mass? I would think
the "increasing mass" would be the increase in mass
of each succeeding frustum with increasing distance
from the origin, due to its larger volume. No?

-- Jeff, in Minneapolis

grapes
2012-Apr-17, 12:59 AM
Does the integral account for...
It accounts for everything I can think of. :)

You can start off with a simple integral that calculates the volume by adding up every single little box, y d\theta dy dx

\int_{x=0}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Then you just multiply that by density to get mass, then multiply by the gravitational constant G and divide by the distance squared to get acceleration, then by the cosine of the angle to get the component parallel to the axis. I think that's everything. :)

xylophobe
2012-Apr-17, 05:38 PM
xylophobe,

I'm glad if you are learning from this (it's your thread,
so that's the first priority), but I just keep getting loster
and loster. Maybe HUb' can explain it to me.

Your diagram showing "exponentially increasing mass"
makes absolutely no sense to me. The density of the
hatching seems to be an indicator of material density.
But you have given no reason for the density to vary.
And you show it increasing radially outward from the
centerline of the cone. What on Earth is that about???
Am I completely misinterpreting the diagram and what
you are saying about increasing mass? I would think
the "increasing mass" would be the increase in mass
of each succeeding frustum with increasing distance
from the origin, due to its larger volume. No?

-- Jeff, in Minneapolis

I guess some of my pictures only make sense to me so I'll try to explain.

In that diagram the density is constant ... the cross-hatching was not intended to imply a changing density.

The mass increases exponentially because the mass of a cylinder is pi()*r2h so as "r" increases the mass increases according to the square of "r".

My increments were not exactly precise (I realized after I posted) because the increments were supposed to indicate rings of equal mass.

Let me know if you have additional questions.

xylophobe
2012-Apr-17, 05:44 PM
It accounts for everything I can think of. :)

You can start off with a simple integral that calculates the volume by adding up every single little box, y d\theta dy dx

\int_{x=0}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Then you just multiply that by density to get mass, then multiply by the gravitational constant G and divide by the distance squared to get acceleration, then by the cosine of the angle to get the component parallel to the axis. I think that's everything. :)

Cut it out .... I'm still trying to figure out the double-integral ..... jk.

After thinking about it a while I think the picture below is an accurate description of what the double-integral is intended to accomplish.

I only show a single slice along the x-axis and a finite number of slices along the y-axis

16748

grapes
2012-Apr-17, 08:21 PM
I dunno if that really depicts what it is trying to accomplish, but it looks like things are labeled correctly.

xylophobe
2012-Apr-18, 05:20 PM
In the picture below the image on the far left is a cylinder divided into 8 sections with equal radial thickness. The numbers are the masses of each ring, using the innermost section (which is a cylinder) as the unit = 1.0

2 times "r" causes the mass to grow by a factor of 4 which is what 1+3 equals
3 times "r" causes the mass to grow by a factor of 9 which is what 1+3+5 equals
4 times "r" causes the mass to grow by a factor of 16 which is what 1+3+5+7 equals
5 times "r" causes the mass to grow by a factor of 25 which is what 1+3+5+7+9 equals
6 times "r" causes the mass to grow by a factor of 36 which is what 1+3+5+7+9+11 equals
7 times "r" causes the mass to grow by a factor of 49 which is what 1+3+5+7+9+11+13 equals
8 times "r" causes the mass to grow by a factor of 64 which is what 1+3+5+7+9+11+13+15 equals
etc.
etc.
etc.

The 5 images to the right of the cylinder are frustum segments that show how the mass of the conic section varies between a beginning at a 1/3 to 2/3 split towards an infinite limit of 1/2 to 1/2 split. By evaluating the cone segments as infinitely small cylinders the double-integral approaches an exact value the farther from point "P' that the segments are but up close the double-integral is inaccurate.

The chart at the bottom of the picture shows how the two conic sections approach equality because as "x" approaches infinity then x+1 approaches "x"

This geometric analysis proves that the cone segments do not have equivalent gravitational influence upon point "P"

Nevertheless, the double-integral is a valuable tool for determining real-life applications because we do not encounter cone-shaped gravitating objects in outer space ... just in thought experiments.


16751

xylophobe
2012-Apr-18, 06:10 PM
Here is a better pictorial explanation of how the double-integral errs from an exact solution.

There is always a constant difference between the approximation and the actual geometry --- adding a constant eliminates the possibility for reaching an exact solution.

16752

Grey
2012-Apr-18, 06:39 PM
Here is a better pictorial explanation of how the double-integral errs from an exact solution.Except that your drawing assumes that dx and dy have finite values, and that you're dividing the cone up into a small finite number of sections. The integral is based on letting the size of the sections go to zero, while simultaneously letting the number of such sections approach infinity. When you do that, the amount of error likewise goes to zero. Indeed, that's the entire point of integral calculus.

grapes
2012-Apr-18, 07:25 PM
This geometric analysis proves that the cone segments do not have equivalent gravitational influence upon point "P"But you haven't even used a distance from the cone segments to the point "P", which is necessary to calculate the gravitational effect.

Your diagrams do not represent how integrals are calculated. The integrals are exact, not approximations, using the same conditions that you assume--that density is constant, and the shapes are perfect cones.

xylophobe
2012-Apr-19, 09:52 PM
But you haven't even used a distance from the cone segments to the point "P", which is necessary to calculate the gravitational effect. ...

If I add the distance from the cone segments as you suggest then everything is fine for the cylindrical portion of the frustum so I will concentrate on just the conic section of the frustum.

The below picture shows the slice as the double-integral estimates which is just a rectangle and because it is wafer thin then its CoG is essentially the same as the centroid for a rectangular area. The "x" and "y" are located, by the integral, in the center and the height of the wafer is "dy" while the width of the wafer is "dx". The distance from point "P" to the CoG for this rectangular wafer solid is shown as a square-root calculation.

The actual geometry has a triangular shaped wafer and because it is a wafer then its CoG matches the location of the centroid (http://en.wikipedia.org/wiki/List_of_centroids) for a triangular area. The distance from point "P" to the CoG for this triangular wafer solid is also shown.

What I show is the very first slice from a cone (or segment 1 from my previous pictures) which clearly shows that the CoG for the actual solid is in a different location than what is used by the double-integral calculation.

The difference in actual location for the CoG and the integral-approximation approaches zero difference but only when the cone reaches infinite dimensions.

The error adds up as the length of the frustum increases ... with the most error being close to point "P" and near zero error at the farthest distance from point "P"

16753

xylophobe
2012-Apr-19, 09:57 PM
Except that your drawing assumes that dx and dy have finite values, and that you're dividing the cone up into a small finite number of sections. The integral is based on letting the size of the sections go to zero, while simultaneously letting the number of such sections approach infinity. When you do that, the amount of error likewise goes to zero. Indeed, that's the entire point of integral calculus.

An integral funtions as you state but only for continuous functions - the frustum is two functions: a cylinder and a conic section.

grapes
2012-Apr-19, 10:01 PM
The integral is not an approximation. It is an exact calculation--that is the point of using calculus.

ETA:
An integral funtions as you state but only for continuous functions - the frustum is two functions: a cylinder and a conic section.I'm not sure exactly what you mean by this, as integrals are applied to multiple functions all the time. **We can illustrate by getting rid of the mass and distance factors and just calculate volume.

\int_{x=0}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Evaluating the theta integral gets us:

\int_{x=0}^{h} \int_{y=0}^{r x/h} 2\pi y dy dx

Evaluating the y integral brings us to:

\int_{x=0}^{h} \pi (r x/h)^2 dx

Which results in \pi r^2 h/3, the exact formula for a cone, no matter what h or r.

ShinAce
2012-Apr-19, 11:31 PM
Integration definitely does give exact answers. Numerical methods, like expanding into a series and integrating, has inherent errors. Be advised that your calculator does many things using series, such as the log function. Integration by hand is not numerical, but it is exact.

I also did the integral for cone volume, and got Pi*r^2*h/3. Exactly.

When I did the integral for force due to gravity, I kept the little m in GmM(it's the test particle, or the mass of your scale) and changed big M to density. I got that F=2*Pi*G*m*density*h / sqrt(1 + r^2/h^2). I also got the curious result that when integrating the third integral dx, x was long gone. It didn't matter if I evaluated x from 1 to 2, or from 3.5 to 4.5. Any slice of fixed height for that cone gives the same gravitational force. Also pretty cool is that I got a negative answer! I integrated with the forces acting from the origin to the infinitesimal 'rings' I was integrating. The answer I get is that the force acts on the test particle, not the other way around like I though I had set it up. Thankfully Newton's forces are equal and opposite pairs.

We can rearrange my expression for F to get:
F = 2*Pi*G*m*density*h^2 / sqrt(h^2 + r^2) , but sqrt(h^2 + r^2) is the length from the cone tip to a point on the base's perimeter. Let's call that s, the length of the slope.
F = 2*Pi*G*m*density*h^2 / s

You can derive r from h and s. It's not lost, just built in.

grapes
2012-Apr-20, 02:22 AM
When I did the integral for force due to gravity, I kept the little m in GmM(it's the test particle, or the mass of your scale) and changed big M to density. I got that F=2*Pi*G*m*density*h / sqrt(1 + r^2/h^2). I also got the curious result that when integrating the third integral dx, x was long gone. It didn't matter if I evaluated x from 1 to 2, or from 3.5 to 4.5. Any slice of fixed height for that cone gives the same gravitational force. Also pretty cool is that I got a negative answer! I integrated with the forces acting from the origin to the infinitesimal 'rings' I was integrating. The answer I get is that the force acts on the test particle, not the other way around like I though I had set it up. Thankfully Newton's forces are equal and opposite pairs.

Thanks for re-doing the integral, it's always nice to have a check. However, your answer differs from mine. You shouldn't have gotten a negative result--all constants and variables are positive. Apparently, you left off the second (positive) value of the interior definite integral, which is lessened by that negative value.



We can rearrange my expression for F to get:
F = 2*Pi*G*m*density*h^2 / sqrt(h^2 + r^2) , but sqrt(h^2 + r^2) is the length from the cone tip to a point on the base's perimeter. Let's call that s, the length of the slope.
F = 2*Pi*G*m*density*h^2 / s
Notice, for a constant h, your F decreases as s increases. That can't be right. As s gets larger, the cone still contains the smaller cones--so the force can't decrease.

ShinAce
2012-Apr-20, 02:59 PM
The only place I can think of where I made a mistake would be the infinitesimal masses of the cone, dM. I'll redo it later as a double integral of rings(aka, drop r*dtheta and replace with 2*Pi*r), instead of the triple integral I tried out.

My expression for the integration looked a lot like yours. All I had to do to get the same orientation of the cone was change rdr for ydy. I had a cos(phi) in the expression for F that comes about due to symmetry arguments. That reduces to x/d where d is the distance between the particles acting on each other. That gave me the same:
xy/(x^2 + y^2)^3/2 dphi dy dx
Integrating that with respect to y gave me the negative. Now you can rearrage the denominator to cancel out x entirely. When I integrate with respect to x, I don't get a second negative to cancel out the first one.

xylophobe
2012-Apr-20, 03:20 PM
In the attached picture I show the integration using horizontal slices in the above image and integration using vertical slices in the bottom image.

The integration using vertical slices yields a correct mass because all the "dx" slices are bounded similarly: "x-axis" on the bottom, "x" on the large end, "x-h" on the small end, and "y=xr/h" on the top. The function is continuous along the entire length.

The integration using the horizontal slices is not over a continuous function because there is a discontinuity at the common point between the boundaries "x-h" and "y=xr/h" (the point) . This violates a fundamental rule of calculus. The left-most boundary of the horizontal slices is "x-h" for the cylinder section and "y=xr/h" for the conic section. The center of gravity for the horizontal slices in the conic section is constantly changing for each "dy" slice.

16754

xylophobe
2012-Apr-20, 03:28 PM
Here are the calculations for the CoGs for the frustums. You can imagine them as being the whole frustum or as three (3) individual "dx" slices ... plus the first cone-shaped "dx" slice.

16755

xylophobe
2012-Apr-20, 04:43 PM
Here are the calculations for the CoGs for the frustums. ...

I was just looking at the picture and if I do not increase the viewing scale on my browser then the divide symbol that I used looks like a "+" symbol .... I guess I should have stuck with the old standby "/" symbol. Hopefully this is only happening on my computer and not everyone else's.

grapes
2012-Apr-20, 04:46 PM
That gave me the same:
xy/(x^2 + y^2)^3/2 dphi dy dx
Integrating that with respect to y gave me the negative.
That should result in -x/(x^2 + y^2)^1/2 dx (dphi?), but now you have to evaluate it at y=rx/h *and* subtract it evaluated at y=0

ShinAce
2012-Apr-20, 05:34 PM
*and* subtract it evaluated at y=0

Oops! There it is.

xylophobe
2012-Apr-21, 08:00 PM
The integral is not an approximation. It is an exact calculation--that is the point of using calculus.

ETA: I'm not sure exactly what you mean by this, as integrals are applied to multiple functions all the time. We can illustrate by getting rid of the mass and distance factors and just calculate volume.

\int_{x=0}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Evaluating the theta integral gets us:

\int_{x=0}^{h} \int_{y=0}^{r x/h} 2\pi y dy dx

Evaluating the y integral brings us to:

\int_{x=0}^{h} \pi (r x/h)^2 dx

Which results in \pi r^2 h/3, the exact formula for a cone, no matter what h or r.

These are excellent confirmations that a properly set up integral will yield a correct answer - now change the limits to get just a frustum volume - either segment 2, segment 3, or segment 4.

As far as applying integrals to multiple functions I think what you are describing is multiple variables, ie: \theta, x, y

What I am referring to is multiple functions --- the function of the sloped cone line y=xr/h, the shape of the horizontal cylinder line y=r (around x-axis for segment 1)

grapes
2012-Apr-21, 11:27 PM
Thank you. It wasn't clear that you agreed with the frustrum results, I was kinda under the impression that you were trying to argue against them.

Grey
2012-Apr-22, 04:17 PM
In the attached picture I show the integration using horizontal slices in the above image and integration using vertical slices in the bottom image.

The integration using vertical slices yields a correct mass because all the "dx" slices are bounded similarly: "x-axis" on the bottom, "x" on the large end, "x-h" on the small end, and "y=xr/h" on the top. The function is continuous along the entire length.

The integration using the horizontal slices is not over a continuous function because there is a discontinuity at the common point between the boundaries "x-h" and "y=xr/h" (the point) . This violates a fundamental rule of calculus. The left-most boundary of the horizontal slices is "x-h" for the cylinder section and "y=xr/h" for the conic section. The center of gravity for the horizontal slices in the conic section is constantly changing for each "dy" slice.It's perfectly possible to use calculus to calculate this in either case. It's true that if you do it using "horizontal slices" as you have it shown here, you'll have to break the integral into two pieces, but that doesn't stop you from using calculus to solve the problem, or from getting an exact answer. As far as I can tell, none of the integrals that grapes has set up would even have this as an issue at all. The integrals he's been using are all the correct method to determine the gravitational force from a cone (or a portion thereof) on a point.

xylophobe
2012-Apr-23, 01:52 PM
It's perfectly possible to use calculus to calculate this in either case. It's true that if you do it using "horizontal slices" as you have it shown here, you'll have to break the integral into two pieces, but that doesn't stop you from using calculus to solve the problem, or from getting an exact answer. As far as I can tell, none of the integrals that grapes has set up would even have this as an issue at all. The integrals he's been using are all the correct method to determine the gravitational force from a cone (or a portion thereof) on a point.

So far I have not seen any integrals that calculate for a frustum. If the integral can not solve for the volume of a frustum then the gravitational force that the integral yields is also not correct for a frustum.

I asked grapes to solve his triple integral for the volume of a frustum because I am sure that he can not because the triple integral is set up to solve for cones not frustums --- I calculated the volumes of the frustums and so far no integral has produced the same result.

Below I will dissect the grapes double-integral into its component parts to show that the volume must match the volume for a frustum.

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}


Constants: G \delta


Distance: {(x^2+y^2)^{1/2}


Inverse Square of Distance: \frac{1}{(x^2+y^2)}


Cosine: \frac{x}{(x^2+y^2)^{1/2}}


Leaves the Volume of a Cone: \int_{x=0}^h \int_{y=0}^{r x/h} 2 \pi y dy dx}


This yields: \frac{\pi r^2h}{3}

This double-integral should solve for the volume of the object being examined which for our discussion is a frustum but it does not solve for the volume of a frustum because it results in the volume of a cone --- I have already shown (post 145 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2007979#post2007979)) that cones follow all the claims: mass squares with distance which cancels out the inverse square of the distance.

This double-integral will not solve for the volume of a frustum because a frustum involves two functions: a cylinder and a conic section. To get the correct solution for a frustum the solver would have to solve individually for the cylinder and then for the conic and then add the two component solutions to receive the correct total solution.

Violating the fundamental requirement (continuity) of integrals results in an integral that yields an incorrect solution --- fix the integral and it will solve for what we are looking for!!

btw: what is the code for creating the brackets around an integral solution, with the limits? I need this to show how the integral is solved. Thanks.

grapes
2012-Apr-23, 02:35 PM
The (interior) y integral stays the same, we just change the limits on x


Evaluating the y integral brings us to:

\int_{x=0}^{h} \pi (r x/h)^2 dx

Which results in \pi r^2 h/3, the exact formula for a cone, no matter what h or r.
So, for a frustrum which has half the height of the cone, the limits would go from h/2 to h, thereby cutting off the top part of the cone:
\int_{x=h/2}^{h} \pi (r x/h)^2 dx

That becomes the evaluation:

\left|_{x=h/2}^{h} \frac{\pi r^2}{h^2} \frac{x^3}{3} \right.

Hmmm, maybe this looks better:

\left. \frac{\pi r^2}{h^2} \frac{x^3}{3} \right|_{x=h/2}^{h}

Which is the same as for the cone, except for the different limits. It evaluates to:

\frac{\pi r^2}{h^2} \frac{h^3}{3} - \frac{\pi r^2}{h^2} \frac{(h/2)^3}{3} = \frac{7 \pi h r^2}{24}

Which is of course 7/8 of the original cone, same answer you got.

Grey
2012-Apr-23, 02:38 PM
I asked grapes to solve his triple integral for the volume of a frustum because I am sure that he can not because the triple integral is set up to solve for cones not frustums --- I calculated the volumes of the frustums and so far no integral has produced the same result.Of course this can be done. In fact, it's pretty trivial. Start with the triple integral for the volume of a cone that grapes provided:


\int_{x=0}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Evaluating the theta integral gets us:

\int_{x=0}^{h} \int_{y=0}^{r x/h} 2\pi y dy dx

Evaluating the y integral brings us to:

\int_{x=0}^{h} \pi (r x/h)^2 dx

Which results in \pi r^2 h/3, the exact formula for a cone, no matter what h or r.
But if we don't want the whole cone, just change the bounds for the x integration. For example, say we want just the frustum that consists of the bottom quarter of a cone of height h. So we change it to:

\int_{x=3h/4}^{h} \int_{y=0}^{r x/h} \int_{\theta=0}^{2\pi} y d\theta dy dx

Evaluating the theta integral gets us:

\int_{x=3h/4}^{h} \int_{y=0}^{r x/h} 2\pi y dy dx

Evaluating the y integral brings us to:

\int_{x=3h/4}^{h} \pi (r x/h)^2 dx

Which gives us (\pi r^2 h/3)(37/64).

The normal formula for the volume of a frustum is generally given as:

V = (\pi h_1/3)(r_1^2 + r_1 r_2 + r_2^2),

where r1 is the radius at the top, r2 is the radius at the bottom, and h1 is the height of just the frustum. Using r for the radius at the bottom, and h for the height of the entire cone, as we did in the integrals above, this gives us (for the bottom quarter):

r_1 = 3r/4
r_2 = r
h_1 = h/4

Substituting those values into the formula above, we get exactly the same result as we got by integration. Changing the bounds of the x integration, we can get any slice of the cone that we want.

Edit to add: Rats! Grapes beat me to it be 3 minutes!

grapes
2012-Apr-23, 02:52 PM
I always take the low road :)

xylophobe
2012-Apr-23, 06:28 PM
For segment 2 then

\int_{x=1h}^{2h} \int_{y=0}^{r x/h} 2 \pi y dy dx = \int_{x=1h}^{2h} \left[2 \pi \frac{1}{2} y^2}\right]_{y=0}^{rx/h}dx \right = \int_{x=1h}^{2h} \left\pi \frac{r^2}{h^2}x^2 dx} \right = \left[\frac{1}{3}\pi \frac{r^2}{h^2} x^3 \right]_{x=1h}^{2h} = \left \frac{1}{3}\pi \frac{r^2}{h^2} (2h)^3 \right - \left \frac{1}{3}\pi \frac{r^2}{h^2} (1h)^3 \right = \left \frac{7}{3}\pi r^2h \right

ShinAce
2012-Apr-23, 07:02 PM
Be advised you're working with the integral for volume only, as defined with the cone having radius r at distance h from the origin.

In four segments, let x go from (0 to 0.25h), (0.25h to 0.5h), (0.5h to 0.75h), (0.75h to h). You get the volume of each frustrum as a result.

In you evaluation of the integral, what is the value of r? It's the radius at the base of which segment? What script are you working with(in response to your post title)?

Grey
2012-Apr-23, 07:18 PM
For segment 2 then...

In you evaluation of the integral, what is the value of r? It's the radius at the base of which segment? What script are you working with(in response to your post title)?I'm curious about exactly what you're intending to calculate with the integral above as well. As you have it set up, it will calculate the volume of a frustum with height h, radius of the smaller circle r, and radius of the larger circle 2r. In other words, the bottom half of a cone of height 2h and radius 2r. And the value that you get is the correct one:
(\frac{7}{3}\pi r^2h)
Maybe you're just not setting up the integrals right for what you intend to be calculating?

grapes
2012-Apr-23, 08:03 PM
Maybe you're just not setting up the integrals right for what you intend to be calculating?No, I think that is what was intended. There was some suspicion that calculus couldn't handle the frustrum, even just for volume, but I think that has been allayed now.

Perhaps now, the gravity calculations will be acceptable as well?

xylophobe
2012-Apr-23, 08:21 PM
I'm curious about exactly what you're intending to calculate with the integral above as well. As you have it set up, it will calculate the volume of a frustum with height h, radius of the smaller circle r, and radius of the larger circle 2r. In other words, the bottom half of a cone of height 2h and radius 2r. And the value that you get is the correct one:
(\frac{7}{3}\pi r^2h)
Maybe you're just not setting up the integrals right for what you intend to be calculating?

The script that I refer to is the way I set up the problem in post 128 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2007695#post2007695) whereby "h" is the individual lengths of the frustums and "r" is the result of which segment we are talking about ... it is just a different way of splitting a cone into 4 equal-length segments.

I could have specified fractions of "h" as ShinAce suggested in post 184 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010835#post2010835) but why mess up a simple representation with fractions? We all know that "r" is directly proportional to "h" so fractions are not needed.

What I would like to see is the full solution (like I just did (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010824#post2010824)) to the grapes double-integral:




\int_{x=0}^h \int_{y=0}^{r x/h} \frac{G \delta 2 \pi y x dy dx}{(x^2+y^2)^{3/2}}



I am sure there are other readers who would also like to see the complete solution, step by step.

grapes
2012-Apr-23, 08:46 PM
What I would like to see is the full solution (like I just did (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010824#post2010824)) to the grapes double-integral:



I am sure there are other readers who would also like to see the complete solution, step by step.
Most of the readers still with us have already done it themselves. I think it is instructive, go ahead and give it a try!

xylophobe
2012-Apr-24, 02:33 PM
Okay, now that I have determined that the double-integral gives the correct volume now I will try to understand how the distance to the CoG is determined for each (dx)(dy) slice .

If x=0 (or the leftmost edge of a segment) then the factor of dx*dy should be zero otherwise there would always be something to subtract in the solution of an integral with a lower limit = 0.

The same reasoning should be true for y=0

So, just using simple areas, the CoG for the rectangular (dx)(dy) slice should be at x-dx/2 and y-dy/2 but the way I currently understand the slicing and dicing fo the double-integral the CoG is in the upper right corner of the rectangular (dx)(dy) area.

So my question is why does the integration ignore the location of the CoG for the rectangular (dx)(dy) slice?

The picture below represents what I am describing.


16766

grapes
2012-Apr-24, 03:13 PM
I'm not sure, but I think the concept that you're missing is that, in the limit (that is, as dx goes to zero), x = x + dx/2 = x + dx

You have to be careful with that, but it's the general idea. Just evaluate the integral!

ShinAce
2012-Apr-24, 03:23 PM
I originally entered this thread showing the steps to find the center of gravity. You must let either the origin move around, or slide the cone around, until it is balanced. Once it is balanced, then you have information about the center of gravity.

To find volume, we integrate the number 1. To find mass, we integrate the density function. For this thread, that is a constant, which can be pulled out of the integral. So you get that mass equals volume*density.

If you want the center of gravity, I would suggest the approach I took way back. Let some portion of the cone be on the left of the origin and then the remaining portion be on the right of the origin. The volume on the left will not necessarily balance the volume on the right. In this case, it won't at all. The mass on the left of the origin will not balance the mass on the right. What will balance is torque!

Torque = force * radial distance
Force = mass * acceleration

Since acceleration appears in both integrals(let's call them the LHS and RHS expressions) and is constant, we can cancel it out. Mass appears on both sides and is simply volume * density. The density can be pulled out of the integrals and cancels for both LHS and RHS.

The way I had written it was that x on the LHS can go from -A to 0. The RHS integral is evaluated in x from 0 to B. Where A + B = height of the cone.

After simplification, it reads that torque goes as the volume of a (vertical) slice times its distance from the origin. The rest is in post 63.

Being in a physics program, I'm quickly learning that simplification of a problem is generally the key. Even if you can nail down all the elements involved, you're better off knowing how to simplify them as opposed to being able to complete those brute force calculations.

xylophobe
2012-Apr-24, 04:21 PM
...Being in a physics program, I'm quickly learning that simplification of a problem is generally the key. Even if you can nail down all the elements involved, you're better off knowing how to simplify them as opposed to being able to complete those brute force calculations.

Hi,

I usually try to simplify things but then questions get asked about the simplification, whether it is valid or not. Like the "script" that I referred to was a great simplification, I thought, but then others either did not recognize it as such or did not understand why it was done. So, I try to include everything and then simplify from there.

Even the picture I just posted in post 189 (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2011083#post2011083) is a simplification because the mass increases by the square of the radius so the CoG for the rectangular (dx)(dy) slices is actually at y-dy*[1-1/(2)1/2] and this is because the rectangular slices are actually a tiny pie-shaped (the pie that we eat) wedge when viewed looking down the x-axis.

ShinAce
2012-Apr-24, 05:30 PM
Look at how complex those coordinates are in your last image. I'm scared to even try integrating something that complex.

I still don't know what the script is. The thread title is COG of a cone. I integrated myself and got 1/4 from the base along the central axis. Then we moved on to frustrums, which is pretty easy once you have the whole cone fingured out. Then we started from the beginning with integration for volume, and integrating for mass. Obviously, there's much more to it than that, but we're talking about covering 3 university calculus courses here.

If the script is for COG, you need to consider torque. First you imagine some mass in a volume, apply a constant acceleration in one direction, and find the point where no torque arises from that acceleration acting at a point. Mass and volume come from straight integration. COG requires some physics in the sense that symmetry arguments and physical formula are first considered, then you integrate.

As grapes is pointing out, dx/2 = dx . The thickness, dx, that you're considering has some value. The dx that we're thinking about is zero. Otherwise, you get all sorts of problems like you're seeing. When working in xyz, you break the space up into tiny cubes. These cubes are so small that you can make ANY shape out of them, even a sphere. The shape you make doesn't have to have a zigzag for a side. It's smooth curves all around.

When I imagine a slice dx of the cone, I forget that the outsides of the cone are actually slanted. The slice dx is so small, that the difference between a super thin frustrum and a super thin disk is zero. If the difference is zero, then I can use disks with thickness dx and get the same answer. Bam! One problem solved.

Since dx=0, then stuff like Xn-dX/2= Xn

So your slices have x = Xn , y = Yn , distance = sqrt(Xn^2 + Yn^2). Therefore, cosine=x/distance.
Make note this is the right way. You will need cosine in the integral, and you can make your life easy by substituting cosine for x/distance.

But which integral? Mass? Gravitational force on a test particle? Cog? Knowing the gravitational field of the cone won't tell you the Cog, you'll need to put in even more work. Start with the one you actually want.

grapes
2012-Apr-24, 06:19 PM
So your slices have x = Xn , y = Yn , distance = sqrt(Xn^2 + Yn^2). Therefore, cosine=x/distance. You might need an angle associated with that cosine. :)


Make note this is the right way. You will need cosine in the integral, and you can make your life easy by substituting cosine for x/distance.
if you're saying that you need that cosine because you make a trig sub to evaluate the integral, be aware that that is the long way! A^2 = y^2 + x^2 is easier, because then A dA = x dx. Actually, I used A = y^2 + x^2

ShinAce
2012-Apr-24, 06:47 PM
Since when does a trig function need an angle? :p I was just sticking to the format of the last image...

The cosine shows up in the expression for force on a test particle at the origin. The substitution is just to replace cosine with x/distance and work in cartesian form for x and y. The way we've had it setup, the angle being integrated, theta, is from the y to the z axis. Working with a second angle from the x to the y axis is not going to make life easy. That's all that was.

When you say trig substitution, you're making me think of half angle formulas. As above, I mean replacing a cosine with something in x and y. Hence why we get:

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{y x dy dx}{(x^2+y^2)^{3/2}}

Instead of:

\int_{x=0}^h \int_{y=0}^{r x/h} \frac{y cos\phi dy dx}{(x^2+y^2)}

Here, \phi is the angle associated with this infamous cosine.

edit: removed stuff from the integral to make the change easier to see

xylophobe
2012-Apr-26, 04:58 PM
In the picture below in the upper image there are two "dx" sizes shown: one large and one significantly smaller. The large "dx" slice essentially represents the entire "Segment 2" from my previous pictures.

By comparing the two slices it can be noted that the number of cylindrical rings ("dy") increases as "dx" becomes smaller and the error of the cylindrical approximation becomes smaller ... but also the number ("n") of slices to fill the large "dx" area increases.

In math terms as "n" approaches infinity then error/n approaches zero ... but the number of errors approaches infinity.

So then the sum is simply n*error/n which still equals error.

Remembering back to my calculus classes there was a methodology whereby (??extinction??) the solution of an integral can be determined to be an exact solution or an approximation. I think it stated that if the error approached zero faster than the quantity of "dx" slices approached infinity then the integral was considered an "exact" solution.

So when considering the volume part of the double-integral the error gets smaller faster than the quantity of "dx" slices approaches infinity because the error is 2/3 of the cylindrical approximation whereas the conical portion of the solid is only 1/3 of the cylindrical approximation. Looking at the bottom image in the picture if we cut the big "dx" in half then we eliminate a cylindrical ring of error, which is double-crosshatched in the picture. So the volume portion of the double-integral delivers an exact solution.

Looking at the distance from point "P" to the individual CoGs for the dxdy rings - the cylindrical portion produces an error that goes extinct (I think) because the location of the CoG for each ring decreases exactly proportional to the decrease in size (and increase in number) of "dx" slices.

Looking at the CoG for the conical portion the error for the distance to the CoG actually increases as the quantity of "dx" slices increases because it moves from its h/4 position in a pure cone to the h/3 for an infinitely small conical ring. This violates the extinction principle which means the double-integral only produces an approximate solution as opposed to an exact solution.

This means that the different segments of the cone produce a different gravitational force upon point "P".

16775

Grey
2012-Apr-26, 05:30 PM
Remembering back to my calculus classes there was a methodology whereby (??extinction??) the solution of an integral can be determined to be an exact solution or an approximation. I think it stated that if the error approached zero faster than the quantity of "dx" slices approached infinity then the integral was considered an "exact" solution.This is incorrect. An integral always produces an exact solution, not an approximate solution. In some cases, it may not be possible to evaluate an integral analytically. In a case like that, we have to use numerical approximation methods, and then (as the name implies) the result is indeed an approximation. But by increasing the number of slices, we can always make that result as arbitrarily close to the correct value as we wish (and it's easy to quantify exactly what the maximum error can be); there's never any inherent error that can't be avoided, and that never applies when you can calculate the integral analytically. I think you must be misremembering your calculus class.

grapes
2012-Apr-26, 08:05 PM
Looking at the CoG for the conical portion the error for the distance to the CoG actually increases as the quantity of "dx" slices increases because it moves from its h/4 position in a pure cone to the h/3 for an infinitely small conical ring. This violates the extinction principle which means the double-integral only produces an approximate solution as opposed to an exact solution.
It would have to grow a lot farther than that! :)

The double integral is an exact solution.

Jeff Root
2012-Apr-26, 10:50 PM
xylophobe,

I don't understand what you are saying in your last post, but it
is clear from the diagrams that the error is cut in half when the
thickness of the slices is cut in half. Reduce the thickness of
the slices to 1/100th and you reduce the error to 1/100th.
Reduce the thickness to 1/1000000000th and you reduce the
error to 1/1000000000th. Reduce the thickness to zero and
you reduce the error to zero.

-- Jeff, in Minneapolis

ShinAce
2012-Apr-27, 02:26 AM
Sounds a lot like the convergence of series. You set some error, E, and then you can find how large of an 'n' to take in order to get the series to within +/-E of the answer. There's also the integral test, where you can take one integral above and one below the function, and then see that as 'n' goes to infinity, there is convergence. You're also using the squeeze theorem, or sandwich theorem here. But assuming it converges, what does it converge to? You still need to complete the integration to get an answer. The integration gives you an exact answer. E = 0.

If you want to break up the cone into n slices, where n is some actual number instead of infinity, then you will have an error. If you want no error, let n be infinity, therefore, you need to integrate infinitesimals. Break up a cone into an infinite number of infinitesimals and you get, the original cone!

edit: the idea of infinite error terms is not correct. The error term gets adjusted as you increase n. Likewise, the set of pieces you're adding, called a series, is not defined by each piece. It is defined as the sum of all the pieces. So as there are more terms calculated, the sum of the series changes, but remains a single sum. The error, if the series converges, goes to 0 as n increases to infinity. Then the error is just 0 and the sum of the series is the answer.

xylophobe
2012-Apr-30, 12:52 PM
xylophobe,

I don't understand what you are saying in your last post, but it
is clear from the diagrams that the error is cut in half when the
thickness of the slices is cut in half. Reduce the thickness of
the slices to 1/100th and you reduce the error to 1/100th.
Reduce the thickness to 1/1000000000th and you reduce the
error to 1/1000000000th. Reduce the thickness to zero and
you reduce the error to zero.

-- Jeff, in Minneapolis

Hi Jeff,

You are correct with your statement that decreasing, by half, the size of "dx" then the error is cut in half .... but realize that the number of errors also increases to a total of two (2) so

error/2 + error/2 = error

In this new picture you can see that as "n" increases the number of errors increases by "n" so

n*error/n = error

In this new picture I also give the mass numbers (I simplified by setting the mass of the smallest ring (with zero I.D.) in the bottom image to equal 1) that show that by eliminating the double-crosshatched ring the error on the mass calculation of the mass integral decreases faster than "n" grows. When "n" increases to the value 2 then the error eliminated = 56 (168/3) whereas the error remaining = 152/3 and thus more error is removed than remains. So I conclude that the integral for mass is exactly accurate.

The argument that the "dx" goes to zero (as well as the error along the x-axis, which is denoted as dx/2) is a not correct because (for the mass integral) if dx=0 then the mass=0 and the sum of an infinite number of zeros is still zero .... which is also disproved by the formula because it produces an actual mass value. You can never get a 3-D object by adding 2-D objects (zero thickness).

Integrating may produce exact solutions but if the formula that is integrated is wrong then you get a solution that is exactly wrong.

btw: the "extinction" descriptor that I used in my previous post was used by my calculus professor who tried to explain things in multiple ways to help the class learn calculus. It was my bad for not remembering the correct terminology.

16804

xylophobe
2012-Apr-30, 01:29 PM
...
edit: the idea of infinite error terms is not correct. The error term gets adjusted as you increase n. Likewise, the set of pieces you're adding, called a series, is not defined by each piece. It is defined as the sum of all the pieces. ...

Every piece has the error in location of the CoG which produces error in the cosine and the distance. I only show the error for each dx slice but for cylindrical rings the error along the x-axis is the same for every rectangle "piece" in the vertical stack. For the bottom image and the right-most stack there are 7 rectangles (1, 3, 5, 7, 9, 11, 13) and each rectangle has the same x-axis error.

Grey
2012-Apr-30, 01:40 PM
You are correct with your statement that decreasing, by half, the size of "dx" then the error is cut in half .... but realize that the number of errors also increases to a total of two (2) so

error/2 + error/2 = [COLOR=#ff0000]errorNo, this is incorrect. It's not the error per slice that drops by half when you double the number of slices, it's the total error. The error per slice drops by a factor of four. It's reasonable to multiply the error per slice by the number of slices to get the total error. But multiplying the total error by the number of slices to try to calculate the total error, as you're attempting to do here, is just wrong.


The argument that the "dx" goes to zero (as well as the error along the x-axis, which is denoted as [B]dx/2) is a not correct because (for the mass integral) if dx=0 then the mass=0 and the sum of an infinite number of zeros is still zero .... which is also disproved by the formula because it produces an actual mass value. You can never get a 3-D object by adding 2-D objects (zero thickness).It really seems like you just don't understand calculus here. The entire point of integral calculus is precisely that by letting the size of the division approach zero, you can get an exact result. Your argument here, if it were correct, should equally say that you can never get the area under the curve by adding up a series of zero-width areas, but that's one of the most straightforward things an integral can be used to calculate.

ShinAce
2012-Apr-30, 01:54 PM
If dx is not zero, then what is it? If you want to know the length of a line that starts at 2 feet and ends at 5 feet you integrate dx from 2 to 5. If dx is not zero, you can pull it out of the integral. If you do that, you'll get nonsense. The length of any dx section is zero. Then you integrate. The integral of dx is x. Now we apply the limits of 2 and 5 feet. We get 3 feet as the length of the line. If dx had a number attached to it, we can pull it out of the integral and then we won't get 3 feet.

In calc, if you can integrate it, you do. And that's the answer. It's only when you can't integrate that you use numerical methods, which is what you're doing.

Gravity due to a cone? Integrate!

Simultaneous orbits of numerous planets? Numerical methods!

Be advised that integration gives exact answers to exact expressions. Different numerical methods can get to the answer more or less quickly. Since you have to dral with straight sides only, you can easily break it down to get zero error. But then you have to solve pieces of the problem, instead of getting a single right answer.

Jeff Root
2012-Apr-30, 08:00 PM
Since you have to dral with straight sides only, ...
Please fix that typo. I'm going nuts trying to figure out
what you meant!


xylophobe,

Again, your diagram makes it very clear that the total error
is cut in half when the thickness of the slices is cut in half.
Each time you double the number of frustums you reduce the
red hatched area to half that of the preceeding diagram.
The red area disappears as you make the frustums thinner
and thinner. The error is obviously proportional to the the
red area, since the red area is the cause of the error. You're
the one making the diagrams so you should see that better
than anyone else.

-- Jeff, in Minneapolis

profloater
2012-Apr-30, 08:10 PM
I think Newton in his "fluctions" sorted this out so that integration could be rigorous! If gravity is tricky how about the optical properties of a cone as opposed to a flat sided prism with incident parallel light rays?

xylophobe
2012-May-01, 02:31 PM
Please fix that typo. I'm going nuts trying to figure out
what you meant!


xylophobe,

Again, your diagram makes it very clear that the total error
is cut in half when the thickness of the slices is cut in half.
Each time you double the number of frustums you reduce the
red hatched area to half that of the preceeding diagram.
The red area disappears as you make the frustums thinner
and thinner. The error is obviously proportional to the the
red area, since the red area is the cause of the error. You're
the one making the diagrams so you should see that better
than anyone else.

-- Jeff, in Minneapolis


In the below picture I put a center-of-mass symbol into each ring - there are 22 rectangular rings and 4 triangular rings.

In my preceding picture/post (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2012744#post2012744) I stated that as "dx" shrinks to half its size then the number of errors doubles but if you count the total number of rings being created then the quantity increases at an increasing rate because there are only two (2) rings in the upper image (http://www.bautforum.com/attachment.php?attachmentid=16804&d=1335788600): one rectangular and one triangular; but in the middle image (http://www.bautforum.com/attachment.php?attachmentid=16804&d=1335788600) the size of "dx" is halved but the number of rings increases to seven (7) with 5 rectangular rings and two triangular rings; when the size of dx is halved again in the lower image (http://www.bautforum.com/attachment.php?attachmentid=16804&d=1335788600) the number of rings increases to twenty-six (26) with 22 rectangular rings and 4 triangular rings.

Each ring has its own CoG location that is used in the cosine calculation and the distance calculation (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010712#post2010712) and the double-integral introduces an error in that location by placing the CoG into the upper right corner which is comprised of an error along the x-axis and an error along the y-axis (I have only been discussing the x-axis error in order to simplify the conversation).

So, yes, Jeff, I was wrong in my statement because the number of errors in the locations of the CoGs increases faster than the size of "dx" decreases so my error formula should look something like this n(1+x)*error/n =more errors


16810

xylophobe
2012-May-01, 03:46 PM
In the below picture I have drawn the first three "dx" segments.

I am just using the innermost ring (with a zero I.D.) so that the mass can be ignored because the comparison is of the same solid but in different locations. The red angled line is the distance to the CoG used by the double-integral while the green angled line is the correct distance to the CoG for the ring.

I put in numerical values for this example so that others can check it.

The length of "dx" is "1" which could be 1 meter, 1 nanometer, or 1/googleplex (http://en.wikipedia.org/wiki/Googolplex) meter ... the scale does not matter because the picture will be the same (its like fractal (http://en.wikipedia.org/wiki/Fractal) geometry).

Since the double-integral uses incorrect CoG locations in its calculation and because this only affects the cosine and distance calculation (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010712#post2010712) I show just the calcs for those two parts.

I have also included calculations for the dx2, dx3, dx4, dx5 to show how the error decreases as "x" increases.

The point that this shows is that the error is real and that it adds up. I was able to arrive at the correct solution for the location of the CoG for a cone (h/4) using this type of "exhaustive" approach so I know that it works.

btw: I cheated and just used 1/2 the radial size as the y-axis location of the CoG (a simplification).

16811

ShinAce
2012-May-01, 05:03 PM
Was it Pauli that said "Not only is it not right, it's not even wrong!".

It looks like you're saying the height of the frustrum is dx. Then you divide that to get? ddx? What the rest of us have been doing is to set the height along the x axis. The height has some difference in x value associated with it. Therefore, the height of the frustrum is x, not dx. Dx is a vertical slice of the frustrum. When you integrate, dx=0. The integral of dx is x. Again, you can't just put a number on dx. That's why it's a symbol and handled with its own rules, like infinity. You can't scale numbers to make infinity=1.

Look at your diagram in post 207. For n=2, the two errors are the areas marked(11, 17/3, 14/3) and (15, 20/3, 23/3).

Now look at n=4. The errors are the areas marked (14/3), (17/3), (20/3), and (23/3).

The number of error terms = n. But all the errors for n=4 are seen included in the errors for n=2. Actually, we can clearly see that by going to n=4, the errors marked 11 and 15 are no longer a part of the total error. Everytime we increase n, the new errors are a smaller part of the old error. We can also easily prove that the error when n doubles is half. That's a fancy way of saying that the triangles marked 14/3 and 17/3 can be rearranged to exactly match the triangle marked 11. The triangles that have parallel lines add up to the same area as the cross hatched square. How can that possibly be seen as the error getting bigger? The error going from n=2 to n=4 has lost the cross hatched area. That is not more error, it's less error. You err in your analysis of error.

So now, when n=1, the error is as large as the calculated value. But everytime we double n, half the error diappears while the calculated value stays the same. Or, error when n=1 is 50%. Error when n=2 is 25%. N=4 gives 12.5%. N=8 gives 6.25%. N=16 gives 3.125% and so on. Which means the amount of error in the total calculated value is 1/2n.

If you want an answer that is good to 1%, then 0.01 = 1/2n. Which gives 0.02n = 1, yielding n=50. I don't know about you, but I'm not interested in repeating the same calculation 50 times only to get within 1% of the answer. I expect better accuracy than that if I'm going to resort to brute force.

If you want zero error, it's pretty obvious that you need to take n=infinity, which means dx=0. You need to integrate!


The point that this shows is that the error is real and that it adds up. I was able to arrive at the correct solution for the location of the CoG for a cone (h/4) using this type of "exhaustive" approach so I know that it works.

When was this? You started this thread because you came up with 0.2063 of height from base while wiki said h/4. I used a two step solution and came up with h/4(btw, error=0). I also realized the same approach can be used for cones or frustrums with any symmetric base without resorting to numerical methods.

You're using numerical methods when you don't have to. You're also coming up with easily falsifiable conclusions like
Since the double-integral uses incorrect CoG locations in its calculation.... What does that mean? The integration doesn't use a COG. It uses things like volume, density, and forces(gravity) to deduce the COG.

Say it with me now! "Not only is it not right, it's not even wrong!". That is not how calculus works, which includes the study of these types of numerical methods.

xylophobe
2012-May-01, 06:22 PM
...

It looks like you're saying the height of the frustrum is dx. Then you divide that to get? ddx? What the rest of us have been doing is to set the height along the x axis. The height has some difference in x value associated with it. Therefore, the height of the frustrum is x, not dx. Dx is a vertical slice of the frustrum. When you integrate, dx=0. The integral of dx is x. Again, you can't just put a number on dx. That's why it's a symbol and handled with its own rules, like infinity. You can't scale numbers to make infinity=1.

...

When we integrate it is applied over an interval and for a cone the limits along the x-axis are x=0 and x=h

As "x" advances from 0 to "h" it goes by increments called "dx" so the first increment is a cone with a height of "dx". All the following increments produce frustums with heights of "dx". This continues until all the increments add up to "h". This is the vertical slicing that professors refer to when discussing integration.

The double-integral also slices horizontal slices which are called "dy" which also happens to equal the "dr" for the very first horizontal slice. So the double-integral cuts the solid into rings which supposedly get progressively thinner along the x-axis and the y-axis until these dimensions are zero length.


...

Look at your diagram in post 207. For n=2, the two errors are the areas marked(11, 17/3, 14/3) and (15, 20/3, 23/3).

Now look at n=4. The errors are the areas marked (14/3), (17/3), (20/3), and (23/3).

The number of error terms = n. But all the errors for n=4 are seen included in the errors for n=2. Actually, we can clearly see that by going to n=4, the errors marked 11 and 15 are no longer a part of the total error. Everytime we increase n, the new errors are a smaller part of the old error. We can also easily prove that the error when n doubles is half. That's a fancy way of saying that the triangles marked 14/3 and 17/3 can be rearranged to exactly match the triangle marked 11. The triangles that have parallel lines add up to the same area as the cross hatched square. How can that possibly be seen as the error getting bigger? The error going from n=2 to n=4 has lost the cross hatched area. That is not more error, it's less error. You err in your analysis of error.
...


You need to carefully read the text that describes these images because I plainly stated that the double-crosshatched areas are eliminated from the solid and represent mass/volume error reduction.

As far as the rest of your post you need to carefully study how the double-integral calculates the gravitational influence upon point "P" because it is composed of a cosine, a distance, and a volume/mass.

xylophobe
2012-May-01, 06:45 PM
I might add that readers should study limits because in every discussion I have read it plainly states something to the effect of "as "x" approaches zero" . When dx=0 it results in a nonsense solution; for instance, the limit as x approaches zero of 1/x is infinity. Mathematics plainly states that when x=0 it is nonsense or undefined.

For the double-integral if dx=0 then there is no volume because volume is a 3-D characteristic of mass and mass produces an effect called gravity.** If dx=0 then there is no volume and there is no mass and there is no gravity.


** - According to Einstein's E=mc2 energy (which according to classical physics does not occupy volume) also produces an effect called gravity.

ShinAce
2012-May-01, 06:47 PM
The double-integral also slices horizontal slices which are called "dy" which also happens to equal the "dr" for the very first horizontal slice. So the double-integral cuts the solid into rings which supposedly get progressively thinner along the x-axis and the y-axis until these dimensions are zero length.

Nope! A slice, dx, is a disk. A slice of that, dy, is a line. A slice of that, dz, is a point. You are not working in cylindrical coordinates with the full shape. Abstracting a cone as a triangle and breaking it up into dx and dy is not the same as what we've done. By the way, I've done the integration, in cylindrical coordinates. It just so happens that there are convenient subsitutions from cylindrical form to make it look cartesian. I don't appreciate being told I need to carefully study how double integrals yields answers. I just got my grade back for multivariable calculus. I got an A. That would imply I did study it, and carefully. Every slice, dx, is a disk. It has straight sides! It cannot be a cone. If it's a cone, it's not an infinitesimal, and you cannot use the notation dx to properly identify its thickness.

Do you even realize what you are calculating? The gravitational influence? That's a force, and involves two masses. What is the mass at point P being influenced? Make it zero and the influence is zero. As you give it a mass, the force calculated grows. If you want acceleration due to gravity at point P, then the value you're trying to find is (force/test mass). That's a field, not an influence.

What I need to do is spend more time in discussions that progress forward.

xylophobe
2012-May-01, 06:57 PM
Nope! A slice, dx, is a disk. A slice of that, dy, is a line. A slice of that, dz, is a point. You are not working in cylindrical coordinates with the full shape. Abstracting a cone as a triangle and breaking it up into dx and dy is not the same as what we've done. By the way, I've done the integration, in cylindrical coordinates. It just so happens that there are convenient subsitutions from cylindrical form to make it look cartesian. I don't appreciate being told I need to carefully study how double integrals yields answers. I just got my grade back for multivariable calculus. I got an A. That would imply I did study it, and carefully. Every slice, dx, is a disk. It has straight sides! It cannot be a cone. If it's a cone, it's not an infinitesimal, and you cannot use the notation dx to properly identify its thickness.

Do you even realize what you are calculating? The gravitational influence? That's a force, and involves two masses. What is the mass at point P being influenced? Make it zero and the influence is zero. As you give it a mass, the force calculated grows. If you want acceleration due to gravity at point P, then the value you're trying to find is (force/test mass). That's a field, not an influence.

What I need to do is spend more time in discussions that progress forward.

I thought it was understood that point "P" represented a point mass.

Sorry.

ShinAce
2012-May-01, 07:18 PM
Point P is a coordinate. The force on a test mass at P is arbitrary. You can make it any number you want by changing the test mass. Are you trying to find the gradient of the gravitational potential at point P? If you use the multivariable calculus of vector fields, then you can get that answer, exactly. Set the gradient to zero and you find the COG, but I imagine you already knew that since you correctly derived the answer. You didn't even have to use calculus and physical principles.

xylophobe
2012-May-01, 07:34 PM
Limit theory states that as a variable approaches zero (0) a function approaches a value and this means that neither "dx" nor "dy" nor "d-theta" ever actually equals zero (0) which means that these three variables define a volume and for our discussion volume=mass which also means that somewhere inside that volume/mass is a CoG.

Because neither "dx" nor "dy" ever equals zero then that means that dx/2 and dy/2 have a non-zero value that affects the outcome of the calculation.

The calculation involves a cosine, a distance, and a volume.

Readers need to ask themselves: what are the endpoints of the distance? One end point is point "P" and the other end point is the CoG of the infinitesimally small mass bounded by "dx" and "dy" and "d-theta"

The hypotenuse of the cosine is that distance so it falls under the same scrutiny.

grapes
2012-May-01, 08:12 PM
Because neither "dx" nor "dy" ever equals zero then that means that dx/2 and dy/2 have a non-zero value that affects the outcome of the calculation.
The integral is a limit itself, as dx goes to zero. As dx goes to zero, the limit of dx/2 is also zero. In the limit.


The calculation involves a cosine, a distance, and a volume.

Readers need to ask themselves: what are the endpoints of the distance? One end point is point "P" and the other end point is the CoG of the infinitesimally small mass bounded by "dx" and "dy" and "d-theta"

The hypotenuse of the cosine is that distance so it falls under the same scrutiny.This thread is in a question and answer forum of BAUT. The original question has been answered, correct?

ShinAce
2012-May-01, 08:20 PM
Limit theory states that as a variable approaches zero (0) a function approaches a value and this means that neither "dx" nor "dy" nor "d-theta" ever actually equals zero (0) which means that these three variables define a volume and for our discussion volume=mass which also means that somewhere inside that volume/mass is a CoG.

Because neither "dx" nor "dy" ever equals zero then that means that dx/2 and dy/2 have a non-zero value that affects the outcome of the calculation.

The calculation involves a cosine, a distance, and a volume.

Readers need to ask themselves: what are the endpoints of the distance? One end point is point "P" and the other end point is the CoG of the infinitesimally small mass bounded by "dx" and "dy" and "d-theta"

The hypotenuse of the cosine is that distance so it falls under the same scrutiny.

Let's not get into a discussion of limits. I'm sure I already mentioned that Newton had the concept of a limit without a rigorous proof of it. The proper definition came much later.

If you're right, then the limit as x -> 0 of sin(x)/x is actually less than 1, since it can never be exactly 1. It would also mean that for a square wave, the fourier series does not converge for all values of x. However, it's not only accepted, but taught at the university level, that this is incorrect.

The limit as x approaches 0 means that if I could plug in 0, that's the answer I would expect. The answer does not need to be defined or real. As long as the function on both sides of x points to the same spot, that's the limit. This limit is not dx. Taking a very small difference, dx, relates to infinitesimals. Infinitesimals are one thing. Limits are another.

I fear that readers are asking themselves what this point P has to do with the center of gravity. They may also be asking themselves why infinitesimals have their own center of gravity. Are we back to the confusion about using the center of mass in gravitational calculations? You're not going to find the COG of a frustrum by manipulating the center of mass of several slices with respect to the origin. You'll either need to move the origin, or move the frustrum.

xylophobe
2012-May-01, 09:12 PM
The integral is a limit itself, as dx goes to zero. As dx goes to zero, the limit of dx/2 is also zero. In the limit.
This thread is in a question and answer forum of BAUT. The original question has been answered, correct?

It is also true that the ratio of (dx/2)/dx = 0.50 even as dx approaches zero, correct? If as dx approaches zero the value of dx/2 approaches the value of dx then the limit of (dx/2)/dx would equal 1.0

This thread was opened to talk about cones.

ShinAce
2012-May-01, 09:35 PM
Then that answers the question of intent. If you don't intend to ask a question, then the thread may be moved to S&T.

We don't need to reinvent differential calculus to talk about cones. You are the only person I've ever seen try to define dx/2.

Jeff Root
2012-May-02, 09:55 AM
Grapes, ShinAce,

This thread was apparently started for xylophobe to understand
how to determine the location of the center of gravity of a cone.
He's still working on that.

Meanwhile, I've learned that the center of gravity of a body is not
(necessarily) the same as its center of mass. Even though I was
already familiar with the idea (my "Which way is down?" web page
depends on it), I didn't realize it. Now I'm getting the equivalent
of the first week of calculus 1001 -- and understanding it!

-- Jeff, in Minneapolis

grapes
2012-May-02, 10:46 AM
I think I've learned something from almost every thread I've ever participated in. The OP asked whether the conventional formula for center of mass for a cone was correct. We've established that it is, and everybody seems to agree on that. However, the bulk of the discussion has been about misconceptions about calculus--and so the thread does not fit into Q&A or S&T. I'm moving it to OTB, we can continue the discussion there.

Jeff Root
2012-May-02, 12:16 PM
Nope! A slice, dx, is a disk. A slice of that, dy, is a line. A slice
of that, dz, is a point.
That isn't at all what xylophobe is doing, and it isn't relevant.

Xylophobe is cutting the cone into frustums of equal height,
and cutting the frustums into cylindrical tubes of equal wall
thickness.

There is no need for him to do that, but it is the method
he has been using since at least post #117, and apparently
you only just became aware of it.

Back in post #150, I said "a hole appears to be a feature of
xylophobe's analysis of frustums". You replied in post #151
"This thread is about solid objects." Apparently you did not
realize he was cutting the cone into cylindrical tubes in order
to analyze it.

Also, all of xylophobe's analysis has been with slices of
finite thickness. Your recent posts seem to assume that the
slices always have zero thickness. You didn't start out that
way. This is what you proposed in post #115:



Imagine staring down at a small pizza. Under it is a medium
and under that, a large. If they're lined up eith matching edges,
like an eclipse, then the force of gravity from any one pizza on
you is a constant. Same thickness, different radius.
At some point you seem to have forgotten that the slices
have thickness and started assuming that they have zero
thickness.

So you and xylophobe have been discussing different things.

-- Jeff, in Minneapolis

Jeff Root
2012-May-02, 12:18 PM
In the below picture I put a center-of-mass symbol into each
ring - there are 22 rectangular rings and 4 triangular rings.
...

In my preceding picture/post I stated that as "dx" shrinks to
half its size then the number of errors doubles but if you
count the total number of rings being created then the
quantity increases at an increasing rate because ...

So, yes, Jeff, I was wrong in my statement because the
number of errors in the locations of the CoGs increases faster
than the size of "dx" decreases so my error formula should
look something like this n(1+x)*error/n =more errors
First, do I understand correctly that you now believe the error
*increases* when you cut the cone into smaller, more numerous
frustums?

If so, I see why you believe that, although ShinAce doesn't seem
to have caught on yet.

This is way, way, way simpler than you think it is.

No exaggeration.

As ShinAce suggests, you don't need to divide the cone with
cuts parallel to its axis. That doesn't accomplish anything useful
to you. Just dividing it into frustums will suffice.

ShinAce apparently only just realized you are doing that. I've
been waiting for somebody to ask you why you are doing it.

Now...

What gives you the idea that the error is "multiplied"? Where
in the calculation of gravitational field strength (force per unit
mass) of the cone on point P do you multiply the error in
each constituent frustum by the number of frustums?

The answer is: You do not multiply the error by the number
of frustums.

Each frustum has an error associated with it, represented
by the red hatched area in your diagrams. When you cut
the frustums in half, the error of each new frustum is 1/4
the error of the previous frustum. Since there are twice as
many 1/4-size frustums as full-size ones, the resulting total
error is 1/2 that of the previous iteration.

When you find the gravitational field strength of a frustum
on point P, you use the distance of the center of mass of
the frustum from point P, as well as the volume of the
frustum. Your neat little diagrams clearly show that
when you divide a frustum in two, the red hatched area
associated with each resulting half-thickness frustum is
only 1/4 that of the parent frustum. So the total error
volume of the two baby frustums is only 1/2 that of the
parent frustum.

This is really, really obvious. So obvious that I am having
a hard time finding a way to explain it. I can understand
your mistake in thinking that the errors multiply, but you
are treating the resulting half-size errors as if they were
full-size errors. When you cut a frustum in two, each part
has less gravitational effect on point P, because the volume
is less. Likewise, each error has less effect, for the same
reason.



... the double-integral introduces an error in that location
by placing the CoG into the upper right corner ...
What do you mean by "the upper right corner"? Where
do you get the idea that that is where the center of mass
is placed, by what operation?

-- Jeff, in Minneapolis

ShinAce
2012-May-02, 01:00 PM
Jeff, a disk is a cylinder. An ideal one. A pizza is shaped like a disk. In calculus terms, a thick one. They're all the same shape. Pipes, disks, cylinders.

The confusion seems to be the outside edge of this frustrum. If we divide it with a finite number of pieces, we have those 'left over' triangular rings. With calculus, you are dividing into an infinite number of zero thickness disks, so there are no ledtovers. The entire cone is covered by disks alone.

Now we are regressing to the abstraction of infinitesimals that you would see in calc1001. For me, it was calc 1002 that covered numerical methods, or numerical integration. We didn't see cylindrical coordinates and 3d integration until calc2001. The integration is best understood after calc2001, but can be done during first year calc.

Why would you want to divide a frustrum into a bunch of 'straws'? That's beyond calc 2001, where we integrate over different types of regions and going into different types of infinitesimals. Besides, the integration grapes and i did starts eith a point, integrated into a ring. The rings are integrated into disks, and the disks are integrated into a cone. No cubes!

Jeff Root
2012-May-02, 01:04 PM
xylophobe,

You have a cone which imposes a gravitational field on point P.
Call the value of the gravitational field strength at point P "1".

As you divide the cone into thinner and thinner frustums, the
gravitational field strength at point P imposed by each frustum
approaches zero. The error in your analysis also approaches
zero.

When you reach the limit, each frustum has zero thickness
and imposes zero gravitational field strength on point P. But
calculus was developed to handle those zeros. As ShinAce
said, that is not how calculus works. It is part of what makes
calculus distinct from the methods you are trying to use. At
all times the total gravitational field strength at point P of all
the frustums which comprise the cone remains 1. It never
changes, whether the cone is a single piece, two pieces, a
trillion pieces, or infinitely many pieces. Your method, using
a trillion pieces, will have an extremely tiny error. Using
calculus, the error is zero.

-- Jeff, in Minneapolis

Jeff Root
2012-May-02, 01:24 PM
Jeff, a disk is a cylinder. An ideal one.
I agree that a disk is a cylinder which tends to be
relativly thin compared to its diameter.

Whether either term refers to an ideal is a separate
question which must be specified separately. It can
usually be understood from the context. Pizzas are
generally not ideal. Canadian bacon helps, though.



A pizza is shaped like a disk.
That's what I said, way back, and you replied with stuff
that seemed non sequiterish.



In calculus terms, a thick one. They're all the same
shape. Pipes, disks, cylinders.
Whoa! A pipe has a hole running through its length.
A pizza should not.

A disk-shaped pipe (or possibly a pipe-shaped disk)
would be an annulus. A disk with a hole.



The confusion seems to be the outside edge of this
frustrum. If we divide it with a finite number of pieces,
we have those 'left over' triangular rings.
Yes, "rings" is the word xylophobe used...



Why would you want to divide a frustrum into a bunch
of 'straws'?
Not straws, actually, but rings. Apparently it seemed like a
good idea to xylophobe at the time. Divide something into
smaller bits, and the analysis should be more accurate.
That seems to have blown up on him.

-- Jeff, in Minneapolis

ShinAce
2012-May-02, 01:43 PM
** - According to Einstein's E=mc2 energy (which according to classical physics does not occupy volume) also produces an effect called gravity.

Watch this! Take a particle with rest mass, m. Then E=mc2. Now, we want to apply the Lorentz transform as a series.

Let's call the transform gamma = 1/(sqrt(1 - v2/c2)). Gamma can be represented as a binomial series (1+x)k = 1 + kx + k(k-1)x2/2! + ... where x=v/c.

Now we expand mc2/(sqrt(1 - x2)) using the binomial series. Which means k= -1/2 but be careful how you plug in x.

Using this numerical approach, we find that E(for that one particle m) = mc2 + (1/2)mv2 + (3/8)mv4/c2 + more terms. Each additional term gets smaller and smaller at low relativistic speeds.

Isn't that neat? Applying gamma to mc2 and expanding into a series gives us the rest mass as the first term(phew! it's not lost), the kinetic energy as the second term, some weird looking terms after that(relativistic terms).

Hmm, if the particle's rest mass produces gravity, then the particle's kinetic energy also adds to gravity. It's getting a little more complicated than you expected now, isn't it?

Jeff: In response to your lightning fast reply(within a few minutes of my earlier post)...Indeed!!! Knowing that each slice has an equal contribution, I would take a very thin slice at the base(since it's the furthest from the origin) and calculate from that one slice. Multiply that answer by the number of slices it would take to make up the full height of the cylinder. Done! It's quick, it's dirty, but it will agree with the previous answers. We're talking about cutting the cone into 100 slices, calculating the force per unit test mass from a single slice, and multiplying by 100. But that would require confidence in the principles of calculus.

edit: Jeff, I think you know full well that the pipe needs to be solid. I'm not comparing plumbing parts to pizzas, ok? Take a circle, make it into a cylinder by giving it height. What is it? A disk? A cylinder? A puck? Pizza shaped? A thick record? They're all different names for the same shape. If we choose different names based on the ratio of radius to height, we are not doing ourselves any favors.

xylophobe
2012-May-02, 02:36 PM
...

Hmm, if the particle's rest mass produces gravity, then the particle's kinetic energy also adds to gravity. It's getting a little more complicated than you expected now, isn't it?

...

Maybe I am reading the "tone" of your replies wrong but you seem to be treating me like an idiot which seems like a violation of forum rules. It does not bother me but I wonder if the moderators care or not.

Anyway, I have know from previous readings on this site that kinetic energy adds to gravity which is why I stated that energy has a gravitational influence: photons and other EM radiation.

Thanks for the reply.

xylophobe
2012-May-02, 02:40 PM
I think I've learned something from almost every thread I've ever participated in. The OP asked whether the conventional formula for center of mass for a cone was correct. We've established that it is, and everybody seems to agree on that. However, the bulk of the discussion has been about misconceptions about calculus--and so the thread does not fit into Q&A or S&T. I'm moving it to OTB, we can continue the discussion there.

I have learned a lot from this thread and I have thought of listing what I have learned but then that would probably start arguments, too, so I hesitate.

Maybe I missed it, but why wasn't a link added to the Q&A section to direct me here?

Thanks.

ShinAce
2012-May-02, 03:25 PM
Maybe I am reading the "tone" of your replies wrong but you seem to be treating me like an idiot which seems like a violation of forum rules. It does not bother me but I wonder if the moderators care or not.

Anyway, I have know from previous readings on this site that kinetic energy adds to gravity which is why I stated that energy has a gravitational influence: photons and other EM radiation.

Thanks for the reply.

I'm not treating you like an idiot. I'm adding to what you said:


I might add that readers should study limits because in every discussion I have read it plainly states something to the effect of "as "x" approaches zero" . When dx=0 it results in a nonsense solution; for instance, the limit as x approaches zero of 1/x is infinity. Mathematics plainly states that when x=0 it is nonsense or undefined.

For the double-integral if dx=0 then there is no volume because volume is a 3-D characteristic of mass and mass produces an effect called gravity.** If dx=0 then there is no volume and there is no mass and there is no gravity.


** - According to Einstein's E=mc2 energy (which according to classical physics does not occupy volume) also produces an effect called gravity.

You clearly wrote that E produces an effect called gravity. I clearly wrote an interesting use of the binomial expansion to show that E includes things like kinetic energy and relativistic corrections when you consider Lorentz invariance. It's a neat application of a series. You've been working with series for a while now.

I'm also addressing your misconceptions about infinitesimals. You ask readers to study limits yet you misuse them. Again, you're working with finite series. It's a crude form of integration, but does not invalidate integral calculus. A major part of my last calculus course was geometry. A point, a line, a vector, a plane, distances between these things, their intersections. We studied it in different ways(cartesian coordinates, polar and cylindrical coordinates) and expressed it in diffferent forms(parametric, symmetric, etc...). We studied arc length, rotated surfaces and their inclosed volumes. We also studied conical sections. Like how cutting a cone at different places can give you: a point, two intersecting lines, a circle, an ellipse, a parabola, a hyperbola.

Everything you need is contained within my first three university calc courses. We saw how to treat error in numerical approaches, how to define various limits, and how to integrate plenty of 3d shapes. We even saw how to find the function's maximum and minimum in 3d. Thank you Lagrange multipliers! So now I can start with one integral, and I get volume. Slap in a density function, I get mass. Add Newton's expression for gravity, and I can find force on a test particle. It all starts with the same integral bounds. To me, it's the same integral.

Another cool thing you see is potential functions. Gravity is a conservative force, which means that the work done by moving around the potential function is independend of the path. So using the same calculus I learned, there's a way to work with potentials directly, or forces directly. You give me a formula that tells me the force at every point in space, and I'll find you the potential at every point. I'll also find where the potential is the lowest, which is the center of gravity.

Yet you've said:


You can never get a 3-D object by adding 2-D objects (zero thickness).

Integrating may produce exact solutions but if the formula that is integrated is wrong then you get a solution that is exactly wrong.

I get 3D objects by adding an infinite number of 2D objects everyday! I use exact formulas to get exact integrals everyday! Are you saying that the expression being integrated is wrong? Can you show this? You're using the same expression for gravitational force, that of Newton. If you show that this expression is wrong, you're showing that your numerical integration is wrong as well.

Calculus gives us several methods to calculate the same thing. The fact that all of us who have done the integration did the same thing tells me this one particular method is the easiest.

xylophobe
2012-May-02, 03:44 PM
I'm not treating you like an idiot. ...

Sorry for the mis-interpretation. I used to be on another forum that had no rules or moderators and it was a real free-for-all. People were always misinterpreting text ... I had a lot of fun there ... until the forum host shut it down. I never got into the habit of using smilies and mood icons so people tend to treat me like a grump but I can take a lot of abuse before it begins to bother me.

Anyhow, I had taken all the courses you speak of but it was a long time ago. I, mostly, do not remember all the tricks for solving integrals which is why I have not written out the complete solution to the double-integral. A week or so ago I dug out my calculus textbook and leafed through it quickly but things are hectic in my life right now so I just think about things in my spare time (like when I am driving).

ShinAce
2012-May-02, 03:49 PM
If you want to pursue a numerical approach to integration, try:
http://digitool.library.mcgill.ca%2Fthesisfile47867.pdf&ei=oVShT8qkNMvG6AHDzN39CA&usg=AFQjCNEn1gzVXfhE_FHRUXSZ79vWrkb5sg

If you still have your textbooks, look over multiple integrals and find the applications in there. Cone volume and center of mass are probably in there as examples.

There's also http://reference.wolfram.com/mathematica/tutorial/NIntegrateIntroduction.html.

If your life is so hectic, why are you spending this much time on a center of gravity problem?

xylophobe
2012-May-02, 04:28 PM
...Yes, "rings" is the word xylophobe used... -- Jeff, in Minneapolis

I use the term "ring" because it is the most descriptive. Essentially when I split a frustum up into increasing numbers of "dx" slices (http://www.bautforum.com/attachment.php?attachmentid=16804&d=1335788600) the number of rings increases faster than the number of "dx" slices. In this picture (http://www.bautforum.com/attachment.php?attachmentid=16804&d=1335788600) I start, in the upper image, with a single "dx" slice which encompasses the entire frustum and then I progress down to the middle image which divides the original frustum into two "dx" slices which I repeat for the bottom image --- the number of slices increases by a factor of 2 whereas the number of rings increases thusly:

1 slice ==> 2 rings = (1+1) = 12+T1
2 slices ==> 7 rings = (2+1)+(2+2)= 22+T2
4 slices ==> 26 rings = (4+1)+(4+2)+(4+3)+(4+4)= 42+T4
8 slices ===> 100 rings = (8+1)+(8+2)+(8+3)+(8+4)+(8+5)+(8+6)+(8+7)+(8+8) = 82+T8

Tn = Triangular number (http://en.wikipedia.org/wiki/Triangular_number) (the sum of numbers 1+2+3+4+...+n)

The integral does not actually cut the frustum into rings but rather pie-chunks of the rings which are dθ in angular size --- see below picture.

So the CoG for these pie-shaped chunks is inside the chunk which is why there is the cosine portion (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010712#post2010712) (post 179) in the formula of the double-integral. The cosine is there to account for the offset mass ... a full ring has a CoG that is in the center of the ring and directly on the x-axis and does not necessitate the use of cosine.

Because the force along the z-axis is canceled by the opposing segment then the fz can be ignored just like the force along the y-axis is also canceled by the opposing segment so the fy can be ignored but fx is calculated by the cosine.

Because fy and fz cancel then really all we have to talk about is the rings.

It is an actual force because the gravitational constant and density converts the units to force units (N).


16815

Jeff Root
2012-May-02, 04:35 PM
edit: Jeff, I think you know full well that the pipe needs to be
solid. I'm not comparing plumbing parts to pizzas, ok? Take
a circle, make it into a cylinder by giving it height. What is it?
A disk? A cylinder? A puck? Pizza shaped? A thick record?
They're all different names for the same shape. If we choose
different names based on the ratio of radius to height, we
are not doing ourselves any favors.
A pipe has a hole in it, by definition. Phonograph records
also always have a hole, in my experience. A hole may be
required by an industry standard, but it really wouldn't be
required by definition.

A disk is a cylinder which is relatively thin compared to its
diameter. A phonograph record is a disk, ignoring the detail
of the actual shape of the edge, which is angled, not square.

A pizza is also a disk, ignoring the detail of the actual shape
of the edge, which is just about anything, and the top surface,
which should be as lumpy as possible.

A hocky puck is close to the limit of where you might want to
stop using the term "disk". But it definitely isn't there.

A "drum shape" is a cylinder which is thicker than a disk.
Of course, some drums don't have parallel sides. I guess
those drums aren't "drum-shaped".

A rod is a cylinder which is relatively long compared to its
diameter. Of course, rods don't necessarily have circular
cross-section or constant diameter, either. But typically...

A tube is the same thing as a pipe, except that pipes are
virtually always rigid, while tubes can be either rigid or
flexible, like hoses. Like rods, they aren't required to have
circular cross-section. And both pipes and tubes can be
bent, whether they are flexible or not.

A ring is a short section of pipe or tube.

Different terms for different things.

-- Jeff, in Minneapolis

ShinAce
2012-May-02, 05:29 PM
So the CoG for these pie-shaped chunks is inside the chunk which is why there is the cosine portion (http://www.bautforum.com/showthread.php/129554-Cones-Center-of-Gravity-Center-of-Mass-etc.?p=2010712#post2010712) (post 179) in the formula of the double-integral. The cosine is there to account for the offset mass ... a full ring has a CoG that is in the center of the ring and directly on the x-axis and does not necessitate the use of cosine.

Because the force along the z-axis is canceled by the opposing segment then the fz can be ignored just like the force along the y-axis is also canceled by the opposing segment so the fy can be ignored but fx is calculated by the cosine.


Then we're not talking about the same cosine.

I'm saying that if you cut up the cone into many many rings, then the cosine appears when calculating the force from a single ring. The distance used in Newton's law of gravity is the distance from the origin to any point on the ring. All points on the ring have the same distance to the origin, so that's cool. Like you're saying, the forces from each point on the ring cancel in the y and z direction. However, the strength of the gravitational force from each ring along the x axis is:
F= cosine(of the angle between x axis and the distance line) * GMm/distance2
G is gravitational constant
M is mass of the ring
m is the test mass
distance as defined above

Start with:
F=GMm/r2, the formula for two particles
Now notice that each point on a ring has its force vector pointing exactly cosine(as defined above) away from the x-axis. Since we already know it cancels in the y and z direction, we want to know how much of that total force vector is pointing in the x direction. From trig, we have Fx = cosine*F.
Therefore,
Fring = cosine(of the angle between x axis and the distance line) * GMm/distance2

This way, the radius, or distance, is easily defined. It's actually sqrt(x2+y2) for that ring. Cosine is x/distance .

Rearrange and you have
Fring = cosine(of the angle between x axis and the distance line) * G * ring mass * test mass/distance2
Fring = (x/distance * G * ring mass * test mass) / x2+y2

But distance = sqrt(x2+y2)
Fring = x * G * ring mass * test mass / (x2+y2)2/2 + 1/2
Fring = x * G * ring mass * test mass / (x2+y2)3/2

Done! Cosine is gone! Distance is gone! We have x and y, which are the x and y coordinates for that ring.

Now you can choose if that ring has a thickness, delta(x) and delta(y), and use numerical methods. Do what you're doing and use the x and y values at the center of each 'square' cross section of the ring for approximation. For the triangles, I would personally pretend that they're full squares and use half of that force value. It's not perfect, but that's why it's called approximation.

Or you can give it an infinitesimal thickness, dxdy, and integrate.

grapes
2012-May-02, 06:28 PM
You give me a formula that tells me the force at every point in space, and I'll find you the potential at every point. I'll also find where the potential is the lowest, which is the center of gravity.
I don't think that is the same center of gravity under discussion in this thread. For two identical spherical masses, the lowest potentials would be on their surfaces, but their center of mass/center of gravity would be halfway between them.


I get 3D objects by adding an infinite number of 2D objects everyday! I use exact formulas to get exact integrals everyday!
Probably, you are adding up infinitessimal 3D objects to get 3D objects.

grapes
2012-May-02, 06:43 PM
Maybe I am reading the "tone" of your replies wrong but you seem to be treating me like an idiot which seems like a violation of forum rules. It does not bother me but I wonder if the moderators care or not.
I didn't notice the tone when I first read it, and I see it's been resolved in your subsequent posts. That's why we encourage you to report posts that you think might be offending, rather than respond in kind in the thread.




Maybe I missed it, but why wasn't a link added to the Q&A section to direct me here?
That was my fault. I never set a re-direct, but I could change if we/I get enough complaints.

ShinAce
2012-May-02, 08:50 PM
I don't think that is the same center of gravity under discussion in this thread. For two identical spherical masses, the lowest potentials would be on their surfaces, but their center of mass/center of gravity would be halfway between them.
Probably, you are adding up infinitessimal 3D objects to get 3D objects.

Wouldn't that be for a spherical shell?

grapes
2012-May-02, 08:58 PM
I don't think that is the same center of gravity under discussion in this thread. For two identical spherical masses, the lowest potentials would be on their surfaces, but their center of mass/center of gravity would be halfway between them.
Probably, you are adding up infinitessimal 3D objects to get 3D objects.

Wouldn't that be for a spherical shell?
What's that? :)

xylophobe
2012-May-02, 09:17 PM
I do not want readers to get the wrong perception (whether by me or someone else) of how integration works so I will direct to:

"By the logic above, a change in x (or Δx) is the sum of the infinitesimal changes dx. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Physical_intuition ) ..."

The above text and other readings on the Fundamental theorem of calculus (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus) seem to agree with my description of how the value of "x" advances by increments of "dx".

Method of exhaustion (http://en.wikipedia.org/wiki/Method_of_exhaustion).

ShinAce
2012-May-02, 10:37 PM
Grapes: looks like we both fumbled that potential function. :)

Xylo: be advised that delta(x) refers to a certain number of slices. Strictly speaking, dx is an infinitesimal. There is not a certain number of dx slices. Dx is used in integration while delta(x) is for numerical approaches. They're not handled the same way.

That quote even says that we sum an infinite number of dx slices to get the integral. Your method can never get to infinite slices. You'll run out of computer processing power and memory trying to get an exact answer.

publiusr
2012-May-05, 04:31 PM
Along the same idea
http://www.rocketreviews.com/shroudtransistion-calculator.html
http://rocketry.newcenturycomputers.net/shroudcalc.html

xylophobe
2012-May-07, 01:10 PM
...

Xylo: be advised that delta(x) refers to a certain number of slices. Strictly speaking, dx is an infinitesimal. There is not a certain number of dx slices. Dx is used in integration while delta(x) is for numerical approaches. They're not handled the same way.

That quote even says that we sum an infinite number of dx slices to get the integral. Your method can never get to infinite slices. You'll run out of computer processing power and memory trying to get an exact answer.

"In common speech, an infinitesimal object is an object which is smaller than any feasible measurement, but not zero in size; or, so small that it cannot be distinguished from zero by any available means. Hence, when used as an adjective, "infinitesimal" in the vernacular means "extremely small". (http://en.wikipedia.org/wiki/Infinitesimal)"

Please notice that it states that the infinitesimal is "not zero in size".

Because "dx" is not a zero value then dx/2 (or any fraction of "dx") is not a zero-valued amount/size.

Leibniz notation uses the fraction "dy/dx" and it should be noted that "dx" is multiplied by the function and if the function is just a constant then 2*dx is acceptable as well as 1/2*dx.

HenrikOlsen
2012-May-07, 01:30 PM
I do not want readers to get the wrong perception (whether by me or someone else) of how integration works so I will direct to:

"By the logic above, a change in x (or Δx) is the sum of the infinitesimal changes dx. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Physical_intuition ) ..."

The above text and other readings on the Fundamental theorem of calculus (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus) seem to agree with my description of how the value of "x" advances by increments of "dx".

Method of exhaustion (http://en.wikipedia.org/wiki/Method_of_exhaustion).
You missed indicating that the quote is from a paragraph on Physical Intuition rather than one of formal definitions.
You're describing an analogy not the thing in itself.

In general it looks like you haven't really understood how the theory of limits is applied to get from sums of slices to integrals, which definitely aren't sums of slices..


"In common speech, an infinitesimal object is an object which is smaller than any feasible measurement, but not zero in size; or, so small that it cannot be distinguished from zero by any available means. Hence, when used as an adjective, "infinitesimal" in the vernacular means "extremely small". (http://en.wikipedia.org/wiki/Infinitesimal)"
"In common speech" is a large red flag being waved frantically to show that this is NOT mathematics.

It approaches mathematics as it vaguely alludes to how limits are defined, but without the formal definition, it's nearly useless to get any insight from.

The mathematical version doesn't talk about measurements and quite easily distinguishes from zero. :)

ShinAce
2012-May-07, 06:26 PM
What Henrik said!

From that same wiki page:
"The method of indivisibles related to geometrical figures as being composed of entities of codimension 1. John Wallis's infinitesimals differed from indivisibles in that he would decompose geometrical figures into infinitely thin building blocks of the same dimension as the figure, preparing the ground for general methods of the integral calculus. He exploited an infinitesimal denoted 1/infinity in area calculations."

You're plugging a number into dx, namely, 1. Try plugging 1/infinity and see what you get. You get 0. That's why the method of infinitesimals applies to integral calculus, not indivisibles related to geometrical figures.

Here's an example.

0.1111111111111periodical is 1/9.
Take that, multiply by 9.
You now have 0.99999999999999999periodical.
But 1/9 times 9 is 9/9. Which is equal to 1.
What's the difference between 0.9999999periodical and 1? We can label it as dx and call it an infinitesimal. Under no circumstances am I allowed to plug any number other than 0 in place of dx. We keep saying it's 0 because you try to make it 1. We're not defending dx as nothing, we're opposing the fact that you continuously replace it with an integer and try to say that this applies to integral calculus. It does not apply to integrals.

As it's own thread, the difference between 9/9 and 1 drew much debate. But the point remains, the difference between 0.9999999... and 1 is determined by how far down the periodical you're willing to go. Go forever, and the difference is an infinitesimal. There's no point trying to pin a number on the difference, it's a concept. Use basic arithmetic and you find that 1 - 9/9 = 0 .

Now we're getting sidetracked again.

Also, make note that everyone that has done integral calculus has used dx/2 at some point. It shows up in a change of variable or even integration by parts.
Let x = 2u. Then dx = 2du , or du = dx/2 .
When you do that, you get rid of dx, and work in du. When you're done, you back substitute 2u for x to get the answer in terms of x. This is something that you would do to make an integral 'doable'. You would use it to reduce the expression into something simpler.

You're not doing a change of variable and getting rid of the original variable. You're trying to redefine an existing variable. X does not equal 2X.
You're not using infinitesimals. You're using plain old differences. Aka, deltas. Writing dx to mean delta(x) is only going to confuse the reader.

Imagine I want to build a staircase. It needs to be 10 feet deep and 10 feet high. Since all edges are right angles, the total length of wood to make the staircase is 20 feet(we're obviously ignoring the width of the staircase). Whether there is 1, 10, 100, or 1000 steps, the total length of wood to make the staircase is 20 feet. But, in calculus, if I let the number of steps be infinity, the right angles disappear and the total length of wood needed is 14.14 feet. That's because it's not even a staircase anymore. It has become a flat ramp. Is your cone like the staircase or the ramp?

Hmmm... even if I had a trillion steps, I would need 20 feet of wood. But if each step was an infinitesimal, I could build the staircase using under 15 feet of wood. It would appear that the two ideas cannot be treated the same way. Yet, by your logic, if the wood I need does not change despite the number of steps, then infinite steps will need 20 feet of wood. This abuse of limits proves nothing.

grapes
2012-May-07, 08:44 PM
Grapes: looks like we both fumbled that potential function. :)
No, we didn't.


Hmmm... even if I had a trillion steps, I would need 20 feet of wood. But if each step was an infinitesimal, I could build the staircase using under 15 feet of wood. It would appear that the two ideas cannot be treated the same way. Yet, by your logic, if the wood I need does not change despite the number of steps, then infinite steps will need 20 feet of wood. This abuse of limits proves nothing.It's still 20. You have one chance to redeem yourself. :)

Analyze his diagrams and show that his error does not increase.

Jeff Root
2012-May-07, 08:54 PM
What should the "d" in "dx" be thought to stand for?

When building a staircase of infinitesimally-thin wood
(like paper! It can fold!), ten feet high and ten feet deep,
why does the length of wood (or paper) needed decrease
from 20 feet to ten times the square root of 2 feet when
the number of steps becomes infinite?

-- Jeff, in Minneapolis

Jeff Root
2012-May-07, 08:58 PM
Oh, wow! That was a mistake by ShinAce? I wondered
how the length could change like that.

-- Jeff, in Minneapolis

ShinAce
2012-May-07, 09:33 PM
Because infinite steps does not make any sense. If you can see that there are steps, there aren't an infinite number of them. If there's an infinite number of them, you can't see if they are steps or slopes. So what does infinite steps imply? If you imagine a bunch of little steps, you say 20 feet of material are required. I ignore that there are any steps at all and just make a ramp. You would need 20 feet because you want to build a staircase with n steps. I would need 14.14 feet because I'm not building steps, I'm making a ramp. I just lay one piece of wood down and call it a day. Don't take the analogy too seriously. I'm not saying that numerical integration gives an answer different than integral calculus. If you really are making little steps, then you need 20 feet of material. If you think infinite steps requires 20 feet, go ahead and buy 20 feet of wood. Your staircase will look and feel the same as my ramp, and I'll save some money on material.

Grapes: the mistake I made was assuming that the center of gravity coincides with the gradient of the potential function being zero. Your mistake was when you said that the potential of a sphere is lowest at the surface. That's true of a spherical shell of mass. But here, the cone is not a shell, it is a solid. Gravitational potential of a solid sphere is lowest at its center. Whether or not you're allowed to go through the sphere's surface is still arbitrary.

Be advised I briefly analyzed the error way back in post 209. I'm not going to redo the work.

dx can be read as "an infinitesimal length along x". It has the same units as the x axis.

HenrikOlsen
2012-May-07, 10:58 PM
Because infinite steps does not make any sense. If you can see that there are steps, there aren't an infinite number of them. If there's an infinite number of them, you can't see if they are steps or slopes. So what does infinite steps imply? If you imagine a bunch of little steps, you say 20 feet of material are required. I ignore that there are any steps at all and just make a ramp. You would need 20 feet because you want to build a staircase with n steps. I would need 14.14 feet because I'm not building steps, I'm making a ramp. I just lay one piece of wood down and call it a day. Don't take the analogy too seriously. I'm not saying that numerical integration gives an answer different than integral calculus. If you really are making little steps, then you need 20 feet of material. If you think infinite steps requires 20 feet, go ahead and buy 20 feet of wood. Your staircase will look and feel the same as my ramp, and I'll save some money on material.
lim_{n \to \infty}20=20, with infinitesimal steps you still need 20 feet of wood.

This time you're the one using intuition rather than formalism to argue your case.