View Full Version : Drawing an accurate crescent Moon

pzkpfw

2009-Feb-15, 08:48 AM

Hi,

I'm looking at making the logo of the company I'm in the process of forming. (Can't pay $$$ to a designer to do it for me.)

I want to include a crescent Moon.

It struck me that there must be "accurate" and "inaccurate" ways to do this.

Is there a (simplified) way to mathematically describe the appearance of the crescent Moon, at it's various stages from new to full?

...I've not yet decided at exactly what point in the cycle I want to draw it.

(I'm in Wellington, New Zealand, Southern Hemisphere).

Cheers,

[I hope this makes at least some sense...]

Look outside when there's a crescent moon? ;)

Seriously, the biggest mistake people make (especially in 3D rendering) is to have the terminator gradually fade to black, and the limb darkening. If you look at the moon you'll see it does no such thing - brightness stays pretty constant on the day side (even at the limb) and the terminator is a sharp line (if you look closely enough you'll see shadows from topography there too)

If you're really accurate you may have to consider earthshine too, in which you'll dimly see the features on the dark side of the moon due to light reflected from the Earth.

I guess from where you are the moon looks upsidedown compared to how it's seen in the northern hemisphere too, which could make it interesting :).

Peter B

2009-Feb-15, 10:45 AM

Hi,

I'm looking at making the logo of the company I'm in the process of forming. (Can't pay $$$ to a designer to do it for me.)

I want to include a crescent Moon.

It struck me that there must be "accurate" and "inaccurate" ways to do this.

Is there a (simplified) way to mathematically describe the appearance of the crescent Moon, at it's various stages from new to full?

...I've not yet decided at exactly what point in the cycle I want to draw it.

(I'm in Wellington, New Zealand, Southern Hemisphere).

Cheers,

[I hope this makes at least some sense...]

In a Full Moon, the relative width of the Moon at any latitude is given by Cosine (Latitude). I think you'll find the same applies to any crescent Moon. That is, the width of the Moon at latitude 60 degrees is half that of the Equator, whether it's a Full Moon, a crescent, or anything in between.

Peter B

2009-Feb-15, 10:46 AM

Or you could take a photo of a crescent Moon and make a tracing!

pzkpfw

2009-Feb-15, 11:15 AM

Thanks.

Perhaps to illustrate what I'm getting at...

Would simply "cutting" a circle of black out of a circle of white make an accurate crescent Moon?

If yes... (*1)

What would the relative sizes of those circles be?

What would be the relative position of the centres of those circles be at different times during the Lunar Month?

(

*1 I'm guessing "no" - that the "cut out" isn't a really a circle as seen from Earth.

Light shining on a ball does make a straight line around that ball - the dark (at that time) half versus the light half... how is that line drawn from a point of view that isn't exactly side-on to the direction the light is coming from?

)

Or you could take a photo of a crescent Moon and make a tracing!

Even easier - steal someone elses photo...

Thanks.

(

*1 I'm guessing "no" - that the "cut out" isn't a really a circle as seen from Earth.

Light shining on a ball does make a straight line around that ball - the dark (at that time) half versus the light half... how is that line drawn from a point of view that isn't exactly side-on to the direction the light is coming from?

)

The line between shadow and light is a circle around the moon, so moon phases are just circle (or half of it) seen from an angle.

Draw a circle, and in the middle draw an ellipse which has semi-major axis same as the radius of the circle. The ellipse inside the circle then defines terminator lines for crescent or gibbuous moon.

Peter B

2009-Feb-15, 12:50 PM

Thanks.

Perhaps to illustrate what I'm getting at...

Would simply "cutting" a circle of black out of a circle of white make an accurate crescent Moon?

No, I don't think so. I think the terminator is the shape of a parabola, but I'm not certain.

If yes... (*1)

What would the relative sizes of those circles be?

What would be the relative position of the centres of those circles be at different times during the Lunar Month?

(

*1 I'm guessing "no" - that the "cut out" isn't a really a circle as seen from Earth.

Light shining on a ball does make a straight line around that ball - the dark (at that time) half versus the light half... how is that line drawn from a point of view that isn't exactly side-on to the direction the light is coming from?

)

Okay, I think I might have got it wrong in my last post. Let's try another way - let's see how well I can dredge up some of my old Tech Drawing skills from *cough - ahem* years ago.

1. Take a blank sheet of paper.

2. Draw a circle on it of 18 cm diameter.

3. Draw horizontal lines across the circle at 1 cm intervals, with the top and bottom lines tangential to the circle. These lines represent latitude lines at 10 degree intervals.

4. Decide how much of a crescent moon you want. This will be a decimal fraction between 0 and 0.5 (a). (Using 0 will give you a New Moon, and using 0.5 will give you a half Moon.)

5. For each latitude line:

5.1 Measure its width (b);

5.2 Multiply b by a, giving you the crescent width at that latitude (c);

5.3 Measure in c cm from the left edge and make a small mark.

6. Connect all the marks with a smooth line. That line is the terminator.

Does that help?

ngc3314

2009-Feb-15, 01:24 PM

The terminator traces a section (half) of an ellipse (in the idealized case of smooth surface and so on). The demonstration is that the boundary of the illuminated hemisphere is a circle in three dimensions, and a circle viewed at a general angle appears as an ellipse. Likewise for a smooth surface lacking topography, the illumination varies as the sine of the sun's angle above the local horizon, but the mapping between that angle and visible location gets complex for crescents and this is overwhelmed by the shadowing from topography (in crescent phases, we are looking at the shadowed side of many features, making it still dimmer than the sine effect suggests).

JohnD

2009-Feb-15, 01:26 PM

The shadow of the earth is a Circle, not an ellipse.

Seen from Earth, it's a circle on the Moon.

It might be an ellipse from somewhere else.

And the shadow circle is the same size as the Moon.

Suggest that the darkened area of the Moon is shown in your logo.

Somehow I think that make sit look more like a 'moon'!

John

hhEb09'1

2009-Feb-15, 02:56 PM

The shadow of the earth is a Circle, not an ellipse.

Seen from Earth, it's a circle on the Moon.That's during a lunar eclipse, but the OP asks about phases of the moon, not eclipses. lek has it right about the shape of the terminator, from our point of view--we're looking at a circle (the terminator) from the side, which means it must be an ellipse, to first order.

pzkpfw

2009-Feb-15, 07:46 PM

Thanks BAUTers.

Next step is: what would be an accurate axis be for the ellipse?

(i.e. angle off vertial (relative to the paper), for the major axis of the ellipse? or the angle between the latitude lines and the top edge of the paper?)

I guess this depends on lat/long of the observer, but perhaps also time of year.

Is simply "vertical" a good enough approximation? What would be a reasonable range of values?

Thanks again.

(I'm amazed at what I've never thought about/noticed. I'm beginning to feel like one of those people who sees Venus for the first time and thinks...)

Thanks BAUTers.

Next step is: what would be an accurate axis be for the ellipse?

(i.e. angle off vertial (relative to the paper), for the major axis of the ellipse? or the angle between the latitude lines and the top edge of the paper?)

I guess this depends on lat/long of the observer, but perhaps also time of year.

Most of all it depends on time of day. If location+date etc is such that the moon can be seen directly overhead, the angle of the semi-major axis to horizon rotates 180 degrees from moon rise to moon set. So defining "accurate" figure would require exact location, date and time.

hhEb09'1

2009-Feb-15, 08:26 PM

Next step is: what would be an accurate axis be for the ellipse?

(i.e. angle off vertial (relative to the paper), for the major axis of the ellipse? or the angle between the latitude lines and the top edge of the paper?)

I guess this depends on lat/long of the observer, but perhaps also time of year.

Is simply "vertical" a good enough approximation? What would be a reasonable range of values?

It does depend upon lat/lon and time of year, but mostly time of night/day :)

A crescent moon that rose as a smile with the points upturned might set in the opposite fashion, with them down--and they'd be straight up and down in between near noon.

Are you going to have the lunar features visible in your logo?

mugaliens

2009-Feb-15, 08:39 PM

While the following method may not be precisely accurate with respect to what we see, 99.5% of the people will not be able to tell the difference:

1. Open Excel

2. Set up two columns, X and Y, which describe a half circle, such that f(x)=(0,-1), (1,0), and (0,1), as well as all the other data points in between.

3. Set up a third column, called Xc, where the value of Xc is a simply, constant percentage (say, 50%) of the value of X.

The two curves, f(x)=f(X,Y) and fc(x)=f(Xc,Y) will give you the outline of your crescent moon.

tony873004

2009-Feb-15, 10:20 PM

If you want to write some code to plot it, these 11 lines of code should work.

XOF = 2000 ' this is the x-coordinate of the middle of your plotted moon

YOF = 2000 ' this is the y-coordinate of the middle of your plotted moon

Size = 1000 'this is the radius in pixels of your plotted moon

PHASE = 0 ' this is the phase of the moon, (-1 = new moon, 0 = half moon, 1 = full moon)

YAD = 1 'this controls the aspect ratio of your plot, so if a circle looks like an oval, adjust this accordingly.

'the following 3 lines draw the terminator

For k = 0 To 3.14 Step 0.01 ' increase or decrease step size for higher or lower resolution plot

Form1.PSet (XOF + Sin(k) * Size * (PHASE), YOF + Cos(k) * Size * YAD)

Next k

'the following 3 lines draw the fixed limb of the moon

For k = 3.15 To 6.28 Step 0.01 ' increase or decrease step size for higher or lower resolution plot

Form1.PSet (XOF + Sin(k) * Size, YOF + Cos(k) * Size * YAD)

Next k

tony873004

2009-Feb-15, 11:28 PM

Here's a combination of my code and mugaliens' idea to use Excel. I'm using Excel 2003 for Windows, so there may be some differences if you're using the new Excel or a Mac.

In cell A1, put 0.01

In cell A2 put =A1+.01

Drag this down to cell A314

(If you want to plot more points, use A1+.001 and drag down to cenn A3142)

In cell B1 put =SIN(A1+PI())

In cell C1 put =COS(A1+PI())

In cell D1 put =SIN(A1)*$F$1

In cell E1 put =COS(A1)

In cell F1 put 0

Highlight B1 through E1 and extend them down to row 314 (just double click the small black square in the bottom right corner of E1 once they are highlighted if your Excel works like mine, and it should extend columns B through E down to the level of column A).

Highlight columns B and C, and press the Graph button. Choose XY(scatter). Press the "Next" button to arrive at "Step 2 of 4", and click the "Series Tab".

Your X-values should be =Sheet1!$B$1:$B$314 and

Your Y-values should be =Sheet1!$C$1:$C$314

Click the button that says "Add" to add a new series.

Make its X-values =Sheet1!$D$1:$D$314

Make its Y-values =Sheet1!$E$1:$E$314

Press "Finish" (or "Next" if you want to label your graph)

You should now have a plot of the moon. Cell F1 controls the phase. -1 is a new moon, 0 is a half moon (quarter phase), and 1 is a full moon. Change its value and it will redraw the plot for you.

The aspect ratio is likely off. In the above example I had you plot a quarter phase moon by setting F1 to 0. The left edge of the moon is the fixed limb. Drag your vertical borders on the graph until it looks like a half-circle. You will probably have to do this each time you change phase, as Excel is constantly calibrating your x and y dimensions based on the data it is plotting.

btw... this is all for fun. I'd have just traced a picture :)

tony873004

2009-Feb-15, 11:44 PM

Here's the excel file if you don't want to do all the above steps:

http://orbitsimulator.com/misc/MoonPhase.xls

JohnD

2009-Feb-15, 11:53 PM

That's during a lunar eclipse, but the OP asks about phases of the moon, not eclipses. lek has it right about the shape of the terminator, from our point of view--we're looking at a circle (the terminator) from the side, which means it must be an ellipse, to first order.

heb,

You've given your opinion, I've given mine.

I've given the reason for my opinion; please explain yours.

To repeat mine:

The shadow of the Earth is a cone.

A horizontal (at right angles to the axis) section through a cone is a circle.

John

pzkpfw

2009-Feb-16, 12:08 AM

Thanks again everyone.

JohnD, I believe hhEb09'1 is correct in the interpretation of my question.

A lunar eclipse is caused by Earths shadow on the Moon, but what I'm after is the phases of the Moon. These are not caused by Earths' shadow, but by seeing the light shone on the Moon by the Sun, from the side.

Sorry if my OP was unclear on what I was after.

Earth

| us viewing "half Moon"

v

Sun ---Light--> Moon

Sun ---Light--X Earth . . Moon (us seeing Eclipse of Moon)

ngc3314

2009-Feb-16, 12:55 AM

heb,

You've given your opinion, I've given mine.

I've given the reason for my opinion; please explain yours.

To repeat mine:

The shadow of the Earth is a cone.

A horizontal (at right angles to the axis) section through a cone is a circle.

John

This is such a common misconception that the point bears emphasis: With the brief exceptions of lunar eclipses, the phases of the Moon are not caused by the Earth's shadow.

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