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mik sawicki
2003-Feb-06, 10:19 PM
This is an excellent and very much needed book, and I sure hope it will sell well.
I spotted some minor problems that I think might need Phil's attention.
Here's the list of 32 items that I think need revisions.
Regards
Mik Sawicki

1. Page 23, 11-th line from the bottom:
"But if they are fired north,…"

Add "on the northern hemisphere"

2. Page 23, third line from the bottom:
"Lets say you are driving a car north at 100 kph (60 mph) in Wiscasset, Maine. The Coriolis effect deflects you…"

A car is not a good example, as the minute Coriolis acceleration there (0.0014 m/s^2) translates for a 1000 kg car into a Coriolis force of 1.4 N (about 0.3 lb). This force can't deflect the car to the right, since it is too small to overcome the sideways force of static friction between tires and the road, that prevents the tires from skidding sideways. This frictional force will easily peak to the value of some 5,000 N -10,000 N before onset of skidding (on a dry roadway).

So I'd suggest something like: "Imagine you rolled a bowling ball on a perfect wooden floor at 100 kph north…" Or "Let's say a hockey puck is sliding north on a perfect frozen lake… "

A bowling ball can roll sideways under the Coriolis force, and it will. The puck is already sliding, so it will deflect under the Coriolis force as well.

3. Page 23, bottom line, continuation of the same sentence:
"…by the teeny amount of 3 millimeters"

According to my calculations, it's about 0.7 mm, even more teeny.

4. Page 24, second line, continuation:
"10 meters (33 feet)"

I got 9 km (kilometers), i.e. about 1000 times more, a reasonable result, considering that in 1 hour you'd move 100 km north, and if you imagine an air molecule moving as a 100 km/h wind near a low pressure system, then this air molecule gets deflected much more than 10 meters over a distance of 100 km - as weather maps on The Weather Channel prove, when you check a curvature of a wind in a vicinity of a low system.

5. Page 29, 11 line from the top:
"No wonder no one could measure it until recently".

It all depends on what one means by "recently".

Back in 1676, Ole Römer, determined it to be about 125,000 miles per second, about
three-quarters of the correct value of 186,300 miles per second, not to good a result.

Then in 1728, astronomer James Bradley found the speed of light to be 185,000 miles per second, with an accuracy of about one percent.

Around 1850 in France, two rivals, Fizeau and Foucault, determined the speed of light with an accuracy of about +/-1,000 miles per second.

Then Albert Michelson in 1877, using Fizeau and Foucault method got 186,355 miles per second, +/- 30 miles per second.

Then in 1931, using new apparatus, he got the speed of light as 299,774 km/s with an average deviation of 11 km/s from the mean.

6. Page 35, line 3 from the top:
"0.00000000001 centimeters"

To many zeroes following the decimal point! (remove two, or say meters instead of centimeters.)

7. Page 37, first line:
"The Earth is a big place. There are 511,209,977 square kilometers of it, give or take a kilometer or two…"

I think this is both an unreasonable accuracy (9 significant figures?) and precision (+/- 1 km).Besides, it's not clear to me what is it meant here - a surface of a land plus the surface of all oceans, or a total surface that would be revealed after draining all oceans? Also, in case of land, is it an actual surface, or a surface as projected onto a two-dimensional map? I doubt that anybody ever calculated the precise area of all the slopes of all these deep and complicated valleys in Himalayas etc. to get a result within +/- 1 sq. km precision as the quoted number implicitly suggests.

Then, if this is a result of some mathematical calculation, was the p number taken to be 3.141592654, as required by the implied precision, or perhaps rounded of to 3.14 ?
For example, if one keeps the 10 digit precision on the p number and use the formula for the surface area of a perfect sphere, the area quoted above corresponds to the radius of the sphere R = 6378.15000 km, i.e. the radius has to be known with a precision of +/- 1 cm. This does not make much sense, as tidal/atmospheric etc. effects change the radius much more than that, and Earth is not nearly a perfect sphere anyway. If an average radius of 6378.15 km was used, then the surface of a perfect sphere would be 511,210 thousands sq. km

In any case, Earth is not a perfect sphere and the best estimate I'm aware of is 510,101 thousands of square kilometers (six significant figures).

8. Page 42.

The problem is with the picture. Since the blue light has a wavelength of some 400 nm, and red about 700 nm, the scattering of blue light is about (700/40)^4 = 9.4 times more effective than that of red light. Moreover, red photons do not scatter much to begin with, and average number of scatterings per red photon is much less than 1. So the average number of scatterings per blue photon is much less than 9.4, and in reality is actually quite close to 1 (single scattering). Yet the picture shows a blue photon that scattered 33 times! The picture needs to be redrawn to show blue photon scattering only once.

9. Page 44.

The problem is with the picture. Perhaps a footnote is needed, explaining that the thickness of the atmosphere is vastly exaggerated in comparison to the Earth's size. (Earth appears to have a radius of 1.25 inch, so the atmosphere (troposphere) should be about 0.003 inch thin!

10. Page 45, line 2 from the top:
"The air bends the light up,…"

The air bends the light down, but our mind does not know it and assumes that the arriving light did not get bent, and hence arrived from a direction at which it entered the eye, i.e. pointing above the horizon.

11. Page 49, line 4 from the top:
Copernicus published his idea in 1543.

12. Page 49, line 7 from bottom:
"…change in distance over the course of the seasons amounts to only a 4-degree Celsius … change in temperature."

Check this number. There are conflicting opinions. Some claim that a Summer in Australia is hotter than Summer in northern hemisphere, but some people claim the opposite.

"Averaged over the globe, sunlight falling on Earth in January [at perihelion] is about 7% more intense than it is in July [at aphelion]," says Roy Spencer of the Global Hydrology and Climate Center in Huntsville, AL." The fact that the northern hemisphere of Earth has more land, while the southern hemisphere has more water, tends to moderate the impact of differences in sunlight between perihelion and aphelion.

Sunlight raises the temperature of continents more than it does oceans. (In other words, land has a lower heat capacity than water.) In July (aphelion) the land-crowded northern half of our planet is tilted toward the Sun. Aphelion sunlight is a little weaker than sunlight at other times of the year, but it nevertheless does a good job warming the continents. In fact, say climate scientists, northern Summer in July when the Sun is more distant than usual is a bit warmer than its southern counterpart in January."

See: http://science.nasa.gov/headlines/y2001/ast04jan_1.htm?list117398

Actually, you make a similar comment on the top of the Page 54, so there's inconsistency between what's on the Page 54 and the 4 degrees C on page 49.

13. Page 52, Figure Caption.

I'd say: "Summer in northern hemisphere and winter in southern hemisphere. In the northern hemisphere, the Sun is higher in the sky. Its light is more concentrated on the Earth's surface there. At the same time, Sun is lower in the southern hemisphere, and the light gets spread out there, heating the Earth less efficiently."

14. Page 66, line 12 from the bottom:
"6 percent"

I've got 7%

15. Page 70, line 12 from top:
"The time of high and low tides changes every day by about a half hour"

Your explanation and calculation are correct, but the wording of the conclusion is a bit confusing. I'd say the morning tide tomorrow will be about 50 minutes later than the morning tide today, etc.

16. Page 73, line 5 from the top in the first paragraph;
"Tides on the Moon are 80 times…"

About 20 times, since diameter of the Moon is about 4 times smaller than Earth's diameter.

17. Page 79, line 20 from the top.
"The air bends the light up,"

See # 10 above.

18. Page 130, line 20 from the top:
"Moon has 20,000 times the tidal force of all the other planets"

It is perhaps worthwhile to add that the tidal force of all the other planets is mostly due to Venus alone, as the tidal force due to Venus is almost 9 times bigger than that generated by Jupiter. Many people think that it's Jupiter that contributes most to the tidal force of all the other planets!

19. Page 167, line 10 from the top:'
"He grabbed a rock weighing roughly a kilogram (two pounds)"

A kilogram is a unit of mass, not weight. A mass of one kilogram weighs about two pounds on Earth's surface, but only one-third of one pound on Moon's surface. Since the sentence refers to what happened on the Moon surface, this needs to be rewritten.

20. Page 178, line 1 through line 11.
"flung outwards by Jupiter's rapid rotation in the same way that a dog shakes its body to spray off water after a bath.
(…)
Jupiter would slow his own rotation every time this happened,"

Not necessarily. If Jupiter had shed some mass from the surface radially outward, it's rotational speed would have not changed. It's just a conservation of angular momentum. Imagine a merry-go-round with a dude riding on a rim. If the dude jumps off pushing away from the platform, the rotational speed of the platform is not affected. (You seem to hint at this on page 198).

On the other hand, if Jupiter ejected a mass radially out from the inside, than the conservation of the angular momentum dictates that its rotational speed would indeed have decreased. But then the dog analogy does not really apply here.

21. Page 181, the last paragraph:
"Worse, the tides from Venus on the Earth would be huge, kilometers high"

Better! Since Venus would be inside the Roche radius of Earth, the tidal force produced by Earth would be stronger than Venus own gravity and the Venus would disintegrate. And Earth would be inside Venus' Roche limit as well, so Earth would disintegrate too, and Velikovsky wouldn't be around to push his bunk.

22. Page 193, first line:
"Something caused this cloud to collapse"

Why not gravity? The Universe probably started as a lumpy, non-uniform mass distribution, so lumpier areas contracted faster than other.

23. Page 193, line 6 from the top:
"flattened due to centrifugal force and friction"

I'd prefer not to use a concept of centrifugal force, as it requires a prior explanation what a noninertial frame of reference is etc. I'd say due to" inertia and friction", or perhaps "due to angular momentum conservation and friction".

24. Page 214, 3-rd paragraph:
"Mars' gravitational influence on the Earth drops by a factor of more than 50 from one side of the sun to the other"

I've got a factor of about 23.

25. Page 214, 4-th paragraph:
"…the Moon has a gravitational effect on the Earth and you that is more than 50 times the combined gravity of planets".

This is correct, but you could push a limit a bit more: According to my calculations Moon's gravity is about 65 times stronger than combined gravity of all other planets even when they are all at the same time at their closest distances from the Earth, and how often does that happen!

26. Page 247, line 13:
"If some atom is sitting around minding its own business and another one comes along moving faster than sound, the first atom is surprised by it. It's literally shocked: it didn't know what was coming. When this happens to a lot of material it's called a shock wave."

I have a problem with that, since for example a speed of sound in air at room temperature (20 degrees C) is about 343 m/s, yet individual molecules comprising air move much faster. O2 molecule has a speed of 480 m/s, N2 molecule 513 m/s and H2O molecule 640 m/s, yet they do not produce a shock wave, obviously.

27. Page 249, line 11:
"This is called the centripetal force,"

This is called inertia. The car accelerates to the left, but the tendency of a passenger is to continue in a straight line. The centripetal force is a force pulling the car towards the center, which on a flat road is provided by a static friction between the roadway and tires. Likewise, for a rock whirled on a string, the centripetal force is provided by the string that pulls towards the center, etc.

"So, if a pilot flying the spaceship banks during the turn, the centripetal force is directed back, pushing the pilot harder against the seat."

Again, the centripetal force is towards the center of the turn. If the spaceship is not banked, the centripetal force on a pilot is provided by a harness (there's no friction, unless his pants are made from Velcro). If he banks the ship, the seat is pushing him at right angle to the seat, and a component of that push towards the center of the turn acts together with his harness and a friction force now present to provide the centripetal force.

28. Page 251, line 12 from top:
"…the nearest planet (is) about 25 million miles away"

The nearest planet periodically comes as close as 25 million miles.

29. Page 251, line 13 from top:
"…the nearest star to the Sun, Alpha Centauri"

The nearest star is Proxima Centauri at 4.22 light years, c.f. page 29, line23. Alpha and Beta Centauri are at 4.35 light years.

30. Page 252, line 3 from the top:
"…Oort cloud, the cometary halo that is almost a light year across."

Do you mean here the inner diameter of the Oort cloud, outer diameter, or its thickness?
Inner diameter is some 0.7 light year, outer diameter is about 3 light years, and thickness, indeed, about 1 light year.

31. Page 252, line 5 from the top:
"…a trillion kilometers from the heat and fierce gravity of the Sun"

An uninitiated reader might erroneously conclude that Sun's gravity does reach there!

32. Page 253, line 13 from the top:
"…even distant Pluto is 8,000 times closer"

Due to large eccentricity of Pluto's orbit, Pluto is between 5330 and 9330 (on average 6780) times closer to Earth than Proxima Centauri.

kilopi
2003-Feb-07, 02:37 AM
On 2003-02-06 17:19, mik sawicki wrote:
I spotted some minor problems that I think might need Phil's attention.
Interesting reading. I agree with a lot of them, at least on first reading. Some of them have been discussed in other threads, I'll see if I can dig them up.

2. Page 23, third line from the bottom:
"Lets say you are driving a car north at 100 kph (60 mph) in Wiscasset, Maine. The Coriolis effect deflects you…"

A car is not a good example, as the minute Coriolis acceleration there (0.0014 m/s^2) translates for a 1000 kg car into a Coriolis force of 1.4 N (about 0.3 lb). This force can't deflect the car to the right, since it is too small to overcome the sideways force of static friction between tires and the road, that prevents the tires from skidding sideways. This frictional force will easily peak to the value of some 5,000 N -10,000 N before onset of skidding (on a dry roadway).
It's not necessary to get anywhere near the 5000 N before the car is deflected. A gentle wind from the side will cause you to steer into it.

3. Page 23, bottom line, continuation of the same sentence:
"…by the teeny amount of 3 millimeters"

According to my calculations, it's about 0.7 mm, even more teeny.
My back of the envelope calculation gave 2 mm, so who knows?


4. Page 24, second line, continuation:
"10 meters (33 feet)"

I got 9 km (kilometers), i.e. about 1000 times more, a reasonable result
I think the 10 meters was just an extrapolation of the 3mm/second times an hour, which does give around 10 meters. If your figure was even more teeny than that, how'd did you get 1000 times more?


9. Page 44.

The problem is with the picture. Perhaps a footnote is needed, explaining that the thickness of the atmosphere is vastly exaggerated in comparison to the Earth's size. (Earth appears to have a radius of 1.25 inch, so the atmosphere (troposphere) should be about 0.003 inch thin!
The vertical is often exaggerated in illustrations--otherwise the diagram would be unreadable. Of course, that's one of the reasons that people imagine a huge cliff at the edge of the continental shelf.


15. Page 70, line 12 from top:
"The time of high and low tides changes every day by about a half hour"

Your explanation and calculation are correct, but the wording of the conclusion is a bit confusing. I'd say the morning tide tomorrow will be about 50 minutes later than the morning tide today, etc.
I remember this one, we hashed it out in Bad "Bad Astronomy: Misconceptions" (http://www.badastronomy.com/phpBB/viewtopic.php?topic=670&forum=9&16)

mik sawicki
2003-Feb-12, 09:19 PM
On 2003-02-06 21:37, kilopi wrote:


On 2003-02-06 17:19, mik sawicki wrote:

4. Page 24, second line, continuation:
"10 meters (33 feet)"

I got 9 km (kilometers), i.e. about 1000 times more, a reasonable result
I think the 10 meters was just an extrapolation of the 3mm/second times an hour, which does give around 10 meters. If your figure was even more teeny than that, how'd did you get 1000 times more?

It's quite simple, actually. The Coriolis force here does not change much over a 100 km distance, so we can treat it as approximately constant. Now, a constant force produces constant acceleration, and distance will be proportional to the time SQUARED. Since the deflection is 0.7 mm after 1 second, it will be (3600)^2 times larger after 1 hour. This gives approximately 9 km.

Likewise, I stand by my other results.
Mik Sawicki

kilopi
2003-Feb-13, 04:24 PM
On 2003-02-12 16:19, mik sawicki wrote:
Likewise, I stand by my other results.

Ah, I think I see the source of the discrepancy. The BA is treating the acceleration as a linear constant. He has calculated the coriolike effect as 2.8mm/s^2 and it appears in his text as 3mm/second. You on the other hand missed a factor of 2, and calculated it as 1.4mm/s^2, and arrived at .7mm displacement in the first second.

That's a guess.

Even using your 1.4mm/s^2 figure, in 3600 seconds, that's a velocity of 5m/second (or double that if the coriolike effect is double that), compared with the automobile velocity of 28m/second--at right angles to the original velocity, constantly. As you point out, such a small force can't do that, overcoming the sideways friction of the tires--the reason you said a car is a bad example.

However, there is a sideways force as I pointed out. If you add it to the rolling resistance and wind resistance on the car, the result is a more-or-less constant force that must be steered against--so in that sense, could be taken out linearly. But the calculations are not as straightforward, and there are tremendous differences depending upon the type of car and atmospheric conditions.

<font size=-1>[ This Message was edited by: kilopi on 2003-02-13 11:59 ]</font>

jimbenet
2003-Sep-08, 02:35 AM
I enjoyed reading Bad Astronomy. However, I found a few minor errors and one major error in it that I would like to point out.

First the minor errors:

p53. Change June to July

p56. The ellipse precesses also with a period of 21,000 years.

p107. Change "brighter" to "dimmer".

The major error is on page 55 in the last paragraph, " ... the Earth's north pole will be pointed away from the Sun in June and toward it in December."

This is not true. The calendar is a tropical calendar that accounts for the precession (with leap years). A year in the tropical calendar is the number of days between successive vernal equinox crossings of the sun. Hence, precession is built into the calendar. That actually makes the calendar year slightly longer than the sidereal year. Hence, spring will always be in March. What this does do is to put the summer constellations in the winter and visa versa.

There are several references on the calendar on the web. Here is just one of them that you might want to check out.
http://astro.nmsu.edu/~lhuber/leaphist.html. I know that websites have been known to be wrong, but I check out several references on this topic and I believe this is correct. It makes a lot of sense. Here is quote from the article:

"The tropical year is defined as the mean interval between vernal equinoxes; it corresponds to the cycle of the seasons. The following expression, based on the orbital elements of Laskar (1986), is used for calculating the length of the tropical year:
365.2421896698 - 0.00000615359 T - 7.29E-10 T^2 + 2.64E-10 T^3 [days]
where T = (JD - 2451545.0)/36525 and JD is the Julian day number. However, the interval from a particular vernal equinox to the next may vary from this mean by several minutes. "

When the Romans produced the Julian calendar, they did not have means to measure star positions. There were no accurate time clocks then and the telescope had not been invented. Hence, they adapted a solar calendar, based on the vernal equinox crossing. This relies only on the accuracy of measuring angles. The Gregorian calendar (circa 1532) modified the Julian calendar to correct for errors by adjusting leap years and adding 13 days in October. Interesting enough, the Europeans did not adopt this calendar for 150 years later (mainly because they did not want to accept anything coming out of the Roman Catholic Church). Still the basis for the calendar did not change. Another quote:

"Gregorian calendar. The third type of calendar, the lunisolar calendar, has a sequence of months based on the lunar phase cycle; but every few years a whole month is intercalated to bring the calendar back in phase with the tropical year."

Jim Benet

kilopi
2003-Sep-11, 06:12 PM
The BA also posted a list of errata in this thread (http://www.badastronomy.com/phpBB/viewtopic.php?t=1070&amp;postdays=0&amp;postorder=asc&amp;high light=edition&amp;start=0). Some other threads also discuss them.


p53. Change June to July

p56. The ellipse precesses also with a period of 21,000 years.

p107. Change "brighter" to "dimmer".

Those don't seem to have been mentioned before, but I don't have my copy handy to check them out.


The major error is on page 55 in the last paragraph, " ... the Earth's north pole will be pointed away from the Sun in June and toward it in December."

This is not true.

That one I brought up before, in this thread (http://www.badastronomy.com/phpBB/viewtopic.php?t=710), but I don't think I've convinced the BA yet. I consider it a major error as well. I appreciate the help. :)


That actually makes the calendar year slightly longer than the sidereal year.

Although the solar day is longer than the sidereal day, the calendar year is about twenty minutes shorter than the sidereal year.


Jim Benet
Welcome to the board!

jimbenet
2003-Sep-12, 03:04 AM
Kilopi-

Thanks for the reply. I stand corrected in the length of the year.

I found it hard to find all the threads pertaining to corrections to the book.

I think the author should have a completely separate section on the website covering errata to the book. He should at least acknowledge the errors and correct them, lest he becomes guilty of propagating Bad Astronomy.

Jim Benet

kilopi
2003-Sep-12, 03:42 AM
I found it hard to find all the threads pertaining to corrections to the book.
H*ck, there's only 102 threads in this forum, that should only take you half an hour to walk through. :)

Seriously, we'll try to help.

I think the author should have a completely separate section on the website covering errata to the book. He should at least acknowledge the errors and correct them, lest he becomes guilty of propagating Bad Astronomy.
I think he does, that was why he started that thread I linked to before, List of errors and typos (http://www.badastronomy.com/phpBB/viewtopic.php?t=1070&amp;postdays=0&amp;postorder=asc). No one is more ready to admit error, fix it, and--this comes close to being unique--keep the errors around to illustrate the progress.

The Bad Astronomer
2003-Sep-12, 08:01 PM
Yup. As a matter of fact, the book is in its fifth (and probably final) printing, and each printing has had some corrections made. I need to find the latest crop and post them here.

Eroica
2003-Sep-24, 11:12 AM
A couple or errors that someone else has probably already pointed out.

page 114 line 12: for "Alpha Ursa Minoris" read "Alpha Ursae Minoris"

page 115 line 5: for "Sigma Octans" read "Sigma Octantis"

Come to think of it, Phil might well be onto something here. Isn't it about time we ditched these old Latin genitives? Who in their right mind could guess that the genitive case of Apus is Apodis, or that the possessive of Canes Venatici is Canum Venaticorum? Alpha Ursa Minor might well be ungrammatical, but it's clear and it sounds much better than the "correct" form. Besides, the IAU already took the first steps in this direction when they decided that the two parts of Serpens should be known as Serpens Caput and Serpens Cauda. Strictly speaking, they ought to be Serpentis Caput and Serpentis Cauda.