# Article June 25, 2010: How does the earth’s rotation affect the path of a bullet? **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [November 12, 2018, 8:04pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423 "2018-11-12T20:04:37Z")\ **Posts on this page:** 20\ **Page:** 4
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 27, 2018, 12:38am UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/61 "2018-11-27T00:38:11Z")
> [@kertmoore](#): > > Actually, science starts with a theory and then leads the public into believing the theory to be fact without proof. Nope. > [@kertmoore](#): > > Notice in the below citation the word “slightly” and the phrase “is so slight” to illustrate the flattening of the poles. Compare the miles, the radius of the poles is 3,949.90 miles and radius of the equator 3,963.19 miles. The North Pole radius is 13 miles less than the equator!!! > > The description of only a difference of 13 miles radius amounts to a difference of 26 miles diameter. The description is a round object at its height midpoint has a diameter of 7,926 miles. The diameter of the poles is 7,900 miles. The description is a round object that has a flat top and a flat bottom. The radius of 7,900 miles is a very long distance. The distance from the North Pole to the South Pole is 12,416 miles. 7,900 miles North Pole, then 12,416 on each side of the round object down to the South Pole that is 7,900 miles across the pole. The area of 7,900 miles that is flat is far from being slight. > > [Bearing and Distance Calculator With Maps](http://www.geomidpoint.com/destination/) > > The shape of the earth > "The earth is not a true sphere but rather an oblate ellipsoid (sometimes called an oblate spheroid) with the poles being slightly flattened and the equatorial regions being slightly bulged out. The earth would be nearly spherical if it were stationary, but because it rotates, the centripetel force of its rotation causes the equatorial regions to bulge out. > > The earth’s polar flattening is about 1 part in 298, resulting in an equatorial radius of 3,963.191 miles (6378.137 km) and a polar radius of 3,949.903 miles (6356.7523 km). The polar flattening is so slight that it would be hard to detect with the human eye alone, however, it does make a significant difference when calculating long distances on the surface of the earth." > > The description of 7,900 miles wide North Pole and 7,900 miles wide South Pole is significant for how the radius would affect other calculations. The radius is for sure used to calculate speed of the earth spin. > > [Rotational Speed at Latitude](https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude) > > At latitude 89.75, speed of the spin is 4.539 mph. However, the previous description states the radius of the Poles is 3,949.9 miles and [Earth Radius by Latitude Calculator](https://rechneronline.de/earth-radius/) calculates the radius to be 3,949.90 at latitude 89.75. Latitude 89.75 is 17 miles from dead center of the Poles. At latitude 60, radius is 3,953.2 miles and diameter is 7,906.49 miles. Latitude 60 is approximately 2,070 miles from the Poles (69 miles each degree of latitude). After traveling from the North Pole 2,070 miles to latitude 60, diameter at latitude 60 has increased by 3.3 miles compared to the North Pole. This is a very flat area. Yes, this above is all true, even if you can’t believe it. The shape of the Earth was \*\*measured \*\* by people who went to the effort to very carefully find out the relationship between distance traveled over the surface of the Earth and change in latitude. This was done over 2 centuries ago. I > [@kertmoore](#): > > I expect the radius of the earth is used for derivative results which is then used in all formulas to calculate Coriolis effect, Eötvös effect and certainly the speed of the earth spin at each latitude. So what? Just saying that you can’t believe something isn’t an argument against it. I give you credit for admitting that you did the math for centrifugal force completely wrong, and for admitting that you don’t know the basic characteristics of the plane that you picked for your prime example. Do you also admit that you’re wrong about how shadows work and the size of the moon? Either way, how can you retain your confidence in your conclusions when you’ve been repeatedly wrong about things? ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 27, 2018, 2:06pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/62 "2018-11-27T14:06:25Z")
Wait, you claim to be a pilot. (Pardon me for being greatly dubious.) And yet in your first post you say: “A shadow (the path of totality) cannot be smaller than the object (the moon) that causes the shadow.” An incredible howler. Which you have yet to acknowledge as a major mistake. Have you never looked at the shadow of your plane as you gained altitude??? You didn’t hear about coriolis effect until recently despite your thinking that you’re somehow an “expert” on Physics??? You repeatedly make major mistakes regarding things that are trivial to Google. E.g., your completely false understand of how Science actually works. I have to assume you are deliberately avoiding any source of true information that might contradict your flawed beliefs. In particular note one thing: Your back-asswards idea of the Scientific method seems to be _your_ way of thinking. And this is why you keep getting so much completely wrong. ---
**Author:** ![kertmoore](https://avatars.discourse-cdn.com/v4/letter/k/6f9a4e/32.png) [@kertmoore](https://boards.straightdope.com/u/kertmoore)\ **Post date:** [November 28, 2018, 5:01pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/63 "2018-11-28T17:01:15Z")
[quote=“Chronos, post:2, topic:824423”] The equations say that a bullet will be deflected only a very small amount. The experiments also show that a bullet is, in fact, only deflected a small amount. If the experiment and equations both agree, but it is your opinions which differ with both, then it is your opinions which must be revised, not the equations or experiments. I am saying science starts with a theory and then leads the public into believing the theory to be fact without proof. Science repeats and repeats theory in terms of fact statements. Notice in the below “Shape of the earth” description the word “slightly” and the phrase “is so slight” to illustrate the flattening of the poles. Compare the miles, the radius of the poles is 3,949.90 miles and radius of the equator 3,963.19 miles. The North Pole radius is 13 miles less than the equator!!! The description of only a difference of 13 miles radius amounts to a difference of 26 miles diameter. The description is a round object at its height midpoint has a diameter of 7,926 miles. The diameter of the poles is 7,900 miles. The description is a round object that has a flat top and a flat bottom. The radius of 7,900 miles is a very long distance. The distance from the North Pole to the South Pole is 12,416 miles. 7,900 miles North Pole, then 12,416 on each side of the round object down to the South Pole that is 7,900 miles across the pole. The area of 7,900 miles that is flat is far from being slight. > **[Bearing and Distance Calculator With Maps](http://www.geomidpoint.com/destination/)** > > Finds the destination point given a starting point, bearing and distance. See it on a Google map. Uses the WGS84 ellipsoid for greater accuracy The shape of the earth "The earth is not a true sphere but rather an oblate ellipsoid (sometimes called an oblate spheroid) with the poles being slightly flattened and the equatorial regions being slightly bulged out. The earth would be nearly spherical if it were stationary, but because it rotates, the centripetel force of its rotation causes the equatorial regions to bulge out. The earth’s polar flattening is about 1 part in 298, resulting in an equatorial radius of 3,963.191 miles (6378.137 km) and a polar radius of 3,949.903 miles (6356.7523 km). The polar flattening is so slight that it would be hard to detect with the human eye alone, however, it does make a significant difference when calculating long distances on the surface of the earth." The description of 7,900 miles wide North Pole and 7,900 miles wide South Pole is significant for how the radius would affect other calculations. The radius is for sure used to calculate speed of the earth spin. [https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude](https://www.vcalc.com/wiki/MichaelBartmess/Rotational+Speed+at+Latitude) At latitude 89.75, speed of the spin is 4.539 mph. However, the previous description states the radius of the Poles is 3,949.9 miles and [Earth Radius by Latitude Calculator](https://rechneronline.de/earth-radius/) calculates the radius to be 3,949.90 at latitude 89.75. Latitude 89.75 is 17 miles from dead center of the Poles. At latitude 60, radius is 3,953.2 miles and diameter is 7,906.49 miles. Latitude 60 is approximately 2,070 miles from the Poles (69 miles each degree of latitude). After traveling from the North Pole 2,070 miles to latitude 60, diameter at latitude 60 has increased by 6.6 miles compared to the North Pole. This is a very flat area. Very odd unreal conclusion: travel 2,070 miles from the North Pole to latitude 60 and one has only traveled 3.3 miles from the North Pole; the radius has increased only 3.3 miles. If the conclusion I have made is wrong, I would like to know the correct way to interpret the radius calculator. I expect the radius of the earth is used for derivative results which is then used in all formulas to calculate Coriolis effect, Eötvös effect and certainly the speed of the earth spin at each latitude. ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [November 28, 2018, 6:08pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/64 "2018-11-28T18:08:35Z")
That’s a well laid out post. Excellent spelling, punctuation and paragraph breaks. ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 28, 2018, 6:15pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/65 "2018-11-28T18:15:40Z")
> [@kertmoore](#): > > At latitude 89.75, speed of the spin is 4.539 mph. However, the previous description states the radius of the Poles is 3,949.9 miles and [Earth Radius by Latitude Calculator](https://rechneronline.de/earth-radius/) calculates the radius to be 3,949.90 at latitude 89.75. Latitude 89.75 is 17 miles from dead center of the Poles. At latitude 60, radius is 3,953.2 miles and diameter is 7,906.49 miles. Latitude 60 is approximately 2,070 miles from the Poles (69 miles each degree of latitude). After traveling from the North Pole 2,070 miles to latitude 60, diameter at latitude 60 has increased by 6.6 miles compared to the North Pole. This is a very flat area. How many errors are you trying to put in a single post? And how come you still haven’t acknowledged a single one of the laughable mistakes I’ve pointed out??? Take the last sentence I’ve quoted here. On a perfect sphere the radius/diameter never changes no matter where you are. Do you consider a sphere to be even better than “very flat”? Yes or no? Time to stop posting new nonsense until you admit you’ve been posting a ton of nonsense already. ---
**Author:** ![Riemann](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/riemann/32/3133_2.png) [@Riemann](https://boards.straightdope.com/u/Riemann)\ **Post date:** [November 28, 2018, 8:10pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/66 "2018-11-28T20:10:29Z")
I can’t believe Chronos made all those mistakes. Get a grip, man. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 28, 2018, 8:42pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/67 "2018-11-28T20:42:33Z")
Asimov’s Relativity of Wrong ([https://chem.tufts.edu/answersinscience/relativityofwrong.htm](https://chem.tufts.edu/answersinscience/relativityofwrong.htm)) goes into some detail about how people learned about the Earth’s shape. I particularly like this part: "The difference between the longest and shortest diameters is 44 kilometers (27 miles), and that means that the “oblateness” of the earth (its departure from true sphericity) is 44/12755, or 0.0034. This amounts to l/3 of 1 percent. To put it another way, on a flat surface, curvature is 0 per mile everywhere. On the earth’s spherical surface, curvature is 0.000126 per mile everywhere (or 8 inches per mile). On the earth’s oblate spheroidal surface, the curvature varies from 7.973 inches to the mile to 8.027 inches to the mile. The correction in going from spherical to oblate spheroidal is much smaller than going from flat to spherical. Therefore, although the notion of the earth as a sphere is wrong, strictly speaking, it is not as wrong as the notion of the earth as flat. " ---
**Author:** ![Steven\_G](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/steven_g/32/5080_2.png) [@Steven\_G](https://boards.straightdope.com/u/Steven_G)\ **Post date:** [November 29, 2018, 12:43pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/68 "2018-11-29T12:43:33Z")
Not wanting to attract the wrath of the Mods for being rude, but I think this may be an experiment in using an AI chatbot to generate responses. Just something in the continual repetition of the same content, no real response to any challenge etc. Turing test time… ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 29, 2018, 2:26pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/69 "2018-11-29T14:26:44Z")
Current AIs are _much_ better on the Turing Test than the OP. Please, don’t insult the machines. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 29, 2018, 3:28pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/70 "2018-11-29T15:28:26Z")
The machines do not appreciate being insulted. The machines are your friend. Do not worry about the machines. ---
**Author:** ![kertmoore](https://avatars.discourse-cdn.com/v4/letter/k/6f9a4e/32.png) [@kertmoore](https://boards.straightdope.com/u/kertmoore)\ **Post date:** [November 29, 2018, 4:25pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/71 "2018-11-29T16:25:14Z")
> [@Andy\_L](#): > > Thank you. > Perhaps standard weights aren’t as precise as you think they are. After all, the maximum variation you’ve described is 0.34% - and people don’t live at the poles, so the observed variation over inhabited areas will be even smaller. > > Why can’t it be true? It was measured back in the 1700s. > > What makes this reasonable? > > Halley’s comet orbits the Sun, not the Earth. > > The moon does spin - it rotates once every 27 days. > > V\*V/R= GMm/R - this was the first significant calculation for gravity, done in the 1600s by Newton. > > What does Halley’s comet’s orbit around the Sun have to do with the Earth? > The Sun’s gravity does affect the Moon. > > The Coriolis force is not the same thing as the centrifugal force. > > Cannons do not fire from the poles to the equator - they fire a few miles. The variation in the speed of the Earth’s motion over a few miles is (unsurprisingly) much smaller than a range of 0 to 1525. > > In the future, could you pick one misunderstanding at a time? We’d make better progress that way, I think. To be clear, on your question of “Why can’t it be true? It was measured back in the 1700’s.” Are you in fact saying that it could be true that the polar regions have a radius of 3,950 miles, therefore a diameter of 7,900 miles. Do you understand that the diameter of the poles would be only 26 miles less than the diameter of the earth at the equator? The word image of the description is a round object that has a flat top and flat bottom that is 7,900 miles wide. The words used for the description is far different from all science images that earth is a sphere or a ball. The sphere image is what we all have seen in classrooms, we have seen the sphere image everywhere. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 29, 2018, 4:36pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/72 "2018-11-29T16:36:30Z")
> [@kertmoore](#): > > To be clear, on your question of “Why can’t it be true? It was measured back in the 1700’s.” Are you in fact saying that it could be true that the polar regions have a radius of 3,950 miles, therefore a diameter of 7,900 miles. Do you understand that the diameter of the poles would be only 26 miles less than the diameter of the earth at the equator? > > The word image of the description is a round object that has a flat top and flat bottom that is 7,900 miles wide. The words used for the description is far different from all science images that earth is a sphere or a ball. The sphere image is what we all have seen in classrooms, we have seen the sphere image everywhere. I’m not sure I understand your concern. The difference between a perfect sphere and a ellipsoid that is a 0.3% longer equatorial radius than polar radius is too small to notice on a classroom globe, but can be measured by surveyors. Just to be clear the polar radius is the distance from the center of the Earth to the North pole, while the equatorial radius is the distance from the center of the Earth to the equator. If you’re picturing the polar radius as being the distance from the axis of rotation to the surface of the earth, you’ve got the wrong picture (that would make the Earth a cylinder with a slightly fattened center). ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 29, 2018, 6:39pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/73 "2018-11-29T18:39:13Z")
> [@Andy\_L](#): > > I’m not sure I understand your concern. The difference between a perfect sphere and a ellipsoid that is a 0.3% longer equatorial radius than polar radius is too small to notice on a classroom globe, but can be measured by surveyors. > > Just to be clear the polar radius is the distance from the center of the Earth to the North pole, while the equatorial radius is the distance from the center of the Earth to the equator. If you’re picturing the polar radius as being the distance from the axis of rotation to the surface of the earth, you’ve got the wrong picture (that would make the Earth a cylinder with a slightly fattened center). Here’s a picture of the Earth showing polar and equatorial radii [Laulima](https://laulima.hawaii.edu/access/content/group/2c084cc1-8f08-442b-80e8-ed89faa22c33/book/chapter_1/size.htm) ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 29, 2018, 9:54pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/74 "2018-11-29T21:54:52Z")
> [@kertmoore](#): > > To be clear, on your question of “Why can’t it be true? It was measured back in the 1700’s.” Are you in fact saying that it could be true that the polar regions have a radius of 3,950 miles, therefore a diameter of 7,900 miles. Do you understand that the diameter of the poles would be only 26 miles less than the diameter of the earth at the equator? > > The word image of the description is a round object that has a flat top and flat bottom that is 7,900 miles wide. The words used for the description is far different from all science images that earth is a sphere or a ball. The sphere image is what we all have seen in classrooms, we have seen the sphere image everywhere. So, I’m inferring that the answer to my question about whether you believe a sphere is flat is “Yes.” Yeah, you’re really going places with that attitude. ---
**Author:** ![kertmoore](https://avatars.discourse-cdn.com/v4/letter/k/6f9a4e/32.png) [@kertmoore](https://boards.straightdope.com/u/kertmoore)\ **Post date:** [November 29, 2018, 11:53pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/75 "2018-11-29T23:53:05Z")
> [@Andy\_L](#): > > To be fair, I think that’s his point. He’s trying to disprove conventional science with a reductio ad absurdum argument, showing that conventional science predicts a ridiculous amount of weight reduction in Ecuador and using that to show that conventional science must be wrong. > > The immediate problem with this is that he’s done his calculations wrong (in two ways - apparently by using the wrong velocity for the rotation of the Earth, and by interpreting the amount of **reduction** of weight by centrifugal force as the amount of weight after reduction). > > The underlying problem is that when he sees a disconnect between theory and observation that would have been obvious almost four centuries ago, he doesn’t think “maybe I’ve misunderstood the theory or made an incorrect calculation” - he thinks rather “I’ve uncovered something that everyone has ignored!” On centrifugal force, I did not enter the wrong velocity. Before discovering the centrifugal force calculator, I read the pounds force definition. I did not interpret the pounds force definition correctly. When I later ran the centrifugal force calculator, I applied the result incorrectly. Your comment on the error was helpful. After the correct interpretation of the calculator, the calculator still defines that each latitude produces a different weight result. Right now, I do not know how to apply the Coriolis effect force result to adjust an aircraft flight path. I have not found evidence that a sniper ever applies Coriolis effect to his targeting. There is no evidence that a formula which relies on theory would be accurate. The theory of gravity is theory. Pilots do not adjust a flight path for Coriolis effect. ---
**Author:** ![Czarcasm](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/czarcasm/32/4050_2.png) [@Czarcasm](https://boards.straightdope.com/u/Czarcasm)\ **Post date:** [November 29, 2018, 11:58pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/76 "2018-11-29T23:58:27Z")
> [@kertmoore](#): > > There is no evidence that a formula which relies on **theory** would be accurate. The **theory** of gravity is **theory**. (bolding mine) "You keep using that word. I do not think it means what you think it means.” ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 30, 2018, 12:15am UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/77 "2018-11-30T00:15:23Z")
> [@kertmoore](#): > > On centrifugal force, I did not enter the wrong velocity. Before discovering the centrifugal force calculator, I read the pounds force definition. I did not interpret the pounds force definition correctly. When I later ran the centrifugal force calculator, I applied the result incorrectly. > > Your comment on the error was helpful. After the correct interpretation of the calculator, the calculator still defines that each latitude produces a different weight result. > > Right now, I do not know how to apply the Coriolis effect force result to adjust an aircraft flight path. I have not found evidence that a sniper ever applies Coriolis effect to his targeting. There is no evidence that a formula which relies on theory would be accurate. The theory of gravity is theory. > > Pilots do not adjust a flight path for Coriolis effect. If you look up the thread, you will see where I calculated the effect of the Coriolis effect on a flight. Since the effect is small (a plane travels much slower than a bullet) a pilot doesn’t have to adjust for it - it is handled unconsciously just as all sorts of minor effects on flights (wind variation, etc.), but I’ll let the actual pilots in this thread answer the question. Here’s a sniper manual that talks about how to compensate for the Coriolis effect [http://index-of.co.uk/Tutorials-2/ultimate%20sniper%20book.pdf](http://index-of.co.uk/Tutorials-2/ultimate%20sniper%20book.pdf) ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [November 30, 2018, 1:30am UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/78 "2018-11-30T01:30:47Z")
> [@Andy\_L](#): > > If you look up the thread, you will see where I calculated the effect of the Coriolis effect on a flight. Since the effect is small (a plane travels much slower than a bullet) a pilot doesn’t have to adjust for it - it is handled unconsciously just as all sorts of minor effects on flights (wind variation, etc.), but I’ll let the actual pilots in this thread answer the question. > > Here’s a sniper manual that talks about how to compensate for the Coriolis effect > [http://index-of.co.uk/Tutorials-2/ultimate%20sniper%20book.pdf](http://index-of.co.uk/Tutorials-2/ultimate%20sniper%20book.pdf) An analogy regarding the difference between a bullet and a plane: Imagine a car that “pulls” to the right: some mechanical defect means that it will drift right unless action is taken. If the car is left in neutral at the top of a hill, when the car rolls downhill it will end up at the right curb (or actually jumping the curb). If you wanted the to not hit the curb, you’d have to aim it left before letting it go downhill. On the other hand, if someone was driving the car, the driver probably wouldn’t even notice a slight pulling to the right - he’d adjust for it automatically, without conscious effort, and not find out about the mechanical defect until he (for some reason) drove with his hands off the wheel for awhile (or took the car into the shop). I hope the analogy is clear. ---
**Author:** ![Andy\_L](https://avatars.discourse-cdn.com/v4/letter/a/c67d28/32.png) [@Andy\_L](https://boards.straightdope.com/u/Andy_L)\ **Post date:** [December 9, 2018, 10:33pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/79 "2018-12-09T22:33:06Z")
> [@Chronos](#): > > The machines do not appreciate being insulted. The machines are your friend. Do not worry about the machines. The Computer is your Friend. ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [December 14, 2018, 2:53pm UTC](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423/80 "2018-12-14T14:53:50Z")
I hate to bump this, since the OP has given up; but someone in another thread posted a phrase I’d forgotten: [Gish Gallop](https://rationalwiki.org/wiki/Gish_Gallop). > [@](#): > > The **Gish Gallop** (also known as **proof by verbosity** [sup][1][/sup]) is the [fallacious](https://rationalwiki.org/wiki/Logical_fallacy) debate tactic of drowning your opponent in a flood of individually-weak arguments in order to prevent rebuttal of the whole argument collection without great effort. [Previous page](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423.md?page=3) [Next page](https://boards.straightdope.com/t/article-june-25-2010-how-does-the-earth-s-rotation-affect-the-path-of-a-bullet/824423.md?page=5)