<blockquote>
  <p>I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.</p>
</blockquote>

<p>You're being sold a bridge with his numbers. If he had done the math correctly he would have obtained a deflection of 33.58 cm at 45° geocentric latitude rather than 17.245 cm. The youtube comments show a number of errors. Two key errors are that the commenter assumes that</p>

<ul>
<li>Local vertical points directly toward the center of the Earth.</li>
<li>Gravitational acceleration due to the Earth's mass points in the same direction.</li>
</ul>

<p>What those youtube comments do show is that the Earth is not quite a sphere. A closer approximation is an oblate spheroid. That said, his comments fall into the wronger than wrong category. Quoting Isaac Asimov,</p>

<blockquote>
  <p>When people thought the Earth was flat, they were wrong. When people thought the Earth was spherical, they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together.</p>
</blockquote>

<p>Note that I qualified latitude in the above as geocentric. Geodesists have multiple concepts of latitude. Geocentric latitude is the angle between a line to the center of the Earth and the Earth's equatorial plane. Geodetic latitude is the angle between the normal to the reference ellipsoid and the Earth's equatorial plane. Yet another concept is astronomical latitude, which is the angle between true vertical and the Earth's equatorial plane.</p>

<p>The difference between geocentric and geodetic latitude is at its greatest at 45°. A geocentric latitude of 45° is equivalent to a geodetic latitude of 45° 11' 32.7". Converting that 11' 32.7" difference to radians and multiplying by 100 meters results in 33.58 cm. The difference between geodetic and astronomical latitude is called the deflection of the vertical (or vertical deflection). The deflection of the vertical is rather small, less than 2' (two minutes of arc) even in extremely mountainous terrain. Typical values are a few tens of arc seconds.</p>

<p>What this means is that an oblate spheroid is a better model of the Earth's shape than a sphere. That does not mean a flat Earth model is correct. The Earth's flattening (how squished the Earth is at the poles versus the at the equator) is rather small. The poles are 22 km closer to the center of the Earth than is the equator. Percentage-wise, that's a 0.345% difference.</p>

<p>The reason an oblate spheroid is a better model of the Earth's shape than a sphere is because the Earth is spinning. That spin does two things: It results in a centrifugal acceleration from the perspective of an observer fixed with respect to the rotating Earth, and it results in an equatorial bulge. If the Earth was a uniform density liquid, the spin would make the Earth naturally relax into an oblate spheroidal shape. "Straight down", the combined effects of gravitational and centrifugal acceleration (geodesists call this "gravity" to distinguish it from gravitation) would be exactly normal to the surface.</p>

CC BY-SA 4.0
2019-10-01T15:11:07.337Z
TBH, I'm shocked that it's as much as 17 *or* 34cm over a 100m drop. I expected an answer in the millimeters at most. But I'm prepared to accept that my intuition is wrong :-)
Semidiurnal Simon
CC BY-SA 4.0
2019-10-01T15:33:31.680Z
@SemidiurnalSimon - It's a consequence of a bad definition of "down" (or "up"). Down is not toward the center of the Earth. It's whatever direction gravity pulls. The difference between "down" and "toward the center of the Earth" is at its greatest at 45° latitude. Except for a few regions (e.g., parts of the Himalaya), the difference between "down" and the inward normal to the reference ellipsoid is over than an order of magnitude smaller than the difference between geocentric and geodetic latitude.
David Hammen
CC BY-SA 4.0
2019-10-01T15:38:15.870Z
Yes, I understand the mechanism - I explained it in my own comment (albeit with less technical language, because I am not a geodesist). I'm just surprised that the effect is so large. That's all.
Semidiurnal Simon
CC BY-SA 4.0
2019-10-02T05:45:57.290Z
Thanks.  Despite the source of the question, it was a great learning opportunity.
Russ Walker
CC BY-SA 4.0
2019-10-01T10:49:32.363Z
2019-10-01T11:24:15.397Z
2019-10-01T11:24:15.397Z
David Hammen
<p>I've been asked this question twice now from our favourite science denying community, and I don't know how to parse it.  I can do centrifugal force and gravity given mass, but have never tried to even see if there should be any deflection of gravity due to centrifugal force at a given latitude.</p>

<p>I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.</p>

<p>Here's the numbers he's given me:</p>

<ul>
<li>Gravitational force (m/s^2) =  9.80576 </li>
<li>Latitude (degrees) = 45</li>
<li>Velocity (mph) = 735.64 </li>
<li>Centrifugal force (m/s^2) = 0.0239809</li>
<li>Gravity    Direction Vector = (0.707107, 0.707107)</li>
<li>Centrifugal force Direction    Vector = (1, 0)</li>
<li>Combined Force = (-6.90974,-6.93372)</li>
<li>Combined Force    (relative to ground) = (0.016957,-9.7888)</li>
<li>Weight reduction (%)    compared to North pole = 0.172929</li>
<li>Projectile drop (time=4.51 seconds)</li>
<li>Deflection south (cm) = 17.2454</li>
<li><p>Drop height 100 m.</p>

<p>How on Earth do I go about double checking what he's telling me? I'll be attempting to share the answer on YT as a reference for handling this supposed "Gotcha!"</p>

<p>From the CF he provides, I'm estimating an object at the surface would have a 712 g mass (from CalcTools), and a 17 cm drift for a 100 m drop for that kind of mass seems like he's off by a factor of 10 (from examples at different lats).</p></li>
</ul>

<p><a href="https://www.youtube.com/channel/UCHZ2fjTguMMT0EDGSCcPoEw" rel="nofollow noreferrer">https://www.youtube.com/channel/UCHZ2fjTguMMT0EDGSCcPoEw</a></p>

CC BY-SA 4.0
2019-09-30T08:54:44.397Z
Please provide some textual context. What is the statement that you want to refute, and where is it exactly? Your link to a Youtube *channel* adds no useful info.
Jan Doggen
CC BY-SA 4.0
2019-09-30T08:59:17.947Z
Excuse me.  The channel is mine where I respond to gotchas, but this one's just a step too far for me.  The claim is that an object at 45 deg N lat will deviate in a 100 m drop by 17 cm due to the vector addition from centrifugal force.  He anticipates that this southwards vector would be measurable on an accelerometer,  My belief is that he's fudged the numbers, after seeing an example of a cannonball in freefall at 32 deg N lat deviate by only 1.9 cm.  My belief is that a typical accelerometer probably wouldn't detect that.
Russ Walker
CC BY-SA 4.0
2019-09-30T09:00:41.997Z
Still: who is *he* and what is he claiming? Please take a step back and look at your question with a new look: will an outsider understand what you are asking here?
Jan Doggen
CC BY-SA 4.0
2019-09-30T09:07:43.793Z
Ok, https://www.youtube.com/watch?v=m2yjiW2STok&lc=z23xjbbrwnn4zjxlvacdp435d34to1p1r1nmon4i2wpw03c010c.1569826217950698 He's a random user on a comment thread.  The moon claim in that thread wasn't a problem.  I happened to have calculated a pretty decent value for that last week based on its angular size.  This claim, however, I don't remember how to tackle at all.  They typically start parroting each other when they think they've got a gotcha.
Russ Walker
CC BY-SA 4.0
2019-09-30T15:58:36.613Z
Why are you bothering with this, @RussWalker? You cannot win a debate with such people; they have been fully inoculated against logic and evidence.
David Hammen
CC BY-SA 4.0
2019-09-30T16:17:58.287Z
Re "Gravitational force (m/s^2) = 9.80576" This is where the crackpot you are trying to debate starts to go wrong. The wrongness continues on down to "Combined Force (relative to ground) = (0.016957,-9.7888)". Uh no. Combined acceleration (relative to ground) = (0.0, -9.806199) at 45 degrees latitude. This guy doesn't know which way up and down point.
David Hammen
CC BY-SA 4.0
2019-09-30T16:46:50.237Z
Thanks.  It's not about winning a debate.  It's about knowing how to check what he's telling me.  Spotting that error is sensible now you show me it.
Russ Walker
CC BY-SA 4.0
2019-09-30T17:35:50.910Z
I think you might have better success at SE Physics
Fred
CC BY-SA 4.0
2019-09-30T18:56:19.900Z
Not a full answer as I lack time, but: Centrifugal force does have an effect on apparent gravity, both in magnitude and direction. This is because the force due to gravity is towards the centre of the earth, but the centrifugal force is away from the earth's axis of rotation - so unless you're on the equator, the two vectors are not exactly opposite in direction. See https://earthscience.stackexchange.com/questions/14514/does-the-coriolis-effect-affect-a-wind-from-east-to-west However, this effect is small, and intuitively the numbers that you give (17cm drift on a 100m drop) sound v unlikely.
Semidiurnal Simon
CC BY-SA 4.0
2019-10-01T06:21:47.253Z
Thanks everyone.  I've an idea how to approach it now.  I'll step through it one piece at a time.
Russ Walker
CC BY-SA 4.0
2019-10-01T11:45:54.560Z
@Friddy - Centrifugal acceleration has nothing to do with the rock *per se*. It is instead a consequence of the observer being in a non-inertial frame, e.g., someone standing still on the surface of the Earth describing the motion of a rock dropped from 100 meters.
David Hammen
CC BY-SA 4.0
2019-10-01T11:56:09.870Z
@SemidiurnalSimon - The youtube poster was off by a factor of two. It's a 34 cm drift rather than 17 cm. The youtube poster intentionally chose a suboptimal definition of "up" (or "down") as being a line from the center of the Earth to the surface. Newton showed that the Earth must deviate a bit from being spherical over 300 years ago. That a spherical model is slightly substandard doesn't mean that the Earth must therefore be flat. That's a non sequitur.
David Hammen
<geophysics><models><planetary-science>
Effects of Centrifugal Force on Gravity Vectors?
Russ Walker
# Effects of Centrifugal Force on Gravity Vectors?

(Asked by: username_0 on 2019-09-30)

I've been asked this question twice now from our favourite science denying community, and I don't know how to parse it. I can do centrifugal force and gravity given mass, but have never tried to even see if there should be any deflection of gravity due to centrifugal force at a given latitude.

I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.

Here's the numbers he's given me:

* Gravitational force (m/s^2) = 9.80576
* Latitude (degrees) = 45
* Velocity (mph) = 735.64
* Centrifugal force (m/s^2) = 0.0239809
* Gravity Direction Vector = (0.707107, 0.707107)
* Centrifugal force Direction Vector = (1, 0)
* Combined Force = (-6.90974,-6.93372)
* Combined Force (relative to ground) = (0.016957,-9.7888)
* Weight reduction (%) compared to North pole = 0.172929
* Projectile drop (time=4.51 seconds)
* Deflection south (cm) = 17.2454
* Drop height 100 m.

  How on Earth do I go about double checking what he's telling me? I'll be attempting to share the answer on YT as a reference for handling this supposed "Gotcha!"

  From the CF he provides, I'm estimating an object at the surface would have a 712 g mass (from CalcTools), and a 17 cm drift for a 100 m drop for that kind of mass seems like he's off by a factor of 10 (from examples at different lats).

<https://www.youtube.com/channel/UCHZ2fjTguMMT0EDGSCcPoEw>

**Tags:** geophysics, models, planetary-science

> **Comments:**
> **username_1**: Please provide some textual context. What is the statement that you want to refute, and where is it exactly? Your link to a Youtube *channel* adds no useful info.  
> **username_0**: Excuse me.  The channel is mine where I respond to gotchas, but this one's just a step too far for me.  The claim is that an object at 45 deg N lat will deviate in a 100 m drop by 17 cm due to the vector addition from centrifugal force.  He anticipates that this southwards vector would be measurable on an accelerometer,  My belief is that he's fudged the numbers, after seeing an example of a cannonball in freefall at 32 deg N lat deviate by only 1.9 cm.  My belief is that a typical accelerometer probably wouldn't detect that.  
> **username_1**: Still: who is *he* and what is he claiming? Please take a step back and look at your question with a new look: will an outsider understand what you are asking here?  
> **username_0**: Ok, https://www.youtube.com/watch?v=m2yjiW2STok&lc=z23xjbbrwnn4zjxlvacdp435d34to1p1r1nmon4i2wpw03c010c.1569826217950698 He's a random user on a comment thread.  The moon claim in that thread wasn't a problem.  I happened to have calculated a pretty decent value for that last week based on its angular size.  This claim, however, I don't remember how to tackle at all.  They typically start parroting each other when they think they've got a gotcha.  
> **username_2**: Why are you bothering with this, @username_0? You cannot win a debate with such people; they have been fully inoculated against logic and evidence.  
> **username_2**: Re "Gravitational force (m/s^2) = 9.80576" This is where the crackpot you are trying to debate starts to go wrong. The wrongness continues on down to "Combined Force (relative to ground) = (0.016957,-9.7888)". Uh no. Combined acceleration (relative to ground) = (0.0, -9.806199) at 45 degrees latitude. This guy doesn't know which way up and down point.  
> **username_0**: Thanks.  It's not about winning a debate.  It's about knowing how to check what he's telling me.  Spotting that error is sensible now you show me it.  
> **username_3**: I think you might have better success at SE Physics  
> **username_4**: Not a full answer as I lack time, but: Centrifugal force does have an effect on apparent gravity, both in magnitude and direction. This is because the force due to gravity is towards the centre of the earth, but the centrifugal force is away from the earth's axis of rotation - so unless you're on the equator, the two vectors are not exactly opposite in direction. See https://earthscience.stackexchange.com/questions/14514/does-the-coriolis-effect-affect-a-wind-from-east-to-west However, this effect is small, and intuitively the numbers that you give (17cm drift on a 100m drop) sound v unlikely.  
> **username_0**: Thanks everyone.  I've an idea how to approach it now.  I'll step through it one piece at a time.  
> **username_2**: @Friddy - Centrifugal acceleration has nothing to do with the rock *per se*. It is instead a consequence of the observer being in a non-inertial frame, e.g., someone standing still on the surface of the Earth describing the motion of a rock dropped from 100 meters.  
> **username_2**: @username_4 - The youtube poster was off by a factor of two. It's a 34 cm drift rather than 17 cm. The youtube poster intentionally chose a suboptimal definition of "up" (or "down") as being a line from the center of the Earth to the surface. Newton showed that the Earth must deviate a bit from being spherical over 300 years ago. That a spherical model is slightly substandard doesn't mean that the Earth must therefore be flat. That's a non sequitur.  

---

## 1 answers

### Answer by username_2 on 2019-10-01. Score: 0 | Accepted answer

> I'm guessing I'm being sold a bridge with his numbers, but can't tell for sure.

You're being sold a bridge with his numbers. If he had done the math correctly he would have obtained a deflection of 33.58 cm at 45° geocentric latitude rather than 17.245 cm. The youtube comments show a number of errors. Two key errors are that the commenter assumes that

* Local vertical points directly toward the center of the Earth.
* Gravitational acceleration due to the Earth's mass points in the same direction.

What those youtube comments do show is that the Earth is not quite a sphere. A closer approximation is an oblate spheroid. That said, his comments fall into the wronger than wrong category. Quoting Isaac Asimov,

> When people thought the Earth was flat, they were wrong. When people thought the Earth was spherical, they were wrong. But if you think that thinking the Earth is spherical is just as wrong as thinking the Earth is flat, then your view is wronger than both of them put together.

Note that I qualified latitude in the above as geocentric. Geodesists have multiple concepts of latitude. Geocentric latitude is the angle between a line to the center of the Earth and the Earth's equatorial plane. Geodetic latitude is the angle between the normal to the reference ellipsoid and the Earth's equatorial plane. Yet another concept is astronomical latitude, which is the angle between true vertical and the Earth's equatorial plane.

The difference between geocentric and geodetic latitude is at its greatest at 45°. A geocentric latitude of 45° is equivalent to a geodetic latitude of 45° 11' 32.7". Converting that 11' 32.7" difference to radians and multiplying by 100 meters results in 33.58 cm. The difference between geodetic and astronomical latitude is called the deflection of the vertical (or vertical deflection). The deflection of the vertical is rather small, less than 2' (two minutes of arc) even in extremely mountainous terrain. Typical values are a few tens of arc seconds.

What this means is that an oblate spheroid is a better model of the Earth's shape than a sphere. That does not mean a flat Earth model is correct. The Earth's flattening (how squished the Earth is at the poles versus the at the equator) is rather small. The poles are 22 km closer to the center of the Earth than is the equator. Percentage-wise, that's a 0.345% difference.

The reason an oblate spheroid is a better model of the Earth's shape than a sphere is because the Earth is spinning. That spin does two things: It results in a centrifugal acceleration from the perspective of an observer fixed with respect to the rotating Earth, and it results in an equatorial bulge. If the Earth was a uniform density liquid, the spin would make the Earth naturally relax into an oblate spheroidal shape. "Straight down", the combined effects of gravitational and centrifugal acceleration (geodesists call this "gravity" to distinguish it from gravitation) would be exactly normal to the surface.

> **Comments:**
> **username_4**: TBH, I'm shocked that it's as much as 17 *or* 34cm over a 100m drop. I expected an answer in the millimeters at most. But I'm prepared to accept that my intuition is wrong :-)  
> **username_2**: @username_4 - It's a consequence of a bad definition of "down" (or "up"). Down is not toward the center of the Earth. It's whatever direction gravity pulls. The difference between "down" and "toward the center of the Earth" is at its greatest at 45° latitude. Except for a few regions (e.g., parts of the Himalaya), the difference between "down" and the inward normal to the reference ellipsoid is over than an order of magnitude smaller than the difference between geocentric and geodetic latitude.  
> **username_4**: Yes, I understand the mechanism - I explained it in my own comment (albeit with less technical language, because I am not a geodesist). I'm just surprised that the effect is so large. That's all.  
> **username_0**: Thanks.  Despite the source of the question, it was a great learning opportunity.  

---

