'average' rotational speed for a point on a rotating sphere?

A sphere is spinning such that a point on the equator has a rotational speed of x cm/sec. I’m dealing with the scalar value here, not the velocity vector.

As you move from the equator to the pole the rotational speed decreases by the cosine of the latitude until you are at the pole where the speed of rotation is zero.

So - if I sampled a trillion random points on the sphere, what would the mean, median, and mode rotational speeds be for that sample?

I’m experiencing a kind of mental hangup that insists that not only is the speed distribution not linear from the pole to the equator, but also that in some vague sense there is more surface area per degree of latitude near the equator, so a function that is based on latitude has to account for this somehow. I may be totally wrong, but I’m stymied. Perhaps I am not formulating the problem correctly.

My gut says that the mean would be something like 0.66x, the median closer to 0.8x, and the mode would be x cm/sec.

This is the key point. By “random points on the sphere”, you mean distributed uniformly by surface area?

The median speed would be the speed at 30º latitude, sqrt(3)/2 times the equatorial speed, since half of the surface area is closer to the equator than that and half is further. The mode would be the equatorial speed. The mean would be pi/4 times the equatorial speed (what latitude that would be is left as an exercise for the reader).

Agree as to mean and median.

Would it?

I agree that considering just e.g. the equator + northern hemisphere (0<= latitude <= +90) we will find that the mode, or most common value, would be the equatorial one, since that is the longest circle containing the most points.

But as soon as we consider both hemispheres, I think the mode would become latitude = ±0.000…0001 degrees. Whose total point count of both circles (north and south) would be just shy of double that of the equator.

Clearly as the sampling gets more extensive eventually we get into the issue of “how many points are on a line of length x?”, and all the rest of the differences between continuous functions vs discrete ones.

Nice.
In the limit as the angle approaches the equator from either direction, the speed approaches the equatorial speed. So the mode is still the equatorial speed. Same logic for sampling. The mode always approaches the equatorial speed with more samples. It doesn’t matter if there is a discontinuity at the equator. Mathematicians cheerfully do such things. Physicists do so without even breaking stride.

The usual argument is of the form, no matter how close a number you choose close to the limit, I can always find one closer.

But with how many zeroes before that 1? No matter how many zeroes you add, you can get better by adding more. In other words, there are an infinite number of zeros before the 1. Which means that the modal latitude is, in fact, exactly equal to 0.

I should probably show my work. For every point on the surface of the Earth, we can define its distance from the equatorial plane (i.e., if we put the equator on the XY Cartesian plane, the Z coordinate of that point, or R*sin(latitude) ). Picking a Z value at random between -R and +R will distribute points uniformly across the surface of the sphere. For simplicity, we can consider just one hemisphere, or Z from 0 to R.

The median is the easiest, since that corresponds to a Z of R/2 (half the points above and half below). Simple trig shows that this corresponds to a latitude of 30º.

For the mode, since all Z values are equally likely, and speed is a function of Z, we want the Z value for which dv/dZ is a minimum. Speed is proportional to the distance from the axis, and so dv/dZ is 0 at the equator (and greater everywhere else). So the modal speed is the equatorial speed.

For the mean, we want the average value of the function v0*sqrt(R^2-Z^2) from 0 to R, which means the integral of that function divided by the width of the interval. The integral is a bit annoying, unless you remember that the area so defined is just a quarter circle, with an area of (pi*R^2)/4.

Thank you both.

Been too long. About time for me to retire from posting to math / calculus logic threads.

There’s some YouTube videos out there, possibly on Numberphile or StandupMaths, that point out that picking such random points is more involved that people think. (YouTube’s search algorithm for channels is terrible.)

If someone showed me some data they got out of picking “random” points on a on a sphere I would seriously wonder if they did it right. And that’s not even getting into the impossibility of picking a number in a “uniform” range of the reals.

Here’s a Wolfram MathWorld article about it.

Here’s a Numberphile video about a paradox you get when you pick two “random” points (a chord) on merely a circle.

Right, that’s why I started by asking for clarification on what the OP meant by “randomly picking points”. But props that it was a lot closer to being unambiguous than I expected on seeing the title.

Which makes the mode meaningless. BTW: I know the OP put it in quote marks but mode is not an average since you can have more than one for a data set. But pedantry aside …
As stated by the OP, the mode would be whichever line of latitude has the most points on it. I agree with Chronos’ figures so far as the limit as the number of points become infinite, but for a large finite number of points while the real results will be close to Cronos’ results for the arithmetic mean and median, I do not believe the same can be said for mode. Look at it this way. Place one-trillion (from the OP) real numbers on the number line from 0 to 1. What is the median of those points? What is the median? Both will be extremely close to 0.5. To calculate the mode, what is the value on the number line with the most points. I believe that because it is a finite subset of an infinite set, that p(2 or more points at any x) = 0. Therefore no mode.

For any real-world problem, when we take the mode, we need to have some sort of binning. If we’re looking at rotational speeds, we might bin it as speeds between 0.9v and v, speeds between 0.8v and 0.9v, speeds between 0.7v and 0.8v, etc. Or if we have enough data points to be able to afford to do so, we might use bins of 0.99v to v, 0.98v to 0.99v, etc., or even finer. In this problem, no matter what width of binning we use, the bin containing the most points will be the top one.

When we talk about the mode of the distribution itself, rather than of the data we collect from the distribution, what we mean is to take the limit as the number of data points approaches infinity and the width of each bin approaches zero. Since with finite-width bins, it was always the top one that had the most points, in the limit, the mode will be the highest number.

It’s true that some distributions don’t have a mode, and some distributions have multiple modes. But the distribution in this problem isn’t one of them: It does, in fact, have one and only one mode.

Except that is not how the problem is stated. I agree with the limit conclusion but the OP stated a finite number of points (one trillion) and I agree that if we set up bins like LSLGuy alludes to, it would be the bin at the equator, but again that’s different than the OP. You’d have to rewrite it (like you did) for the discussion of mode to work.

Yes. It’s always worthwhile to poke at how “random” things are selected.

To clarify, when standing on the North pole, you’re still rotating (in the sense that the sun will go around your front from your left to your right and then behind you), but your linear speed around the axis is zero.

One assumes, when someone asks the question “What is the mode?”, that the mode is defined. Any definition of “mode” in this case will give the same answer. Even if someone asks something like “What is the mode of a uniform distribution?”, “mode” is still defined (and the definition leads to the answer “the mode does not exist”).

And I claim that given the original conditions there is no mode. Yes mode is always defined and what I was referring to was the discussion of mode like LSLGuy’s use of bins or your use of the limit.

With a finite number of points, there is a mode. With an infinite number of points, there’s still a mode. With any number of points, the mode is always the same. I’m still not seeing any argument by which there isn’t a mode.

If so, then the mode would be all points

No, the mode would be the bin at the equator edge.

Just to be clear:

We are discussing point sampling on the sphere’s surface, and not points within the volume…

Correct?

Yes, surface points only.

I understand your arguments for the median and mode, but how do you get pi/4 as the mean?