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I and my friends periodically go on a trip. This is a "round" trip as we always end up back in our original location.

I noticed something curious:

Even though...

  • We all take the same vehicle,
  • We all leave simultaneously,
  • If we were to use stopwatches, the trip would last exactly the same amount of time,

...there is a measurable difference in how far we've traveled.

To the best of my ability to measure this discrepancy,

  • Friend "A" travels the farthest, we'll call it 100% of the possible distance (>= than any of the others).
  • Friend "B" travels the smallest distance (<= others), which is ±98.79% of the distance traveled by "A".

Can you tell me what vehicle we are using and what trip we are taking?

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    $\begingroup$ '±98.79% of the distance traveled by "A".' What does the ± sign there mean? It seems like you mean B's distance is 1.21 % to 198.79 % of that traveled by A, but you said that B travels the smallest distance. $\endgroup$ – JiK Sep 26 at 9:34
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    $\begingroup$ @JiK I'm sure OP meant "about/approximately" 98.79% of A, a little more or a little less than 98.79%. $\endgroup$ – HeadhunterKev Sep 26 at 10:12
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I believe, you two are using:

A normal car...

And the trip will be:

A normal trip...

What, why?

For simplicity, assume that the trip is in a full circular route. But it can be generalized to any route although harder to prove. Also assume that the trip is in anti-clockwise direction.

Still, why?

Again, assume that the car driver is on the right seat. And again, assume that the left side and the right side are separated just $60.5$ cm.

So?

A is always the driver because B can't drive!

Huh?

The distance they take is different, because of... MATH!

Oh!

If the diameter of the trip is exactly $100$ meters, then the distance of A will be $\pi \times 100$ while B will be $\pi \times (100-2 \times 0.605)$, so they are $100:98.79$!

And yup, both get a window seat! :)

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  • $\begingroup$ Even though I intended the larger, spherical vehicle shown in other answers, this was faster, correct, and very well explained. $\endgroup$ – Chowzen Sep 25 at 19:29
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Perhaps

Your "vehicle" is the planet earth, and your "friends" are other people at your longitude. Your "trip" starts at sunrise on one of the equinoxes, and ends 24 hours later. You're all traveling on the same vehicle (planet earth), and your trip takes the same amount of time (24 hours), but depending on your latitude, you will travel different distances. A friend on the equator will travel 24,901 miles, while a friend at the poles will travel 0 miles. Select the latitudes of friend A and friend B as you like so that B travels 98.79% of the distance of A.

And of course, everyone has a window seat for this trip.

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  • $\begingroup$ I guess we're ignoring the distance travelled by the vehicle on its own, longer journey here? $\endgroup$ – Toby Speight Sep 26 at 10:54
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    $\begingroup$ @TobySpeight That distance will be very nearly identical for everyone on the planet, so it doesn't really factor in. The orbital path distance around the sun will vary very slightly with latitude, although that variation will be minimized at the equinox. I'm quite certain it's possible to pick two latitudes such that one travels 98.79% the distance of the other, even accounting for the distance traveled by the earth around the sun. Two orbits separated by an earth-width differ in their path by only ~137 miles in the course of a day, and nearly all of that will be averaged out by rotation. $\endgroup$ – Nuclear Wang Sep 26 at 13:17
  • $\begingroup$ @TobySpeight Actually, I have to take back my previous comment. The earth travels somewhere around 1.6M miles on its path around the sun in one day. If that's taken into account in the "distance traveled", it swamps any variation due to latitude/rotation, which can only be ~25k miles at most. There would be no way to vary by by more than 1%, so the 98.79% figure wouldn't work out. But you could pick a different reference frame for the car in the accepted answer and have the exact same "problem", so using the earth itself as reference is justified. $\endgroup$ – Nuclear Wang Sep 27 at 13:15
  • $\begingroup$ Ah, yes, it's all reference-frame dependent. And a rotating reference frame could make even "stationary" vehicle and passengers travel different distances! BTW, in my original comment, I was thinking that the time period was a calendar day, which would make a difference to passengers on the sides facing toward/away from the sun, but that's much less so if we use a sidereal day. $\endgroup$ – Toby Speight Sep 27 at 13:30
  • $\begingroup$ @TobySpeight (I'm gonna regret this) Here is how I got 98.79%: One year is 365.2425 days, making 9095071.83 miles at equator in one year and 1897056.83 miles at 78°.. BTW, this is how I picked 78°. Add to each of these a trip around the sun (584002364.58 miles). Equator trip: 593097436.41 mi. 78° trip: 585899421.41 mi. Hence, 98.79%. $\endgroup$ – Chowzen Sep 28 at 11:13
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I'm thinking

It's a normal day, on the earth. The vehicle is the earth itself, and after 24 hours, you all wind up at the same place as you started, but those friends further from the equator don't travel as far.

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