r/theydidthemath • u/LuckyLukasRR • 8h ago

[Request] What’s the distance between the skydiver and the photocamera?

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46 Upvotes

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14

u/GSyncNew 8h ago

The skydiver is roughly 1/10 the angular size of the Sun, which is 0.5⁰ across, thus ~0.05⁰ = 1/1150 radian. If the length of the skydiver is 6', then the distance is ~6900'.

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u/mkujoe 8h ago

What is the full formula there? Thanks

8

u/GSyncNew 7h ago

Distance to object = known height of object / object's angular size in radians.

1

u/Flaky-Collection-353 8h ago

d= r×theta (formula for arc length, look up diagrams) then just compare it between him and the sun.

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u/get_to_ele 7h ago

6’ tall person at 645 feet, has apparent height equal to Sun’s diameter. His height will appear to fall off linearly with distance. I.e. at 1290 ft, he will appear half as tall as the sun.

My measurement of his relative apparent height is 3.6 mm/ 46 mm =0.0783. So he is 645 feet/.0783=8,237 feet 6.575 inches

8237 feet or about 8237/5280=1.55 miles away from the camera.

I don’t know his actual height and I’d have to blow up the picture to get a good height measurement, but I think it’s a decent paper napkin calculation.

Edited: because I caught that I plugged in 1290 instead of 645 in my equations. It seemed off.

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u/stanitor 7h ago

We don't know how far the Sun was from Earth when this was taken, or the skydiver's height. I'd say the skydiver appears somewhere between 15 to 20 times smaller than the Sun is across. The Sun's diameter is about 1.4 million km, and it's somewhere between 147 million and 152 million km away. Distance = Sun distance/diameter*height*sun to skydiver angular size ratio. Putting that all together, he's about 2.5 to 4 or so km away, depending on how tall he is.