While watching the first 4 seconds of driving at 745 km/h is ludicrous from any angle wondered

  1. If we knew the curvature of the Earth in a "flat" desert, what would be the speed of the car?

  2. Assuming we don't know the curvature of the Earth, how long would it take for the car to go around the Earth ( assuming a "flat" desert all the way travelling on a great arc)?

Using my ninja pause skills it seemed it took 2 seconds from the time the car appeared ( as a dot ) in horizon to the time it passed by the camera, although 2 second window of observation seems to lead to a great error in overall estimates, using the 745 as the value for 1 and 2 the error of human observation could be calculated (?)

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    $\begingroup$ You need to know the height of the camera to know how far the horizon is on a flat surface. $\endgroup$ – Ron Maimon Nov 13 '11 at 6:14
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    $\begingroup$ Additionally, one cannot assume that the car popped over the horizon in the video, as the point at which it appears could have simply been the point where its angular diameter fills a single pixel of the camera. $\endgroup$ – Richard Terrett Nov 13 '11 at 7:37
  • $\begingroup$ @RichardTerrett : true, it seems there is nothing more can be done $\endgroup$ – Arjang Nov 14 '11 at 6:38
  • $\begingroup$ well, if you know the height of the car, even assuming the camera is at equal height can you give a good estimation. However, my guess is that there might be refraction atmospheric effects that will make the car visible much later than pure straight line optics would predict $\endgroup$ – lurscher Nov 14 '11 at 12:31
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    $\begingroup$ I hate to say "it depends", but it depends on how far away you can see the other car. If it's the size of an ant, not very far. If it's the size of a planet, really far. (As a pilot, this is not academic. You train yourself to inspect the sky for specs that appear not to be moving, because any other aircraft on a collision course with you looks exactly like that.) Also those awesome trains in France appear out of nowhere and flash past you. $\endgroup$ – Mike Dunlavey Dec 13 '11 at 22:55

In order to solve this question, we need to determine the distance from the eyes of the observer to the horizon, as this will be the relevant distance the car travels within those 2 seconds. Let's define a few variables:

$d =$ distance between observer and horizon
$v =$ velocity of the car
$t =$ time for car to travel from horizon to observer
$h =$ distance of observer's eyes above ground
$R =$ radius of Earth

This calculation makes a few considerable approximations. First, it assumes that the Earth is spherical, and that the region between the observer and the horizon is a part of that sphere. Light is assumed to move in perfectly straight lines, neglecting any effects such as diffusion or refraction.

Here is a highly exaggerated diagram showing the Earth, and a 'not-to-scale' observer: enter image description here

We can now create a right angle triangle, the three vertices located at the observer's eyes, the horizon (the end of the dotted line furthest from the observer), and the centre of the Earth. The sides of the triangle are $h+R$, $R$ and $d$. By a bit of Pythagoras' Theorem, it turns out that:

$$d = \sqrt{h^2+2hR}$$

The mean radius of the Earth, $R$, according to Wikipedia, is $(6.371 \times 10^6)m$, and let's assume that $h\approx 2m$. Therefore, $d\approx 5048m$.

Then, given that $t = 2s$, we can calculate the velocity of the car with:

$$v = \frac{d}{t} \approx 2524ms^{-1}$$

$$= 9086kmh^{-1}$$

So, with a velocity of around $9000kmh^{-1}$, and a Mach Number of around $7.4$, it goes without saying that we have stumbled upon a pretty absurd result. Which makes me question whether this deserves be an answer or a comment. Anyways, this value gives us a few possible scenarios to work with:

  • The radius of curvature of the desert may be a lot smaller than that of the Earth, i.e. there might have been a hill
  • The car can go ridiculously fast without becoming scrap metal (unlikely)
  • The camera resolution/human observation cannot detect something crossing the literal horizon
  • The camera is a mere few centimetres above the ground
  • The effects of light diffusion and refraction are significant to the distance of the horizon
  • The above theory and calculations are wrong
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    $\begingroup$ If the camera was 3 cm from the ground (0.2m), then you get about 784 km/hr. If you just laid the camera on the ground (unmounted, that is), I imagine a few cm would be appropriate. Another consideration: the car is accelerating during the few frames, in which case the constant-velocity method fails. $\endgroup$ – Kyle Kanos Dec 20 '14 at 21:46
  • $\begingroup$ and the height of the car is needed also. Remember ships appear on the horizon mast first. $\endgroup$ – ja72 Dec 21 '14 at 1:25

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