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I was watching the first episode of the Genius documentary series by Dr. Hawking. At the beginning of the documentary, the people prove that the Earth is round and not flat by going to a lake, and seeing how a laser beam moves in height as they move in the lake. Then they calculate the circumference based on that.

The "data" would: first they go into the lake 3 miles from the laser beam, and the laser beam is at a height of 6 feet. Afterward, they go farther (to do that they use a helicopter and a telescope to see when the chopper disappeared from the view of the telescope, where the laser beam was before), they say that they went 6 miles and chopper disappeared when they were 24 feet above the ground.

So our data basically is: at 3 miles the laser is at 6 feet, at 6 miles it would be at 24 feet. I tried to do the calculations, and my Earth has a circumference twice that of our known planet... I'm using I made some stupid error but I can't find it.

I attach an image with my calculations (I changed everything into km). Image with calculations I apologize in advance because the error is probably quite absurd, but I'm going crazy.

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  • $\begingroup$ Did the laser beam start at ground level? $\endgroup$
    – Mark H
    Commented Aug 13, 2023 at 22:00
  • $\begingroup$ I can't quite figure it out but it seems to be a laser which is floating on a bed so that it is always horizontal, and then 3 miles or 6 miles away they have some sort of yardstick which measures where the laser beam is hitting, and it's six feet then twenty-four feet above the base level when the laser was adjacent to the yardstick. $\endgroup$ Commented Aug 13, 2023 at 22:25
  • $\begingroup$ The laser was at the bottom of the lake, and it is supposed to be always horizontal in the same position. You can check the experiment here: pbslearningmedia.org/resource/hawking_genius_ep06_clip01/… $\endgroup$ Commented Aug 14, 2023 at 9:33

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The way that you're doing the calculation using similar triangles, what you have as the "opposite" in the first equation is actually approximately twice the height of the laser beam, which you will see if you draw your diagram more carefully. If you use a factor of two before it you'll get the correct height. Here is a computer program in Perl which does it for you:

use Math::Trig;
my $hyp = sqrt (4800**2+(2*1.8)**2);
my $theta = acos (4800/$hyp);
print 4800/tan ($theta);

But it would be smarter to use the equations to eliminate that angle completely.

I don't have time to write all the equations out and draw the diagram more carefully for you, but there is a very nice web page here which gives all the details of how to calculate the horizon. Here is the diagram from that reference, copyright Andrew T. Young.

dip of the horizon, by Andrew T. Young

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  • $\begingroup$ Using this answer I found a useful video. Basically from the video we get R = D^2/(2*C) = 9.6km^2/(2*7.3m) and we get the Radius of the Earth and from that the circumference :) $\endgroup$ Commented Aug 14, 2023 at 17:17
  • $\begingroup$ @TaelaShanthi I'm glad you were able to solve the problem using the answer and the video. Keep up the good work. $\endgroup$ Commented Aug 14, 2023 at 21:32

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