Distance to the HorizonDate: 09/18/97 at 23:08:17 From: Tony Turner Subject: Geometry/Trig If a 6-foot man is standing on the beach at sea level looking straight out to sea, how far can he see - i.e. what is the distance from the man to the horizon? The radius of the Earth is approximately 3963.21 miles. Also, this question assumes the Earth is perfect circle. We have been arguing over this question for two months at work. Using a squared + b squared = c squared, we came up with 3 miles. We know this is wrong because you can see farther than 3 miles. Any help would be greatly appreciated. Tony Date: 09/22/97 at 16:35:32 From: Doctor Ken Subject: Re: Geometry/Trig Hi Tony. You're right, you need to use the Pythagorean Theorem to solve this problem. If you let the radius of the Earth be R, and the distance your 6-foot man can see be D, then you get this formula (I assume you got this far): R^2 + D^2 = (R + 6)^2 Which simplifies to: D^2 = 12R + 36 D^2 = 12*20925748.8 + 36 (plug in radius of earth in feet) D^2 = 251109021.6 (do the arithmetic) D = Sqrt(251109021.6) D = 15846.4 feet D = about 3 miles So you were right. But note that this actually tells you how far away the farthest point _on_the_surface_ of the Earth is. If you're trying to see some other 6-foot person standing on a raft way out in the ocean, he/she can actually be 6 miles away (you'll only be able to see the top of his/her head, though). Can you draw a picture to justify this? An interesting exercise for you might be to come up with a formula that tells you how far away you can see another object - given your height and the height of another object, how far away can the other object be while you can still see it? Again, your solution would use the Pythagorean Theorem. Another interesting thing to try would be to come up with a good approximation to your formula that's easier to compute in your head. If you can come up with something like this, it will come in handy in mountainous regions - if two mountains are 9 miles apart and you can see the top of one from the top of the other, what can you say about their height? -Doctor Ken, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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