The Top of the Tower
Date: 07/20/2001 at 15:02:38 From: Ashley Sherman Subject: Word problem Given a 300-mile-tall tower on the earth, how far away from it can you go before the top disappears below the horizon? (Assume unlimited visibility.) We've drawn diagrams and triangles to try to figure it out using trigonometry, but we're looking for an equation(s) that will help us get the answer.
Date: 07/23/2001 at 11:41:00 From: Doctor Ian Subject: Re: Word problem Hi Ashley, It's hard to draw a circle with a keyboard, but let's pretend this is the earth: . . . . . . . . . . . . Also, let's pretend the earth is a sphere, and that it's perfectly smooth, even though neither of those things is true. If your eye is at ground level at some point, then you can look out tangent to the earth: <---------------- . . . . . . . . . . . If the top of a tower of height h just intersects your line of sight, <---------.------ . . . . . h . . . . . . . then you can form a right triangle with you, the center of the earth, and the tip of the tower at the corners: A C <---------.------ . . . . . . . h . . . . . B . . . . . If the radius of the earth (in miles) is r, then AB = r BC = r + h and r cos(ABC) = ----- r + h On a sphere, the distance along the surface (s) corresponding to a central angle theta is s = r * theta So the distance you'd have to walk from point A to the base of the tower would be s = r * theta r = r * arccos( ----- ) r + h For r = 3950 miles and h = 300 miles, 3950 s = 3950 * arccos( ---- ) 4250 = 1493 miles That seems like a pretty long distance! Is it reasonable? It's tough to visualize something 300 miles tall. Let's think about something more realistic, like the Sears Tower. The Sears tower is 1450 feet tall, which works out to be about 0.275 miles. So our equation says that we should be able to see it up to 3950 s = 3950 * arccos( -------- ) 3950.275 = 46.6 miles which means that you would just be able to see the Sears tower from the Indiana Dunes State park - if you were lying on the ground. I know for a fact that you can see it from there standing up, because I grew up near there. Okay, so what if you're _not_ lying on the ground? Well, think about this problem. If your eye is height h' above the ground, what's the farthest point you can see? A' | h' | . . . . . . . C' . . . . . B . . . . . Now we have a different right triangle (A'C'B is the right angle), and we know that r cos(A'BC') = ------ r + h' This looks familiar, doesn't it? Now extend the line A'C', and you have the same problem we just solved. The central angle is now (A'BC' + ABC), so s = r * (A'BC' + ABC) r r = r *[arcos( ------ ) + arccos( ----- )] r + h' r + h Let's check that again for the Sears Tower. If you're 6 feet tall, that's about 0.0011 miles tall, so 3950 3950 s = 3950 * [arccos( --------- ) + arccos( -------- )] 3950.0011 3950.275 = 49.6 miles which isn't a lot of difference. Now, here's the interesting thing. From Indiana, it seems as though you can see, not just the top of the building, but the top half. The greatest distance from which that should be possible is 3950 3950 s = 3950 * [arccos( --------- ) + arccos( --------- )] 3950.0011 3950.1375 = 35.9 miles I just checked with maps.yahoo.com, and the driving distance from the house where I grew up in Indiana (just outside the Dunes State Park) to the Sears Tower is 49.7 miles, and the distance from the Dunes State Park to the Sears Tower is a little less than 40 miles (as measured on the map that accompanies the driving instructions). So this appears to work out quite well. I hope this helps. Write back if you'd like to talk about this some more, or if you have other questions. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum