Date: 07/07/2003 at 22:30:20 From: James Subject: Earth's Curvature My son and I wondered what would happen if there were no gravity. I jokingly said that you'd take a step and simply float in a straight line forever. That got me to thinking: How far would you have to "walk" this way to get 1 foot off the ground? Scientifically, how far would you have to follow a tangent of the earth to be one foot above the surface of the curvature? Could you use triangulation somehow? Could you use the formula for a circle to calculate the slope? In another discussion: Earth's Curvature http://mathforum.org/library/drmath/view/54904.html you say that the formula for a circle is x^2 + (y+R)^2 = R^2 and that, "if 2 points are separated by a distance L, they might be point A at x = -L/2 and point B at x = +L/2. At both A and B we have: y = - L^2 / 8R" That leads me off in the general direction, but I'm not finishing the connection.
Date: 07/08/2003 at 02:23:31 From: Doctor Jeremiah Subject: Re: Earth's Curvature Hi James, Imagine a perfectly spherical world with a tangent: A B +++++ -------+ +++ | +++ / +++ | +++ + | / + + | / + + | / + + R R+1 + + | / + + | / + + | / + + |/ + + + + You travel from A to B. Your distance from the center of the Earth at A is R, and your distance from the center of the Earth at B is R+1. This is a simple right angle triangle and can be solved for the distance from A to B with the Pythagorean formula: (R+1)^2 = R^2 + d^2 where d = the distance from A to B d^2 = (R+1)^2 - R^2 d^2 = R^2+2R+1 - R^2 d^2 = 2R+1 d = sqrt(2R+1) The radius of the Earth (R) is 3963.19 statute miles or, more importantly for us, 20925643.2 feet. That means: d = sqrt(2R+1) where R = 20925643.2 feet d = sqrt(2*20925643.2+1) d = sqrt(41851287.4) d = 6469.26 feet So you would have to travel along the tangent for over a mile to get one foot off the ground. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/
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