Distance to the HorizonDate: 05/23/99 at 01:09:50 From: Robert Ziering Subject: Line of sight How far is the horizon? Date: 05/26/99 at 14:49:59 From: Doctor Fwg Subject: Re: Line of sight Dear Robert, Thanks for sending this interesting question. However, it is difficult to answer exactly because certain conditions must be specified, whether they happen to exist or not. For example, one estimate of the distance to "the" horizon can be made if one assumes a perfectly smooth surface for the earth and an observer looking toward the horizon with his/her eye level exactly 5.28 ft (or "h") above the earth's surface (this height is arbitrary but the value used for it will affect the final answer). The nearest approximation of this condition might exist for a short person standing in a row boat floating in the middle of a very large (calm) lake. If you now assume that the distance from the center of the earth to the earth's "perfectly spherical" surface is exactly 4,000 mi (or "R") and that the distance from the center of the earth to the short person's eye level is 4,000 mi plus 5.28 ft (or R + h), you can estimate the distance to the horizon (assuming no other problems, such as refraction or bending of light coming from the horizon to the observer's eyes). Let "s" stand for the actual distance between the observer's eyes and the horizon. You also need to assume that rays of light coming from the horizon to the observer's eyes make a right angle with the earth's radius vector at the horizon. In other words, the light rays coming from the horizon point to the observer's eyes are tangent to the earth's "surface" at the horizon point. You may ignore the actual curvature of the earth and assume that "s" and the actual distance between the observer and the horizon, along the curved part of the earth's surface, are equivalent. From these conditions and approximations, you can draw a "right" triangle and then use the Theorem of Pythagoras to estimate the distance to the apparent horizon. Using the Theorem of Pythagoras and the conditions above, you can write: R^2 + s^2 = (R+h)^2 = R^2 + 2Rh + h^2, or s^2 = 2Rh + h^2 = h(2R + h). But 2R is much greater than h, so: s^2 = h(2R), approximately, or s^2 = 2Rh. Therefore: s = Sqrt(2Rh) = Sqrt[2(4000 mi)(5.28 ft/5280 ft/mi)] = Sqrt[(8000)(0.001)] = Sqrt[8] = Sqrt[(4)(2)] = 2[Sqrt(2)] = 2(1.414) = 2.8 mi. So, for these conditions, the distance to the horizon is about 2.8 mi. I hope this has been helpful. With best wishes, - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
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