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### Distance to the Horizon

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Date: 05/23/99 at 01:09:50
From: Robert Ziering
Subject: Line of sight

How far is the horizon?
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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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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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