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Distance to the Horizon

Date: 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 

Using the Theorem of Pythagoras and the conditions above, you can 

     R^2 + s^2 = (R+h)^2 = R^2 + 2Rh + h^2,


     s^2 = 2Rh + h^2 = h(2R + h).

But 2R is much greater than h, so:

     s^2 = h(2R),  approximately,


     s^2 = 2Rh.


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   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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