Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Line of Sight Distances between Two Objects

Date: 01/27/2005 at 12:03:42
From: Kenneth
Subject: Line of sight distances on the Earth

An offshore drilling rig is being towed out to sea.  What is the 
maximum distance that the navigation lights can still be seen by an 
observer standing at the shoreline?  The observer's eye height is 
5'9", and the uppermost navigation light is 225 feet above the water. 
Give your answer in miles.

I'm confused because I believe there is a lack of direct information
to find out the exact answer, like the curvature of earth.  That
should play a role, but that's what I'm unsure of.  How does it play a
role, and is there a formula to determine when a line of sight will be
lost?  I know that the radius of the earth is 3959 miles.



Date: 01/27/2005 at 12:48:41
From: Doctor Jerry
Subject: Re: Line of sight distances on the Earth

Hi Kenneth--

Thanks for writing to Dr. Math.

Imagine a sector of a circle with represents the Earth (like a slice
of pie).  At one end of the sector's arc erect a tower of height H and
at the other end erect a tower of height h.  The tops of these are the
points between which we want to decide upon the longest line-of-sight
distance.  Draw a line connecting the tops.  We want to adjust the
size of the sector so that this sight line is tangent to the circle at
a point between the two bases.  If the two towers were further apart,
they could not see each other.

Now draw a radius from the center of the Earth to the point of
tangency between the sight line and the circle, and recall that the
radius and the tangent line will be perpendicular.  Thus, we have two
right triangles, one on either side of the radius to the tangent point.

One of these triangles has hypotenuse H + r and the other h + r.  So
we see that

  (H + r)^2 = x^2 + r^2

  (h + r)^2 = y^2 + r^2

You can see this all set up here:

    

The maximum line-of-sight distance is x + y.  Solving the first
equation above for x:

  x^2 + r^2 = (H + r)^2
  x^2 + r^2 = H^2 + 2Hr + r^2
        x^2 = H^2 + 2Hr
          x = SQRT(H^2 + 2Hr)

Following the same steps to solve the second equation for y, we can
then express x + y as:

  x + y = SQRT(H^2 + 2Hr) + SQRT(h^2 + 2hr)

Using the values of r, h, and H you gave before, I find the maximum
line of sight between the rig and the person to be x + y = 21.3 miles.

This formula will work for any line of sight problem between two
objects of known heights (H and h) above the Earth.

Does this help?

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 01/27/2005 at 21:51:47
From: Kenneth
Subject: Thank you (Line of sight distances on the Earth)

Thank you very much, Doctor!  I really appreciate your timely reply.  
I was completely stumped and now I can get some sleep.  I REALLY
appreciate it, this is an excellent service you supply.
 
Thank you greatly,

Kenneth
Associated Topics:
High School Higher-Dimensional Geometry

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/