Line of Sight Distances between Two ObjectsDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/