# Homework Help: Distance to the horizon

1. May 28, 2006

### Alec

My problem is to determine how far one can see to the horizon, depending on how far above the water the person is and to create a formula that determines the distance as a function of the height.

To start with, the hypotenuse is the distance from the core to the persons eyes. (x+e). The cathetus being the distance from the core to the waterlevel (x) and the 2nd cathetus being the distance to the horizon (d).
This is easily calculated with Pythagoras (x+e)^2 - (x)^2 = (d)^2.
With the fact that a cathetus to a spheric object has an angle of 90degrees.

How shall I proceed to create this formula?

2. May 29, 2006

### Tide

You're almost there. x is the radius of the Earth so just solve for d. Noting that $a^2-b^2 = (a-b)(a+b)$ may help.

3. May 30, 2006

### Alec

Okay, to put this formula into situations depending on the height above the water level in meters:
(x+e)^2 - (x) = "the distance to the horizon" ^2
When I keep my eyes just above the water:
(6378150+0.02)^2 - (6378150)^2 = 255126.
I can then theoretically see 505.1 meters untill I can only see the horizon.

A person of 1.85 meters have his eyes at a height of 1.65 meters above the water. (Roughly)
(6378150+1,65)^2 - (6378150)^2 = 21047898
The person is capable of seeing 4587.8 meters until the horizon.

A person in a lighthouse looking at the horizon from a height of 20 meters:
(6378150+20)^2 - (6378150)^2 = 255107020,1
The person is then looking at a distance of 15972 meters to the horizon.

Are these numbers probable?
How do you advice me to solve the problem to make a formula giving the distance as a function of e (the height from my eyes to the water) ?

4. May 31, 2006

### Tide

Alec,

Those numbers look pretty good!