Naked Science Forum
General Science => General Science => Topic started by: chris on 14/12/2004 16:21:39

Assuming perfect conditions, what is the limit of visibility, i.e. how far away is the horizon ?
And now considering an aeroplane flying at cruising altitude (35,000 feet)  how far away is the horizon for them, and how far away can they been seen from the Earth's surface ?
Chris
"I never forget a face, but in your case I'll make an exception"
 Groucho Marx

I'm posting this without having checked it in a reference, so I'm prepared to be full of s.
This can be simplified to a problem in plane geometry, which I studied more years ago than I will ever admit. Here's the construct: Draw a circle of radius r. At the top of the circle, draw a tangent line. Draw a radius line between the center of the circle and the tangent point. Label this line "r". Draw another line from the center, out at an acute angle (30 degrees in my drawing) to intercept the tangent line. Label this line "h". Label the line "d" that goes from the tangent point to the intercept between h and the tangent line. You now have a right triangle rdh defined. The Pythagorean theorem tells us that h^2 = d^2 + r^2. Break the line h into two segements, r, between the center and the circle, and a, between the circle and tangent line. h = r + a. The altitude above the circle, where we observe the horizon is a. The distance to the horizon is d.
Substitute r + a for h, and solve the Pythagorean formula, and find:
d = square root (a^2 + 2ra).
The answer for r = 3900 miles and a = 35,000 feet is d = 228 miles.

Bump...
Yo Chris, is this right or wrong? What's my grade on the quizz?

Your calculation would give you the lineofsight distance to the horizon, i.e. straight line from the observer's eyeball to the horizon. The actual ground distance along the curved surface of the earth would be a little more.

quote:
Originally posted by gsmollin
Bump...
Yo Chris, is this right or wrong? What's my grade on the quizz?
I've no idea, I was asking the question because I was curious to know the answer, rather than to test anyone !!
Chris
"I never forget a face, but in your case I'll make an exception"
 Groucho Marx

OK, so I broke down and Googled it. Here is a web page where the guy uses my derivation: http://www.wolfram.demon.co.uk/rp_horizon_distance.html
He gets the same formula for the distance to the horizon that I did.
DrPhil, for your edification, method 2 on this web page derives the distance along the surface of the earth. I interpreted Chris's question as straightline distance. It's pretty simple to calculate the fraction of the circumference of a circle you would travel, once you know the central angle, using some trig.