The Curve of the Earth
I was sitting on a sailboat with a friend and far off at the edge of the horizon we could see some thunderheads. My friend asked, “How far away do you think those are?”
I thought about it for 2 years. I didn’t Google it – I wanted to think it out myself. Here is what I’ve come up with.
The surface of the earth, taken in 2 dimensions, is a circle with radius R. In Cartesian coordinates, a circle centered on 0 with radius R has a very simple equation:
Equation 1
In the image above a is the observer and b is where the thundercloud is. The cloud is y tall and x away from the observer. We can see that with a horizontal line-of-sight we won’t be able to see something shorter than y. (There is some approximating going on here but because the angles are so small, we can get away with it.)
In the picture above, the point a is at x=0=x0, y=R=y0, which we write (0,R) and point b is at (x0 +x, y0–y). In both of these cases we know that x2+y2=R2. Since we know x0, y0, y and R, the only unknown in our equation is x, the maximum distance which we can see something y tall.
So calling the coordinates for point a x0 and y0 and the coordinates at point b x and y, plugging them both in Equation 1 and setting them equal, we can solve for the distance x.
You’ll recall that x0=0, y0=R (the radius of the earth) and y is the height of the object.
If we assume the thunderclouds were 30,000 feet high (y = 9000 meters) and use the radius of the earth (y0 = 6,378,100 meters) the right hand side of the last equation above gives us a distance of about 200 miles (300 km).
There are a few assumptions in here but that answer should be right to within a factor of two or better.
Saturday, April 28, 2007
