Thread: How far away is the horizon??

Originally Posted by mike_ie

No. The equation is based on a line tangental to a circle, i.e. the Tangent Secant Theorem, and won't work for negative values. Place the point D somewhere between C and F and you will see what I mean when you try to draw a line to the horizon point. ...........

That is why you need to calculate a new diameter to work with, that way both C and B stay on the circle and you can still work with a tangent.

Originally Posted by mike_ie

.......

however for an accurate answer that means you need to know the height above sea level of your horizon line to factor it into the equation, i.e. you need to know the answer before you can calculate the answer Which obviously, no equation can do. That's why these equations are only really used to describe a standing-in-a-lighthouse-staring-out-to-sea type problem.



...........

For a lake it is simple to determine the altitude of any other point on the lake.

What I was trying to point out was that a couple of thousand m in 12576km makes very little difference. The approximation of the distance to the horizon at sea level is more or less the same distance over any body of water irrespective of altitude, particularly when you start with the assumption that the earth is a sphere.

Originally Posted by slug_burner

That is why you need to calculate a new diameter to work with, that way both C and B stay on the circle and you can still work with a tangent.



For a lake it is simple to determine the altitude of any other point on the lake.

What I was trying to point out was that a couple of thousand m in 12576km makes very little difference. The approximation of the distance to the horizon at sea level is more or less the same distance over any body of water irrespective of altitude, particularly when you start with the assumption that the earth is a sphere.

ASSUMING.... that the lake you're looking out over is longer in that direction than the horizon point is far, then yes Which is a pretty safe assumption if you're looking out over the lake and all you can see is water meeting sky.

PERSON STANDING AT THE SEA SHORE

d = sqrt (h(D+h))
D = 12756200m (avg diameter of earth)
h = 2m (observer height ASL)

d = sqrt (2(12756200+2)
d = sqrt (2(12756202))
d = sqrt(25512404)
d = 5050.98m or 5.05km



PERSON STANDING AT LAKE TITICACA:

d = sqrt (h-x(D+x+h))
D = 12756200m (avg diameter of earth)
h = 3814m (observer height ASL)
x = 3812m (lake height ASL)

d = sqrt (3814-3812(12756200+3812+2)
d = sqrt (2(12760014)
d = sqrt (25520028)
d = 5051.73m or 5.05km

Difference is negligible....

Wow, I've learned a few valuable things here:

1, the maths teaching system (smile), that we had in the UK in the late 80s was woefully inadequate!

3, my brain hurts.

Thanks anyway, made interesting reading

Rohan.