Thread: How far away is the horizon??

Oooooooh!!! My brain hurts.

Originally Posted by slug_burner

Craig are you saying that on Lake Eyre at approximately 15 m (49 ft) below sea level the horizon is at a different distance to that at Lake Titicaca with a surface elevation of 3,812 m?

I would imagine it's not that straightforward. Distance to horizon equations are all based on what's known as the Tangent Secant Theorem, and therefore count on the horizon being at sea level. Both lakes being inland lakes, far from the sea, the distance to the horizon would be completely dependant on the terrain surrounding the lakes.

what about if your standing on the edge of a black hole looking in how far to the horizin then

You would be stretched so much by the warped space time that the electrical impulses traveling along your optic nerve would never reach your brain....

Originally Posted by slug_burner

Craig are you saying that on Lake Eyre at approximately 15 m (49 ft) below sea level the horizon is at a different distance to that at Lake Titicaca with a surface elevation of 3,812 m?

Originally Posted by mike_ie

Distance to horizon equations are all based on what's known as the Tangent Secant Theorem, and therefore count on the horizon being at sea level.

So given that at best we are approximating the earth with a sphere of approximate diameter of 12756.2 km the Tangent Secant Theorem gives us a horizon for an observation point of 2m above the surface of Kati Thanda – Lake Eyre of 5050.977m and on Lake Titicaca 5051.735. For the 0.75m difference in 5km it is hardly worth the effort of doing the calculations, the error in the approximation of the earth to a sphere probably accounts for any error introduced by the altitude of the surface for which you are working out the distance to the horizon. But according to the theorem yes there is a difference

Originally Posted by slug_burner

So given that at best we are approximating the earth with a sphere of approximate diameter of 12756.2 km the Tangent Secant Theorem gives us a horizon for an observation point of 2m above the surface of Kati Thanda – Lake Eyre of 5050.977m and on Lake Titicaca 5051.735. For the 0.75m difference in 5km it is hardly worth the effort of doing the calculations, the error in the approximation of the earth to a sphere probably accounts for any error introduced by the altitude of the surface for which you are working out the distance to the horizon. But according to the theorem yes there is a difference

As has been said the calculations are based on visible horizon, with no interference (mountains, trees, buildings etc etc). Hence why if we ever need to see the horizon you find a high point with unobstructed views. With GPS it is fairly easy as you will have your elevation and some GPS units or weather units will even calculate your distance to horizon. Generally we will only use the horizon for locating incoming aircraft, ships - boats etc. Fairly mute now with the advent of GPS for anything other than spotting.

We have that F bisects AC and that CD is added to it.
Apply Square of Sum less Square:

AD⋅DC+FC^2=FD
But FC=FB and so:
AD⋅DC+FB^2=FD^2

Applying Pythagoras's Theorem, FD^2=FB^2+DB^2 therefore:

AD⋅DC+FB^2=FB^2+DB^2
AD⋅DC=DB^2
DB = sqrt (AD.DC)

Make the following substitutions:
d = DB = distance to the horizon
D = AC = diameter of the Earth
h = DC = height of the observer above sea level
D+h = AD = diameter of the Earth plus height of the observer above sea level

d = sqrt (h(D+h))

PERSON STANDING AT LAKE TITICACA:
D = 12756200m
h = 3812m

d = sqrt(48641165744)
d = 220547.42m or 22.5km


PERSON STANDING AT THE SEA SHORE
D = 12756200m
h = 2m

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

...so yes, there's a big difference in what a person standing at Lake Titicaca can see, vs a person standing on the beach.

Below sea level values throw the formula into disarray, due to having to get the square root of a negative value, but it applies to all values above zero.

Of course, this is the theoretical, and assumes that the horizon you are looking at is at sea level. Easiest real world scenario to prove effects of terrain - stand on a beach that has a mountain range behind you. Looking out to sea, the horizon id about 5km away. Looking behind you, the horizon i wherever the top of the mountain is, probably 6-700m away....

Originally Posted by mike_ie

We have that F bisects AC and that CD is added to it.
Apply Square of Sum less Square:

AD⋅DC+FC^2=FD
But FC=FB and so:
AD⋅DC+FB^2=FD^2

Applying Pythagoras's Theorem, FD^2=FB^2+DB^2 therefore:

AD⋅DC+FB^2=FB^2+DB^2
AD⋅DC=DB^2
DB = sqrt (AD.DC)

Make the following substitutions:
d = DB = distance to the horizon
D = AC = diameter of the Earth
h = DC = height of the observer above sea level
D+h = AD = diameter of the Earth plus height of the observer above sea level

d = sqrt (h(D+h))

PERSON STANDING AT LAKE TITICACA:
D = 12756200m
h = 3812m

d = sqrt(48641165744)
d = 220547.42m or 22.5km


PERSON STANDING AT THE SEA SHORE
D = 12756200m
h = 2m

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


Below sea level values throw the formula into disarray, due to having to get the square root of a negative value, but it applies to all values above zero.

Of course, this is the theoretical, and assumes that the horizon you are looking at is at sea level. Easiest real world scenario to prove effects of terrain - stand on a beach that has a mountain range behind you. Looking out to sea, the horizon id about 5km away. Looking behind you, the horizon i wherever the top of the mountain is, probably 6-700m away....

Simple, really, cant work out how those hopeless Mayans got it all wrong, actually. Bob