Diesel

Jan 17 2004, 07:18 AM

Assuming perfectly clear weather, one can see to the horizon. My goal is to outdo that through a small blimp, fiberoptic line, and a mage.

Does anyone have information on how the range to the horizon is increased with elevation, preferably in increments? My google-fu is weak, I got a bunch of exercise equipment.

Thanks in advance!

RedmondLarry

Jan 17 2004, 07:27 AM

The horizon is farther away for a Troll than it is for a Dwarf ...

Link to page with formulaIf the blimp is 50 meters over your head, the formulas should let you calculate how far someone on the blimp could see, assuming a spherical earth with no trees, buildings, or mountains. And assuming clear air.

Diesel

Jan 17 2004, 07:28 AM

It's at sea, so it should work out.

Diesel

Jan 17 2004, 07:31 AM

Hmm, that doesn't seem right. At two meters off the ground I was under the impression I could see farther than two miles.

John Campbell

Jan 17 2004, 09:08 AM

Whee, math time again.

If we assume that the Earth is a perfect sphere with a radius of 6400km, then the dimensions we're interested in should form a right triangle, with one leg (d) being the distance from your eye to a given point on the horizon, the other leg being the distance from the center of the Earth to the same point on the horizon (which is just the radius (r ) of the planet) and the distance from the center of the Earth to your eye forming the hypotenuse, being the radius (r ) of the planet plus your height (h).

So, applying Pythagoras, that gives us:

r² + d² = (r+h)²

r and h we know... we're interested in finding d. With a little shuffling terms around, we get:

d = sqrt((r+h)² - r²)

Or, with the radius of Earth plugged in:

d = sqrt((6400000m + h)² - (6400000m)²)

If we plug in 2m for the height, that works out to:

d = sqrt((6400000m + 2m)² - (6400000m)²)

d = 5060m

... or about 3.14 miles.

That's distance as the eye-laser fires... distance following the curve of the surface could be figured by applying some basic trig to determine the portion of the Earth's circumference that represents, but it's probably not worth the effort. The difference is going to be fairly small... I'm fairly sure that it should always be less than h.

I think the formula given by the site OurTeam linked is approximating the planetary radius out from under the radical to simplify the math, which is reasonable given the multiple orders of magnitude difference that're likely between the height and radius terms. I believe mine's more accurate, but the other is easier to calculate, and should be close enough for government work, especially since the margin of error is going to be swallowed by the fact that Earth is *not* a perfect sphere.

Sunday_Gamer

Jan 18 2004, 01:15 AM

Uh, it was my understanding that there would be no math?

Kong

Mal-2

Jan 18 2004, 03:57 AM

Just running a few numbers through John's equation:

**CODE** |

Height (m) Horizon (km) 2 m 5 km 10 m 11 km 20 m 16 km 50 m 25 km 100 m 35 km 250 m 56 km 500 m 80 km 750 m 98 km 1000 m 113 km |

I have no idea what limits the atmosphere plays in sighting distances at these kinds of ranges though.

Mal-2

Crusher Bob

Jan 18 2004, 04:02 AM

Note that this distance also covers your radar horizon. Bonus points for a new equasion that determins the dead zone altitude for any given distance.

zephir

Jan 18 2004, 08:10 AM

The formula give the horizon on a perfectly spherical planet, so if you don't actually want to target the ground, your reach lengthens: if you target another human (2m tall) your range doubles. If the target stands on a small hill, if there's a small valley between you ...

This is a "lo-fi" version of our main content. To view the full version with more information, formatting and images, please

click here.