# The Horizon

Tonight, as I drove along the long stretch of road from Longview to Dallas, Texas, I couldn’t help but wonder how far away the horizon is.

Now, I’m not completely math illiterate, so I decided to figure it out. The problem is… My answer doesn’t seem right. Perhaps I have done something wrong. See if any of you find a mistake in my method:

Assumptions:
The Earth is perfectly round
Our test subject is 5.8 feet tall.
Because the Earth is so big, the distance ‘x’ (shown below) is a close enough approximation of the arc length and actual driving distance to the horizon.

Okay! Calculation time!
According to Wikipedia, the radius of the earth is an average of about 3956.545 miles * 5280 ft/mile=
20890557.6 ft

Add 5.8 feet to that and we get a hypotenuse of 20890563.4 ft

according to Pythagoras
20890557.6^2 + x^2 = 20890563.4^2

or

436415396838917.76
+
x^2
=
436415639169419.56

so x^2 = 24330501.8

x is about equal to 15566.96 ft

Divide by 5280 and we get… 2.948 miles. That’s so… short… did I do something wrong?

### 8 thoughts on “The Horizon”

• 12/6/2007 at 12:04 am
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I have no idea, but it sounds ok to me. I just love that the stick figure in the diagram is holding a guitar.

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• 12/6/2007 at 12:59 am
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you know what this problem needs?

more sig figs

also this site’s calculator jives with the 3 mile ball park http://www.boatsafe.com/tools/horizon.htm

Another way to do the problem is to set up that same triangle and then find acos(r/h). If you do this in radians just multiply the radius by the angle to get the arc length rather than the other side of the triangle. which incidentally yields 2.948 miles as well.

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• 12/6/2007 at 9:30 am
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Having spend a large part of my life in extremely flat, treeless areas, that sounds about right to me.

But wait! There’s a problem! The dude’s eyes aren’t at the top of his head! Make your observer 6.1 feet tall, and your math is still probably accurate though.

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• 12/6/2007 at 11:17 am
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That sounds right, as a general rule I learned that the horizon is usually around 3 miles away if you are looking out over a flat area.

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• 12/6/2007 at 12:19 pm
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Yeah, it’s a short three miles if you’re in a flat field. If you get up onto some higher ground — say, 3500 feet from the top of Mount Greylock — you can see a lot farther. 72 miles by the above calculation, which is clear to Connecticut.

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• 12/24/2007 at 11:09 am
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Hi Mike – Wikpedia deals with Horizon Distance Directly. As can be seen from their analysis, you get less than 3 miles visual horizon (your answer) if you’re a dwarf. You get 12 miles at a 100 feet vantage point.

See Below:

Distance to the horizon

The straight line of sight distance d in kilometers to the true horizon on earth is approximately

d = \sqrt{13h},

where h is the height above ground or sea level (in meters) of the eye of the observer. Examples:

* For an observer standing on the ground with h = 1.70 m (average eye-level height), the horizon appears at a distance of 4.7 km.
* For an observer standing on a hill or tower of 100 m in height, the horizon appears at a distance of 36 km.

To compute the height of a tower, the mast of a ship or a hilltop visible above the horizon, add the horizon distance for that height. For example, standing on the ground with h = 1.70 m, one can see, weather permitting, the tip of a tower of 100 m height at a distance of 4.7+36 â‰ˆ 41 km.

In the Imperial version of the formula, 13 is replaced by 1.5, h is in feet and d is in miles. Examples:

* For observers on the ground with eye-level at h = 5 ft 7 in (5.583 ft), the horizon appears at a distance of 2.89 miles.
* For observers standing on a hill or tower 100 ft in height, the horizon appears at a distance of 12.25 miles.

The metric formula is reasonable (and the Imperial one is actually quite precise) when h is much smaller than the radius of the Earth (6371 km). The exact formula for distance from the viewpoint to the horizon, applicable even for satellites, is

d = \sqrt{2Rh + h^2},

where R is the radius of the Earth (note: both R and h in this equation must be given in the same units (e.g. kilometers), but any consistent units will work).

Another relationship involves the arc length distance s along the curved surface of the Earth to the bottom of object:

\cos\frac{s}{R}=\frac{R}{R+h}.

Solving for s gives the formula

s=R\cos^{-1}\frac{R}{R+h}.

The distances d and s are nearly the same when the height of the object is negligible compared to the radius (that is, h

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