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# Fun with Math ... on the Moon!

03/26/2018 7:05 PM

Something to fill your lonely hours: you are standing on the Moon (radius 1079 miles). How far away is the horizon? What other information is needed to get the answer? Extra point: can you give an expression for the horizon distance as a function of one factor?

This is the kind of thing that comes to me in bed. See what you can do in your head without paper and pen. I figured out a method, but had to write it out to get the actual number.

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#1

### Re: Fun with Math..on the Moon!

03/26/2018 7:39 PM

The horizon is the distance where your line of sight is tangent to the surface. Thus you have a right triangle where one side is the radius of the surface, the other side is the distance to the horizon (the unknown), and the hypotenuse is the sum of the radius and the height of your eye above the surface. Thus the distance can be calculated using the Pythagorean theorem. Given the radius is known, the only variable is the height of your eye

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#2

### Re: Fun with Math..on the Moon!

03/26/2018 9:03 PM

Even simpler, your height = h, distance to horizon = d, and radius of the moon = r.

By similar triangles:

h/d = d/r

d=sqrt(h*r)

This is a good approximation unless you are very tall...

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#3

### Re: Fun with Math..on the Moon!

03/26/2018 9:09 PM

d = [(h/63,360 + 1079)2 - 10792]1/2; where h is eyeball height in feet and d is horizon distnce in miles. I believe there is an approximation equal to a number x √h.

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#4

### Re: Fun with Math..on the Moon!

03/26/2018 9:44 PM

Sorry, my dreams are about the girls I didn't get a chance to date

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#6

### Re: Fun with Math..on the Moon!

03/27/2018 2:43 AM

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#7

### Re: Fun with Math..on the Moon!

03/27/2018 1:04 PM

Oh my, she is my dream girl too. Burt Reynolds was a lucky man who really screwed up!

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#5

### Re: Fun with Math..on the Moon!

03/27/2018 2:41 AM

This is an opportunity to put the versine formula, used in railway track radius calculations, into action.

• versine = (chord)2/(8*radius), which is derived from Pythagoras and very nearly correct, and probably satisfactory given that the Moon isn't completely spherical.

One's height is the versine. The chord is twice the distance to the horizon. The radius is given. Into the calculator and "Robert is your mother's brother" - Anonymous poster #0.

Beware, though. Anyone building a railway on the Moon needs to put on about 6 times as much cant into a curve than for the same curve on Earth, otherwise the train will probably derail to the outside of the curve.

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#8

### Re: Fun with Math..on the Moon!

03/27/2018 1:22 PM

I am having a senior moment. Using Gringogreg and Pwslack's approachs I get d=sqrt(2h*r), but Rixter's answer of d=sqrt(h*r) also seems correct. What am I missing?!?!

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#9

### Re: Fun with Math..on the Moon!

03/27/2018 1:38 PM

The internet answer of 8124 ft appears to support d=sqrt(2h*r). Somehow Rixter's approach has a flaw, but damn if I can see it.

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#14

### Re: Fun with Math..on the Moon!

03/28/2018 5:48 AM

Assume the angle subtended by d is ß then the two angles at the chord are (90-½ß)

tan(ß) = d/R

and tan(½ß) = h/d

For these small angles tan(2x) = 2tan(x)

so d/R = 2h/d

d² = 2Rh

In other words his triangles aren't similar.

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#10

### Re: Fun with Math..on the Moon!

03/27/2018 2:31 PM

Just as an aside: By "distance" I assumed that SSCpal was referring to the linear distance as the line of sight. However we can also take the "distance" as the distance along the circumference of the surface, the distance that one would have to travel to the horizon.

Knowing the 3 sides of the right triangle, the inverse cos of r/(r+h) expressed in radians would give this dimension

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#11

### Re: Fun with Math..on the Moon!

03/27/2018 6:01 PM

How 'approximate' do you want?

Typically on Earth at sea level the horizon is said to be 7 miles away. Since we're dealing with small angles, linear approximations are reasonably good. So since the Moon's radius is about 1/4th the Earth's radius, then the horizon on the Moon would be about 1/4 the distance as it is on Earth.

Thus, 1/4 of 7 miles is 7/4ths = 1.75 miles. If I want it in feet, then doing a rough approximation in my head 3/4ths of 5000 feet (rounding a miles to the nearest 1000s) is 3250; so 1 mile + 3/4 miles would be about 5000 + 3250 = 8250 feet, approximately. Maybe a bit more since I rounded the feet down; call it 8500 feet.

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#12

### Re: Fun with Math..on the Moon!

03/27/2018 6:05 PM

Computing the distance to the horizon on the Moon could avoid or minimize some of the influences affecting similar calculations for horozons

Nothing like looming to be concerned with, as there is no atmosphere to speak of (or with for that matter).

Also, far less equatorial bulge to contend with.

However there is no body of water ob the moon to occassionally provide a smooth surface.

Anyone know a simple model for relating surface roughness and scale to the horizon distance calc?

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#13

### Re: Fun with Math..on the Moon!

03/27/2018 6:05 PM

I would just walk, count my paces, until I fell off. Wait, this isn't a flat earth post? Damn....

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#15

### Re: Fun with Math..on the Moon!

03/28/2018 8:46 AM

fast calc: D=1.17 * sqrt(h)

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#16

### Re: Fun with Math..on the Moon!

03/28/2018 10:05 AM

Don, What are the units in your equation? How did you derive it?

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#17

### Re: Fun with Math ... on the Moon!

03/28/2018 1:03 PM

Lots of answers based on math formula and all of them useless! Calculating horizon distance assumes that the horizon is at mean sea level and you know the observer's height above mean sea level. The mean radius of the moon is 1079 miles but actually over 98% of the surface is above or below that height. For instance if you were standing on Mons Huygens, the highest mountain on the moon, with an elevation of 3.4 miles, and you did the calculations based on 3.4 miles high you would not get the correct answer because it stands on a plateau which itself is over 1 mile high. If you were standing in the centre of the crater Picard Y which is approximately round with a diameter of 2.48 miles your horizon would be 1.24 miles away. But if you were standing at the edge of Picard Y your horizon in one direction may be less than 20 yards and in the opposite direction 2.48 miles.

So the correct answer is INDETERMINATE as the formulae cannot be applied in this instance.

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