Covering the Earth's Surface With Circles (1 Viewer)

[braintic]

This is a question I have been thinking about which is not really related to the Ext 2 course, but it should involve work from only HS maths.

Lets say I want to write a program that randomly creates circles of radius 1 km on the earths surface.

To do this, lets say I simply pick a random longitude and a random latitude. If I ran this program many times, it would pick as many circles outside the 45 degree latitudes as inside. But there is less surface area outside the 45 degree latitudes than inside, so the effect would be to cover the polar regions with circles more densely than equatorial regions.

I am looking for a function which would cover all regions of the earth evenly by area. To do this, you might use a random number generator Rnd(x) which picks a continuous random number between 0 and x. For example, if I wanted to pick a random longitude, I might use Rnd(360)-180 to create a random number between -180 and 180 (say -123.371).

Is there someone who understands the problem who would be able to create such a function. Or perhaps there is another coordinate system which would allow one to randomly choose a point on the earth's surface without adjustment.

 

[asianese]

Are you talking about the Earth as a sphere? If so, would you be talking about great circles (circles whose centres is also the centre of the sphere)?

What exactly do you mean by 'cover all regions of the Earth evenly by area'?

 

[braintic]

Are you talking about the Earth as a sphere? If so, would you be talking about great circles (circles whose centres is also the centre of the sphere)?

What exactly do you mean by 'cover all regions of the Earth evenly by area'?

A great circle would not have a radius of 1 km. As previously stated, I am talking about circles of radius 1 km on the SURFACE of the earth. As though you were covering the earth with circular stickers of radius 1 km.

I explained in the 3rd paragraph that picking a perfectly random latitude and longitude would result in a greater coverage of the surface near the poles, ie. there would be more circles per unit area of the earth's surface near the poles than near the equator. I am looking for a function that would even out this discrepancy.

 

[asianese]

Ok I'm still trying to understand.

I think you are thinking of equal area maps. http://en.wikipedia.org/wiki/Cylindrical_equal-area_projection is this what you are thinking of?

If you have a parameterisation of the Earth's sphere into spherical coordinates, you can make it so that its 2-D representation (a map) has area preserving properties. Indeed at the poles you have more area than at the equator for the same 'radius' 1km circle. A common map we see in everyday lives is this one: https://en.wikipedia.org/wiki/Mollweide_projection which preserves area.

There are many more maps that have such interesting properties!


(*Studied a module on map projections for a 1st year special studies program*)

 

[braintic]

I might be wrong, but I don't think I need to know any uni maths to answer the question. To give you another perspective on the question, imagine a right-angled triangle sitting on one of its short sides, and imagine a line parallel to the base and bisecting the altitude. You repeatedly toss a coin. Every time you toss a head, you draw a dot somewhere above the line, and when you toss a tail you draw a dot below the line. As there is three times as much area below the line, the density of dots by area will be three times as great above the line. This is what will happen in my apparently random experiment. Someone observing the earth from space will see a greater concentration of circles near the poles, as there is less area between successive parallels of latitude as you approach the poles.

I am wondering if the answer is as simple as factoring a cosine of the latitude somewhere into the calculation, as cos(latitude) determines the radius of each of the parallels of latitude.

 

[asianese]

I am still having trouble understanding your question. Someone else should try and interpret what you're saying.

 

[seanieg89]

Such a construction has been known since Archimedes. Imagine that you have a transparent sphere with an internal linear light source joining the north and south poles. Wrap this sphere in a cyclinder whose axis of symmetry contains this light source.

Now the "projection" from points on the sphere to points on this circumscribing cylinder is an area preserving map from the sphere (excluding north and south poles) to a cylinder. (this is the main thing to prove and is not difficult). Finding an area preserving map between a cylinder and a planar domain is trivial, which then gives us a coordinate system with the desired properties.

Interestingly this yields a quick way of computing the surface area of a sphere. (The cylinder is of radius r and height 2r, so its surface area is .)

 

[braintic]

Such a construction has been known since Archimedes. Imagine that you have a transparent sphere with an internal linear light source joining the north and south poles. Wrap this sphere in a cyclinder whose axis of symmetry contains this light source.

Now the "projection" from points on the sphere to points on this circumscribing cylinder is an area preserving map from the sphere (excluding north and south poles) to a cylinder. (this is the main thing to prove and is not difficult). Finding an area preserving map between a cylinder and a planar domain is trivial, which then gives us a coordinate system with the desired properties.

Interestingly this yields a quick way of computing the surface area of a sphere. (The cylinder is of radius r and height 2r, so its surface area is .)

So for the sake of my question, are you saying that I would just need to pick a random distance between the poles corresponding to a point in this light source?

What I don't get - since the light source is not a point source, the projection of any point on the earth's surface onto this cylinder is also not a point.

 

[seanieg89]

So for the sake of my question, are you saying that I would just need to pick a random distance between the poles corresponding to a point in this light source?

What I don't get - since the light source is not a point source, the projection of any point on the earth's surface onto this cylinder is also not a point.

Yeah, a random z coordinate and a random line of longitude would do.

Yeah a light source was perhaps a poor choice to explain, the projection is perpendicular to the line joining the N and S poles.

 

[braintic]

Yeah, a random z coordinate and a random line of longitude would do.

Yeah a light source was perhaps a poor choice to explain, the projection is perpendicular to the line joining the N and S poles.

OK, thanks. I think that answers my question. So I would just need to pick a random number between -1 and 1 and find the arcsin of this number to get my random latitude, is that correct?

 

[seanieg89]

OK, thanks. I think that answers my question. So I would just need to pick a random number between -1 and 1 and find the arcsin of this number to get my random latitude, is that correct?

Yep.