Basic spherical geometry/math question

[Junghalli]

Probably a stupid question, I know, sorry:

Say I have a sphere with a radius of 10 (inches, miles, whatever).

Now I draw a square 4 (inches, miles, whatever) on a side on that sphere

Now say I have a sphere with a radius of 5

It will have 1/4 the surface area of the first sphere

And I draw a 2 X 2 square on it.

The square on the smaller sphere should have 1/4 the surface area of the square on the bigger sphere, right?

And this should be generalizeable to shapes of different kinds (triangles, polygons etc.) drawn on spheres, right?

I'm trying to work out geography on a fictional planet with ~.9X Earth's radius. I figure I could just take terrestrial continents like (e.g.) South America and knock about 20% off its area while keeping the latitudinal and longitudinal extent relative to the planet the same and work from there. That's geometrically/mathematically correct, right?

[Terralthra]

This question is more complex than you give it credit for. Before beginning to answer, you'll need to give a more precise definition of "square". Specifically, the two interpretations I can see would involve either an area with corners that are not tau/4, or an area with curved sides, neither of which are "squares" as one might easily calculate the area of.

Girard's Theorem might help you here, it's for calculating the area of a triangle projected onto a sphere (the "curved sides" interpretation), but doing it twice for equilateral (sorta) triangles whose hypontenuses coincide will give you more or less a useful answer.

[Simon_Jester]

On the one hand the question is more complex than you give credit for.

On the other hand, I am PRETTY sure the answer is still simple: if you take South America, reduce its dimensions by 50%, and put it on a sphere with 50% the Earth's radius, yes it will fit and you'll get the desired result. Inverse square law in action.

Rigorously defining a 'square' on a curved surface is tricky. Knowing how it scales as a function of the radius of the sphere not so much.

[Darth Holbytlan]

What Simon_Jester said. Terralthra, you're making this far more complex than it needs to be. As long as we are talking Euclidean space[*] and proportionate scaling by a factor f, then lengths all scale by f, areas all scale by f2, and volumes all scale by f3. So if you scale a planet by .9x its radius, then all lengths will be .9x the original length, all areas .92=.81x the original area, and all volumes .93=.729x the original volume.

[*] We are, for all intents and purposes.

[madd0ct0r]

but we're talking euclidean space are we? you can draw a triangle on the surface of a sphere that has 2 right angles.

[Zeropoint]

Yes, but the spheres in question are embedded in a euclidian 3-space, so the simple scaling relationships still apply. If you were scaling down a figure drawn on a sphere of fixed radius, then the non-euclidian nature of the surface would be an issue.

[Simon_Jester]

Yes. Spherical geometry happens in Euclidean space, even if it doesn't happen on a Euclidean surface. Things like "what does it mean to draw two similar triangles on a curved surface" or "what does a square look like on a curved surface" have screwy answers because of the non-Euclidean character of the surface. But basic scaling relationships (so long as the objects are scaling up or down proportionate to the size of the sphere) will persist as normal in Euclidean space, simply because the sphere itself and everything described on it CAN be described in terms of Euclidean geometry... in three dimensions.

[Darth Holbytlan]

This is worth explaining several times because it can be hard to understand.

If we scale up or down the entire planet and everything on or in it, then we get nice, simple scaling for all the measurements because we are performing a Euclidean scaling operation in what is Euclidean 3-dimentional space.

If we keep the planet the same size but try to scale up or down a continent on the surface, things get screwy. In fact, scaling that way isn't even well-defined: If you try to scale it by 3D Euclidean rules, the result won't even lay flat on the surface because the curvature of the scaled object will be different from the original surface. If you try to scale it by 2D rules, you'll find that different methods produce different results.

For example, pick a point O on the surface of a sphere and represent every point in the object by coordinates (θ,r) where θ is the direction of the point from O and r is the surface distance of the point from O. Then define the scaling operation as taking point (θ,r) -> (θ,rf). On a Euclidean surface this would define a normal scaling operation and would have the same result for any point O, modulo a translation.

On the surface of a sphere, different points O result in different shapes: Take a triangle[*] 1 unit to a side on the surface of a sphere 20000 units circumference and scale it up by f = 5000. If you pick O to be at the center of the circle (all three corners equidistant from it) then the scaled up object will be symmetrical but not a triangle—the sides will curve inwards. If you pick O to be one of the corners, then the angle at that corner will obviously be preserved and the two line segments connecting to it will remain straight[**], but the third segment will still be curved inwards.

[*] Defined as a shape composed of three straight[**] line segments.
[**] A straight line is a great circle on the sphere.

The good news for the OP is that he seems to be asking about the simple scaling case where the entire planet and everything on it are scaled.

[Terralthra]

Holbytian, just for my own understanding, can you clarify what you mean by **? It seems ambiguously phrased.

Ie, do you mean to say that "all straight lines on a sphere are great circles"? In which case, I question that, because e.g. lines of latitude are (from some points of view) "straight". If you instead mean that "by definition, any great circle is a straight line, and any non-great circle is, when talking about spherical geometry, not 'straight'," then I understand your meaning, but it seems to be stated in an imprecise manner.

[Esquire]

Hang on - I thought a great circle was just the equatorial circle for a given sphere, and by extension that any circle could be said to be the great circle of some particular sphere. Unless I missed something, while any cutting plane applied to sphere will produce a circle, only one of those will be the great circle of the sphere.

[Terralthra]

Hang on - I thought a great circle was just the equatorial circle for a given sphere, and by extension that any circle could be said to be the great circle of some particular sphere. Unless I missed something, while any cutting plane applied to sphere will produce a circle, only one of those will be the great circle of the sphere.

Spheres have an infinite number of great circles. If you pick any two points on a sphere and draw a circle which connects them using the shortest distance between them when projected on the surface of the sphere, the circle you've just drawn is a great circle. E.g. the route from SF, California to Japan, naively drawn, goes straight west across the Pacific, but the shortest route actually ends up being to go, from SF, northwest up along the west coast and back down southwest along parts of the east coast of Asia, forming an arc of a great circle, like so.

[Esquire]

Ah, right, regular figures, my bad. You still have to cut through the center, though, so although there are infinite great circles there are infinitely more non-great circular cuts.

Which is, of course, not what we're talking about. My brain's stuck in Apollonian conics-land for some reason.

[Darth Holbytlan]

Holbytian, just for my own understanding, can you clarify what you mean by **? It seems ambiguously phrased.

Ie, do you mean to say that "all straight lines on a sphere are great circles"? In which case, I question that, because e.g. lines of latitude are (from some points of view) "straight". If you instead mean that "by definition, any great circle is a straight line, and any non-great circle is, when talking about spherical geometry, not 'straight'," then I understand your meaning, but it seems to be stated in an imprecise manner.

I'm defining how I am using the term "straight line" when describing the triangle, which is by a great circle on the sphere.

Now that I think about it, I was using this same definition of "straight line" when I described the transformation: When measuring the distance and angle to a point from O, that is implicitly using a straight line measurement and I didn't specify what that was. But it is also based on a great circle.

[Kuroneko]

The essential property of a square is that it is a regular quadrilateral. This definition is not dependent on whether the geometry is elliptic, Euclidean, or hyperbolic, and is sensible in each. But of those, similar polygons are only possible in Euclidean geometry.

Ah, right, regular figures, my bad. You still have to cut through the center, though, so although there are infinite great circles there are infinitely more non-great circular cuts.

Yes, in the sense that you have two degrees of freedom for the great circles: starting from the center of the sphere, pick any direction (unit vector), and let the plane be the one normal to the vector and passing through the center. Equivalently, this picks out the direction to the point on the sphere that to serve as the 'north pole'.

[Junghalli]

As far as calculating the area of made-up continents go, I think I've found a quick and dirty method.

This site says:

Suppose we divide a hemisphere into q equal lunes by drawing great cirles all from one point on the great circle which froms the boundary of the hemisphere. The lunar angle of each is TT/q, and the area of each lune is 2TTR^2/q.

The latter equation, if I understand this right, can be used to get the surface area of a triangle with the equator as its base, the pole as its vertex, and parallels of longitude as its sides. Taking a planet the size of Earth and a lune 10 degrees wide:

2TT(6371^2)/36 = 7,084,228.8 km^2

Which about fits which that the Earth's surface area is 510 million km^2 and this triangle should have (510/2)/36 = 7.084 million km^2

This pdf gives a table of the percent of Earth's surface area between different parallels of latitude. For instance, 70.56% of Earth's surface is between 45 N/S, and 17.3% is between 10 N/S. We can apply this to the triangle which's area we just figured to get, say, an area for an island that fills a box 10 W to 20 W and 10 N to 45 N

7,084,000 X .7056 = 4,998,470 km^2
7,084,000 X . 173 = 1,225,532 km^2
4,998,470 - 1,225,532 = 3,772,938 km^2 (bit bigger than India...)

We can try the same for my fictional planet with .9X Earth's radius, which is 5733.9 km:

2TT(5734^2)/36 = 5,738,425.5 km^2

Noteably
7,084,228.8 / 5,738,425.5 = 1.2345
1/1.2345 = .81 => 81% of the area
Which is also what I get for comparing the surface areas of the two spheres.

And the same 10 W/ 20 W / 10 N / 45 N box on the smaller planet:

5,738,425.5 X .7056 = 4,049,033 km^2
5,738,425.5 X .173 = 992,747.6 km^2
4,049,033 - 992,747.7 = 3,056,285.3 km^2

Comparing this with the equivalent area on Earth:
3,772,938 / 3,056,285.3 = 1.2345
1 / 1.2345 = .81 => 81% the area (again, fits with comparing total surface area of the two spheres)

Whew! Hope I didn't make any silly errors typing that all out.

And you can use different great circles, and model a continent as a number of these partial triangles.

This sound right to you?

[Darth Holbytlan]

I found the presentation in that article to be convoluted. It looks like they are trying to explain calculus without using calculus and failing. Also, I don't know why they chose to do their calculation in terms of hemispheres instead of the full sphere. I think that is more confusing, and it has lead you into a math error. There are only 18 lunes 10° wide in a hemisphere, not 36, so your calculated areas are all half the size they should be.

A sphere has the area A = 4πr 2 and that a lune takes a proportionate slice of the sphere. So if a lune is 10° wide, or 1/36th of the whole sphere, then the area is 4πr 2/36. For the Earth, where r = 6371 km, that is about 14.2M km3. If you use a lune angle in radians (θ), the formula is just A = 2θr 2.

A very useful trick is that there is a nice relationship between the area of a sphere and a cylinder enscribed around it—projecting the surface perpendicular to the axis from the surface of the sphere to the cylinder is area-preserving. This is used for certain equal-area map projections, most notably the Lambert projection. So you can draw your continents on the cylinder and know that you will still get the right area. See this image for reference. Note that the area of an open cylinder is height*circumference, or 2r*2πr = 4πr 2, so the full areas match as expected.

Note also that you don't need a look-up table to get the proportion of area between parallels of latitude. The conversion is just proportion = sin φ, where φ is the latitude. So sin 45° = sin π/4 = .7071 (70.71%) and sin 10° = sin π/18 = .1736 (17.36%). If you want to get the area between two parallels of latitude for a lune (α and β), you can combine the formulas to get A = (sin β - sin α)θr 2.

NB: There is a slight difference between my calculated proportions and the looked up versions. This appears to be because they are correcting for the fact that the Earth is not quite spherical. I recommend not worrying about the difference.

[Junghalli]

Ah, thanks, let me try that again:

Let me try a continent 40 degrees wide in longitude and stretching from 10 N to 60 N, first on a planet with Earth's radius and then on a planet with .9 times Earth's radius:

Earth:

4TT(6371^2)/9 = 56,673,830.21 km^2

sin10 = .1736
sin60 = .866
.866 - .1736 = .6924
.6924 X 56,673,830.21 = 39,240,960 km^2
39,240,960 / 2 = 19,620,480 km^2

.9X Earth:

4TT(5733.9^2)/9 = 45,805,802.47 km^2
.6924 X 45,805,802.47 = 31,715,937.63 km^2
31,715,937 / 2 = 15,857,968.82 km^2

Comparing the two
19,620,480 / 15,857,968.82 = 1.237
1/1.237 = .808 => the land mass on the smaller planet is ~81% of the size of the one of the same longitudinal/latitudinal dimensions on the bigger planet, which fits with a comparison of the area of the two spheres
4TT(6371^2) = 510 million km^2
4TT(5733.9^2) = 413 million km^2
510/413 = 1.2348
1/1.2348 = .8098 => 81%

Right, now?

[Darth Holbytlan]

That checks out with my calculations.