OK, so I actually have been googling this for a while before I ended up here. I have an unproven formula for this (with good reasoning behind it though) that I came up with myself (or discovered independently without knowledge of anyone else's version of the same).
Here it is:
the solid angle (represented by units of one great circle worth of area, or pi steradians) = (Number of Sides on the Polygon)*Average of internal degrees (in a unit of pi radians ie. 180 degrees) -((Number of sides on the polygon)-2)
A little neater:
Let N be the number of sides of the polygon in question.
Let A be the average internal angle of the shape (Ie, the total internal angle sum divided by N).
Let Z be the number of pi steradians the shape occupies on the surface of a sphere.
Then: Z = N*A-(N-2).
I'll just post this and do the reasoning and examples in the next post.
OK, so here are the examples.
First of all, any shape with the internal degrees the amount they should be on a flat plane, will occupy "no space" on the sphere.
this is because the flat plane is an "infinitely" zoomed in version of the sphere, so the sphere would be infinately large, despite for
example a flat square having an area of a billion acres, it is still negligible when compared to the infinite (on a sphere, flat shapes don't exist!).
Therefore the table below shows nill Solid angle values.
N A Z
3 1/3 0 (any flat triangle)
4 2/4=1/2 0 (any flat quadrilateral)
5 3/5 0 (any flat pentagon)
6 4/6=2/3 0 (any flat hexagon)
N N-2/N 0 (any flat polygon)
Now, this also tells us something interesting about the same polygon that has those figures as it's external angles...
The OUTSIDE of the shape doesn't exist on the sphere, an thus the polygon takes up the entire sphere's worth of solid Angle!
N A Z
3 5/3 4 (any sphere enclosing triangle)
4 6/4=3/2 4 (any sphere enclosing quadrilateral)
5 7/5 4 (any sphere enclosing pentagon)
6 8/6=4/3 4 (any sphere enclosing hexagon)
N (N+2)/N 4 (any sphere enclosing polygon)
Ok, so we have NO Solid Angle for A = (N-2)/N and FULL Solid Angle for A=(N+2)/N
So do we have HALF the sphere (Z=2) when we get A= N/N, that is, A=1?
YES!
A=1 means the average internal angle is pi rads or 180 degrees. this means the polygon's sides (or single side in this case) is a great circle,
one which encompasses exactly half of the sphere
So now we have the extremedies and centre of this formula working out, is that it? Does everyone just believe it's a linear function of solid angle for each polygon and hope it works?
No. we can use the platonic solids to see further evidence of a linear relationship, though there's only 5 shapes to work from, they are easily incorporated into this formula.
The tetrahedron:
Join 4 equidistant points by arcs of a great circle.
Each point has 3 congruent sides coming from it, so when we 'zoom' right in until the sphere becomes flat, the lines go off into oblivion at an angle of 2/3
(or 2/3 pi rads or 120 degrees).
This is true of all the angles of all for spherical triangles.
the area of each said triangle is 1 (or 1/4 of the sphere or pi steradians)
So:
Z = N*A-(N-2)
1 = 3*2/3 - (3-2)
1 = 2 - 1
1 = 1 (so this checks out!)
You can check the following out yourself (by thinking about the shapes), I'll just give the figures in my terms.
Cube: Z = 2/3, N = 4, A = 2/3
Octahedron: Z = 1/2, N = 3, A = 1/2
Dodecahedron: Z = 1/3, N = 5, A = 2/3
Icosahedron: Z = 1/5, N = 3, A = 2/5
They all check out!
You'll notice though that in each case the "Average" angle, is the only angle, the same throughout each platonic shape's surface shapes.
Next I'll post up some irregular shapes, derived from the platonic solids.
Please let me know what you guys think, and if this is what you were looking for before!
OK, so now for some irregular shapes.
first, lets turn the square on the side of the cube into 2 triangles via the use of a diagonal,
so one 2/3 angles remain and the other 2 diagonally opposite angles become 1/3.
2/3 + 1/3 + 1/3 = 4/3
therefore A = (4/3)/3 = 4/9
Z is halved, Z = (2/3)/2 = 1/3
N is now 3.
Z = N*A - (N-2)
= 3*(4/9) - (3-2)
= 4/3 - 1
= 1/3
How about the same sqaure as 2 rectangles?
Angles are: 2/3, 2/3 1/2, 1/2 (think about it)
A = (4/6 + 4/6 + 3/6 + 3/6)/4
= (14/6)/4
= 7/12
Z = 1/3 (halved)
N = 4 (still)
Z = N*A - (N-2)
= 4*7/12 - (4-2)
= 28/12 - 2
= 7/3 - 2
= 1/3 (Awesome!)
Lets try a harder one now.
We will halve the Dodecagon's pentagonal surface.
we start with Angles 2/3, 2/3, 2/3, 2/3, 2/3
now we divide the top angle in half, and join the bottom of the pentagon halfway along at angle 1/2 (90 degrees)
now we have 1/3, 2/3, 2/3, 1/2 as our angles.
A = (2/6 + 4/6 + 4/6 + 3/6)/4
= (13/6)/4
= 13/24
the Z of 1/3 becomes 1/6 now.
And N drops from 5 to 4.
Z = N*A-(N-2)
= 4*13/24 - (4-2)
= 13/6 - 2
= 1/6 (Pretty good formula if you ask me!)
This may have already been published, it seems quite basic when you think of the sort of stuff done in topology to do with surfaces of a torus etc.
but if not, and anyone can come up with a proof, please call it:
"your name here"'s proof of the Andrews conjecture on linear relationships between internal angles of geodesic polygons and their solid angle value.
thanks! lol.
now, if you can work out the relationship between the 3 angles at the centre of the sphere between 3 radius vectors, AND the three internal andles they produce upon the surface, you could tell quite easily how many steradians or degrees squared or whatever was in each polyhedra and find a pattern perhaps. when you compare triangle and tetrahedron, then circle with sphere, it makes you think there definately is a relationship there.
