Unimpressive and quirky

[UsuallyPresent]

You know what’s super interesting about that question? The sum of the inside angle at each vertex and the outside angle at each vertex is still only 360 degrees, no matter what the curvature of the surface is (spherical or hyperbolic).

That means that the total for all 3 angles, summing interior and exterior angles, is always 1080 degrees (360 x 3) no matter what the requirements of the “inside” angles are in that particular geometry.

That property should remain consistent in any 'smooth' realm - IE, no spikes running to plus/minus infinity, no disjointed space (gaps), etc - if I recall my advanced geometry correctly.

 

[Jackie.Hikaru]

You know what’s super interesting about that question? The sum of the inside angle at each vertex and the outside angle at each vertex is still only 360 degrees, no matter what the curvature of the surface is (spherical or hyperbolic).

That means that the total for all 3 angles, summing interior and exterior angles, is always 1080 degrees (360 x 3) no matter what the requirements of the “inside” angles are in that particular geometry.

Unimpressive Police here! This thread is focused on the unimpressive Y'all need to tone down this impressive math shit alright? Good. Carry on!

 

[Britva415]

Yes, notifications.

This happens to me when I miss the notification and don’t open the thread right away.

It’s like the site thinks, “I already notified you” and stops doing it until I take it upon myself to catch up with the thread.

Which I don’t know to do because I missed the notification.

Vicious cycle!

But once I catch up with the thread, I’ll start seeing notifications again.

I’ve also somehow mistakenly “in-watched” a thread.

 

[Britva415]

Unimpressive Police here! This thread is focused on the unimpressive Y'all need to tone down this impressive math shit alright? Good. Carry on!

Math nerds aren’t quirky?

Ah, there’s an AND in that title.

 

[Bramblethorn]

If you place the vertexes at the maxima of each axis, covering 1/8th of the sphere, you get 90 degree angles.

Pulling the vertexes in closer and the angles shrink down towards the planar normal of 60 degrees as the curvature flattens towards a plane.

I can't figure out the max angle if you go past the maxima of the angles and put the bulk of the sphere inside the triangle instead of just one eighth of the whole, not that it really matters to this conversation.

edit: argh, didn't see @Britva415 's more succinct answer until I'd posted this one. At least we agree!

That's an easy one when you see the trick to it.

Take a little equilateral triangle, small enough that the sphere is approximately flat over that region. As you've said, the angles will be just over 60 degrees, for a total of (180+ε)°.

Expand it until it covers one-eighth of the sphere, and again as you've said, the angles are 90° for a total of 270°.
Expand it further again, until the triangle covers one-half of the sphere, with our three points evenly spaced along an equator. (It's now a perfect circle, but by our definitions it's still a triangle.) Now the angle at each vertex is 180°, for a total of 540°.

But which half of the sphere is the "inside" and which is the "outside" of that triangle?

Any time we draw a triangle on a sphere, we're actually creating two triangles: one "inside", one "outside" the lines, for whichever side we designate as "inside". Normally we're thinking about the smaller triangle, but the larger one is just as valid by spherical-geometry definitions (even though it's a long way from the plane triangles we're used to).

At any vertex, the sum of the interior and exterior angles is 360°. So for any set of three lines that split the sphere into two triangles, the total angle sum of the two is 3*360° = 1080°.

Since the minimum angle sum possible for a triangle on the sphere is 180°, the maximum is (1080-180)° = 900°. Going back to our original tiny equilateral triangle with angles each just over 60°, the partner triangle is an equilateral triangle with angles each just under 300°.

 

[UsuallyPresent]

Math nerds aren’t quirky?

Ah, there’s an AND in that title.

We're not iterating over polydimensional space, fractal spaces, or other truly impressive math. Carry on, we fit.

 

[Kitty_so_frisky]

Since I’ve been in the author’s hangout I’ve learned that my fellow lit authors are an incredibly impressive bunch who have lived incredibly impressive lives and to simply be here among you I feel so fortunate and honored. That said I would love to learn more about you all! Specifically how about we share something entirely unimpressive yet perhaps a bit quirky about ourselves.

I’ll go first! The only Cheetos I eat are hot Cheetos and I always eat them with chopsticks. I guess that’s two facts…

Quirkiness, a Cowboy Bebop reference in tagline AND you eat Cheetos with chopsticks?! You can come to the Sex & Shenanigans thread and sit by me and @UnquietDreams , we were just talking about eating Cheetos and other snacks with chopsticks yesterday!

 

[Jackie.Hikaru]

You can come to the Sex & Shenanigans thread and sit by me and @UnquietDreams , we were just talking about eating Cheetos and other snacks with chopsticks yesterday!

such classy people in this joint!

 

[Bramblethorn]

That property should remain consistent in any 'smooth' realm - IE, no spikes running to plus/minus infinity, no disjointed space (gaps), etc - if I recall my advanced geometry correctly.

It's been a while, but I think the smoothness condition needs to be a little tighter than that: not just "no spikes running to plus/minus infinity", but no spikes at all.

For instance, replace our sphere with an icosahedron (d20), made up of twenty flat equilateral triangles glued together in a ball. For any point not at one of the vertices, the space is locally flat so the sum of angles around a point is exactly 360°. But for a point at one of the vertices, the angle sum is only 300°.

For a more obnoxious example, consider the surface defined by x = r cos θ, y = r sin θ, z = r sin nθ, for some positive integer n. This is continuous, and for any point not at (0,0,0) the angle sum is 360° as usual, or 2π radians. But at (0,0,0)?

The angle sum at a point is equivalent to the limit of (circle circumference/circle radius) as the radius tends to zero.

For a circle of radius r with centre at (0,0,0), the circumference is equal to:

integral from 0 to 2π of sqrt( (dx/dθ)^2 + (dy/dθ)^2 + (dz/dθ)^2)) dθ
= integral from 0 to 2π of sqrt( (r^2 sin^2 θ + r^2 cos^2 θ + r^2 n^2 cos^2 (nθ) ) dθ
= r * integral from 0 to 2n of sqrt(1 + n^2 cos^2 (nθ)) dθ

So the angle sum is equal to:

integral from 0 to 2n of sqrt(1 + n^2 cos^2 (nθ)) dθ

which is clearly larger than 2π, and can be made arbitrarily large by increasing n.

 

[Bramblethorn]

Unimpressive fact about me: when I was a kid I used to eat apples core, seeds and all. I may have thought I was building up a resistance to cyanide, I don't remember for sure.

I loved apricots and cherries, and I would do a thing where I'd carefully split an apricot in half, remove the kernel, and stick a cherry in the hollow before eating it.