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Behold, a square (sopuli.xyz)

submitted 2 months ago by hydroptic@sopuli.xyz to c/science_memes@mander.xyz

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[-] spankmonkey@lemmy.world 160 points 2 months ago

As long as we ignore the parallel sides requirement, sure.

[-] kryptonianCodeMonkey@lemmy.world 97 points 2 months ago

And that the 90 degree angles should be interior angles.

[-] ninja@lemmy.world 49 points 2 months ago

And that polygons should only consist of straight lines.

[-] bstix@feddit.dk 38 points 2 months ago

Yes sure, in Euclidean geometry, but this is clearly keyhole shaped geometry.

[-] rain_worl@lemmy.world 1 points 2 months ago

geodesics

[-] wolfpack86@lemmy.world 11 points 2 months ago

They're also not actually right angles, as the curvature starts departing from the angles origin. They may be approximately 90, down to many many small decimal places, but they are not 90.

[-] kryptonianCodeMonkey@lemmy.world 29 points 2 months ago* (last edited 2 months ago)

That's not accurate. If you are measuring the angle of a line intersecting with a curved surface, you measure against the tangent at the point of contact/intersection. It can be and still is exactly 90 degrees.

[-] Prunebutt@slrpnk.net 36 points 2 months ago

Take shitposts seriously and point out their obvious errors

-Carl Friedrich Gauss, probably

[-] Eheran@lemmy.world 22 points 2 months ago

Science memes is not r/shitposting? I would assume the person is serious when posting here.

[-] rain_worl@lemmy.world 4 points 2 months ago

gasp!!! it is c/!!!

[-] hydroptic@sopuli.xyz -5 points 2 months ago

I would assume the person is serious when posting here.

This sounds like a "you" problem

[-] Tja@programming.dev 4 points 2 months ago

The name of that Gauss?

Ampere

[-] lugal@sopuli.xyz 22 points 2 months ago

c/gatekeeping squares

[-] hydroptic@sopuli.xyz 10 points 2 months ago

You're no fun

[-] Tolookah@discuss.tchncs.de 5 points 2 months ago

Polar coordinate square?

[-] Lemjukes@lemm.ee 32 points 2 months ago

I remember enough from geometry to know this is horseshit and be annoyed at it but not enough to actually prove why

[-] Tja@programming.dev 11 points 2 months ago

Sides must be straight and parallel two and two.

[-] OldChicoAle@lemmy.world 3 points 2 months ago

QED

[-] Doll_Tow_Jet-ski@fedia.io 11 points 2 months ago

What are the 4 sides?

[-] Assman@sh.itjust.works 17 points 2 months ago

The black lines

[-] CodexArcanum@lemmy.world 8 points 2 months ago

The semi-circle is one side, then the 2 straight edges, and the arc between them is the 4th side.

[-] Doll_Tow_Jet-ski@fedia.io 7 points 2 months ago

That's what I thought. The only way on which this has four sides is if the semi -circle is a side. But if that's the case, then I don't know wha the definition of "side" is

[-] bstix@feddit.dk 10 points 2 months ago

Knock knock. Do you have a moment to discuss non-euclidean geometry?

[-] Doll_Tow_Jet-ski@fedia.io 10 points 2 months ago

/slams door

[-] UrLogicFails@beehaw.org 5 points 2 months ago

Someone may want to double-check my math on this one, but the length of the sides will be dependant on the radius of the smaller circle

ϴ=π+1-√(π^2+1), l=(2π-ϴ)r_1, l is the length of the sides. r_1 is the radius of the smaller circle

[-] m0darn@lemmy.ca 4 points 2 months ago* (last edited 2 months ago)

I look at your diagram and see:

ϴ= L/(L+R)

And

2π-ϴ = L/R

I solved those (using substitution, then the quadratic formula) and got

L= π-1 ± √(1+π²) ~= 5.44 or -1.16

Whether or not a negative length is meaningful in this context is an exercise left to the reader

Giving (for L=5.44):

ϴ~= 0.845 ~~48.4°

I'm surprised that it solved to a single number, maybe I made a mistake.

[-] UrLogicFails@beehaw.org 3 points 2 months ago

That lines up pretty similarly with what I found also. The angle should be a constant since there is only one angle where the relationship would be true. I just left it in terms of π because I try to avoid rounding.

Having said that, L would be a ratio of r; which I think lines up with what you found as well.

[-] affiliate@lemmy.world 11 points 2 months ago

it’s homeomorphic to a square, so why not

[-] hydroptic@sopuli.xyz 4 points 2 months ago

See, you get it

[-] davidagain@lemmy.world 2 points 2 months ago

I'll tell you why not! You hippie homeopaths are all the same! Science has scienced the evidence that there's no evidence for homopathic medicines otter than the libido effect.