this post was submitted on 07 Oct 2024
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Science Memes

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[–] spankmonkey@lemmy.world 158 points 1 week ago (5 children)

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

[–] kryptonianCodeMonkey@lemmy.world 95 points 1 week ago (2 children)

And that the 90 degree angles should be interior angles.

[–] ninja@lemmy.world 49 points 1 week ago (2 children)

And that polygons should only consist of straight lines.

[–] rain_worl@lemmy.world 0 points 4 days ago
[–] bstix@feddit.dk 37 points 1 week ago

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

[–] wolfpack86@lemmy.world 12 points 1 week ago (1 children)

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 27 points 1 week ago* (last edited 1 week 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 34 points 1 week ago (2 children)

Take shitposts seriously and point out their obvious errors

-Carl Friedrich Gauss, probably

[–] Eheran@lemmy.world 21 points 1 week ago (2 children)

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

[–] rain_worl@lemmy.world 3 points 1 week ago

gasp!!! it is c/!!!

[–] Tja@programming.dev 4 points 1 week ago

The name of that Gauss?

Ampere

[–] lugal@sopuli.xyz 21 points 1 week ago

c/gatekeeping squares

[–] hydroptic@sopuli.xyz 10 points 1 week ago

You're no fun

[–] Tolookah@discuss.tchncs.de 5 points 1 week ago

Polar coordinate square?

[–] Lemjukes@lemm.ee 31 points 1 week ago (1 children)

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 1 week ago (1 children)

Sides must be straight and parallel two and two.

[–] affiliate@lemmy.world 11 points 1 week ago (1 children)

it’s homeomorphic to a square, so why not

[–] hydroptic@sopuli.xyz 3 points 1 week ago

See, you get it

[–] Doll_Tow_Jet-ski@fedia.io 11 points 1 week ago (3 children)
[–] Assman@sh.itjust.works 16 points 1 week ago

The black lines

[–] CodexArcanum@lemmy.world 8 points 1 week ago (1 children)

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 1 week ago (1 children)

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 8 points 1 week ago (1 children)

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

[–] Doll_Tow_Jet-ski@fedia.io 9 points 1 week ago

/slams door

[–] UrLogicFails@beehaw.org 5 points 1 week ago (1 children)

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 1 week ago* (last edited 1 week ago) (1 children)

I look at your diagram and see:

ϴ= L/(L+R)

And

2π-ϴ = L/R

I solved those 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 1 week 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.