Fools!
…limiting themselves to Euclidean geometry…
this post was submitted on 15 May 2025
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Only one of them is limiting himself to Euclidean geometry. The others are perfectly calm.
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(2 replies)
Saddle shaped universe confirmed
Looks like a tetrahedron to me.
Exactly! The diagram is simply a schematic.
Just wanted to … nevermind.
Too late is too is too late is …
That implies one person is observing 3 other people from the above (or flying over), which is not exactly trivial.
Nobody said it would be easy
exactly what I came here to say
ikr? It's like some people don't even recognize a tetrahedron
Ah, D&D rules
Pffft, Dnd had the 'first diagonal 5, second diagonal 10' rule. It worked well enough, aye?
It doesn't anymore =(
5e uses diagonal = 5'
Alternating diagonals is in the (2014) DMG as an optional rule at least
Oh good! Octagons are a much better approximation of a circle than squares
Well anything after 3.5e is a watered-down, bastardized version of the game anyway.
4e just used "squares" instead of 5 feet, but it, like 5e, used chebyshev distances.
Pathfinder 2e uses alternating diagonals though.
D&D still doesn't have hexagons?
Well you see, space isn't flat in this very localized area!
No one said right angles.
Equal sides in a triangle are only possible if the corners are equal. So, 60⁰ each.
But its height cannot be half of base because of the same Pythagorean theorem
(1,5)²+(1,5/2)²=2,8125
sqrt(2,8125) ≈ 1,677, which is half of a diagonal
So, we get 4 sides that are 1,5 in a parallelogram, but diagonals are 1,5 and 3,354, as opposed to both being 1,5 as shown on the picture
TL;DR: Won't work because Pythagorean theorem
It's funny how we say "because of such and such theorem" as if if some greek dude didn't come up with his little story, the height could totally be half of base.
Pyramid?
Yes, it is possible with a 3-sided pyramid, i.e. tetrahedron. If we dont look at all 4 points as being on the same plane but 2 opposite corners being offset above or below the other two, this could totally be a tatrahedron.
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calm down, they're constraints on distance, not distance
if the people were aranged in 3d in the shape of a tetrahedron (triangular pyramid) this would work out fine
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Tetrahedrons man, tetrahedrons.
So this makes me wonder if one could force a move into a higher dimension by somehow constraining a set of connected distances in this way.
Sort of like protein folding as a way to bootstrap a dimensional jump.
You might like And He Built a Crooked House by Robert Heinlein - the story of a tesseract-shaped house that folds itself into a real tesseract during an earthquake.
Middle one should be the square root of 4.5 meters, or 2.12 meters
You're all thinking too two-dimensionally. Clearly the people are being instructed to arrange themselves into a tetrahedron.
What? Everytime I meet other people we always arange ourselves in the shape of a simplex of the appropriate dimension. Doesn't everyone?
So the fifth person to arrive moves to the centre of the tetrahedron and shifts roughly 1.299m into the past or future.
I have a few questions.
- How do you attain time offset?
- Doesn't that make conversation difficult?
- What even is the fifth dimension?
- How do you convert a distance in metres into a distance in time? You would surely then have a universal m/s? Oh, wait, there is a universal speed, it's the speed of light, which means 1.299m is equivalent to about 4.3 billionths of a second, which is considerably less impressive for question 1 and just not at all problematic for question 2.
- If you're using very fast motion for your time offset, doesn't that make conversation even more difficult? How fast would you need to be going to dilate time for a few billionths of a second? Doesn't Heisenberg uncertainty start to have an impact here? How can you be sure you got it right?
If you have to ask, you wouldn't understand.
If I understood, I wouldn't have to ask.
- So, the diagram doesn't represent it well, but the 1.5m is a minimum. So, I just delay myself by half a heartbeat which is well over 4.3^e-9s.
Yes, but if they're just minimums, there's no need for even using the third dimension, let alone the fourth.
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This is so good hahaha
Thank you
Explanation:
So, Theres the sentence of Pythagoras. It says that c^2^ = a^2^ + b^2^ when the triangle has a 90° corner
Since a square is just 2 triangles, it applies. That means c (the distance from Person a to Person c) should be √(2×1.90^2^). But that is 2.7m.
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