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hi, me again sorry; puzzle 391 in book 3 doesn't appear to have a label number!

2 months ago 1 0 1 0
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2 months ago 1 0 0 0
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you might have changed it in the wrong place, i don't know much about website backend stuff; on hover it says the correct link but the bottom left is where clicking it directs you

2 months ago 1 0 2 0

@lukathemouse.gay sorry to bother you but on the alblune site your bsky link is outdated since you changed to .gay, love your games <3

2 months ago 3 0 1 0

idk about you but mine was more or less modelled on who i was when i made it, so it isnt surprising to me that it has threads leading back into my past because i have those threads too

1 year ago 0 0 0 0

it takes pi arcs of length d to cover the circumference of the cross section but since you can't have a fraction of a pass it has to round up

1 year ago 2 0 0 0

to perfectly return to the original position it would be an infinite number of rotations because pi is irrational, but if you want it to be close enough to not notice it skipping back when restarting, you'd need a good rational approximation of pi that you can get close to, like 355/113

1 year ago 1 0 1 0

actually someone posted the question in a discord im in, i made this account just to reply :p

1 year ago 1 0 0 0

just saw this after writing out an answer, you don't need to consider the whole surface to calculate, just the cross section at one point; each time the arc passes that point on the torus it covers "d" of the circumference, so it has to pass circ/d times, which is pi (but it would really be 4)

1 year ago 1 0 0 0

if i misunderstood the question lemme know where i got it wrong and ill try to help out, otherwise gl with your sound donut

1 year ago 1 0 1 0
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apologies for the bad ms paint rendering but each colour indicates a single revolution around the major axis, the blue part that doesn't overlap with red will be 2pi-6 minor radii long (~0.283r)

1 year ago 1 0 1 0

with no overlap, the arc length would need to be (circumference/k) where k is an integer and it would take k sweeps to do it

if you want to fully cover the torus with a length equal to the diameter, you'd need 4 passes of the wand you described to cover it with overlap, would look like this image

1 year ago 1 0 1 0

right rate for the start of the next arc to meet the end of the previous arc, it will take pi arcs to cover the torus. The arc length being equal to the diameter means it covers an angle around the cross section of 2 radians, and the full rotation is 2 pi radians. If you want it to cover the surface

1 year ago 1 0 1 0

if im reading this correctly you want an arc that traces around the surface of the torus to cover it entirely, the length of the arc is the minor diameter, and want to know how many times the arc would have to sweep around to cover the whole thing; assuming it's rotating around the minor axis at the

1 year ago 1 0 1 0