Orienting a hexagon around another hexagon without 60-degree turns

CombatHammie · 2020-07-13T15:02:14

To simplify the question, I'm boiling down one of the mechanics of my game into two hexagons. The mechanic comes down to rotating the black hexagon to align with the red hexagon. The black hexagon will encounter red hexagons that are oriented around an arbitrary angle. The black hexagon will rotate smoothly to match the orientation of the next selected red hexagon. The red hexagons might not be perfect hexagons. That is, they might be slightly deformed and the angles of the corners might be bigger or smaller than 60 degrees. The alignment portion of the rotation (getting all vertices to align with other vertices) is straightforward: Choose a reference vector (e.g. Vector2.UnitY) Determine the topmost vertex and calculate the angle between it and the reference vector Apply rotation with the given angle The problem is that eventually, the orientation will have a full 60/-60 rotation in it. The reason is simple, depending on which hexagon selected next, the new “topmost” vertex might now be on the other side (e.g. shift from +x to -x). At first, I figured that this is just a matter of selecting the topmost vertex with the smallest angle by comparing the absolutely value of the topmost and second topmost angles. But this makes no difference, eventually, the 60/-60 degree rotation will still be applied, because either side will have a new topmost vertex. I tried a number of approaches including constraining the maximum orientation to 30 degrees (and then flipping to the other angle). But with every approach I try, at some point, I once again hit the 60 degree rotation, either in cases where the orientation reaching a flat top, or the angles are being exchanged. Am I hitting some sort of a mathematic constraint and/or this simply not possible? Is there no way to orient one hexagon with another with an arbitrary angle without eventually doing a full 60-degree rotation?

Math and Physics Programming

Started by CombatHammie July 07, 2020 04:59 PM

12 comments, last by alvaro 4 years, 6 months ago

alvaro

21,610

July 09, 2020 01:53 PM

#include <complex>

typedef std::complex<double> C;

C rotation_to_align_hex(C original, C target) {
  C o6 = std::pow(original, 6);
  C t6 = std::pow(target, 6);
  C r6 = t6 / o6;
  return std::pow(r6, 1.0/6.0);
}

Following the initial suggestion of using complex numbers, this works beautifully. You can plug in the orientation of any of the vertices of the two hexagons (as length-1 complex numbers) and it will return the smallest rotation needed to line up the hexagons.

JoeJ

4,406

July 09, 2020 03:15 PM

alvaro said:
Following the initial suggestion

It was your introductions from years ago that made me using complex numbers for 2D rotations, and it's really convenient. : )
Though, i wonder why division gives difference of angle, pow multiplies angle, arg gives angle, conj gives dot product, etc. There might be some deeper sense for the mathematician, but i'm still missing that :D

Your example assumes regular hexagons. I thought this is not the case and they can be irregular. If i got this wrong there is no need to iterate over all 6 points ofc, and no need for weights either.

alvaro

21,610

July 13, 2020 03:02 PM

Oh, I'm glad to hear my years-long propaganda campaign for complex numbers is having an effect! ?

It's too hard to explain in this forum why division gives difference of angles, etc. Maybe this might help: https://www.youtube.com/watch?v=mvmuCPvRoWQ

I don't know even know what the desired behavior is in the case of irregular polygons, so I just went with the case I understood, and I demonstrated how using complex numbers in the way you indicated does work beautifully.

Orienting a hexagon around another hexagon without 60-degree turns

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Orienting a hexagon around another hexagon without 60-degree turns

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