In my last post I was happy to announce that I had gotten the vector equations for orthogonal refraction sorted out. I was even able to plot the locations of dots with an exact distance of one between each neighboring dot:
But what this equation didn’t yet produce was an explanation for the secondary refraction that makes the entangled dots bend.
While I can’t be 100 % certain, I think I know what the known unknown is. Already quite some time ago, I knew (or at least had a very strong hunch) that refraction would be split into two components that I think I called tori (or toruses) of refraction. One of the most visual illustrations of this is in the image below.
However, while the image itself might be correct, the problem with it is that I wasn’t really able to show how the single torus with no refraction could be split into two tori.
But then it hit me. While you cannot connect the two refracted dots (or ends of vectors) with a single circular arc, you can do this with an elliptical arc. Like this:
The image is quite cluttered, so its contents need to be to be clarified. The green vectors are all non-refracted, as is the green circular arc. The semi-transparent red tori are the ‘conventional’ tori of refraction, whereas the semi-transparent red elliptical torus is the new ellipse of refraction. The important thing to understand is that the ellipse connects the two tori, while still grazing the surface of both large transparent spheres (that indicates all possible locations of the vectors of a neighboring connected dot)
However, viewed from this angle the Eureka moment of the ellipse of refraction isn’t too visible. Viewed from a slightly different angle we can see that if an orthogonal circular arc is drawn from the non-refracted point to the ellipse of refraction, the point of connection is not on the torus of refraction!
This means that while the initial model was very close to reality, it wasn’t the whole truth.
Now that I know where I was incorrect, I should be able to correct the equations of refraction. What these equations will be, I don’t know yet. But my guess is that they won’t be that much more difficult than the ones that I’ve already found.
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