The Easiest Angle of Refraction

For the past two weeks I’ve been attempting to introduce refraction to the otherwise perfectly fine vector equations of everything. Or if I look at things from a longer perspective, I’ve been attempting to do this for months.

While I’ve had the simple shapes for quite some time from which these equations should be deduced, actually coming up with the equations isn’t straightforward. The problem is that you have the ellipse of refraction, which governs the shape of the orthogonal circle of refraction, which is the actual shape that governs refraction. However, my attempt at tilting the refracted shape so that the orthogonal circle of refraction would be on a single plane produced equations where three angles were dependent on each other in a way that I just couldn’t solve.

So, I once again decided to leave the tilted projections alone for a bit and first come up with an equation that describes the ellipse of refraction. If I choose the primary angle that defines all the other angles to be the angle by which the ellipse of refraction tilted to the non-refracted state, this should make the equations really easy. In the Below image this angle is φ, but it isn’t exactly the same φ as in the tilted projections. 

To cut a long story short, using this image and the law of cosines, it is possible to determine the radius L of the semi-major axis. Without boring you with the details:

Also, without boring you with the details, when we know that the radius of the semi-minor axis is √3r, we can plot the projections of the ellipse in all dimensions. Below we see the ellipse plotted in y-z plane, making it look like a line.

Just looking at the image above one might be fooled to think that the solution to the orthogonal circle of refraction was just around the corner. However, without the nice tilts the orthogonal circle of refraction looks like a nightmare:

However, there might be a few tricks up my sleeve that might make it easier to handle. So, next I’ll play around with the three projections and look at what the ellipse looks like in each of them and with luck I should be able to come up with the equation for that as well.