​@Ivair.Gontijo 

Its a good question. I’ve always wanted to write the equations down for people to easily have access to this information but never found the time to do it…

For me the closest description of this algorithm is found in the book Aberrations Of Optical Systems by W T Welford in Chapter 4 Finite Raytracing. Eventually, you’ll want to pay close attention to Section 4.7 Raytracing through quadrics of revolution.

Regarding question 1. Yes, for a standard surface with only a radius of curvature and a conic coefficient there is a closed form solution and it is what’s implemented in case 5.

Regarding question 2. There’s a relatively good match between the variable names in case 5 and the equations in the book. The transfer (finding the ray intercept with the surface) involves solving a quadratic equation. a, b, and c are the coefficients of the quadratic equation, and rad is basically the discriminant (they removed a factor of 4, but once you have the equations in front of you it’ll make sense). So what is this quadratic equation you need to solve?

In short (because I don’t have the time to go into the details now), its the equation of the sag (of the standard surface):

where r * r = x * x + y * y. And you can express the intercept coordinates (x, y, z) as a function of the vertex (you call it tangent plane) coordinates (x0, y0, z0=0) and the unknown (called Delta in the book) distance between those two points. This is visible in the code in the lines:

            (UD->x) = (UD->l) * t + (UD->x);             (UD->y) = (UD->m) * t + (UD->y);             (UD->z) = (UD->n) * t + (UD->z);

Where t is the unknown distance. At this point in the code, they’ve solved the quadratic equation already, but its just to show you how you’d derive the quadratic equation to be solved. You plug the:

x = l * t + x_0

y = m * t + y_0

z = n * t + z_0

Into the equation of the sag, and solve for t.

PS: in the book, they use an asphericity parameter epsilon, and as far as I remember it isn’t the conic coefficient but they give the equation to do the transformation (4.26). Double check this because I studied this a long time ago.

Regarding question 3. Its very unlikely, but it all depends on the equation of your sag (as you just saw). What kind of surface are you trying to model exactly? Did you not find something suitable among the existing surfaces?

I apologize for being brief on 2. but I really don’t have much time now. In the future, I’ll try to edit or make a clean version of this with details.

I hope it still helps though.

Take care,

David

Thank you David for the detailed answer above and also for the reference to Welford’s book, which was very helpful! Yes, I noticed that in the book they use epsilon, which is connected to the eccentricity of the conic, which is in turn connected to the conic constant k by k=-e^2.

Unfortunately none of the surfaces in Zemax are sufficient to model the surface I need to work with. So, I am going to study that chapter in the book carefully, but you are right that with a more complex surface, an iterative solution will be needed.

Thanks again for taking the time to explain the details above!

Ivair

By the way, equation (4.25) in Welford’s book is z = 0.5c(x^2+y^2+ez^2), which is in implicit form. It can be manipulated into the standard surface equation used in Zemax by making r2 = x^2+y^2 and rearranging it into 

cez^2 -2z + cr^2 = 0

 Then solve this quadratic equation, as r^2 is constant with respect to z. Choose the minus sign because when r = 0, z should also be zero.  Multiply above and below by the expression with the positive sign, to move the square root to the denominator. Then make e=1-k and you are done! :-)

There is a mistake in my comments above, regarding the connection between epsilon and k. Epsilon is the constant used in Welford’s book implicit formula for the conic (I called it ‘e’ in my comment) and the conic constant ‘k’. The correct relation between these two quantities is epsilon = (1+k).

It was fun to derive all the formulas used in the standard surface dll code. Once ​@David.Nguyen pointed me in the right direction, to use the implicit equation for the conic, it was easy. Thank you David!

Note that there is also a third constant related to these two, which is the eccentricity. However, we don’t use it, so there is no need to complicate things further. :-)