The knowledge base article ‘Understanding the geometry in OpticStudio Curvature Cross-Section analysis’ authored by Shawn Gay doesn’t explain this discrepancy (I mean sphere case above).

Eric

Hi Eric,

I think I see where the confusion lies, and I have some comments that may help you. 

Firstly, the value given by the SCUR operand isn’t measuring circular radius of the cross-sectional cut of the surface in x or y. 

SCUR is reporting the surface curvature of your surface, measured at a certain (x,y) coordinate along your surface, and you can specify whether you want to know just the curvature in the x direction, just curvature in the y direction, the x-y curvature, just the tangential curvature, and so on, with the data column. 

So as an example, let’s take the spherical surface. Since it is spherical, by definition it has the same curvature along the entire surface, let’s say, I’ll use a 20mm radius of curvature (aka a curvature of 1/20 = 0.05). I use the SCUR operand to report the surface curvature at multiple points along the surface, and it reports back the same 0.05 curvature, since the spherical curvature is consistent across the entire surface.

As an example of the surface curvature is not being consistent across the surface, I set the conic constant of Surface 1 to -1, making a parabolic surface. Now, the x curvature and y curvature are not necessarily the same at a given (x,y) coordinate. So, here you can see that I used the Data column to specify if I wanted the x or y curvature reported per each coordinate point, and that they differ slightly. However, I also wanted to see the effect of axial symmetry, so you can see that at points equidistant from the origin s (3,0) & (-3,0) ], it reports the same curvature. 

Lastly, as a more dramatic example, I tried out a toroidal surface with a curvature in y but no curvature in x, and you can see the more obvious difference between curvature in x and y per each point. 

So, I hope this explains the use of the SCUR operand, and what exactly is meant by ‘surface curvature in x and y’ but let me know if I’ve been unclear or if I misunderstood your question. 

Best,

Nikki 

Hello Nikki,

Thank you for your nice detailed explaination.

But the answer still didn ‘t clarify my confusion. Still take  sphere  case as example, because  curvature calculation depends on direction, I think the statement ‘Since it is spherical, by definition it has the same curvature along the entire surface’ is not strict correct. For sphere case, the curvature is constant only along local radial direction, but not in a global cartesian coordinate system(sag defined coordinate system).

In sag defined cartesian coordinate system(I think this is what coordinate system SCUR data4/5 used), the curvature only constant in x and y direction for vertex point(0,0). Any depature (x,y) point will make curvature in x and y direction different. However, SCUR(data4/5) still give same constant value for any depature point. This doesn’t make sense.

Best regards, 

Eric

Hi Eric,

I’d recommend looking at this page on how curvature is derived- essentially as the second derivative of the arc/surface at the desired point x,y (rather than the sag of a cross-section)  2.2 Principal normal and curvature (mit.edu).

I hope this clarifies things,

Nikki 

Hey Eric,

I think this part is not quite correct:

But the answer still didn ‘t clarify my confusion. Still take  sphere  case as example, because  curvature calculation depends on direction, I think the statement ‘Since it is spherical, by definition it has the same curvature along the entire surface’ is not strict correct. For sphere case, the curvature is constant only along local radial direction, but not in a global cartesian coordinate system(sag defined coordinate system).

I think the bit that is missing is that in ray-tracing, we always imagine a ray hitting some x,y sag point and create a surface normal at that location. In the infinitesimal area that normal vector covers, the gradient is always zero in that coordinate system. Hence the curvature (second derivative) is given by 1/R at all places, in a coordinate system that is located normal to the ray-surface intersection point.

So the curvature that matters from the perspective of ray-tracing is truly independent of where you look on a sphere.

• Mark

Thank you Mark, Nikki

I made a mistake and both your explaination make sense. Nikki,the article you gave is very clear.

Great support! and Many thanks, again!

Eric