If I have a radius of curvature R = 0.865” and a surface semi-diameter of 0.75” and I apply a TAPR decenter DY of 0.25”, I’ll get a range of r going from -1.0” to 0.5”, which then gives sin(alpha) = abs(-1/0.865) = 1.156. Taking the asin of this gives me an error (in Excel). Am I doing this correctly? (All “x” components are “0” for this question, e.g. no DX involved). Thanks.
Alex
Best answer by Jeff.Wilde
The radial coordinate can only be positive:
Note that a hemispherical surface has a semi-diameter = radius of curvature R. Each point on this surface has a radial value (in the xy-plane) that must physically be less than the radius of curvature. Based on the way that a cosine taper is defined in OpticStudio, the same constraint applies to the coating’s local radial coordinate.
In your case, the surface is close to being hemispherical, and your coating decenter value causes the value of the local coating radial coordinate r to exceed R. This in turn causes alpha to be undefined. You need to reduce the decenter value or perhaps try using a decentered radial polynomial taper.
+3The radial coordinate can only be positive:
Note that a hemispherical surface has a semi-diameter = radius of curvature R. Each point on this surface has a radial value (in the xy-plane) that must physically be less than the radius of curvature. Based on the way that a cosine taper is defined in OpticStudio, the same constraint applies to the coating’s local radial coordinate.
In your case, the surface is close to being hemispherical, and your coating decenter value causes the value of the local coating radial coordinate r to exceed R. This in turn causes alpha to be undefined. You need to reduce the decenter value or perhaps try using a decentered radial polynomial taper.
Thanks for bringing some clarity to the subject, Jeff. I wasn’t seeing the forest for the trees! Makes sense now. I appreciate the speedy feedback.
+3No problem, you’re very welcome.
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