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
View original
+3
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.
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.
+3
No problem, you’re very welcome.
Enter your username or e-mail address. We'll send you an e-mail with instructions to reset your password.
Do not provide any information or data that is restricted by applicable law, including by the People’s Republic of China’s Cybersecurity and Data Security Laws ( e.g., Important Data, National Core Data, etc.).
不要提供任何受适用法律，包括中华人民共和国的网络安全和数据安全法限制的信息或数据（如重要数据、国家核心数据等）。