imk
New member
- Joined
- Apr 7, 2023
- Messages
- 13
Hello, so my math is poor but i do try!
Now I am interested in calculating the width of a beam when projected at an angle over a distance and I have found the following which tests correctly:
w = 2*d*tan(a/2) where w=width, a=angle and d=distance.
Now I need to transform this to find d or a, so I found the site https://www.symbolab.com/ which has a function transformation page, so I entered
w = 2*d*tan(a/2) to be transformed and have two solutions one of which seems odd.
Solution for d=w/(2*tan(a/2) looks and tests correctly.
Solution for a = 2arctan(w/2*d)+2PIn. Why +2PIn I have no idea what n is and what part it plays in the transformation?
I have however tested a = 2arctan(w/2*d) and it seems correct.
Insight most gracefully received please![Smile :) :)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Now I am interested in calculating the width of a beam when projected at an angle over a distance and I have found the following which tests correctly:
w = 2*d*tan(a/2) where w=width, a=angle and d=distance.
Now I need to transform this to find d or a, so I found the site https://www.symbolab.com/ which has a function transformation page, so I entered
w = 2*d*tan(a/2) to be transformed and have two solutions one of which seems odd.
Solution for d=w/(2*tan(a/2) looks and tests correctly.
Solution for a = 2arctan(w/2*d)+2PIn. Why +2PIn I have no idea what n is and what part it plays in the transformation?
I have however tested a = 2arctan(w/2*d) and it seems correct.
Insight most gracefully received please