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Anonymous Thu 16 Sep 17:09:40 2021 No.13649139 View Reply Original Report

Quoted By: >>13649553 >>13650168

Guys how the fuck does one get the solid angle of a point projected on a plane?

Lets say I have a single point of illuminating light. Then, I have a rectangular detector of a certain size sat a certain distance away from the point. How one earth can I calculate the angle/percent (im guessing it would be out of 4pi) of the light that the detector covers?

Very confusion here...

Anonymous

Anonymous Thu 16 Sep 2021 17:33:23 No.13649241 Report

Quoted By: >>13649430

I don't know what the red is, or a "solid angle", and your descriptions are so poor I want to vomit a bit. I will assume the rectangle is normal to the vector pointing back to the point source of light, implying it is also centered. I would then "inscribe" the rectangle inside of a spherical shell so the four corners are in contact. I think you are asking what percentage of area of the spherical shell (of this size or larger) would be covered in shadow.

Anonymous

Anonymous Thu 16 Sep 2021 18:24:11 No.13649430 Report

Quoted By: >>13649588 >>13649653 >>13649816

>>13649241
Holy shit are you retarded? Why would you assume the plane is normal to the vector to the center? DId I say it was? It's a fairly simple problem. A point of light/particles emits radiation in all directions (4*pi). I have rectangular detectors that I need to mount in specifiic places. I need to know what percentage of that 4*pi each detector covers. I'm guessing its as simple as working out the are legnths in theta and phi but it might also require integration, I'm unsure.

Anonymous

Anonymous Thu 16 Sep 2021 19:13:36 No.13649553 Report

Quoted By: >>13649643

>>13649139
Very cool paper with algorithms you can use. If you project the 4 vertices of your rectangle onto a sphere with radius 1 you can calculate the area of the projection which is also the solid angle you are looking for.

https://trs.jpl.nasa.gov/handle/2014/40409

Anonymous

Anonymous Thu 16 Sep 2021 19:28:21 No.13649588 Report

Quoted By: >>13649643

>>13649430
>Why would you assume the plane is normal
If it wasn't, treating it like a small part of a bigger one that was would make it easier.

Anonymous

Anonymous Thu 16 Sep 2021 19:44:57 No.13649643 Report

Quoted By: >>13649852

>>13649588
I thought about this, would that really help though? I'm not conviced it becomes easier if its normal, and even if so once you have that area, you are faced with the anagolous problem of how to reduce it to a smaller area that is no longer normal.

It seems >>13649553
is the best approach, project on to a sphere and measure the area from there

Anonymous

Anonymous Thu 16 Sep 2021 19:47:24 No.13649653 Report

Quoted By:

>>13649430
youre a stupid little cocksucker

Anonymous

Anonymous Thu 16 Sep 2021 19:52:44 No.13649667 Report

Quoted By:

>>>13644707

Anonymous

Anonymous Thu 16 Sep 2021 20:45:55 No.13649816 Report

Quoted By:

>>13649430
>Begs for help on a Mongolian basket-weaving forum
>"Holy shit are you retarded?"
I definitely am, I just tried to help an asshole out. Good luck on your own.

Anonymous

Anonymous Thu 16 Sep 2021 20:59:52 No.13649852 Report

Quoted By:

>>13649643
>how to reduce it to a smaller area that is no longer normal
Subtract from it parts you don't want, which are also normal/coplanar. So it's easy to go from the normal solution to an abnormal.
>not conviced it becomes easier if its normal
Might not. It would at least be symmetric. Neater equations from zeros and right angles.

Anonymous

Anonymous Thu 16 Sep 2021 22:40:21 No.13650168 Report

Quoted By:

>>13649139
1)look up how to compute the solid angle/projected area of a triangle onto a sphere.
2)split your rectangle into two triangles.
3) profit

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