A little geometry problem

[Howard]

Basically, let's say I have a flashlight that projects a cone of light whose angle is 90 degrees (included, i.e from side to side) onto a hemispherical wall. The vertex of the cone (the position of the light source) and the center of the hemisphere are at the same point.

What is the ratio of the area covered by the light to the area of the wall?

 

[TheoPetro]

wouldnt it depend on how far away it was?

 

[Mo0o]

are you looking for an equation? since the exact ratio woudl change depending on how far you are from the wall

 

[dullard]

Originally posted by: TheoPetro
wouldnt it depend on how far away it was?

Originally posted by: Mo0o
are you looking for an equation? since the exact ratio woudl change depending on how far you are from the wall

He says clearly how far away it is in the OP.

 

[BrownTown]

Maybe not the way its meant to be done, but I would just integrate the concentric circles coming up from the bottom to the 90 degree mark. Taht probably doesn't make sense to say it in words, but I think it would werk.

 

[Mo0o]

Originally posted by: dullard

Originally posted by: TheoPetro
wouldnt it depend on how far away it was?

Originally posted by: Mo0o
are you looking for an equation? since the exact ratio woudl change depending on how far you are from the wall

He says clearly how far away it is in the OP.

oh i thought he meant the vertex of the light cone is aligned with the center of the hemisphere.

 

[dullard]

To calculate the answer, look at this and this.

r-h = r*cos(45°) = r* 2^0.5/2

h = r*(2-2^0.5)/2

Acap = 2*pi*r*h = pi*r^2*(2-2^0.5)

Ahemisphere = 2*pi*r^2

Acap/Ahemisphere = (2-2^0.5)/2 = 29.29%

Unless I did a stupid error. Note: I assume the light is all hitting the wall (you could be shining the light on the edge of the hemisphere and some of it goes outside the hemisphere). Otherwise, this problem is unanserable without more information.

 

[TheoPetro]

Originally posted by: Mo0o

Originally posted by: dullard

Originally posted by: TheoPetro
wouldnt it depend on how far away it was?

Originally posted by: Mo0o
are you looking for an equation? since the exact ratio woudl change depending on how far you are from the wall

He says clearly how far away it is in the OP.

oh i thought he meant the vertex of the light cone is aligned with the center of the hemisphere.

thats how I read it too

 

[Howard]

Originally posted by: dullard
To calculate the answer, look at this and this.

r-h = r*cos(45°) = r* 2^0.5/2

h = r*(2-2^0.5)/2

Acap = 2*pi*r*h = pi*r^2*(2-2^0.5)

Ahemisphere = 2*pi*r^2

Acap/Ahemisphere = (2-2^0.5)/2 = 29.29%

Unless I did a stupid error. Note: I assume the light is all hitting the wall (you could be shining the light on the edge of the hemisphere and some of it goes outside the hemisphere). Otherwise, this problem is unanserable without more information.

Sorry, the cone would be aimed at the center of the hemisphere, so there would be no light loss.

Thanks for the answer, sounds pretty reasonable.