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rumble

Questions about radiance

9 posts in this topic

Radiance is: L = dPhi/(dw*cos(theta)*dA) Phi = flux w = solid angle theta = angle of incident with surface normal A = area 1) For the cos(theta) term. As theta increases, cos(theta) decreases, thus L increases. Does this imply radiance at a grazing angle is larger than radiance aligned with the surface normal? And at theta = PI/2, L blows up? 2) I am still unsure how to relate the differential solid angle and differential area in the same formula. I can accept that dPhi/dw is the radiant intensity. But further divide it by the projected differential area I can not visualize as easily. Thanks
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> 1) For the cos(theta) term. As theta increases, cos(theta) decreases,

Yes.

> thus L increases.

No. It is divided by the cosine term, so as cos(theta) decreases, L increases.

> Does this imply radiance at a grazing angle is larger than radiance aligned with the surface normal?

The opposite - I hope this helps you visualize it better.
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Quote:
Original post by crowley9
> 1) For the cos(theta) term. As theta increases, cos(theta) decreases,

Yes.

> thus L increases.

No. It is divided by the cosine term, so as cos(theta) decreases, L increases.


I think that's the same as my post, that L increases as theta increases.

Quote:

> Does this imply radiance at a grazing angle is larger than radiance aligned with the surface normal?

The opposite - I hope this helps you visualize it better.

[/quote]

Well, that's not what the formula for radiance predicts? Like we both mentioned, since the cosine term is being divided, L increases as theta increases to approach a grazing angle.

Hence my confusion.
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Hi!

Quote:
Original post by rumble
1) For the cos(theta) term. As theta increases, cos(theta) decreases, thus L increases. Does this imply radiance at a grazing angle is larger than radiance aligned with the surface normal? And at theta = PI/2, L blows up?


That's right. Radiance gives the transmitted radiant power through the given solid angle per unit of projected area. That is why you need to divide by the projected area given by cos(theta)*dA. At a grazing angle the projected area approaches zero, hence the radiant power per unit of projected area approaches infinity.

Quote:

2) I am still unsure how to relate the differential solid angle and differential area in the same formula. I can accept that dPhi/dw is the radiant intensity. But further divide it by the projected differential area I can not visualize as easily.


The reason why you need to divide by the projected differential area is because it's size can vary for the same solid angle based on distance. Once again, you want to get transmitted radiant power through the given solid angle per unit of projected area.

Hope that made some sense :)
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Quote:

That's right. Radiance gives the transmitted radiant power through the given solid angle per unit of projected area. That is why you need to divide by the projected area given by cos(theta)*dA. At a grazing angle the projected area approaches zero, hence the radiant power per unit of projected area approaches infinity.


Thanks that makes sense. In practice however, the radiance sampled from the scene/environment at grazing angles are not significantly larger? I.e., the sampled texel data from a texCUBE() would not become higher even if treated as radiance, regardless of angle with surface normal.

Quote:

The reason why you need to divide by the projected differential area is because it's size can vary for the same solid angle based on [bold]distance[/bold].


Instead of distance, I think you mean the orientation of the solid angle relative to the surface normal? The projected area should be independent of distance.
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Quote:
Original post by rumble
1) For the cos(theta) term. As theta increases, cos(theta) decreases, thus L increases. Does this imply radiance at a grazing angle is larger than radiance aligned with the surface normal? And at theta = PI/2, L blows up?


Imagine this. You grab a cardboard paper towel tube and look straight down at a book. You can see a very small circular area of the book. Power is being emitted from the book through the tube and into your eye. Lets say you moved close enough to the book so the area you see perhaps is just one letter of the books title name. That one letter is emitting power into your eye. Now you should be able to tilt the book so instead of just seeing the one letter you can see the whole book cover. Now that you can see the whole book cover more power is coming through the tube into your eye.

I believe that is how you are suppose to interpret it.

I found this link http://omlc.ogi.edu/classroom/ece532/class1/radiance.html

-= Dave
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Quote:
Original post by rumble
Well, that's not what the formula for radiance predicts? Like we both mentioned, since the cosine term is being divided, L increases as theta increases to approach a grazing angle.

Hence my confusion.


Urgh, I read your post incorrectly, and thought you said the opposite. It was very late at night :-)
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This is much easier to understand if you read the formula backwards:

dPhi = L * cos(theta) * dw * dA

So we have a small patch of a surface, radiance L and area dA. We want to calculate the flux dPhi passing through a small hoop of solid angle dw, at an angle theta to the normal.

Well obviously if we increase the radiance of the patch, the flux will go up linearly. That accounts for L. Also a larger emitting area will have a higher, hence the dA. Also if we increase our hoop's solid angle we will catch more flux, so that accounts for dw.

All that's left is the cos(theta) term. But it's not hard to see that a patch with constant radiance has lower flux at grazing angles. That's just because from the point of view of the hoop, the patch appears to have area dA * cos(theta) instead of dA.

Hope that helps.
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Thanks all with your varied interpretations.

I was thinking about the sampling of radiance in the scene and how come they are not larger at grazing angles. Perhaps the cosine term in the integrand is ommitted because it cancels with the cosine term from the incoming radiance?

And David Neubelt, how did you find that link. I searched 'radiance', 'radiance optics', and similar terms but this nice page does not show up in the first and many pages of the results.
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Quote:
Original post by rumbleI was thinking about the sampling of radiance in the scene and how come they are not larger at grazing angles. Perhaps the cosine term in the integrand is ommitted because it cancels with the cosine term from the incoming radiance?


In computer graphics when you use an environment map it represents the radiance. To work with light equations you want to convert it to irradiance. The relationship between irradiance and radiance is this

E = Integral [L * cos(theta)] dOmega (where E is irradiance and L is radiance)

That says the hemisphere irradiance is the sum of the radiance from all directions multiplied by cos(theta). This pretty much cancels out the cos term on the bottom. The cos(theta) is a specific example of a transfer function. Different transfer functions map to different material types.

Lighting equations take irradiance and convert it to radiant exitance (or radiosity). With Lambertian surfaces the radiosity is just the irradiance multiplied by an albedo term rho. Since your environment map is in terms of radiance you want to first convert it to irradiance via cos(theta).

So to sum up,

Environment map = radiance
Irradiance (needed for lighting equations) at a specific angle = radiance * transfer function
Radiant exitance (out going light) = rho * Irradiance

So usually you just sample your environment (radiance) and multiply it by N dot L * color map.

Quote:
And David Neubelt, how did you find that link. I searched 'radiance', 'radiance optics', and similar terms but this nice page does not show up in the first and many pages of the results.
I searched for radiant intensity

More reading
http://www.helios32.com/Measuring%20Light.pdf
http://graphics.stanford.edu/papers/invlamb/josa.pdf
http://graphics.stanford.edu/papers/envmap/envmap.pdf


-= Dave
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