Off-angle irradiance from Radiant Intensity

[SunThief]

For a light source with a given radiant intensity, I'm confused about the calculation of irradiance at off-angles. I've uploaded a drawing that shows a few spots and their location relative to the source, along with what I understand to be standard calculations. Then at the bottom of the drawing, I pose two alternate means for obtaining the irradiance from the perspective of a surface. Hopefully one of the two is correct. Would someone take a look for me?

 

[Andy Resnick]

For a light source with a given radiant intensity, I'm confused about the calculation of irradiance at off-angles. I've uploaded a drawing that shows a few spots and their location relative to the source, along with what I understand to be standard calculations. Then at the bottom of the drawing, I pose two alternate means for obtaining the irradiance from the perspective of a surface. Hopefully one of the two is correct. Would someone take a look for me?

I'm not sure I understand your figure: for example, is the source a point isotropic source, or does the source emit an intensity I(α)? If the emitted intensity depends on an angle, what is the reference direction? If the source is isotropic, the irradiance at any point A, B or C is simply I/r^2 ,and the incidance is I cos(θ)/r^2 where θ is the angle between surface normal and direction of the source- so for point 'A', θ=0 since it's not associated with a surface. Point 'D' is in the interior of the object?

 

[SunThief]

I'm not sure I understand your figure: for example, is the source a point isotropic source, or does the source emit an intensity I(α)? If the emitted intensity depends on an angle, what is the reference direction?

No, it's not isotropic--it's an LED. I meant I(α) to indicate that the radiant intensity was a function of angle. I was provided a radiant intensity graph that displays (W/sr) as a function of angle. I simplified the curve and turned it into an equation. Sorry I wasn't more clear, I meant the orange line to indicate the reference (zero angle) direction from the LED to the perpendicular lines CA and DB.

Point 'D' is in the interior of the object?

Pretend it's inside a transparent box. I've been struggling with this stuff for a bit, I drew this diagram to try to view points that are common to different scenarios, to help isolate my main difficulties. The first four calculations were meant to solidify what I think is supposed to be basic:

• Points A and B are along the zero-angle axis from the LED, so the angles for both the radiant intensity equation and the cosine argument are zero.

• Point C is at an angle to the LED on the line where point A is directly underneath, and point D is at an angle to the LED on the line where point B is directly underneath. Both of these have radiant intensities that are functions of α. Likewise, their cosine arguments are also α. Although these points are at an angle to the LED, they are each on their respective lines coming from the perpendicular. So, the irradiances at points C and D are referenced in the same direction as those at points A and B (perpendicular to their lines).

Where I am at a loss is when one tries to determine the irradiance at point C, but from a different standpoint. My last 2 equations were two different attempts to calculate the component of irradiance at point C, but perpendicular to the box instead of relative to the LED axis.

• The first attempt simply takes the irradiance already calculated (referenced to the axis), and multiplies it by the cosine of the angle to the box perpendicular.

• The second attempt takes the original equation, using angle α for the radiant intensity argument, and angle β for the cosine argument.

The latter approach was inspired by a piece I read on foreshortening that talked about using the angle between the surface normal and the line back to the source for the cosine argument. I may have misunderstood the context for that approach.

I appreciate your feedback, thanks.

 

[SunThief]

Ok, I clarified and streamlined the figure. I'm looking to verify the irradiance calculation referenced to the flat surface instead of the Light's axis. Would someone take a peak?

 

[SunThief]

Follow-up... this is the crux of my overall confusion:

For calculating irradiance, everyting I read instructs me to use the angle between the line of sight (from the light source) and the surface normal. In point P of the attached figure, this corresponds to angle α. The calculation returns an irradiance value that is referenced perpendicular to the surface. That is, it "points up" at the same angle as for the irradiance directly underneath the light source (i.e. along D1). This makes sense to me.

But if I consider point P as part of the tabletop (extending to the left), the "surface" is different now. And this is where I'm confused: the angle between the line of sight and the surface normal now becomes (α + β). If I use this combination angle as the argument for the cosine in an irradiance calculation, I get one result. However, if I simply multiply the result of the original irradiance calculation by the cosine of the angle between its perpendicular and the surface normal (i.e. cos(β)), I get a different result. This is not surprising, because cos(α)*cos(β) is not the same as cos(α + β). Which is correct?

 

[Andy Resnick]

Follow-up... this is the crux of my overall confusion:

But if I consider point P as part of the tabletop (extending to the left), the "surface" is different now. And this is where I'm confused: the angle between the line of sight and the surface normal now becomes (α + β). If I use this combination angle as the argument for the cosine in an irradiance calculation, I get one result. However, if I simply multiply the result of the original irradiance calculation by the cosine of the angle between its perpendicular and the surface normal (i.e. cos(β)), I get a different result. This is not surprising, because cos(α)*cos(β) is not the same as cos(α + β). Which is correct?

I think you are getting closer. First, I(α) = mα+b is an odd expression and may not be physical. More typically, I(α) = I₀cos²(α) or I₀cos⁴(α). In any case, let's start with the fundamental equation of radiative transfer between two surfaces:

dΦ=L/r² dA₁cosθ₁dA₂cosθ₂

where A is the area of source and surface, r the distance between surfaces, θ the angle between surface normal and line of sight between the two, L the radiance (units W/m²*sr) from one surface to the other, and Φ the radiant flux (units W). Now, for a point source, you don't have radiance, you have intensity (units W/sr), the units still work out because of the dA₁cosθ₁ term in the numerator, so for your point source you have:

dΦ=I/r²dAcosθ

You state I = mα+b, so

dΦ=(mα+b)/r²dAcos(α+β)
and thus the incidance is

E = dΦ/dA = (mα+b)/r² cos(α+β)Does that help?

 

[SunThief]

First, I(α) = mα+b is an odd expression and may not be physical.

[I just uploaded the graph from which I made the (simplified) equation. I know it's not precisely linear... but for now I assume both halves are symmetrical, and I "integrate" one side at a time.]

E = dΦ/dA = (mα+b)/r2 cos(α+β)

So then it is the original equation with the combined angle. I was hoping it was the other one. When converting the sun's Direct Normal irradiance to Beam Horizontal irradiance, you just multiply the Direct irradiance by the sine of the elevation angle (or divide by the cosine of its complement).

I have a lot of difficulty with the language that they use in the radiometry references--I have trouble visualizing the ideas. I did buy that Wolfe book, but it seemed more suited as a review handbook than as an introductory text. Would you recommend any other sources that are more basic? I've seen the Hecht book referenced, but I've been hesitant to get it without some idea about it.

Thank you so much for your help!

 

[SunThief]

Does that help?

I knew I was having tunnel vision about something basic, so I feel pretty silly.

With the sun, Direct Normal Irradiance (DNI) is independent of the surface on which it falls--it doesn't change. So for example, the sun might exhibit a DNI value of 800 W/m2 at some angle. That value does not depend on whether the recipient* of the light is a wall or a floor, etc. So while radiometry texts define irradiance (or incidence) relative to a surface, in my context it seemed more an attribute of the light source itself. And while I'm trying to incorporate the angular anomalies of the LED light over a surface, I want to ultimately maintain representative DNI values for that light.

Anyway, I think I got it now, thank you! You helped clear the mud away.

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*Implicitly, it strikes an imaginary surface perpendicular to the line of sight. The real surface of course matters when transposing the light.