Integrating candela values from IES files

1

(Not exactly Radiance, but I have nowhere else to ask.)

I’ve got fixtures for which the manufacturer refuses to divulge photometric data. It is, unfortunately, a poor country, and the vendor is the only one I can find, so I’m trying to generate one based on data about fixture geometry and output. I have then run into a second problem: I seem unable to correctly go from an array of proportionate candela values to a valid IES file; my attempts at integrating the candela values in existing IES files lead to total lumen outputs of about 1.7-1.8 times the correct value. I suspect I am missing a factor of √π somewhere. My procedure (in Julia) is the following. Anyone have any thoughts on what is going wrong here?

cv = pho.candelavalues
va = pho.verticalangles
ha = pho.horizontalangles
rows,cols = size(cv)
total = 0
for i = 1:rows-1
    for j = 1:cols-1
        Iᵥ  = sum(cv[i,j] + cv[i+1,j] + cv[i,j+1] + cv[i+1,j+1]) / 4
        Ω = 4 * asin(
            sind((ha[i+1]-ha[i])/2) * 
            sind((va[j+1]-va[j])/2))
        total += Iᵥ * Ω
    end
end
total
2

Hi Randolph,

this looks familiar, and I think I once ran down that rabbit hole too. :wink:

The solid angle calculation is more involved, as it requires calculating the spherical excess of the quadrilateral spanned by the cv array entries:

Ω = ∑ α_i - 2π

where α_i is the angle (in radians) between the two adjacent planes passing through the hemisphere’s origin and the i-th and ((i+1) mod 4)-th quadrilateral edge. There’s a nice illustration here for a spherical triangle: Spherical trigonometry - Wikipedia

Incidentally, evalglare does the same calculation under the hood for pixels.

Hope this helps. Let me know if you need hints calculating the plane normals. I wrote this stuff in Python a few years back.

–Roland

1 Like
3

Ouch! Thanks. Yes, I certainly got that wrong. To be continued…

BTW, I have put in a request to have equation support added to the Radiance Discourse server. This is one of those “Duh” things that seems to have been overlooked.

4

Hi Randolph,

no worries. Been there, done that. Of course I’m not ruling out other issues with the calculation, but the spherical excess is a common issue here.

I do agree equation support would be very helpful here. Punching in Unicodes is no fun. Had to revert to TeX notation with the subscripts.

–Roland

1 Like
5

Lo! I have found it!

Turns out that the horizontal angles in type C photometry describe great circles - lines of longitude, but the vertical angles describe circles in planes perpendicular to the axis of horizontal angles, which are not great circles - lines of latitude. With that in hand, and a formula found in Wikipedia’s “Solid Angles” article, this code produces a total approximating the total lumen output of the luminaire.

Thank you for the advice.

cv = pho.candelavalues
va = pho.verticalangles
ha = pho.horizontalangles
rows,cols = size(cv)
total = 0
for i = 1:rows-1
    for j = 1:cols-1
        Iᵥ  = (cv[i,j] + cv[i+1,j] + cv[i,j+1] + cv[i+1,j+1]) / 4
        Ω = (sind(va[j+1]-90) - sind(va[j]-90)) * (deg2rad(ha[i+1]) - deg2rad(ha[i]))
        total += Iᵥ * Ω
    end
end
total