>> The swamp I'm trying to drain, is how much energy is a 
>> given LED
>> painting on the scene at 30' (or other distance)

Do you have specified candella at beam centre and 
distribution patterns?
Or just gross lumens and distribution patterns?

If the former then it's very easy [tm].
And probably wrong :-).

Given what you said about guaranteed minimum and half power 
points.

- Select beam orientation with NARROWEST angle.
- Note angle for half power or power at desired angle.
- Candella are lumens/m^2 at 1m. so
If radius = 30 feet = 9.144m.
At 30 feet brightness will be down 1/9.144^2 over 1m =
1.196% of level at 1m.

Minimum guaranteed irradiance =

Cd x k_a x 0.01196

    k_a = 0.5 for half power point
              or whatever for selected angle in step 2 above


QED ???

if you have gross lumens and no other info then you will 
need to integrate the lumens with angle. 'Untidy' for a non 
rotationally symmetrical pattern. Easy enough with squared 
paper for a rotationally symmetric pattern.

We still really don't know exactly what you have available 
and what you are trying to achieve.
Does your Golden Dragon have a custom lens on it or is it an 
OTS unit? If the latter can you point us to a data sheet. If 
the former is the lens spec and LED spec available? Or the 
combined radiation pattern?


        Russell McMahon



        Russell



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