Inspired by a discussion from this thread, I present a method to estimate total output of a reflector-based light with a single intensity measurement at the edge of the spill.
The idea is simple: at the boundary of the spill, the light comes directly from the LED without any transformation by the reflector. Since a bare LED follows a Lambertian emission profile, a single intensity measurement at a known angle gives you enough information to calculate the emitter’s total output.
What you need:
- A wall 1m from the LED
- An estimate of the diametrical spill cone angle (2θ)
- A lux meter
What you do: [EDIT: this procedure will be replaced by one more robust to measurement errors, to be updated later today.]
- Position the light such that the LED is 1m away pointing directly at the wall.
- Place the phone just to the interior of the spill boundary, flat against the wall, and take an intensity measurement, say L lux.
Schematic:
The computation:
- The total OTF output of the bare emitter (post reflection losses from glass lens) is given by Ω = πL / (cos(θ))^4 lumens.
- For a more accurate result, perform the measurement at a farther distance of d meters, and the final output is Ω = πLd^2 / (cos(θ))^4 lumens.
This method does not yet account for losses at the reflector (which should be easily to estimate, I will update this post with the calculations later), but should already give a +/-10%-accurate output estimate, assuming a correct intensity measurement.
Quick proof sketch for the formula
At point (1), the illuminance is [1]=Ω/π lux; this is a standard result about Lambertian emitters, which follows by considering the sphere of diameter 1m in front of the emitter–this sphere’s surface has constant illuminance and area π.
Point (2) is also 1m away from the emitter, but offset by angle θ. Thus, a lux meter positioned at (2), facing the emitter, picks up [2]:=[1]*cos(θ) lux.
Point (4) is farther than point (2) by a factor of 1/cos(θ); for this reason, a lux meter positioned here picks up [3]=[2]*cos(θ)^2 lux.
By laying the lux meter flat against the wall at point (4), it is now tilted θ away from the emitter, and thus picks up an intensity of [3]*cos(θ) = (Ω/π)(cos(θ))^4 lux.


