I found myself routinely repeating the same calculations trying to quantify the performance of throwers, so I decided to condense it all into a more convenient formula to save myself some work in the future.
Suppose you have a light that emits a beam of uniform** intensity; let
- L denote the total output (lumens),
- θ the beam’s full divergence angle (degrees), and
- C the intensity of the beam (candelas).
Then, the relation
C θ^2 = 4180 L
holds with accuracy within ±0.3%, for beam angles θ ≤ 20 degrees (which is not a restrictive condition at all for throwers).
Sketch of derivation
- Express the total flux as L = CΩ where Ω is solid angle subtended by the beam
- Solid angle is the area of a unit-spherical cap subtending a diametrical angle of θ, which is 2π(1-cos(θπ/360)) for θ in degrees
- Simplify using the small-angle approximation 1-cos(x) ≈ x^2/2
In the case that the beam is not uniform in brightness, the above formula, with C re-interpreted as peak intensity, and θ interpreted as widest angle spanned by the beam, turns into
C θ^2 ≥ 4180 L, and gives you
- A lower bound for peak intensity, in terms of output and beam angle;
- A lower bound for beam angle, in terms of output and peak intensity;
- An upper bound for output, in terms of peak intensity and beam angle.
Example: the Acebeam W35 has been measured with 733lm of output and 1.89Mcd of intensity. By the above formula, the beam divergence angle is at least \sqrt{4180*733/1,890,000} = 1.27 degrees, which is about 2.5 times the width of the moon’s disk.
** Uniform meaning that a very large sphere centered at the emitter receives constant illuminance (lux) on the part of the surface covered by the beam.
1 Thank
I think that if you use the hotspot cutoff of p ≈ 0.37×Max (if you keep it Gaussian with s = 2) in the mini-calc at the bottom of this page you will get the same values for beam spread.
The Map uses p = 0.5 for hotspot size comparison.
There is a good reason for this: the intensity cutoff 0.37*max corresponds to a radial/angular cutoff of \sqrt{-2\ln(0.37)} ≈ 1.41σ, where σ is the standard deviation of the Gaussian. This is very close to \sqrt{2}σ. It turns out that a uniform beam of this radius contains exactly the same total output as a Gaussian beam of standard deviation σ: a cylinder of radius \sqrt{2}σ and height 1/(2πσ^2) (max value of the 2D Gaussian density) contains volume 1, the same volume enclosed under any probability density function.
A similar constant (not necessarily equal to 0.37) exists for pretty much any finite-variance distribution you choose–the Gaussian is not special in this regard.
One thing that is not immediately clear to me, however, is how to interpret the angular distribution as Gaussian, because the Gaussian is supported on an unbounded domain while the angular domain (the surface of a (hemi)sphere) is very much bounded. Perhaps the distribution you really want is the planar Gaussian distribution mapped onto the sphere by rays from the sphere’s center? That is, a Gaussian illuminance profile on an infinite white wall, mapped back to angle-space.
We’ve discussed in the past. Seems to work as intended.
I remember asking why the Gaussian profile is chosen, but don’t remember receiving a response. Could you explain again?
I remember having a couple concerns about the Gaussian model, and would appreciate hearing your take. I made a comment in the other thread to continue the discussion, because it seems more appropriate for the subject.