Originally Posted by

**Walterk**
Surface brightness of the source; measure the candlepower-output of the bare source first, divided by the surface-area of source. The only way to do this without much uncertainty, is to do a lux-measurement on the bare source at one meter (with a calibrated lux-meter) and determain the source size, then divide the lux-measurement by the mm2 surface of the source.

There is indeed one other way to accurately estimate the axial intensity of a domeless LED, which is the preferred LED type for throw.

The peak axial intensity of the bare LED at 1 meter is simply the maximum lumen output divided by π divided by the area of the emitter. We divide by the area of the die to get surface brightness.

In the case of an Osram Oslon Flat Black tested at Taschenlampen Forums, cooled with a fan and copper heat sink within the operating temperature range of 25-35C, having a peak lumen output of 937 lm and a die area of 1.122 mm^{2}:

SB=L_{TOT}/π/s^{2}, where L_{TOT} is total lumens (usually the peak lumens but it can be whatever value you are designing around) and s^{2} is die area in mm^{2}

SB=937/π/1.122 mm^{2
}SB= 266 cd/mm^{2
}(Peak intensity is 298 cd)

And this result is consistent with the surface brightness of 260 cd/mm^{2 }that is the measured peak this LED. It is derived here from the data in the same link and experiment mentioned here.

**Proof:**

Approximating the domeless LED as a lambertian emmiter, the intensity is I(**θ**)=I(0)*cos **θ **from -90 to 90 degrees.

One can then multiply this intensity with an infinitesimal area element on the hemisphere of radius 1 meter to get lumens in that element. Summation of these elements within a boundary defined by the half angle of the beam determines total lumens in that given field. (Integration).

Long story short, the percentage of total lumens from the emitter in the boundary of the beam angle is sin^{2}**θ** where theta is the half angle of the beam.

The area of this sector is different, and at 1 meter radius, the area of the sector is given as 2 π * (1-cos **θ** ) square meters.

So the average intensity in a defined sector is

I_{AVG }(**θ**)= [L_{TOT} * sin^{2}**θ** ] / [ 2 π * (1-cos**θ**) ] ......(lumens / square meters)

Where L_{TOT} is total lumen output of the emitter across teh hemisphere.

To get the intensity of the axial vector at 1 meter (candella), you take the limit of the function as **θ** approaches zero. This is indeterminate at first due to dividing zero by zero when **θ**=0 is plugged in. So the numerator and denominator are independently solved for the corresponding derivatives until the ratio become determinant. (From this point on, I will solve for surface brightness, then correct to solve for axial intensity.)
SB=L

_{TOT}*(1/2

π)/s

^{2}*

_{lim θ--->0} [(2* sin

**θ***cos

**θ)** / sin

**θ] **(dividing sines here is a big mathematical no-no

)

SB=L_{TOT}*(1/π)/s^{2}* _{lim θ--->0} [(sin**θ***cos**θ**) / sin **θ**]

Again, you get an indeterminate ratio, so you take the derivatives *one more time*.

SB=L_{TOT}*(1/π)*_{ lim θ --->0} [(cos^{2}**θ** -sin^{2}**θ**) / cos **θ**]

Taking the limit of theta approaches zero, we enter **θ**=0, L_{TOT }=937 lm and die area =1.122 mm^{2}.
SB=937*(1/

π) [(1-0)/1]/1.122

**SB=266 cd/mm**^{2 }
(Peak intensity on the axial is SB * s

^{2 }= 298 cd)

One can easily check this result by placing a very small half angle (say 1 degree) into the formula and finding a value that is very close to the limit.

SB=L_{TOT}/s^{2}* lim **θ** --->0 [ sin^{2}**θ** ] / [ 2 π * (1-cos**θ**) ]

SB=937/1.122* [ sin^{2}**1**°] / [ 2 π * (1-cos**1**°) ]

SB=937/1.122* [ 3.04586e-4 ] / [ 9.56960e-4 ]

SB=266 cd/mm^{2} (rounded to three significant digits)

I(1°)=266 cd/mm^{2} * 1.122 mm^{2} = 298 cd

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