Formula for calculating throw using aspheric lens

[Ra]

Thanks, I didn't know what a cd was

I know what a cd is.... I'm listening to one now..!!

Sorry, couldn't resist...

 

[Dr.Jones]

@saabluster: Ah, I see.

Hm, somehow we seem to agree now.

It had a few rough words, but was an otherwise interesting discussion.

I'll hit the bed now... Have fun

 

[bshanahan14rulz]

So a pre-collimator increases the final image size, while keeping the lux the same?

So would it also cut down on the divergence of the beam then? the image is larger, would the divergence be smaller?

 

[Walterk]

I put some theory to the test about beam diameter, and look for verification.
I did a maseurement using a XRE and a 66mmm lens, and measured the spot. Set it out in Cad, with the advantage of zooming in- and out without loosing resolution.
It appears to my findings that the spotsize is plainly resulting from the absolute diesize and the distance of the plane side of the aspheric lens:

- Is it correct that the line from the widest part of the source (here 1mm) through the center of the plane of the lens forms the half beam cone ? (wouldn't that be easy)
- Is it correct that: beam diameter = source size / focal length ?
- Using the apparent die size for calculating candlepower has to do with units and light, and not the geometric ray-path ? (just throwing in interesting words here)
- The dome makes that the theoretical focus length is different form the empirical witnessed focal length ?

 

[Dr.Jones]

So a pre-collimator increases the final image size, while keeping the lux the same?

Basically yes. Minus some additional losses, and maybe plus a small gain from reduced aberration.

So would it also cut down on the divergence of the beam then? the image is larger, would the divergence be smaller?

No. A bigger final image size means a wider beam angle and thus a bigger beam divergence.

It appears to my findings that the spotsize is plainly resulting from the absolute diesize and the distance of the plane side of the aspheric lens:

- Is it correct that the line form the widest part of the source through the center of the plane of the lens forms the half beam cone ? (wouldn't that be easy)
- Using the apparent die size for calculating candlepower has to do with units and light, and not the geometric ray-path ? (just throwing in intersting words here)
- The dome makes that the theoretical focus length is different form the empirical witnessed focal length ?

I'll start from the end:

- The dome creates a virtual image of the actual die. This virtual light source can be seen as the effective light source for all following optics. This virtual light source is magnified and sits a bit behind the real light source (in your picture: below).
- Thus in your drawing, you should use the apparent die size.
- The line from the die edges of the virtual light source through the lens center forms indeed the half beam cone (effective at larger distances). However, the lens center is not at the plane side, but somewhere in the middle of the lens.
- Thus it should be more like "the spot size is plainly resulting from the apparent die size and the distance to the effective center of the aspheric lens."

 

[bigterk]

After reading this thread I have determined that I need 2 things, opiates and some pie. :shrug:

 

[Genzod]

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 mm2​:

SB=LTOT​/π/s2​, where LTOT​ is total lumens (usually the peak lumens but it can be whatever value you are designing around) and s2​ is die area in mm2​

SB=937/π/1.122 mm2

​SB= 266 cd/mm2

​(Peak intensity is 298 cd)​

And this result is consistent with the surface brightness of 260 cd/mm2 ​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 emitter, 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 sin2​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

IAVG ​(θ)= [LTOT​ * sin2​θ ] / [ 2 π * (1-cosθ) ] ......(lumens / square meters)

Where LTOT​ is total lumen output of the emitter across teh hemisphere.​

To get the intensity of the axial vector at 1 meter (cd), 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=LTOT​*(1/2π)/s2​* lim θ--->0​ [(2* sinθ*cosθ) / sin θ] (dividing sines here is a big mathematical no-no

)​

SB=LTOT​*(1/π)/s2​* lim θ--->0​ [(sinθ*cosθ) / sin θ]​

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

SB=LTOT​*(1/π)* lim θ --->0​ [(cos2​θ -sin2​θ) / cos θ]​

Taking the limit of theta approaches zero, we enter θ=0, LTOT ​=937 lm and die area =1.122 mm2​.

SB=937*(1/π) [(1-0)/1]/1.122

SB=266 cd/mm2 ​

(Peak intensity on the axial is SB * s2 ​= 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=LTOT​/s2​* lim θ --->0 [ sin2​θ ] / [ 2 π * (1-cosθ) ]

SB=937/1.122* [ sin2​1°] / [ 2 π * (1-cos1°) ]​

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

SB=266 cd/mm2​ (rounded to three significant digits)

I(1°)=266 cd/mm2​ * 1.122 mm2​ = 298 cd​