[PART 1] Official BLF GT Group Buy thread. Group buy officially closed! Lights shipping.

Without minimizing or diminishing anything you guys are doing, all of this is theoretical. Who is it that has the signature: “Theory sounds like a nice place, I’d like to go there one day, I hear everything works there.”?

Thank you! :smiley:

It is almost exactly that, but If I had used cos then it would have overestimated the intensity a lot, the curve I overlayed is a lot closer except for the tiny bit at 80-90 degrees. It is basically just cos but a bit skinnier :slight_smile:

Yes, just need to integrate it by spherical coords and we should be able to determine what % of total light is being captured and reflected.

That’s pretty much what I just added to the graph yesterday :stuck_out_tongue: except instead of polar it is just a regular graph with degrees on the X axis.
Although I think this may not be necessary if I use spherical coordinates to calculate the intensity…so it’s just an intermediate step.

Pity all the math is beyond me, but maybe this is helpful to visualise things futher:

The question being, is it worthwhile to have more than 93mm reflector height?
Or does it cost the beloved focal distance at the base too much?

I belive we would go for 112mm at focus point.

Why do you want to integrate? As far as I understood it the diagram in the cree datasheet is already accounting for that.

Good job on that pic, that will help a lot of people understand what’s going on here :slight_smile:
The question isn’t really “is it worthwhile” rather, “at what point does it stop being worthwhile”
Since the more you increase the depth, the less difference in makes :confused:

Because that is only a horizontal slice…?

If you have intensity 50%, at 45 degrees, for a unit sphere you have 2*pi*r circumference which is 2*pi*cos(45)
You then multiply that by 50 to get 222

If you have intensity 90% at 10 degrees, for a unit sphere you have 2*pi*cos(80) , remember that 10 degrees from vertical is 80 from horizontal, and multiply that by 90 to get 98

So as you can see the total amount of light coming out of the LED at 10 degrees is less than half of what you get at 45 degrees.

If you look at this image, you can see the blue circle circumference changes depending on the angle of R.

Yeah, but you can integrate and then plot it on a 2D graph again. I thought that’s what cree has done, but I guess not. I have to admit I’ve never looked at this part of the datasheet before.

[quote=Enderman]

But the focal distance at the base will decrease as the height increases.
This is not what we want to happen.
It’s a 14% decrease when you go from 93mm to 112mm height (while keeping the width 120mm).

In addition, here is why I need to integrate over the surface area of a sphere:

So the graph of intensity from cree is measured at different angles, all with the same radius, like this:

Except instead of those numbers, it goes from –90 to +90 with 0 being directly above the LED.

Looking at the sphere at distance R from the LED, we see this:

Point A has 100% intensity, point B has~80% intensity, point D has 0% intensity, etc……
So if we integrate the function of intensity over the surface area of a sphere we get the area * intensity which will be the total amount of light coming out of the LED. This number will not mean anything in real life, because cree decided to only state the %, BUT you can use this formula to compare two different reflectors that collect light with different angles of R:

If we take the angle of R, you can calculate the total light of the half sphere, subtract the total light of the cap, and you get the total light that will be reflected.

Again, this will only let us compare between different angles, and say “this reflector collects 32% more light than this other one”, we cannot know how many lumens will actually be collected.

Realistic way of thinking about it:
Lux is the measure of intensity, which is 1 lumen per square meter. Then you have a light that emits in a half circle, with 1 lux measured at every point on that half circle.
Take the surface area of the half circle, which is 4*pi*r^2 (assume unit sphere) and you get ~ 12.6 square meters.
12.6 square meters * 1 lux (lumens per quare meter) and you get a total of 12.6 lumens! :smiley:

Cree’s graph goes from –90 to 90 degrees and has 100% intensity at the center.
If you imagine the light emitted around an LED, the amount of area at 0 degrees is 0, so the total lumens should be 0 there.
Intensity is 100%, but area is 0, so lumens is 0.

Yes, these calculations are only to calculate the total % of light that will be collected, aka lumens, except measured in percent.
The calculations about dispersion will need to be calculated after this.
This is the first steep of the process to calculating the throw.

First we see how much light is being collected, then we can calculate the dispersion of the total light (by integrating over a parabola) and we can find out how intense the light will be at a certain distance.

One step at a time!

Now this is what I call a BLF design thread.

I probably sound like a broken record, but our friend, I=LA, is all that’s needed to predict the maximum intensity of the beam. I is the luminous intensity (cd) of the beam, L is the luminance (cd/mm^2) of the LED, and A is the frontal area of the reflector. This is the max possible throw of the reflector; when the LED is properly focused in the reflector.

It seems like the calculation you are proposing would give you the beam profile, which would be nice to know.

You can imagine it as a hemisphere over the LED full of holes.
At around 0° there is just one hole, but it will let through the highest intensity light.
At 45° you have a circle of many holes, that each lets through less intense light, but there are many more holes so it adds up to much more than the light from the single top hole.
Going towards 90° the holes are not that many more in number as the increase from 1 hole to the many holes at 45°, plus the LED loses more intensity in that direction, so the sum of light around 90° is comparable to the light at 0°.
The sweet spot seems to be very close to 45°.

Hasn’t time for family dinner…but I’ll think I’m divorced if I dont go…

Catch you guys later…
Hahaha :innocent:

This will be the last time you see me here in this thread.
Al those expensive words, hard to understand graphs and those equations, are giving me a headache… :rage:

Kidding of course, haha :smiley: :smiley:
Keep it going guys! :beer:

Maybe that would be all that was needed to predict throw if all of the following were possible:

  • The reflective surface was 100% true (no imperfections).
  • The reflective surface was 100% reflective of all frequencies of light emitted from the LED. This would require both the right materials and proper execution in the application of the finish.
  • The LED was a true point source of light.
  • No light reflected back by the lens or lost in the lens that is in front of the reflector.

EDIT: Sorry for the multiple edits. Kept missing typo's.

You are right about the non-perfect reflectivity. There is typically 10-20% loss from the reflector and glass. A couple approximate examples of LED luminances: dedomed XPG2 S4 2B has a luminance of ~180cd/mm^2 (at typical direct drive current ~4.4A) after reflector losses and a typical XPL HI has a luminance of ~110cd/mm^2 (at typical direct drive current ~5.5A) after reflector losses. In an EE X6, for example, with reflector active area ~630mm^2, the dedomed XPG2 does ~113Kcd, and a typical XPL HI does ~69Kcd.

The LED doesn’t have to be a point source; in fact I don’t think the concept of luminance makes sense for a point source of light.

Is there a mathematical proof for that?
Or is this just empirical evidence?
I have trouble believing that you can just calculate the area of the front reflector and have that tell you the throw…