Spatial Distribution vs...

So I've been hanging around here for 4.5yrs and still don't know what I'm talking about! :duh2:
OK, taking the "Typical Spatial Distribution" graphs from Cree etc datasheets. Am I correct in interpreting it as, if you hold a light meter at a set distance directly in front of the LED (say an XM-L2) you'll get x reading, then if you go to y angle you'll get a corresponding % of the reading? eg. if you read "100" at 0degrees, at 30degrees you should read about "90", and at 70degrees you should read about "30"?
That's how I assume it works. Now comes the bits I'm less confident on. And if I got the above wrong, then I'm waaaaay off.

- say you're driving this XM-L2 at a current that would result in a theoretical 1000 lumens. Now put an aspheric lens in front, capturing a cone from -30degrees to +30 degrees (is this called a 30degree or a 60degree cone?). Theoretically, how much of the emitted lumens are available to the lens? ie., is there a direct correlation between the Spatial Distribution graph, with it's Relative Luminous Intensity %, and lumens? Do you maybe have to work out the volume under the curve, then see what % of volume this -30 to +30 slice contains? (hadn't thought of that until just this second)

- brightness-wise; If you were to isolate 10degree bands, and focus the output of that band into one spot - from the graph, you could say(?) that 0-10 is roughly an average of 100%. From 70-80 is roughly an average of say 25% (70 = ~30%, 80 = ~20%). If I were to focus just this 0-10 band, and compare it to the 70-80 band, would it be roughly 4x as bright? And if you add the 70-80 spot to the 0-10 spot, how bright is it now, 125% of 0-10?

I'll chime in on the first paragraph to get things started. I suppose you could use the data chart that way but I can't say as I have known anyone to do so. The "Typical Spacial Distribution" chart would be used to characterize the output pattern or output angle of an LED. If you take the peak of the chart as 100% then go to the 50% point on the curve that angle times two is the "full width, half max" of the LED. For the XM-L2 example that would be about 65 degrees X 2 = 130 degrees. Having a defined method for measuring the output angle is important so that everyone knows what to expect from a product.

Second paragraph: it's the total area (not volume but you're on the right track) that would give you an estimate of the output not including Fresnel loss, absorption, scatter, etc.. So for the XM-L2 the chart has about 30ish squares and for +/- 30 degrees you have about 14 so just under 50% or less than 500lm. As a guess....

You're definitely on the right track. But the math is a little more complex than that.

Sorry, that's being euphemistic. The math is actually a lot more complex than that.

First two paragraphs, dead on. If you read 100 at 0 degrees, you should read 90 at 30 degrees (if that's what the spacial distribution graph shows) and 30 at 70 degrees (repeat).

You are absolutely correct to think about an aspheric lens in front of an LED collecting only the light that's within the proper angle. However, some light is reflected off the flat surface of the back of the lens. At wider angles this increases. By the time you consider intensity vs. angle, reflection vs. angle, and interpolation/integration of illumination/steradian, the calculation is far beyond my capabilities, as well as those of most others. There are people who can do this calculation, but most of us prefer to estimate or guess, or make comparisons to the results of those who have gone to the trouble to make measurements or have taken on the enormous task of calculating the expected result.

Hi Harold, my main thoughts are on how to get a handle on how exactly reflectors / TIRs work with regards to capturing and redirecting the light; without having access to or knowledge of raytracing systems etc., I'm doing hand calculations to try to figure out how this bit of light coming out here can be redirected to this spot, and how bright it can roughly be expected to be...

Also thought it might be a handy way to get across ballpark ways of knowing what to expect OTF of a torch; if for example said XM-L is being driven at 3Amps, and it's got an aspheric only catching +/- 30degrees, then at the extreme most you can only possibly expect 500lm OTF, rather than the 1000lm at the emitter. And then take out some standard rough percentage for losses. So you can buy your "1600lm" ebay special safely guestimating that it'll be more like 400lm OTF, and be quite happy with that.

Essentially I'm talking about "back of napkin" workings that will get you within a very rough ballpark of what a scientifically calculated figure would, but may be "good enough" for the average hobbyist.

Harold_B said:

it's the total area (not volume)

Click to expand...

yep I knew what I meant
Had a passing thought and after some investigation I've found an "easy" method;
- take a screenshot of the graph, crop it to just the graph (no labels)
- select (freehand, bezier, etc) the curve and the area below it
- find your graphics program's Histogram; in The GIMP it's colors/info/histogram, in Photoshop it's window/histogram. This will tell you how many pixels in the current selection!! Note the figure down.
- deselect the portions you are not concerned with, eg. outside of +/- 30degrees, and again use the histogram to count the selected pixels.
- divide, and you have a proportional guestimate of the lumens living within the angle! :thumbsup:

For example, I just did the above as a selection, the rectangle of the graph had 148292px, the curve contained 96614px, and the +/-30 part of the curve contained 48134px. 48134 / 96614 = 0.4982. Pretty damn close to what Harold said...

Close, but you are missing the point that the graph is a cross-section of a three-dimensional phenomenon. A pixel at 30 degrees carries much more weight than one at 10 degrees.

You must integrate not the intensity times the width of the pixel, but the intensity times the width of the pixel times the circumference of the annular ring it represents.

See that's one of the things I haven't figured out, and there doesn't seem to be much in the way of laymans info on interpreting. I've contemplated it as a ring before but I couldn't nut it out. This is where the third part of my badly-written query eluded to, being, what happens when you gather all of the light from one "ring" and gather it together into a focused spot; will the "20% intensity ring" spot be 1/4 as bright as the "80% intensity ring", or as you say do you bring in circumference as a factor?
eg. say for the XM-L graph, if I measure the 80% diameter as 260 pixels, and the 20% diameter as 480 pixels. As circumference is pie x D, we can discard pie and just divide one by the other and say that the 20% ring is 1.85 times the circumference of the 80% ring. So if we focus these rings to spots, is the 20% ring spot in actuality 1/4 x 1.85 = 46.25% the brightness of the 80% ring spot?

Yeah I know, clear as mud... just trying to get some sort of idea of, for example, how various parts of the spatial distribution add into the spot produced by a reflector, why/how the reflected part is brighter than the spill when the spill is coming directly from the high-intensity parts of the LED, etc... etc...

Yes it a profile of a volume although I'm missing the point of weighting the data at various points on the curve. Without a model we're making a lot of assumptions so my thought is keep it simple. If you really like doing the math there's this somewhat older white paper to use as a guide: http://www.researchgate.net/profile..._LED_radiation/links/02e7e51cef65da5168000000

Imagine the emitter at the center of a sphere. You measure intensity of illumination on the sphere at different angles from the axis of the emitter (note this is not an integrating sphere, but is designed to absorb all light, not reflect it, so all the light hitting the sphere comes directly from the emitter).

If you took a 2-dimensional slice of this sphere through the axis of the emitter, and graphed the illuminance vs angle, you'd get the spacial distribution graph in the data sheet.

For argument's sake, let's say the sphere is 1 meter radius, and we measure exactly 100 candela (100 candlepower) illuminance on axis. Instead of a continuous integration (since differential equations was so long ago for me, and many of us never took it anyway), we'll use a piecewise approximation, using a 1 square cm pixel size. The pixel centered on axis has an illuminance of 100 candela, which is 100 lm/m^2. Since 1 cm^2 is equal to 1/10000 m^2, the number of lumens falling on this pixel is 100 lm/m^2 * 1/10000 m^2 = 0.01 lm.

Now let's skip a bunch and go out to 30 degrees off axis. There's a ring on the sphere that's all 30 degrees off axis. This ring is illuminated at 70 candela. 30 degrees is a great number to use because the radius of this ring is exactly half the radius of the sphere, or 0.5m. If we were to peel off a strip of the sphere 1 cm wide, centered on this ring, then flatten and straighten it (getting a bit deeper into approximation territory here, but not too bad), we'd have a strip 1 cm wide by 3.1415927 (pi) meters, or 314 cm, long.

The number of lumens falling on one pixel is 70 cm/m^2 * 1/10000 m^2 = 0.007 lm. But this area is 314 cm^2, so the lumens on the whole area at 30 degrees off axis is 70 lm/m^2 * 314/10000 m^2 = 2.2 lm.

That's what I meant when I said "A pixel at 30 degrees carries much more weight..." It's 30% less bright, but 220 times more lumens.

I'm not interested in doing the math to complete this calculation, but hopefully you get the idea.

To complicate matters, very little light gets reflected off the lens on axis. As you get farther off axis, more of the light reflects off the back of the lens and never enters the lens to get included in the output. At some point, ALL the light reflects off the lens. That's for flat (plano) back lenses. The transmittance (is that the right term?) vs angle curve of the lens must be multiplied by the radiance vs angle curve of the emitter. Of course the transmittance vs angle curve depends on the lens material, surface finish, and shape. AFAIK, TIR lenses are designed to have nearly 100% transmittance at all angles by having a spherical concave back surface.

Even if you can do the math, you still need to know a lot about the lens before you can even begin to make a decent approximation of its output characteristics when coupled with a particular emitter at a particular distance. I think that's why most of us DIY hacks depend more on tests than calculations. We lack the skills and/or the information to calculate decent approximations.

Thanks for the thorough and well thought out reply DIWdiver. Perhaps I've gotten lazy with so many tools at my disposal that I don't think these things through like I should. I'm just glad we're not translating between polar and Cartesian plots too. The paper I linked has a lot of the math you mention wanting to avoid.
i was thinking about this thread today while looking at the simulation results for an LED and reflector system. Perhaps I can take advantage of having the system modeled to illustrate RoGue_Streak's question about energy distribution and spill.

Thanks guys, sorry for the late reply, for some reason work has now blocked this thread as "Weapons" (happens a lot with CPF threads), though I can't figure out the logic that triggered the new classification...

Harold, yeah from what I saw of that white paper (I think I got the right one) it seemed a lot more mathy than I intended going into. I find it kinda fun to figure out refractions and reflections when light's coming into a medium at x degree and hitting a plane angled at y, but when things start going into inverse trig then I'm outta there

DIW, I think I follow, but I'll try tomorrow to see if I can come up with another angle as an another example (0degrees is a bit of an anomoly in the comparison) to see if I'm tracking right?

No, zero degrees isn't an anomaly, just the extreme end. If you were to compare 10 or 20 degreees to 30 degrees, you'd find the difference to be less dramatic, but not fundamentally different.

Sorry I meant with regards to the calcs and it just being a single "pixel" compared to a strip of them