OSRAM CSLNM1.TG 1mm, CSLPM1.TG 2mm, CULNM1.TG 1mm

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EasyB
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LouieAtienza, I think one concept you are missing is that of luminance. This is a measure of the surface brightness (of the LED) and it is important in the determination of beam intensity or throw (intensity in cd is equal to luminance times apparent reflector area). The luminance of LEDs does not vary with angle; the reason the intensity goes down as you go to the side of the LED is because the apparent area decreases. See here for some explanation.

http://budgetlightforum.com/node/15818

Enderman
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LouieAtienza wrote:

Well, the lux meter can be 30deg. to either side of the central axis of the LED and still read 90% luminous intensity of dead center. I found a calculator online and found the surface area of a parabolic reflector for a front-facing LED whose diameter is the approximately the same as the distance from opening to focal point about four times that of the surface area of the opening. But even though the surface area is four times as large, the light intensity I would estimate to be about 2/3 that of the light emitted from a 30deg. to 0. And the efficacy of an electroformed parabolic reflector surface with rhodium coating is around 90%; I’m sure the cast, machined, and polished/coated surface of say a BLF GT to be slightly less. So maybe the ratio of light from the reflector versus light directly from 0-30deg. is more like 2:1 rather than 3:1.

But the reason for using these emitters is to effect a longer throw, not to put the light on an integrating sphere to get a lux reading. Your very own OptoFire uses that 30deg. half angle since that’s the aperture size of the RLT you used,. And the lens is approximately the same size as the reflector on the GT – yet it throws farther than a Black Flat installed in a GT with a reflector – despite the GT’s reflector having over four times the surface area of the RLT collar you used. A collar which you noted is about 33% efficient at collecting light from 30deg – 90deg. So using your theory a larger collar (with the same aperture angle) would be more efficient – but I don’t think so, because any errors on the surface would be greatly magnified dur the distance of the reflector from the LES. But for a larger emitter it may be beneficial.


The intensity of a flashlight depends on two things, the emitter’s luminance (cd/mm^2) and the front area of the optic.
Just ignore reflectivity for now, to simplify things.
The luminous flux output of a flashlight depends on the angle of light collection, because the angle tells us both the area on the hemisphere as well as the intensity.

Try playing around with these calculators, you might see how things work when you adjust the value sliders:
http://budgetlightforum.com/node/57049

LouieAtienza
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EasyB wrote:
LouieAtienza, I think one concept you are missing is that of luminance. This is a measure of the surface brightness (of the LED) and it is important in the determination of beam intensity or throw (intensity in cd is equal to luminance times apparent reflector area). The luminance of LEDs does not vary with angle; the reason the intensity goes down as you go to the side of the LED is because the apparent area decreases. See here for some explanation.

http://budgetlightforum.com/node/15818

I do have some understanding of the importance of luminance… which is why I take interest in this thread in the first place! I actually have a few of the Black Flat and White Flat 1mm^2 and 2mm^2; and have already done a couple “learning” builds with them (which I still tinker with). I modded my GT Mini with a White Flat 1mm^2 not too long ago; my Uncle liked it so much he’s likely the only guy in the Philippines who has a GT Mini with a White Flat! I thought Enderman’s canon-style layout was great because it gives me a test bed for swapping emitters and optics and electronics. I’ll have to wait however until it gets warm enough outside to get more accurate long-range lux readings without me shaking from the cold.

The recent comments stemmed how to achieve additional luminance to a standard layout reflector-based flashlight in a retrofit scenario – that is, after de-doming (which I haven’t done yet to my XHP70.2). The question was as to how much benefit there’d be if a spherical reflector with a width coinciding with the 30deg. half angle to take the “spill light” and reflect it back onto the die. Enderman estimates about an 11% gain, and how that’s probably not worthwhile. Coming from someone who chases a 1 percent gain to obtain the maximum out of every component just seemed a bit off to me, when other ideas to gain 11% would be far costlier. So I probably defended my position with some half-true assumptions. Still, I think even half, a 5.5% gain, for $3 in parts from the box-‘o’-parts, and a bit of sweat equity from a man that barely watches TV, and otherwise doesn’t have a clue, is not too horrible. Sure I could just buy a precision aspheric lens, get a MarineBeam Illuminator RLT to harvest the Wavien collar (because that’s probably the only way I could get one now) and spend hours like_ pscal_ to make a 3.508Mcd GT. But since I’m trying to stick to the “budget” part of BLF just trying to come up with ways to get more performance on the cheap…

Thanks for the link – it reinforced some notions I intuitively had as well as gave insight on some that were not-so-intuitive…

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Enderman wrote:
LouieAtienza wrote:

Well, the lux meter can be 30deg. to either side of the central axis of the LED and still read 90% luminous intensity of dead center. I found a calculator online and found the surface area of a parabolic reflector for a front-facing LED whose diameter is the approximately the same as the distance from opening to focal point about four times that of the surface area of the opening. But even though the surface area is four times as large, the light intensity I would estimate to be about 2/3 that of the light emitted from a 30deg. to 0. And the efficacy of an electroformed parabolic reflector surface with rhodium coating is around 90%; I’m sure the cast, machined, and polished/coated surface of say a BLF GT to be slightly less. So maybe the ratio of light from the reflector versus light directly from 0-30deg. is more like 2:1 rather than 3:1.

But the reason for using these emitters is to effect a longer throw, not to put the light on an integrating sphere to get a lux reading. Your very own OptoFire uses that 30deg. half angle since that’s the aperture size of the RLT you used,. And the lens is approximately the same size as the reflector on the GT – yet it throws farther than a Black Flat installed in a GT with a reflector – despite the GT’s reflector having over four times the surface area of the RLT collar you used. A collar which you noted is about 33% efficient at collecting light from 30deg – 90deg. So using your theory a larger collar (with the same aperture angle) would be more efficient – but I don’t think so, because any errors on the surface would be greatly magnified dur the distance of the reflector from the LES. But for a larger emitter it may be beneficial.


The intensity of a flashlight depends on two things, the emitter’s luminance (cd/mm^2) and the front area of the optic.
Just ignore reflectivity for now, to simplify things.
The luminous flux output of a flashlight depends on the angle of light collection, because the angle tells us both the area on the hemisphere as well as the intensity.

Try playing around with these calculators, you might see how things work when you adjust the value sliders:
http://budgetlightforum.com/node/57049

I’ll check that out later, thanks for the link…

Barkuti
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Mmm, suggested maximum continuous driving current for KW CSLPM1.TG in a C8S?

Gonna pimp an LD-29 for 9 - 10A Silly of output LoL, check the update in HKJ's 2014/02/16 LD-29 review. Wish me luck. 

 

Cheers Party 

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Interesting two lens design (LEDiL Seanna) with an Osram 1mm^2. The first wall shot is with out the exterior Fresnel lens. Not the most practical of lights but it was fun to build. I like to try a center gasket with different thicknesses to see the effect. Maybe removing the first lens and taking a shot. Early stages of this build.



LouieAtienza
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Inkidu wrote:
Interesting two lens design (LEDiL Seanna) with an Osram 1mm^2. The first wall shot is with out the exterior Fresnel lens. Not the most practical of lights but it was fun to build. I like to try a center gasket with different thicknesses to see the effect. Maybe removing the first lens and taking a shot. Early stages of this build. !{width:100%}http://i807.photobucket.com/albums/yy353/inkidu2007/DSC_6422.jpg!

Cool! I have two of these… I fitted one with a 2mm*2, have to take some beam shots. But yes, the interior TIR lens produces a weird beam, like an axicon lens for a laser.

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I remember seeing some automotive presentation which recommended conical light pipes before aspherics to improve light collection. So….axicon.

Lexel
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I modded for a german Forum member an Utorch UT02 I measured 440k he 480k with the 2mm2 at 7A with 2S buck, about 1500 lumens

It was the Other Osram package with the die, on white flat the centering ring needs to be sanded down

netprince
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I upgraded my maxtoch shooter 2x with the CSLNM1.TG. I never cared much for the stock dedomed xml2, very green. I had a bit of trouble at first getting the LED onto an XPL MCPCB, then had a LOT of trouble getting the focus, but I finally got it.

 


IMG_20190214_220943_crop.jpg



IMG_20190215_120207.jpg

EasyB
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Lexel wrote:
I modded for a german Forum member an Utorch UT02 I measured 440k he 480k with the 2mm2 at 7A with 2S buck, about 1500 lumens

It was the Other Osram package with the die, on white flat the centering ring needs to be sanded down

Awesome result! Is that with 2×26350 cells? So is this the white flat 2mm^2 or not? I’m not sure what you mean by “other osram package with the die”.

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EasyB wrote:
LouieAtienza, I think one concept you are missing is that of luminance. This is a measure of the surface brightness (of the LED) and it is important in the determination of beam intensity or throw (intensity in cd is equal to luminance times apparent reflector area). The luminance of LEDs does not vary with angle; the reason the intensity goes down as you go to the side of the LED is because the apparent area decreases. See here for some explanation.

http://budgetlightforum.com/node/15818

So I had time to digest this particular tidbit of info and now it’s bugging me. Because if the intensity remained the same, while the apparent area decreased, then that would be the luminous intensity would increase. But I intuitively guess that’s not the case, so I’m missing something here.

Thanks all for your patience and explanations. I do love reading but most of what I’ve learn has been through experimenting and trial-and-error – meaning I am an LED executioner of sorts. But one never knows the limits unless they’re exceeded…

DavidEF
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LouieAtienza wrote:
EasyB wrote:
LouieAtienza, I think one concept you are missing is that of luminance. This is a measure of the surface brightness (of the LED) and it is important in the determination of beam intensity or throw (intensity in cd is equal to luminance times apparent reflector area). The luminance of LEDs does not vary with angle; the reason the intensity goes down as you go to the side of the LED is because the apparent area decreases. See here for some explanation.

http://budgetlightforum.com/node/15818

So I had time to digest this particular tidbit of info and now it’s bugging me. Because if the intensity remained the same, while the apparent area decreased, then that would be the luminous intensity would increase. But I intuitively guess that’s not the case, so I’m missing something here.

Thanks all for your patience and explanations. I do love reading but most of what I’ve learn has been through experimenting and trial-and-error – meaning I am an LED executioner of sorts. But one never knows the limits unless they’re exceeded…


It’s really easy to overthink it. When you think of the intensity being constant over a given area then it makes more sense that the bigger the area, the more light you will see coming from it.

The Cycle of Goodness: “No one prospers without rendering benefit to others”
- The YKK Philosophy

LouieAtienza
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DavidEF wrote:
LouieAtienza wrote:
EasyB wrote:
LouieAtienza, I think one concept you are missing is that of luminance. This is a measure of the surface brightness (of the LED) and it is important in the determination of beam intensity or throw (intensity in cd is equal to luminance times apparent reflector area). The luminance of LEDs does not vary with angle; the reason the intensity goes down as you go to the side of the LED is because the apparent area decreases. See here for some explanation.

http://budgetlightforum.com/node/15818

So I had time to digest this particular tidbit of info and now it’s bugging me. Because if the intensity remained the same, while the apparent area decreased, then that would be the luminous intensity would increase. But I intuitively guess that’s not the case, so I’m missing something here.

Thanks all for your patience and explanations. I do love reading but most of what I’ve learn has been through experimenting and trial-and-error – meaning I am an LED executioner of sorts. But one never knows the limits unless they’re exceeded…


It’s really easy to overthink it. When you think of the intensity being constant over a given area then it makes more sense that the bigger the area, the more light you will see coming from it.

Yes… I understand that. But, as I mentioned, if the intensity of the surface remained the same, and its apparent area became smaller (because we’re looking at it from an angle), then by the definition above the luminance would increase as you moved further away from center. I figure intuitively that is not the case, but I could may well be mistaken. I don’t think it’s overthinking things; We could easily take a lux meter and orbit 180 degrees around an LED and know. But we don’t need to – the manufacturers provide us that data. So something’s amiss in the explanation; either there’s a correction factor based on an angle from the axis of the LED, or something.

DavidEF
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LouieAtienza wrote:
DavidEF wrote:
It’s really easy to overthink it. When you think of the intensity being constant over a given area then it makes more sense that the bigger the area, the more light you will see coming from it.

Yes… I understand that. But, as I mentioned, if the intensity of the surface remained the same, and its apparent area became smaller (because we’re looking at it from an angle), then by the definition above the luminance would increase as you moved further away from center. I figure intuitively that is not the case, but I could may well be mistaken. I don’t think it’s overthinking things; We could easily take a lux meter and orbit 180 degrees around an LED and know. But we don’t need to – the manufacturers provide us that data. So something’s amiss in the explanation; either there’s a correction factor based on an angle from the axis of the LED, or something.


I guess we’re thinking of “intensity” differently then. I think of intensity as a constant, like one unit of brightness. More units take up more space. So, if you have less space, you have less brightness. To get more brightness from the same space (or from less space), you have to increase the intensity itself. Like a building has a specific size footprint. You can get more space by adding more buildings, but you need more land to do that. If your land is limited, the only way to get more space is to make the building taller.

The Cycle of Goodness: “No one prospers without rendering benefit to others”
- The YKK Philosophy

EasyB
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LouieAtienza wrote:
Yes… I understand that. But, as I mentioned, if the intensity of the surface remained the same, and its apparent area became smaller (because we’re looking at it from an angle), then by the definition above the luminance would increase as you moved further away from center. I figure intuitively that is not the case, but I could may well be mistaken. I don’t think it’s overthinking things; We could easily take a lux meter and orbit 180 degrees around an LED and know. But we don’t need to – the manufacturers provide us that data. So something’s amiss in the explanation; either there’s a correction factor based on an angle from the axis of the LED, or something.

I think there might be some wording confusion. In my quote above, when I said “intensity goes down as you go to the side of the LED” I meant go down in angle, as the angular distribution graph in the datasheet shows.

That graph shows the luminous intensity (cd) as a function of angle. It basically is a graph of the apparent area as a function of angle. The apparent area goes like cos(theta) which is the approximate shape of the graph. The luminance of LED (cd/mm^2) is a constant. Right above the LED (zero degrees) you have the full die area, say 2mm^2, then at an angle of say 60 degrees you have an apparent area of half that, 1mm^2 which gives you half the intensity.

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DavidEF wrote:
LouieAtienza wrote:
DavidEF wrote:
It’s really easy to overthink it. When you think of the intensity being constant over a given area then it makes more sense that the bigger the area, the more light you will see coming from it.

Yes… I understand that. But, as I mentioned, if the intensity of the surface remained the same, and its apparent area became smaller (because we’re looking at it from an angle), then by the definition above the luminance would increase as you moved further away from center. I figure intuitively that is not the case, but I could may well be mistaken. I don’t think it’s overthinking things; We could easily take a lux meter and orbit 180 degrees around an LED and know. But we don’t need to – the manufacturers provide us that data. So something’s amiss in the explanation; either there’s a correction factor based on an angle from the axis of the LED, or something.

I guess we’re thinking of “intensity” differently then. I think of intensity as a constant, like one unit of brightness. More units take up more space. So, if you have less space, you have less brightness. To get more brightness from the same space (or from less space), you have to increase the intensity itself. Like a building has a specific size footprint. You can get more space by adding more buildings, but you need more land to do that. If your land is limited, the only way to get more space is to make the building taller.

Yes, I understand that too. I was referring to the definition given to me above, that luminous intensity is the product of the surface area and candela. But since the apparent LES will reduce with increased viewing angle from center, then a) if the luminous intensity is constant, and the apparent surface area is reduced, that means the candela increases with the angle, which doesn’t make sense, or b) the luminous intensity decreases as one moves away from the center; because if the candela remained constant, and the surface area became smaller, that has to be. In other words, the intensity of the LED light at a certain angle is directly proportional to that angle it’s being viewed from in relation to head-on. And the reason I bring this up is because I’m trying to understand exactly how_ much_ of the total output of an LED goes into a typical parabolic reflector (using a GT/GT70 as an example), and how much goes into spill. And I postulate that the light that goes into the spill is actually more intense as a percentage than described to me in relation to the light reflected.

EasyB wrote:
That graph shows the luminous intensity (cd) as a function of angle. It basically is a graph of the apparent area as a function of angle. The apparent area goes like cos(theta) which is the approximate shape of the graph. The luminance of LED (cd/mm^2) is a constant. Right above the LED (zero degrees) you have the full die area, say 2mm^2, then at an angle of say 60 degrees you have an apparent area of half that, 1mm^2 which gives you half the intensity.

That’s what I’m saying all along. If the luminous intensity from 60 degrees to 90 degrees drops off a cliff, then how does 66% of the beam half-angle contain 75% of the lumens as claimed above?

luminarium iaculator
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netprince wrote:

I upgraded my maxtoch shooter 2x with the CSLNM1.TG. I never cared much for the stock dedomed xml2, very green. I had a bit of trouble at first getting the LED onto an XPL MCPCB, then had a LOT of trouble getting the focus, but I finally got it.


 



IMG_20190214_220943_crop.jpg



IMG_20190215_120207.jpg

Whoa Bro!

That is is serious reflector thrower… 550-650 kcd?

Modding is making something how you want it to be, not how it comes stock...

Old-Lumens

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Luminance is cd/mm^2 (well technically it is per m^2 but we mostly deal with small LEDs here)
Intensity is cd
The luminance of the LED stays the same, but when you look at the LED from off-axis the area is smaller and therefore the intensity (cd) is smaller because luminance * smaller area = less intensity.
That is why a lux meter at 90 degrees from an LED will show low lux even though the LED’s luminance (cd/mm^2) is constant.

EasyB
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LouieAtienza wrote:

That’s what I’m saying all along. If the luminous intensity from 60 degrees to 90 degrees drops off a cliff, then how does 66% of the beam half-angle contain 75% of the lumens as claimed above?

Oh, that is another geometric effect. The intensity vs angle graph in the datasheet is not the correct graph to integrate in order to get the total lumens. You have to also integrate as you rotate around the axis normal to the LED surface. Think about how all the light from an LED fills a half hemisphere. There is a lot more area and light at around 45 degrees compared to close to 0 degrees even though it is less intense. See here for discussion and a graph of the resulting light output vs angle.
http://budgetlightforum.com/comment/1047573#comment-1047573

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Enderman wrote:
Luminance is cd/mm^2 (well technically it is per m^2 but we mostly deal with small LEDs here) Intensity is cd The luminance of the LED stays the same, but when you look at the LED from off-axis the area is smaller and therefore the intensity (cd) is smaller because luminance * smaller area = less intensity. That is why a lux meter at 90 degrees from an LED will show low lux even though the LED’s luminance (cd/mm^2) is constant.

Bingo! That’s the tidbit of info I overlooked. Thanks…

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LouieAtienza wrote:

Bingo! That’s the tidbit of info I overlooked. Thanks…

No prob Smile
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I have it about nailed now. Speaking of full angles, the topmost 60° is (πr/3)² = (π²/9)r² volumetric area, which is just (100(π²/9)/2π)% = ≈17,4533% of the total half sphere volumetric area since a half sphere's area is 2πr². It now becomes clear why ≈25% of the light comes out of the topmost 60° cone.

Uuuugh! Some steam is coming out of my brain right now. Big Smile

 

Cheers Party 

Enderman
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Barkuti wrote:

I have it about nailed now. Speaking of full angles, the topmost 60° is (πr/3)² = (π²/9)r² volumetric area, which is just (100(π²/9)/2π)% = ≈17,4533% of the total half sphere volumetric area since a half sphere’s area is 2πr². It now becomes clear why ≈25% of the light comes out of the topmost 60° cone.


Uuuugh! Some steam is coming out of my brain right now. Big Smile


 


Cheers Party 


This is what was literally explained in the beginning of this post I linked already.
http://budgetlightforum.com/node/54696

You need to take into account area because there is more surface area on a hemisphere at larger angles.
A lux meter used to measure the intensity on the datasheet is kept at a constant 1m distance from the LED.
So the total luminous flux output of the LED is the integral of the candela over the whole hemisphere.

When flattened to 2d the result is the lobes graph:

Then you can take the area of one portion of the lobe to determine the fraction of lumens collected or lost.

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Here, have this, it makes calculating what % of light you collect easy:
https://www.desmos.com/calculator/rvazvbuo3u

Lexel
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EasyB wrote:
Lexel wrote:
I modded for a german Forum member an Utorch UT02 I measured 440k he 480k with the 2mm2 at 7A with 2S buck, about 1500 lumens

It was the Other Osram package with the die, on white flat the centering ring needs to be sanded down

Awesome result! Is that with 2×26350 cells? So is this the white flat 2mm^2 or not? I’m not sure what you mean by “other osram package with the die”.

yes 2×26650

I will have for the 1mm ² a CC FET soon,
maybe can drive at 7A with proper board layout and special cooling layout adding 0.5mm copper sheet and fill tons of viases with solder as well

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Lexel wrote:

maybe can drive at 7A with proper board layout and special cooling layout adding 0.5mm copper sheet and fill tons of viases with solder as well

That refers to the CSLPM1.TG (2mm²) emitter, doesn't it?

What host are you speaking of here, Utorch UT02? 0,5mm copper sheet, is that for focusing?

Special cooling? 

 

Party 

Lexel
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Barkuti wrote:

Lexel wrote:

maybe can drive at 7A with proper board layout and special cooling layout adding 0.5mm copper sheet and fill tons of viases with solder as well

That refers to the CSLPM1.TG (2mm²) emitter, doesn’t it?


What host are you speaking of here, Utorch UT02? 0,5mm copper sheet, is that for focusing?


Special cooling? 


 


Party 

sized are planned for 20,22 and 30mm, possibly also 17mm
for high current on CC FET regulation a proper heatsinking is needed, so I am massively enhancing the PCBs thermal properties with solder filled viases and copper sheet,
if there is enough interest and prototypes tested I may produce a 2 layer MCPCB driver (PCB cost for that is very high about 200$ for a bout 100 drivers)
I am not planning on using MOSLED for the FET cooling as this should work in as many hosts as possible

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Barkuti wrote:

I have it about nailed now. Speaking of full angles, the topmost 60° is (πr/3)² = (π²/9)r² volumetric area, which is just (100(π²/9)/2π)% = ≈17,4533% of the total half sphere volumetric area since a half sphere’s area is 2πr². It now becomes clear why ≈25% of the light comes out of the topmost 60° cone.


Uuuugh! Some steam is coming out of my brain right now. Big Smile


 


Cheers Party 

The surface area of said spherical cone comprises about 13.4% of the total surface area of the hemisphere, yet accounts for ~25% of the light. So ~75% of the light comes from the remaining ~86.6%. So if we “weighed” the luminous intensity coming from said 60 deg. cone, which is between .9-1 relative luminous intensity (I’ll guesstimate on average .96 at any point on that surface) and multiply that by .134 I get a result of .1286. Guesstimating the average luminous intensity of the remainder, using a typical chart, between 0 and .9, I’d say about .6. So multiplying .6 by .866 gives me .5196, which is just about 4 times the result I got from the top section. I’m sure the actual equation is a bit more complicated than this (and I hadn’t done calculus or differential equations in about 26 years) but intuitively it seems to jibe now.

It also makes sense why an RLT collar at 33% efficiency would potentially double the luminous intensity of the light coming out its aperture (given a 30 deg. half angle aperture).

While impractical, it also seems clear to me that given a luminous intensity chart for a given LED, there would be an optimal aperture angle for the RLT factoring in the luminous intensity graph, meaning there is an ideal focal length (ratio).

Of course now for a reflector, even a precision electroformed one with the most reflective coating is only approaching 90% efficiency, so that boosts the percentage of light from said 60 deg. cone to about 27% of total light output, and likely more given a typical flashlight reflector. Which makes me ponder that there is an “equilibrium angle” where the light coming from the top cone is equal to that of the lower section with the included efficiency factor. Not clear to me yet what use this would be of, but it should generate the most OTF lumens.

Enderman
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LouieAtienza wrote:

The surface area of said spherical cone comprises about 13.4% of the total surface area of the hemisphere, yet accounts for ~25% of the light. So ~75% of the light comes from the remaining ~86.6%. So if we “weighed” the luminous intensity coming from said 60 deg. cone, which is between .9-1 relative luminous intensity (I’ll guesstimate on average .96 at any point on that surface) and multiply that by .134 I get a result of .1286. Guesstimating the average luminous intensity of the remainder, using a typical chart, between 0 and .9, I’d say about .6. So multiplying .6 by .866 gives me .5196, which is just about 4 times the result I got from the top section. I’m sure the actual equation is a bit more complicated than this (and I hadn’t done calculus or differential equations in about 26 years) but intuitively it seems to jibe now.

It also makes sense why an RLT collar at 33% efficiency would potentially double the luminous intensity of the light coming out its aperture (given a 30 deg. half angle aperture).

While impractical, it also seems clear to me that given a luminous intensity chart for a given LED, there would be an optimal aperture angle for the RLT factoring in the luminous intensity graph, meaning there is an ideal focal length (ratio).

Of course now for a reflector, even a precision electroformed one with the most reflective coating is only approaching 90% efficiency, so that boosts the percentage of light from said 60 deg. cone to about 27% of total light output, and likely more given a typical flashlight reflector. Which makes me ponder that there is an “equilibrium angle” where the light coming from the top cone is equal to that of the lower section with the included efficiency factor. Not clear to me yet what use this would be of, but it should generate the most OTF lumens.


It’s just integration, it’s easy.
http://budgetlightforum.com/node/65922

2) there is no optimal aperture, the more light that a wavien collar collects the less lumens are output and the more intense the light becomes.
For highest efficiency you use no collar, for highest candela/mm^2 you use a collar that covers almost 100% of the hemisphere.
60 degree opening was just chosen as a good middle ground by wavien.

3) silver coatings are 98%, aluminum 90-95, and dielectric coatings can get to 99%+

4) the highest OTF lumens are with a mule and no optic. For even highest efficiency, remove the front protection lens too, that will get another 1-3% light output.

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