1. ## Optic theory

Newbie's ~2.5"dia ~1.25"fl aspheric lens with AR coating measured 37k lux with a cree xre at 3.some watts, IIRC. Way back I remember some mag luxeonIII measuring 10-12k lux. Point is, we have a lot of data points on this.
What I am missing in these boards, is guidance towards a theoretical and numerical approach to this as well. So that's what I am fishing for here. I have the feeling that ballpark numbers could be found without expensive and complicated raytracing.

To start with a simple question, if I were to scale an aspheric to half the dimensions, would I get 1/4 as many lux? I'm thinking that degrees in=degrees out; If I am sitting on the flat surface of the lens, the LED is covering x degrees of my view; and that is the resulting beam angle as well. If I and the lens gets two times closer to the LED, then the beam angle is doubled - resulting in the light being spread over an area four times as large.

There are lots of factors which may or may not be relevant to a napkin-class calculation - I would very much welcome the discussion of these.

....

3. ## Re: Optic theory

edited out some wrong stuff.

...

5. ## Re: Optic theory

Let me try again. Here is a chart showing how many percent of the total
emitted lumens of an xr-e which falls within any angle.

It shows that newbie's lens catches about the same as the hd45 and other

It is made with the following method: The beam is divided into rings, and
each ring's area on a sphere is calculated, and multiplied by the intensity as shown in the xre datasheet:

For instance, between 20 and 40 degrees included (10-20 deg half angle),
is a ring that projects upon 0,0225576 parts of a sphere. The average
intensity for that part of the beam is then ogled from the datasheet graph -
I'd say about 90 percent. 0,0225576 is multiplied with 90, to
find how much of the total output of the led falls within this ring. Then the rings' light amount are converted to percent of the total light amount and added from 0-180deg and tossed into the chart. apparently in an inexplicable way. Hope it works.

keep in mind the difference between half angle and included angle: 1:2

6. ## Re: Optic theory

Here is an image that shows what I think happens with the cree and a true parabolic ~45x45mm reflector. The apparent die image moves around a bit.

the angles on the side should have a "-" in front of them.

7. ## Re: Optic theory

Hi AilSnail,

If you use a f/1 aspheric optical lens with lets say 50mm diameter, the f/1 determines the angle of light that is collected from the led.. So the f/1 determines the amount of lumens you collect.

Now if you would use a f/1 lens with 25mm diameter, the amount of lumens you collect stays the same, because it grabbes the same angle of light.

BUT: (And this is a fact! If you measure the lux output at a distance of lets say 10 metres, with the 50mm lens, you'll get a value 4 times higher than with the 25mm lens! (If both are properly collimated..)

You are right: with the smaller lens closer to the led the emitting surface seems 4 times bigger to that lens: Conclusion is: with the laws of light the 25mm lens 'sees' a 4 times bigger surface, so it illuminates a 4 times bigger surface at 10 metres distance than the 50mm lens does!!

So with the 25mm lens the surface brightness in the beam at 10 metres is 1/4 of the surface brightness of the beam comming from the 50mm lens..

That is why the lens- or reflector-diameter determines the throw of a torch. (together with the surface brightness of the source)

Fact is that the Cree and Luxeon emitters are front-emitters, so don't work very well with conventional reflectors, indeed if you use conventional reflectors you'll need them very deep !!

The acrylic optics, especially designed for these emitters do a much better job: They grab the light where it is most emitted: directly at the front. They are also much more efficient because they use the 100% internal reflection law of light ! A conventional reflector reflects about 84% of the light at the most!

Regards,

Ra.

8. ## Re: Optic theory

Hey, nice graphs! I haven’t seen people here use those graphs of ray tracing. I agree that it is very complex.

It is too bad that nobody has replied yet. I have a post in this "general lighting discussion" area and only people who happened to wander on would give any help. Sometimes you have to search around to some of the "old" and experienced members that know their stuff and invite them over. It sucks that you have to post important lighting questions here only because they are not directly flashlight related. I was thrown in here because I was trying to discuss my LED headlight project. Oh well.

Now you asked some questions, and I am probably not qualified to answer them, but I do have some understanding of optics.
Originally Posted by AilSnail
… if I were to scale an aspheric to half the dimensions, would I get 1/4 as many lux?
Is this when the lens stays the same distance away from the light source? Wouldn’t you get the same lux if you move the aspheric lens (scaled to half the size) to half of that distance? That may be because it is at the correct focal point of the lens. If this is true, I see what you mean that the beam angle would appear to be bigger. That would happen because to scale (with this smaller and closer lens), the LED die size would appear to the lens as being “bigger”. Then you said:
Originally Posted by AilSnail
If I and the lens gets two times closer to the LED, then the beam angle is doubled…
Is the lens the same size as the original you were talking about in this statement (not the half size)? If so, then the beam angle coming out should be bigger (I don’t really know if it would be a pretty and exact double size). This should be because the lens is moved closer past the focal point, right? If the lens was half of the size and half of the distance, then the beam angle coming out would be only slightly larger, right?

I hope that you know what I mean when I talked about scaling down the lenses and distances, as well as the apparent “size” of the LED die compared to the lens at this scaled down size.

Now I see that you changed your focus from the aspheric lens to a parabolic reflector. Do you have a question about this, or are you just showing what you had found? It is interesting using ray tracing. I tried doing this by hand to find the intensity of my LED headlights at different angles, trying to pretend that I do not have a cluster of 18 lights, but instead, one continuous 9 degree optic and light source. I wanted to find the beam angle of the projected light, and then consider how bright the light would be projected on the ground with the light tilted down at 1.5 degrees. It is tough and I will just rely on testing the finished product.

Anyway, what do you need to know now? Are you still trying to achieve that awesome 37K lux? That will be tough. That is like a 230 cd/lm on-axis efficiency with the optic (with the XR-E at 160lm). I mean the standard Fraen FHS optic has a 21 cd/lm on-axis efficiency, as well as the typical 27mm IMS reflectors. The optic that Cree makes has a supposed 46cd/lm on-axis efficiency (according to the specs sheet). These are all tiny optics, so you might have to stick with a larger reflector or lens. I don’t know. Did you need to achieve this 37Klux with a smaller optic or something, or are you trying to enhance the lux with the same size 2.5” by 1.25” optic?

Also, with this claimed 37K lux, was this measured at 1 meter, or even closer? You can blow that number out of the water with using those concentrator optics made by Polymer Optics, but this measurement is at 14.5mm. You can get up to 5 million lux at this distance, all in a 6mm diameter circle! This is with the LED at 160lm. This would be useless as a flashlight because the 14.5mm is the focal point, and the beam spreads out further from that, BUT, what if you could incorporate a second lens to turn this intensely concentrated light to a parallel beam? This would take some experimentation with different optics, but it is doable. I thought about getting a concentrator optic from an online store (used in projectors) to take this spot and project it forward at a narrow angle, but never did so. It even had the perfect focal point (of light coming in). A aspheric lens would do the same (they are similarly shaped), all you would have to do is turn it so the convex end would be facing the concentrator optic. Then just focus it in.

I remember another way to project the light forward at huge distances, but I don’t think that the lux rating is that extreme. I managed to clamp in my old V-binned LuxV flashlight with 27mm IMS reflector into our 6in diameter reflector telescope where the eyepiece would have went. The thing probably was not efficient, but I could project a beam that is 8-12 inches across at huge distances (+100yds). The beam isn’t smooth at all, but it was cool because it felt like you were holding a light cannon.

Oh, I just remembered that I seen somewhere in CPF where someone used these XR-E’s in their D-sized Maglites with stock reflectors. This person made a mistake (I think) and they took off the dome lens from the XR-E. There was some silicone inside, but the wires were intact. They turned their XR-E with the odd 70 degree half angle intensity into a Lambertian beam pattern LED. The light was better focused with the stock Maglite reflector and more lux was produced! The beam was far more usable. If you can do this, then it would be just as easy to focus the light of the XR-E as the standard Lambertian Luxeon LED. As long as the reflector or optic doesn’t press against the delicate die and die wires (like an optic or reflector with a holder or legs), then you would be good! I will try to find that thread where this guy modified his XR-E and post a link.

What do you think? Sorry if my posts are too long. I cant help it. I hope you get stuff figured out. I think that what you are doing is just as important to the LED and flashlight world as what some do with optimizing batteries and driver circuits. Please keep us updated with graphs and pictures of your findings…

-Tony

EDIT:
Ha! I just realized that this was posted 5 minutes after the last post by Ra. It took me like 15 minutes to type this thing, so I missed that. Ra was right on the dot. Nice. Some of the things that I said was similar to what Ra said, but he said it in a shorter and more clearer way. If it sounds like what I said was wrong, well, I meant the right thing, but I just didn't say it right. Does that make sense? Lol...

9. ## Re: Optic theory

Thanks! Nice to have some theories confirmed and seen from slightly different angles.

It is not real raytrace, but a rather long rhino3d session. The apparent die placement in the drawing is only what it looks like in the real world, and the base of the package is placed at the fp of the parabola.

I think it was 3rd shift who beheaded his cree and put it in a light.
Newbie's lens was 37k at 1m. Ra, I like how you are always talking about lux at 10m and more! I suspect that those readings can give a different picture than the 1m readings - which I believe can be skewed a lot from both the effect of aperture diameter and from crossover points in the beam.

I do have a question: I should like to be able to have excel (or openoffice) return an y value from the first graph when i feed it an x value. So that if I put 70degrees into one cell, another cell will show 50%. I think it would be called a look-up table.

Tony, I like it here in the slow moving backwaters.

10. ## Re: Optic theory

Thanks. Well, I found another. Two have done this and took pictures so far (that I have seen), but several others claimed that they have done it. I guess that this mod makes the tint shift to a warmer color, but in a good way (as some has said). I guess that the lens can be removed easily, but the metal ring is more difficult. You have to use a cutting disk (like a dremel) to cut a small notch in the ring for grip, then set a sharp flathead screwdriver into the notch and bend and twist to pry it off. According to others findings, the hemispherical lens is made of a solid "glass"-like material (it isnt hollow). Between the solid lens and the die (in the region surrounded by the metal ring), it is full of a soft silicone material. So you can safely remove the lens (responsible for the 70 degree angle), and get a more lambertian beam. Make sure you cut or score the silicone on the inside edge of the metal ring before you pry it of. They say that there is a chance that if you dont, all the silicone may go with it, messing up the bond wires possibly.

For my next project, I might take the lens off of four of these in my Quad 2D Maglite upgrade. This way, I can use the existing IMS20XA reflectors that sat atop the LuxeonIII's. I may have to modify them a little to focus them, but that wont be difficult. I thought about using some clear silicone glue to attach old lambertian domes from some dead K2's I had to protect the dies even further, as well as make the beam pattern nicer. If I scrape a little silicone out, the surface will then be very rough, throwing light in random ways. If I use a dome over top filled with silicone glue, then those rough spots would be smoothed out (as long as I dont trap any air bubbles). This will be weird for the XR-E because the original dome was slid and glass-like, while the K2 dome is softer and filled with a little silicone.

Anyway, sorry. Here are those links:
This was the first one I seen by 3rd-shift:
http://www.candlepowerforums.com/vb/...d.php?t=143195
If you look at the first post in this thread, he made a link to the previous thread where there are better explanations.

Then I stumbled upon this one by Download. He shows it a little ways down:
http://www.candlepowerforums.com/vb/...d.php?t=144141

AilSnail, I wish I can help you with the Exel spreadsheet. I learned how to add in formulas, and I have used it to measure out beam angles of my project before, but it is really time consuming. I tried today for a simple task, but it just didnt like me and I had to enter data in manually. I have no clue how to set up "programs" where you enter a number and it gives you a response automatically (like a calculator). I have seen it done with an elaborite database program made in Exel, but I am clueless how it was made.

That first graph of yours is confusing me. I am having a hard time gripping what you are trying to achieve with that one. It seems tough to read it and put it to use. Maybe if it turn it upside-down. LOL...

-Tony

11. ## Re: Optic theory

Yep, you are right Tony...

But perhaps I can enlighten you a bit: In that graph you can see how much of the lotal lumens-output you grab with a sertain lens: If you take a F/1 lens: It has a focal-angle of about 53 degrees, you can see in the graph that it will collect a little over 30% of the total lumens generated..

If you take a F/0.5 lens you would grab a 90 degree angle: 70% of the lumens output according to the graph..

Hope this clears things up for you..

Regards,

Ra.

12. ## Re: Optic theory

Tony - Yes! It is "right" for lenses only. For reflectors you have to look at it the other way! Does it make sense now?
In the second graph, you can see that the beam has 50% intensity at 80 degrees. I believe this is what is usually referred to as "beam angle"? Maybe not.
In the first graph, you can see that half of the light falls within a 70 degree cone.
edit: you beat me to it Ra. thanks.

13. ## Re: Optic theory

Here is the same data presented differently; It shows how much of the light of the XR-E that will hit a lens of a given F/#. The F/# is the same as focal length divided by diameter.

14. ## Re: Optic theory

Some wrong stuff deleted

15. ## Re: Optic theory

I think this drawing from wikipedia would be nice in this thread:

16. ## Re: Optic theory

Keep up the good work guys! This is one of the best discussions I read here on CPF in awhile.

17. ## Re: Optic theory

......They are also much more efficient because they use the 100% internal reflection law of light ! A conventional reflector reflects about 84% of the light at the most!
I am not familiar with the 100% internal reflection law of light? The acrylic lenses I have seen which for the most part are a combination of TIR and refractive lens element certainly are not 100% efficient in their TIR aspect as I see much light leaving from the optic when viewed from the side? I understand that SureFire uses a very expensive grade of optical plastic and I believe they do see transmision efficiencies greater than those of an external reflective surface.

In terms of total efficiency, in the case of a reflector, we get at best your 84% transmision of that light actually encountering the reflector but we get 100% of the light which leaves the optic unaltered in direction. In the acrylic optic, you have all of the light traveling through the acrylic and its absorption loss will be levied on 100% of the source's output. Additionally, you will loose some light out the side due to the T in TIR not being Total but some fraction there of. Yes?

I guess the only point I am trying to make here is that there will be losses regardless of the type of optic used. If the primary goal is minimal loss in flux but with a secondary optic required, then one can identify the type of optic which results in minimal loss. Typically one is more concerned with more specific photon management and say a tight collimation of light. In such a case, the delivery of light on target is of primary concern and the nature of loss of light is not as important as the loss itself.

19. ## Re: Optic theory

This is at approx 1m.

20. ## Re: Optic theory

Hi McGizmo, welcome to this dicussion...

I said that the acrylic optics use the 100% internal reflection law of light. I didn't say that the acrylic optics have 100% efficiency !!

All optics have losses, or else you would not see any light comming from the side!

Earlier in this discussion we agreed on the fact that the Cree emitters put out most of their lumens within a 30 degees or so at the front!

Conventional reflectors are designed to work best with light-sources that emit their lumens sideways (Halogen with axial fiament, automotive HID, short arc.) So not with light-sources that emit at the front !

The acrylic optics use a combination of internal refraction and internal reflection, that means that these optics grab almost the entire lumens output of the emitter: Even the center-hole is covered by a collimating lens !

If they are kept very clean and undamaged, they have much higher total efficiencies compared to conventional reflectors with the same diameter (ofcource only if they are used with emitters like Cree and Luxeon)

AND NOW FOR SOMETHING COMPLETELY DIFFERENT : THROW :

And this will be the hard part of this post...!

Let me try it this way: Lets assume you are an object 100 yards away, that wants to be illuminated by a torch. Which are the things that determine the amount of light you actually receive 100 yards away ?? Or, in other words, what determines the apparent brightness of a torch, seen from a distance??

Well, two major things: Surface brightness of the source, and the dimensions of the source ! NOTHING ELSE ! (well,, atmospheric conditions, ofcource,, but lets say they are perfect...)

A reflector not only reflects lumens, it also reflects the sources surface brightness. FACT: The apparent surface brightness of the reflector cannot ever be higher than the surface brightness of the source! It always is lower due to reflection losses in the reflector.

Now its time for a picture:

What does it show? Two operating torches, directly photographed in the hottest part of the beam (through a type 13 welding-filter)
A 35watt halogen torch at the left, a 10watt HID torch at the right.
You see the reflectors, lit by the emitting surfaces (filament and arc)

These two torches have the same throw !! The low surface brightness of the halogen is compensated by the bigger reflector!

FACT: A torch has max throw when the entire reflector is lit by the source, seen from a distance.

You are propably not going to beleve what I'm going to tell you now: THROW IS ABSOLUTELY NOT AFFECTED BY THE FOCAL LENGTH OF THE REFLECTOR OR LENS !! ONLY LUMENS OUTPUT IS !!

From a distance the object 'sees' only a two-dimensional surface, no matter how deep or how shallow the reflector.

If you have a deep reflector, you collect more lumens from the source, that results in a wider beam (more sidespill) But you'll have the same throw with a shallow reflector, but because you send less lumens towards the object, the beam will be tighter (more laserlike if visible)

So, bottom line: With a lens diameter of 50mm and a focal length of 1 metre you'll have exactly the same throw as with a 50mm lens with a focal length of 30mm !! Only the beam of the first will be useless because of the pathetic amount of lumens its made of..

I hope this isn't too long for you.. any questions, please ask..

Regards,

Ra.

21. ## Re: Optic theory

Doug, I like it too - Ra's last post has me confused though.

The green filtered picture illustrates how lux at 1m gets skewed in favor of small reflectors.

Ra, I'm not sure I am following you. You seem to be opposing your earlier post. A short focal length will not compete with a long focal length lens when it comes to illuminate something far away, especially if f/# is kept constant.

I'm going to think a bit more on what you are saying.

22. ## Re: Optic theory

This one illustrates a long and a short fl, with the same diameter.
It shows that the light is spread over a larger area with the shorter fl, like we concluded earlier. It also illustrates that in a perfect lens, the only light beams which are going straight ahead, are the ones coming from the direction of the infinately small focal point.

Now, I happen to have a 21.5mm aperture x 6.9mm bfl aspheric, and another lens with a focal like 50mm or so, which I masked to to get the apertures the same.
The latter is much better at lighting up distant targets.

So if I understand correctly what you are saying, Ra, either your theory is flawed, or my equipment, method or eyes skews my observation in favor of the longer fl.

23. ## Re: Optic theory

That is what I meant with "This will be the hard part of this post !"

On these forums there is a lot of confusion and mis understanding about what determines the throw of a light !

But there is hope: Take a good look at the last picture AilSnail posted:

Now concentrate only on the green beams: That beam, comming out of the lens, towards the object is exactly the same with both lenses !!

That means that the center of each beam has exactly the same amount of light (lux at 1m), This also means: THE THROW OF THE CENTER BEAM IS THE SAME WITH BOTH LENSES !! And that is that I've been trying to tell you..

For the entire beam, I hope the earlier posts were clear enough about that: If the lens "sees" a larger surface, it projects a larger surface, but if the lens diameter stayes the same, the surface brightness at the object stayes the same!

So, back to the pic AilSnail posted: The lens with the short focus projects a bigger spot at a sertain distance compared to the lens with the longer focal length. But in the middle of both spots the lux-reading will be the same, so they'll have the same throw.

Now what happens is: The lens with the short focal length grabes much more lumens from the emitter, thats why the beam is wider. Now if you illuminate an object with this wider beam objects in the neighborhood light up as well by the wider beam, That is the main reason the narrow beam seems to throw further: You have a clearer wiew to the object, not disturbed by illuminated objects nearby !!

AilSnail,, Try to do the same experiment, but now measuring the lux at a distance of about 10 metres, at the center of each beam. The brightness of a illuminated object is directly related to the amount of lux it receives, so the amount of lux you measure at a distance is directly related to throw !!

Hope this helps you to understand..

Regards,

Ra.

24. ## Re: Optic theory

Thanks for explaining more. I don't have a lux meter, but maybe I can set up something with a photoresistor.

Meanwhile I need to think some more about whether your theory holds up, in theory.

What a cliffhanger.

Imagine that you put a reflector behind a star, a reflector so immensely large that the star is almost not measurable in comparison - I'm talking about a sincerely huge construction, almost half the universe. The reflector is parabolic, and focused towards the earth. That was just for fun, I'll get back to the theory shortly.

25. ## Re: Optic theory

Originally Posted by AilSnail
The green filtered picture illustrates how lux at 1m gets skewed in favor of small reflectors.
Sorry, but no.. It does not !! the size of the reflector isn't a issue here, only the difference in surface brightness is: If I would put a halogen bulb in the smaller reflector the image would look like this:

The other thing you can see is that the surface brightness reflected by the reflector is mostly of the same intensity, no matter the source-reflector surface distance: The outer rim of the reflector is further away from the filament than the central part, yes ?? But it has the same intensity !! So no matter how small or big or deep or shallow your reflector is,, if its parabolic (with the source in focus..) it would look the same when you look into it from a distance, evenly lit over its entire surface !

The HID-arc has a much higher surface brightness than halogen: Thats why the picture with the small HID-torch looks like this: Not because of the smaller reflector!!:

Regards,

Ra.

26. ## Re: Optic theory

The HID-arc has a much higher surface brightness than halogen: Thats why the picture with the small HID-torch looks like this: Not because of the smaller reflector!!:
But the smaller one with the short arc reads a higher lux at 1m than the large halogen doesn't it? I guess it would depend on which sort of measurement equipment you use?

27. ## Re: Optic theory

Originally Posted by AilSnail
Here is the same data presented differently; It shows how much of the light of the XR-E that will hit a lens of a given F/#. The F/# is the same as focal length divided by diameter.

A very useful graph! Optics is a bit far from my formal training but I can offer one practical comment on using this graph. It becomes very difficult to design lenses with very low f numbers because at some point the rays from the source will strike the lens material at less than the critical angle and thus never enter the lens.

28. ## Re: Optic theory

Nice of you to point that out, we should delve into that a bit also.

29. ## Re: Optic theory

Every time I try to quote someone the link to CPF crashes!!!

So I'll try it this way:

QUOTE: (AilSnail But the smaller one with the short arc reads a higher lux at 1m than the large halogen doesn't it? I guess it would depend on which sort of measurement equipment you use? END QUOTE.

No, it doesn't !! That is, not if the entire reflector of the halogen torch is lit by the filament!

The amount of lux you measure comming from a torch is a combination of the surface brightness and the dimensions of that surface. So with a lower surface brightness (halogen) you need a bigger reflector to get the same lux-reading at a distance, than with higher surface brightness (HID).

Every lightsource has its typical surface brightness (within a sertain range)

You can overdrive a halogen bulb to get a higher surface brightness, but physics of nature determine the limits: Going too far will melt the filament!
With halogen you'll never reach the surface brightness of HID!!

As a result: With a halogen torch with a 4 inch reflector, you'll never reach the throw of a HID-torch with the same reflector diameter!

AND: You'll never see military searchlights with a CCFL-tube as a lightsource !!

Regards,

Ra.

30. ## Re: Optic theory

And...

QUOTE (Doug_S): A very useful graph! Optics is a bit far from my formal training but I can offer one practical comment on using this graph. It becomes very difficult to design lenses with very low f numbers because at some point the rays from the source will strike the lens material at less than the critical angle and thus never enter the lens. END QUOTE.

You are absolutely right Doug..

With short focal lenghts the lens absolutely needs to be aspheric, and even then there indeed is a limit of angle of the incomming rays at the edge of the lens.

F/0.7 is about the shortest possible focal length of a single aspheric lens!

A lens is the best way to grab light from an emitter that emits at the front.
A conventional reflector is the best way to grab light from emitters that emit to the side and towards the back.

The acrylic optics for use with emitters like Cree and Luxeon are a combination of the workings of a conventional reflector with a lens: They grab almost 95% of the lumens output of the emitter: more than ever is possible with only a reflector or a single lens!

Regards,

ra.

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