I’m glad you guys found it interesting. One more interesting thing is now that so much light is moving straight forward, I could put a set of lenses on it and create sort of high power white laser.
Outside the box, I like it. Very interested to see what a lens will do. Thank you for sharing 
No, the reflector basically amplifies the size of the light source as seen from the hotspot.
Such a cool idea/build. Thanks for reporting your results.
This is a common misconception. It doesn't work like that. The "tightness" you mention is not actual throw (range), but just the shape of the hotspot. A smaller hotspot does not neccessarily mean that it actually goes further. The luminous intensity (measured in candela[cd] or lux @ 1m) determines the throw.
You can actually calculate it:
luminous_intensity [cd] = luminance_of_the_light_source [cd/mm2] x area_of_the_reflector_or_optic_as_seen_from_the_hotspot[mm2]
Imagine that you are the hotspot some distance away from the flashlight. You can't see how deep the reflector is (you see a flat circle lit up by the LED).
area_of_circle = pi x radius2
radius = diameter / 2
As you can see the area is mainly determined by the outer diameter of a circle. This has by far the largest influence on the area.
The depth of a reflector (for a given outer diameter) only has a tiny influence (a deeper reflector for a given diameter usually has a smaller hole for the lightsource => this increases the area of the reflector slightly). The depth mainly influences the shape of the beam. A deeper reflector decreases the size of the hotspot, increases the size of the corona around the spot and decreases the emission angle of the outer spill.
You can test this yourself by doing a few measurements or by using the calculators here.
Surely divergence is a factor in throw too, it can’t be purely down to luminosity lets say for arguments sake i have two identical emitters with identical luminosity / surface brightness I setup one as a flood light and the other a tightly focused beam they are not going to go the same distance.
The hotspot from the reflector could have a very high luminosity at 1 meter but have terrible divergence at distance , where as another reflector giving the same luminosity at 1 meter may have a more collomated beam
I dont understand how you can exactly determine Distance from Lux without taking into account divergence? maybe someone can explain?
Some reflectors may look similar to my example but not actually be a parabola. Some reflectors use circle geometry. Divergence is a huge deal with that type.
Id be very interested to see how it stands up against an Aspheric setup with the same emitter and drive , if it is indeed collimated as you suggest then it should harness all of the light from the emitter and give similar performance to an Aspheric setup with a wavien collar
Well I don’t have an aspheric lens or wavien collar. But donations are welcome 
Ah ha, I have an aspheric lens I unscrewed from a zoomie and held in front of the E21 at the same power, distance. It’s a much larger hotspot area. Not to mention the gaps in the LED show up terrible 
With lens:
With reflector:
I am only talking about parabolic reflectors of differents sizes. Luminous intensity (throw) only depends on the frontal surface area (a flat circle with a hole in the middle) and the luminance of the light source (and some losses because of reflectivity, transmission etc.). The divergence doesn't affect it in a meaningful way. Try the calculators I linked to above.
We have had this discussion multiple times over the years here in the forum and in other forums. Here's a German explanation (use google translate).
Also, you don't need to consider any special equations for specific LEDs (unless they have a dome and a very untypical emission angle like 80° etc.). All LEDs without a dome are lambertian emitters and thus all work the same in conjuntion with reflectors.
The maximum luminance values of many LEDs can be found here. You can use those to calculate what is possible with a given reflector (or optic).
All of the intensity measurements that this community has done support what the driver says. The beam intensity is equal to the emitter luminance multiplied by the frontal area of the reflector or lens.
There have been many discussions of this. Maybe the first thread introducing this concept to the community is the one by Dr Jones.
JoshK, are you saying that you can get more beam intensity from a smaller diameter reflector than a significant larger one just by making the reflector longer? If so take some measurements and see.
Try out the formula above. Read the threads. Build some lights and measure them. It will all work out. What you are stating is a common misconception, like I said.
Why do you mention a toothpick sized reflector? It obviously needs to be wider than the die of the LED as you also mention.
Example:
Lets say you have two identical LEDs with a luminance of 200cd/mm2. Each has a die with an area of 1mm2.
Lets also say that you have two parabolic reflectors. Both have an LED hole with a diameter of 10mm. Both have a center depth of 20mm. Both have an aluminium coating with 90% reflectance of visible light. Both have an ar-coated glass lens with 96% transmittance (typical chinese flashlight quality) on top.
Reflector A has an outer diameter of 20mm.
Reflector B has an outer diameter of 50mm.
Now you want to find out how far you can see at night with these two lights (the ANSI range of a flashlight). To calculate this distance you need the luminous intensity [candela]. To calculate luminous intensity you need to calculate the frontal surface area of both reflectors.
Reflector A:
refl_a_area = area_circle - area_led_hole
= (20mm / 2)2 * pi - (10mm / 2)2 x pi
= 314.2mm2 - 78.5mm2
= 235.7mm2
refl_a_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity
= 200cd/mm2 * 235.7mm2 * 0.96 * 0.9
= 40,729cd
Note here that cd is the same as lux@1m.
refl_a_ansi_range = sqrt ( refl_a_lum_intensity / 0.25 Lux )
= sqrt (40,729lux / 0.25lux)
= 403.6m
Reflector B:
refl_a_area = area_circle - area_led_hole
= (50mm / 2)2 * pi - (10mm / 2)2 * pi
= 1963.5mm2 - 78.5mm2
= 1885mm2
refl_b_lum_intensity = luminance * refl_a_area * transmission_losses * reflectivity
= 200cd/mm2 * 1885mm2 * 0.96 * 0.9
= 325,728cd
refl_b_ansi_range = sqrt ( refl_b_lum_intensity / 0.25 Lux )
= sqrt (325,728lux / 0.25lux)
= 1141.5m
As you can see, the depth of the reflector has no effect on the results.
I think I see where our points of view differ. Let me explain it using your calculator instead of my modelling software.
Here’s a simple 4” reflector. We will use this as the base. It will be visible in the background of the other pictures.
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Now when I say “a wider function” or “wider mathematically”, I mean this:
All other things equal, it becomes shorter, and reduces the “degrees of light captured”. This is why the diameter (and front surface area) of the reflector is unreliable as a measure of throw.
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Now when you say wider, you mean this:
I hope that all makes the sense I intended it to.
I learned this stuff in my college class Non-linear Equations. So I think about it starting from the equation. I know that’s not the most intuitive.
You are correct it’s not just the outer diameter that matters, it’s the frontal area. To find this you subtract the area of the small inner circle from the area of the large circle. But because of the way area grows with diameter the inner circle area is always relatively small. So making the reflector shallow, within reason, doesn’t have a huge effect on the frontal area.
This is not correct. Most flashlights use reflectors with similar collection angles. So would you expect, for example, that the BLF GT reflector would make a larger less intense spot than a C8 reflector, with the same LED? This is obviously not the case.
Please put your current understanding of this subject aside for a moment and read this:
Ok, I have been working on this and discovered an error on my part.
When I was making changes I was accidentally allowing the focus point to shift. But because it didn’t shift much, it went unnoticed. The error was even there in the reflector I printed.
So I came back to correct my mistake and hopefully explain this clearly.
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For parabolic reflectors, you only have 1 thing you can change about the geometry. That’s the vertex location. (The vertex is a theoretical point behind the LED). The effect of moving the vertex away from the LED is more total offset of the reflector surface.
For thrower flashlights, you want enough offset to properly fit the reflector around the LED, and that’s it. Then the only choice that’s left is where to chop off the infinite geometry. You could call this the “diameter”, like The_Driver does… But this causes confusion, because as you see in the pic above, diameter isn’t forced to increase when you fiddle with the vertex. Diameter can be kept the same just by chopping off that infinite geometry sooner. Again, see pic.
You need to mathematically show how you derive the luminous intensity (= throw). That's what I did (even accounting for typical losses!). The size of the inner opening in a reflector has a very small impact on the overall frontal surface area (because the area of a circle increases with the radius squared). The radius (and thus outer diameter) is the deciding factor.
You have switched to using the term "performance" instead of "throw". How can you quantify that?
Do you own a lux meter? You could try doing some measurements with it.
I think the real issue here might be your understanding of what throw is?
We are talking about throw, so I don’t see an issue with the word performance, but I edited the word as you wish.
I didn’t say your math was bad. It looks good to me because you used “area” and subtracted the flat area around the emitter. It’s the fact you say front diameter is all you need to know that causes confusion. IRL it works well, but it causes confusion in discussions.
Saying throw is based on “area_circle - area_led_hole” sounds good to me. Saying it’s based on the front diameter only, confuses people. It confused me.