6" Long Reflector on E21

Flashlight Modding and DIY Parts
1

Just for fun, I modeled and 3D printed a 6” long x 1.8” wide Parabolic Reflector for my E21A emitter.

I did not use the best print quality because I didn’t want to waste too much time on an odd project. And then I didn’t have any Mirror Chrome paint on hand either, so I used a regular Metallic Silver. The inside seemed good enough for some fun.

I put the camera 10 feet from the wall, but ended up backing the light up to 20 feet for the pic.

What I really noticed is how I could see the 4 individual emitters in the hotspot! I was not expecting that from a hollow tube. It was easier to see when I waved it around. These camera settings capture it a little, but it is harder to see without the motion. (Try scrolling the page as you stare at the hotspot.)

Not too bad for a rushed project :smiley:

2

That’s cool. That form factor could be interesting with the cell mounted alongside instead of behind it, but might be impractical to produce. Getting tooling to the base of the reflector would be difficult.

3

Printing a reflector seems like a cool idea!

Just a note here:

If you want throw you need to make a reflector with a large diameter. The front facing surface area (as if it were a flat circle) determines the throw, not the depth.

4

Looks like fun!
The effect is more like a zoomie than a thrower, and I think with the narrow reflector it works better than a bigger diameter one - but only because there’s no real mirror, if there was a mirror then the larger diameter option would be a better for sure.
If I had to guess, if you made a wide one with no mirror it wouldn’t do much at all, and have next to no throw. I’m probably wrong though lol! :person_facepalming:
Would like to see what happens!

5

Neat project, thanks for posting.

6

so you mean no reflector is the best, because the area in front of the led is infinity large

7

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.

8

Outside the box, I like it. Very interested to see what a lens will do. Thank you for sharing :+1:

9

No, the reflector basically amplifies the size of the light source as seen from the hotspot.

10

Such a cool idea/build. Thanks for reporting your results.

11

:sunglasses: :+1:

12

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.

13

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?

14

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.

15

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

16

Well I don’t have an aspheric lens or wavien collar. But donations are welcome :slight_smile:

17

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 :confounded:

With lens:

With reflector:

18

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).

19

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.

20

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.