Aspherical Lens for pocket lights

[EDIT] Summary of conclusions from this thread:
click here -> https://www.candlepowerforums.com/posts/1892834&postcount=30

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I really didn't wanna hi-jack another thread so I figured I would post this in a new post.

I see more and more nice custom lights being made with optics, and every one of them is rather big/not quite pocketable. I'd love to have a 1xcr123a CREE led light with an aspherical lens for major throw. I know mac got one of his lens from surplusshed.com so I checked it out.

I found 4 that looked rather small with a little focal length.
Here they are in order, from smallest to bigger:


12mm diameter, 17mm focal length
http://www.surplusshed.com/pages/item/pl1099.html


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18mm diameter, 22mm focal length
http://www.surplusshed.com/pages/item/l3694.html


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30mm diameter, 15mm focal length
http://www.surplusshed.com/pages/item/l3461.html


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30mm diameter, 20mm focal length
http://www.surplusshed.com/pages/item/l3787.html


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So what do y'all think?
Could I make a nice pocket-size aspherical cree x-re flashlight with one of these?
Besides obvious diameter and focal length, is there anything I have to watch for?

I assume that since the actual surface is smaller than on a big flashlight, the light will not be AS concentrated in the center, but will it still be concentrated enough to make it a major thrower like we see in bigger aspherical flashlight mods?

Thanks!

Libbs

Unfortunately I don't have any answers to your questions but I am very interested to see what beam characteristics you get with those. If you have a chance to put an led behind I'd love to see some beamshots.

Cypher said:

Unfortunately I don't have any answers to your questions but I am very interested to see what beam characteristics you get with those. If you have a chance to put an led behind I'd love to see some beamshots.

Click to expand...

+1 :rock:

Very best regards

_____
Tom

I purchased some of the 30mm x 15mm focal length model for my bike lighting system. From my initial try they seem to have a slightly longer focal length than 15mm though, although I don't have the fittings completed yet (My fabrication skills don't quite match my ambitions) .
In any case I'll be very interested in whatever you can come up with.

tspoon said:

My fabrication skills don't quite match my ambitions

Click to expand...

You just described my curse as a newborn flashaholic...

There are currently 4 lights I want to custom build but I have no idea how to do half of it. Got a week off so I'm gonna read about some stuff and as soon as I feel confident enough I'm gonna give a shot at the pocket aspherical light.

All the body machining/thread making/electrical circuits and boards is just hard to pick up out of nowhere without a course.

Libbs, please read post #66 and #83 at this page!

Small lenses will not give a narrow beam!

tl

Small lenses are as capable of giving a tight beam as a larger lense, as the critical ratio is that of focal length to diameter. Unfortunately it is true that most of the lenses in this thread are less than suitable, however the 30d x 15f actually has the best ratio available of the aspheric lenses from surplus shed (if the focal length is 15mm, as posted earlier).

tspoon said:

Small lenses are as capable of giving a tight beam as a larger lense

Click to expand...

I tryed some small ones and big ones and it makes a big difference, it's like reflectors - bigger equals tighter beam.
But it mostly depends on size of the emiter. Take for example laser diode, the size of the emiter is like diameter of hair and 5mm lense (collimator) is enough to make the beam parallel. Led emiter size is probably 100 times larger...

its obvious that bigger will give a tighter beam because a bigger portion of the light coming directly from the led will go into the lens, that is at equal focal length

if you put the led at the focal length of the lens, all the beams going towards the lens will go through the lens and come out in a straight pattern...only thing is the light reflected by the reflector and then sent to the lens will not go in a straight pattern because it does not come from the focal point...bottom line is the light thats usually the hot spot (light directed from the reflector) is now the spill while the light thats usually the spill (mostly direct light from the led) now becomes the hot spot

if the lens is small, it will throw as much as a big one as long as the proportion of light that go directly into it from the led is the same

ideally you want a small focal length compared to the diameter.

out of memory, this could be incorrect...but say you want to calculate the proportion of light caught by the lens when the led is at focal length it should be approximatly

light caught by lens / surface area of the sphere

light caught by lens = pi*(Diameter/2)² [surface of the lens]
surface area of the sphere = 4*pi*(FocalLength)² [led is at focal length]

NOTE: THIS IS WRONG BECAUSE IT DOES NOT TAKE IN CONSIDERATION THE CONE OF LIGHT PROVIDED BY THE LED RATHER THAN A FULL SPHERE...SEE POSTS BELOW

This is an approximation because you would need to use integrals on parametric equations to get the exact result. You would also have to consider thickness of the lens for the calculations and the surface of light caught by the lens would be bigger because you would obtain a portion of a sphere as opposed to a disk like we do now.


Example 1: diameter=30mm and focal length=15mm

pi*(Diameter/2)² = 225pi
4*pi*(FocalLength)² = 900pi

225pi / 900pi = 1/4 = 0.25

25% of the light emitted by the LED will directly go into the lens, everything else that goes into the lens will be from the reflector inside the head of the light
a good % like this results in great throw, less spill

Example 2:

12mm diameter, 17mm focal length

pi*(Diameter/2)² = 36pi
4*pi*(FocalLength)² = 1156pi

36pi/1156pi = 0,0311

3,11% of the light emitted by the LED will directly go into the lens, everything else that goes into the lens will be from the reflector inside the head of the light
a low % like this results in low throw, much spill


example 3:

https://www.candlepowerforums.com/threads/154760
here, mac used a lens that has these characteristics:

52mm dia, 37mm focal length

pi*(Diameter/2)² = 676pi
4*pi*(FocalLength)² = 5476pi

676pi/5476pi = 0,1234

12,34% of the light emitted by the LED will directly go into the lens, everything else that goes into the lens will be from the reflector inside the head of the light
he is getting good results with that kind of %



Like I said, this is an approximation but it should give you a rough estimate of how much throw you can get if you try the formula with aspherical lens mods that have already been done
This does not consider exact geometry, reflector pattern inside the head of the light, thickness of the lens.


Please tell me what I just said makes sense, I've been pulling some optic physics notions I remember from college.


Libbs

You have the basic idea. I want to point out that you're assuming LED's emit light in all directions, e.g. 360°!!! some emit at 180° and according to some on this forum, the CREE emits at 70°, making 100% absorption into the lens possible.

also, you wont need any parametric equations, its basic geometry.

I worked it out, and for 100% absorption I get:

fl = d / (2 * tan(theta / 2))

where fl = the focal length of the lens, d = diameter of the lens, theta = emission angle/spread of the LED

one can clearly see that for a 180 degree LED, its physically impossible to get 100% absorption, but for a 70°?? LED like the CREE its quite possible!!

yeah ur definitly right about LEDs emitting in only a certain angle, that entirely slipped my mind. doh.

In my situation, I suggested to use parametric equations to obtain the exact top surface area of the cone part of the sphere because I was using the surface area of a sphere. I thought calculating disks was incorrect because light travels as far in all directions, so I figured we should calculate cones imbricated into spheres rather than plain disks.


I got same result as you doing your logic:

radius of emission = fl*tan(theta/2)
radius caught by lens = d/2

pi*(d/2)²
------------------ = proportion
pi*(fl*tan(theta/2))²


d²
---------------- = proportion (1 for 100%)
4*fl²*tan²(theta/2)


d = 2*fl*tan(theta/2)


fl = d / (2 * tan(theta/2)) <--- same as yours

another problem with small focal length is that the led is not 1 very small source of light, its actually 5mm wide or so
so if its perfectly centered, you have 2.5mm on each side popping out of the focal point, that gives a more diffuse hot spot
the longer the focal point, the less this problem will occur


oh yeah, also...

can anyone backup the 70 degree angle for CREEs?
might be nice to test it out if someone has a bare cree led available and a battery
if it is small enough, theres a possibility that the 30mm dia 15mm fl lens would make an excellent pocket size thrower...
in the event where the emission angle was very small, you could go for 30mm dia 20mm fl lens for better results if the beam still all fits in the lens, because you would have the 2.5mm popping on each side issue a tad less since the focal length would be increased

anyhow, a 30mm wide short cr123a light sounds like something that fits well in coat pockets without being noticed

Thanks,

Libbs

I see what you mean by surface area of a 3-d cone. you could use parametric eq. and spherical coordinants and it would be a pain. there is probably a shortcut.

but we dont need to do that cause we're assuming the base of the lens is flat (or rather, using the diameter of the lens as the 'cut off zone,' for lack of a better word.)

let's make it easier.

for 100% absorption we need the diameter/focal length ratio to be,

D/FL > 2*tan(theta/2)

so, if the CREE infact emits at 70°, look for a lens with a Diameter/Focal Lenth ratio greater then 1.54.

e.g. D/FL > 1.54 for nearly 100% beam absorption

I wish I'd payed more attention in school, my head hurts



Trey

http://ledlightingsupply.com/ledlightingsupply/Reference%20Materials/XLampXR-ESecondaryOptics%20(2).pdf
75 degrees.

So look for an aspherical lense with a D/FL ratio of 1.54 or higher, for near 100% beam collection.

Out of curiosity I did the parametric equations to calculate the proportion...

The englobing sphere is x²+y²+z² = fl² fl being the focal length

in spherical coordinates: (p = fl)

x = fl*sin(phi)*cos(theta)
y = fl*sin(phi)*sin(theta)
z = fl*cos(phi)


r = [fl*sin(phi)*cos(theta)]i + [fl*sin(phi)*sin(theta)]j + [fl*cos(phi)]k

dr/d(phi) = [fl*cos(phi)*cos(theta)]i + [fl*cos(phi)*sin(theta)]j - [fl*sin(phi)]k

dr/d(theta) = -[fl*sin(phi)*sin(theta)]i + [fl*sin(phi)*cos(theta)]j


for this part I wont copy everything I wrote, just the results:

dr/d(phi) ^ dr/d(theta) = [fl²*sin²(phi)*cos(theta)]i + [fl²*sin²(phi)*sin(theta)]j + [fl²*sin(phi)*cos(phi)]k

||dr/d(phi) ^ dr/d(theta)|| = fl²*sin(phi)


now for the surface area of the sphere:

A = //||dr/d(phi) ^ dr/d(theta)||dA

A = //fl²*sin(phi) d(phi) d(theta)

boundaries (very important):
for the emitting cone that the led produces:
0 <= theta <= 2*pi
0 <= phi <= (emit/2)
where emit is the emitting angle, for crees here its 75 degrees

for the absorbing cone that the lens catches:
0 <= theta <= 2*pi
0 <= phi <= arctan(d/(2*fl)) where d is the diameter of the lens


For the emitting cone after doing the integrals you should obtain:

surface = 2*pi * fl²(1 - cos(emit/2))

For the absorbing cone:

surface = 2*pi * fl² (1 - cos(arctan(d/(2*fl))))

however we know cos(arctan(x)) = 1/sqrt(1+x²)



absorbing cone
--------------- = R
emitting cone

(1 - 1/sqrt(1+d²/4fl²))
--------------------- = R
(1 - cos(emit/2))

if we take R = 1, for complete beam absorption with a 75 degree emitting cree led:


d/fl = sqrt[4/(cos²(emit/2)) - 4]
d/fl = 2*sqrt[1/(cos²(emit/2)) - 1]

this is actually the SAME equation as d/fl = 2*tan(emit/2)

put emit = 75 in there, you obtain a ratio of d/fl equal to

1.54

muy bien!

My totally unscientific opinion without any calculation:


The angle of the cone - built between the outer diameter of the lens and the
focal point- has to match the angle of the cone of light emitted from the LED.


This will lead to high efficiency of light collecting.


Given this - the bigger the lens (or smaller the emitter), the narrower will be
the beam …


tl

now, to calculate the actual error you get because the led is not a infinitely small source of light we do:

angle = arctan(LEDwidth/(2*fl))

tan(angle) = spill radius / distance

LEDwidth*distance
------------------ = spill radius
2*fl


we assume the center of the led is directly on the focal point
so say we have a led with a width of 5mm, a focal length of 15mm and a distance of 100m, we get a spill radius of 16,7 meters

I ran the numbers again to compare the 30mm dia 15mm fl lens against mac's aspherical

Mac gets 83,9% absorption and worst spill of 6,76m at 100m distance
we get 100% absorption and worst spill of 16,7m at 100m distance

lets say we only want a spill of 1 meter radius at 100m distance for sick throw
with Mac's light: LEDwidth = 0,64mm
with the 30/15 lens = 0,3mm

lighting surfaces:
0,64² = 0,41mm² for Mac

0,3² * 100/83,9 = 0,107mm²


0,107 / 0,41 = 0,261 = 26,1%

It seems the throw from the 30/15 lens would only be 26% as strong as Mac's light

Your scientific and unscientific explanations are all true, not far from true is "practical" explanation from DM51...