Formula for calculating throw using aspheric lens

[gcbryan]

Throw can be defined. A common standard for search lights is the distance at which luminance is 1 lux.



Here is a key relation:

Relative Candlepower = Relative Source Intensity x Relative Light Gather x Optic Efficiency x (Relative Optic Focal Length)^2

Relative Throw = sqrt(Relative Candlepower)

Relative Throw = sqrt(Relative Source Intensity) x sqrt(Relative Light Gather) x sqrt(Relative Optic Efficiency) x Relative Optic Focal Length

For this relation, light gather is the percent of source light utilized by the optic. For a lens, optic efficiency is the light transmittance of the lens as a percent, and also the light reflectance of a reflector as a percent.

You will see that optic diameter is not a part of the relation, but optic diameter is involved because optic diameter provides for increases in light gather and focal length, but optic diameter alone does not indicate the relative amounts of light gather and focal length. Optic diameter can be used for general comparisons, but it's not as accurate as using the actual light gather and focal lengths.

When considering throw should focal length really be in the computations? My understanding is that two optics of the same diameter will throw the same distance regardless of focal length.

Focal length will only affect the amount of light not the intensity (which is all that affects throw).

 

[Walterk]

@GetLit:

Here is a key relation:

Most interesting posting ! Thx. Finally a leap forward. And reading your NightSword thread I think you know pretty well how it works. I will need some time to get related with this.

@TH232: Thanx for the input. Measuring and testing surely will follow,(waiting for meter) but probably not as elaborate as suggested .
Can you explain more about 'The bigger lens will throw further, because the LED more closely approximates a point source.' ?
Is that because the actual led is relatively smaller for the big lens, and thus the beam more narrow and thus more intense/concentrated ?

@GcBryan: It may be 'just' another approach. You are right in that you propably can't just mix principles.

Just a quote from GetLit:
Longer focal distances enlarge the size of the reflector in relation to the luminance area, which in turn decreases the overall angle of incidence to the target, thus increasing collimation.

Focal length is just the name for a certain set of parameters.

 

[get-lit]

Most interesting posting ! Thx. Finally a leap forward. And reading your NightSword thread I think you know pretty well how it works. I will need some time to get related with this.

Thanks. I put a lot of work into deriving that relation. It was most useful for me in determining optimal design of the Nightsword, which is on hold due to lack of funds for now.

When considering throw should focal length really be in the computations? My understanding is that two optics of the same diameter will throw the same distance regardless of focal length.

Focal length will only affect the amount of light not the intensity (which is all that affects throw).

No. Lens diameter alone is useless without focal length. With no focal length, there is absolutely no gain in throw, you would have a light bulb with a just a flat piece of glass that does no good for throw, no matter how big it is. As you increase focal length, the etendue relation is increased for more throw, and with the same diameter optic, light gather diminishes, until a point is reached at which further gains in throw due to increased focal length are no longer achieved due to losses in light gather. The optimal utlization of focal length and light gather is then reached.

Increased diameter allows for increased light gather and focal length. For the same diameter, you can increase focal length by lessening light gather and vice-versa. For any given diameter there is always an optimal utilization of focal length and light gather, based solely upon the light source angular intensity profile.

Focal length has a square relation to candlepower and thus a direct relation to throw.

Light gather has a direct relation to candlepower and thus a square root relation to throw.

Optic diameter provides for increased focal length and light gather. When the utilization of focal length and light gather remain relatively the same, optic diameter has a square relation to candlepower and thus a direct relation to throw.

 

[get-lit]

Just a quote from GetLit:
Longer focal distances enlarge the size of the reflector in relation to the luminance area, which in turn decreases the overall angle of incidence to the target, thus increasing collimation.

I'm glad you located that quote. You've really been looking into this

Taken from an earlier post I made on the subject:

Here's a decent article for learning a bit more about the inverse relation of luminance area to focal length to produce collimated light output, and in this article Etendue is the key term you want to review:
http://thomann.net/uhp.pdf

In that article, there are many sections describing the Etendue effect, and section 3.4 about arc length is particularly useful.

Here's more about Etendue:
http://en.wikipedia.org/wiki/Etendue

Although the formulas become complex, the concept is really quite simple. To produce the most candlepower, we want the the optical system to produce the most photons going the same direction. If we reflect an infinitely single point source of light with a perfectly parabolic reflector, we would have perfectly collimated light, and candlepower would simply be the amount of light emanating from the point light source. But since an infinite point source of light is impossible, we always have imperfect collimation because photons emitted further from the focal point are never perfectly collimated, and it is impossible for any reflector shape to perfectly collimate photons emitted from different points. As the size of the luminance area continues to increase, more and more photons emit further from the focal point and those photons divert further and further from collimation. If we then increase the focal length of the parabolic reflector, we increase the size of the parabola, which proportionately decreases the diversion of the photons that are further from the focal point.

Please refer to the following diagram:

In the diagram, the blue rectangles are the luminance area. The orange lines are the collimated light from photons emitted at the focal points of the parabolic reflectors. The blue lines are the diverted light from photons emitted at the opposite end of the luminance areas furthest from the focal points. The green lines are the extent of the angles from the luminance area captured by the reflector.

With optical system "A" as a reference, Optical system "B" has double the luminance area, and with the same focal length reflector, double the diversion.

Optical system "C" also has double the luminance area of system "A", but system "C" has also double the focal length of system "A", so the net effect on collimation is nil, and they collimate with exactly the same amount of diversion. System "C" is simply larger in relation to system "A", both the luminance area and the focal length are double in size. Divergence is the same because scaling size does not affect divergence angles.

But if you look at the green lines for system "C", you will also notice that it is capturing less of the angles from the luminance area. It now has less "light gathering" than both "A" and "B". The extent of this amount is dependent upon the luminance distribution pattern of the luminance area.

--- The above is from a discussion about reflectors, but the same relation applies to lenses.

 

[gcbryan]

Get-lit, I wasn't saying "no focal length" as in a flat pane of glass!

I was referring to comparing, for instance, a 28 mm diameter aspheric with a fl of 24 mm with a 28 mm diameter aspheric with a fl of 48 mm.

The center beam in the focal point is the same with either lens. It's only the beam off center that is larger or smaller and that doesn't contribute to max throw. It contributes to lumen.

https://www.candlepowerforums.com/threads/143017

Read post #23 for the point I was trying to make.

 

[Th232]

The center beam in the focal point is the same with either lens. It's only the beam off center that is larger or smaller and that doesn't contribute to max throw. It contributes to lumen.
.

While you and Ra are correct that only the light that's perfectly on the focal point will go parallel to the lens' axis, I think I know what's missing.

Take an image of the die that is a few microns off from the axis. While almost none (or none at all?) of this light will go exactly parallel to the axis, that light is still going to be projected roughly forward. This overlaps with the on-axis beam, but as we go down range this beam gets wider and moves further away from the on-axis beam. However, at a distance of, say, 100 metres the beam will still cover most of the area that the original beam did.

The definition of lux (as per my previous post) is lumens/m^2. You're now putting more lumens on roughly the same surface area, therefore lux increases, therefore throw increases.

At a range of infinity, then yes I'd agree that only the beam at the focal point matters, all other die images will have widened to the point of uselessness. That said, I can't see out to infinity with any reasonable degree of resolution.:shrug:

Get-lit, I wasn't saying "no focal length" as in a flat pane of glass!

I was referring to comparing, for instance, a 28 mm diameter aspheric with a fl of 24 mm with a 28 mm diameter apheric with a fl of 48 mm.

Practically speaking, he is right. Why not instead compare aspherics with focal lengths of 24 and 72 mm, or 24 and 1200 mm, or 24 and 10^6 mm? By that last example, you're practically at a pane of glass.

I've often found taking extreme cases of something to be good, not just for finding upper and lower bounds, but also as a sanity check.

 

[gcbryan]

TH232, I disagree (by the way, I'm no expert and I'm hear to learn and discuss).

For instance, let's say a flashlight with a 28 mm aspheric lens with a fl of 15 mm puts a spot 5 feet wide on a building 300 feet away

Let's say a flashlight with a 28 mm aspheric lens with a fl of 20 mm puts a spot 4 1/2 feet wide on that same building 300 feet away.

Is one throwing further than the other? No, the spot is just larger with one.

You're agreeing that at infinity throw isn't affected by focal length and you (I think) would agree that at the 300 foot example above it's not about throw as you can see both spots (one is just bigger) so to me it's not about throw.

I assume you would argue that at some point the dynamic would change and it would be like a laser. I have a smaller green pointer laser that I guess you could say does project further than I can see it. However, that doesn't fit the definition of throw either. If I can't see it...that's where throw stops.

The point-like beam of my laser does diverge and therefore I can see it for a long way (it makes the same size image on my retina). When it no longer does that its reached the limits of throw. If the beam had sidespill and was wider for part of it's distance that wouldn't be throw anymore than it is with a flashlight.

If you can see it without the less intense beam then it's just lumens and not increased lux (intensity).

 

[Walterk]

@GetLit; Can you correct me or give an example in a way that we can follow where the numbers and units come from ?
The F-number dictates light-gathering (viewing-angle) and collimation, so really keen on understanding how to work the numbers.

Relative Candlepower = Relative Source Intensity x Relative Light Gather x Optic Efficiency x (Relative Optic Focal Length)^2

Source Intensity :250 Lux/mm2 ( Suface brightness of Cree XRE 250Lumen/1 mm2 Led-die Area)
Light Gather : 40 (40% from FluxvsAngle Led-datasheet following viewingangle lens. )
Optic Efficiency : 90 ( 90% for standard condensor lens made of B270 glass.)
Optical Focal Length: 0.7 (For a lens with f# 0.7 , focal length 70mm and a diameter of 100mm )

250 x 40 x 90 x (0.7x0.7) = 441.000 Relative Candlepower

How do you get from the calculation (Relative) to real life measurements (Absolute), do you use a correction factor, interpolated from measurements of other lights?
As in the end I want to be to predict the beam intensity, and relate/verify the outcome to real-life Lux-measurements at distance.
(FYI: Relative Divergence = Luminance Area / Focal Length showed me to be close to my own estimates. )

 

[Th232]

I'd have to disagree that the above example isn't about throw, the reason one isn't throwing further than 300 feet than the other is that there's (obviously) a building in the way, nothing more. You're failing to take into account the brightness of each spot, which is how we can demonstrate which one throws better (note how everyone measures flashlight throw in terms of lux at a certain distance, from Saabluster to Selfbuilt).

Let's extend your example. Suppose I swap out the aspheric in your second light with an aspheric that has a focal length of infinity (yep, that's actually just a flat piece of glass and yes, it's still actually an aspheric lens). Given the focal length, it doesn't matter where we place it, so we'll do a straight swap with the original lens It lights up the whole building, but to such a low degree that you can't actually tell any difference. All I've done is changed the focal length of the aspheric, yet that one will definitely not throw far at all, it's now a mule.

By your same example, I can take a DEFT and a bare XR-E at the end of a black cardboard tube to make a similar size spot, shine them both at a wall 1 metre away. Using your logic, because I can see the same sized spot in both, I can therefore safely conclude that having an aspheric in place or not doesn't affect throw. Evidently this is wrong, because the intensity of the spot isn't being taken into account.

In the explanation I gave in my previous post, the dynamic doesn't suddenly change. The further the target is, the less important focal length becomes, until you're at infinity, where focal length is irrelevant. For much shorter distances (say, closer to 0 than to infinity), focal length remains very important.

I think we have a fundamental difference of the definition "throw" (didn't this happen in Saabluster's thread as well?). You mention that:

If I can't see it...that's where throw stops.

I think the issue with this is that it and takes into account usefulness (a subjective quality) as well as throw.

Suppose you can only see out to 100 metres for some medical reason. You have two flashlights, the first manages to put out a spot 2 metres in diameter with an intensity of 10 lux. The second manages to put out a spot, again 2 metres in diameter, but with an intensity of 100,000 lux. By your subjective definition, because you can't see past 100 metres anyway, both of those lights throw equally far.

For a person who can see further than 100 metres, suppose they take those two lights and shine them out to 600 metres, and sees only the spot from the second light. By his definition, the second one throws better. From what you've said above, both of these definitions are right?

For reference, my personal definition of throw is in terms of on-axis intensity, and makes no reference whatsoever of its usefulness. That includes lasers. For comparing usefulness, throw and flood beams, then I take FWHM into account.

If you can see it without the less intense beam then it's just lumens and not increased lux (intensity).

Was that supposed to be "with" the less intense beam? If it is, then all I have to say is that lux = lm/m^2, for a given size spot, more lumens = more lux.

Out of curiosity, how much experimentation have you done with aspherics? I'm not trying to say I've got better credentials than you or anything, I just want to see if I can describe things in terms of your past experience with them.

 

[gcbryan]

I'd have to disagree that the above example isn't about throw, the reason one isn't throwing further than 300 feet than the other is that there's (obviously) a building in the way, nothing more. You're failing to take into account the brightness of each spot, which is how we can demonstrate which one throws better (note how everyone measures flashlight throw in terms of lux at a certain distance, from Saabluster to Selfbuilt).

You're missing my point (my fault). Yes, I'm aware that the building is keeping the beams from throwing and I'm aware that throw is measured in terms of lux.

My point was that the difference in the two beam diameters in that (obvious) case was just a difference in size (lumens). So if you moved the building out of the way the one with the greatest intensity (collimated beam) would throw the furthest and the size of the off center beam wouldn't have an effect.

Out of curiosity, how much experimentation have you done with aspherics? I'm not trying to say I've got better credentials than you or anything, I just want to see if I can describe things in terms of your past experience with them.

I have some experience and I'm not claiming to have more than you or anyone else (I'm not saying that is what you are saying either).

I understand that there is a fine line between Z axis collimated max intensity lux and the contribution of off Z axis lumen output. I posted a question in the RA Optic thread to this effect as I still have a few questions along these lines.

In general I've found there to not be many "experts" around here on this subject and therefore some misinformation. So I try to work things out for myself through experimentation and by asking questions. I still have a few questions.

 

[Th232]

You're missing my point (my fault). Yes, I'm aware that the building is keeping the beams from throwing and I'm aware that throw is measured in terms of lux.

My point was that the difference in the two beam diameters in that (obvious) case was just a difference in size (lumens). So if you moved the building out of the way the one with the greatest intensity (collimated beam) would throw the furthest and the size of the off center beam wouldn't have an effect.

Thanks for the clarification, I think I can what you're trying to get at now.

Just to further clarify terminology, when I'm talking about the on-axis beam, I'm talking about just that one projected image that is collimated perfectly (i.e. all the rays are 100% parallel to each other and the aspheric's axis), this coming from the part of the die that is directly below the aspheric's axis. The off-axis beam for me is every other image that doesn't come from that spot. Are those your definitions?

Here's an optics applet:
http://webphysics.davidson.edu/applets/Optics/intro.html

It's rather basic, only has point sources and so on, but the laws of refraction are simple for point sources, and it helps to illustrate my point. Here's what I've done with it:

Fairly simple, one point source, one lens (note that refraction happens at the centre, but that's irrelevant for the purposes of this exercise) and one aperture, since the applet doesn't allow us to change the lens diameter. Same effect though.

Top image is the bit of the die that's directly under the lens' axis. Obviously, everything is parallel to the lens' axis. Second image is for a bit of the die that's a bit off-axis. Note how at short distances, the second beam still covers the first, but as we reach the right side of the image, we've almost got two completely separate projections.

Observe here that in real life, you've got much smaller "steps" in the off-axis distance, and this happens in 360 degrees around the axis. Lots of overlapping.

In the third image, I've extended the focal length. Obviously, since the lens (well, aperture in this case) is still the same size, the beam coming out of the lens is tighter. Fourth image is the same offset that I've introduced in the second image.

Observe the differences between the 2nd and 4th images and where the rays cross the centreline. (Side note: the bottom ray of both has been cut off, note that while this affects both, it makes the 4th image look a fair bit worse than it actually is). In the second, let's put an object where the bottom ray crosses the axis. It's now lit up by both projections, and there's a 50% overlap between them, leading to a doubling of the brightness in that region.

In the 4th image, put an object at the same distance. Now observe that although the beam is tighter, the off-axis projection covers a greater proportion of the on-axis projection. Also note that the angle those rays make with the lens axis is much less than that created by the rays in the second image, confirming that this second setup results in a more collimated beam. That bit can be done by visual inspection or just by using trig.

As I've said before, as the target approaches infinity, the focal length becomes less important. This is due to the fact that for the light from a certain point to still light up the target and not completely miss it, it must be ever closer to the axis, until you hit infinity and no other projection ever hits the target.

Therefore we can conclude that by only changing the focal length, while the first setup will gather more light (which I think we all knew), the second will result in a more collimated beam. Hence the throw will increase, but with the reduction in light being gathered the usefulness of such a beam will decrease.

Finite element analysis would be great here, since it would allow us to more accurately quantify the effects of focal length.


Walter, my apologies, I only just noticed this:

Can you explain more about 'The bigger lens will throw further, because the LED more closely approximates a point source.' ?
Is that because the actual led is relatively smaller for the big lens, and thus the beam more narrow and thus more intense/concentrated ?

That is correct. :thumbsup: It can be described either way, depending on whether you're looking at things from the point of the lens or the LED.

 

[gcbryan]

TH232, can you tell me how you define "infinity"? Thanks.

(I'm asking as in one sense (to me) it means an infinite distance never reached and in another sense as marked on camera lenses it can be quite close.).

 

[Th232]

In this case, I'm defining it as a distance never reached.

 

[gcbryan]

I guess my question regarding the charts you've provided is whether that's the way it works regarding the off axis beam.

When you talk about a subject being at a point where the off axis beams overlap you say the throw (intensity) increases.

I would ask...does it? For intensity to increase I would think that you would have to add collimated beams to collimated beams.

 

[Th232]

I would say so, since lux is defined as lumens/m^2 on a given area. In both examples, say we have each projection putting out 1 lumen*, the overlapping areas will have the same intensity, but in the second example we'll get an overlapping area for a much greater distance when compared to the first.

Shine two dim lights onto a surface from an equal distance, and the intensity increases, regardless of where the lights are positioned. The catch there is that the beams are nowhere near parallel, so with a focus that's decidedly not at infinity, it won't throw. You can do similar with a magnifying glass and the sun, the output from the magnifying glass most definitely isn't collimated, but it gets intense enough to burn paper, leaves, small insects &c. The lumens captured is identical, but it gets concentrated into a smaller area, resulting in a higher lux.





*For the physicists, engineers, mathematicians and others who know integration, yes, I know I should be using an "infinitesimally small number" instead.

 

[gcbryan]

It only happens at a point in time and one in which the beam isn't at max throw anyway. By the time the beam needs any "extra throw" the beams have already diverged haven't they?

 

[Th232]

Bit uncertain about what you mean by "extra throw" and "max throw", could you define those terms?

 

[gcbryan]

Bit uncertain about what you mean by "extra throw" and "max throw", could you define those terms?

Not very scientific I'll admit...my point is that we were talking about focal length not effecting throw.

You say in effect that although at the limits of throw fl may not have much effect at practical ranges it does.

Your chart shows a cross over of less collimated beams at a certain point. You say that throw is improved at this point. That would be at a certain distance. After that distance throw would not be improved because the beams would have diverged.

So for the throw to be increased this cross over would need to occur close to the limits that we can perceive throw otherwise it would not be increasing throw at a point where it would make much difference.

In other words this cross over doesn't by definition continue on to infinity so it doesn't extend throw in that sense.

 

[Th232]

Ahh, right. Got it.:thumbsup:

You've got to remember that this is two points out of what is practically an infinite number (or at least the number of molecules in the phosphor). I wish the applet was able to show non-point sources, because it would clear things up with much greater ease.

You're indeed correct that the particular off-axis point in my example will be useless at certain points, namely before and after it overlaps with the central beam. But there're also points closer and further from the axis, and when our example point fails to overlap, there will be other points that do overlap. If you work out the integral, I'm pretty that increasing the distance will cause the intensity to fall off in line with the inverse square law.

When comparing the focal lengths, as mentioned previously, for a given point the projection through the lens with the shorter focal length will go off at a greater angle. If we break this down and take the molecule right next to the one that's under the lens' axis, that projection will continue on to "nearly infinity", but for the lens with the longer focal length, the projection will be "closer" to infinity.

In other words this cross over doesn't by definition continue on to infinity so it doesn't extend throw in that sense.

And that's why I've been saying that at infinity, focal length doesn't matter. More and more projections diverge until at infinity, none of them overlap. The focal length determines the rate at which they diverge.

 

[gcbryan]

I'm sure you've seen the RA thread on Optics.

What is your argument to his description of a viewer standing in front of a reflector not being able to tell if it's deep or not (which is focal length)?

It's only by moving off of the Z axis that someone can tell if it's a deeper reflector or not which shows that it produces a wider beam but not a brighter one.