Formula for calculating throw using aspheric lens

[Ra]

Hi guyzzz,

I think it is about time for me to step in..

Reading this and other threads where the throw discussion starts (or is the main issue..) I cannot help getting the thought that, propably unintentionally, these discussions end up in very long threads that incorporate some confusing remarks, definitely when they are not true! Don't get me wrong here, but this is not very helpfull for the ones that are not quite into this, but still want to know how things work..

Question is: Do you want to know the facts about throw, and live with them without understanding the detailed law's behind them, or not..? (most do propably not..)

We have to face that some members do not have the abillity to understand it all, and when those members are not satisfied with this, long, uninformative threads start to grow..

I sinceraly hope you're not offended by the above, it's absolutely not my intention to lose CPF-friends, but to some, these are maters which are quite demanding to the brain, and so don't allow space for complicated discussions..



Maybe it's a good idea for the OP to edit the first post with a section in which the short conclusions we all agree about are summed up, so members can find a short summary with all the facts about this matter, and can decide for themselves if they want to read how we came up with those facts..



That said,

It's time for a little input from myself:

Definition of throw: The abillity to enlighten distant objects.. It's as simple as that!

But: Different objects have different levels of reflection when enlightened, so we have a problem.

Ok, then lets agree on specify the throw of a torch by stating the distance from the torch at which 1 lux is measured (like most do..)

Don't forget to post the simple square law formula for people who want to recalculate the throw-distance when they need for example 3 lux.


-Throw is not lumens related: A laser pointer throws far but has very poor lumens output!

Years ago, I experimented with lenses and led's on a testbench, and came to the following hard conclusions: (all theoretical, based on actual tests)

-Within the range of aviable lenses: For all lenses with the same diameter: The focus length does absolutely not affect throw!-

If it does, something else is wrong, like difference in quality, or transmittance of the glass,
or the entre surface of the lens doesn't play along. Those are only a few things that can be wrong.

-For all lenses with the same diameter: Focus length does affect the amount of lumens, collimated into the beam, affecting the wideness of the beam.

So what is important about focal length: Angle of emittance that is grabbed:

Most sources emit their lumens in a wide area (for most led's about 140 degrees), so the more you cover that area with a lens or reflector, the more lumens you collimate into the main beam. But we're not talking about lumens on this thread, we're talking about throw.

-Lenses more easily give high throw: With led's, which are front-emitting ofcource, lenses are best suited as they grab the light in front of the source. Conventional reflectors are designed mostly for use with side-emitting sources and are less efficient with led's (but still work to certain extend, when you accept the lower efficiency..)

Theoretical:

If a omnidirectional source emits 250 lumens, the lux measurement at 1 meter should give 20 lux (250 divided by 4 times pi) This is called MSCP (Mean Spherical Candle Power)
That also means, that when you know the size of the source, for example led-die 1x1mm, you can calculate the surface brightness: 1x1x20 equals 20 lux/mm2

Then you simply need to know the surface of the collimator in use: For example, 30mm diameter aspherical lens.. 15x15xpi=706.85 mm2

A high quality lens copies the surface brightness of the source, minus the losses caused by surface reflections, absorbtions and stuff like that. So we need to know the effective transmission of the lens: Uncoated, that will be about 90%..

So here we are:

-Source: 20 lux at one meter comming from a 1x1mm source size

-Effective lens surface (always 2-D, seen from a distance..): 706mm2

-Lens efficiency 90% (note that this is the efficiancy for surface brightness, not for lumens output!!)

Source has 1mm2 surface, measures 20 lux at 1 meter: Source + lens will give:

(Lenssurface divided by sourcesurface) x lux @ 1 meter x lensefficiency:

(706/1)x20x0.9= 12708 lux at one meter ! So there is your formula...well not quite..

with the inverse square law, you now can calculate the throw:

Taking the square root from 12708 (which is the actual CP-output, as this already is at 1 meter) gives 113 meters as the distance at which 1 lux should be the measurement result.

Another formula for throw: Take a calibrated measurement at any distance from the source, but far enough to be sure that the entire lens- or reflector-surface plays along,
and multiply that measurement with the quadratic of the distance:

Example:
Torch-luxmeter distance is 100 meters reading 24 lux: Beam-CP output is 100x100x24= 240,000 B-cp


Now there is a catch: Omnidirectional means emitting in all directions.. (like the sun does)
Led's are not omnidirectional. So IMO, the only way to do this without much uncertainty, is to do a lux-measurement on the bare source at one meter, with a calibrated lux-meter,
and determain the source size
The remaining of the theoretics do have less uncertainties.


Notes: When a torch does not give the theoretically calculated lux reading at one meter, chances are that, among other things, the entire surface of the lens does not play along: You are too close to the torch, or you need to focus the torch.

When does one know, he (or she) has enough distance for an accurate measurement:

Double the distance: according to the inverse square law, you should measure 1/4 of the measurement at half the distance. If this doesn't apply to your results, something is wrong in the way you measure..Propably too close with the first measurement.


Like I said, this is matter that is not easy to understand for some. This is my effort to be clear about the facts in a post that hopefully isn't too long..

Any question's?


Regards,

Ra.

 

[Th232]

My reflector theory isn't quite up to scratch compared to my optic theory. Truth be told, I fail to see the logic in his argument, it seems to go:

* The object sees a 2D object
* Deep reflectors collect more lumens, resulting in a wider beam (I personally dispute this bit, see below), and vice versa for shallow reflectors.
* Therefore focal length doesn't matter

Could you try explaining it to me if you think I'm completely missing his argument (which I think I am)? Because as is, I just flat out can't see the logic.

Ok, deep reflectors collecting more lumens. While depth is one function, width also plays a role. Here are two parabolae I plotted.

The green one is obviously deeper than the blue one. Focal points for both are at 0,0. The green one is obviously deeper than the blue one (also wider and with a greater surface area), but as is obvious, the blue one will collect more light due to its width:depth ratio, which is better than that of the green parabola.

In case anyone's curious, that's y = x^2 and y = 3x^2, adjusted so the focal point is at (0,0).

The blue reflector will, however, result in a wider hotspot and less throw for the same reasons that a shorter FL aspheric will result in less throw as in my previous posts.

As for the object seeing a 2D surface. I'm not quite sure what Ra is trying to get at, but think about this for a minute. Take an object with a reflector + LED illuminating it from 100 metres. Now take a single point on the object. All light striking it comes from the reflector or the LED. These rays are evidently not parallel to each other, regardless of their source, or whether this point is on the axis of the reflector or not. Draw the lines from each point on the reflector/LED to the point on the object. As the rays are not parallel, this is the only point where every one of those rays will coincide.

Move the object back 1 mm. Now we have an almost completely new set of rays striking the object. The sole exception is one ray parallel to the reflector's axis. It's either coming LED or the reflector, and note that some points won't have this ray at all.

Move the object back another mm. Same process repeats itself. Note the similarities between this process and the aspherics mentioned previously. Same deal, new projections shine on that point, but eventually we're out at infinity and the only rays are from the point of the die that is perfectly in focus with the reflector. Again as with aspherics, focal length determines the rate at which the other projections diverge from the axis for the same reasoning detailed in my previous post.

In short, as with aspherics, focal length is irrelevant if we go out to infinity, otherwise it plays a role. Also like aspherics, the longer the focal length the less light gathered, and the usefulness of the beam diminishes.

 

[Th232]

And in the meantime Ra posts! Thank you very much for that.

One small question though. Regarding your calculations, I agree with most of it, but you seem to be making the assumption that the whole LED is in focus, when (as we know) a reflector or aspheric has a focal point, not a focal plane, and hence while one point on the LED will be in focus, the other projections diverge and/or will not be parallel to the reflector/aspheric's axis as per the diagram in my earlier post.

How do you account for the fact that nearly none of the LED die will be in perfect focus? In my experience the focal length determines the rate of this divergence (again, as with the raytracing diagrams in my previous post), and hence the intensity of the beam over long distances. At short ranges the difference is quite small, but the effect becomes more pronounced over longer distances.

 

[Ra]

@Th232,

As for post #42:

Yup, you're missing the point: When I talk about the deepness of a reflector, then I talk about a reflector with the same diameter! So now draw two reflectors with the same diameter and different focal lengts: You'll see that there is a difference in angle from the source that is covered by each reflector..

And about the 2-D approach: Look at the sun! (use a filter!!) It's a giant ball right? WRONG!!
it's a flat disk!! Due to astronomics, you know that the sun is a ball. Now forget everything you know about the sun and look again.. Can you tell that it's a ball? It's not brighter at the center (not that much that we'd notice it) No, it's a 2-D disk with a certain (very high!) surface brightness..

It's the same with reflectors! look at them from a distance, and you cannot tell wether they have depth at all !

Lets post a picture:

Two operating torches seen from a distance: Can you see the deepness, no silly, it's a picture!! But it's not different when you look at the torch from a distance, with one eye, right? Same as the sun: From a distance, the enlightened object only 'sees' a 2-dimensional disk or surface, with a certain diameter and a certain surface brightness!

The amount of lux that is receved by the object is only determined by the (apparent !!) diameter and surface brightness of the light-source (sun, torch, candle...)
When you have lower surface brightness (halogen at the left) you need a larger reflector diameter: Both torches displayed have the same throw! At the right my Mini-HID..


Hope this more clearly shines some light at the subject..


Regards,

Ra.

 

[Ra]

And in the meantime Ra posts! Thank you very much for that.

One small question though. Regarding your calculations, I agree with most of it, but you seem to be making the assumption that the whole LED is in focus, when (as we know) a reflector or aspheric has a focal point, not a focal plane, and hence while one point on the LED will be in focus, the other projections diverge and/or will not be parallel to the reflector/aspheric's axis as per the diagram in my earlier post.

How do you account for the fact that nearly none of the LED die will be in perfect focus? In my experience the focal length determines the rate of this divergence (again, as with the raytracing diagrams in my previous post), and hence the intensity of the beam over long distances. At short ranges the difference is quite small, but the effect becomes more pronounced over longer distances.

You posted just before me...

Yep, when you want to have optimal throw, the one thing you should make, is the assumption that the led is in focus!! It is: With a high quality lens (in focus) an image of the light-chip (or die) is projected at the wall further away, proving that the entire die is in focus! The size of the projection at the wall is (only!!) exactly determined by the size of the die itself, the focal length of the lens, and the distance to the wall !

And:

Every lens or reflector has a focal plane!!
An infinitely small lightsource doesn't exist, so with a led die of 1x1mm, you already are using a focal plane, right?
Indeed: One point of the led-die will be in focus, but the other parts are in focus as well, but will be projected away from the optical axis, forming the image on the wall.
I now assume we have a high quality lens, when we use a low quality lens, you will not get a distinct image of the die, but a blurry spot with a bright center, more like when you use a conventional reflector.
Now the of-axis part of the source (die, arc or filament) causes what we call sidespill..


Regards,

Ra.

 

[Th232]

I think we're getting somewhere now!

Every lens or reflector has a focal plane!!
An infinitely small lightsource doesn't exist, so with a led die of 1x1mm, you already are using a focal plane, right?
Indeed: One point of the led-die will be in focus, but the other parts are in focus as well, but will be projected away from the optical axis, forming the image on the wall.

Let's just make sure we're on the same page.

Focal plane: The plane along which the rays from a point source will be projected, parallel to each other, in a certain direction, as per the second and fourth images in my previous post. Would that be your definition?

For points on the focal plane but not directly on the axis, they will be projected at an angle to the axis. Would you agree that the angle they make with the axis is dependent on their distance from the axis and the focal length? Specifically that with a shorter focal length, the projections from a certain point on the focal plane will diverge at a greater angle.

Ah-ha! I think we have a definition problem.

I am defining as the "hotspot" of a beam as the on-axis part. I note that you're defining the off-axis part as the sidespill. Let's use that definition for this next bit. At short distances, the sidespill will still overlap with the hotspot, as per the raytracing diagram in my earlier post, hence increasing its brightness. After a certain distance however, it will no longer overlap, and instead illuminate an area outside the hotspot. Yes/No?

 

[Ra]

For points on the focal plane but not directly on the axis, they will be projected at an angle to the axis. Would you agree that the angle they make with the axis is dependent on their distance from the axis and the focal length? Specifically that with a shorter focal length, the projections from a certain point on the focal plane will diverge at a greater angle.

Yep, you got it !!!


Regards,

Ra.

 

[Ra]

I am defining as the "hotspot" of a beam as the on-axis part. I note that you're defining the off-axis part as the sidespill. Let's use that definition for this next bit. At short distances, the sidespill will still overlap with the hotspot, as per the raytracing diagram in my earlier post, hence increasing its brightness. After a certain distance however, it will no longer overlap, and instead illuminate an area outside the hotspot. Yes/No?

Yep.

Edit: As most of the sidespill is created at the reflector part close to the source (inner rim), it needs some projection distance to free itself from the central spot..

Regards,

Ra.

 

[Th232]

And so when we have a longer focal length, the off-axis projection will overlap with the on-axis projection for a greater distance, increasing the intensity of the hotspot over that distance.

In all cases though, the off-axis projections will move off the on-axis projection as we approach infinity and become irrelevant.

Funny, that's what I said back in post 26.

Ahh well, I think we've cleared all the confusion up?

 

[Ra]

And so when we have a longer focal length, the off-axis projection will overlap with the on-axis projection for a greater distance, increasing the intensity of the hotspot over that distance.

No, not quite... There is another matter you don't include: When jou project over a great distance, you focus your setup at that distance, so closer to the torch, depending on the source dimensions, only part of the reflector will play along!

Edit: To be more clear: When focused at great distance, the source is not in the focus-plane for the shorter distance, so does not use the entire lens for that shorter distance!
For that, the lens is too close to the source!

When you focus at infinity, the reflector or lens is not fully lit when you look at it from close by!

This causes the spot not change much in surface brightness, when you increase the projection distance (to prevent more confusion: Here I mean the surface brightness of the projected spot)

That's what I meant when I said one could be too close to the torch for a reliable lux-measurement. That can only be done at a distance at which the entire reflector or lens plays along. And from that point further away, the lux-readings should follow the inverse square law... (note that the atmospheric conditions must play along as well)

With Maxablaster (and it's very small source..) I need a distance of several hundered feet to obtain a decent lux-measurement! And for lux-measurements close to the end of the throw of the beast, at 6-7 miles, Sure atmospheric conditions are going to interfere!!
Mostly Maxablaster doesn't reach much over 4-5 miles (if that's not enough...!)
I only once reached 9 miles, under perfect conditions !!


Regards,

Ra.

 

[Th232]

Good point, didn't consider focussing the source for dedicated long distance beams.

 

[Ra]

I found some pics from an early throw-test I did with lenses with various focal lengths..

Lens with short focal length:

Longer focal length:

Even longer focal length:

Notice that the lux-readings are (almost..) the same! (plus or minus 0.2... imperfections in the setup..)

And notice the difference in projected spots: The shorter focal lengts grab more lumens from the source, not resulting in higher throw, but bigger spot's !

I needed to cover the sensor, leaving a tiny hole, small enough to make a possible point-measurement within the smallest projected spot (longest focal length..)

Edit: In all situations, the lenses were placed at exactly the same distance from the sensor (within 1-2 mm.). And for each situation, the same diafragm was used
to exactly obtain the same diameter during each measurement..


Regards,

Ra.

 

[Walterk]

Thank you all for your contributions these last productive days!
And welcome Ra, glad you helped me out.

One question: You say;

-Source: 20 lux at one meter comming from a 1x1mm source size
-Effective lens surface (always 2-D, seen from a distance..): 706mm2
-Lens efficiency 90% (note that this is the efficiancy for surface brightness, not for lumens output!!)

Source has 1mm2 surface, measures 20 lux at 1 meter: Source + lens will give:

Lenssurface x sourcesurface x lensefficiancy:

706x1x0.9= 635.4 lux at one meter ! So there is your formula...

I wonder, is it correct that I do not see the surface brightness in the calculation?
You take for sourcesurface the figure of 1.
Why not 20?
Would it be 9 in case of a 3x3mm Led-die?

 

[Ra]

I wonder, is it correct that I do not see the surface brightness in the calculation?
You take for sourcesurface the figure of 1.
Why not 20?
Would it be 9 in case of a 3x3mm Led-die?

If a 1x1mm source gives 20 lux at 1 meter distance, the surface brightness is automatically embedded in that fact..
I take 1 because most led-die's are one square mm..

EDIT: Uhh... :thinking: Wait a minute.. Walter, I think you're right. I forgot something very important !! It's 706x20x0.9 !!!!

EDIT2: Maybe indeed a bit confusing, taking a 1x1mm source because it doesn't change anything and so can be left out... The right formula: (706/1)x20x0.9) = 12.708 lux @ 1 meter

Now when the led-die is bigger, lets say 2x2mm (4 square mm), with the same 20 lux output at one meter, then the torch would give: (706/4)x20x0.9) = 3.177 lux @ 1 meter



Wait, I'm going to correct that right away !

Thanks for pointng this out !



Regards,

Ra.

 

[Walterk]

Thank you Ra ! Finally this thread brought me the answer to my question.

I am pretty sure I've read the same in the Optic theory-thread, but not clear enough to trust as only source.
I found al sort of hints here and there, but no one could state me a satisfactorily and watertight answer.
Although most people (38.3%) claim to know few care to share.
Many thanks to the people that did help, by contributing to this thread or by PM !
Finally I will accept the formula is as simple as that, and nothing more. Stubborn me.

Just wonder: How much variation is there in Lux within the complete spot ? I suppose close to none at near infinity?

NB In time I might rephrase some of the content in the first posting.
I leave the thread open for further contributions on light, rays, lenses and the distribution and properties of all this.

 

[gcbryan]

Thanks WalterK for starting this thread! TH232 I'm picturing focal length and how it affects the sidespill better now and thanks especially to RA for contributing and especially for bringing up the focus issue.

That seems to clear up the focal length aspect.

I have a couple of practical questions that many of you can answer I'm sure.

1). The formula for determining lux at 1 meter is based on a lens so pi R ^2. How would this formula be changed to work with reflectors?

2). In many threads I see lux at 1 meter vastly overstated according to the formula above.

Would the explanation for this be not moving away from the source to the point where the sidespill is starting to diverge from the hotspot before taking the reading?

 

[gcbryan]

1). The formula for determining lux at 1 meter is based on a lens so pi R ^2. How would this formula be changed to work with reflectors?

Anyone?

 

[Walterk]

1). The formula is the same for lenses as reflectors. Only difference is lenses are more efficient. The shape of the reflector works the same as F-# for lenses. (See also post44)

2) Inaccurate measurements, to close by.
Note that High power Leds are not omnidirectional.
In the White led test-thread, the Cree XRE shows to have 170Candela surface brightness on axis, not 250 or 340.

 

[gcbryan]

1). The formula is the same for lenses as reflectors. Only difference is lenses are more efficient. The shape of the reflector works the same as F-# for lenses. (See also post44)

2) Inaccurate measurements, to close by.
Note that High power Leds are not omnidirectional.
In the White led test-thread, the Cree XRE shows to have 170Candela surface brightness on axis, not 250 or 340.

Since led's aren't omni directional is that the reason for the 4 * pi part of the equation? 360 degrees/90 degrees (XR-E beam angle) =4 ?

 

[Ra]

Since led's aren't omni directional is that the reason for the 4 * pi part of the equation? 360 degrees/90 degrees (XR-E beam angle) =4 ?

The 4 * pi part is only part the optical law that states how many lux at 1 meter an omnidirectional lightsource would give based on the total lumens output of that source.
Like I said earlier, led's are not omnidirectional! But it is too difficult to make a formula that overcomes this problem.. So that is why I ommited that part, and took the road of measuring the lux@1meter for the bare source first and calculate the surface brightness using the (apparent..) source-dimensions.


And for the throw formula for conventional reflectors:

Walterk already said it but forgot to complete his remark: Lenses are more efficient than reflectors... on throw!!...




This is important: We're still talking throw here ! So the things below are about the reflector and lens efficiencies for throw.. If we do this for lumens-output, the results will be different (and TIR will be the king of all !!! Only with led's ofcource..)




Formula is the same, indeed you must incoporate the lower efficiency of reflectors: It has a central hole that does not play along, and the reflectivity of the reflective layer can be lower as well.
Lenses more easily have high effectivity (for throw) because they have less different factors that can ruin things.

Example: You only have to polish a lens to make it reach a transmission of 90%. Aplying a simple antireflex coating can increase the transmission to 95%.. That's what you need for throw..
Note that this is the same for glass based TIR optics: In a TIR there is no reflection layer, the angle of the lightrays inside causes 100% internal reflection at the inner-side of the reflector.

Now with conventional reflectors: The best reflection layer is silver at 97% minimum, but this degrades very quickly by oxidation of the silver. Second best is aluminium at 88%, but you have to work hard to reach that! More likely is a surface reflection of about 80-85%, which already is much lower than the transmission of an uncoated lens.

And... for throw, the entere surface must play along: Conventional reflectors have this black hole in the center, which only further decreases throw-efficiency!

Cleaning conventional reflectors almost always cause a decrease in efficiency, by damaging the reflective layer.


All these factors and their inconsistencies make it impossible to calculate the throw of all types conventional reflectors by only one formula !! The solution for that would be an efficiency-spec, delivered with the reflector... But then again, how many manufacturers lie about the specs of their products (30 Million CP Thor for example ??)
I wouldn't be surprised by reading a claim of 150% reflector-efficiency some day.. :thinking:

BTW: When conventional reflectors are used with led's you can somewhat increase throw by glueing a lens on the front window which coveres the 'black hole' of the reflector and brings some of the lost light in collimation as well.. Could mean 5-7% improvement on actual 'throw-surface'..



Regards,

Ra.