Formula for calculating throw using aspheric lens

Do you understand the physics of throw

  • I don't but want to

    Votes: 16 18.8%
  • I want to understand better how throw works

    Votes: 27 31.8%
  • I understand enough to decide what's working

    Votes: 17 20.0%
  • I think or am pretty certain I know how it works

    Votes: 25 29.4%

  • Total voters
    85

Walterk

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

Summary: (from posts in this thread )

- 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 .

- There is a simple ' inverse square law' formula to recalculate distance and lux to go from candlepower to throw-distance and vice versa from Lux-measurement to candlepower.
Rephrased by Walter this comes to: Source intensity (Candlepower) = Intensity (Lux measurement at chosen distance ) / Distance x distance (in meters, of chosen point where the Lux-measurement was or will be).

- Throw is determined only by three things: Lensdiameter (or reflector diameter), surface brightness of the bare source, and the (throw-) efficiency of the lens (or reflector)
Basic formula for calculating throw:

(lenssurface divided by apparant
sourcesurface (all in mm2)) x [email protected] meter (bare source) x Throw-efficiency lens (or reflector).

- Surface brightness of the source; measure the candlepower-output of the bare source first, divided by the surface-area of source. 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, then divide the lux-measurement by the mm2 surface of the source.


- We're talking about throw: Reflectors and lenses have two types of efficiency: Efficiency for throw and efficiency for lumens output (torchlumens).
Throw-efficiency of lenses is almost always (sometimes much..) higher than of reflectors with the same diameter. (on lumens-efficiency it's mostly the other way around !)

For throw-efficiency: 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%, when coated with anti-reflective coating this can be 95%.

- F-ratio of the lens (or depth of reflector) does not affect throw, it only affects total lumens output (torchlumens, wideness of the beam and sidespill).
- Within the range of aviable lenses: For all lenses with the same diameter: The focus length does absolutely not affect throw!
- 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..)
- Throw is not lumens related: A laser pointer throws far but has very poor lumens output !

When putting this to the test: Always use a stable power supply, and the same source (led or bulb). Never use batteries !
When comparing collimators on throw: Always check that the entire surface of the collimator plays along at the test-distance !
It's best to use calibrated equipment (lux-meter).

Theoretical example of calculations:

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

(Although usually the source will be a unidrectional source like a high power starmounted Led. Then calculate surface brightness from Lux-measurements/mm2).
Then you simply need to know the surface of the collimator in use: For example, 30mm diameter aspherical lens.. 15x15xpi=706.85 mm2

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 apparant 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.


- 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.
Most likely flaw; second measurement too close with the first measurement.
The amount of lux that is received by the object is only determined by the (apparent !!) diameter and surface brightness of the light-source (sun, torch, candle...)
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 on very long distances atmospheric conditions play along as well)


Edit 24.06.2010: My thread-starting-post was :
I really would like to see a formula that predicts beam intensity and throw, and then calculated back from real life measurements corresponds (for the larger part ) to the assumptions.
I am happy with the results, I think we 've got it pretty much covered ! It sure helped me understanding light and lenses. Now will start experimenting and build me a light !
 
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Batou00159

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:eek: somone else with numerical dislexia

If this thred is as good as some of the others have beenit will make my life so much easier cpf rules

sorry this dosent help much dose it
 
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gcbryan

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There is info on here regarding throw. The only problem is that the threads get so long with definitions and theory that it quickly loses its usefulness.

As far as I know it comes down to surface brightness of the led and the diameter of a reflector or optic and depth of a reflector plays a part as well as to it's usefulness.

Everything else is related to surface brightness and reflector/optic diameter.

The XR-E R2 has the greatest surface brightness. The other bright emitters that have greater output are larger but don't have greater surface brightness.

The larger the diameter of a reflector the more light collected but the main factor is that the further from the emitter the more the light is like a point source and so it will be brighter and more tightly focused (collimated).

With optics there are some factors to be considered (I don't have as much practical experience here) but practically speaking I think closer with a greater diameter is better for capturing the light and a longer focal length is better for focusing the light so I guess there is some optimal balance here.

I don't think in terms of formulas so I'll leave that to others.

Good thread. I hope we all learn something.
 
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gcbryan

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I should add if someone understands throw they should be able to briefly and concisely describe it. In the past no one did this.

I'd hate to see this thread turn into a long illustrated textbook of confusion.

Throw is how far a light will shine. The individual has to decide after what point the light isn't useful. A laser throws a long way but the spot isn't large enough to be useful but it does throw.

Some people will choose a thrower with a reflector because the spot of light may be larger and more useful and some will chose an optic which may result in a smaller spotspot and may be less useful to some.

This is all throw however and doesn't need to be defined beyond this as the rest is all individual preferences.

I hesitated to even post all this but I'm hoping it will eliminate the thread from going in this direction (as all other such threads have).
 

kengps

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What you're gonna find is that "Throw" is a subjective thing. Look at Saablusters post about "the mechanics of throw" Or something like that. Maybe the title changed since I seen it last.
 

Walterk

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Thanks for your input. Hope this thread will not get muddy. Indeed let us not talk definitions and throw.

I want to be able to predict the Lux measurent within the spot at a known distance and spot-size, so I am about halfway:
Surface-area of the lens x Geometric correction factor x Surface brightness of the Led-die gives Beam-intensity expressed in Lux.
Inverse-square law gives the light fall-off at a distance.

How do you do it ?
 
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gcbryan

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I'm not a numbers guy but in chatting with someone who is I've been playing around with the following (just to get a framework for how all this fits together in my mind).

To predict potentially how far a particular light will throw you take a lux reading at 3 meters or so and convert it back to 1 meter.

(by the way does anyone have a lux reading for the Uniquefire HS-802?)

You also have to come up with a minimum lux figure so let's say 1 lux.

If the lux reading from 1 meter is 30,000 (for instance) the math would be squared(30,000/1) or 173 meter would be the throw at which your subject would still be receiving 1 lux.

If we standardize on an emitter (XR-E R2) driven at 1A then the only changes would be in the reflector or optics.

I think 1 lux might be a reasonable figure to use for the minimum illumination.

I'd like to know what the optical laws say regarding increasing diameter as it relates to lux. It's not linear I don't think so what is it?

If you have an aspherical lens of 30 mm with a lux reading of 15,000 lux what is the predicted lux with a 60 mm lens?
 

Walterk

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Throw is determined by surface brightness (lux/mm2) x diameter (mm2) x efficiency.

So if efficiency and surface brightness is the same, then:
Diameter lens 30mm, then surface is 706mm2.
Diameter lens 60mm, then surface area is 2827mm2.
So, two times the diameter is 4 times the surface area.

30mm Lens then 1 x 15.000Lux.
For 15.000Lux inverse square gives 1 Lux reading at 122 meter.
60mm Lens then 4 x 15.000 => 60.000Lux.
For 60.000Lux inverse square gives 1 Lux reading at 244 meter.
So, two times bigger lens gives two times effective throw-length.

That the beamshape might be different, makes no difference for inverse square.
I don't know if the focal length/Geometric correction factor makes a difference, I think it should.
 
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gcbryan

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Throw is determined by surface brightness (lux/mm2) x diameter (mm2) x efficiency.

So if efficiency and surface brightness is the same, then:
Diameter lens 30mm, then surface is 706mm2.
Diameter lens 60mm, then surface area is 2827mm2.
So, two times the diameter is 4 times the surface area.

30mm Lens then 1 x 15.000Lux.
For 15.000Lux inverse square gives 1 Lux reading at 122 meter.
60mm Lens then 4 x 15.000 => 60.000Lux.
For 60.000Lux inverse square gives 1 Lux reading at 244 meter.
So, two times bigger lens gives two times effective throw-length.

That the beamshape might be different, makes no difference for inverse square.
I don't know if the focal length/Geometric correction factor makes a difference, I think it should.

I don't think it does make a difference in throw. It's more of a lumen or spot size thing rather than an intensity thing (focal length that is).

I think it's interesting to be able to adjust for environment as well. 1 lux at 122 meters may allow for subject identification if the subject is a white building and 2 lux may be more reasonable if the subject is non-reflective such as trees in the forest so throw then becomes 86 meters at 2 lux.

I'm running into this in testing some of my lights. People are listing throw figures that are more than I'm getting but in looking at their beamshots I see that their subject is a white building while my local test environment is (non-reflective) trees.
 
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gcbryan

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But what about the GAP between calculation and measurement!!

Surface lens 706mm2 x 250Lux/mm2 x 60% efficiency = 123.550 Lux

That is 10 times more the Lux then you expected from measurements.....

How to explain this?

I'm sorry but I don't understand what you are trying to say here?
 

gcbryan

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I've edited-ed the post, hope it is more clear.
Anyway, still working on it, and waiting for my Lux-meter.


Where are you getting this figure...250Lux/mm2.

I guess I don't understand what you are referring to when you end up with 123,550 lux from a 30 mm diameter lens? That's not happening anywhere that I know of.

Is the 250 Lux really meant to be 250 lumen (max output for a XR-E)?
 
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gcbryan

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Sorry, got the formula screwed up: have corrected it as follows:

706mm2 x 250 Lux/mm2 x 0.6 efficiency = 105.900 [email protected]

Where indeed 250lux/mm2 is the surface brightness of the Led.
For [email protected] at typical Ampere:
250 Lumen / Surface area of the die 1 mm2 = 250 Lux / mm2

I don't think it works like that. You are using lumen figures for lux figures and then wondering why there is a gap between calculated and observed values.

250 lumen/surface area of 1 mm2 = 250 lumen/mm2

It's still a lumen figure and therefore doesn't belong in the formula for throw (unless I'm missing something obvious).

The 250 lumen /mm2 figure is a way to compare XR-E to other emitters to see which has the greatest surface brightness but the number itself isn't a brightness (lux) figure. It's simply figuring lumen output as a ratio to size.
 
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Th232

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Concur with gcbryan, you're mixing up lux and lumens, all that bit is good for is the relative surface brightness between different LEDs.

1 lux is defined as 1 lm/m^2, and is a measure of intensity. Don't forget that 1 m^2 = 1,000,000 mm^2.

BIG catch though. Lux is the light that hits a surface, not the light emitted.

Further thoughts:
The problem with this is that the physics are simple for a single photon, but when you have to take into account every angle it gets rather messy, especially since an aspheric lens may not have a nice geometric formula. The jump from a single case to making a simple formula is quite tricky if you want it to be right. How much error are you willing to tolerate?

I personally can't be bothered doing all this myself, but try this:

* Find several lenses, different focal lengths, same diameter. If you can't do that, tape off the lenses from the flat side so they have roughly the same effective diameter. Note that this will break down at shorter focal lengths.
* Test each of them and plot the width of the beam vs the focal length. Find a relationship (if any).
* Find several lenses, same focal length, different diameter. Put them in and test flux. Find a relationship (if any).
--- Optional: Make readings at points in, say, a 10x10 grid. See how different they are. This could be very useful later on.
* Go through the datasheet for the LED you're using, calculate the total output over a set angle from the axis. Find out how many lumens are being collected by each aspheric.

By that stage you'll have a good idea of how focal length and diameter actually influence both the lumens and lux. I'm not trying to say this in a condescending manner, but as long and tedious as the above procedure is, it'll really give you a much better idea of how it all works.

That'd be a good start, and at least get you a fair formula for the LED you're using. Then repeat the process with other LEDs. To be honest, I'd start off this whole process with LEDs that have a Lambertian emission pattern (i.e. not XR-Es), so then you can easily determine differences due to apparent die size. They obviously won't throw as well as an XR-E, but at this point you're just looking for a formula, aren't you?

On focal length, here's an additional step in the above experiment you might try. The following aspherics have different focal lengths, but both collect the same amount of light:
aspherics.png


See what happens. The bigger one will throw further, because the LED more closely approximates a point source. How much further it'll throw is something I leave to you.
 
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Launch Mini

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Without getting into numbers , I think the following makes sense.

I would think that the tighter & smaller the beam, from any light source, the further your throw will be. Yes a larger lense can collect more light, but there is only so much light being emitted from the LED.

Think of this in terms of a garden hose with and without a nozzle.
How far does your water spray without anything? ie no lens
Then restrict that same flow into a narrow opening ( lens) and you get greater distance.
Disperse that flow over a larger area and less distance.
Now gravity takes effect with water, but this is my thought on the matter.

You still need the math to determine ulitimate throw but I'll leave that to the experts.
 

get-lit

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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.
 
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