I have seen a need in recent times to create a thread that will help people to understand the physics behind throw. It can be a very difficult thing to understand as intuition would tell you that a 2000 lumen light would out-throw a 300 lumen light every time. This is not necessarily the case for a variety of reasons. The recently released Olight SR90 seems to have shown light on a lack of true understanding among the CPF populace as to what makes a light throw. The reason for this confusion seems to stem from its extraordinary performance in the throw category while at the same time being an ultra-large die LED. People assume that it out throws the smaller die lights like the DBS because it is just putting out so much more light. There is no such thing as just using brute force to overpower the weaker small die LED lights. So what is it?
Let's try and understand the fundamentals and the reasons that one light has more throw than another. We will examine HID, incandescent, and LED flashlights and the various collimation methods used to create throw.
I am starting this thread even though I don't have the time to have it be fully featured like I would like but we can get the conversation rolling and I would appreciate any suggestions from those of you out there that might have a good way to illustrate it in a way that is easily understandable. Please refrain from saying what you think is the truth. This thread is intended to be a resource for CPF and to clear up confusion not create it so please think and read before you post. Questions are good and can help to refine the information or layout of the first post if some have a hard time understanding with the given information or layout. I will work as best as I can as time permits to refine this thread's first post.
Reading notes-The intent of this is to enlighten the CPF layperson and therefore strays away from uber technical terms such as luminance or the adherence to literal or arcane understandings of such words as throw. I use terms that are in common use here on CPF but the scientific principles that this "argument" is based on is intended to be sound. I do not wish to hear from those who are trying to be overly literal or picky in the minutia. If there is a fundamental flaw in the arguments presented here then please speak up.
A request has been made to include a comprehensive definition of what "throw", as it relates to our usage in the flashlight word, means.
As with most things I do there was a great deal of thought that went behind it so before making suggestions otherwise think it through and have a well reasoned "argument" for its change if you see the need.
Definitions
Emitter- A device from which photonic energy emanates(such as a bulb filament or LED die).
Surface brightness- A characteristic of a light-source by which emitted light at the emitter's surface can be ranked from light to dark in correlation with its intensity.
Throw-The ability of a flashlight to illuminate and enable identification of distant objects.
Surface Brightness
It all starts with this one basic thing. This is the foundation. The higher the surface brightness the more the throw potential. This does not guarantee the most throw however as there are a lot of other factors that can ruin all that potential.
So what do we mean by surface brightness?
It can be defined as a characteristic of a light-source by which emitted light at the emitter's surface can be ranked from light to dark in correlation with its intensity.
Surface brightness is not dependent on the area of the emitter but the saturation of light within the area. The use of area in respect to surface brightness is only to quantify what the actual relative surface brightness is.
The emitter in the second picture can be said to have higher surface brightness. The LED die in this instance is aprx. 1mmx1mm. If the first emitter is putting out 10 lumens and the second one 100lumens you can calculate that the second one has 10X the surface brightness of the first one.
The very same thing applies to incandescent lights. The filament is the same for both pictures but one has more power going to it and therefore it is putting out more lumens. Since the area that emits light has not increased, only the amount of lumens in that area, we know that it has higher surface brightness.
If you have two lights of exactly the same spec in every way except the LED in one has higher surface brightness, the one with higher surface brightness will throw farther every single time. There are no exceptions to this unless you are God and can change the laws of the universe.
How can I say this with such certainty? Grab one of your multi-mode lights. Which setting does the light throw farther on? High or low? The only variable you have changed in this equation is surface brightness. You have increased the amount of lumens created without making the size of the LED die any larger thereby increasing the intensity. That means that on high there is now more light in the same amount of space(at the die level) and therefore it(the die) must have a higher surface brightness.
How do we figure out the relative surface brightness of say an LED of 9 square mm(SST90) vs an LED with only one square mm of emitting space(XR-E)? You need two crucial bits of information to make a calculation. Total lumen output and surface area of the emitter.
The top bin XR-E@1A(max per spec) puts out 270lumens inside of its 1mmx1mm box. 270 divided by the area(1mm square) gives us a figure of 270. Now let's do that with the SST90.
The top bin SST90 at maximum current is putting out 2250lumens inside of its 3mmx3mm box(9 square mm). To get the surface brightness calculation we take the lumens(2250) and divide that by the area(9) which gives us a figure of 250.
So at maximum spec the XR-E's surface brightness is 270 vs the SST90's 250. That means the XR-E has more throw potential than the SST90.
This same formula can be done with any type of light emitter such as HID or incandescent sources but it is far more difficult to figure out what your surface area is with these types of lights due to their 3 dimensional nature. Since LEDs are very easy to find the surface area of I have used those as examples.
Distance to Reflective/Refractive surfaces
This area of concern is a huge determining factor in how concentrated the beam from a light is. First there are some things we need to know to set this up so it will take a bit before you realize how this all relates to the distance of the surfaces away from the emitter.
Collimated light is light that is parallel or close to parallel when leaving the light. This means the beam does not spread very fast. Collimation allows a concentration and transmission of light farther than it would otherwise be capable of. This collimated light forms what is known as the "hotspot" of the beam.
Collimation in the truest sense of the word is not possible in the real world as it would require a point of light infinitely small. Assuming you did have the impossible point source of light, and this point was at the reflector or len's focal point, the resultant beam would never grow any larger than when it left the light. If I shone a light with a diameter of 80mm towards the moon there would be an 80mm beam of light on the moon.
Here in the real world we make due with collimation that we might call "good enough". Collimation in the real world is a relative term as there are degrees in the amount of collimization from light to light. All of our light sources in the universe regardless of what kind they are(xenon, LED, incandescence, stars) emit their light from an area not a point. Because the light is being emitted from an area and not a point we now have to contend with divergence.
What do we mean by divergence? At the root of that word it means it is diverging from something. From what? It is diverging from perfect collimation. The light emitted from the very center of the source, say an LED die, does come close to perfect collimation. The light from the edge of the LED however is not at the exact focal point(notice point not area) of the lens or reflector. (Keep in mind that while we can never have a point source light emitter, all of our collimation methods "require" a point source. NO EXCEPTIONS!) Only the center of the emitter can be there. Sure you "could" move the edge of the LED die to be in the perfect focal point of the lens but then the center of the LED is no longer at the focal point.
Distance as it relates to reflectors
The beam from a reflector consists of an infinite amount of images of the source laid one over the other. Imagine for a second we were only dealing with one teeny tiny spot on that reflector-say the middle. The beam that is projected will be an image of the emitter albeit very dim. Now move to a spot directly to the side(not closer or farther to the emitter). The same thing happens here. Light from points all across the emitter shine at that tiny spot on the reflector and the beam sent out will be another image. However since you have moved to the side, essentially circling the emitter it(LED's image) will be rotated slightly from the other one. You can repeat this infinitely going around the reflector so that you have image over image over image. This repeated overlapping eventually creates a circle instead of a single image as you would have with an optic.
Now while the basics of what I just described do not change when you move closer or farther to the emitter one thing does change. The size of image that teeny tiny spot projects. The closer that tiny spot is to the emitter the larger the emitter appears just as someone far away appears tiny to you even though your brain knows otherwise. So that tiny spot close in throws an image of the emitter but larger than when we were in the middle of the reflector. And the opposite is true when we move to the outer rim of the reflector. From this vantage point the emitter looks much smaller. Therefore the image thrown is much smaller. Now remember that this whole process is just laying image over image.
So you want to know what part of the projected beam consists of what images? Let's start in the center of the projected beam or the beam you would see on a white wall at 20 feet or so(beam has to have space to mature). The most intense round spot you see is mostly what has been reflected by the outer area of the reflector. It also consists of the less intense and larger images created by the lower portions of the reflector. So all of the reflector contributes to the center of the projected beam but the outer portions of the reflector do not add any intensity to the outer areas of the projected beam.
So to recap. The images from the outer portions of the reflector are the smallest. As we move closer the images get larger.
For more information see this video.
Optical System Efficiency
This is subdivided into two categories.
1-Efficiency of Reflective/Refractive elements
2-Emitter's Spatial Distribution vs Type and Diameter of collimating element.
This page is still under construction. Sorry I am such a slow typist.
Let's try and understand the fundamentals and the reasons that one light has more throw than another. We will examine HID, incandescent, and LED flashlights and the various collimation methods used to create throw.
I am starting this thread even though I don't have the time to have it be fully featured like I would like but we can get the conversation rolling and I would appreciate any suggestions from those of you out there that might have a good way to illustrate it in a way that is easily understandable. Please refrain from saying what you think is the truth. This thread is intended to be a resource for CPF and to clear up confusion not create it so please think and read before you post. Questions are good and can help to refine the information or layout of the first post if some have a hard time understanding with the given information or layout. I will work as best as I can as time permits to refine this thread's first post.
Reading notes-The intent of this is to enlighten the CPF layperson and therefore strays away from uber technical terms such as luminance or the adherence to literal or arcane understandings of such words as throw. I use terms that are in common use here on CPF but the scientific principles that this "argument" is based on is intended to be sound. I do not wish to hear from those who are trying to be overly literal or picky in the minutia. If there is a fundamental flaw in the arguments presented here then please speak up.
A request has been made to include a comprehensive definition of what "throw", as it relates to our usage in the flashlight word, means.
As with most things I do there was a great deal of thought that went behind it so before making suggestions otherwise think it through and have a well reasoned "argument" for its change if you see the need.
Definitions
Emitter- A device from which photonic energy emanates(such as a bulb filament or LED die).
Surface brightness- A characteristic of a light-source by which emitted light at the emitter's surface can be ranked from light to dark in correlation with its intensity.
Throw-The ability of a flashlight to illuminate and enable identification of distant objects.
Surface Brightness
It all starts with this one basic thing. This is the foundation. The higher the surface brightness the more the throw potential. This does not guarantee the most throw however as there are a lot of other factors that can ruin all that potential.
So what do we mean by surface brightness?
It can be defined as a characteristic of a light-source by which emitted light at the emitter's surface can be ranked from light to dark in correlation with its intensity.
Surface brightness is not dependent on the area of the emitter but the saturation of light within the area. The use of area in respect to surface brightness is only to quantify what the actual relative surface brightness is.
The emitter in the second picture can be said to have higher surface brightness. The LED die in this instance is aprx. 1mmx1mm. If the first emitter is putting out 10 lumens and the second one 100lumens you can calculate that the second one has 10X the surface brightness of the first one.
The very same thing applies to incandescent lights. The filament is the same for both pictures but one has more power going to it and therefore it is putting out more lumens. Since the area that emits light has not increased, only the amount of lumens in that area, we know that it has higher surface brightness.
If you have two lights of exactly the same spec in every way except the LED in one has higher surface brightness, the one with higher surface brightness will throw farther every single time. There are no exceptions to this unless you are God and can change the laws of the universe.
How can I say this with such certainty? Grab one of your multi-mode lights. Which setting does the light throw farther on? High or low? The only variable you have changed in this equation is surface brightness. You have increased the amount of lumens created without making the size of the LED die any larger thereby increasing the intensity. That means that on high there is now more light in the same amount of space(at the die level) and therefore it(the die) must have a higher surface brightness.
How do we figure out the relative surface brightness of say an LED of 9 square mm(SST90) vs an LED with only one square mm of emitting space(XR-E)? You need two crucial bits of information to make a calculation. Total lumen output and surface area of the emitter.
The top bin XR-E@1A(max per spec) puts out 270lumens inside of its 1mmx1mm box. 270 divided by the area(1mm square) gives us a figure of 270. Now let's do that with the SST90.
The top bin SST90 at maximum current is putting out 2250lumens inside of its 3mmx3mm box(9 square mm). To get the surface brightness calculation we take the lumens(2250) and divide that by the area(9) which gives us a figure of 250.
So at maximum spec the XR-E's surface brightness is 270 vs the SST90's 250. That means the XR-E has more throw potential than the SST90.
This same formula can be done with any type of light emitter such as HID or incandescent sources but it is far more difficult to figure out what your surface area is with these types of lights due to their 3 dimensional nature. Since LEDs are very easy to find the surface area of I have used those as examples.
Distance to Reflective/Refractive surfaces
This area of concern is a huge determining factor in how concentrated the beam from a light is. First there are some things we need to know to set this up so it will take a bit before you realize how this all relates to the distance of the surfaces away from the emitter.
Collimated light is light that is parallel or close to parallel when leaving the light. This means the beam does not spread very fast. Collimation allows a concentration and transmission of light farther than it would otherwise be capable of. This collimated light forms what is known as the "hotspot" of the beam.
Collimation in the truest sense of the word is not possible in the real world as it would require a point of light infinitely small. Assuming you did have the impossible point source of light, and this point was at the reflector or len's focal point, the resultant beam would never grow any larger than when it left the light. If I shone a light with a diameter of 80mm towards the moon there would be an 80mm beam of light on the moon.
Here in the real world we make due with collimation that we might call "good enough". Collimation in the real world is a relative term as there are degrees in the amount of collimization from light to light. All of our light sources in the universe regardless of what kind they are(xenon, LED, incandescence, stars) emit their light from an area not a point. Because the light is being emitted from an area and not a point we now have to contend with divergence.
What do we mean by divergence? At the root of that word it means it is diverging from something. From what? It is diverging from perfect collimation. The light emitted from the very center of the source, say an LED die, does come close to perfect collimation. The light from the edge of the LED however is not at the exact focal point(notice point not area) of the lens or reflector. (Keep in mind that while we can never have a point source light emitter, all of our collimation methods "require" a point source. NO EXCEPTIONS!) Only the center of the emitter can be there. Sure you "could" move the edge of the LED die to be in the perfect focal point of the lens but then the center of the LED is no longer at the focal point.
Distance as it relates to reflectors
The beam from a reflector consists of an infinite amount of images of the source laid one over the other. Imagine for a second we were only dealing with one teeny tiny spot on that reflector-say the middle. The beam that is projected will be an image of the emitter albeit very dim. Now move to a spot directly to the side(not closer or farther to the emitter). The same thing happens here. Light from points all across the emitter shine at that tiny spot on the reflector and the beam sent out will be another image. However since you have moved to the side, essentially circling the emitter it(LED's image) will be rotated slightly from the other one. You can repeat this infinitely going around the reflector so that you have image over image over image. This repeated overlapping eventually creates a circle instead of a single image as you would have with an optic.
Now while the basics of what I just described do not change when you move closer or farther to the emitter one thing does change. The size of image that teeny tiny spot projects. The closer that tiny spot is to the emitter the larger the emitter appears just as someone far away appears tiny to you even though your brain knows otherwise. So that tiny spot close in throws an image of the emitter but larger than when we were in the middle of the reflector. And the opposite is true when we move to the outer rim of the reflector. From this vantage point the emitter looks much smaller. Therefore the image thrown is much smaller. Now remember that this whole process is just laying image over image.
So you want to know what part of the projected beam consists of what images? Let's start in the center of the projected beam or the beam you would see on a white wall at 20 feet or so(beam has to have space to mature). The most intense round spot you see is mostly what has been reflected by the outer area of the reflector. It also consists of the less intense and larger images created by the lower portions of the reflector. So all of the reflector contributes to the center of the projected beam but the outer portions of the reflector do not add any intensity to the outer areas of the projected beam.
So to recap. The images from the outer portions of the reflector are the smallest. As we move closer the images get larger.
For more information see this video.
Optical System Efficiency
This is subdivided into two categories.
1-Efficiency of Reflective/Refractive elements
2-Emitter's Spatial Distribution vs Type and Diameter of collimating element.
This page is still under construction. Sorry I am such a slow typist.
Last edited:
oo: That's not even a word! 
