smokinshogun
Member
.....
Inverse-Square 'LAW' of Light
- Thread starter smokinshogun
- Start date
This may help you understand the concept better:
http://en.wikipedia.org/wiki/Inverse-square_law
All energy in the entire universe follows this law. Life as we know it would be totally different were you to change these rules. It definitely refers to the diffusion of a ray based on the observer. While it is true that the radiation we calculate is actually curved(that is, if you made all the rays the same exact length), we simplify that to a flat surface like the floor of a grow room. In distances as short as those from a grow light to the floor, it doesn't make much difference to adjust. When you're looking at stellar scale though, it is very important to remember sometimes.
Think of it this way, we're basically defining a square in space, arbitrarily, because it helps us calculate these things. So if your light is a certain intensity at 1 foot, at 2 feet, it WILL be 1/4 that intensity. You can use reflective surfaces to focus more light back into the same area, but all the energy involved still radiates under this same principal, still gets 1/4 as intense each time you double the distance. Just because your reflective surfaces cause these rays to overlap each other, does not mean the law isn't true. Also, the reflector on your lamp adds more light that would have gone the other direction, but that light still obeys the inverse square law.
In your example of the prisms, you are not using an equilateral prism, and so some of the light from your original prism would radiate past the edges of your longer prism. You need to increase the size of the prism in all dimensions, or you need to adjust your calculations. You'll definitely find that for the surface area of your longer prism, it's exactly 1/4 as intense. If you made the longer one 2mx2mx2m, you'd have 4 square meters of cross-section, so your 1mx1mx2m prism still has only 1 square meter of cross-section, 1/4 of your original. See?
http://en.wikipedia.org/wiki/Inverse-square_law
All energy in the entire universe follows this law. Life as we know it would be totally different were you to change these rules. It definitely refers to the diffusion of a ray based on the observer. While it is true that the radiation we calculate is actually curved(that is, if you made all the rays the same exact length), we simplify that to a flat surface like the floor of a grow room. In distances as short as those from a grow light to the floor, it doesn't make much difference to adjust. When you're looking at stellar scale though, it is very important to remember sometimes.
Think of it this way, we're basically defining a square in space, arbitrarily, because it helps us calculate these things. So if your light is a certain intensity at 1 foot, at 2 feet, it WILL be 1/4 that intensity. You can use reflective surfaces to focus more light back into the same area, but all the energy involved still radiates under this same principal, still gets 1/4 as intense each time you double the distance. Just because your reflective surfaces cause these rays to overlap each other, does not mean the law isn't true. Also, the reflector on your lamp adds more light that would have gone the other direction, but that light still obeys the inverse square law.
In your example of the prisms, you are not using an equilateral prism, and so some of the light from your original prism would radiate past the edges of your longer prism. You need to increase the size of the prism in all dimensions, or you need to adjust your calculations. You'll definitely find that for the surface area of your longer prism, it's exactly 1/4 as intense. If you made the longer one 2mx2mx2m, you'd have 4 square meters of cross-section, so your 1mx1mx2m prism still has only 1 square meter of cross-section, 1/4 of your original. See?
excellent post smoking. Ive been sitting here pondering the duality of light : is it a wave, is it a particle lol. Waves diminish in intensity as they travel, but is that as a result of dissipation or the consumption/transferance/transformation of the energy as it travels, whereas particles can lose energy quickly but tend to still exist, so is a photon a particle or a wave and is the energy it contains reduced at all by its travels? I follow the arguement about the redirection of the photons from the source point to the destination point, and it's a absolutely right, I'm just not sure about how much energy it will still contain. After all, light bouncing from different points will interfere with itself, and if bouncing from 2 points may well cancel itself out.
river rat01
Member
math makes me
i'm so glad that the sun is my light source and i dont have to worry about this crap.
i'm so glad that the sun is my light source and i dont have to worry about this crap.
You're really not listening. I wish you'd not be so confrontational about this idea. This "law" you mock so much is the fundamental law governing all radiant energy.
What you're not hearing is that I'm saying if you have 2 times the distance, but the same space, IN AN OPEN AREA, you'd have 1/4 the light. That's not my opinion. That's the way the universe works! When you bounce that light back off the walls, you're condensing that light which has spread out, back into the same space. With 100% reflective materials on the wall, you'd lose nothing. Life doesn't work that way though.
That chart you show does not account for reflective walls. It never claims to! It's just determining the sunburn point for cannabis plants. The chart shows the same surface area at all distances, so reflective walls are not part of the picture. If you have reflective walls, and IF your light is focused in such a way as to hit them, then they have to be factored in as well. The light doesn't vanish, it just spreads out. Even the light bouncing back from the walls is spreading out as it does so. It's not inverse square that is ruined by your walls, it's your over-simplification of it which is ruined.
The reason you can see stars billions of light years away is because they are intense on a level that HID doesn't even begin to approach. For every star you can see on Earth, there's hundreds of trillions(at least), which you cannot see. Why? Inverse square! All radiation does this. Gravity does this, sound does this. It's fundamental to energy.
What you're not hearing is that I'm saying if you have 2 times the distance, but the same space, IN AN OPEN AREA, you'd have 1/4 the light. That's not my opinion. That's the way the universe works! When you bounce that light back off the walls, you're condensing that light which has spread out, back into the same space. With 100% reflective materials on the wall, you'd lose nothing. Life doesn't work that way though.
That chart you show does not account for reflective walls. It never claims to! It's just determining the sunburn point for cannabis plants. The chart shows the same surface area at all distances, so reflective walls are not part of the picture. If you have reflective walls, and IF your light is focused in such a way as to hit them, then they have to be factored in as well. The light doesn't vanish, it just spreads out. Even the light bouncing back from the walls is spreading out as it does so. It's not inverse square that is ruined by your walls, it's your over-simplification of it which is ruined.
The reason you can see stars billions of light years away is because they are intense on a level that HID doesn't even begin to approach. For every star you can see on Earth, there's hundreds of trillions(at least), which you cannot see. Why? Inverse square! All radiation does this. Gravity does this, sound does this. It's fundamental to energy.
Philosophelon
Member
I'm with Magic -- the inverse square law is pretty widely accepted in the field of physics.
I think what is throwing you might be the use of reflectors and reflective wall covering. With reflectors and poly/paint/mylar/etc., all the light that would otherwise be radiated to most of the sphere is reflected back to the canopy, but the inverse square law still applies to that reflected light.
So even though most of the sphere of light is shining on one spot (your canopy) on the sphere (some directly and most reflected), if you raise your reflector, all of the sphere of light is now that much farther away from your canopy. I realize I'm not explaining this very well, and I got a C in physics, but I promise that the inverse square law applies even with reflectors and poly.
I think what is throwing you might be the use of reflectors and reflective wall covering. With reflectors and poly/paint/mylar/etc., all the light that would otherwise be radiated to most of the sphere is reflected back to the canopy, but the inverse square law still applies to that reflected light.
So even though most of the sphere of light is shining on one spot (your canopy) on the sphere (some directly and most reflected), if you raise your reflector, all of the sphere of light is now that much farther away from your canopy. I realize I'm not explaining this very well, and I got a C in physics, but I promise that the inverse square law applies even with reflectors and poly.
Let me try it this way....
The DIRECT light is 1/4 the original intensity. If you didn't have reflectors, or your light was focused so that it did not need them, then the rule applies completely.
If you want to really get a good answer, go to science.com and ask there. I can tell you right now they won't be as nice as I am about it. Nerds get pretty worked up about science. I've even been threatened with death on science boards before(because I dared to suggest Einstein was wrong on special relativity and Lorentz was right).
For some simple math lets assume you have a source with 10,000 lumens(rated at 12"). We'll also assume this lamp is focused to cover the exact area of the grow space. At 24" that same light would be a DIRECT 2500 lumens in the same space. That means that your four walls have to redirect the remaining 7500 lumens to the grow. Assuming even dispersion, the reflected light from each wall(90% reflectivity), would be:
7500 lumens of scattered light divided by 4 (4 walls), 1875 lumens per wall, 1687.5 lumens net from each wall. So recombining that into the same space, that's 6750 lumens, plus the 2500 direct lumens, giving you 9250 lumens, or a bit over 92% of your original light. That's also assuming a nice uniform reflection from each wall. Realistically, there's bright spots, especially with HID, so some areas would be brighter than that, some would be more dim, but that would be the average.
You're not violating the law of inverse squared by doing this, you're just minimizing the dispersion inherent to it.
The DIRECT light is 1/4 the original intensity. If you didn't have reflectors, or your light was focused so that it did not need them, then the rule applies completely.
If you want to really get a good answer, go to science.com and ask there. I can tell you right now they won't be as nice as I am about it. Nerds get pretty worked up about science. I've even been threatened with death on science boards before(because I dared to suggest Einstein was wrong on special relativity and Lorentz was right).
For some simple math lets assume you have a source with 10,000 lumens(rated at 12"). We'll also assume this lamp is focused to cover the exact area of the grow space. At 24" that same light would be a DIRECT 2500 lumens in the same space. That means that your four walls have to redirect the remaining 7500 lumens to the grow. Assuming even dispersion, the reflected light from each wall(90% reflectivity), would be:
7500 lumens of scattered light divided by 4 (4 walls), 1875 lumens per wall, 1687.5 lumens net from each wall. So recombining that into the same space, that's 6750 lumens, plus the 2500 direct lumens, giving you 9250 lumens, or a bit over 92% of your original light. That's also assuming a nice uniform reflection from each wall. Realistically, there's bright spots, especially with HID, so some areas would be brighter than that, some would be more dim, but that would be the average.
You're not violating the law of inverse squared by doing this, you're just minimizing the dispersion inherent to it.
Philosophelon
Member
Don't think of it as distance from the canopy, think of it as total distance the light has to travel. It's easier to imagine if you think of light as particles. Imagine light particles shooting out of your arc tube at all angles and then bouncing off reflective surfaces until they reach the canopy and are largely absorbed. The inverse square law applies to the total distance that a given particle travels.
If you want to independently verify that the law applies, you could run a test with a light meter.
If you want to independently verify that the law applies, you could run a test with a light meter.
niacinamide
Member
I really still think I'm right...
It'll be easy to prove one way or the other. Get a cheap used light meter off eBay and record the values received at different heights within your grow area.
I do agree with you that the inverse square law does not apply -exactly- when dealing with small enclosed areas due to the presence of reflected light as you suggest. However, it is still a good guide to keep in mind when dealing with light sources.
Another area to consider is exactly how useful reflected light is if the majority of it reaches the underside of the leaves where there is comparatively fewer chloroplasts than the top of leaves.
I think that is smoky's whole point. He believes that the law is true, but does not really directly apply to your growroom because we use reflectors and walls to minimize the dispersion inherent to it.
So basically you both agree on the same law's but smokey is taking into consideration reflectors and walls. There is alot of marketing hype and people in general that like to quote the inverse square law but fail to take into consideration reflectors and walls, thus making the situation sound alot worst than it really is.
Someone needs to go out and recruit a physicist to clear this and other issues up IMO. This question would actually be pretty easy to resolve empirically with a light meter. I tend to believe that the inverse square law doesn't apply as light at 1ft away that hits a reflective surface is only diminished by 90% or so.
Some other long standing questions-
(i) do lumens stack? what is the difference between 4 5k lumen bulbs and single source 20k lumens
(ii) if you put a fan on the back of a sailboat would it move forward? Hmm.
Pine
Some other long standing questions-
(i) do lumens stack? what is the difference between 4 5k lumen bulbs and single source 20k lumens
(ii) if you put a fan on the back of a sailboat would it move forward? Hmm.
Pine
Lt. Herb
Member
Your primary point may be valid (doubling the height of your light does not result in 25% intensity) but just because you don't understand how the Inverse Square Law applies to your grow does not make it false.
In it's simplest form [the Inverse Square Law] describes the dispersion of light in the vacuum of space from a point source. When you add anything else to it (atmosphere, reflectors, walls, ect.) you must make accomodations for this in the math. This is the part you're missing, the part where you have to account for the real world in your equations. The reason you won't find that math anywhere is every grow room will have a different equation. To know exactly what losses you are getting, you must understand the math involved and make measurements and calculations yourself. Other than that, you use the ISL as a guideline, and know that the farther away your light is, the less intense it will be, regardless of how well your walls/reflector work.
In it's simplest form [the Inverse Square Law] describes the dispersion of light in the vacuum of space from a point source. When you add anything else to it (atmosphere, reflectors, walls, ect.) you must make accomodations for this in the math. This is the part you're missing, the part where you have to account for the real world in your equations. The reason you won't find that math anywhere is every grow room will have a different equation. To know exactly what losses you are getting, you must understand the math involved and make measurements and calculations yourself. Other than that, you use the ISL as a guideline, and know that the farther away your light is, the less intense it will be, regardless of how well your walls/reflector work.
Philosophelon
Member
OK, here's another way to think about it.
Say your arc tube is 4" from the reflector. Say that two inches from the arc tube there is X amount of radiant energy in one square inch. In this example, X remains constant. Four inches (twice the distance) away from the arc tube (at the surface of the reflector) X amount of energy covers four square inches. After the light reflects off the surface of the reflector, it travels another two inches back towards the bulb -- that same energy is now covering 9 square inches. Then the light travels the final two inches back to the arc tube. At this point X amount of energy is now covering 16 square inches. And so on, until it reaches the canopy, or all/some reflects off some other surface.
Another way to think about it:
You have a set of 3 parabolic mirrors, angled such that an intense beam of light will be continuously reflected off of all three mirrors. The beam of light does not get brighter the longer you leave it on -- even though light is being continuously reflected, it still rapidly loses energy as it travels through space. Or visualize it like this - if you put a bulb inside a mirrored sphere, so that all of the light is reflected back into the sphere, the amount of radiant energy in the sphere would not keep increasing indefinitely. Again, because light loses energy whenever it travels through space, no matter how effectively it is reflected, etc.
Say your arc tube is 4" from the reflector. Say that two inches from the arc tube there is X amount of radiant energy in one square inch. In this example, X remains constant. Four inches (twice the distance) away from the arc tube (at the surface of the reflector) X amount of energy covers four square inches. After the light reflects off the surface of the reflector, it travels another two inches back towards the bulb -- that same energy is now covering 9 square inches. Then the light travels the final two inches back to the arc tube. At this point X amount of energy is now covering 16 square inches. And so on, until it reaches the canopy, or all/some reflects off some other surface.
Another way to think about it:
You have a set of 3 parabolic mirrors, angled such that an intense beam of light will be continuously reflected off of all three mirrors. The beam of light does not get brighter the longer you leave it on -- even though light is being continuously reflected, it still rapidly loses energy as it travels through space. Or visualize it like this - if you put a bulb inside a mirrored sphere, so that all of the light is reflected back into the sphere, the amount of radiant energy in the sphere would not keep increasing indefinitely. Again, because light loses energy whenever it travels through space, no matter how effectively it is reflected, etc.
Philosophelon
Member
Well, never mind, there is a reason I'm not a physics professor. Several actually 
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T
T.J
Great reading guys!
It's easier to misunderstand something than actually understand it fully, a point quite proven in this thread
It's easier to misunderstand something than actually understand it fully, a point quite proven in this thread
F
fully baked
First off, let's clarify, it's not the lights energy we're discussing, it's the lights intensity. The energy is simply a product of the wavelength and constants.
Let me see if I understand this right, but what smokinshogun is trying to say, is the inverse square law, as it is written for a particle in space, does not accurately estimate the intensity in our grow rooms. I support him 100%, and that simple fact is the reason we use reflective material. It is an attempt to raise the lighting intensity on our canopy.
I believe that is the basis of the chart. It shows the intensity without the use of reflectors and can be estimated as a point source emitting light in all directions. My only problem with your statements is your prism argument, the light is still losing it's intensity according the inverse square but because the light that should have shot out the sides and top of the tube is redirected, our total intensity does not drop off accordingly.
But, each and every photon is still losing intensity according to the inverse square, it's only our manipulation of reflectors that allows us to increase intensity so that it doesn't drop off at the predicted level.
So my final conclusion, yes the inverse square law still applies to each individual photon, but because of reflectivity, our TOTAL intensity doesn't follow this law because of the redirection of light that should have gone elsewhere...
Then you have to consider interference and the loss of light due to absorption and imperfect reflectors and shit will start to get ugly real fast with the mathematics!! And I thought I was done studying quantum...
My
Cheers,
FB
Let me see if I understand this right, but what smokinshogun is trying to say, is the inverse square law, as it is written for a particle in space, does not accurately estimate the intensity in our grow rooms. I support him 100%, and that simple fact is the reason we use reflective material. It is an attempt to raise the lighting intensity on our canopy.
I believe that is the basis of the chart. It shows the intensity without the use of reflectors and can be estimated as a point source emitting light in all directions. My only problem with your statements is your prism argument, the light is still losing it's intensity according the inverse square but because the light that should have shot out the sides and top of the tube is redirected, our total intensity does not drop off accordingly.
But, each and every photon is still losing intensity according to the inverse square, it's only our manipulation of reflectors that allows us to increase intensity so that it doesn't drop off at the predicted level.
So my final conclusion, yes the inverse square law still applies to each individual photon, but because of reflectivity, our TOTAL intensity doesn't follow this law because of the redirection of light that should have gone elsewhere...
Then you have to consider interference and the loss of light due to absorption and imperfect reflectors and shit will start to get ugly real fast with the mathematics!! And I thought I was done studying quantum...
My
Cheers,
FB
centralcoastbud
Member
So after tossing the question at some #physics heads the official reply is....
Q:i'm basically trying to figure out "how" the inverse square law applies to light as it pertains to a bulb being held from a fixture in a closed space with a high amount of reflectivity
A:It doesn't.
Reason:the inverse square law only applies to undisturbed spherical emmission. the inverse square law comes from the fact that the energy (or power) distributes over a spherical surface that grows with radius squared.so if you have a different geometry, a good approximation for the law of intensity fall-off is to find the cross sectional wavefront area and divide the emitted power by that.
If we are referring to the intensity drop off -- then that is "the inverse square law in an isotropic case".
I'm fairly stoned off some killer black domina but theres a response if that helps!
Q:i'm basically trying to figure out "how" the inverse square law applies to light as it pertains to a bulb being held from a fixture in a closed space with a high amount of reflectivity
A:It doesn't.
Reason:the inverse square law only applies to undisturbed spherical emmission. the inverse square law comes from the fact that the energy (or power) distributes over a spherical surface that grows with radius squared.so if you have a different geometry, a good approximation for the law of intensity fall-off is to find the cross sectional wavefront area and divide the emitted power by that.
If we are referring to the intensity drop off -- then that is "the inverse square law in an isotropic case".
I'm fairly stoned off some killer black domina but theres a response if that helps!
