Inverse-Square 'LAW' of Light

[GMT]

The reflector above the light doesn't affect the measurements, since the light measured 4 " below the bulb and 8" below the bulb are both recording the reflected light, in effect the reflector just ups the starting values, its the reflective walls that make the difference to the measurements. If your light reflector aims the light at the width of the canopy, the walls will make little difference to the measurements, but if your reflector is high enough to allow some of the light to be hitting the walls before the plants, the measurements will change. To do this accurately though, the measurements need to be taken without walls and with walls for a comparrison. If someone with a light meter could measure half way from the bulb to the walls, then take readings at the half way point and then at the wall, (providing the light is hitting the wall at a higher point than the measurement is taken at), then that would be a good measurement to have.

 

[devilgoob]

GMT, your studies are appreciated..have you looked at my mspaint drawing http://icmag.com/ic/showpost.php?p=2386170#post2386170

who knew changing thread in "showthread.php" to post would've allowed me to do that.

 

[SativaDreams]

you guys made it really complicated.


in simple terms

the formula describes (in IDEAL conditions.. ill explain after) a relationship between distance (distance the light travels for the calculation) and lumens recorded at that point.

this formula DOES work in the growroom scenarios. let's explore

firstly, what are ideal conditions. ideal conditions are assuming the light is traveling in a vacuum ( a non-existent completely theoretical but mathematically sound place where the light is not inhibited by real world factors I will mention in a moment).

SOME Real world factors which are probably the most relevant to light dispersion in an applied setting of the growroom(including its constructive and deconstructive interferences due to reflection/refraction)
minor:
-Medium the light is traveling through- aka the air in your grow room(including its moisture content or humidity, its % makeup of different compounds and molecules, etc)
-Temperature of the medium -aka temp air in grow room
major:
-quality of the reflector cone used around the light (the angles it was constructed at, the thickness of it, the material it is made out of(or the surface is made of), its surface area, distance from the bulb, etc.)
-reflective properties of all other surfaces the light will bounce off of.. walls most importantly (although the light will to some extent bounce off everything like the plant containers, soil, plant itself, floor, parts of the bulb itself, but different percentages of the light will be lost as heat depending on the chemical/physical makeup of the surface the light is reflecting off of etc etc.) -
- the thickness of the reflective surface of the walls is also important (if it is mylar, and it is very thin then light is passing through it.. this light MAY be trapped behind it and lost as heat energy, MAY bounce off the wall and back through the mylar (it could be constructive then - which amplifies intensity (lumens) or it could be deconstructive - lowers the intensity (lumens) )

all of these factors and many more would only be considered if one knew advanced physics and calculus.

The base formula, although excluding factors that should be calculated in (like those mentioned above), does apply even though it is not 100% accurate. for most scenarios even in grow room it could be assumed quite accurate because most of these things don't make too significant a difference (Save the mylar and reflective cone.)


The reason this formula is fine to use if used properly is you must treat EVERY surface light bounces off of as a source of light.

So the filaments in the bulb are the origin, you must calculate the distance from the filaments to each part of the plant to determine intensity at those points.

then you must consider the reflector cone (at any of its near infinite points on the surface of it) as sources of light and first calculate the lumens at a point on the cone (use distance from filaments of bulb to the point on the reflector cone) then you must calculate distance from that point on the reflector cone to each part of the plant you wish to find intensity for by using the lumens found at that point as initial. you must repeatedly do this for every wall and surface which light is reflected off of, because that surface becomes a source of light so to speak.
we have not even included the fact that all of these sources of light (reflector cone/wall/floor/etc) will absorb certain parts of the spectrum and in simple terms the lumens will be lost as heat.


this is best done using advanced calculus by taking integrals about different parts of reflective surfaces.

obviously since laymen cannot do such a task, the general formula is used to calculate a fairly accurate number, its a general rule for use by non-physics doctorates like me and probably most of you! most people say "double the distance 1/4 the lumen output".. get some sort of lumen measuring device for yourself, and figure it out!

BTW to the guy who started this whole thing the inverse "square" law doesnt mean light is a square. it means the square function in the formula (LOL)

 

[GMT]

Hi devilgoob, yes I looked at your diagram, and then did one of my own, but it got far to complicated and messy as I wanted to keep adding more and more to it lol. In your diagram I suspect that for the blue line, the ISL will hold true, however if the walls intersect the green line, which then has to bounce to the tops of the plants, then those photons can be added to the green line that hits the plant tops directly, increasing the measurement that would be taken, which without the walls would equal the ISL but with them = ISL+reflected ISL photons > ISL on its own.

 

[devilgoob]

Why is everyone saying lumens? Lumens are the perceived power of the intensity of the light by the human eye. Humans have a sensitivity between colors with the same amount of photon densities. So if green light was being emitted onto a white surface with 100u/mols/m^2/s density and red light with 100u/mols/m^2/s density, the green would appear brighter...because your a human! To a plant, lumens don't matter...it's the frequency of the light and the photometric density. Basically it's something you stoner know already. Photosynthetically Active Radition (PAR) Pretty self-explanatory.

A real world use of this is a 6500K light spectrum (cool white) at 42 watts is using it's power to emit blue light...and it could be 3800 lumens even, but you know you want the 42 watt 2700K bulb with only 3400 lumens because it has red light. Even at a lumens advantage..the red light is key. Therefore ultimately the lumen is dead and frequencies and the amount of photons per surface area unit/second.

If you look at the mspaint diagram I drew, you can see the surface area reasoning of the inverse-square law vs real life.....which was the real argument here.

http://icmag.com/ic/showpost.php?p=2386170#post2386170


Oh yea GMT? Started adding tons of things onto it? LOL thats exactly what I did..most of the time spent making it was screwing with text boxes.

 

[T.J]

Why is everyone saying lumens?

Now you opened another Pandora's box

 

[GMT]

I guess the best representation of what is being said here is this:
In 2 lengths of fibre optical cable, one being twice the length of the other, the output at the end of the longer one isn't 1/4 of the shorter one.
Clearly our grow boxes aren't as efficient at reflecting light as the example, but the principle holds true.

Devil, I started to play with the number of reflections above the bulb, and which photons cancelled out which other photons, before I realised that all that was a red herring.

 

[GMT]

It definately does matter how reflective the surface areas are where the light that is being measured is light that is reflected. Picture the fibre optic cable as a black electrical cable, how much light will come out of the other end then?

 

[GMT]

Unless there are no walls, or the walls completely absorb the light, then I concur.
The reflector above the light is completely irrelevant to your objection, the only thing that matters is the walls.

 

[GMT]

I don't get 25% smokin, I agree with you.

 

[CannabisSativa]

Plants don't reflect AS MUCH light as these which means they are essentially GATHERING THE LIGHT!

That is common knowledge. Plants absorb the light that they need and reflect what they don't. Who cares what happens to the light that is 'reflected' from the plants? If they reflect it then they don't need it. It's that simple.

It doesnt matter if you have 1/4 the total light since plants 'learn' to absorb MORE light when there is less of it. The percentage of light they absorb WILL be greater as compared to when they have plenty of light. Its also 4 times the space which obviously means more yield, since there would be more plants which are also more efficient..

Dude, you have to understand how a plant works first before tackling other problems. Read up on the Calvin Cycle and other topics such as those. And please tell me how they "learn" to absorb more light.

You have to understand that if it doesnt sound right it doesnt mean it wrong. That's where math comes in to prove it.

This should answer your question:

Here is a Java applete where you can play with this:
http://csep10.phys.utk.edu/astr162/lect/light/intensity.html


Try this analogy since this also works with the ISL:
Lets say someone screams in your ear. Now someone walks about 20 feet away from you and screams...Do you hear it the same? Sound waves are going to bounce off of you and only some make it to your ear. Now have that same person yell into a long piece of tube and it is also next to your ear. What do you hear? And why?

It definately does matter how reflective the surface areas are where the light that is being measured is light that is reflected. Picture the fibre optic cable as a black electrical cable, how much light will come out of the other end then?

True. They use very reflective material and base it using Snell's law to get the most optimal reflectivity. Granted, Optics use either LED's or Laser depending on the application and the distance.

 

[CannabisSativa]

Alright, given this situation with a small piece of paper, which one will have the most intense light to the given surface to the paper?

Now what if you add more pieces of papers around that area where the light is shown...will it equal the same amount of intensity as used on one sheet of paper?

I am not taking either sides of the argument, just posting facts and different analogies. This is probably one of the most interesting threads I have read in a while.



EDIT: I am asking a panel of experts on this subject...I am awaiting to hear back soon. I will post the question and answers when I get them. Like I said, very interesting thread! K+ OP!

 

[Gold123]

From what I gather from this discussion is smokinshogun is trying to simply say you do not loose 75% of your light by doubleing the distance in a small grow room. the ISL is measuring from a single point light source emiting light in 3 dimensions. In our grows we use a parabolic reflector to focus the light downward basically 2 dimensions. Yes there is light lost when you raise the light but not 4 times as much at double the distance as in the ISL. We are raising inches or at most a foot or so.
Thus the loss of light is there but not as much, unless you have a large grow and start raising the bulbs much higher. This of coarse doesn't apply to hanging cfl's without reflectors.

 

[CannabisSativa]

From what I gather from this discussion is smokinshogun is trying to simply say you do not loose 75% of your light by doubleing the distance in a small grow room. the ISL is measuring from a single point light source emiting light in 3 dimensions. In our grows we use a parabolic reflector to focus the light downward basically 2 dimensions. Yes there is light lost when you raise the light but not 4 times as much at double the distance as in the ISL. We are raising inches or at most a foot or so.
Thus the loss of light is there but not as much, unless you have a large grow and start raising the bulbs much higher. This of coarse doesn't apply to hanging cfl's without reflectors.

Exactly! Like I said, I am trying to confirm this.

 

[GMT]

actually no. You are still working in 3 spacial dimensions. The point is that the walls reflect light and so by raising the lights, you get more reflected rather than direct light. The direct light is reduced by the ISL but you then add the reflected light to that light increasing the total light available and so 75% is not lost, a much smaller percentage is lost.

 

[CannabisSativa]

The law is about surface area, thats it. I just had to provide this to back up my wild 'claim' of plants being able to learn.

http://www.scielo.br/scielo.php?pid=S0100-84042006000200013&script=sci_arttext
"The effects of shade conditions caused changes in the leaf biophysical characteristics. C. parviflora leaves showed a significant absorption increase in the 400-750 nm range under 1.5% PFD when compared to full sun. This increase was achieved through reductions in transmittance and reflectance."

but "Reduction of absorption is generally due to epidermics modifications. Structures like wax or trichomes are associated with higher leaf reflectance and lower absorptance (Ehleringer 1981). It has been suggested that in sunny environments, such modifications helped to limit water loss by lowering the amount of absorbed light and the consequential heat load (Vogelmann 1993, Poorter et al. 1995). The reflectance increase in both species in full sun was probably due to the presence of a dense layer of cuticle folds in C. parviflora leaves and to a high density of trichomes in G. virgata. Moreover, in full sun and 40% PFD, G. virgata leaves stood erected to reduce light penetration by 50%. "

since plants 'learn' to absorb MORE light when there is less of it.

I was assuming you meant that they are able to 'learn' to produce more sugars with less light. You need more light to produce more sugars, thus getting bigger plants. Look up the Calvin Cycle.

and you have our common knowledge wrong...
Plants don't somehow 'absorb the light they need and reflect light because they don't need it'. Plants use green light and reflect blue and red. Its all in the percentages, really has nothing to do with if it 'wants' the light or not. Light reflected by plants COULD be re-absorbed!

Wrong. You have it backwards. Chlorophyll reflects green light and absorbs some blue and red light. And yes, they do absorb the light (spectrum) they need. It's basic plant biology.

 

[CannabisSativa]

S
Its louder when they are yelling into the tube, this is because its ALL about surface area. You can't spread things out when they don't have the space. That is the 'law"....
Twice the distance of a prism doesn't equal 4 times the surface area. Its imposible to get 25% according to the law...

Dude, the ISL law assumed to be in optimal conditions with no type of walls/reflection. You use that base formula in conjugation with other variables and it will change.

Lets say you add walls..well you have to factor that in...know what I mean? Its not some magical formula that gets applied directly to all situations. Its like using the generic speed of sound formula (c = sqrt(C/P)) to calculate the speed of sound when its traveling down a hallway or anywhere. You have to factor the walls, size, if there is any wind, etc.
I am not calling you an idiot or assuming...I just think that everyone is NOT on the same page.

 

[CannabisSativa]

The ONLY variable that can effect the ISL is SURFACE AREA. In real life, other things can effect the actual intensity. Do not try to say its the ISL causing the decrease though, if you don't have a certain increase in surface area...That is what your answer will be.

Good read:

Because of this fall of in light intensity with distance, the intrinsic energy output of stars (also called their luminosity can only be known if their distances are known. Later on this term we will be measuring distances to nearby stars using the Parallax applet. Thus, if you have a known distance to a star and are able to measure the flux at that distance, the total energy output of the star can easily be ascertained (this will be shown in class).
A simple way to think about this, is the following:

• Suppose you are in a large, darkened room and in the center of that room is a 100 watt light bulb. When you turn on the light bulb that is the only source of light there is. The light fills the room so the entire room contains all of the energy.
• Now let's suppose that you stand 1 meter away from the light bulb. Let's say you have a 1 square centimeter detector and you hold it up and you measure the energy incident on it. For the sake of this argument, let's assume that energy is equivelent to 4 million photons per second per square centimeter.
• Now you move twice as far away or 2 meters away. Since the surface area of a sphere (or the area of a 2D circle) goes as the square of the radius, the same amount of energy (or number of photons per second from the lightbulb) now has to go through an area 4 times larger. As a result, the flux of the light at 2 meters is down by a factor of 4, so you measure 1 million photons per second. If you were to move out to 4 meters, you would see 250,000 photons per sq. cm per second.
• The instrinsic energy output of the lightbulb (its total number of emitted photons per second) can be found by taking your measured flux and multiplying by the total number of square centimeters corresponding to the surface area of a sphere at your distance. The surface area of a sphere is:
  4pR2 This means the total energy output is your measured flux multiplied by the surface area:
  So, at R = 1 meter
  
  f = 4 million photons per square cm per second
  
  ; the total surface area is 4p square meters (or 40,000 square centimeters)
  
  so the total Energy emitted is 40,000 x 4 million photons per second.
  Now notice, at R = 2 meters f is down to 1 million photons per square cm per second but the total surface are has increased to 16p square meters or 160,000 square centimeters. So the total Energy is 160,000 x 1 million photons per second, which is the same number as before.
  Thus if you can measure f at a known distance, then you know the intinsic energy output. The ability to measure f, however, depends upon the DETECTOR!

http://zebu.uoregon.edu/2001/ph122/isq.html

 

[CannabisSativa]

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.

Actually, he summed up everything right there. Everyone is partially correct in this thread. The ISL isnt wrong, its just that it has to be USED correctly depending on the circumstances. I understand that you are not trying to prove it wrong, but trying to prove that those "light" chart distances are wrong. Right? I wish we could get a hold of the people who made those charts and ask them what variables they used...but I do know that some of the distances there are somewhat correct because I have seen them tested with light meters. If you have the bulb illuminating in an open space, then yea it will be accurate via ISL, but if not, then it wont be. Do you agree with this?

"Light waves are spread out over a larger and larger surface (as it travels outward). As the radius increases, the photons are spread further and further apart, and the intensity (# of photons) decreases. So lower intensity means less photons, which means fainter light."

you can't have 1/4 the intensity without 4 times the surface area. Thats the law

If you move the light higher you can get 1/4 the intensity at a certain point even if you don't have the surface area being 4 time the size.

 

[CannabisSativa]

Here are two crappy pictures I just draw in paint:

Light is closer

Light is farther

As you can see in the first picture, the light source is close, this the intensity is better because you have more photons closer together.

In the second picture the light source if further, and the photons spread apart more thus having less intensity. You can see less "photos". Here is another good example:

See the A? See what happens once the source is farther away? Think of that "A" square as your grow room, a leaf, a plant, or anything.