Photon-Photon Scattering [Archive]

kochab

10-04-2008, 06:26 PM

well I would believe that it is possible for the photons to be as of a too strong of a nature...
I had a 3foot plant outdoors this year get fucking light poisioning dude...
That is How MrsMcGreggor and I met, Lb introduced us because of her vast knowledge of plants.

The issue was that I had let the plant veg for about a month and a half to 2 months, then used a trashcan to start force flowering her to determine its sex, she was a female and so was the other counterpart that I did in that same trashcan.
One was an indica dominant hjack and one having issues was a sativa dominant hijack. The indica never experienced this @ all...

When I moved them both back they got light poisoning, we deferred this from many readings on what happens to greenhouse plants occasionally. The fans from the center of the plant started having a big yellow spot in the center finger of the leaf in the center of the blade..
It started in the center of the plant and was working its way through most of her Large fans from the center to the bottom and top both...
It went on to look like something had sucked all the juice out of the leaves there where they yellowed. The spot was not brittle but looked like a dry oak leaf in texture and color, the area felt really leathery to touch if that makes sense... Some extreme TLC ang she finally made it back to full health

The chances of this happening indoors are pretty minimal I would assume but Im not capable of remembering what exact strength the sun has ATM... Im not sure how the photons would react if they were to be able to hit each other...Light travels in straight beams supposedly, so do those photons travel in straight beams as well by chance?

skunkushybrid

10-04-2008, 06:43 PM

I don't know kochab... i started out reading papers on biology, and now here I am moving into physics. I hate physics.

kochab

10-04-2008, 06:58 PM

lmao. I feel you there.

Trashed

10-05-2008, 03:54 PM

Just had a brain fart here. if light, being made of molecules of matter, does indeed scatter differently under different circumstances, wouldn't that be yet another reason for keeping temps down in a grow room? I mean, if the air temperature does affect the movement. colder air slowing down the molecules and letting space between them "stay" there longer would have the same effect on light particles as it would on sound waves/pressure waves, on te premise that they're both presure waves, just of different types. If so, then cooler temps would allow the light particles to move more freely and with less interference, because they wouldn't be bouncing off each other so often.

hell, it's sunday morning and i've got this full-on karma-in-a-vacumn thing going on here and need to go outside or soemthing.

TetraHyC

10-05-2008, 04:04 PM

A photon has 0 mass.

What do you mean by 'scattering'

"Light travels through space in a straight line as long as nothing disturbs it. As light moves through the atmosphere, it continues to go straight until it bumps into a bit of dust or a gas molecule. Then what happens to the light depends on its wave length and the size of the thing it hits.

Dust particles and water droplets are much larger than the wavelength of visible light. When light hits these large particles, it gets reflected, or bounced off, in different directions. The different colors of light are all reflected by the particle in the same way. The reflected light appears white because it still contains all of the same colors.

Gas molecules are smaller than the wavelength of visible light. If light bumps into them, it acts differently. When light hits a gas molecule, some of it may get absorbed. After awhile, the molecule radiates (releases, or gives off) the light in a different direction. The color that is radiated is the same color that was absorbed. The different colors of light are affected differently. All of the colors can be absorbed. But the higher frequencies (blues) are absorbed more often than the lower frequencies (reds). This process is called Rayleigh scattering. (It is named after Lord John Rayleigh, an English physicist, who first described it in the 1870's.) "

This is an easy READ (http://www.sciencemadesimple.com/sky_blue.html)

seamaiden

10-05-2008, 04:47 PM

Well.. it has zero mass, yet it behaves both as a particle and as a wave. However, light is not "molecules" of any sort.

Light poisoning is a new one to me, kochab, at least with plants that have evolved to grow out in the open rather than within or under a thick canopy. That's interesting.

I think what you're getting at, skunky, is something almost akin to what CERN was made to do (although they're making a much more violent reaction), in that I think you're asking about photons bouncing off of each other... once light gets to a certain density. Do I understand you correctly?

ceestyle

10-05-2008, 06:44 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 4 2008, 04:50 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38318)</div>
In trying to understand a little more as to how light interacts with our plants... i came across photon-photon scattering... meaning that when different photons collide we get a scattering effect.

However, what i want to know is... do photons of the same type also bounce off each other, or do they pour down in an orderly stream?[/b]

Photon-photon scattering is a current area of research. These phenomena only exist in the confines of extremely small geometries and extremely high intensity light - at least that is what current theory predicts. This is why - as SM pointed out - experiments designed to expose this type of scattering may only be carried out in a few places in the world.

So the short answer is that the effects of PPS are not relevant at the energy/length scales observed in the real world.

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 4 2008, 05:29 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38336)</div>
Yes and our plants are that something else... although i'm actually talking about photon-photon scattering, what happens when a red photon meets a blue photon?

But a light wave is full of photons, and as these photons travel through the wave... are they orderly, or do they bounce off each other?[/b]

As THC accurately pointed out, photons have zero mass - they are quasiparticles. Although a quantum mechanical description of light would be necessary to describe this in a way that (sortof) makes sense, they can 'interfere' with one another constructively and destructively. This is best understood in the context of a wave: if you take a sinusoidal wave that represents one photon and superpose it atop another that represents another particle, as you shift one relative to another you can see that the maxima and minima take different positions relative to one another. If you were to add the magnitude of the waves (the square of the waves' magnitude, to be more accurate), the sum would yield a pattern with maxima and minima.

You can observe this rather easily at home if you like. Take a bright flashlight and a piece of paper, and put two very narrow (~1mm) slits in it an inch or two apart. Turn off the lights. Turn on the flashlight from behind the piece of paper, with the piece of paper a foot or so from a blank wall. You should see light and dark patches on the wall. These arise from the interference of the light particles with one another.

All of that said, these are effects that are irrelevant at the length-scale and with the type of light source that we use.

<div class='quotetop'>QUOTE </div>
Also what happens with our plants? is a single photon absorbed as a whole, or could it be that only a certain amount of energy from the photon is needed, and thus only a certain percentage of the photon used?

Th photon must be of a certain intensity to ionise the plant cell... but can it be that a photon can be too intense?[/b]

Yes. If you remember when I was pointing out that blue light is not always better, especially if you have a fixed total light energy, a photon can certainly be too intense.

Light acts as a catalyst for chemical reaction. A very easy way to think about it is this. There are a number of reactions that need to occur in order to produce food for the plant, in the form of sugars and such. Say that one such reaction involves the conversion of the reactants "A" and "B" into "C", denoted by A+B->C. A fixed amount of energy is required to carry out this reaction. A good analogy is this: you have a ball on one side of a hill, and you need to get it to the other side. Even if where you need to get it to on the other side is at the same altitude, you need energy to push it over the hill - a fixed amount of energy. Too little will not get the ball over, and too much won't be any more effective at getting the ball over - it will just be moving faster when it gets there. So is the case for the light-catalyzed chemical reaction. It takes a fixed amount of energy, which corresponds to a certain wavelength (color) of light (you can also think of this as a photon energy). If the light is too far red-shifted, it cannot catalyze the reaction. If it is too blue-shifted, it also may not catalyze the reaction (photoactive compounds within the plant are only designed to absorb energy at certain wavelengths), but even if it does, a large amount of energy is wasted. This excess energy (the input energy minus the energy required to initiate the reaction) is either converted to heat, or may in fact damage plant tissue or compounds such as THC.

To a very good approximation - based on light intensity and the time-scales of chemical reactions - one photon at a time interacts with a chemical process within the plant. What this means is that two photons will not help initiate the same reaction. So what you want is as many photons at the correct energy as possible. To see what the effect of having too highly energetic photons on overall conversion efficiency is, let me give you an example:

Say that you have light source A, which produces highly energetic photons, and source B, which produces perfectly energetic photons for a given photosynthetic process. Say (optimistically) that the higher energy photons each induce one chemical reaction, as do the perfectly energetic photons. Say also that each source has the same energy efficiency (lumens/watt, PAR watt/electricity watt, whatever). What you get is the following for the number of chemical reactions induced by each source, assuming N is the energy provided per watt input, W is the number of watts of the light source, and Ea and Eb are the energies of the photons from each bulb:

Source A:

# initiated reactions = W * N / Ea

Source B:

# initiated reactions = W * N / Eb

You can see from this simple example that if you have the same light energy from two bulbs, having photons of energy exceeding that required is simply a waste. Source B, despite having lower energy photons, initiates more reactions by a ratio of Eb/Ea.

Another intuitive way to look at this is to think that at one extreme we could use all UV, XRay, or gamma rays, but that would simply be too much for the plant to digest.

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 5 2008, 08:47 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38911)</div>
Well.. it has zero mass, yet it behaves both as a particle and as a wave. However, light is not "molecules" of any sort.

I think what you're getting at, skunky, is something almost akin to what CERN was made to do (although they're making a much more violent reaction), in that I think you're asking about photons bouncing off of each other... once light gets to a certain density. Do I understand you correctly?[/b]

Yes, an EXTREMELY high density

skunkushybrid

10-05-2008, 07:15 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 5 2008, 07:44 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38957)</div>
Another intuitive way to look at this is to think that at one extreme we could use all UV, XRay, or gamma rays, but that would simply be too much for the plant to digest.[/b]

UV is not too much for the plant to digest... i have grown under a pure UV light source, and the plants grew normally.

thanks for answering my question. i was in interested in the power of the scattering.

skunkushybrid

10-05-2008, 07:21 PM

so now... back to the breakdown of or semi-useability of the photon. although it does make sense that only a certain pertion of the photon is needed... why do we assume the rest of the photon is discarded? In fact couldn't there be secondary cells specifically used to scavenge the left over light?

Goddamn this plant... why does shit have to be so complicated?

ceestyle

10-05-2008, 07:49 PM

<div class='quotetop'>QUOTE (Honkeytown @ Sep 30 2008, 01:22 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=36963)</div>
Radiohead's "in rainbows" album...its great!
also some Mars Volta..."de loused in the comatorium" album[/b]

<div class='quotetop'>QUOTE (max @ Oct 1 2008, 09:28 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=37438)</div>
Tiesto-suburban train[/b]

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 5 2008, 11:15 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38971)</div>
UV is not too much for the plant to digest... i have grown under a pure UV light source, and the plants grew normally.

thanks for answering my question. i was in interested in the power of the scattering.[/b]

That's interesting. Do you have a thread on that? I'd bet, however, that if your UV light and a full-spec light had the same output efficiency, they would grow much better under the full-spec. Plants just don't evolve to most efficiently absorb UV.

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 5 2008, 11:21 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=38976)</div>
so now... back to the breakdown of or semi-useability of the photon. although it does make sense that only a certain pertion of the photon is needed... why do we assume the rest of the photon is discarded? In fact couldn't there be secondary cells specifically used to scavenge the left over light?

Goddamn this plant... why does shit have to be so complicated?[/b]

See, there's not left over light. There's left-over energy. A dye within the molecule - such as chlorophyll - absorbs the photon. The absorption of the photon excites an electron within the molecule. The transition of this electron from a ground to an excited state has a fixed energy, as dictated by quantum mechanics. The energy levels within the molecule have discrete energy levels, and the gap between the ground and excited state has a set energy. This is why you need a different molecule to absorb different wavelengths of light (in general). Although it's complicated to describe exactly how this works, the excess energy is converted to other forms, called rotational and vibrational energies, and also into translational energy, which we observe as heat.

As I mentioned earlier, this happens so quickly that the absorptions and reactions effectively occur in isolation. If many occurred in the same place, it is possible (but improbable) that the sum energy could be harvested, but this is not the case. Otherwise, it would have to be stored chemically and harvested collectively. To the best of my knowledge, this does not occur. Practically, you would again have to overcome the fact that these reactions are few and far between (relatively), loss mechanisms due to entropy and otherwise, and the simple fact that the wasted energy is relatively small compared to the energy harvested and the other energy modes of the molecule.

There are, however, systems within plants that are designed to absorb higher-energy wavelengths and effectively waste some of the energy and then transfer the correct energy quanta to the chemical process. This is still not any more efficient per photon, but can use energies not typically thought of as being photosynthetically active.

skunkushybrid

10-05-2008, 07:59 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 5 2008, 08:49 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=39004)</div>
That's interesting. Do you have a thread on that? I'd bet, however, that if your UV light and a full-spec light had the same output efficiency, they would grow much better under the full-spec. Plants just don't evolve to most efficiently absorb UV.[/b]

Mine did exactly that... i used a 300W OSRAM... and hit them with it 24/0 before the seed casing had even come off. The plants got rid of the casing, but there were no true leaves on either plant. Only cotyledons. It took 3 days for the true leaves to even look like they were going to appear... as they looked burnt. Then they came through... anyway, here's a link: http://www.rollitup.org/advanced-marijuana...-radiation.html (http://www.rollitup.org/advanced-marijuana-cultivation/48409-death-radiation.html)

I didn't manage to complete the whole thread as i had to leave the site.

Trashed

10-06-2008, 02:39 AM

energy is energy. It doesn't dissapear, it merely changes form, whether it be 13 megs of juic blowing someones socks off, or light scattering.l it's still there in our case for what we want, turned into a form of energy used by these funny green weeds. I'm not really worried about a photon bouncing off something in my box here. But I am learning a little more. In the meantime, I'm going into town and buy a photon producing device. ;P

ceestyle

10-06-2008, 03:24 AM

<div class='quotetop'>QUOTE (Trashed @ Oct 5 2008, 06:39 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=39379)</div>
energy is energy.[/b]

yabbut try and grow some weed with a space heater!

<div class='quotetop'>QUOTE </div>
It doesn't dissapear, it merely changes form, whether it be 13 megs of juic blowing someones socks off, or light scattering.l it's still there in our case for what we want, turned into a form of energy used by these funny green weeds. I'm not really worried about a photon bouncing off something in my box here. But I am learning a little more. In the meantime, I'm going into town and buy a photon producing device. ;P[/b]

you do that. make sure it be bright.

skunkushybrid

10-06-2008, 08:43 AM

the only evidence i can find that actually says anything specific and sure... is that the energy must be powerful enough to create the reaction with whatever substance it comes into contact with. I can't find anything that actually says the light can be too powerful.

Indeed if that were the case, then wouldn't it be that a 400w was enough to grow bud, so a 1000w is a waste of energy? How does light work in that sense? Does a richer light source merely contain more photons, or does it supply a richer more powerful photon?

I apologise if i seem a little ignorant, I didn't go to school http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif

StealthPanda

10-06-2008, 08:59 AM

This may seem off topic, but is it possible to ionize a quasiparticle? I was theorising about the effects of mass amounts of ionized particles the sun blasts at the earth, well, only some of it penetrates the atmosphere. But i wonder what the effects of ions are to plants. And how it would be possible to create this synthetically.

Trashed

10-07-2008, 10:48 AM

i just bought a photon producing device last night before coming to work and think im going to use it for a space heater. my plants might not like it but i will. but i might use it to scatter some quasiparticles into the same area of my box as the quarks that are already there. they're sneaky. th' quarks i mean. the metal halide ones to be specific. Sylvania has a small space heater imbedded in the new 360w ones disquised as an igniter, making people think it's an hps. but then again, im sitting here in this control room, it's a quarter til six a.m. and still have two hours left of a thirteen hour shift. my photons are scattered. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/tongue.gif

seamaiden

10-07-2008, 02:31 PM

Oh.. Man, I fell fast asleep with my browser open and trying to wrap my head around the Second Law of Thermodynamics (because I'm trying to understand entropy in relation to thermodynamics). Good monring! http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/laugh.gif

skunkushybrid

10-07-2008, 03:53 PM

why's everything got to be so difficult?

seamaiden

10-07-2008, 04:03 PM

Well, see, for me it's like this. Use words and I can usually get it. Start using EQUATIONS, though, make it look like math, and I am lost. I do not get cee's equations on the earlier page, and completely fail to see that they are a demonstration that shows that the perfectly energetic photons are more efficient (i.e. "better") than the highly energetic photons. Unless 'a' is inherently just a better letter than 'b', which is news to me. I like my 'b's' about as much as I like my 'a's'.

tom crudes

10-07-2008, 04:03 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 7 2008, 03:53 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=40593)</div>
why's everything got to be so difficult?[/b]

LOL it's only as hard as you make it, or so they say.

You guys are all super nerds, I hate learning about light...

skunkushybrid

10-07-2008, 04:08 PM

I was ok with it until I needed to start studying physics...

Trashed

10-09-2008, 01:19 AM

if you take a 12 guage pump-gun and shoot a lightbulb, everything will scatter, regardless of if it has mass or not. i prefer #4 high-brass loads. leaves very few hot-spots. double-0 buck doesn't have enough actual pellets in it to make much difference. helps aerate the leaves also, opening up the stomato to recieve scattered stuff. ;P

ceestyle

10-09-2008, 11:37 PM

<div class='quotetop'>QUOTE (Trashed @ Oct 8 2008, 05:19 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=41342)</div>
if you take a 12 guage pump-gun and shoot a lightbulb, everything will scatter, regardless of if it has mass or not. i prefer #4 high-brass loads. leaves very few hot-spots. double-0 buck doesn't have enough actual pellets in it to make much difference. helps aerate the leaves also, opening up the stomato to recieve scattered stuff. ;P[/b]

i now have everyone in the room staring at me for explosively laughing when i read this.

TetraHyC

10-09-2008, 11:41 PM

I like 000 for making holes in the pots.

ceestyle

10-10-2008, 12:02 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 6 2008, 12:43 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=39512)</div>
the only evidence i can find that actually says anything specific and sure... is that the energy must be powerful enough to create the reaction with whatever substance it comes into contact with. I can't find anything that actually says the light can be too powerful.

Indeed if that were the case, then wouldn't it be that a 400w was enough to grow bud, so a 1000w is a waste of energy? How does light work in that sense? Does a richer light source merely contain more photons, or does it supply a richer more powerful photon?

I apologise if i seem a little ignorant, I didn't go to school http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif[/b]

no man, you got it. 1000W just produces more photons. Approximately 1000/400 = 2.5x as many photons, to be more precise.

for a really simple analogy to the too powerful photon: return to the case where you're trying to push a ball over a hill, and you need a certain amount of energy to do that. Think about trying to hit the ball too hard, and you might "whiff" or "shank" it and not get it over the hill. It's not really the same, but the same end effect. Now think about if you hit it super hard, you'll just pop the ball. That's what a gamma ray would do to plant material. Toast.

<div class='quotetop'>QUOTE (StealthPanda @ Oct 6 2008, 12:59 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=39519)</div>
This may seem off topic, but is it possible to ionize a quasiparticle? I was theorising about the effects of mass amounts of ionized particles the sun blasts at the earth, well, only some of it penetrates the atmosphere. But i wonder what the effects of ions are to plants. And how it would be possible to create this synthetically.[/b]

No. Ionization requires at least two particles that are balanced in charge (before you ionize them: an ion is a charged particle). The simplest one (with whole, rather than fractional charges) is the Hydrogen atom, containing an electron and a proton. Both electrons and protons are particles (real ones).

<div class='quotetop'>QUOTE (Trashed @ Oct 7 2008, 02:48 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=40461)</div>
i just bought a photon producing device last night before coming to work and think im going to use it for a space heater. my plants might not like it but i will. but i might use it to scatter some quasiparticles into the same area of my box as the quarks that are already there. they're sneaky. th' quarks i mean. the metal halide ones to be specific. Sylvania has a small space heater imbedded in the new 360w ones disquised as an igniter, making people think it's an hps. but then again, im sitting here in this control room, it's a quarter til six a.m. and still have two hours left of a thirteen hour shift. my photons are scattered. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/tongue.gif[/b]

dude, you are seriously funny. my photons are now also scattered.

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 7 2008, 06:31 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=40549)</div>
Oh.. Man, I fell fast asleep with my browser open and trying to wrap my head around the Second Law of Thermodynamics (because I'm trying to understand entropy in relation to thermodynamics). Good monring! http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/laugh.gif[/b]

Okay, you need an example. Say you have some hot water and some cold water. The second law of thermodynamics says you can't make the hot water hotter by making the cold water colder, even though the system could have the same total energy. Entropy is a measure of order. When you have hot water and cold water, you have all the hot water molecules separated from the cold water molecules: an ordered system. By letting them mix themselves all up, it's totally disordered and all mixed up. Obviously without some other energy input, you couldn't just go backwards and turn lukewarm water into separate cold and hot water again. Entropy must always increase or stay the same in a closed system!

Wrap your head around this: using the above principle, the entropy of our universe is constantly increasing.

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 7 2008, 08:03 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=40596)</div>
Well, see, for me it's like this. Use words and I can usually get it. Start using EQUATIONS, though, make it look like math, and I am lost. I do not get cee's equations on the earlier page, and completely fail to see that they are a demonstration that shows that the perfectly energetic photons are more efficient (i.e. "better") than the highly energetic photons. Unless 'a' is inherently just a better letter than 'b', which is news to me. I like my 'b's' about as much as I like my 'a's'.[/b]

Yeah, that equation didn't really help. Let me again use the ball example.

Say you (personally) have a limited amount of energy and a stack of balls, and you'd like to get as many as you can over the hill. It takes a certain amount of your energy to get them over the hill. Are you going to push them as hard as you can, or just hard enough to get them over the hill? Say in one case you push just hard enough to get them over the hill, and you end up pushing 10 over the hill. If you were to have pushed them, for example, twice as hard, you would have only gotten 5 over, right?

So your fixed amount of energy is the amount of total energy the light puts out, the "just hard enough" push is a perfectly energetic photon, and the "twice as hard" push is the too energetic photon. With a fixed amount of energy, there are twice as many "just hard enough" pushes or "just energetic enough" photons, and you get twice as many balls over the hill, which corresponds to twice as many photoinitiated chemical reactions. Does that make more sense?

seamaiden

10-10-2008, 04:30 PM

Yes, that makes sense, after I had to post a thread on another forum asking for help with the laws.
I've got those two prof friends, one's a physics prof at SFSU, and the other's a chem prof (the researcher W. Scott I told you about), so I posted on the forum where I met them asking for help understanding the Second Law and entropy as a measure of unavailable energy in thermodynamics. The SFSU guy helped me get it, I think. The chem prof guy muddled it all to fuck. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/laugh.gif He started using equations and it went all to hell.

I understand the basic principals of entropy, but when it was mixed with the thermodynamics I hadn't quite gotten it. BUT! With regards to these different types of photons (perfectly energetic versus highly energetic) I now think I can grasp how that entropy can be measured, because it essentially (usually?) becomes heat. Which can be measured. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/biggrin.gif

Go perfect photons! Boo super-energetic photons.

skunkushybrid

10-17-2008, 08:32 AM

<div class='quotetop'>QUOTE (ceestyle @ Oct 10 2008, 01:02 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=41835)</div>
no man, you got it. 1000W just produces more photons. Approximately 1000/400 = 2.5x as many photons, to be more precise.

for a really simple analogy to the too powerful photon: return to the case where you're trying to push a ball over a hill, and you need a certain amount of energy to do that. Think about trying to hit the ball too hard, and you might "whiff" or "shank" it and not get it over the hill. It's not really the same, but the same end effect. Now think about if you hit it super hard, you'll just pop the ball. That's what a gamma ray would do to plant material. Toast.[/b]

Yes... but i'm thinking of things on a much lower level... as in say, a 400w blue photon against a 400w red photon... i cannot understand why only a part of the blue photon would be used and the other unnecessary. for example, if i have a 400w and a 1000w HPS, the 1000W will grow bigger plants and provide more of a yield. so back to the MH with a richer photon, would this not then be more comparible to a larger HPS, rather than the plant deciding the MH is only 400w so it must take an equivalent of light to a 400w HPS?

It makes more sense (to me) that the richer blue photon would be totally used. it may only take a certain amount of light to create the ionisation, but a 400w hps v a 1000w hps proves that you've got ionisation then you've got IONISATION.

ceestyle

10-17-2008, 09:36 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 17 2008, 12:32 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45059)</div>
Yes... but i'm thinking of things on a much lower level... as in say, a 400w blue photon against a 400w red photon...[/b]

Red photons are red photons and blue photons are blue photons. More watts just means more photons

<div class='quotetop'>QUOTE </div>
i cannot understand why only a part of the blue photon would be used and the other unnecessary. for example, if i have a 400w and a 1000w HPS, the 1000W will grow bigger plants and provide more of a yield.[/b]

That's because there are more photons with more watts.

So with the ball / hill analogy, if you push the ball just hard enough to go over the hill, and you measure success with how many balls you push over the hill, if you push it harder than you need to, you're wasting that extra energy. The ball will be going faster when it gets to the other side, but you don't care.

<div class='quotetop'>QUOTE </div>
so back to the MH with a richer photon, would this not then be more comparible to a larger HPS, rather than the plant deciding the MH is only 400w so it must take an equivalent of light to a 400w HPS?[/b]

So let's simplify things and say that MH makes only blue photons and HPS makes only red photons. It takes more of that 400W to make each blue photon than it does for each red photon, so if MH and HPS are the same efficiency (they're fairly close), than you would get more red photons for 400W than you would blue photons. Does that make sense? So if you get one chemical reaction per photon (regardless of whether they're blue or red), then you get more chemical reactions from the HPS, since you get more red photons than blue. I hope that run-on sentence was clear http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif

<div class='quotetop'>QUOTE </div>
It makes more sense (to me) that the richer blue photon would be totally used. it may only take a certain amount of light to create the ionisation, but a 400w hps v a 1000w hps proves that you've got ionisation then you've got IONISATION.[/b]

If it takes an energy input to initiate a chemical reaction that corresponds to the energy of the red photon, there is nothing left to do with the extra energy of the blue photon but waste it somehow. In fact, the bigger difference between the photon energy and the energy required to initiate the chemical reaction, the less efficient that photon will be at initiating it (even if it has more energy). This is a quantum mechanical effect known as resonance, and explains why you need more than one wavelength of light to get optimum growth. The fact that LEDs are nearly a single wavelength (monochromatic light, like lasers) may explain why they are so inefficient at growing plants. The fact is that the photosynthetic pathways are not fully understood.

skunkushybrid

10-17-2008, 10:16 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 17 2008, 10:36 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45350)</div>
Red photons are red photons and blue photons are blue photons. More watts just means more photons.[/b]

This doesn't sound right to me... the light from a 400W diminishes much more quickly than from a 1000w. so more watts also means a greater penetration. which also suggests that the power is sufficient to push the light further away. If we could slow down time we would see the pulses of light, the stutters as it starts and fires. so light has to travel, it's fast... but the more power it has the faster it gets and the further it can travel. so we are merely given the perception of consistency... and especially with artificial light. a fly lives life so fast that it can actually see the flickers from a light bulb as the energy is pulsed through it. so there has to be more to it than a higher wattage simply meaning more photons. are you sure that more wattage doesn't equal a more powerful photon? i don't think a photon has any mass, so maybe the velocity thing doesn't work... but it could be more powerful in the sense that it is moving faster from a 1000w than it is a 400w.

<div class='quotetop'>QUOTE </div>
That's because there are more photons with more watts.

So with the ball / hill analogy, if you push the ball just hard enough to go over the hill, and you measure success with how many balls you push over the hill, if you push it harder than you need to, you're wasting that extra energy. The ball will be going faster when it gets to the other side, but you don't care.[/b]

yes i got this the first time... only, a 400w (the ballast the kicker, the light the ball, the hill the distance to travel) is good enough to grow a plant. so a 1000w is a waste of energy?

<div class='quotetop'>QUOTE </div>
So let's simplify things and say that MH makes only blue photons and HPS makes only red photons. It takes more of that 400W to make each blue photon than it does for each red photon, so if MH and HPS are the same efficiency (they're fairly close), than you would get more red photons for 400W than you would blue photons. Does that make sense? So if you get one chemical reaction per photon (regardless of whether they're blue or red), then you get more chemical reactions from the HPS, since you get more red photons than blue. I hope that run-on sentence was clear http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif[/b]

MH is clearly a better light to veg' with... and there are clearly different growth patterns. plants veg'd under MH are stronger and bushier. so, at least during veg' there are clear differences in a plant's growth patterns. so there is also a difference in the chemical reactions that take place once the photon hits. all of the light sensitive hormones react to the blue photon.... carotenoids and the chlorophylls... yet only the chlorophylls react to red.

<div class='quotetop'>QUOTE </div>
If it takes an energy input to initiate a chemical reaction that corresponds to the energy of the red photon, there is nothing left to do with the extra energy of the blue photon but waste it somehow.[/b]

Not necessarily so... the left over energy could be scavenged by another cell.

<div class='quotetop'>QUOTE </div>
The fact that LEDs are nearly a single wavelength (monochromatic light, like lasers) may explain why they are so inefficient at growing plants. The fact is that the photosynthetic pathways are not fully understood.[/b]

LEDs are no good because they have no power behind them... the light doesn't reach far enough.

ceestyle

10-17-2008, 11:38 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 17 2008, 02:16 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45379)</div>
This doesn't sound right to me... the light from a 400W diminishes much more quickly than from a 1000w. so more watts also means a greater penetration. which also suggests that the power is sufficient to push the light further away. If we could slow down time we would see the pulses of light, the stutters as it starts and fires. so light has to travel, it's fast... but the more power it has the faster it gets and the further it can travel.[/b]

Trust me; it's right. Take a look at this: http://www.rollitup.org/indoor-growing/831...html#post968899 (http://www.rollitup.org/indoor-growing/83128-lumens-lux-adding-all-up.html#post968899)

All light travels the same speed : 3x10^8 m/s .. in any reference frame. This is the theory of relativity.

Each photon is a light quantum. More watts means more quanta. Again, there is no difference between the photons coming out of a 1000W bulb and those coming out of a 400W bulb; there are simply more of them.

Think about it like this: If you took the filament of a 1000W HID and cut it in half, it would put out the light of a 500W. You can duplicate this experiment by taking a 1000W and covering half of it with something that blocks light. You now have 500W output. There is nothing more to making a higher power bulb. In fact, this is how they make higher-power CFLs. You take the same tube that you make a 23W CFL with, add more coils, and you have a 42W CFL. It's just a matter of light output per unit length.

This train of thought is demonstrates that lumens and therefore lux add.

The reason that a higher power bulb seems to offer more penetration is simply that in order to have the same power, you have to separate the bulbs. By doing so, less light reaches lower branches, but only in some places. Let me show you what I mean with a little text example.

Say you have one 1000W bulb over a canopy like this:

-----------O------------
------------------------
------------------------
-----/\----------/\-----
----/--\--------/--\----
---/----\------/----\---
--/------\----/------\--
y/------- \x-/-------\y

Or you have two 500W (I know, they don't exist, but follow me) bulbs spaced like this:

----o-------------o----
------------------------
------------------------
-----/\----------/\-----
----/--\--------/--\----
---/----\------/----\---
--/------\----/------\--
y/------- \x-/-------\y

The reason you get more penetration is that light at position "x" sees little or no light from either bulb in the case of two lights, but with one large light, light reaches it very well. Even if light from both lights reached point 'x', it's farther from each light, and the sum is still weaker than the one light. On the other hand, points "y" see better coverage with two smaller lights.

As you move those two small lights closer and closer together, you approach the equivalent of the larger bulb. If you could put them next to each other and keep the same reflector efficiency, you would have identical lighting with the two smaller bulbs.

To see that this is in fact the case, see the example in the thread I posted above.

<div class='quotetop'>QUOTE </div>
yes i got this the first time... only, a 400w (the ballast the kicker, the light the ball, the hill the distance to travel) is good enough to grow a plant. so a 1000w is a waste of energy?[/b]

No, it's definitely not a waste. In order to see that it's not a waste, you have to trust me that the photons from each source are the same, however. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif

If you don't believe me and have a copy of Jorge's bible, look at the section on light coverage and how he shows in the FIgure the difference between using one 1000W, a couple 600W, or a few 400W (if I'm remembering it correctly). That figure would not make any sense if photons quantities didn't add.

<div class='quotetop'>QUOTE </div>
MH is clearly a better light to veg' with... and there are clearly different growth patterns. plants veg'd under MH are stronger and bushier. so, at least during veg' there are clear differences in a plant's growth patterns. so there is also a difference in the chemical reactions that take place once the photon hits. all of the light sensitive hormones react to the blue photon.... carotenoids and the chlorophylls... yet only the chlorophylls react to red.[/b]

It is clear from the plant response curves that you cannot describe photosynthesis by the sum of carotenoids and chlorophylls only. I feel like we've talked about this before.

<div class='quotetop'>QUOTE </div>
Not necessarily so... the left over energy could be scavenged by another cell.[/b]

You'll just have to take my word for the fact that this is not how photoinitiated reactions work. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/wink.gif The energy is wasted.

<div class='quotetop'>QUOTE </div>
LEDs are no good because they have no power behind them... the light doesn't reach far enough.[/b]

I'm not sure I know what you mean - in terms of how light works. Power does not determine how far photons go. Photons go until they run into something to either reflect off of or absorb into. This is another reason the penetration argument does not fly. They don't go anywhere until they hit something - they can't just disappear. This is why we can see light from the stars - there is nothing (vacuum) between us and them.

ceestyle

10-17-2008, 11:54 PM

Okay, let me draw an analogy I have found helpful in describing how light works, and how the d^2 rule works.

Say you have a sprinkler that puts out 1 gallon per minute, and say it does so in a perfectly spherical fashion and is unaffected by gravity. The area of a sphere is 4*pi*r^2, where r is the radius of the sphere. What this means is that if you are 1 foot from the sprinkler, you get hit with:

1 gallon per (4*pi*(1')^2) square feet per minute, or about 1/12 gallon per square foot per minute

If you're 2 feet away from the sprinkler, you get hit with:

1 gallon per (4*pi*(2')^2) square feet per minute, or about 1/48 gallon per square foot per minute

So if you had an area that was 1' x 1', you would collect 1/12 gallon in a minute at one foot, and 1/48 gallon in a minute at two feet.

You can see that as you doubled the distance from 1' to 2', you divided the water collected by four, hence the d^2 rule.

The droplets are all the same in this example. They don't go anywhere. The d^2 rule has nothing to do with the "power" of the droplets. It's just that the same amount of water coming out of the sprinkler is distributed over more area as you move away from the sprinkler.

Now, imagine in this analogy you either take another sprinkler putting out 1 gallon per minute and put it immediately next to the other sprinkler (assuming they don't interfere with one another, of course). Now imagine you simply increase the output of the first sprinkler to 2 gallons per minute.

The amount of water received at 1 and 2 feet away are simply doubled for either case, right? That's because it's the exact same thing, as far as you are concerned.

This analogy holds for one color of light: each droplet of water is a photon, and each sprinkler is a light.

seamaiden

10-18-2008, 08:06 PM

And here I was going to suggest making a plumbing analogy.

TetraHyC

10-18-2008, 10:05 PM

I think something about RGB and CMY. The primary and secondary colors of light.
C-MY-RGB (http://www.glenbrook.k12.il.us/gbssci/phys/Class/light/u12l2d.html)

skunkushybrid

10-18-2008, 10:15 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 18 2008, 12:38 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45410)</div>
Trust me; it's right. Take a look at this: http://www.rollitup.org/indoor-growing/831...html#post968899 (http://www.rollitup.org/indoor-growing/83128-lumens-lux-adding-all-up.html#post968899)

All light travels the same speed : 3x10^8 m/s .. in any reference frame. This is the theory of relativity.

Each photon is a light quantum. More watts means more quanta. Again, there is no difference between the photons coming out of a 1000W bulb and those coming out of a 400W bulb; there are simply more of them.

Think about it like this: If you took the filament of a 1000W HID and cut it in half, it would put out the light of a 500W. You can duplicate this experiment by taking a 1000W and covering half of it with something that blocks light. You now have 500W output. There is nothing more to making a higher power bulb. In fact, this is how they make higher-power CFLs. You take the same tube that you make a 23W CFL with, add more coils, and you have a 42W CFL. It's just a matter of light output per unit length.

This train of thought is demonstrates that lumens and therefore lux add.

The reason that a higher power bulb seems to offer more penetration is simply that in order to have the same power, you have to separate the bulbs. By doing so, less light reaches lower branches, but only in some places. Let me show you what I mean with a little text example.

Say you have one 1000W bulb over a canopy like this:

-----------O------------
------------------------
------------------------
-----/\----------/\-----
----/--\--------/--\----
---/----\------/----\---
--/------\----/------\--
y/------- \x-/-------\y

Or you have two 500W (I know, they don't exist, but follow me) bulbs spaced like this:

----o-------------o----
------------------------
------------------------
-----/\----------/\-----
----/--\--------/--\----
---/----\------/----\---
--/------\----/------\--
y/------- \x-/-------\y

The reason you get more penetration is that light at position "x" sees little or no light from either bulb in the case of two lights, but with one large light, light reaches it very well. Even if light from both lights reached point 'x', it's farther from each light, and the sum is still weaker than the one light. On the other hand, points "y" see better coverage with two smaller lights.

As you move those two small lights closer and closer together, you approach the equivalent of the larger bulb. If you could put them next to each other and keep the same reflector efficiency, you would have identical lighting with the two smaller bulbs.

To see that this is in fact the case, see the example in the thread I posted above.

No, it's definitely not a waste. In order to see that it's not a waste, you have to trust me that the photons from each source are the same, however. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif

If you don't believe me and have a copy of Jorge's bible, look at the section on light coverage and how he shows in the FIgure the difference between using one 1000W, a couple 600W, or a few 400W (if I'm remembering it correctly). That figure would not make any sense if photons quantities didn't add.

It is clear from the plant response curves that you cannot describe photosynthesis by the sum of carotenoids and chlorophylls only. I feel like we've talked about this before.

You'll just have to take my word for the fact that this is not how photoinitiated reactions work. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/wink.gif The energy is wasted.

I'm not sure I know what you mean - in terms of how light works. Power does not determine how far photons go. Photons go until they run into something to either reflect off of or absorb into. This is another reason the penetration argument does not fly. They don't go anywhere until they hit something - they can't just disappear. This is why we can see light from the stars - there is nothing (vacuum) between us and them.[/b]

But i have seen penetration for my own eyes... if i hold a 250w light 12" above my plants they will stretch... yet a 600w at 12" and they don't. also, i'm not a fan of cervantes, but i'm sure that what he was talking about is that 2 400's could outdo a 1000 by width coverage. the 400w's would need to be lower and you would get shorter colas. i know this as I have used varying sizes of lights. You do not get any extra penetration, believe me... i have 3 lights all same wattage, they do not create any extra penetration... i know this.

sgtpeppr

10-18-2008, 10:26 PM

What exactly is the question?

is it still......."do photons of the same type also bounce off each other, or do they pour down in an orderly stream?"

skunkushybrid

10-18-2008, 10:31 PM

so if i shined up a flashlight from earth and pointed it at the moon on a clear day, you would see that flash from the moon?

now a big bomb going off... lots of power... you could probably see that. light must lose energy as it travels, this is the only thing that makes any sense to me. yes a weak light source can easily disappear quite quickly... a small LED in a very dark open field.... you'd see it for a few meters, but gradually the light would fade...

with a 400w light even with 2 together your plants will do no better than an 18" main cola... yet a 600w will grow a 24-30" cola. this is a fact, and something i have witnessed over and over again... not just on forums but with friends too.

TetraHyC

10-18-2008, 10:34 PM

What if you use the Hubble with and 11 day exposure.

http://msnbcmedia3.msn.com/i/msnbc/Components/Photos/040923/hudf.jpg

skunkushybrid

10-18-2008, 11:06 PM

ceestyle... is this the current scientific thinking?

It has to be wrong... of course light gets weaker as it travels, we can see it with our own eyes. the more power it has to push it out the further it can travel.

let's take the ball and the hill analogy... the more power there is behind the ball the further over the hill it will go... put enough force into it and you can bury it into the ground on the other side.... so what if you lose the ball, you got another one to kick and all the power you need to kick it.

if you place a light meter on the floor beneath a 400w you will get a low reading, yet as you move the meter up uou get a higher one.

seamaiden

10-19-2008, 02:59 AM

You could see the flashlight IF you could filter out all the interference, including brighter light, skunk, and with a long enough exposure (i.e. not with the unaided human eye).

ceestyle

10-19-2008, 03:03 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 18 2008, 02:15 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45835)</div>
But i have seen penetration for my own eyes... if i hold a 250w light 12" above my plants they will stretch... yet a 600w at 12" and they don't. also, i'm not a fan of cervantes, but i'm sure that what he was talking about is that 2 400's could outdo a 1000 by width coverage. the 400w's would need to be lower and you would get shorter colas. i know this as I have used varying sizes of lights. You do not get any extra penetration, believe me... i have 3 lights all same wattage, they do not create any extra penetration... i know this.[/b]

Well yes, if you replace a bulb with a larger one, the plants are getting more light. If you replaced two 250s right next to each other with a 400, they might stretch.

What I meant about the Cervantes diagram is that if lighting didn't add, then the diagram wouldn't make any sense.

ceestyle

10-19-2008, 03:10 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 18 2008, 02:31 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45846)</div>
so if i shined up a flashlight from earth and pointed it at the moon on a clear day, you would see that flash from the moon?[/b]

Nope, but a laser would. The reason it's different is that while each photon travels in a straight line, each one has a slightly different trajectory from a point source (like a light bulb). In other words, they're not traveling on parallel trajectories, whereas with a laser they are. The reason this effect isn't noticeable from light rays from the sun is that by the time they get to the earth, the ones that arrive here are by that time essentially traveling parallel to one another. Back to the sprinkler analogy: think about the path of each droplet of water. None of the droplets go anywhere, but as you move away, fewer hit you per unit area.

<div class='quotetop'>QUOTE </div>
now a big bomb going off... lots of power... you could probably see that. light must lose energy as it travels, this is the only thing that makes any sense to me.[/b]

Only if things in the atmosphere such as dust and elements in our atmosphere absorb it. At the length-scales we're dealing with, these absorption effects are totally irrelevant.

What I don't think you understand is that I really do know what I'm talking about. No offense, and I'm not trying to be a know-it-all, but I'm not reasoning this out as I go along. These are simply the laws of physics.

<div class='quotetop'>QUOTE </div>
yes a weak light source can easily disappear quite quickly... a small LED in a very dark open field.... you'd see it for a few meters, but gradually the light would fade...

with a 400w light even with 2 together your plants will do no better than an 18" main cola... yet a 600w will grow a 24-30" cola. this is a fact, and something i have witnessed over and over again... not just on forums but with friends too.[/b]

I don't know what to tell you, except that the experimental conditions were flawed, such as having problems with the bulbs physically interfering with one another.

ceestyle

10-19-2008, 03:16 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 18 2008, 03:06 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45863)</div>
ceestyle... is this the current scientific thinking?

It has to be wrong... of course light gets weaker as it travels, we can see it with our own eyes. the more power it has to push it out the further it can travel.

let's take the ball and the hill analogy... the more power there is behind the ball the further over the hill it will go... put enough force into it and you can bury it into the ground on the other side.... so what if you lose the ball, you got another one to kick and all the power you need to kick it.

if you place a light meter on the floor beneath a 400w you will get a low reading, yet as you move the meter up uou get a higher one.[/b]

The analogy doesn't apply to photons.

Yes, the reading you get will increase as the inverse square of the distance. This is for the reason I outlined in my previous post where I derived the d^2 law.

Think about a sprinkler. Seriously, it's an essentially exact analogy.

These are all basic properties of light that were the foundation of the theory of relativity, the photoelectric effect, and any other light-based phenomenon that has been studied since, including quantum mechanics. If what I have said is wrong, then so are Einstein, Newton, and Maxwell.

ceestyle

10-19-2008, 03:17 AM

.. and Bohr, Heisenberg, Schrodinger ...

seamaiden

10-19-2008, 03:20 AM

<div class='quotetop'>QUOTE (ceestyle @ Oct 18 2008, 07:17 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45955)</div>
.. and Bohr, Heisenberg, Schrodinger ...[/b]
http://icanhascheezburger.files.wordpress.com/2007/06/schrodingers-lolcat1.jpg

sgtpeppr

10-19-2008, 04:51 AM

I think the problem here is that light is both a wave and a particle.....until you get to the Planck scale of course.

ceestyle

10-19-2008, 05:40 AM

<div class='quotetop'>QUOTE (sgtpeppr @ Oct 18 2008, 08:51 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45996)</div>
I think the problem here is that light is both a wave and a particle.....until you get to the Planck scale of course.[/b]

yes, an interesting conundrum, to be sure. this is one (along with fractional realities) that led me to a deep study and research in quantum theory.

you can couch all the relevant arguments in terms of waves as well, but they're more intuitive when you describe them in terms of particles.

ceestyle

10-19-2008, 05:41 AM

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 18 2008, 07:20 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45957)</div>
http://icanhascheezburger.files.wordpress.com/2007/06/schrodingers-lolcat1.jpg[/b]

that is too fucking funny. that was obvious written by someone who has both an understanding of quantum mechanics and a sense of humor.

i assume that means you've heard of schrodinger's cat? an interesting metaphysical mind-puzzle.

skunkushybrid

10-19-2008, 08:10 AM

<div class='quotetop'>QUOTE (ceestyle @ Oct 19 2008, 04:16 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=45952)</div>
The analogy doesn't apply to photons.

Yes, the reading you get will increase as the inverse square of the distance. This is for the reason I outlined in my previous post where I derived the d^2 law.

Think about a sprinkler. Seriously, it's an essentially exact analogy.

These are all basic properties of light that were the foundation of the theory of relativity, the photoelectric effect, and any other light-based phenomenon that has been studied since, including quantum mechanics. If what I have said is wrong, then so are Einstein, Newton, and Maxwell.[/b]

I'm thinking of a sprinkler... and i'm thinking that if i turn down the power on the sprinkler, the droplets will lose their power too... sure i'll still get a little wet, but i'd get much wetter if you turned the power up... and a laser is far more powerful than a flashlight. takes far more energy to produce a laser beam than it does to power a flashlight.... and if photons are all the same then cfl's wouldn't be so crap at growing. and likewise 2 400w's should produce 2 foot colas, but they don't... a 600 does though... i do not need to understand science to KNOW what is best for my plants. i have seen it for my own eyes. also if you put a light meter right next to a 1000, and put the meter right next to a 400... right next to the bulb, so there's no scattering, the 1000w will read higher.

sgtpeppr

10-19-2008, 08:34 AM

skunkushybrid

10-19-2008, 12:21 PM

thanks SP...i think this is more along the lines: http://lyndonashmore.com/tired_light_front_page.htm

ceestyle

10-19-2008, 05:34 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 19 2008, 01:10 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46025)</div>
I'm thinking of a sprinkler... and i'm thinking that if i turn down the power on the sprinkler, the droplets will lose their power too... sure i'll still get a little wet, but i'd get much wetter if you turned the power up...[/b]

That's only because gravity pulls them down. If there weren't gravity, there would be fewer droplets.

Ok, you're right in that I left an assumption out of the sprinkler analogy as-stated. A garden sprinkler operates on a pressure drop that is a function of how open the valve is, meaning the droplets do move faster as you turn it up.

So ... you have to assume that regardless of the output, the drops are moving the same velocity (the speed of water? http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif ) as they exit the sprinkler. All light moves the same speed.

<div class='quotetop'>QUOTE </div>
and a laser is far more powerful than a flashlight. takes far more energy to produce a laser beam than it does to power a flashlight....[/b]

Only because of the optics and materials involved in pumping a laser to get coherent light. It has nothing to do with the power of the photons as they are emitted from the laser. The reason laser light travels to far is that all of the photons are moving exactly parallel to one another.

Imagine you place an explosive charge in a mass of projectiles, like bbs or slugs and detonate it some distance away from a target. Now imagine you load those projectiles into several firearms and point them all at the same target, from the same distance. Do you think more projectiles will hit the target from the explosion or from the firearms? This is the difference between a point light source and a laser. Does that make sense?

<div class='quotetop'>QUOTE </div>
and if photons are all the same then cfl's wouldn't be so crap at growing. and likewise 2 400w's should produce 2 foot colas, but they don't... a 600 does though... i do not need to understand science to KNOW what is best for my plants. i have seen it for my own eyes.[/b]

You know, that's fine, but I've explained all of your observations to you in terms of what light is and what the science explains.

Two 400s are more light than a 600. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/mega_shok.gif0 / 600 => 1/3rd more light than the 600 alone. Your observations are a result of the bulbs being spread. I guarantee if you did a proper side-by-side with the two 400s vs. the 600 you would yield more with the two smaller lights.

CFLs simply don't have the raw output that HIDs do. In order to get the same output, you have to have (a lot) more bulbs, which means they must be spaced, reducing center penetration but often bettering coverage. It also reduces the efficiency of any reflectors you might have. For the last time, I am going to post this link: http://www.rollitup.org/indoor-growing/833...html#post972220 (http://www.rollitup.org/indoor-growing/83378-16k-lumens-hps-vs-cfl.html#post972220)

This should prove to you that there is no difference between CFL light and HPS light. If you don't believe me, either I fabricated the data, my experiment is flawed, or my light meter is broken.

<div class='quotetop'>QUOTE </div>
also if you put a light meter right next to a 1000, and put the meter right next to a 400... right next to the bulb, so there's no scattering, the 1000w will read higher.[/b]

Of course the 1000W will read a higher light intensity. It is emitting more photons. A light meter measures the number of photons per unit area per unit time (adjusted for their energy). Did you click the links I put up earlier of the experiments I've done that prove all this stuff for the average user?

I'm actually getting annoyed/offended that you can't believe me, do some research of your own, or even look through the experiments that I've linked to find that what I'm posting here is truth. This is at the heart of what I do for a living. I've spent the better part of ten years learning this stuff. Instead of believing me, you're countering with intuitive arguments that just don't reflect the truth of the science.

ceestyle

10-19-2008, 05:43 PM

<div class='quotetop'>QUOTE (sgtpeppr @ Oct 19 2008, 01:34 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46028)</div>
It sounds to me like you are talking about light intensity...
Several measures of light are commonly known as intensity:

Radiant intensity (http://en.wikipedia.org/wiki/Radiant_intensity), a radiometric quantity measured in watts per steradian (W/sr) Luminous intensity (http://en.wikipedia.org/wiki/Luminous_intensity), a photometric quantity measured in lumens per steradian (lm/sr), or candela (cd) Irradiance (http://en.wikipedia.org/wiki/Irradiance), a radiometric quantity, measured in watts per meter squared (W/m<sup>2</sup>) Intensity_(physics) (http://en.wikipedia.org/wiki/Intensity_%28physics%29), the name for irradiance used in other branches of physics (W/m<sup>2</sup>) Radiance (http://en.wikipedia.org/wiki/Radiance), commonly called "intensity" in astronomy and astrophysics (W·sr<sup>-1</sup>·m<sup>-2</sup>)For more info, I got this from here (http://en.wikipedia.org/wiki/Light_intensity)[/b]

Yes, but they all measure the same thing in different units.

You see that first unit there in all of those? The "W" ? That is the unit for power, the watt. It has units of energy per second.

This corresponds to the product of the number of photons times their energy per second :

Power = Number of photons per second x energy per photon = total energy per second

The unit is the denominator in all of those is a "per unit area" or steradian. One is per unit area, one is a fraction of the angular distribution , so what you have is the energy per second per unit area (or per angular fraction) of the light.

The way this is measured in a light meter is the following:

There is a photodiode that converts a fraction of the photons at each wavelength (energy) into an electrical charge that is collected. The current is the number of these charges that is collected per second (the Ampere). Knowing the area of the photodiode and its quantum efficiency (the fraction of light at each energy that produces an electrical charge), you can easily compute the power (W) per unit area. It's that simple.

ceestyle

10-19-2008, 07:47 PM

i thought of another good way to explain a laser. It's like a sprinkler vs. a squirt gun (again ignoring gravity).

skunkushybrid

10-19-2008, 10:32 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 19 2008, 06:34 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46229)</div>
Two 400s are more light than a 600. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/mega_shok.gif 0 / 600 => 1/3rd more light than the 600 alone. Your observations are a result of the bulbs being spread. I guarantee if you did a proper side-by-side with the two 400s vs. the 600 you would yield more with the two smaller lights.[/b]

I actually said that 2 400's together are not enough to grow a 2ft cola per plant... but a single 600 is... this has nothing to do with overall yield, but proves a greater penetration of the 600 in comparison to 2 400's.

<div class='quotetop'>QUOTE </div>
CFLs simply don't have the raw output that HIDs do. In order to get the same output, you have to have (a lot) more bulbs, which means they must be spaced, reducing center penetration but often bettering coverage. It also reduces the efficiency of any reflectors you might have. For the last time, I am going to post this link: http://www.rollitup.org/indoor-growing/833...html#post972220 (http://www.rollitup.org/indoor-growing/83378-16k-lumens-hps-vs-cfl.html#post972220)

This should prove to you that there is no difference between CFL light and HPS light. If you don't believe me, either I fabricated the data, my experiment is flawed, or my light meter is broken.[/b]

Well I see no plants in that experiment, and i cannot see how your graphs prove anything.

<div class='quotetop'>QUOTE </div>
Of course the 1000W will read a higher light intensity. It is emitting more photons. A light meter measures the number of photons per unit area per unit time (adjusted for their energy). Did you click the links I put up earlier of the experiments I've done that prove all this stuff for the average user?

I'm actually getting annoyed/offended that you can't believe me, do some research of your own, or even look through the experiments that I've linked to find that what I'm posting here is truth. This is at the heart of what I do for a living. I've spent the better part of ten years learning this stuff. Instead of believing me, you're countering with intuitive arguments that just don't reflect the truth of the science.[/b]

But i have experimented... i have been growing for a long time now... and now when i try to find out about the science behind what i know, it seems to be conflicting.

Do the experiment again, only set up two areas with a plant in each of them. see what happens then... i am only concerned with light in as much as it interacts with our plants. for it is this that i truly want to understand. this is not personal, i don't have to believe you... as i'd much rather believe myself. my experience comes from actually growing plants... and everything you're saying does not fit in with what i know about growing weed.

ceestyle

10-19-2008, 11:13 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 19 2008, 02:32 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46322)</div>
I actually said that 2 400's together are not enough to grow a 2ft cola per plant... but a single 600 is... this has nothing to do with overall yield, but proves a greater penetration of the 600 in comparison to 2 400's.[/b]

Sorry, I used the wrong word. If you had two 400s right next to each other, it would be superior to a 600W in every way. If that means bigger colas, higher yield, whatever. It's more light.

<div class='quotetop'>QUOTE </div>
Well I see no plants in that experiment, and i cannot see how your graphs prove anything.[/b]

They prove that multiple weak light sources can sum to higher intensity than one stronger light source, which you apparently don't believe.

I'm not talking about plants. I'm talking about the physics of light, which you will not accept as true.

<div class='quotetop'>QUOTE </div>
But i have experimented... i have been growing for a long time now... and now when i try to find out about the science behind what i know, it seems to be conflicting.

Do the experiment again, only set up two areas with a plant in each of them. see what happens then... i am only concerned with light in as much as it interacts with our plants. for it is this that i truly want to understand. this is not personal, i don't have to believe you... as i'd much rather believe myself. my experience comes from actually growing plants... and everything you're saying does not fit in with what i know about growing weed.[/b]

Well, you can choose to not believe me if you like. I have focused on the physical nature of light, not the effect it has on plants. My statements are just reiterations of the commonly accepted physics of light. If you choose not to accept them, that is up to you.

My point is this: when an electrician comes onto a thread - which has happened previously - anyone who does not have that expertise would be poorly served by trying to intuitively reason through the fundamentals of circuits while the electrician attempts to explain to you how things actually work. The reason that is the case is because he has received a great deal of training in the fundamentals of circuit elements, their application, and the relevant science. So, while you or I may inquire *why* our observations don't seem to jive with a particular description, it would just be ludicrous to tell him that he is wrong based on my personal intuition and experience. I mean, it's just rude, in my opinion.

What I am saying is that I can describe everything you've observed (in your grows) in terms of the physics of light, yet you continue your disbelief.

skunkushybrid

10-19-2008, 11:30 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 20 2008, 12:13 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46333)</div>
So, while you or I may inquire *why* our observations don't seem to jive with a particular description, it would just be ludicrous to tell him that he is wrong based on my personal intuition and experience. I mean, it's just rude, in my opinion.[/b]

Ludicrous? lol... if something doesn't make sense to me i will say so... and i will keep saying so till the cows come home. I gave you a link of my own which you must've missed. http://lyndonashmore.com/tired_light_front_page.htm

<div class='quotetop'>QUOTE </div>
What I am saying is that I can describe everything you've observed (in your grows) in terms of the physics of light, yet you continue your disbelief.[/b]

As can I... light loses power as it travels http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/wink.gif

ceestyle

10-20-2008, 01:59 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 19 2008, 03:30 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46341)</div>
Ludicrous? lol... if something doesn't make sense to me i will say so... and i will keep saying so till the cows come home. I gave you a link of my own which you must've missed. http://lyndonashmore.com/tired_light_front_page.htm[/b]

Yes, ridiculous. It is one thing to say that something doesn't make sense, in terms of your observations - your personal experience doesn't seem to make sense in the context of the theory. It is another to take science facts that I am simply conveying to you and say they are false.

<div class='quotetop'>QUOTE </div>
As can I... light loses power as it travels http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/wink.gif[/b]

Well, in addition to this being debunked and not accepted by today's scientists (http://en.wikipedia.org/wiki/Tired_light), there are a few main points that make this concept irrelevant:

This refers to red-shifts of light, meaning that the wavelength changes; it doesn't mean that it changes speed or doesn't reach as far.

The phenomenon that is described occurs over distances on the magnitude of light-years, not meters. Furthermore, the redshift is in fractions of nanometers, undetectable to anything but the most sensitive scientific instruments.

Here. I think the earth is flat: http://www.alaska.net/~clund/e_djublonskop...arthsociety.htm (http://www.alaska.net/~clund/e_djublonskopf/Flatearthsociety.htm)

skunkushybrid

10-20-2008, 07:53 AM

You are not conveying science fact... but science theory...

as i said... grow some plants and see what the diffeence is. if i had the choice between those cfl's or 150w of hps, i'd pick the hps... guess you'd pick the flouro's.

so many people make these claims with the cfl's.. but we never get to see the finished grow. just pic's from the top of the bud, shit like that. fluoro's do not produce full colas... if you were a grower you'd know that.

the science theory may say one thing, but in practice it doesn't work... so that means the science is flawed somewhere. where as light losing power as it travels (as blue photons are more powerful than red ones) makes the most sense.

You can come up with as many equations as you like... show me some plants http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/smile.gif

ceestyle

10-20-2008, 04:39 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 20 2008, 12:53 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46524)</div>
You are not conveying science fact... but science theory...

as i said... grow some plants and see what the diffeence is. if i had the choice between those cfl's or 150w of hps, i'd pick the hps... guess you'd pick the flouro's.[/b]

Well, it depends what you think is fact or theory. It is accepted as fact that gravity will keep you attached to the ground following F=ma. If you discount that as theory rather than fact, then I guess I am. Let's just say everything I've stated is accepted the world over and describes every experiment that has ever been run to test it, and it has been amply. If you're implying that you have insight to a new theory to describe light, by all means lay it on me. From what I can see of your experiences, these are all effects that can be explained by the theory of light as written in textbooks that teach physics the world over ...

In the size box that 150W was in, I would take the CFLs over the HPS for veg absolutely. The coverage is just better, as the data show. For flower, I would use both. If I had to choose one, I'm not sure which I'd choose ... the coverage on the corners is pretty poor, and they are a lot heavier in IR. It started to cook my tops if I didn't LST them low.

<div class='quotetop'>QUOTE </div>
so many people make these claims with the cfl's.. but we never get to see the finished grow. just pic's from the top of the bud, shit like that. fluoro's do not produce full colas... if you were a grower you'd know that.[/b]

First of all, I am a grower. Did you not see the plants in the posts? That was unnecessary.

I already explained why CFLs don't have the same penetration as HID. Were you not reading? All of a sudden I'm a CFL advocate? WTF?

<div class='quotetop'>QUOTE </div>
the science theory may say one thing, but in practice it doesn't work... so that means the science is flawed somewhere. where as light losing power as it travels (as blue photons are more powerful than red ones) makes the most sense.[/b]

Dude, you are out of your mind ... the theory you linked has been proven incorrect .. not to mention it doesn't apply at the lengthscales we are talking about. Unless there are LIGHT YEARS between your bulbs and you plants and they can detect the difference of FRACTIONS of a nanometer wavelength of light, it makes absolutely no sense to invoke this outdated theory.

Give me one example of where the science doesn't work. I've explained everything you've brought up here.

The science is why you have your fancy grow systems in the first place .. I don't think you have a very healthy respect for it.

seamaiden

10-21-2008, 05:11 PM

So, last night I was communicating by flashlight with this guy on the moon....

ceestyle

10-21-2008, 05:31 PM

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 21 2008, 10:11 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=47128)</div>
So, last night I was communicating by flashlight with this guy on the moon....[/b]

and what did he say? I'm telling dave.

wait ... was it Ron Paul?

seamaiden

10-21-2008, 06:02 PM

Had to be, because I've yet to meet the resident Moonbat. http://www.skunkskool.com/style_emoticons/<#EMO_DIR#>/LMAO.gif

sgtpeppr

10-21-2008, 06:38 PM

<div class='quotetop'>QUOTE (Maiden of the Sea @ Oct 21 2008, 10:11 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=47128)</div>
So, last night I was communicating by flashlight with this guy on the moon....[/b]

Hey wait.....last night I was communicating with someone on the earth by flashlight!!!

seamaiden

10-22-2008, 03:20 PM

<div class='quotetop'>QUOTE (sgtpeppr @ Oct 21 2008, 10:38 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=47168)</div>
Hey wait.....last night I was communicating with someone on the earth by flashlight!!![/b]
Dude! That was YOU? Ok, so, are three fast clicks yes, or no? I was going as fast as I could.

skunkushybrid

10-23-2008, 01:27 PM

<div class='quotetop'>QUOTE (ceestyle @ Oct 20 2008, 05:39 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=46658)</div>
Well, it depends what you think is fact or theory. It is accepted as fact that gravity will keep you attached to the ground following F=ma. If you discount that as theory rather than fact, then I guess I am. Let's just say everything I've stated is accepted the world over and describes every experiment that has ever been run to test it, and it has been amply. If you're implying that you have insight to a new theory to describe light, by all means lay it on me. From what I can see of your experiences, these are all effects that can be explained by the theory of light as written in textbooks that teach physics the world over ...

In the size box that 150W was in, I would take the CFLs over the HPS for veg absolutely. The coverage is just better, as the data show. For flower, I would use both. If I had to choose one, I'm not sure which I'd choose ... the coverage on the corners is pretty poor, and they are a lot heavier in IR. It started to cook my tops if I didn't LST them low.

First of all, I am a grower. Did you not see the plants in the posts? That was unnecessary.

I already explained why CFLs don't have the same penetration as HID. Were you not reading? All of a sudden I'm a CFL advocate? WTF?

Dude, you are out of your mind ... the theory you linked has been proven incorrect .. not to mention it doesn't apply at the lengthscales we are talking about. Unless there are LIGHT YEARS between your bulbs and you plants and they can detect the difference of FRACTIONS of a nanometer wavelength of light, it makes absolutely no sense to invoke this outdated theory.

Give me one example of where the science doesn't work. I've explained everything you've brought up here.

The science is why you have your fancy grow systems in the first place .. I don't think you have a very healthy respect for it.[/b]

were not talking about light from the sun, but light from an electrical appliance... why would there need to be light years between the plants? Obviously the more powerful the light then the longer it will take to dissipate.

You've explained that the spread of the photons means there are less photons as you get toward the bottom of the plant, or even so far either side, etc.. and i can picture what you mean.

However, photons can change with the slightest interaction... gravity for example.

No, I didn't see any plants in that experiment.. although i would like to. I really need to see the cfl's match the HPS.

skunkushybrid

10-23-2008, 01:28 PM

and i'd really love to see them match the HPS in flower...

TetraHyC

10-23-2008, 01:46 PM

IDK if you'll ever match HPS, but I grew some nice CFL bud myself, just to see first hand how they came out. 12 (100w equiv) CFLs.

StealthPanda

10-23-2008, 03:05 PM

I think it is very possible to get similar results with a CFL setup as with an HID... Its just all about what you got, really, what you can honestly put together to get the results, And, in my personal experience, the buds grown by CFLs are honestly some of the densest buds i have EVER felt in my life, this may have to do with some one the factors of yeild.

I mean, yeah dude, HPS lights are bright, and grow huge buds. But its not like its the tip top notch stuff, its just the most efficient way to get top notch stuff.

ceestyle

10-23-2008, 04:30 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 23 2008, 05:27 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=47912)</div>
were not talking about light from the sun, but light from an electrical appliance... why would there need to be light years between the plants? Obviously the more powerful the light then the longer it will take to dissipate.[/b]

To see the 'tired light' phenomenon he mentioned, you'd need ridiculous amounts of distance over which to see this fractional effect.

<div class='quotetop'>QUOTE </div>
You've explained that the spread of the photons means there are less photons as you get toward the bottom of the plant, or even so far either side, etc.. and i can picture what you mean.[/b]

Good. Once you can see that, you can envision how lights add, and how coverage changes.

<div class='quotetop'>QUOTE </div>
However, photons can change with the slightest interaction... gravity for example.[/b]

Not on our length-scale. Light is affected by planetary gravity when traveling through space according to the general theory of relativity, but not in the context of your plants.

<div class='quotetop'>QUOTE </div>
No, I didn't see any plants in that experiment.. although i would like to. I really need to see the cfl's match the HPS.[/b]

If you know what the spectral distribution is and the areal distribution is, you don't need the plants. Same spectrum = same growth.

skunkushybrid

10-23-2008, 06:16 PM

well if only shit was that easy... i really need the plants.

I have heard this before you know... a guy called babygro and a couple of others that have come along with pretty much the same idea.

yes they make it sound great, and like it would work... but in practise it doesn't. grow some plants in there and see.

ceestyle

10-24-2008, 12:45 AM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 23 2008, 11:16 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=48001)</div>
well if only shit was that easy... i really need the plants.

I have heard this before you know... a guy called babygro and a couple of others that have come along with pretty much the same idea.

yes they make it sound great, and like it would work... but in practise it doesn't. grow some plants in there and see.[/b]

well, like i said, you can't physically get as many CFLs in an area to replicate an HID, so you can't really do the experiment ... and that's why you don't get the same results from CFLs that you do from HIDs.

skunkushybrid

10-24-2008, 07:26 AM

...but you just matched them... you used 150w of hps and those cfl's...

You said that you proved it... surely all that is left now is the final test?

ceestyle

10-24-2008, 03:43 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 23 2008, 11:26 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=48407)</div>
...but you just matched them... you used 150w of hps and those cfl's...

You said that you proved it... surely all that is left now is the final test?[/b]

I proved they were different!

skunkushybrid

10-24-2008, 03:51 PM

I already know they are different... but you also said that you were unsure whuich you would choose for flower. this must mean you believe those cfl's to be the equal of the HPS.

You've also said that photons are all the same, and blue ones are only as much use as red as the photons merely invoke a chemical reaction. Yet blue and red will clearly show you different growth patterns. indeed the plant under red will stretch more, which is similar to a shade response to my mind...

so this difference you speak of... what is the exact point of your experiment?

I sincerely apologise for missing it.

ceestyle

10-24-2008, 09:04 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 24 2008, 08:51 AM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=48522)</div>
I already know they are different... but you also said that you were unsure whuich you would choose for flower. this must mean you believe those cfl's to be the equal of the HPS.

You've also said that photons are all the same, and blue ones are only as much use as red as the photons merely invoke a chemical reaction. Yet blue and red will clearly show you different growth patterns. indeed the plant under red will stretch more, which is similar to a shade response to my mind...

so this difference you speak of... what is the exact point of your experiment?

I sincerely apologise for missing it.[/b]

The CFLs are only a possible replacement for 150W HPS equivalent and less. Otherwise, you can get the coverage you need with the power efficiency and penetration superiority of HID. At the power I compared, however, the coverage of HPS sucked. Granted, that was partially due to my homemade reflector, but the coverage was just really bad ... so much so that I would probably choose the CFLs alone if I had to choose.

skunkushybrid

10-24-2008, 10:53 PM

yes so with the coverage you would get a nice layer of scrawny bud... with the 150 HPS you'd at least get half a cola.

it also depends on what you term coverage... i mean you could consider the penetration aspect to be like coverage too.

The only way the coverage (widthways) would suck is if the light were too close to the plants. light emanates out until it hits something yes? if nothing was in the way it would form a perfect sphere. did you make accounts for the distance a HPS would need to be from the plants than the CFL's would?

the cfl's would need to be far closer to the plants, so this would negate some of the coverage. The great thing about cannabis is that it's a light loving plant, and like all plants relies on light for food. so they are ideal test subjects.

sometimes all the sums can add up, but when it comes to the test something is missing. I can understand fluoro for veg'... but for flower I've never seen them cut the mustard.

ceestyle

10-24-2008, 11:58 PM

<div class='quotetop'>QUOTE (skunkushybrid @ Oct 24 2008, 03:53 PM) <{POST_SNAPBACK}> (index.php?act=findpost&pid=48840)</div>
yes so with the coverage you would get a nice layer of scrawny bud... with the 150 HPS you'd at least get half a cola.[/b]

I don't know about the scrawny bud part, but the sides that didn't get covered by the HPS at all wouldn't give you shit. If you're running SOG, then the penetration is not an issue.

<div class='quotetop'>QUOTE </div>
The only way the coverage (widthways) would suck is if the light were too close to the plants. light emanates out until it hits something yes? if nothing was in the way it would form a perfect sphere. did you make accounts for the distance a HPS would need to be from the plants than the CFL's would?[/b]

I mean, you can look at those coverage plots and make that determination for yourself. If you have the HPS the right distance for good intensity, the coverage is crap.

<div class='quotetop'>QUOTE </div>
the cfl's would need to be far closer to the plants, so this would negate some of the coverage. The great thing about cannabis is that it's a light loving plant, and like all plants relies on light for food. so they are ideal test subjects.[/b]

Yeah, but the thing is that if you have six bulbs, you can ram them down on the plants and still get way better coverage.

<div class='quotetop'>QUOTE </div>
sometimes all the sums can add up, but when it comes to the test something is missing. I can understand fluoro for veg'... but for flower I've never seen them cut the mustard.[/b]

I mean, 99% of the time it's because people are just using not nearly enough of them.