14x1750=1750 WTF, Small Cfl Array, Input Appreciated

[Al B. Fuct]

Thanks much.

No worries

What do you mean by 'foliar penetration'?

Do you mean light getting past the canopy to lower leaves? Or do you mean light penetrating outer leaf tissues or something to be used as energy by the plant?

All of the above. Energetic photons have a better chance of doing the job than more of the less energetic ones.

 

[Maccabee]

Oh, DAYUM, he went there. Well, OK then. Let's get it on!

From Wikipedia:

Photosynthesis uses light energy and carbon dioxide to make triose phospates (G3P). G3P is generally considered the prime end-product of photosynthesis. It can be used as an immediate food nutrient, or combined and rearranged to form disaccharide sugars, such as sucrose and fructose, which can be transported to other cells, or packaged for storage as insoluble polysaccharides such as starch.

A commonly used but slightly simplified equation for photosynthesis is:

    6 CO2(g) + 12 H2O(l) + [B]photons[/B] → C6H12O6(aq) + 6 O2(g) + 6 H2O(l)
    carbon dioxide + water + [I]light energy[/I]→ glucose + oxygen + water

When written as a word equation the light energy appears above the arrow as it is required for photosynthesis but it is not actually a reactant. Here the monosaccharide glucose is shown as a product, although the actual processes in plants produce disaccharides.

The equation is often presented in introductory chemistry texts in an even more simplified form as:[2]

    6 CO2(g) + 6 H2O(l) +[B] photons[/B] → C6H12O6(aq) + 6 O2(g)

Photosynthesis occurs in two stages. In the first phase, light-dependent reactions or photosynthetic reactions (also called the Light reactions) capture the energy of light and use it to make high-energy molecules. During the second phase, the light-independent reactions (also called the Calvin-Benson Cycle, and formerly known as the Dark Reactions) use the high-energy molecules to capture carbon dioxide (CO2) and make the precursors of carbohydrates.

In the light reactions, one molecule of the pigment chlorophyll absorbs one photon and loses one electron. This electron is passed to a modified form of chlorophyll called pheophytin, which passes the electron to a quinone molecule, allowing the start of a flow of electrons down an electron transport chain that leads to the ultimate reduction of NADP into NADPH. In addition, it serves to create a proton gradient across the chloroplast membrane; its dissipation is used by ATP Synthase for the concomitant synthesis of ATP. The chlorophyll molecule regains the lost electron by taking one from a water molecule through a process called photolysis, that releases oxygen gas.


In the Light-independent or dark reactions the enzyme RuBisCO captures CO2 from the atmosphere and in a process that requires the newly-formed NADPH, called the Calvin-Benson cycle releases three-carbon sugars, which are later combined to form sucrose and starch.

Photosynthesis may simply be defined as the conversion of light energy into chemical energy by living organisms. It is affected by its surroundings and the rate of photosynthesis is affected by the concentration of carbon dioxide, the intensity of light , and the temperature.

 

[Al B. Fuct]

One more note on foliar penetration.

Tissues that make up leaves and seed bracts (the main component of buds) are not quite opaque. Some light will pass through them. The greater the oomph of the photons, the better the chance they will penetrate seed bract tissues to support formation of big, dense buds on the shaded, inner parts the bud.

Fluoros by their nature can't make the intensity needed to make really large, dense buds all over a plant. However, I've seen some neat, dense colas produced with CFLs in clip-type bell reflectors on a pole beside a plant, one CFL per bud cluster. Plants tend to elongate under fluoros, so the CFLs have to be moved daily. It's really heroic stuff but quickly becomes impractical for much more than toy grow ops.

 

[Al B. Fuct]

ZOMFG... not the CALVIN CYCLE!!

aieeeeeeeeeeee......

 

[Maccabee]

ZOMFG... not the CALVIN CYCLE!!

aieeeeeeeeeeee......

Growing pot chains you to it, like a galley slave.


PEDAL FASTER! WE'RE IN FLOWERING!

 

[Tanuvan]

Ok, in its simplest sense...if you have 1sq ft, and you stick one 42 watt cfl rated at 2700 lumens...and measure...and then you stick 8 42 watt 2700 lumen cfl's in that same space...you are telling me that the light in that area of 1 CFL bulb is just as bright as 8 bulbs in that 1 sqft area?

Something is getting added because there are 8 times the number of photons hitting that area. As I have stated before, lumens are a measure of perceived brightness. That area appears brighter than just with 1 bulb. Also you can't use single point formula's that apply to a light like HID or incandescent to a Fluorescent bulb.

Also I love how people just start mixing everything up...we have people quoting lumens, luminous intensity, light intensity, and lux (even using lux meters) all of which are different things.

Lux versus lumen

The difference between the lux and the lumen is that the lux takes into account the area over which the luminous flux is spread. 1000 lumens, concentrated into an area of one square metre, lights up that square metre with an illuminance of 1000 lux. The same 1000 lumens, spread out over ten square metres, produces a dimmer illuminance of only 100 lux.

It seems that the math here is far more complicated than people want it to be. It is not as simple as whether lumens add or not. It is also a function of perceived brightness and total area.


I'd like to see some lux meter readings at different areas...ie (by adding more bulbs inside of a 1 foot box) If you have 100 CFL's and measure them at 1 inch, it will read the same because the number of photons hitting that one spot will be equal due to it's surface area.

What you are not taking in account is the area being covered. Light intensity can't be added, but luminous output can be increased per given area. If that were not the case...how does adding more lights make a room brighter?

What do those charts floating around that show how many Flouro's are needed to approximate an HID? (We've all seen those charts)

What are Flouro manufacturers quoting when they say they have a 40K lumen 8 bulb light fixture?

For anyone with any serious photography background...this would be readily obvious. When you want to increase the illumination on a subject...we point more light at it. How does this happen? Why does the subject get blown highlights if you keep adding more light to the subject? Film is measuring light isn't it?

 

[Tanuvan]

All of this bulb moving crap can be mediated with a Scrog setup. I've not experienced having to reposition bulbs around anymore more so than with my HPS in a Scrog. And yield has been about the same as well given similar Lumens. 28K lumens on my HPS vs 2700 lumens a whole bunch of times with my CFL's..since we are saying that somehow the lumens aren't adding. The results even out.

No one has yet to say why dense buds are better than the supposed airy buds of CFL. 8 oz of dense bud vs 8 oz of airy bud...is still 8 oz of bud...it still smokes...still makes hash...still makes oil...Am I missing something?

Now, for a standard non LST/Scrog over a large area...HPS is King for now.

 

[Maccabee]

Film is measuring light isn't it?

It's measuring light, but not necessarily its intensity. Look at the chemical equation for photosynthesis: that reactions happens for each photon, not once the photon bucket fills to a certain point. When you carry out the reaction with a higher energy photon, you get more out of it. Or maybe it costs less, making it more efficient. I'm a little vague there.

Furthermore, using higher energy photons helps deliver light to more areas, as adding point sources only goes so far. Every square inch of photosynthetically active tissue can't be brought to within an inch of the CFL tube.


More intense light is better. Increasing the amount of light delivered also helps, up to a point. Increasing the number of lights can increase the amount of light delivered, but there is also a logistical limit there. So you're working within the limit of benefit possible to derive from increased light volume, and then within the smaller yet limit of how how to practically approach that theoretical limit.

 

[Tanuvan]

"Right. The luminous output of a lamp is measured in lumens, the amount of light energy hitting a particular area is quantified in lux (lumens per m2).

Since intensity reduces with distance from the light source by an inverse square function, the greatest lux figure available will be right up against the lamp's tube." Quoted from AL

This is only the case for point light sources like HID. The greatest lux is closest to the bulb because the light is emanating from one spot.

 

[Maccabee]

All of this bulb moving crap can be mediated with a Scrog setup....The results even out.

I tend to agree from my limited experience. But people who have been at it longer than I maintain that there will always be a quantitative and qualitative difference in product between relying on many weaker sources and HID lighting.

>shrug<

I say, do both. Everybody has room to experiment with CFLs if curious.

No one has yet to say why dense buds are better than the supposed airy buds of CFL. 8 oz of dense bud vs 8 oz of airy bud...is still 8 oz of bud...it still smokes...still makes hash...still makes oil...Am I missing something?

No, not really. It's a taste thing, and there's a vast preference for the dense bud associated with HIDs (not that people want pellets either.) I think those who discourage CFL grows feel that the airer buds are in fact lighter. You don't get the same weight, or it's not the same potency.

My growing experiences don't really amount to enough data for me to have a strong opinion. Maybe a little ways down the road.

 

[Tanuvan]

It's measuring light, but not necessarily its intensity. Look at the chemical equation for photosynthesis: that reactions happens for each photon, not once the photon bucket fills to a certain point. When you carry out the reaction with a higher energy photon, you get more out of it. Or maybe it costs less, making it more efficient. I'm a little vague there.

Furthermore, using higher energy photons helps deliver light to more areas, as adding point sources only goes so far. Every square inch of photosynthetically active tissue can't be brought to within an inch of the CFL tube.


More intense light is better. Increasing the amount of light delivered also helps, up to a point. Increasing the number of lights can increase the amount of light delivered, but there is also a logistical limit there. So you're working within the limit of benefit possible to derive from increased light volume, and then within the smaller yet limit of how how to practically approach that theoretical limit.

I beg to differ. Film does register different light intensity. Try taking a picture of the Sun, vs a Candle.

 

[Maccabee]

"

This is only the case for point light sources like HID. The greatest lux is closest to the bulb because the light is emanating from one spot.

No, AFAIK it's the case for all light sources, it's just that the emitter has a larger surface area in a tube or CFL so there's more complex 'shape' to the pattern of radiated energy that drops off according to the inverse square law rather than a straightforward cone of rays. Think about an HID and a reflector for a minute and ask yourself if that doesn't seem like a distinction w/o much of a difference.

We need a physics kid in the thread, to set us straight and then make fun of us.

 

[Tanuvan]

It isn't the case for all sources. Just point sources of light. HID has photons radiating from a central localized area. The light distribution is vastly different than a flouro tube. CFL including straight tubes radiate light along the entire surface area...which is why they give the impression of being physically cooler lights.

People are also forgetting when they call CFL "Toy" grows that T5 is also CFL technology and people are having great results with those as well. I have BOTH technologies and have no stake in either. I just want people who downplay Flouros to play fair.

It seems the only real argument I hear about is "airy" bud...which to me really doesn't matter...especially when we are talking the same weight. The only real drawback I have seen is when people attempt to use Flouro lights like it has the intensity of HID and try and grow tall plants...then bash Flouros as inferior when in reality it is there technique which accounts for their dismal yields.

GK has proven without a doubt as well as many other growers that Flouros can produce very well over small grows which a good majority of people on here do anyway. Just don't use flouros like an HID.

 

[Maccabee]

GK has proven without a doubt as well as many other growers that Flouros can produce very well over small grows which a good majority of people on here do anyway. Just don't use flouros like an HID.

That's true, but GK defers to Al when it comes time to explain the whys and hows of lumens, lux, etc.

Don't use flouros like an HID....and don't expect the same growth profile.

 

[Tanuvan]

That's true, but GK defers to Al when it comes time to explain the whys and hows of lumens, lux, etc.

Don't use flouros like an HID....and don't expect the same growth profile.

Indeed Sir.

 

[Maccabee]

It isn't the case for all sources. Just point sources of light. HID has photons radiating from a central localized area. The light distribution is vastly different than a flouro tube. CFL including straight tubes radiate light along the entire surface area...which is why they give the impression of being physically cooler lights.

How does that inhibit the operation of the inverse square law as the observation point is moved away from the emitter surface? The inverse square law still operates when HID light is diffused by a reflector, altering the patter of distribution of the radiated energy from that of a naked bulb hanging on a cord.

I'm no physics genius, but I don't think what you're saying is quite on point. It's true--but not really on point. Meh. I feel like calling someone up from college and hassling them about it, but then they'd want to know why.

Maybe we should go try and rope somebody from the XKCD forums in.

 

[Tanuvan]

How does that inhibit the operation of the inverse square law as the observation point is moved away from the emitter surface? The inverse square law still operates when HID light is diffused by a reflector, altering the patter of distribution of the radiated energy from that of a naked bulb hanging on a cord.

I'm no physics genius, but I don't think what you're saying is quite on point. It's true--but not really on point. Meh. I feel like calling someone up from college and hassling them about it, but then they'd want to know why.

Maybe we should go try and rope somebody from the XKCD forums in.

I was not referring to the measure of intensity. I was referring to how the area affects the Lux meter reading in particular. But I was sure that the the inverse square law was for point light sources. Even still, you can have a flouro tube much closer to a plant and thus negate a good portion of the inverse square law...which mind you... is the whole point of the way GK setup his lights...and the whole point of a SCROG.

The real problem here is that no one knows just how much intensity or lumens are required before diminishing returns sets in. Here is some more reference :

As light intensity increases, the photosynthetic rate increases until a point is reached where the rate begins to level off. At low light intensity, photosynthesis occurs slowly because only a small quantity of ATP and NADPH is created by the light dependent reactions. As light intensity increases, more ATP and NADPH are created, thus increasing the photosynthetic rate. At high light intensity, photosynthetic rate levels out, not due to light intensity but due to other limiting factors, including competition between oxygen and carbon dioxide for the active site on RUBP carboxylase.

 

[Maccabee]

That's interesting, especially the part about not knowing when diminishing returns occur.

Looking at Inverse-square law - Wikipedia, the free encyclopedia, I found this:

Light and other electromagnetic radiation

The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only ¼ the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

For example, the intensity of radiation from the Sun is 9140 watts per square meter at the distance of Mercury (0.387AU); but only 1370 watts per square meter at the distance of Earth (1AU)&#8212;a threefold increase in distance results in a ninefold decrease in intensity of radiation.

Photographers and theatrical lighting professionals use the inverse-square law to determine optimal location of the light source for proper illumination of the subject.

The fractional reduction in electromagnetic fluence (&#934 for indirectly ionizing radiation with increasing distance from a point source can be calculated using the inverse square law. Since emissions from a point source have radial directions, they intercept at a perpendicular incidence. The area of such a shell is 4&#960;r2 where r is the radial distance from the center.

The law is particularly important in diagnostic radiography and radiotherapy treatment planing, though this proportionality does not hold in practical situations unless source dimensions are much smaller than the distance r.

Also, this:

Field theory interpretation

For an irrotational vector field in three-dimensional space the law corresponds to the property that the divergence is zero outside the source. Generally, for irrotational vector field in n-dimensional Euclidean space, inverse (n &#8722; 1)th potention law corresponds to the property of zero divergence outside the source.

I'm still tempted to think that a flouro tube operates more or less like a point emitter so far as this is concerned. Or at least in keeping with teh spherical wavefront thing. Photons are emitted from the tube, from a point on its surface (many points at the same time, in fact) and travel outwards. HID's have arc tubes too! They're just sealed in a glass envelope for safety. So are they also not point sources? And, again, the reflector.

I wish I had more science. Growing pot makes you want to know more about physics, chemistry, botany.....

Oh well. Just as soon as that IP over NPK RFC comes out, I'll be ready to help.

EDIT:
The articles on illuminance and irradiance are interesting. Illuminance mentions the following measurements:

  Quantity     Symbol     SI unit     Abbr.     Notes
Luminous energy     Qv     lumen second     lm·s     units are sometimes called talbots
Luminous flux           F     lumen (= cd·sr)     lm     also called luminous power
Luminous intensity     Iv     candela (= lm/sr)     cd     an SI base unit
Luminance         Lv     candela per square metre     cd/m2     units are sometimes called nits
Illuminance         Ev     lux (= lm/m2)     lx     Used for light incident on a surface
Luminous emittance     Mv     lux (= lm/m2)     lx     Used for light emitted from a surface
Luminous efficacy           lumen per watt     lm/W     ratio of luminous flux to radiant flux; maximum possible is 683.002 lm/W

And irradiance these:

 Quantity     Symbol     SI unit     Abbr.     Notes
Radiant energy     Q     joule     J     energy
Radiant flux     Φ     watt     W     radiant energy per unit time, also called radiant power
Radiant intensity     I     watt per steradian     W·sr−1     power per unit solid angle
Radiance     L     watt per steradian per square metre     W·sr−1·m−2     power per unit solid angle per unit projected source area  Sometimes confusingly called "intensity".
Irradiance     E     watt per square metre     W·m−2     power incident on a surface. Sometimes confusingly called "intensity".
Radiant exitance/emittance     M     watt per square metre     W·m−2     power emitted from a surface.
Radiosity     J or Jλ      watt per square metre     W·m−2     emitted plus reflected power leaving a surface
Spectral radiance  Lλ o rLν watt per steradian per metre3 or watt per steradian per square metre per hertz    W·sr−1·m−3 or W·sr−1·m−2·Hz−1 commonly measured in W·sr−1·m−2·nm−1
Spectral irradiance Eλ or Eν     watt per metre3 or watt per square metre per hertz  W·m−3 or W·m−2·Hz−1 commonly measured in W·m−2·nm

Also, the article on photometry might be informative. I'm going to try and make some more sense of these. Condsidering all the different possible light measurements and trying to understand how their units interrelate is going to take a little while.

 

[Tanuvan]

I agree with you in that respect Maccabee. Any photon loses intensity with distance as it travels unless in vacuum. However as far as light intensity per given area...I was saying that an HID focuses light like this:
____
/ \ vs a Flouro which is like ================= Thus the brightness or Lux per area is different.

I still believe that for some reason in these debates...people love to start arguing over the addition of lumens...and comparing them to HID. This is almost as bad as debating religion or politics. Regardless of whether lumens add or not, we know that the key to getting the yields that GK experienced lies in adding more CFL's and keeping them close(Whichever means you choose in keeping them close). Who really cares if the light is adding or not...other than for a mental debate or theoretical exercise. What we want are end results. GK gave us a simple formula to work with to yield half a pound...which is nothing to sneeze at. (That formula being 2700 lumens * 8 42watt bulbs...or for those who believe they add...that is 21600 lumens)

Again quality and potency seems to not be affected. "Airy" buds is subjective...and used in a negative way for some reason as if airy buds don't contain resin or something. I can't smoke "density" anyway...so I really am perplexed by that.

The real questions always go unanswered...the questions that matter most...optimum spectrum and when is enough...enough? i.e. how much light do we need to be the most efficient. If adding another 2-3 CFL's don't increase yield...then it is best to not add them.

 

[Maccabee]

It get's worse! My mind is going to leak out my nose.

The lumen is the photometric unit of light output. Although most consumers still think of light in terms of power consumed by the bulb, in the U.S. it has been a trade requirement for several decades that light bulb packaging give the output in lumens. The package of a 60 watt incandescent bulb indicates that it provides about 900 lumens, as does the package of the 15 watt compact fluorescent.


The lumen is defined as amount of light given into one steradian by a point source of one candela strength; while the candela, a base SI unit, is defined as the luminous intensity of a source of monochromatic radiation, of frequency 540 terahertz, and a radiant intensity of 1/683 watts per steradian. (540 THz corresponds to about 555 nanometres, the wavelength, in the green, to which the human eye is most sensitive. The number 1/683 was chosen to make the candela about equal to the standard candle, the unit which it superseded).


Combining these definitions, we see that 1/683 watt of 555 nanometre green light provides one lumen. [Thus, the lumen scale is calibrated to a point in the spectrum that photosynthesis doesn't use -- Mac]


The relation between watts and lumens is not just a simple scaling factor. We know this already, because the 60 watt incandescent bulb and the 15 watt compact fluorescent can both provide 900 lumens.


The definition tells us that 1 watt of pure green 555 nm light is "worth" 683 lumens. It does not say anything about other wavelengths. Because lumens are photometric units, their relationship to watts depends on the wavelength according to how visible the wavelength is. Infrared and ultraviolet radiation, for example, are invisible and do not count. One watt of infrared radiation (which is where most of the radiation from an incandescent bulb falls) is worth zero lumens. Within the visible spectrum, wavelengths of light are weighted according to a function called the "photopic spectral luminous efficiency." According to this function, 700 nm red light is only about 4% as efficient as 555 nm green light. Thus, one watt of 700 nm red light is "worth" only 27 lumens. [This suggests to me that lumen ratings would be misleading for comparing, say, HPS to CMH. CMH is much greener light, and will appear to be 'worth' more lumens, even though it should be less useful to flowering Cannabis.-Mac]

Because of the summation over the visual portion of the EM spectrum that is part of this weighting, the unit of "lumen" is color-blind: there is no way to tell what color a lumen will appear. This is equivalent to evaluating groceries by number of bags: there is no information about the specific content, just a number that refers to the total weighted quantity.

It's like someone just told me that Santa Claus and the Easter Bunny aren't real. Plus, the grocery bag analogy is now fighting for space in my head with the 'lumens don't add' concept.

There aren't any reliable measurements! Ladies and Gentlemen, we are floating in Outer Space.