Light attenuation/penetratability for HPS vs. CFL
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growingpeace
Member
The light intensity coming from HPS or MH lights is much stronger than that of CFLs I wanted to go CFL all the way through my grow but I'm very glad I decided on a HPS for flowering. You gotta get one to see the difference. My HPS is 250 watts and my CFL reflector has about 268 watts worth of bulb and its still not as brights as my HPS system. Not to mention I have to keep them 1-3 inches away to keep that light nice and strong. My HPS is hanging 7-10 inches above my plants and I get and 25-30 thousands lumens as oppose to 17,200 from the CFL setup
tyler.durden
Well-Known Member
The lumen differential between your cfl and hps set up is much greater than that. If you're getting 35k lumens from the hps and your plant is 10 away, you're probably getting close to what you posted, but you weren't getting anywhere near that with the cfl set up. Lumens don't add, much like decibels don't add: If a trumpet is playing at 90 decibels and another trumpeter starts playing next to him also at 90 decibels, you have two trumpets playing at 90, the decibels don't add to make it 180 decibels (could you imagine?) Same with lights, if you have a cfl putting out 5000 lumens close to other 5000 lumen bulbs, that's multiple sources of 5000 lumens, they don't add to make 10,000. There's no comparison in light intensity between cfls and hps.
As to the OP, you're right; the light intensity from the bulb follows the same formula of degradation over distance regardless of the type of light. Unless by degradation you mean that flouros decay slower than hps, that is true...
growingpeace
Member
Truth because the HPS is coming from once source and the CFL light is coming from multiple places. except where the light from two bulbs intersect then it may be a bit higher on the lumen rating.
heckler73
Well-Known Member
Assuming those two trumpets are being played with a zero phase shift, the (ideal) output combined would be 93dB.
At least that's what my intuition is telling me.
Lumens also sum, otherwise there would be no advantage to running two lights instead of one.
Furthermore, comparing HPS to CFL seems incorrect, since they do NOT have identical spectra.
To properly test the hypothesis, one could use filters at the ideal wavelengths for Chlorophyll A&B along with the carotenoid wavelengths, and then see what the relative intensities of the bulbs are.
This is an interesting puzzle though. I should sit down with the math later, and see if I can figure something out, unless someone else does so before I get a chance.
Anyone know the math off-hand?
EDIT: Thought I would add a little vid discussing Superposition of waves
[video=youtube;CAe3lkYNKt8]http://www.youtube.com/watch?v=CAe3lkYNKt8[/video]
tyler.durden
Well-Known Member
The reason for using multiple lights is to cover more area. Take a PAR meter or a lumens meter to 2 1000w lights that are touching rated at 150k, and you'll get the same reading below them as if there were only one. It won't be 300k lumens, simply 2 sources of 150k. Same with decibles, if you have two fire alarms at 90 decibels next to one another, you won't get a sound twice as loud, just multiple sources. The decibel level may be slightly more where the sound waves intersect, but you can round to 90 for all practical purposes. This is the reason you can hear a violin soloist above 30 other violins in the orchestra, if you added the decibels of the 30 violins playing, we'd never be able to hear the soloist. I can't believe how difficult it is to find scientific info about sound and light not adding, but I did find something on decibels -
Does one add decibels?
For example, if I have two computer fans each making 30 dB, how much noise do I hear? It's not 60, is it?
No, two 30 dB fans would equal 33 dB.
A 30 dB fan and a 10 dB fan would equal 30 dB. You wouldn't hear the much quieter fan.
================
Large increases in Intensity level of sound are not perceived as much "louder". In general a ten times increase in acoustic intensity is perceived as twice as loud. The decibel scale was invented to comply with this perceived measurement.
================
Decibels are a "relative measure", not an absolute measure. Each decibel measurement is related to a linear quantity.
A=10*log(a)
B=10*log(b)
To add the two signals you get 10*log(a+b) where you add the linear values.
================
In your original case a=b, and 30 dB = 10*log(a)
The sum of the two fans is:
10*log(a+b) or 10*log(2*a) = 10*log(a)+10*log(2) = 10*log(a)+3= 33 dB
You can calculate "a" if you want, but it is not necessary.
10*log(2)=3.010299957 but most people round it off to 3.
heckler73
Well-Known Member
You're not summing them correctly, as shown below.
So my original intuition was correct
3dB = a doubling of power (i.e. Watts), hence 90dB + 90dB = 93dB
Of course the ideal is difficult (if not impossible) to achieve, simply due to effects from destructive interference as the double-slit experiment demonstrates. Walter Lewin does an experiment in his 8.03SC (Physics III) lectures demonstrating that principle using sound waves.
You should be able to prove this to yourself by vector superposition.
Back to the light issue, what does a PAR meter measure?
For reference, a quantum refers to the amount of energy carried by a photon. Quantum meters approximate the quantity of photons between 400 and 700 nanometers. Photosynthesis is largely driven by the number of photons between these wavelengths. Photosynthetically Active Radiation (PAR) is also called the Photosynthetic Photon Flux (PPF) or Photosynthetic Photon Flux Density (PPFD) and the units are μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP] (micromoles of photons per meters squared per second).
http://www.apogeeinstruments.co.uk/quantum/
They are measuring the quanta of light in a specific bandwidth. If one has a source that puts out 100 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP] and one doubles that (by adding a 2nd source in phase), ideally one will get 200 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP]
http://physics.info/intensity/
The equation derived there shows all the factors which affect intensity.
If you are trying to argue this from the perspective of destructive interference, then sure, under a specific scenario, what you are implying could result, but that is HIGHLY unlikely.
In general addition of light sources will increase the intensity, although it may not double perfectly. After all, intensity does fall off at r^-2 for point sources.
I think Stardustsailor had a thread exploring this topic in gory detail in the LED section.
tyler.durden
Well-Known Member
Your intuition is good, Heckler. Thanks for clearing the decibel thing up, I understand that two 90 decibel sources in close proximity raises the level 30%. Not doubling, but not insignificant. I still don't know if this is valid:You're not summing them correctly, as shown below.
So my original intuition was correct
3dB = a doubling of power (i.e. Watts), hence 90dB + 90dB = 93dB
Of course the ideal is difficult (if not impossible) to achieve, simply due to effects from destructive interference as the double-slit experiment demonstrates. Walter Lewin does an experiment in his 8.03SC (Physics III) lectures demonstrating that principle using sound waves.
You should be able to prove this to yourself by vector superposition.
Back to the light issue, what does a PAR meter measure?
For reference, a quantum refers to the amount of energy carried by a photon. Quantum meters approximate the quantity of photons between 400 and 700 nanometers. Photosynthesis is largely driven by the number of photons between these wavelengths. Photosynthetically Active Radiation (PAR) is also called the Photosynthetic Photon Flux (PPF) or Photosynthetic Photon Flux Density (PPFD) and the units are μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP] (micromoles of photons per meters squared per second).
http://www.apogeeinstruments.co.uk/quantum/
They are measuring the quanta of light in a specific bandwidth. If one has a source that puts out 100 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP] and one doubles that (by adding a 2nd source in phase), ideally one will get 200 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP]
http://physics.info/intensity/
The equation derived there shows all the factors which affect intensity.
If you are trying to argue this from the perspective of destructive interference, then sure, under a specific scenario, what you are implying could result, but that is HIGHLY unlikely.
In general addition of light sources will increase the intensity, although it may not double perfectly. After all, intensity does fall off at r^-2 for point sources.
I think Stardustsailor had a thread exploring this topic in gory detail in the LED section.
They are measuring the quanta of light in a specific bandwidth. If one has a source that puts out 100 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP] and one doubles that (by adding a 2nd source in phase), ideally one will get 200 μmol m[SUP]-2[/SUP] s[SUP]-1[/SUP]
I'm not sure this demonstrates the doubling of lumens with two light sources in close proximity. You're good at looking these things up, and I will search around to try to support your hypothesis...