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Old 06-19-2008, 05:21 PM
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Default Lumens, lux, and adding it all up.
1. What is light?

Light is electromagnetic radiation, and may be interpreted as either a wave or a particle: thus the phrase wave/particle duality. It exhibits phenomena that may be explained by its wavelike characteristics - interference, for example - while also demonstrating some that are best explained by its particle-like properties - such as the photoelectric effect.

Treating light like a particle, a light source emits many light particles called photons. These are distinguishable by ONE unique property - their energy. When treating light like a wave, this is referred to as the wavelength, and is visible to us as a difference in color.

The main point is this: photons of the same energy (wavelength, color) from different sources, for example MH, HPS, CFL, T5, etc. are THE SAME.

2. What is light intensity?

Light intensity is simply the flux of photons per unit area per unit time. More photons means increased intensity. That's it.

3. What do lumens, lux, and foot-candles mean?

*Lumen:

This is the unit that defines "luminous flux", which is "radiant flux" (energy emission) adjusted for the sensitivity of the human eye. In the context of a light bulb, it is a measure of the total number of photons being emitted by the bulb (summed over all directions), adjusted for the sensitivity of the human eye to different photon energies.

What this means is that different colors are weighted differently according to a "luminosity function" that describes the sensitivity of our eyes. You can see this function and a more detailed description here.

*Lux, foot-candles:

The lux is simply defined as the number of lumens per second incident upon one square meter. In other words, it is the number of photons striking a finite area per unit time, weighted by the luminosity function described above.

The foot-candle is a derivative unit that is equal to 10.76 lux. It was defined based on the square foot, so the conversion is simply based on an area conversion.

4. What does this mean when using multiple light sources?

So we've established that lux are the number of photons striking a unit area per unit time, weighted by a luminosity function.

We've also established that photons from the same light source are indistinguishable, as long as they have the same energy/wavelength/color.

What this means is that if you put two lights the same distance from a point, and each light provides N photons per unit area at the point, with two lights you will have 2N photons per unit area at the point. Because intensity is a measure of the number of photons per unit area, the light is twice as intense, whatever unit you choose to use. Twice the lumens, twice the lux, twice the footcandles.

An obvious practical caveat to this point comes when using multiple low-intensity light sources. Notice how I stated that the lights were at the same distance? A practical problem with CFLs, for example, is that while you can get 27000 lumens from 10 x 42W CFLs, it's difficult to get them close enough to make them useful. If you have them in a line, for example, as I've shown below, each successive light is further from the meter, and the effective increase will be reduced. They still add, but according to the 1/d^2 rule, so having a bulb 2cm farther away will yield diminishing returns. On the other hand, this can be an effective way of distributing light, whereas with HID you need to distribute the plants around your single point light source.

5. Seriously? Prove it.

Here is a simple experiment that demonstrates this point. Below are some 42W bulbs that are a part of a flowering cabinet. I have suspended it from some pots for the sake of this experiment. You can see there is also a 150W HPS in there; I'll do some HID vs. CFL comparisons at some point as well.

The point is simple. Lights that are equidistant from a point contribute additively.

Ambient light = 0



1 x 42W = 6500 ftc ~ 65000 lux



2 x 42W = 13100 ftc ~ 131000 lux



Any questions?

More examples and experiments to come.

Last edited by ceestyle; 06-21-2008 at 04:14 PM..
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Old 06-19-2008, 05:55 PM
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Good post.

okay, so saying the lux doubles, but what does your meter say in lumens rather then lux?
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Old 06-19-2008, 06:44 PM
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Quote:
Originally Posted by firebullet View Post
Good post.

okay, so saying the lux doubles, but what does your meter say in lumens rather then lux?
Lumens aren't really meaningful in this case. To get the absolute number of lumens the meter reads, you'd just have to know the area of the photodiode in m^2 and multiply by the lux, which is lumens per m^2.

What you can do, however, is figure out how far away the meter is from the light source, figure out the total area that the lumen output is being distributed to, and backcalculate to get lumens.

The problem with such a calculation is that it assumes that the light is radiating equally in a spherical manner, which is certainly not the case unless the bulb is spherical.

Let me give you an example:

When I've got one bulb on at what is about 7", I get the reading of 65000 lumens / m^2 .

At 7", the area the light is being distributed to is:

4 * pi * r^2 = 4 * pi * (7")^2 = 615 in^2 ~ 3971 cm^2 = 0.397 m^2

So for 65000 lux = 65000 lumens / m^2 , we have:

65000 lumens / m^2 * 0.397 m^2 = 25805 lumens

Obviously, the bulb - rated at 2700 lumens, is not putting out ten times that. It is simply that the distribution of light is not spherical. At either end of the bulb, the lux reading is much lower. You can see how important it is to properly place your bulbs.

If instead you assume that the bulb is distributing light roughly in a cylindrical pattern the length of the bulb - approximately 3" - you get the following calculation.

Area of distribution:

2 * pi * r * L = 2 * pi * 7" * 3" ~ 132 in^2 ~ 850 cm^2 = 0.0850 m^2

So for 65000 lux = 65000 lumens / m^2 , we have:

65000 lumens / m^2 * 0.0992 m^2 = 5525 lumens

This is still not quite right, but the order of magnitude is right. The difference can be explained by the rapid falloff in intensity as you move away from the center of the coil. In addition, the panel above the lights is covered in mylar, which reflects rather effectively.
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Old 06-19-2008, 06:52 PM
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i want to see you put 3 and 4 cfls over that meter... and then compare it to the 150 watt HPS,, that would be very interesting
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Old 06-19-2008, 06:59 PM
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It will. What will be especially interesting is to compare the light distribution.

By the numbers, each should provide about the same lumen count.

6 x 42W = 6 x 2700 lumens ~ 16200 lumens

1 x 150W HPS ~ 16000 lumens

There will be errors due to my cheapass HPS and the different spectra, age, etc., but we should be able to get the point.

What is especially interesting is that at the corners we should see better coverage from the CFLs, as they're closer, while at the center, the HPS should be better. And of course all of this will change with the distance from the canopy.

Let experimentation begin! (but after dark, 'cause it's hot as shit in this room atm)
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Old 06-19-2008, 08:23 PM
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pics man,, PICS!
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Old 06-19-2008, 08:31 PM
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This should be stickied.
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Old 06-19-2008, 08:43 PM
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This should be stickied.
I don't know what that means, but ok.
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Old 06-19-2008, 09:00 PM
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Ask fdd. He has the powah.

Stickied means that it's permanently "pasted" at the top of a given forum. It sticks on top no matter what.
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Old 06-19-2008, 09:15 PM
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Ask fdd. He has the powah.

Stickied means that it's permanently "pasted" at the top of a given forum. It sticks on top no matter what.
we'll see how it's received. i'm getting a lot of "CFLs suck, so we don't care", but it's applicable to HID too...

man, i should take a picture of the burn on my arm i just got moving my light meter under that HPS. *sizzle*
 

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