1550 watts of cfl s

ceestyle

Well-Known Member
You're not very good at admitting when you're wrong. It's not becoming. All the science, anecdotal, and experimental evidence is contrary to your assertion, which - from what I can make of your attempt at explanation - is based on your flawed understanding of the way that light works.

If you take a look at the thread in my sig about mapping intensity of lights, it is completely obvious to all but those practiced in cognitive dissonance that you are absolutely wrong.

Get over it. You could man up and admit it, but I never expected that.
 

ceestyle

Well-Known Member
Nope, wrong answer! Get a lux meter and try it yourself. I did.
I'm sure there are others here with lux meters. Please do try it and post your results. Even a three-way meter is better than nothing. ABF clearly won't listen to me, or the body of evidence to the contrary of his intuition.

ABF, why don't you try the experiment again, putting the lights equidistant from the meter, say a foot, rather than 2", and turn them on one at a time?
 
Last edited:

ceestyle

Well-Known Member
I'm not wrong. I did the experiment, got the result I reported, the science backs me up.
What science? Read this thread . This is the science. You can shout spaghetti monster until you're blue in the face, but it doesn't make it so.

I've pointed out the flaws in your explanation, and how they are inconsistent both with the laws of physics, and how the physics is inconsistent with your experiment. I challenge you to point out a flaw in the science I have outlined, which is consistent with both my experiment, the results of people growing with multiple CFLs, and ... did I mention physics?
 
Last edited:

Al B. Fuct

once had a dog named
The science- lumens are a measure of intensity. No lamp gets brighter by putting one lamp next to another.
 

ganjagoddess

Well-Known Member
Im not arguing intensity of light.

We are arguing lux. and lumens are lux/m2

intesity/par is completely different...
 

ceestyle

Well-Known Member
The science- lumens are a measure of intensity. No lamp gets brighter by putting one lamp next to another.
Read my post. Lumens are proportional to the number of photons. Flux is the number of incident photons per unit area per unit time. Where do you think the photons from a second light are going?

Add a light, add photons. Add photons, add intensity. What don't you get?

If you took your thumbs out of your ears for long enough, you might understand that. Try actually reading my other thread.
 
Last edited:

ceestyle

Well-Known Member
Im not arguing intensity of light.

We are arguing lux. and lumens are lux/m2

intesity/par is completely different...
Lumens is essentially the raw amount of photons emitted in total from the bulb, adjusted for the sensitivity of our eyes. The number of lumens is therefore not a function of where your target is - plants, whatever.

Lux is an intensity - lumens / m2 - that is the number of photons striking a unit area per second. This therefore depends on where you make the measurement.

So intensity is therefore the relevant quantity for plants. The measurements I have made are of intensity, and the most meaningful quantity.
 

ceestyle

Well-Known Member
Here. Maybe a picture will help. You see the intensity map below? It's in lux - intensity, right? It's a map of the intensity below two rows of three CFLs, also shown below.



If intensity didn't add, then why would the intensity be highest in the middle of three bulbs?

IF they did not add, there would be a constant intensity underneath all three bulbs. Obviously this is not the case.

Denying this any further is just being obstinate.
 

ganjagoddess

Well-Known Member
This explanation makes it easy to understand why it would be impossible to replicate a 600 watt HPS or MH using alternate lighting sources.

and how it might, be possible to come within the distance of a 400 using HO- CFL's
 

ceestyle

Well-Known Member
Yes. If you could increase the lumens/bulb area of a fluoro, you could do it. With CFLs, however, it's much easier to simply increase the area of the bulb. The stuff inside a fluorescent bulb just makes a certain amount of light per unit volume, and it's not as dense, power-wise, as an HID. Hence, High Intensity Discharge, where intensity means photon flux per bulb area. With a fluoro, you just have to add more bulb area to get the same amount of light. Light is light - at a given color, anyway.
 

bobtokes

Well-Known Member
GG just checked out your journal i will be following it. my grow area is about half the size of yours and hope to get some tips.
i toke cos ive got ms fair play to ya for helping ppl hope every thing goes well for ya good luck.
 

BigBudBalls

Well-Known Member
Hey cee, just inquiring, in your experiment and photos, lux/lumens are a direct computation. Now has the multiple lamp lux number ever exceeded the rating of a single? You have used the sensor further away from the bulb (ABF had his right up against them) Now plants are usually a little bit away, but under the lumens add up, try this, put the meter right up against the bulb, get a reading, then turn on 5 or so of them and drop the sensor down a foot and see if the number *rises*.
 

ceestyle

Well-Known Member
Hey cee, just inquiring, in your experiment and photos, lux/lumens are a direct computation. Now has the multiple lamp lux number ever exceeded the rating of a single? You have used the sensor further away from the bulb (ABF had his right up against them) Now plants are usually a little bit away, but under the lumens add up, try this, put the meter right up against the bulb, get a reading, then turn on 5 or so of them and drop the sensor down a foot and see if the number *rises*.
There really is not a lux rating for a bulb. There is only a lumen rating, and a lux reading from the meter. Elsewhere (in this thread I believe) I made a back-calculation to compute the lumens using an assumed light distribution and the measured lux. It was fairly close, but it's just too difficult to know how exactly the light is distributed well enough to make it just right.

In any case, you can take a known reading for a distance from the bulb, and calculate the anticipated change in intensity when adding bulbs based on that. Obviously, the simplest case of that is the one that I've demonstrated, where both lights are equidistant. You just multiply times two. If you throw in geometry, it gets more complicated, and you see why it's practically difficult to add all those lumens in a finite space. Here goes.

Basically, if the sensor is "D" inches directly under the bulb, and the bulbs are spaced laterally at a distance of "S", for three bulbs you'd get a combined intensity of:

I = Io * (1/D^2 + 2/(D^2 + S^2))

where Io is some constant that defines the intensity of whatever bulb you're using as a function of the distance D. This will be a function obviously of the wattage and type of bulb.

If you're using five bulbs spaced laterally, you get:

I = Io * (1/D^2 + 2/(D^2 + S^2) + 2/(D^2 + (2S)^2))

etc...

I can draw a diagram that would help explain that.

What this means is that at any distance D from the bulb, as you pull the meter away, the intensity will decrease. This is simply because if you are centered below the bulb, as long as light is not obscured from the side bulbs, you are always increasing the distance from all of the bulbs.

Without care, this result might be obscured slightly by the experimental condition that the meter remains flat, and as you pull it away, a wider angle of light reaches the meter from the side bulbs. This is not practically relevant, as the plant will adjust its foliage to maximize this angle. This would be experimentally observed by tilting the meter so that it is perpendicular to the bulb.

It might be interesting to compare the theoretical to predicted results, but I"m satisfied with it thus far. About to watch Superhighme. Cheers!
 

ceestyle

Well-Known Member
Okay, so check this out. If you have bulbs in an array separated by a distance S, you center a target below the center bulb and slowly draw it away, the intensity is affected as shown below, for the number of bulbs indicated. The x-axis on the plot is normalized by the separation, so that at D/S = 1, the target is as far from the center bulb as they are from each other.





You can see that at close distances, the other bulbs don't even have an effect, are especially effective at medium distances, and at long range they are additive as essentially one point source. Make sense?
 
Top