# Question about lumens and lamps.



## HomerSimpson (Mar 16, 2009)

So I'm just wondering how lumen ratings work with regard to bulbs.

For example, using the surface area of a sphere, measures using the radius: 40,000 lumens becomes 3200 lumens at 1 ft. 2 ft is 800 lumens. 18,000 lm is 5 inches away. Correct?

Where as a 1600 lumen CFL emits 130 lm at one foot. 2 inches is 4600 lm. Once inch is 18,000 lm. Correct?

Thanks.


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## asdalton (Mar 16, 2009)

No, lumens is a measurement of total light output, which is independent of distance. A lamp has a certain lumen output, period.


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## HomerSimpson (Mar 16, 2009)

Ok, so then at 1 foot the lamp would have that lumen rating per sq ft?

What about getting closer to the lamp than one foot?

Since I = P / A or i = p / r ^ 2.

Would, say, a 10,000 lumen lamp, @ 6" be 40,000 lumens?


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## StarHalo (Mar 16, 2009)

asdalton said:


> lumens is a measurement of total light output, which is independent of distance.



A 10,000 lumen lamp viewed at a distance of one mile is...10,000 lumens.

The lumens-per-meter measurement you're looking for is Lux: http://en.wikipedia.org/wiki/Lux


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## HomerSimpson (Mar 17, 2009)

Ok. So say I have two flashlights.

And I point their beams at a single point one foot away from each of them. This would double the lux, correct?


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## StarHalo (Mar 17, 2009)

If both lights have the same lux score individually, then yes.


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## ponygt65 (Mar 17, 2009)

ALso research "Foot Candles".

That may give you more understanding.


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## Ken_McE (Mar 17, 2009)

"Lumens" is a measure of the total amount of light emitted by an object. you can change how much of that light falls on you, but you are not changing how much light it emits. 

What does change is the amount of light that falls on a given surface, and this does of course shrink as you get further way. We use different units to measure how much light is recieved by a surface such as lux and foot candles. 

You may be interested in: http://en.wikipedia.org/wiki/Lux


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## HomerSimpson (Mar 18, 2009)

So, If I have, say four 25 watt bulbs, and compare it to a 100 watt bulb, and for the sake of this, they produce the same lumen/watt output.

If I take those 4 bulbs, and put them in a tight grid(like a square), I'd have very nearly the same lux, using those four 25W bulbs compared to one 100W, when measuring lux at identical distances between the two setups, correct?


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## StarHalo (Mar 18, 2009)

HomerSimpson said:


> So, If I have, say four 25 watt bulbs, and compare it to a 100 watt bulb, and for the sake of this, they produce the same lumen/watt output.
> 
> If I take those 4 bulbs, and put them in a tight grid(like a square), I'd have very nearly the same lux, using those four 25W bulbs compared to one 100W, when measuring lux at identical distances between the two setups, correct?



Yes.

The two setups would also have the same lumen output.

And how you direct the light can then alter those numbers; if we put a mirrored bowl over the 4x25w setup, directing the light in one general direction, that setup would then have a much higher lux rating than the 1x100w. Still the same lumen output, but differing lux throw - This is a very important concept when dealing with flashlights.


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## ponygt65 (Mar 18, 2009)

StarHalo said:


> Yes.
> 
> The two setups would also have the same lumen output.
> 
> And how you direct the light can then alter those numbers; if we put a mirrored bowl over the 4x25w setup, directing the light in one general direction, that setup would then have a much higher lux rating than the 1x100w. Still the same lumen output, but differing lux throw - This is a very important concept when dealing with flashlights.


+1.

It's VERY important when dealing with any type of lighting. Optics/reflectance is key in some lighting setups.

Edit: Just for reference, 10lux = 1fc.


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## HomerSimpson (Mar 18, 2009)

Ok, so back to one of my previous questions.

Would moving closer than 1' would increase the lux, like 6" would be 4*lumens in lm/ft^2. 3" would be 16*lumens in lm/ft^2, etc. Correct?

Thanks for the replies!


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## asdalton (Mar 18, 2009)

Only if the light source is very small.

The inverse square law can be extrapolated only when the distance from the light source remains very large compared to the size of the light source. When the two sizes are comparable (for example, 1 foot away from a standard household light bulb), then you will not get such a strong increase in lux by shortening the distance.

If you think about it more, it has to be this way, or else the lux would become infinite at the surface of the light source.


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## HomerSimpson (Mar 18, 2009)

Well, I was thinking along the lines of... like a candle, if you touch it you get burned... it'd seem to take a lot of lux to actually burn.


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## StarHalo (Mar 18, 2009)

Lux is just a unit/expression of how much light is being emitted, not heat. A very efficient light source will emit more of the energy going into it as light and not heat.

A candle flame is not very efficient, it expels almost all its energy (~50-100 watts) as heat. The hottest point in a candle flame (the blue part) is over 1000 degrees Fahrenheit.

The image in my sigline is of a High Intensity Discharge bulb cooling down; the orange part (the arc tube, where the light is created) sees temperatures over 2000 degrees Fahrenheit; if you were to put an aluminum-bodied flashlight in that reaction, it would melt like ice cream.


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## HomerSimpson (Mar 18, 2009)

What about IR 'light'?


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## Morelite (Mar 18, 2009)

HomerSimpson said:


> What about IR 'light'?


 IR is measured in watts by using a infrared detector and converting the received radiation into an electrical signal. The detector can be a thermocouple, or any number of solid state devices.


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## HomerSimpson (Mar 18, 2009)

I'm mainly wondering like how much the intensity can increase, I suppose.

Like I found this chart I guess it could be called a lm/sq ft(non SI lux) chart:






Is it more or less accurate? It seems to obey I = P/r^2.


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## StarHalo (Mar 19, 2009)

The chart doesn't make any sense to me; For one, there's no correlation between watts and lumens. And the numbers as they're displayed there seem to imply that a 300 lumen light source would be completely invisible at a distance of just over six inches.. 

What "intensity" is it you're wanting to increase, lumens/output, lux/throw, or.. ?


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## HomerSimpson (Mar 19, 2009)

The chart seems to have minimums of 1000 lm and 1.2". Yea... 900 lm @ 1' doesn't even make the chart... :laughing:. As far as I know the watts/lumens relate to typical CFL bulbs.

Lux/throw, I think.

I heard those Aero Garden have small CFLs in them. I was thinking of using ?? CFLs for a small indoor herb garden of basil, cilantro, thyme, maybe some others. My question is sort of how much light do I need(and how close). They don't seem to publish any PAR(µmol/area/time) ratings on these type of bulbs.

I'd like to do better than pure trial and error. 

I've just heard you need them *really* close, like within a couple inches. Most bulbs seem to be over 40,000 lm/sq ft within 2 inches.

I'm just wondering, if it doesn't increase between 1' and 1".... then would 1 foot be the same as 1", in lm/sq ft(or lux)? Is this what you're telling me?

I like your signature picture. It's pretty.


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## asdalton (Mar 19, 2009)

I've grown plants indoors using fluorescent tubes. If you have more than a couple of small plants, I recommend using a full sized shop light fixture with standard cool-white tubes. Get good quality tubes but don't pay for the expensive "full spectrum" tubes. Quantity of light matters more than quality.

You can even use two shop fixtures side by side, for a total of 4 tubes. More light is better here, and using more tubes makes the whole setup more forgiving of the exact distance between the plants and the lights.


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## StarHalo (Mar 19, 2009)

+1 for the full-sized fluorescent shop light fixture; I've done indoor gardening before and there's really no math involved, just set aside a desk/table for your plants and hang the fixture using chain or some other adjustable method so you can raise or lower it as needed. 

The exception is if you're doing the "medicinal"/Grateful Dead kind of herb garden, those folks use very elaborate HID setups that require a lot of money..



HomerSimpson said:


> I like your signature picture. It's pretty.



Thanks, the full image and thread/info about it is here: https://www.candlepowerforums.com/threads/225844&page=5


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## HomerSimpson (Mar 19, 2009)

Well I was looking at spiral bulbs, and they usually have several spirals, so it seems kind of like multiple tubes, really close to each other.

Apparently a 26 watt CFL has around a 2.5' long 3/8" tube, just spiraled(4x). Those T8 tubes are really long and distribute the light over a much greater area with less intensity.

I saw this video on youtube, where a guy just takes a HVAC duct, a lamp dual socket fixture, drills a hole to attach it to the center, in a sort of U shape. Then drill a couple more holes for chains. The whole set-up costs under $20. I don't really have a lot of room. I was just going to use some of those mushroom containers(the blue ones), poke a few holes, and put them on a tray, just 2 square feet, really. In the video he uses two 42 watt, which is around 5600 lumens total.

I don't really want to have to fuss too much with adjusting them daily. And I have a *big* tree on the south side of the house.... Which makes for little light coming in the windows.


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## StarHalo (Mar 19, 2009)

You could just use a basic utility/"brooder" lamp, those usually sell for ~$10 (here's one for $6: http://www.thehardwarecity.com/?sku=5914130 ). Since these usually have clamps and swivels, there's all kinds of mounting possibilities.. A decent $4 26-watt CFL bulb and you've got a complete light assembly with ~1200 directed lumens for ~$10..


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## HomerSimpson (Mar 20, 2009)

Interesting.

So to get kind of back to my original question.

To predict(calculate) the lm/sq ft, would I take the total lumens divided by 4*pi*r^2(surface area of sphere, defined by radius r in feet)? Or just r^2(steradian, r=radius in feet)?


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## asdalton (Mar 20, 2009)

Illuminance (commonly measured in lux, or lumens/m^2) is a measure of incident luminous flux per area. So you need to do calculations based on area, not solid angle.

But the premise of your question assumes a point source of light, or a source that is point-like at the distances being measured. For a compact fluorescent light, this means something like 10-20 ft away. Obviously, you won't be growing plants this way.

Forget about the inverse square law; it's NOT a fundamental law of nature (conservation of flux is more generalized), and it can can be used for calculations in only a limited set of circumstances, none of which are relevant for growing plants under artificial light.

Just set up a compact fluorescent lamp with a reflector as StarHalo suggested, and keep the light a few inches away from the plants. There is no need to do any calculations.


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