? watts per plant?

Plant-specific MH are excellent.

HPS is not as good as MH, but I suppose has earned its place.

Stare at green for ten to thirity seconds. Then look at a white surface. You should see a magenta after-image. This would be the ideal color of light for plants.

PAR is based on the photosynthetic absorption and utilization which has an upper limit where more light simply does not matter if it's only in a narrow wavelength within 300 nm to 700 nm.

TeaTreeOil said:

Plant-specific MH are excellent.

HPS is not as good as MH, but I suppose has earned its place.

Stare at green for ten to thirity seconds. Then look at a white surface. You should see a magenta after-image. This would be the ideal color of light for plants.

PAR is based on the photosynthetic absorption and utilization which has an upper limit where more light simply does not matter if it's only in a narrow wavelength within 300 nm to 700 nm.

Click to expand...

my point exactly plants use every color of the light wave lenth. why do plants grow better in the summer than in winter?= intensity. o.k other factors come in to play, but if you had a heated green house the limiting factor is (depending on plant) is the the suns intensity. the sun gives out full spectrum of light so why does it not matter as long as there are blue and red light avalabe to the plant?

Despite the fact that chlorophyll absorbs maximally in the blue and red,
photosynthesis (and thus growth) is almost equally efficient across the
entire visible spectrum from 400 to 700 nm. Therefore, sunlight is good
stuff, HID's would be good except for the large heat load if you are
using it in a closed area, and the fact that the heat load represents watts
you are paying for without results. Fluorescents, halides, mixed halides
and all those give usable light, depending on what your specific plant
requirements are and you budget.

Click to expand...

TeaTreeOil said:

It looks like you're doing a^2 + b^2 = c^3, which is how a right triangle's sides(lengths) are related.

I don't see how it's related to lumens.

Adding lumens is as simple as adding lumens.

Click to expand...

no

it is

sq rt of (a^2 + b^2)

so if:

light a = 1700 lumens
light b = 2000 lumens

sq rt (1700^2+2000^2)

sq rt (2890000+4000000)

sq rt 6890000 = 2625 lumans


not 3700

. . . . . . .. . . . . . . . . . . . . . .. . . .. . . .. . . . . . . . . .
area = L x W

units that are for area need squared (2 dim)

units that are distance can be add. (1 dim)

units volumn need cubed (3 dim)

. . . . . . . . . . . . .. .. . . . .. . . .. . .. . . . .. . . . .. . .. ..

also use comon sence

if 1 cfl 26 watt = 1700 lumanes
and 1 mh 400 watt =32000

then you your way would use 19 cfls would = 1 mh ( we know this is not true)

but the truth is it would take 353 cfls to = 1 mh

sq rt (1700^2 * 353)

sq rt (2890000 * 353)

sq rt 1020170000 = 31940

Yes, 19 CFLs would output more lumens than that single MH.

Light is typically 3D. You make no sense. I don't know where to even correct you... because I don't see anything even close to being correct or used properly.

Your math would make sense if both lights were in a 2 dimensional set up, they're directional. You then take a measure of a plane @45 degrees.

Lumens is just an intensity measure. Lux takes into account spatial flux variance.

I've designed a 3d model(it's even animated) to represent your math.



Attached is another rendering. If you can check/compare a couple pixels(you seem smart), you'll see the intensity doubles. Thumbnail attachment is a view, top-down, at the setup. A&B are lights. C is the plane.

Cheers.

if that were true the the sun is 100000 lumens

so would 588 cfl = the sun?

or

346000 cfl = the sun?

. . . . . . . . . . . . .. . . . . .. . .

this may help

1 foot = 12 inches

1 foot square = 144 inches sq

1 foot cube = 1728 inches cu.

. . . . . . . . . . . . . . . . . .

lumens measure the light in a one foot square.

so even if you combine them correctlly it won't all focus on the same 1 foot square anyway.

Sunlight, by the time it reaches a square foot on Earth, is around 10,000(average summer day @ noon) to 25,000(extreme) lumens.

This is pretty interesting: http://www.norcalblogs.com/watts/2007/04/post_3.html