Am I ready to put the pot under light? (Picture)

I start my seedlings under a 13W cfl. When they show the first set of true leaves I move them to higher wattage cfls.

yep that is ready...been ready. Good Luck!!!

Distiller said:

Using the inverse square law of I = 1/d^2 which is Intensity = Light Output/Distance^2. A 400 w HID has 32,000 lumens. The end result is 8,000 at 2 feet away. Not only is the actual light output too high for seedlings, the heat given off could fry your plants.

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This is confused. I = 1/d^2 applies to the Irradiance of light. The power per unit area. Irradiance is watts per sq meter. The equivalent for lumens is Illuminance which is lumens per sq meter. The bulb's rating in watts or lumens gives the total power out output or luminous power output.

http://en.wikipedia.org/wiki/Radiant_flux
http://en.wikipedia.org/wiki/Luminous_flux

Also, the Inverse Square Law only applies to point sources or perfectly uniform, perfectly spherical sources. Like the stars and the sun. An HID lamp is really a bar or strip emitter. But this is a quible. 1/d^2 is still useful for thinking about HID lamps. A more important consideration is the presence of a reflector which complicates things.

If we ignore the point source issue and think about the surface of a sphere with a 2' radius ( 0.6m) and the 400W HID at it's center, then the surface area of sphere is about 4.5m^2. Which gives an irradience of 88W/m^2.

However light is also being reflected by the hood and this does not look anything like a point source and doesnt really follow the inv. sq law. But let's simplify and just say that under the lamp, the irradience is boosted by 50% by the presence of the reflector, ie at 2' away we have 132W/m^2


If we do the same calculation for a distance of 4', we have a sphere who's surface area is now 18m^2 but the radiant flux of the light is still 400W so the irradience is 22W/M62 or 33W/m^2 with the reflector. Only 1/4 of that intensity at a 2' distance.

This little example shows why light from a point source follows the inv sq law. It's because of the relationship between the radius of a sphere and it's area. Doubling the radius makes a sphere with 4x the area. It also shows that if you are illuminating a 1m^2 canopy with a 400W light, you are wasting 80% of the light's output depending on your reflector efficiency.

For illuminance the calculation is exactly the same using lumens in place of watts.

If the bulb emits a total of 32000 lumens, then at 2' you have 32000/4.5 = 7111 lumens/m^2 and at 4' 1,777 lumens. Or 1000 lm/m^2 & 27000lm/m^2 with the reflector.

I dont like using lumens to rate a grow lamp's effectiveness. Lumens fall in exactly that part of the spectrum that the plant doesnt use. From the point of view of an MJ plant, lumens are a unit to measure wasted light output. If you check out the PAR curve, which shows what parts of the spectrum a plant uses, you will see there's a big hole in the middle right where the photopic response goes.

Finally, from a practical experience, I run a 600W HPS and keep my plant tops less than 1' away.