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figuring WPSF in a vertical system

river rat01

Member
if you were to try and figure out how many wpsf were in the various vert sys., how would you measure it?

lets go with a 1000w light in its optimun footprint.

lets say that you have a cylinder that is 3 feet tall, and has a circumfrence of 10 feet. = 30 square feet.
33wpsf?


or, would you count the square footage of the volume inside the cylinder?
3 feet dia.-3 feet tall = 9 square feet.
111wpsf?
 

C21H30O2

I have ridden the mighty sandworm.
Veteran
growing area will be the circumference of the growing cylinder 2*pi*r
 

magiccannabus

Next Stop: Outer Space!
Veteran
Yeah imagine the cylinder turned on it's side and rolled out flat. That's your surface area. It can be a real pain to get it tweaked though. Your best bet is to try to calculate how many foot-candles intensity you have at the various points around your growing area. As long as you don't exceed 70,000 anywhere you should be fine. If it's getting higher than that you may sunburn them no matter how much air you blow across them. Realistically you should aim for 30-50k as a maximum intensity at your expected canopy top. Aim to distribute about that intensity over as much space as evenly as you can. It's easier to reach those intensities with HID, but it's easier to have even distribution with fluorescents. There's always compromises to be made in this.
 
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river rat01

Member
mmm..pi.

well i'm not very good at spherical geometry so you "mathy" guys will have to help me along.

ok, so with a 1000w that has 140,000 lumens divided by the surface area of 30 feet gives me about 5000 lumens psf. thats no where near the 50,000 your saying that i need, magiccannabus.

so what size cylinder would be best for a 1000w sodium to be placed in the center?
 

00420

full time daddy
Veteran
Pi = 3.14
r = 3'
H = 4'

2 x 3.14 = 6.28 X 3 = 18.84 x 4 = 75.36sqft /1k = 13.26wpsqft

I Belive this is right all double check when I get home ....
This would be for a 6 foot round 4 foot tall colisum type grow area I would cut it down to 3 foot on the hight n get up too 22-24wpsqft that's about where I'm at n I have no prob's
 
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purpledomgoddes

volume=the number of units that can be arranged to completely fill the space within the object.

volume of a cylinder:
v=bh
=(pi*radius sq)(h) edit: pi times radius squared times height=volume

1k w/ out cool tubes w/ blower/10-18" fan underneath, directing heat upwards+out.
vegetables will grow in to (very close) to light. tomatoes permitted to grow as close to illumination as possible. w/in 3-6".
take into account lateral heat dissipation if no cool tubes used.
higher temps accelerate metabolism of plants - as long as other environmentals are w/in parameters, cylindrical-type air flow are points to take note of.
important (debatable) to measure par (photosynthetically active radiation) as well.
practice run(s) w/out vegetables+mock routine maintenance activities w/in area, w/ temp+rh readings more considerations.
1k, whether mh/hps will illuminate approx. nine cu ft thoroughly from one ft away.
draw 1 ft circumference around bulb(s) as reference point/line of perimeter w/ greatest illumination points around light(s). moves veggies back as maturation progresses (stretch) and attempt to create circular canopy using 1 ft circumference line as target imaginary line.
 
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purpledomgoddes

actual formula:

n=desired light level in lux*surface area/effective flux[lumens]

where n is the number of fixtures required to achieve desired lux

example: 9'x128'(252 feet of area) table-area over which 400w hps' will hang. light intensity of 8600 lux(800 foot candles) is desired.
effective flux(lumens) is ~38,400 lumens.
the number(n) of fixtures is equal to the light level times the surface area to be lighted divided by the effective flux, or

(800*9*128)/38,400=24 fixtures

the fixture pattern is defined by the horizontal spacing and the height above the crop surface (or, in this case, the concave curvature of the imaginary 1 ft circumference around the 1k hung vert).

the light level is equal to the effective flux divided by the area, or,

e=f/a [and, therefor, a=f/e]

where e is the light level, f the effective flux (lumens), and a the area
 
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purpledomgoddes

full sunlight=~2000 umol/sq m/sec - ~450 par watts/sq m (400-700nm)=5000 lumens=50,000 lux ['umol' is micro-moles. mole is quantity of substance that contains same number of ultimate particles as is contained in 12g of carbon-12, or avogadro's number, 6.0228x10^23]

~photon (par) saturation point: 300 par watts upon leaf surface/3000 lumens of full spectrum(mercury vapor+incandescent+uvb+cmh+hps+sulphur/etc).

1k hps ~ 5000 lumens @ 1 ft/~250 par watts

1k mh ~ 133 par watts @ 1ft (due to lack of par colors, standing alone)

pyranometer-measures photons between 400-1100nm.

~ 20 photons required to make/store 1 molecule of sugar.

note: sunmaster warm deluxe states 345 par watts.
 
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river rat01

Member
00420, a cylinder with a 6' radius would only have aprox. an 18 - 20'' diameter.

if a 1000w sodium were in the center, would that be too much?
 
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purpledomgoddes

mmm..pi.

well i'm not very good at spherical geometry so you "mathy" guys will have to help me along.
above raw data from several greenhouse engineering/plant physiology/research facility/etc. texts. if inapplicable, don't apply.

5,000 lumens=50,000 lux. just different measurement name. may be the target sought per other poster.

1 lumen=1 candle from 1 foot away shining on 1 square foot=1 foot-candle=1fc

1 lux=1 candle from 1 foot away shining on 1 square meter.

1 lumen=10 lux.

1kmh/hps 1ft away from plant=5,000 lumens/=50,000 lux/=~133-250 par watts(depending on bulb spectrum(s))

exponentially declining/point of deminishing returns further away from 1 ft optimum space between light source and canopy

edit: just foot/meter conversion for clarity
 
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magiccannabus

Next Stop: Outer Space!
Veteran
My bad I was too high when I wrote that and put a zero on the end of a couple numbers there. The range you're aiming for is 3-5K not 30-50K. Oops. Still take those as maximums, but remember that 30K at 1 foot is much more intense than at 1 yard, and the bottom of your plants are going to start much further from the lights than the tops will be when they finish. This is a big problem with growing with light shining from the inside-out. I think the future of vertical is outside-in. Surround the plants with light, not the other way around. Time will tell though.
 

00420

full time daddy
Veteran
00420, a cylinder with a 6' radius would only have aprox. an 18 - 20'' diameter.

if a 1000w sodium were in the center, would that be too much?

6' R = 12' D
your thinking 6' circumference

Surface Area = (2 • π • r²) + (2 • π • r • height)
Where (2 • π • r²) is the surface area of the "ends" and (2 • π • r • height) is the lateral area (the area of the "side").
 
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river rat01

Member
it would be cool if there were a chart that showed a 1k light from the side and displayed the light intensity levels at various distances.
 

river rat01

Member
i know what you mean.
i'm a visual person.
i wish there were a graph or some purdy colorful chart that would explain it all instead of these headachey words.

i guess i'll have to make one.
 
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purpledomgoddes

keep it simple.
target distance from 1k hps/mh bulb is approximately 1 foot.
place temp monitor on/by closest foliage.
max temp approximately 86f.
move garden/bulb away/closer depending on temp max+minimum distance.
 
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