greatbranch
Well-Known Member
I'm using a RW75 + RW150 in a 32"x32", and would guess 3-4 W90s would be one heck of an upgrade. I defer to the technical folks, but your conclusion seems sound to me.
"Area 51 LED" Information and discussion
- Thread starter captainmorgan
- Start date
1.21 Gigawatts!
Well-Known Member
5x5 is 60"x60". If the recommended coverage is 20"x20", wouldn't that mean you'd need 3 rows of 3, or 9 total to cover the 5x5?
captainmorgan
Well-Known Member
I did my quick calculations based on the power going to the COB,not the total power draw of the light. My CXAs are very close to the Veros in performance. Based on my CXAs at 30 watts a sq/ft I came up with 2.66 sq/ft of coverage for the A51 light ,A51 says 20"x20" which is around 2.77 sq/ft based on the COB draw. These are rough numbers with rounding but should be in the ball park. What someone needs can go up or down depending on his set up. A scrog with plants that don't stretch much could possible use less with good results and if your growing tall plants you may want more.
vitamin_green_inc
Well-Known Member
How do you end up with 3-4 units following the stats of one covers 20" x20" to a room that is 60x60? Your math is waayyyyyy off, must have been an early morning lol
Humanrob
Well-Known Member
Good point. This seems to be a gray area. Minimum, maximum, ample, optimal... There is the amount of light that will reach a total surface area of the canopy sufficiently to grow evenly, and then there is the additional amount of light you can add to the same area to encourage maximum growth. Sellers tend to want to promote the largest possible footprint, growers seem to want to figure out how much light they can throw down before the plants are over saturated. Somewhere in the middle is probably a sweet spot... but I doubt there is one single right answer, just guidelines and references.
JimmyIndica
Well-Known Member
110chained? or would you have to go 220? w/ 9 those?
JimmyIndica
Well-Known Member
Its all preference?IM sure will some run 6 some will run 9 in the 5x5. It would be cool to see what #s would look like with 9 those in a 5x5?
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captainmorgan
Well-Known Member
Yep it's all gray,you can add environment (temp,RH,CO2 levels),strain and even grow medium to your variables.
JimmyIndica
Well-Known Member
Hey Captain, Show us what 9 those will do in that 5x5?
Greengenes707
Well-Known Member
It's a vero29 at 2.15a. We know exactly what it is putting out and thus how many we need for any given space.
At ~39.5% efficient and 79w= 31.21 par watts and at 4.94umols/PARwatt =154.18umols.
*yes, 4.94 is total possible photons, not just 400-700nm.
20"x20" = .258m2
154.18 divided by .258m2 gives the w90 a beat case scenario of 597umols ppfd in a 20"x20"
You want 700-1000umols over each square meter of grow space. Smaller spaces can get away with lower intensities, but I would say closer to 19" diameter to get in the prime range. Diameter as in circular coverage and smaller area. Not a square.
At ~39.5% efficient and 79w= 31.21 par watts and at 4.94umols/PARwatt =154.18umols.
*yes, 4.94 is total possible photons, not just 400-700nm.
20"x20" = .258m2
154.18 divided by .258m2 gives the w90 a beat case scenario of 597umols ppfd in a 20"x20"
You want 700-1000umols over each square meter of grow space. Smaller spaces can get away with lower intensities, but I would say closer to 19" diameter to get in the prime range. Diameter as in circular coverage and smaller area. Not a square.
captainmorgan
Well-Known Member
That's why I said ball park and based on my results with a closely performing CXA3070. I'm getting big numbers from my CXA set up but that might not be the most common way they're used.
I'm running around 529 PPFD with my CXA's.
OneHitDone
Well-Known Member
Sounds like a good fit for my setup
JimmyIndica
Well-Known Member
Be following that run!Sounds like a good fit for my setup
greatbranch
Well-Known Member
I could swear he said you'd need 6-8. Edit: Nevermind - you mean the first part. Why is it looking like the answer is as many as you can physically fit in the space?
vitamin_green_inc
Well-Known Member
Well if 5x5=25sq foot and 1fixture will cover optimally the manufacturers stated claim which is about 2.7 sq feet then if we multiply that by 9 and we get 24~ sq feet, I would do 10 of these but then that is a lot of freaking power, 855w+ and a lot of heat from individual units............. so I'm sorry man, I understand what you're saying now, that doesn't make sense...maybe I'm just confusing myself now, I'm dry so not in the right frame of mind
captainmorgan
Well-Known Member
I think 9 in a 5x5 would be the bare minimum I would use.
JimmyIndica
Well-Known Member
agree with ya! 3 rows 3? spaced even? lotta ratchets, but worth it!
Greengenes707
Well-Known Member
8 units(7.7 technically) per 4x4 gets just about 800umols ppfd.
Some will try and work with less. Some may opt for more(like 3 rows of 3).
Some will try and work with less. Some may opt for more(like 3 rows of 3).
JimmyIndica
Well-Known Member
Hey I just glad its somewhat peaceful !That other shit just sucks