onthedl0008
Well-Known Member
PLEASE READ THE FOLLOWING
Ok heres where im at....
I know that the reccomended watts/sq.ft. is 50 and the way U figure this is to take the total amount of grow space In a grow area (LXW) and times that # by 50 to achieve the size of Lamp U need.
FOR EXAMPLE: My grow space is (2X4=8sq.ftX50=400 watts is ideal)?
WHERE IM stuck @ is In my experience more light is better i guess and in my particular grow space I am maxed out @ 66.7777 watts/sq.ft....
IF Possible Would it not be better to try and achieve 50 watts/CUBIC FT.?
What formula could I use to attain total watts/cubic ft.?
PLEASE if there are any super stoner math dudes here tell me if this looks right...
(LXWXD IN FT.=TOTAL CUBIC.FT CONTAINED IN A GROW AREA)
WOULD YOU SIMPLY THEN JUST TAKE ur # of total cu.ft and times that x50 to attain 50 watts/cubic ft.?
For example: My area is (2x4X4=32x50=1600 total watts/cubic ft.)
I have seen monsterous grows with extraordinary harvest here...
Ok heres where im at....
I know that the reccomended watts/sq.ft. is 50 and the way U figure this is to take the total amount of grow space In a grow area (LXW) and times that # by 50 to achieve the size of Lamp U need.
FOR EXAMPLE: My grow space is (2X4=8sq.ftX50=400 watts is ideal)?
WHERE IM stuck @ is In my experience more light is better i guess and in my particular grow space I am maxed out @ 66.7777 watts/sq.ft....
IF Possible Would it not be better to try and achieve 50 watts/CUBIC FT.?
What formula could I use to attain total watts/cubic ft.?
PLEASE if there are any super stoner math dudes here tell me if this looks right...
(LXWXD IN FT.=TOTAL CUBIC.FT CONTAINED IN A GROW AREA)
WOULD YOU SIMPLY THEN JUST TAKE ur # of total cu.ft and times that x50 to attain 50 watts/cubic ft.?
For example: My area is (2x4X4=32x50=1600 total watts/cubic ft.)
I have seen monsterous grows with extraordinary harvest here...