Realistic Yield per plant

The HornyCulturist

The HornyCulturist

Active Member
moparhomo said:
That is a good yield for a 400w hps. Back in the day I would be lucky to get 250gram with my 400w in the closet. It was a small closet though. Ty for the watt to gram ratio. Very interesting.
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Sorry for the delay in response! Internet went down!

Cheers for your comment, Yes I am always happy with my yield. I find square pots tend to give a wee bit more yield as well. More space for soil and roots just i guess!?
I normally have a proper greenhouse set up in my attic 8x8x6 with a proper fresh air cycling and filter system and all the rest. I normally grow 4 plants that I veg for 2 months under 200w and finish in 21 inch square pots under 400w. They always finish about 5 feet tall including pot.
I did something a bit different this time as i had only one seed left (aurora indica with shockingly bad genetics so no cloning, turned sativa and was a mental grow. All 5 seeds the same!!).
This time i squeezed the top to try and get 4 colas but just got no main cola and 4 sort of medium ones at best. Harvesting tonight as long as the soil is dry. I'm expecting to still get AT LEAST 3 ounces off it. Ill take before and after pics tonight and let you know the dry weight when I'm done drying and curing if your interested?
 
moparhomo

moparhomo

Well-Known Member
The HornyCulturist said:
Sorry for the delay in response! Internet went down!

Cheers for your comment, Yes I am always happy with my yield. I find square pots tend to give a wee bit more yield as well. More space for soil and roots just i guess!?
I normally have a proper greenhouse set up in my attic 8x8x6 with a proper fresh air cycling and filter system and all the rest. I normally grow 4 plants that I veg for 2 months under 200w and finish in 21 inch square pots under 400w. They always finish about 5 feet tall including pot.
I did something a bit different this time as i had only one seed left (aurora indica with shockingly bad genetics so no cloning, turned sativa and was a mental grow. All 5 seeds the same!!).
This time i squeezed the top to try and get 4 colas but just got no main cola and 4 sort of medium ones at best. Harvesting tonight as long as the soil is dry. I'm expecting to still get AT LEAST 3 ounces off it. Ill take before and after pics tonight and let you know the dry weight when I'm done drying and curing if your interested?
Click to expand...
I would!
 
The HornyCulturist

The HornyCulturist

Active Member
moparhomo said:
I would!
Click to expand...
So here is some pics of the Aurora (pain in my ass!) Indica. Didn't really put as much effort into this plant as normal due to the weird genetics.

This plant was mostly chopped 2 days ago with the remainder done yesterday. Threw in a pic of a wee bit of the GOLD I have left from harvesting as well! ¦0)
I will let you know the end weight when i get it all dried and cured properly.

1.jpg 2.jpg 3.jpg 4.jpg 5.jpg 6.jpg
 
jacksthc

jacksthc

Well-Known Member
nuevo said:
It would be interesting to see if there is some kind of equation that could be used to help estimate yield. Something like this - Yield in grams (Y) = Total wattage (W) x total area of plants in square feet (A) x veg time in days plus flower time in days(T1 + T2). You would need some kind of conversion factor to convert the watts x ft^2 x days to grams. The time factor is always missing when I see growers do their yield estimates.

For my set up, I typically veg the plants to about 18 inches in about 40 days, with another 70 days in flower. The plants are in 8 inch round pots, so I can fit about 40 into a 5 x 5 space. I use a 1000W light. My typical yield is about 30 grams per plant. So in my case W = 1000, A = 25, and T = 110, and Y = 1200. So the conversion factor for this set up would be 0.000436.

Plugging your numbers into the equation and solving for Y would be as follows:

Y = 800W x 16 ft^2 x 84 days x 0.000436
Y = 469 grams

I assumed you flower for about the same time as me - 10 weeks or so. Also did a guesstimate for the total wattage with a composite light system.

Would like to see some other growers use there own numbers to see if there is a fairly constant conversion factor. Obviously, there is going to be lots of variation, but it would be cool to have some idea if an equation like this could be useful.
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I am trying a grow with the min amout of electria from start to finsh but finshing off under a 600w Aiming for 10 oz's

sg = seeds germinating under 16watts * 20hours (hours) * 4 days (room size 12" L* 10" W * 12" H)
veg = veg lights t5's 40w * 24hours *28 days (room size 12" L* 10" W * 12" H)
earlyFlower = early flower lights 400w hps * 12 hours * 14 days (room size 100cm * 90cm)
flowerRoom = flower lights 600w * 12 hours * 42 days (room size 110cm * 90cm)

sFan= 4" InTake fan 60w * 12 * 56 days (only on when the lights on)
Fan= 6" outTake fan 100w * 24 * 56 days

sg = 1.28kw
veg = 26.88kw
earlyFlower = 67.2kw
flowerRoom = 302.4kw
sFan = 40.32kw
Fan= 134.4kw

total = 572.48kw
cost amount of eletria = total * (price per kw)
(price per kw) = 0.09p

cost amount of eletria = £51.52p

baic soil and nutes cost about £30

so I want to know what i can get out of a £81.52p grow :)
 
ryan s

ryan s

Well-Known Member
Bugeye said:
It is my observation that people who need to ask about their yields are not very experienced growers. Less experience equals greater chance of fuck up. So realistically, I would expect you to get very little per plant. Maybe even zero. Hope this helps.
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Yeah man but before I started this grow I'm working now I wanted to know the same exact thing and 2 months into it I'd like to see you do better.

OP, defoliate well and dial in every variable you can and you can expect maybe 10 grams a plant, give or take. If you get 7 grams a plant that is one pound. I would suggest flood and drain in 5" or 6" rockwool cubes. Start them in 2" cubes and when the roots are all healthy at the bottom of the 2" cube stick that cube into the 5" or 6" cube
 
mc130p

mc130p

Well-Known Member
nuevo said:
It would be interesting to see if there is some kind of equation that could be used to help estimate yield. Something like this - Yield in grams (Y) = Total wattage (W) x total area of plants in square feet (A) x veg time in days plus flower time in days(T1 + T2). You would need some kind of conversion factor to convert the watts x ft^2 x days to grams. The time factor is always missing when I see growers do their yield estimates.

For my set up, I typically veg the plants to about 18 inches in about 40 days, with another 70 days in flower. The plants are in 8 inch round pots, so I can fit about 40 into a 5 x 5 space. I use a 1000W light. My typical yield is about 30 grams per plant. So in my case W = 1000, A = 25, and T = 110, and Y = 1200. So the conversion factor for this set up would be 0.000436.

Plugging your numbers into the equation and solving for Y would be as follows:

Y = 800W x 16 ft^2 x 84 days x 0.000436
Y = 469 grams

I assumed you flower for about the same time as me - 10 weeks or so. Also did a guesstimate for the total wattage with a composite light system.

Would like to see some other growers use there own numbers to see if there is a fairly constant conversion factor. Obviously, there is going to be lots of variation, but it would be cool to have some idea if an equation like this could be useful.
Click to expand...
The units don't work out in your equation. Watts is J/s and you multiplied it by time in days, and then multiplied it by the area, so
your final units for Y in the equation are J/s*m^2*days...that's not grams of yield.

After smoking several bowls, and a big doobie, I decided I would redesign the equation like this:

Define a maximum attainable yield per unit area, g/m^2, call this variable G. G is a function of your grow space coordinates, such that G = G(x,y) and it might take considerable effort to map G over the entire area of the room. For simplicity, we will assume constant light intensity, temperature, etc. and make G(x,y) is a constant over the area. The maximum attainable yield, Ymax is equal to the integral of G(x,y) over the area of the grow space utilized.

We also need the probability density function that describes the probability that the grower can achieve the maximum G, P(G,x,y) (some areas of the room might be easier to grow in than others). Perhaps choose a Gaussian distribution with a mean and variance based on past grows of the grower.

Then, your estimated yield is the triple integral of G(x,y) *P(G,x,y)dx*dy*dG over the area of your grow space and over G = (0,inf). In the OP's case, the integration works out to about a pound, amazingly.

:weed:
 
Gquebed

Gquebed

Well-Known Member
64 plants in a 4x4 area is pretty crowded. Light penetration would be a problem. Choke lots of them out.

In another thread a guy was saying he did 25 plants per 4x4 table and had two tables. He claimed 1.5 lbs under 1000w each table for a total of 3lbs. I believe him. I was doing 5x5 area under 1000w and 4 plants for 1.5 lbs every run. So his math seems right....
 
nuevo

nuevo

Well-Known Member
mc130p said:
The units don't work out in your equation. Watts is J/s and you multiplied it by time in days, and then multiplied it by the area, so
your final units for Y in the equation are J/s*m^2*days...that's not grams of yield.

After smoking several bowls, and a big doobie, I decided I would redesign the equation like this:

Define a maximum attainable yield per unit area, g/m^2, call this variable G. G is a function of your grow space coordinates, such that G = G(x,y) and it might take considerable effort to map G over the entire area of the room. For simplicity, we will assume constant light intensity, temperature, etc. and make G(x,y) is a constant over the area. The maximum attainable yield, Ymax is equal to the integral of G(x,y) over the area of the grow space utilized.

We also need the probability density function that describes the probability that the grower can achieve the maximum G, P(G,x,y) (some areas of the room might be easier to grow in than others). Perhaps choose a Gaussian distribution with a mean and variance based on past grows of the grower.

Then, your estimated yield is the triple integral of G(x,y) *P(G,x,y)dx*dy*dG over the area of your grow space and over G = (0,inf). In the OP's case, the integration works out to about a pound, amazingly.

:weed:
Click to expand...
The conversion factor I utilized accounts for the units. In my case, the conversion factor is a new unit that I call grams per square foot-watt-days, or grafootfootwadaze. Works perfectly, that is, works after a couple or twenty puffs of good smoke.
 
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