Lumens per SF according to the inverse square law HELP

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blueX

Active Member
I just cant get thing right mathematically, maybe someone can help me.
If i put a lamp 1.5f above the canopy, according to the inverse square law a 600w HSP(~90K lumens) will make 40K FC(90000/1.5*1.5).
If i want to have 7K lumens per SF it means that i need 600w HSP on every ~5.7 SF area(40000/7000).
Its a lot more then the recommendations here and more then what ive seen on the journals...
What did i calculate wrong ?

Also i want to know if i put the lamp lesser then 1f closer to the canopy(with cool tube), theoretically i should get more lumens then what the bulb originally should "give"(because the original number measured from 1f distance). Is this correct ? If so, how do i calculate things when im closer then 1f?

thx
 
mindphuk

mindphuk

Well-Known Member
blueX said:
I just cant get thing right mathematically, maybe someone can help me.
If i put a lamp 1.5f above the canopy, according to the inverse square law a 600w HSP(~90K lumens) will make 40K FC(90000/1.5*1.5).
If i want to have 7K lumens per SF it means that i need 600w HSP on every ~5.7 SF area(40000/7000).
Its a lot more then the recommendations here and more then what ive seen on the journals...
What did i calculate wrong ?

Also i want to know if i put the lamp lesser then 1f closer to the canopy(with cool tube), theoretically i should get more lumens then what the bulb originally should "give"(because the original number measured from 1f distance). Is this correct ? If so, how do i calculate things when im closer then 1f?

thx
Click to expand...
It isn't the math but your concept of trying to convert lumens to fc. A 90,000 lumen bulb can illuminate 1.5 square meters (close to 5'x5' or 25sf) at about 60,000 lux. You can only use the inverse square law on lux or foot-candle. Since lumen is distance zero, doubling that is still zero. There are charts that will give you lux or fc based on your area of illumination and distance.
 
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blueX

Active Member
mindphuk said:
It isn't the math but your concept of trying to convert lumens to fc. A 90,000 lumen bulb can illuminate 1.5 square meters (close to 5'x5' or 25sf) at about 60,000 lux. You can only use the inverse square law on lux or foot-candle. Since lumen is distance zero, doubling that is still zero. There are charts that will give you lux or fc based on your area of illumination and distance.
Click to expand...
Oh ok. I thought it was a problem with lumens and fc analogy.
Anyway can you(or anyone) help me understand how to do the math correctly ? I dont trust any charts(since a lot of them are showing different things) and anyway i want to
understand things better.
How do i calculate how much SF(or SM if its easyer) can a 90K lumens lamp(600w hsp) illuminate at 7K light intensity(at the top of the canopy!)? The lamp distance is 1.5f(18i/45cm).

thx again :)
 
haze2

haze2

Well-Known Member
Remember that the sun gives out 10,000 lumens per square foot on a sunny summer day. If you had 100,000 lumens that would be highly sufficient for 10 ft by 10 ft
 
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blueX

Active Member
haze2 said:
Remember that the sun gives out 10,000 lumens per square foot on a sunny summer day. If you had 100,000 lumens that would be highly sufficient for 10 ft by 10 ft
Click to expand...
But again, there is the distance thing. The sun gives 10k lumens per sf on the ground.
90K lumens are from the bulb - and the distance is a key factor.
Any good mathematician here ? :)
 
TeaTreeOil

TeaTreeOil

Well-Known Member
haze2 said:
Remember that the sun gives out 10,000 lumens per square foot on a sunny summer day. If you had 100,000 lumens that would be highly sufficient for 10 ft by 10 ft
Click to expand...
That's 10' *10' = 100 sq ft. Which is 1000 lm/sq ft.

blueX said:
But again, there is the distance thing. The sun gives 10k lumens per sf on the ground.
90K lumens are from the bulb - and the distance is a key factor.
Any good mathematician here ? :)
Click to expand...
Yes. Hello.

3' * 3' = 9 sq ft, 90,000 lm / 9 sq ft = 10k lm/sq ft.

A fc(foot candle) is defined as 1 lm/sq ft.
A mc(meter candle) is defined as 1 lux(lm/sq meter).
 
B

blueX

Active Member
TeaTreeOil said:
Yes. Hello.

3' * 3' = 9 sq ft, 90,000 lm / 9 sq ft = 10k lm/sq ft.

A fc(foot candle) is defined as 1 lm/sq ft.
A mc(meter candle) is defined as 1 lux(lm/sq meter).
Click to expand...
Well if so my math was correct ? In order to give my plants 7K lumens per sf on the canopy(=fc) i need 600w hsp on every ~5.7 sf area (see the calculation above)?
 
B

blueX

Active Member
Twistyman said:
Check these lighting charts....
Click to expand...
I know those charts. According to it a 600w hsp give only 13K lumens per sf at 1f(18i) distance. It means that it can only cover 1.85 sf area at 7K lumens per sf - there is no way for that .
Maybe i dont understand things right ?
 
jigfresh

jigfresh

Well-Known Member
I am not entirely sure I can help but here goes.

You talk about wanting 7k per square foot. Well, unless you have one of those circular rotating garden things, the plants are gong to be varying distances from your light. So it will not be that you have 1 square meter with exactly 7k per square meter. You can make it so that the very center is 7k and goes down from there.

Do you mean a 7k average over an area, that would make more sense to me.

If you wanted an area with a minimum of 7k per square foot then according to the chart above with a 600w HPS, the maximum distance the light could be from the 'edge' of the area you are trying to light is ~24". So depending on how close your light is to your canopy, and the configuration of your reflector you could light varying areas with a minimum of 7k using your 600w.

Are you going for 7k average, 7k minimum, 7k maximum? If you want a perfect spread you should either drop huge money on a rotating garden thing, or next best thing would be a scrog, you could even bend the thing around to hug the light. I do one, because I like the idea of spreading the light out more effeciently.

I know I didn't answer your question, but there you go, hope I helped a little.
 
Twistyman

Twistyman

Well-Known Member
blueX said:
I know those charts. According to it a 600w hsp give only 13K lumens per sf at 1f(18i) distance. It means that it can only cover 1.85 sf area at 7K lumens per sf - there is no way for that .
Maybe i dont understand things right ?
Click to expand...
I have a book here lets see.....
P = primary coverage... S = Supplementary lighting
1000W 6 x 6 ft P (2m) 10 x 10 ft S(2.5m).............................
600W 4 x 4 ' P (1.3m) 6 x 6' S (2m)................................
400W 4 x 4' P " " ".............................................
250W 3 x 3" P (90cm) 4 x 4 S (1.3)................................
 
E

easygrinder

New Member
i tend not to go with lumens so much, but your calculations would give you excellet light coverage.

600w over 6 sq foot is giving you 100w psf, the min you wanna roll with is 50 psf.

i grow in a 4x4 area so thats 16 sq foot and i use 2 x 600 for that space so i get about 75 per square foot

180,000 lumens divided by 16 sq foot = 11250 lumens psf
 
TeaTreeOil

TeaTreeOil

Well-Known Member
3.6 feet away from the bulb is 7k lumens/sq ft. This is without a reflector.

3.6'*3.6' is 12.96 sq feet. This shape, without a reflector, resembles the shape of a contact lense(it is a, 3D, circular section of sphere).
 
Gastanker

Gastanker

Well-Known Member
I don't know much about growing but alittle about lighting. - just want to make sure you dont forget about overlap. Using that chart - which I understand you do not like - a 3' distance on a 600w bulb produces 3k lumens. Well if you have 2 lights next to eachother - you mentioned multiple lamps - the overlap doubles. A plant inbetween two lamps, 3' from each lamps (lamps being 6' apart) would be recieving 6k lumens.
 
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easygrinder

New Member
Gastanker said:
I don't know much about growing but alittle about lighting. - just want to make sure you dont forget about overlap. Using that chart - which I understand you do not like - a 3' distance on a 600w bulb produces 3k lumens. Well if you have 2 lights next to eachother - you mentioned multiple lamps - the overlap doubles. A plant inbetween two lamps, 3' from each lamps (lamps being 6' apart) would be recieving 6k lumens.
Click to expand...
if the lights were 3 foot from the plant tops and the plants were a reasonable distance apart, say 2 foot then the plant inbetween would be 3.6 foot away, so by time you divide the intensity by that extra foot your no where near doubling 3000k
 
B

blueX

Active Member
Im still VERY confused. A lot of people here are saing different things. This is way i want to know the math behind what you say so ill know everything is correct .
Lets say i have:
4'*8' grow space, i use 600w hsp's(~90K lumens each), 1.5f distance between the bulb and the canopy and i want to have 7K lumens per sf. How do i calculate this ? If someone can show me the way so i can understand and apply this on different variables it can be great !

thx for your replys .
 
jigfresh

jigfresh

Well-Known Member
easygrinder said:
if the lights were 3 foot from the plant tops and the plants were a reasonable distance apart, say 2 foot then the plant inbetween would be 3.6 foot away, so by time you divide the intensity by that extra foot your no where near doubling 3000k
Click to expand...
For your math to work the plants would be 4 feet apart.

2 feet apart would be 3.16 feet away (see pic)

Plus, I think my plants are a reasonable distance apart for what I am doing and they are 1 per square foot. I think 'reasonable' depends on what kind of grow you are doing.

I also don't know the worth of discussing two lights right next to each other, as people would either usually have two lights spread out over an area, or just one big light.

What's the verdict BlueX, has anyone come close to helping you, or are you more confused?
 

Attachments

B

blueX

Active Member
jigfresh said:
I am not entirely sure I can help but here goes.

You talk about wanting 7k per square foot. Well, unless you have one of those circular rotating garden things, the plants are gong to be varying distances from your light. So it will not be that you have 1 square meter with exactly 7k per square meter. You can make it so that the very center is 7k and goes down from there.

Do you mean a 7k average over an area, that would make more sense to me.

If you wanted an area with a minimum of 7k per square foot then according to the chart above with a 600w HPS, the maximum distance the light could be from the 'edge' of the area you are trying to light is ~24". So depending on how close your light is to your canopy, and the configuration of your reflector you could light varying areas with a minimum of 7k using your 600w.

Are you going for 7k average, 7k minimum, 7k maximum? If you want a perfect spread you should either drop huge money on a rotating garden thing, or next best thing would be a scrog, you could even bend the thing around to hug the light. I do one, because I like the idea of spreading the light out more effeciently.

I know I didn't answer your question, but there you go, hope I helped a little.
Click to expand...
I meant average. I want the light intensity to be 7K per sf at the canopy.
 
TeaTreeOil

TeaTreeOil

Well-Known Member
TeaTreeOil said:
3.6 feet away from the bulb is 7k lumens/sq ft. This is without a reflector.

3.6'*3.6' is 12.96 sq feet. This shape, without a reflector, resembles the shape of a contact lense(it is a, 3D, circular section of sphere).
Click to expand...
90,000 lm / 12.96 sq ft = 6944 lm/sq ft

lm / area = lm/area

90000 / x = 7000
Multiply x by each side & cancel X on the left:
90000 = 7000x
Divide each side by 7000 & simplify:
12.86 sq ft = x

Square root 12.86.
3.59 feet away from the bulb is 7000 lm.

blueX said:
Im still VERY confused. A lot of people here are saing different things. This is way i want to know the math behind what you say so ill know everything is correct .
Lets say i have:
4'*8' grow space, i use 600w hsp's(~90K lumens each), 1.5f distance between the bulb and the canopy and i want to have 7K lumens per sf. How do i calculate this ? If someone can show me the way so i can understand and apply this on different variables it can be great !

thx for your replys .
Click to expand...
Hope that helps.
 
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