Lumens per SF according to the inverse square law HELP
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easygrinder
New Member
3.6 foot will get you about 7k lumens per square foot
In your equation you dont include the distance from the bulb to the canopy and it does make a different(or i just dont understand the "3.59 feet away from the bulb" thing).
I the distance is 1f(12i) or 1.5f(18i), how does things change? Can you show me ?
Sorry im pain in the a$$ but i realy want to be able to understand it fully(the math equation are the best way imo)
jigfresh
Well-Known Member
Hey math major. He is figuring out the distance you need.In your equation you dont include the distance from the bulb to the canopy and it does make a different(or i just dont understand the "3.59 feet away from the bulb" thing).
I the distance is 1f(12i) or 1.5f(18i), how does things change? Can you show me ?
Sorry im pain in the a$$ but i realy want to be able to understand it fully(the math equation are the best way imo)
He started with your light: 90k. He used x as your distance (x means unknown at this point), then he used your 7k lum you want (that is what you wanted right?)
Then he solved for x. Is that 8th grade or 10th.
He did that math for you. It is in easy to understand steps. If you don't get that, you are going to have problems totally understanding the situation I think.
TeaTreeOil
Well-Known Member
Well, you want to always use the correct units for whatever problem you're trying to solve. Inches should always be converted to feet(divide by 12).
Light decays according to Iavg = P / A. That's: average intensity = power / area.
Most bulbs produce a omni-directional, more or less, luminous flux. So your area is more circular than a square. With a good reflector this area becomes flatter, broader, more even, and is pushed out horizontally.
If you're really concerned about it, I'd suggest a light meter.
Technically, with a point, omni source the equation is: Iavg = P / r^2. That's: average intensity = power / radius * radius. And the area is a section of a sphere, extended off a cone(cone/sphere intersection). Basically the shape of a contact lens.
1 is the cone. 2 is the area that Iavg covers. This is just a side-shot, it's really 3 dimensional. Imagine a second one of these perpendicular to the first to kind of get the idea. Top to bottom would be the height. And left to right would be the width and length.
Examples
One foot:
Iavg = 90000 lm / 1 ^ 2
Iavg = 90,000
Two feet:
Iavg = 90000 lm / 2 ^ 2
Iavg = 22,500
One & a half feet:
Iavg = 90000 lm / 1.5 ^ 2
Iavg = 40,000
None of these assume a reflector is being used.
Light decays according to Iavg = P / A. That's: average intensity = power / area.
Most bulbs produce a omni-directional, more or less, luminous flux. So your area is more circular than a square. With a good reflector this area becomes flatter, broader, more even, and is pushed out horizontally.
If you're really concerned about it, I'd suggest a light meter.
Technically, with a point, omni source the equation is: Iavg = P / r^2. That's: average intensity = power / radius * radius. And the area is a section of a sphere, extended off a cone(cone/sphere intersection). Basically the shape of a contact lens.
1 is the cone. 2 is the area that Iavg covers. This is just a side-shot, it's really 3 dimensional. Imagine a second one of these perpendicular to the first to kind of get the idea. Top to bottom would be the height. And left to right would be the width and length.
Examples
One foot:
Iavg = 90000 lm / 1 ^ 2
Iavg = 90,000
Two feet:
Iavg = 90000 lm / 2 ^ 2
Iavg = 22,500
One & a half feet:
Iavg = 90000 lm / 1.5 ^ 2
Iavg = 40,000
None of these assume a reflector is being used.
mindphuk
Well-Known Member
Please don't listen to Teatreeoil, he is misapplying the formulas. You cannot take a point source with x lumens and calculate how much lux or footcandles you will get by dividing it over a specified area. The reason is specifically due to the inverse square law. You need a lux or FC measurement to start your calculation, because you need distance in the equation. You can't calculate the distance without starting with a distance.
If the sun is 10,000 FC (lumen/sf), it is not going to get weaker over a larger area as he claimed. If you have a source that is 10,000 fc at the surface of the earth, then that same light does not get divided over 9 sf, it's still 10,000 FC at falling on every square foot since they are all the same distance from the sun. You can add, and get 90,000 fc total over 9sf. The correct way to apply the inverse square law in this instance is to say the sun will be 2500 fc at double the earth orbit.Originally Posted by TeaTreeOil said:
You have your answer, look at the chart. If you want 7000 lumen/sf, then refer to the chart under the 600w column, it's between 24 and 25" away. Of course, that intensity is going to be directly under the bulb and fall off toward the sides since those are longer distances so you need to be closer than 24" so that you can get 7000 FC to the sides as well. Take 18" for example, with 13,000 lumen/sf directly under the bulb. in a 3x3 area, the furthest part away from the bulb is about 25.46" (it is the hypotenuse of a right angle triangle with both legs 18", so you see why?), go to your chart and you will see that at 25", the intensity is 6747 FC, just under the amount you need, so 3'x3' area is going to be about the max at 18". With an air-cooled hood, you can bring the lamp closer to the canopy covering more area on the fringes too.
Thx a lot that's exactly what i wanted .
Only one thing - can you tell me how you calculate the furthest part away from the bulb ? Its been a long time since my last math lesson
Anyway TeaTreeOil thx a lot for your help too. I learned some things from you too .
mindphuk
Well-Known Member
He did no such thing. You cannot figure out intensity from the area alone, you need a measurement first before you can start the process. Lux and foot-candle is a measured, not calculated value. You can make calculations for various distances and areas once you have that initial number, but until then, the lumen number is meaningless without a distance.
If you take TeaTreeOil's figure and work backwards, the intensity should be 28,000 lumen/sf at 1.8 feet (half the distance), or 112,000 lumen/sf at 0.9ft! How can a 90,000 lumen lamp produce 112,000 lumen/sf at any distance?
jigfresh
Well-Known Member
It seems like that is what he did to me.
I never said the equation answered all questions asked and I never said the light figures are correct. I was just explaining what he did because the op didn't understand.
Sorry.
mindphuk
Well-Known Member
Just remember that the sum of the 2 sides of a right angle triangle sqaured, equals the square of the hypotenuse. In the 3sf example, I took 1/2 of 3ft or 18" to determine the distance to the perimeter from the centerpoint, and of course the other leg of the triangle is the distance from the bulb.Thx a lot that's exactly what i wanted .
Only one thing - can you tell me how you calculate the furthest part away from the bulb ? Its been a long time since my last math lesson
Anyway TeaTreeOil thx a lot for your help too. I learned some things from you too .
Let's try 4sf area with the bulb 12" away, so you can see the math.
The two legs of the triangle will be 12" x 24". The distance the edge of this area from the bulb is the hypotenuse, (12^2)+ (24^2)= 144+576=720 sqrt(720)=26.83 inches. Go to the chart, 26" is 6238 fc, under 7000, but still sufficient.
mindphuk
Well-Known Member
Yes, but you were a bit condescending to the OP with your math major comment. My point was that even if the math was correct, TTO used an inappropriate formula to make the calculation. It's no wonder the OP was confused with these strange and vastly different answers he's getting.
I did understand the equation - i even did the same thing at the start of the post.
I just didnt understand the distance factor in the equation but now i do.
easygrinder
New Member
3.6 foot from 90,000 lumens is 7000 lumens
7000 lumens is 1/13 of 90,000
90,000 lumens / 3.6x3.6=6944 lumens per square foot
so you would get 12.96 sq foot of canopy area or however that was focused with your reflector
7000 lumens is 1/13 of 90,000
90,000 lumens / 3.6x3.6=6944 lumens per square foot
so you would get 12.96 sq foot of canopy area or however that was focused with your reflector
TeaTreeOil
Well-Known Member
1.8'*1.8' = 3.24 sq ft
90,000 lm / 3.24 sq ft = 27,778 lm/ sq ft. Yup. Close enough.
Because 90,000 lumens isn't being distributed upon an entire square foot. What do you don't understand??? That's like saying you can't convert Lux to Lumens/sq ft.
Mindphuk is clueless.
I never said the sun's 10,000 lm/sq ft(on Earth) decays over more area. It's already in terms of area. Because really, 10,000 lm/sq ft is derived from the sun's output / distance from sun to earth squared - atmospheric shielding. There's no increase in distance, only area, as per your example.
You dumbfound me. Just astonished, man.
According to you, CFLs are just as effective 1 foot away as 1 inch.
90,000 lm / 3.24 sq ft = 27,778 lm/ sq ft. Yup. Close enough.
Because 90,000 lumens isn't being distributed upon an entire square foot. What do you don't understand??? That's like saying you can't convert Lux to Lumens/sq ft.
Mindphuk is clueless.
I never said the sun's 10,000 lm/sq ft(on Earth) decays over more area. It's already in terms of area. Because really, 10,000 lm/sq ft is derived from the sun's output / distance from sun to earth squared - atmospheric shielding. There's no increase in distance, only area, as per your example.
You dumbfound me. Just astonished, man.
According to you, CFLs are just as effective 1 foot away as 1 inch.
jigfresh
Well-Known Member
My condescension come from the fact the op asked an extremely vague question, which leads (unavoidably) to confused answers and a big chase by everyone involved just to dig to the REAL question the person MEANT to ask in the first place.
Seems like a lot of work by a lot of people who are really trying to help, when it would have been great to be clear to begin with.
Next time, think a little more about what exactly you are trying to ask. It will save MindPhuk having to clear everything up for all of us.
Seems like a lot of work by a lot of people who are really trying to help, when it would have been great to be clear to begin with.
That's a little like the people who post "I have two seeds, a 1000w light, and some tin foil. Does anyone see a problem with this?"
Next time, think a little more about what exactly you are trying to ask. It will save MindPhuk having to clear everything up for all of us.
I dont see the problem with my question. I mixed things up a bit with light intensity, fc, distance an so on when trying to use things together. It was wrong and i knew it - THIS IS WAY I OPENED THIS THREAD. Im not the only one here who confused things and from there came misunderstanding.
Im sure im not the only one who didnt realy understand things fully(newbie and more advanced growers), Im also sure that in the future when someone will want to understand this subject - he will fully understand it after reading this thread because it is so detailed. So it wast bad at all .
Im sure im not the only one who didnt realy understand things fully(newbie and more advanced growers), Im also sure that in the future when someone will want to understand this subject - he will fully understand it after reading this thread because it is so detailed. So it wast bad at all .
mindphuk
Well-Known Member
I'm clueless but came up with the correct answer for the OP, while your answer is still wildly inaccurate.1.8'*1.8' = 3.24 sq ft
90,000 lm / 3.24 sq ft = 27,778 lm/ sq ft. Yup. Close enough.
Because 90,000 lumens isn't being distributed upon an entire square foot. What do you don't understand??? That's like saying you can't convert Lux to Lumens/sq ft.
Mindphuk is clueless.
You cannot convert lumen to lux or foot-candle. You can convert between f-c and lux, and I never said or implied otherwise.
You did absolutely claim that the 90,000lum/sf would be diluted over a larger area:
No, if the sun is 90,000lum/sf then over 9sf is still 90,000 lum/sf, not 10,000. And you are trying to say I don't understand? This seems like a pattern with you. You bring out all of this scientific knowledge but then apply it incorrectly. Other people have called you out on this and you still refuse to listen.
You are the perfect model for the saying, "a little knowledge is a dangerous thing."
At what distance?
See this is where you keep fucking up. Taking a point source of x lumens and dividing it over a specified area does not equal foot-candles. You don't have distance in the equation. Yes, you have 27K lumen/sf at zero distance from the point source. How does that help you calculating at 1 foot, 2 feet, etc.?
How so? I'm the only one using distance in my calculations. Where did I imply that CFLs don't adhere to the inverse square law?
TeaTreeOil
Well-Known Member
Because light is based on waves. Waves propagate is spheres/circle. Not squares.
The distance is the radius. The radius squared is the area of a conical intersection of the sphere defined by radius r. Got it? This is how light decays.
http://searchcio-midmarket.techtarget.com/sDefinition/0,,sid183_gci528813,00.html
Lumen is a unit of power. So it satisfies I = Power / Area.
We agree on the last part.
Distance is tied to the area. Like a wave that ripples outward.
1 candela / steradian = 1 lumen
The distance is the radius. The radius squared is the area of a conical intersection of the sphere defined by radius r. Got it? This is how light decays.
http://searchcio-midmarket.techtarget.com/sDefinition/0,,sid183_gci528813,00.html
Lumen is a unit of power. So it satisfies I = Power / Area.
I agree with the first sentence. Uhm. I'm not applying it incorrectly. You're just misunderstanding. Because other people are wrong, and they try to educate others with their 'Internet education' when I learned physics in college and got a 4.0.
We agree on the last part.
Distance is tied to the area. Like a wave that ripples outward.
1 candela / steradian = 1 lumen
mindphuk
Well-Known Member
No one is claiming anything different. Why must you obfuscate the issue with irrelevant information?
Yes, and you can use this to approximate the lux or FC, but it will be for a lamp with no reflector so your measured value will be different.
For example, take that 90,000 lumen lamp and say that the light will radiate away from that point and at one foot, the energy will have dissipated in a sphere. That calculation is 4/3 Pi x r^3= 4.188 recurring. 90,000 lumens divided by 4.188= 21,489 F-C, lower than the chart but that didn't take into account a reflector, so the higher value in the chart is probably more accurate if using a reflector.
At 2 feet, the F-C would be 5372 F-C, much lower than you 7,000 F-C @ 3.24 feet as you claim.
You absolutely did apply it incorrectly. Did you or did you not claim that the sun would be 10,000 F-C over 9sf or not?
mindphuk
Well-Known Member
Did you forget this thread?
https://www.rollitup.org/indoor-growing/173926-yet-another-light-quiz.html
O TeaTreeOil, O Tea Tree!
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!
Not only in the summertime,
even under CFL's is thy are so shiney.
O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
Much pleasure your threads bring me!
O TeaTreeOil, O Tea Tree,
Much pleasure doth thou rendering bring me!
For every year the TeaTreeOil,
Brings to us all both the bullshit and glee.
O TeaTreeOil, O TeaTree,
Much your bullshit is boaring me!
O TeaTreeOil, O TeaTree,
Thy CFL's shine out so brightly!
O TeaTreeOil, O TeaTree,
Thy candle power burn so powerfully!
Each tiny lumen doth hold its PAR light,
That makes my toy HPS to sparkle like a fairy light.
O TeaTreeOil, O TeaTreeOil,
Thy shit shine out so brightly!
(thank you 9inch Bigbud)
https://www.rollitup.org/indoor-growing/173926-yet-another-light-quiz.html
O TeaTreeOil, O Tea Tree!
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!
Not only in the summertime,
even under CFL's is thy are so shiney.
O TeaTreeOil, O Tea Tree,
How are thy leaves so sticky!
O TeaTreeOil, O Tea Tree,
Much pleasure your threads bring me!
O TeaTreeOil, O Tea Tree,
Much pleasure doth thou rendering bring me!
For every year the TeaTreeOil,
Brings to us all both the bullshit and glee.
O TeaTreeOil, O TeaTree,
Much your bullshit is boaring me!
O TeaTreeOil, O TeaTree,
Thy CFL's shine out so brightly!
O TeaTreeOil, O TeaTree,
Thy candle power burn so powerfully!
Each tiny lumen doth hold its PAR light,
That makes my toy HPS to sparkle like a fairy light.
O TeaTreeOil, O TeaTreeOil,
Thy shit shine out so brightly!
(thank you 9inch Bigbud)