InTheShed Grows Inside & Out: Jump In Any Time
All that home time has paid off. Not just nice, but very nice!
In other news, there are already roots coming out of the @Weedseedsexpress Chocolope pot, four days after transplant.
I remember when we all met in SD, I was expecting y'all to talk me into a new fertilizer or additive I should use. Nope. If you grow good weed then keep on doin whatchur doin. Thanks for that!
Hey, Shed, hope all is well. Since this is where you and the other folk with gifted smartitudes tend to hang out I was wondering if you could check my math?
I am keeping my MegaLight at 180 W 24 inches from my canopy top. It is just about impossible to keep a canopy from one plant perfectly on the same plane so even at that distance I noticed the light intensity falls off pretty considerably on the parts of my plant that are 6 inches to 12 inches below the main canopy.
So I was thinking if I increased the distance to 36 inches between light and canopy and adjusted intensity accordingly there would be less of a drastic fall off of light to the parts of the canopy 6 to 12 inches below the main canopy since light falls off with the square of the distance.
But using that formula I would need a lot more light and I wonder if people agree with my math. Using feet 2x2 is 4 and 3x3 is 9 so dividing 9 x 4 equals 2.25 so instead of 180 W at 2 feet I would need 405 Watts at 3 feet opposed to 2 feet distance from my canopy.
I do have a light meter and a power draw meter I can hook up to the light so I can verify this experimentally but does this sound right to you guys?
I am keeping my MegaLight at 180 W 24 inches from my canopy top. It is just about impossible to keep a canopy from one plant perfectly on the same plane so even at that distance I noticed the light intensity falls off pretty considerably on the parts of my plant that are 6 inches to 12 inches below the main canopy.
So I was thinking if I increased the distance to 36 inches between light and canopy and adjusted intensity accordingly there would be less of a drastic fall off of light to the parts of the canopy 6 to 12 inches below the main canopy since light falls off with the square of the distance.
But using that formula I would need a lot more light and I wonder if people agree with my math. Using feet 2x2 is 4 and 3x3 is 9 so dividing 9 x 4 equals 2.25 so instead of 180 W at 2 feet I would need 405 Watts at 3 feet opposed to 2 feet distance from my canopy.
I do have a light meter and a power draw meter I can hook up to the light so I can verify this experimentally but does this sound right to you guys?
- Thread starter
- #20,965
Why yes I am!
As I carry the AF-47 around (still!) I have memories of last summer's IIP, so I think only autos or short-veg photos will be LST'd.
Keeping it simple is a heck of a lot easier. It's one of the reasons I switched to MegaCrop as well.
We were a pretty easy-going bunch down there, weren't we
.Raising the light and increasing the wattage will increase your penetration, but light is subject to the inverse square law, which while I can't explain it, I can certainly link to it.
It states: "The intensity (or illuminance or irradiance) of light ... radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period). "
The basic takeaway is that you can't use straight multiplication and division to sort out your change.
A light meter will give you your answer with no math needed!
The inverse square law refers to the light intensity and not the wattage. Simply stated, when you increase the distance to the light source by "x", the intensity decreases by x2Raising the light and increasing the wattage will increase your penetration, but light is subject to the inverse square law, which while I can't explain it, I can certainly link to it.
It states: "The intensity (or illuminance or irradiance) of light ... radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period). "
The basic takeaway is that you can't use straight multiplication and division to sort out your change.
A light meter will give you your answer with no math needed!
@Homer Simpson, what are the dimensions of your tent?
Why yes I am!
As I carry the AF-47 around (still!) I have memories of last summer's IIP, so I think only autos or short-veg photos will be LST'd.
Keeping it simple is a heck of a lot easier. It's one of the reasons I switched to MegaCrop as well.
We were a pretty easy-going bunch down there, weren't we
Raising the light and increasing the wattage will increase your penetration, but light is subject to the inverse square law , which while I can't explain it, I can certainly link to it.
It states: "The intensity (or illuminance or irradiance) of light ... radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period). "
The basic takeaway is that you can't use straight multiplication and division to sort out your change.
A light meter will give you your answer with no math needed!
Actually my calculations were based on the inverse square law giving the result I will need 2.25 as much light to go from 2 to 3 feet but I wasn't sure I did it right. Seems a lot more power hence money for a bit more penetration so not sure I will do it. Thanks anyway. Hope you and yours are well.
Hey Mr. Krip. Yes, I know the inverse square law refers to the light intensity and not the wattage but I assumed (perhaps incorrectly) the intensity and wattage are directly proportional. And yes I know what the inverse law says as you and Shed have quoted it because I remember it from high school but what I wonder is if my calculations based on it assuming wattage and light intensity are directly proportional are correct. Thanks for the input. I think I got it.
I don't have a tent but moveable reflectors. My MegaLight is 37"x 37" but my plant is getting so big I have them a few inches beyond that.
So, you hit the nail on the head! Light intensity is not, as a rule, proportional to wattage. It would only apply, as far as I know, if you were comparing the same light at different distances. You want about 350w actual power draw from the wall in a 10 sqft space for your lights. and 500w would be even better
I really don’t want to get in a whole argument but I find it bewildering you could say light intensity is not proportional to wattage. I could understand someone saying it’s not perfectly directly proportional because I assume lights have points where they are more efficient at certain power applications just like engines have peak horsepower at certain RPMs. But to say light intensity is not as a rule proportional to wattage is disputed by the mere existence of a dimmer switch. That would be like saying stepping on a gas pedal applying more gas doesn’t make an engine spin faster.
I think you are looking for some kind of exactitude when all I was wondering if it would be worth moving my light a foot further away from 2 to 3 feet for more penetration but at 2.25 times as much power I don’t see it being practical. But if light intensity is not directly and absolutely proportional to wattage with LEDs and I am out by 10 to 20% it still close enough to tell me whether it’s doable or not. So I appreciate the input but I think I got this.
Many LED lights these days, like Mars, provide “PAR” graphics that show the PPFD over various square areas at different distances. It’s a good illustration that might address what you are asking, Homer.
We don't need to argue, but take, for example, two 100w light bulbs, one 3000k and one 6000k. At the SAME distance, they would have two different intensities.
Damn, I'm hitting the like button for shit I don't even understand!We don't need to argue, but take, for example, two 100w light bulbs, one 3000k and one 6000k. At the SAME distance, they would have two different intensities.
I am happy we agree there is no need to argue.We don't need to argue, but take, for example, two 100w light bulbs, one 3000k and one 6000k. At the SAME distance, they would have two different intensities.
My issue is that I asked a pretty specific question but admittedly probably in such a roundabout way no one knew what I was really asking. The question I am asking is that when applying the illumination inverse square law if my calculations were correct. And I calculated if I move from 2 feet to 3 feet my illumination difference would be a factor of 2.25. That is what I was really asking.
And this is the formula for the inverse square law of a luminance.
And using this formula at 2 feet I would have one quarter the illumination and at 3 feet I would have one-ninth the illumination and 4÷9 = 2 .25 which means with my light at 3 feet I would need 2.25 times as much light intensity to have the same illumination as I did at 2 feet.
Which I assume would mean I need somewhere in the neighborhood of 2.25 times as much power at 3 feet as I do at 2 feet assuming which I admit is probably incorrect that wattage and illumination are perfectly directly proportional but close enough to give me an idea of whether it’s worth it.
And with your to 100 W light bulbs one at 3000 K and won it 6000 K I would think they would have the same intensity but just in different spectrums of light. I honestly don’t know how that would work because the higher temperature I assume would have a higher wavelength and I don’t know how that would relate to intensity but honestly I don’t understand how that is pertinent to the discussion.
This has escalated way beyond what I anticipated because I do have a light meter and a power draw meter so I can equalize the intensity at 3 feet and tell the difference I just wondered if my math was right before I started out of curiosity. All is well.
I like it!
I checked one of them and they did not agree with my calculations so either they are bogus or I am totally wrong, lol.
You are the smart one Grand Daddy Black, I should have never opened this can of worms, lol.Damn, I'm hitting the like button for shit I don't even understand!
I just had a thought why perhaps the PAR graphics I checked didn't seem to follow the inverse square law for light intensity and neither did my light meter just now.
That law applies to a pinpoint source of illumination and my light panels are 37" x 37" so if I put my light meter in the middle of my light right against it I am already over 18 inches away from some of the lights. So moving 1 foot away would not be moving the same proportional distance to all the LEDs. So the formula cannot really apply to multi-light sources like I have, my bad, never mind. Sorry, everyone.
OK, technically you're correct but for our purposes, your really not!
We've been using the term "intensity" to describe the "power" of the lights hitting the plants. Really, intensity measures the amount of energy photons hitting the plants in a given amount of time. In the LED world, this is the PPFD (Photosynthetic Photon Flux Density). However, the wavelength (PAR) determines the amount of energy per photon.
I understand we're getting way deeper than you originally intended. My point is simply that the spectrum effects the PPFD and inverse square law is only accurate when looking at the same light at two distances and does not necessarily apply to two different lights.
And since most LEDs illuminate over a 120 angle, there’s a lot of overlap from multiple LEDs.
It gets confusing, but it’s real world.
It gets confusing, but it’s real world.