Grow light height

[Tazard]

I read it and it says unless you are using a focused light source such as a laser ISL applies. Like with a CFL or MH or HPS or other "radiant " light.

[Tazard]

I did at about 3" I have 1100 lumens at about 4" I have 700.

[DrPhoton]

At least read those threads, i have measured light at various levels from 12" to as much as 68". It does not follow ISL.

[Tazard]

Place a lumen meter at canopy level and note the change as you change the distance. I=W/d^2

[DrPhoton]

Heres some discussions on the matter, hopefully this will help clear it up for you. Also my physics101 had very little in the form of calculas and mathematics.

http://www.cinematography.com/index.php?showtopic=63448

https://www.photo.net/discuss/threads/inverse-square-law-light-modifiers.467889/

https://physics.stackexchange.com/questions/230367/can-focused-light-be-treated-as-a-point-source

https://www.quora.com/Is-the-light-from-lasers-reduced-by-the-inverse-square-law-as-distance-grows-similar-to-other-light-sources

[DrPhoton]

Tazard, im not one to get into arguments and i ment no disrespect if felt given. I assure you nothing is copy and paste except for what is quoted. What i have provided is based on my "own" understanding, which i believe is correct. Iv seen many people get hung up on this problem but i assure you myself and others agree with what i have suggested. I am all about the convincing argument so rest assured if there are errors i will always rationally accept any.

Your suggested source i did not apply any calculation, i merely looked at the reference and pointed out the suggested protocols needed for calculating light intensity. I do not know at what distance ISL starts to apply, it may be after a metre. But if you are wanting references, heres wikipedia detailing it.

Under justification it details the requirements needed for ISL to apply

https://en.m.wikipedia.org/wiki/Inverse-square_law

"The inverse-square law generally applies when some force, energy, or other conserved quantity is evenly radiated outward from apoint source in three-dimensional space. Since the surface area of a sphere (which is 4πr2 ) is proportional to the square of the radius, as the emitted radiation gets farther from the source, it is spread out over an area that is increasing in proportion to the square of the distance from the source. Hence, the intensity of radiation passing through any unit area (directly facing the point source) is inversely proportional to the square of the distance from the point source."

Please do not comment on my integrity, i am all for a healthy discussion. Im here to learn, just like you.

[Tazard]

I read your posts. I can tell what is copy paste and what is your opinion (interpretation) of the data supplied. Did you understand the example given? That's a rhetorical question since the example was based on 81cm (81x2.54=31.89") using a 60w bulb. I do believe you are well read and even possibly an electrician However, you are reacthing conclusions not supported by the references and posting these as facts. 🤔

[DrPhoton]

Yep this is correct, as you can see the calculations here take into equation the geometry of the point source.

"The light intensity at a particular point depends on the configuration of the light source and the directions in which it radiates light."

The geometry and direction of the light source affects the math for calculating ISL, so much infact that ISL does not apply until the geometry of the point source starts acting like a point source from an observers POV. Standard incandescent light bulbs will act like a point source after a couple of meters. But at close distance ISL will not apply.
This is how fibre optics go for such long distances, their energy is reflected and does not diverge, allowing only reflective losses to occur.

As for most other lights, at close distances (talking metres) the light will fall off less than ISL. You can measure this with a meter, done it myself.
Take care, good luck

[Tazard]

I also will rarely provide facts I cannot reference.

[Tazard]

http://sciencing.com/calculate-light...y-7240676.html Light electricity optics magnetism polarization and heat were covered in my phy 202 class phy 201 was static and dynamic motion. Phy101 (I have never taken) is algebra based physics as opposed to calculus based. I guess the only real difference is we learn to derive the algebraic equations you were given in phy101. I have a BSCE and a Meng.

[Tazard]

Thanks. I have 3 strains growing with excellent coverage from the CFL's. 2 of the strains handle 1100-1200 lumens at canopy level very well (see pics seeds were started on the 23rd of May). Some sprouted 24h later others took 2-3days. The other 2 strains continue to curl with lightning at 700 lumens. I'm about to just give that strain shade. 🤢

[DrPhoton]

Yep thats correct, higher intensity will penetrate deeper. But doesnt follow ISL (read my last post). Also the total amount of photon energy received by the plant, is unchanged no matter what intensity or height the light is (excluding variables such as reflective losses, light footprint, coverage etc). Only the distribution characteristics of the photons to the plant. In other words, for the most part. Yeild is the same, but the uniformity of the developed produce will by some degree be dictated from the distribution of light. Saying that, increased penetration could offer better quality and slightly more yeild.

Im well read, electrician by trade. Have read several books on numerous science subjects. Have covered PHY101.

[Tazard]

Point being a higher intensity will penetrate the canopy further but over a smaller area (solve ISL for d). I can tell you are educated, well read, or a great copy and paster or a combination. I would suggest phy202 will cover light.

[DrPhoton]

Yep i understand, i was just interpreting the weight given as either pressure or weight, which i believe doesnt really matter either way. Comes to the same conclusion.
The rest of the information given was just to hopefully supply some information to acertain the right idea on light manipulation.

[Tazard]

DrPhoton I only read the first sentence or two but pressure does not equal weight or force. pressure is force per area so it would be given in like pounds per square inch not pounds or should you be British it would be given in kilograms per centimeter squared as opposed to kilograms. I'm not here to correct anyone's physics so I won't bother reading any further.