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Inverse-Square 'LAW' of Light

devilgoob

Active member
Veteran
a little thingy I made
untitled.PNG
 
P

purpledomgoddes

just for clarity...

there are industries that have worked all of this out, is nothing else, for cost savings, and efficiency of the business. namely, agri-buisness; specifically, greenhouse engineering and management. they itemize every coin invested, every cubic inch of c02 produced, every photon received by plants, and most importantly, the profit they derive each square foot of greenhouse space.

of course, ancient physicists worked it all out back when great pyramid was build in exact center of land mass of earth, and placed it there so they could observe great year - or, procession of the equinoxes; approximately 25,000 year cycle. sun worship, and dependence on food crop made the observance of light paramount. they mapped the sky, and all light shining therefrom meticulously.

modern physics makes it clear for all to comprehend - and reproduce.

photometry/radiometry definitions

*(vital importance of par, leaf saturation, uvb, ir, not discussed here, but relevant)
the inverse square law:
this law applies to the illumination of a surface due to the luminous intensity of a point source of light.
the illumination of a surface is called illuminance and is defined as the amount of light falling on a unit area.
illuminance is proportional to the luminous intensity of the light source and is inversely proportional to the square of the distance.

e=i/d^2 ("^" is symbol for exponents)

where

i= the luminous intensity of a light source
d= the distance to the illuminated surface in feet, and
e= foot candles

the flow of light energy is called light flux and is measured in lumens

the lumen is defined as the amount of visible light flux which falling normally on one square foot of area will produce an illuminace of one foot-candle.

example: the illuminace on the road directly under a street light suspended 25 feet above the ground is 1.2 foot-candles. calculate the power of the lamp.

solution:
e=1.2 foot-candles, and d=25 feet, gives

i=ed^2=1.2(25)^2=750 candle power.

photometric quantity
luminous flux:
the rate of light from a source with respect to the sensitivity of the human eye, measured in lumens (lm).

illuminance:
the luminous flux density incident on a surface, in lumens per square meter expressed in lux (lx) or lumens per square foot (lm/ft^2) expressed as foot-candles (fc). one lux equals 10.76 foot candles.

radiometric quantity
irradiance:
the radiant flux density incident on a surface; that is, the radiant energy received by a surface. the units are milliwatts per square meter (mw/m^2) or in watts or milliwatts per square foot (w/ft^2/mw/ft^2).

lumiunous efficacy
the ratio of total lumious flux emitted by a lamp to the energy consumed expressed in lumens per watt (lm/w).

the best unit for expressing the requirement of a particular plant species is the irradiance expressed in milliwatts per sqaure meter (mw/m^2) or milliwatts per square foot (mw/ft^2).
it is a measure of the quantity of energy in wavelength band 380-850 nanometers (nm) received by the plants.

the light intensity at any point distant from the source varies inversely as the square of the distance; that is, if the distance between plant and light source is doubled, the light intensity is reduced to one-quarter of its original value.
however, reflectors may make inverse square law inapplicable; as same luminaire in different reflectors, or bare bulbed, will provide different results, both radiometrically, and photometrically, due to the exponential factors involved.
also, reflectivity of the surface on which the light is incident is important in rendering readings.
in short, two different 400w bulbs may not give same, concrete 1/4 results when variables such as reflectors, glass casings, hoods, angles of discharge, etc are placed into equation.
law mainly applies in vaccum.

light measurement
luminous intensity
the total output of a light source measured in candle power. measurement made by manufacturer.

illumination
the amount of light falling on a unit area, and is measured in foot-candles.

luminance
(brightness), measured in foot-lamberts, refers to the light that a surface gives off in the direction of an observer. it can be either self-luminous, or reflective. some surfaces maight reflect more light than others. also, as you go further from a surface, the apprent brightness decreases.

an accurate measurement of the light the plants actually recieve can be done w/ a meter that measures par (photosynthetically active radiation).

types of light meters
photometers
measure illumination

quantum meters
measure the number of electromagnetic units (photons) available on the leaf surface in units of microeinsteins per meter per second (um/m^2s).

radiometers
measure amount of radiant energy available on the leaf surface in units of watts per square meter (w/m^2).

pyranometer
measures light in par watts per meter, between 400-1100 nm.

this may also be helpful:

lighting conversion data
http://www.icmag.com/ic/showthread.php?t=108993

enjoy your garden!

note: fwiw, run 1k's (mh+hps conversion). 6-8" seedlings/clones start off no less than 1 foot away. after 1 week of fruiting, they are placed within 2-3 inches of 1k's; or as close as plants can possibly get to light.
air flow+exchange a must. condensation gone 1 min after foliar. no lighter/match can stay lit for 5 seconds.
also have 400 in middle that plant gets 1/2-1 inch from.
all bare bulb, vertically suspended.
w/ 400-600, plant can get as close as 1/2"-1". want full par saturation.
 

Maj.PotHead

End Cannibis Prohibition Now Realize Legalize !!
Mentor
Veteran
nice informative thread kudos but i can say this for smoken your 400w 1ft away from plants tops is about optimal distance for that light. i run dual 400 w lights 1 hps 1 mh anything gets 6inch or closer to my hps gets bleached the mh they can touch the glass and not burn. all i know is when i went from the whole closet w/35sqft floor space 7x5ft to my current 30in D x 50 in W x 73 T jumped from 25w per sqft to 80 w per sqft my yields increased also. my cab isnt perdy just boxed in a corner of the closet 2x4's and blk/wht poly for 2 walls plywood n poly for the top. 1 coolube n 400w hps 1 aircooled hood n 400w mh when i extend the Depth of the cab another 6inch. i can hang the lights lookn into the flower chamber and make optimum use of the lighting, and hit my target yields n/p

all in all very nice thread
 
T

T.J

I
Do you really believe that walls have any effect on the "spread"? Or, do you just like to:dueling:?

Gotta give ya points for holding your ground, I guess.

Wait, did I get this right? You are claiming that walls do NOT have any effect on the spreading of light? :yoinks:

Give him points for being right,
The law is correct but doesn't apply to growing. Is it so hard to grasp?
 

Weezard

Hawaiian Inebriatti
Veteran
Too um, off, to grasp

Too um, off, to grasp

Wait, did I get this right? You are claiming that walls do NOT have any effect on the spreading of light? :yoinks:

Less than 10% by my measurements.
I have a tiny space, lined with Reflectix ,and expected to get the same results as fluorescent tubes seem to have from the chart, (2X D = 50%)
But, I don't have tubes so no can measure.

Give him points for being right,
The law is correct but doesn't apply to growing. Is it so hard to grasp?

I'll give him more points when he simply measures it and shows us, all, yah?

Regards,
Weezard
 

Weezard

Hawaiian Inebriatti
Veteran
"i dont need to prove it" Hmmmmm.

"i dont need to prove it" Hmmmmm.

i dont need to prove it, its simple.

Im folling this ISL, which says that everytime you have 4 times the surface area, you wil have 1/4 the intensity. Thats all it says. there not even a need to mention anything else involving spheres, radius, twice as far away, point source, anything. Thats the law, when things disperse they get less dense. You are saying that you achieve a cetain percentage density (25% for every 2x distance) decrease, without a certain percentage of increase in surface area(4 TIMES). NOT POSSIBLE...

This law is true, and which is widely accepted. Im asking how do you end up with 1/4 the intensity (twice as far away) without 4 times the surface area. It doesnt work if you follow that law....

It simply means that you are missing something, brah.
There seems to be a flaw in your logic.
I suggest you re-examine the hi-lighted statement.
Every now and then, it's a good idea to play devils advocate with one's self.
Go back over your arguements with a different perspective.

You actually had me doubting what I thought I knew to be true.
I actually like that. Keeps the mind open.

When thought experiments failed me, I fell back on physical experiments.

So...
I measure, I take new picture, I post.

You look at dem, den maybe you can 'splain it to dis dumb lizard, yah?

Mahalo
Weeze
 
T

T.J

You got to many variables in your measurements!

Take out the plants, and close the door, then you at least get more "accurate" readings.

And please if you do experiments do them like scientists, do not only take two measurements, take many many and then the average of those.


EDIT: Anyone that went farther in school than highschool can clearly see and read that you Wizzard are the one that has got the flaw in your logics.

It's like arguing with some one who thinks the world is flat, whatever you say the returning argument will be; "Well where I stand the earth is flat, just look!"
 
T

T.J

smokin.. Can't we apply that law to only one half of the sphere? I'm thinking like double the height = 1/2 intensity minus reflection losses (the reflector and walls)

And sure the bulb isn't a point source but lets simplify ;)
 
T

T.J

You are right ! Totally forgot that the cab doesn't grow in size when you raise your light :D

Its harder to come up with a formula for a rectangular space when it's just getting "longer" and not wider too... I need to be sober for that brainstorming I think :p

EDIT: Well it wasn't hard to pinpoint the variables I need in the formula, but for now I only have equations that you do and summarize the sums of them. I would like to get them together in the same equation/formula. But that is not a thing to do at 02:00 hours.
 
Great Thread

Great Thread

Wow Smokin

This is a great thread.

I always had ISL in mind an keep the light as close as poss. But, ISL totally does not account for the fact that in a well designed grow space, any light that does not directly hit the plant should be re-refllected back toward the plants.

ISL does not apply to focused light. Think of a laser beam, like a laser pointer. That light spot does NOT get twice the diameter when you increase the distance 2X. When LED grow lamps become useful, ISL will not apply to them. Most LEDs have a built in lens to make the light directional.
 
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purpledomgoddes

actually provided the relevant definitions and data in post 88.

important distinction must be made between just providing light, and the light the plants receive and use.

in any event, this should provide the raw math data used by greenhouse engineers:

example
a gardener has a 9x128 garden.
gardener wants to suspend 400w lamps over table.
the desired light intensity is 8600 lux (800 foot-candles; remember, 1 lux=10.8 foot candles).
the effective flux from a 400w lamp is 38,400 lumens.

the number of fixtures needed is equal to the light level times the surface area to be lighted divided by the effective flux, or,

n=[light level(800 fc)*surface area (9*128)/effective flux (38,400)]

this works out to

800*9*128/38,400=24 fixtures.

thus, 24 fixtures are required to provide 800 foot-candles, or 8600 lux, or 86,000 lumens per square foot.
*note: adjust foot-candles and re-work formula/equation, w/ conversions in mind, to derive desired levels in own garden...

fixture pattern
the fixture patern is defined as the horizontal spacing and the height above the crop surface. if h is the height, l the distance between the fixtures along the line, and b the distance between the lines, then for a 400w hps light, l should not exceed 1.55 feet l, and b should not exceed 2.7 feet h, for a uniformity of at least 80%.
these relationshipships will be different for each light, depending on manufacturer and type.

the light level is equal to the effective flux divided by the area, or

e=f/a [and, therefore, a=f/e]

the area a=l*b=1.55h*2.7h=4.2h^2

therefore, h (the height above the table)=

square root of a/4.2

=square root of f/4.2*e

=square root of 38,400/4.2*800

=3.38 feet...

l=1.55*h=1.55*3.38=5.24 feet

b=2.7*h=2.7*3.338=9.13 feet

the fixtures should be placed with the long axis of the fixtures across the table and placed 5'3" apart down the length of the table.

again, the best unit for expressing the energy requirement of a specific plant species is the irradiance expressed in milliwatts per sqaure meter (mw/m^2), or milliwatts per sqare foot (mw/ft^2).
it is a measure of the quantity of energy in wavelength band 400-700 nanometers (nm) received by the plants.

quantum meter: measures the number of electromagnetic energy units (photons) available on the leaf surface in units of microeinsteins per square meter per second (um/m^2/s)

the sun provides ~450 par watts.

sunmaster warm deluxe states ~345 par watts, allegedly.

best possible situation is to get 1k/600/400w light ~2-12" away from bulb/reflector glass/etc.

@ 1 foot, a 1k will provide ~5000 lumens/50000 lux/150-350 par watts. nowhere near 450 par watts of full spectrum sunlight.

also, leaves reach saturation point @ around 300 par watts, allegedly. after this, point of diminishing returns triggered, relevant to spectral (400-700nm) reception. they get saturated in light and intenstity prevails over specificity.

leaves also assimilate uvb and ir frequencies of light. these are generally outside of conventional par readings (380 nm uvb being effecictive reproductive gland stimulator).

keep in mind that intensity is not all either. mimic aurora borealis (purple sunset) and see how they love it @ certain times during maturity. purple+red party lights may influence plant as well, even if not very intense. plants sense as change of seasons, etc.
hope this helps. enjoy your garden!
 

GMT

The Tri Guy
Veteran
We havent factored in the fact that a photon that has reflected off 2 surfaces and hitting one travelling in the opposite direction that hasnt bounced at all will cancel each other out emitting a combination of heat and other radiating wavelengths of energy, one having bounced off 3 will behave the same way meeting one that has bounced off 1 surface. Which is probably why if you look directly at a bulb and reflector, the reflector surface isnt as bright as the bulb less reflection loss. Therefore while the reflector will be redirecting an great deal of photons, it wont be half of all emitted. So reflector design will have a huge impact on the brightness beneath the bulb.

Light emitting from the bulb will be lost in accordance with the ISL but light reflecting from the reflector will also be lost through various methods. Therefore what we measure at any point below the bulb will contain both the light emitted from the bulb and the light reflected from the reflector that has survived.

I can't see how a photon bouncing off a surface can be said to keep its initial energy twice as efficiently as one being generated at source. I'd like to propose something that I'm just making up. If a photon travels 6", then the wave front around it has a radius of 6" in all directions. If it travels 3" and then bounces 3" back in the other direction, it will still have travelled 6" and have a wave front 6" in all directions other than the direction of its bounce point 3" behind its current direction, and an interference pattern inbetween itself and its rear wave front. In this way the ISL would remain constant, even though what we are measuring has a higher intensity than it would without the reflector. The same would be said of the walls. So I guess what I'm saying is that I'm not sure if I want to change my mind or not. What does anyone else think of that idea? Remember photons are waves as well as particles and can't be considered ping pong balls.
 
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purpledomgoddes

We havent factored in the fact that a photon that has reflected off 2 surfaces and hitting one travelling in the opposite direction that hasnt bounced at all will cancel each other out emitting a combination of heat and other radiating wavelengths of energy, one having bounced off 3 will behave the same way meeting one that has bounced off 1 surface. Which is probably why if you look directly at a bulb and reflector, the reflector surface isnt as bright as the bulb less reflection loss. Therefore while the reflector will be redirecting an great deal of photons, it wont be half of all emitted. So reflector design will have a huge impact on the brightness beneath the bulb.

Light emitting from the bulb will be lost in accordance with the ISL but light reflecting from the reflector will also be lost through various methods. Therefore what we measure at any point below the bulb will contain both the light emitted from the bulb and the light reflected from the reflector that has survived.

I can't see how a photon bouncing off a surface can be said to keep its initial energy twice as efficiently as one being generated at source. I'd like to propose something that I'm just making up. If a photon travels 6", then the wave front around it has a radius of 6" in all directions. If it travels 3" and then bounces 3" back in the other direction, it will still have travelled 6" and have a wave front 6" in all directions other than the direction of its bounce point 3" behind its current direction, and an interference pattern inbetween itself and its rear wave front. In this way the ISL would remain constant, even though what we are measuring has a higher intensity than it would without the reflector. The same would be said of the walls. So I guess what I'm saying is that I'm not sure if I want to change my mind or not. What does anyone else think of that idea? Remember photons are waves as well as particles and can't be considered ping pong balls.

yes, friction is involved w/ wavelengths traveling thru space. and particles do collide. heat is eithe lost/gained, etc. first law of thermodymanics: no energy ever destroyed; just converted.

the inverse square law is a constant, in a vaccum. not w/ every garden having many variables, and no two the same.

best photovoltaics by bp (the orbic type; not flat panels), was @ best 30% efficient. the regular flat panels are only 15% efficient.

this means that out of every 100 photons that are incident upon a surface of a photovoltaic cell, only 15 are converted into usable energy. the remainder are lost in the process.

dont know how efficient leaves are @ converting photons. the meters listed above will provide data on how many photons are striking the leaves. will require brix, or something, noting distances from the light w/ sugar samples, to see @ what distance from the light saturation occurs.

i.e., @ a certain distance, the plant will be acquiring+processing its maximum physical quantity.

to make it simple, place lamp as close as possible and move on to other areas of garden...

enjoy your veggies!!
 

pigeonpie

Member
health and safety should be made aware of this thread!

i'm afraid someones head might just Pop

Please think cautiously

Mmmmm...*pop*
 

GMT

The Tri Guy
Veteran
Smokin, yes but the surface area of the wave front, not the surface area of the destination.
 
P

purpledomgoddes

The ISL law only applies to SURFACE AREA, it has nothing to do energy losses/gains from reflection/abosrption. That is a completely different 'law' than we are discussing right now. The only one I'm talking about says that THINGS (not just light) will spread out. When its spread across 4 times the space, it will be 1/4 times as much. Thats as far as the law goes. You start off with a 100% 'relative' value and when its spreads across 4 times the area its 25% of htat original value...You guys misunderstood this law, and tried to say the 25% was because of then ISL EVEN WHEN THE SURFACE AREA ISNT 4 TIMES! Thats against the law...

important surface area of interest in an open forum about gardening and plants is the surface area of:

PLANTS...

the MOST IMPORTANT SURFACE AREA re a gardening forum are surfaces of leaves, no?
no point in shining light on walls. walls dont photosynthesize the light, leaves do.
your thread+discussion; however, is open public forum w/ many opinion, facts, data sets. other bright individuals have considered these questions previously. some at length.

in applied physics forum/thread, pure question of light and surfaces would be fascinating. but here, fascinating aspect is how these living creatures actually capture, assimilate, and process light into usable energy.

the inverse square law has not been disproved. it is just that gardeners have just stumbled upon the fact that the law is flexible, may be influenced by external variables, and depends upon both source of light and surface light is incident upon.

not quite sure what the general discussion goal is? please clarify what is sought/hope to convey?

enjoy your garden!
 

GMT

The Tri Guy
Veteran
The way I'm thinking atm, is this, if the photon travels for 1 cm, then it has a wave front of 1cm in all directions, in effect the photon itself does not move, only its wave front. Think of dropping a pebble into the centre of a pond, what we see on the surface of the pond is in effect a 2 dimensional wave front of ripples circulating outwards, and becoming less intense as they travel outwards. A light photon does this in 3 dimensions. In practice this wave front collapses at the point of measurement, (quantum mechanics), and can be recorded as a particle rather than a wave. How it behaves on its travels though relies upon it being a wave up until its point of measurement. In effect, the photon stays stationery until it decides it's destination point, then vanishes from where it was and simultaniously appears at its destination point. Only the wave front takes time to travel, the photon itself doesn't. As the wave front travels, it dissipates. This dissipation is what I was refering to. Like I said, I'm trying to work this out with everyone else, I only half know what I'm talking about. What I don't know is how the energy contained in a single photon is measured. We can count the photons, but I suspect that that is not the same thing as measuring intensity of light or total energy reaching the destination.
 
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purpledomgoddes

The way I'm thinking atm, is this, if the photon travels for 1 cm, then it has a wave front of 1cm in all directions, in effect the photon itself does not move, only its wave front. Think of dropping a pebble into the centre of a pond, what we see on the surface of the pond is in effect a 2 dimensional wave front of ripples circulating outwards, and becoming less intense as they travel outwards. A light photon does this in 3 dimensions. In practice this wave front collapses at the point of measurement, (quantum mechanics), and can be recorded as a particle rather than a wave. How it behaves on its travels though relies upon it being a wave up until its point of measurement. In effect, the photon stays stationery until it decides it's destination point, then vanishes from where it was and simultaniously appears at its destination point. Only the wave front takes time to travel, the photon itself doesn't. As the wave front travels, it dissipates. This dissipation is what I was refering to. Like I said, I'm trying to work this out with everyone else, I only half know what I'm talking about. What I don't know is how the energy contained in a single photon is measured. We can count the photons, but I suspect that that is not the same thing as measuring intensity of light or total energy reaching the destination.
the best unit for expressing the energy requirement of a specific plant species is the irradiance expressed in milliwatts per sqaure meter (mw/m^2), or milliwatts per sqare foot (mw/ft^2).
it is a measure of the quantity of energy in wavelength band 400-700 nanometers (nm) received by the plants.

quantum meter: measures the number of electromagnetic energy units (photons) available on the leaf surface in units of microeinsteins per square meter per second (um/m^2/s).

see posts 87 & 100.

do not know of any meter than can attach to plants to determine ratio of photons incident on surface to photons assimilated and converted into simple sugars.

brix measures sugars...

if more/specific nm wavelength/proximity of light stimulates greater rate of photosynthesis, can then measure sugars based upon control plant @ set distance/known light irradiance upon plant. adjust data set #'s for plant growth. should be recordable if highly motivated to do study.

most of this data is readily available from various light manufacturers.

other resources gardeners interested in plant light:

horticultural lighting, by phillips lighting co. write them and ask for literature, w/ specific q's. they'll reply.
 

GMT

The Tri Guy
Veteran
ok now I know a little more, I'm going to back smokin again lol. We will still lose a lot of photons due to the reflection cancelling some of the photons out, (see the first video below) however since intensity is measured by the number of photons and that the photons dont lose energy as they travel, unlike the water ripples, (see the second video below) the ISL cant be used in an enclosed reflective area. The basics of the ISL are shown very clearly with the 3rd video, but the first 2 are the important ones.

http://www.youtube.com/watch?v=qpQABLRCU_0&feature=related superposition of wave function of a single light photon
http://www.youtube.com/watch?v=USP0NSO8yUw measuring the intensity of light.
http://www.youtube.com/watch?v=JW3tT0L2gpc&feature=related inverse square law for beginners.
 
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