do photons loose energy as they travel through air or do they just get spread out?
- Thread starter Dako
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Tym
Well-Known Member
They refract (get scattered). It takes miles of air for any absorption to take place. But light does slow down.
So the reason that you want to put lights as close as possible to your plants is that the light will spread out, and not because traveling through air reduces the energy contained in light. Is this correct?
In other words, the relationship between light intensity(lux) and distance would be the same in a vacuum.
In other words, the relationship between light intensity(lux) and distance would be the same in a vacuum.
Mr.BigsBud
Member
Sick sick sick puppy.
Tym
Well-Known Member
That's right, there will be less scattering the closer you get the light, this is obvious since the farther you move the light away, the bigger the light foot print gets.. You want it nice and close to focus the most amount of light possible on to the plants.. Absorption does happen in air, but it take miles of air to absorb lots of light. You lose more to scattering than you do absorption. The 12 - 18 inches of air your light travels will have almost 0 loss to absorption, we are talking small numbers like (for example cause I don't know the exact numbers off the top of my head) 0.0001 percent.
This leads to my next question. It is impossible to answer but I want guesses. If I put my 400w hps at the top of my 2x4x7 Mylar grow tent and measured the lux at the bottom, (6 feet away say) what would be the loss compared to any other distance say 3,2, and 1 feet away respectively? I realize there is going to be a huge loss but it can't be as much as the inverse square law for open spaces. I know that every time a "light beam"(every angle) hits a wall there will be a 5% loss due to Mylar. I realize that the light may be coming at an angle that is hard for the plant to use(compared to the unreflected light coming relatively strait down). But still it seams like there wouldn't be that much of a difference in lux at the bottom and from I'm hearing on my other thread is that there will be a huge difference in the amount of light the plants can use at the bottom (6 feet) compared to say 2 feet. Where is the energy going and what is it's form?
Tym
Well-Known Member
The loss will be much larger, cause it's %5 per reflection. At 6 ft, the light will most likely reflect more than once or twice, so you;re talking about %5 - %10 (PER SURFACE, assuming you have 4 walls) for a minimum of %20 loss, just to reflection. But the larger problem is stretching. The plant can tell what direction the more intense light is comming from, and it will head right for it, spending all it's energy trying to get closer to it and no much on root production and bulking out.. Even with a %20 loss from the walls the plant knows that the best light is coming from above and reaches for it.
The main reason you use reflection on the walls is to support growth below the top of the plant where the light from above doesn't penetrate as well, This can be seen better when people put CFLs under the and beside the plants to give it more light under the canopy of the plants.. Otherwise they won't get much light, and won't produce much bud on the lower extremities.
So to answer the question, your loss of light would be minimum %20, to a max of %40 (and if it's far enough away to reflect 3 times off the mylar %60) %5 per wall per reflection. measure the angle of your light reflector (bat wing or whatever you use), and chart out how many times it will reflect off the mylar, say 2 times before it hits the plants.
5 (loss of light do to mylar) x 2 (times it reflects before it hits the plant) x 4 (number of walls).
5 x 2 x 4 = 40
This is just an estimate, to get an exact number, you would need the exact measurements of the surface areas that are reflecting light, the reflectivity of your reflecting hood (yes you loose light there too) and the exact angle of the light coming off the reflective hood. But it would be close to this calculation. That's why when you move it closer, the plant does better, it gets more light, it doesn't have to stretch for it, and the light scatters less..
Hope this helps.
I still don't understand as well as I would like. Tym, your answer was still very helpful, thanks. I think making an equation for this would require some nasty calculus but I looked at it pretty much like you did. The majority of the(lost/indirect) light would bounce less than 4 times.
NLXSK1
Well-Known Member
I understand what you are trying to figure out but it is far more complex and really wont affect the grow that much.
Just look at it this way... The useable light has to hit the tops of the leaves to provide energy that the plant can use. That light is most powerful hitting the plant leaves at a perpendicular angle. The chances of a photon missing the leaves and bouncing back around and hitting the leaf at a perpendicular angle are almost 0.
So, the mylar and other reflective surfaces help somewhat but are by no means perfect at recycling the light for the plant.
Tym
Well-Known Member
Yup that's what I said, Also he's talking about having the light 6ft away from the plants. He will lose light (in the form of heat) from the reflections off the walls. The more it has to reflect off a surface, the more you lose. They do slow down, and they do so measurably, just not significantly in such a short distance. You lose light when it interacts with matter. Usually in the form of heat. but in the plants case, in the form of photosynthesis.photons never loose energy... they ARE energy
are you taking lux readings under your canopy?
if so... the light is lost from being absorbed and/or reflected off of the leaves above it.
and photons do NOT slow down... sorry. not measurably anyways.
Nice Ol Bud
Well-Known Member
This question is..
very confusing..
Makes no sense brother..
NoB
very confusing..
Makes no sense brother..
NoB
IAm5toned
Well-Known Member
actually, the laws of physics dictate that photons travel at the speed of light, which is a universal constant.
the only way light 'slows down' is when it is passing through a lens... and it doesnt slow... there is a unit of time that occurs from when a proton strikes an object and another photon is emmited from the same object... but a photon, always travels @ at [SIZE=-1]3 to the 8th power, meters per sec.[/SIZE]
[SIZE=-1]"When light enters a material, photons are absorbed by the atoms in that material, increasing the energy of the atom. The atom will then lose energy after some tiny fraction of time, emitting a photon in the process. This photon, which is identical to the first, travels at the speed of light until it is absorbed by another atom and the process repeats. The delay between the time that the atom absorbs the photon and the excited atom releases as photon causes it to appear that light is slowing down. [/SIZE]" -Gary Russel, M.S,
If photons or light slowed down, E=MC2 wouldnt work......
and we know it works.
the only way light 'slows down' is when it is passing through a lens... and it doesnt slow... there is a unit of time that occurs from when a proton strikes an object and another photon is emmited from the same object... but a photon, always travels @ at [SIZE=-1]3 to the 8th power, meters per sec.[/SIZE]
[SIZE=-1]"When light enters a material, photons are absorbed by the atoms in that material, increasing the energy of the atom. The atom will then lose energy after some tiny fraction of time, emitting a photon in the process. This photon, which is identical to the first, travels at the speed of light until it is absorbed by another atom and the process repeats. The delay between the time that the atom absorbs the photon and the excited atom releases as photon causes it to appear that light is slowing down. [/SIZE]" -Gary Russel, M.S,
If photons or light slowed down, E=MC2 wouldnt work......
and we know it works.
Tym
Well-Known Member
Sorry, but no..
Straight from the pages of wikipedia.
Photons in matter
See also: Group velocity and Photochemistry
(Visible) light that travels through transparent matter does so at a lower speed than c, the speed of light in a vacuum. X-rays, on the other hand, usually have a phase velocity above c, as evidenced by total external reflection. In addition, light can also undergo scattering and absorption. There are circumstances in which heat transfer through a material is mostly radiative, involving emission and absorption of photons within it.
Light does slow down. The speed of light is only constant in a vacuum.
Some instances light can be slowed down to a speed we can actually watch. We can slow it down so much, you can watch light travel.. Check out this link from Harvard university.
http://www.news.harvard.edu/gazette/1999/02.18/light.html
But in the short distance we use our lights in the air we have them in, the slow is negligible.
Tym
Well-Known Member
Are you kidding? I love this stuff! lol.
Science is my life. Literally, it's what I do for a living
robert 14617
Well-Known Member
View attachment 1319505this is what i use
Brick Top
New Member
all you really need to look at is inverse square law and some trig for angles, everything else is academic and more than required to understand what is going on in your grow op
Inverse Square Law, Light
As one of the fields which obey the general inverse square law, the light from a point source can be put in the form
