MH lighting is a myth

[bmw]

Hi Andy,
Thanks for the response. I have seen Harker's articles. But he was measuring differences in the icecap ballast and tar ballasts. Do you think you could dig up some figures on on light intensities Icecap measured for vho and MH, in terms of lumens(or lux)and the parameters involved in the measurement? Would it help to say I am a customer
Thanks,
b.

[TheDonger]

"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality."--Albert Einstein

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Visit Keith's Reef

[bmw]

"Mathematics is the queen of sciences" --Karl Gauss, reputed to be a better mathematician

[Tadashi]

What units are the inverse law dealing with? Some of you guys are using feet (meters) some inches (centimeters). I have a book that states the earth's surface receives 2000 muE/m^2/s (metric) or PAR which is roughly equivalent to 100,000 LUX or 10,000 foot candles. If the formula uses meters (feet) then the light lost in air above the water is negligible from 4-6" if it uses cm or inches then the light lost at 2" is great. How does lumens fit into the picture? (been 10 years since I did college physics) Evan and I have been arguing about this for weeks someone please help us.

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"Honey, put the bleach down and step away from the reef tank. I promise we will spend more quality time together."
My Bonsai-reef site

http://www.intl-light.com/handbook/ch06.html

this url explains the inverse square law. L8r mega

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[Cookoopod]

Mega,

Does inverse law account for the medium that the lighting has to penetrate? Is it assuming distance through air? Seems like there should be some type of coefficient to account for this. Also, the units you used with Inverse's Law do not match the units used at the website you referred to. Can you please explain this?

TIA BUD

Unit of measure doesnt matter, see the "are there any physics major's in the house post" Yes, medium the light passes though does matter, but at the depths we are dealing with, not very much, short of a algae bloom. That said, Bretton WAde sp? has a java program that will calculate the K in water as depth drops & effects of water clarity on this. First things first though, the formula is a BEST case for dropoff. L8r mega

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[ATJ]

Mega,

How are you calculating the lumens quoted above? Shouldn't the light intensity be in lux?


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ATJ
http://atj777.tripod.com/

If I had lux I would use it, I now have par for 250/400 halides as I said above, but I do not for vho/pc/no so for compairson it wont work. L8r mega

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[ATJ]

Mega,

lux = lumens per m². As I said in the other thread, lumens don't make sense. Like measuring pressure in pounds, it doesn't work.

[ATJ]

Tadashi,

I'm not sure what the relevance is of the amount of light that hits the earth. The sun is so far from the earth, relatively, that the distances we are talking about will make no difference in the intensity of light from the sun.

However, the inverse square law still holds for the sun. If you could get twice the distance from the sun, as the earth is, you'd only have 25% as much light.

[Tadashi]

I used the sun measurement as a reference. PAR has metric values in it. I am not sure about lumens or foot-candles. I know that the inverse law is geometric but to get the right units you must measure correctly. TO determine the PAR from 1 cm versus 1 m would make a difference in the formula.

The next chapter in the handbook of the previous link explains how the different units play into the measurements.

Has anyone seen this: http://www.masla.com/dana01.htm
They state that PAR is a more relavent measurement since it measures the entire band of light versus Lux.

I am still tyring to read all this and understand everything.

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"Honey, put the bleach down and step away from the reef tank. I promise we will spend more quality time together."
My Bonsai-reef site

Dont worry atj, I completely agree with you, unfortunatly I am unable to find lux or better yet par levels for vho, pc, no, and 175w halide bulbs, so I used lumens for the posts here, in my database, I have par for the bulbs that I use. L8r mega

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[ATJ]

Tadashi,

The inverse square rule is a relative thing and the units are irrelevent. Much like specific gravity (now theres a discussion we could start).

If you double the distance you quarter the light. It doesn't matter if you go from 1m to 2m or 1" to 2" or 1cm to 2cm. In all cases the light intensity will be reduced by 4.

Look at the second formula in the handbook. E1.D1**2 = E2.D2**2 as long as the units on both sides of the equation are the same it doesn't matter what they are as long as they are a unit of length.

BTW, lumens and lux are derived SI units and so are considered "metric". lux is lumens per square metre.


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ATJ
http://atj777.tripod.com/

[ATJ]

Mega,

lumens is used to measure light output. lux is used to measure light intensity. Lamps are rated in lumens because this is how much light they output - it is an absolute number. You won't find lamps rated in lux because it doesn't make sense. The best you could hope for is a rating like 800 lux at 1m from the light source.

That's why I'm worried about your calculations. While they may be useful for comparing one distance from a light source to another distance, the values you have calculated are meaningless.

How are you going from the rated output of the lamp, to the values at a certain distance?


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ATJ
http://atj777.tripod.com/

"While they may be useful for comparing one distance from a light source to another distance" That's all I'm doing, the only numbers that have any usefullness in documenting coral requiremnts are par, I wouldnt and wont use lux, as par is MUCH better. ALL I am doing is showing how much difference a increase in distance make, I have par for my bulbs, which is a much more usefull number as I said above 5 times. L8r mega

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[markrb]

This post made my head hurt!

I never had any idea that there was so much to this. I never even considered such things. I guess in my naive way I always assumed that watts=watts give or take. If I hadn't already purchased my VHO ballasts I would have reconsidered.

Thanks for the entirely confusing and way above my head info. Anyone want to know what how many cpu's can go into a Sun 450?

Mark

[jameso]

Just FYI guys, the inverse square law does not apply if you use a parabolic (or anything BUT a flat) reflector.

The inverse square function is derived from the surface of a spherical shell (integral of a circle over 2pi).

In any case, the inverse square law is based on the fact that light emmitted from a point source "spreads out" in ever increasing spheres from the point.

Put a reflector next to the lamp, and the rule does not apply.

In fact, if 100% of the light (ideal of course) is directed into the tank, then 1/R^2 goes to 1/1.

Cheers
James Wiseman www.reefs.org

[ATJ]

Mega,

I guess one of the points I was trying to make was this: your calculations are OK for comparing the different light initensities for one light source at different distances, however, I'm not sure if the comparisons from VHO to MH are accurate. That is, what you are saying is perfect to get the point across that the closer the lights are to the water surface the better. I don't think you can use it to compare one light source to a different light source.


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ATJ
http://atj777.tripod.com/

[ATJ]

James,

The inverse square rule applies whenever the light rays are not parallel. Only when they are parallel will there be no dispersion. If the rays are near parallel, the intensity may be brighter and any point away from the light source but it will still be one quarter as bright at twice the distance.

Even if you had a perfectly parabolic reflector, with the light source at its focus, the light emitted from the near side of the light source would disperse.

With any other reflector all the light disperses. It doesn't matter that the light does not disperse evenly.

It is important to note that the distance from the light source to the reflector will make some differences to the calculations, but if the reflector is fairly close, the differences quickly become insignificant.

In your ideal case, unless you have 100% of the light directed to a spot that is exactly the same size as the light source (i.e. a point), there will be dispersion and so drop of in intensity as per the inverse square rule.


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ATJ
http://atj777.tripod.com/

[jameso]

A reflector can and is designed to take 1/2 the photons emmitted from a pt source, and send them all out on parallel paths. This is how reflecting telescopes work, only backwards. I'm pretty sure the function is parabolic.

I agree w/ you that half the photons will still disperse in a hemisphere away from the pt source.

Now that I think about it, I think that you may be right about the dispersion relation for light waves, just because of the way we model them. We assume that a wavefront is an infinite line of emmitters...Ok, I don't want to try to explain it here, but the inverse square law may still apply, but w/ modifications due to the reflector. Interesting.

Cheers
James


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James Wiseman
www.reefs.org

[Gannet]

The inverse-square law still applies, even with a reflector. Nominally, all a reflector does is maximize the amount of light available by redirecting light that would otherwise be lost off the "back" of the light. Yes, you could have a focused reflector that does a lot more than that, but is it still a "reflector" at that point, or a lens? If I could focus the output of an EYE 400 to an infinitely small point would it be infinitely bright?

In any case, the reflector has no effect on inverse square. Imagine the emitter, instead of being an assumed point or line, was instead a plane parallel to the plane of measurement. It doesn't matter how the photons got to the backside of the plane...once they emit from the front, inverse-square applies.

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when in doubt, change water

[bmw]

The function is hyperbolic.
The so called inverse square law is a model for the dispersion of light in a three dimensional space. It assumes linear direction at infinite angles. It works well for short(say under a few light years) distances. Einstein developed some models, though the mathematical ground work had preceded him by 40 years, that showed a different geometry--where parallel lines do in fact meet.

It (inverse square law)incidentally coincides with the geometry of a sphere if you consider a point source. Doesn't have to.

I am not aware of anyone actually demonstrating parallel lines of light from an artificial source. A laser is a closer approximation that had previously been achieved--but even in the best applications it will show dispersion over enough distance, indicating non parallel lines. The more finite the angles, the less the rate of dispersion,

b.

[ATJ]

quote:

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Originally posted by bmw:
It (inverse square law)incidentally coincides with the geometry of a sphere if you consider a point source. Doesn't have to.

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bmw,

This is a very good point. The inverse square rule is more corelated to the geometry of isosceles triangles and areas being a function of the square of the sides or radius.

You could just as easily represent it with a cone or even a pyramid.

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ATJ
http://atj777.tripod.com/

[Martyn]

WOW
Only just found this how come these sorts of discussion I have not come across.
just had to refresh printed it all
Martyn