View Full Version : MH lighting is a myth

Well--I am not sure.

But I thought that would get a few peoples' attention.

I cannot figure why Mh bulbs(250 watt) at 16000 lumens at 8" from the water surface are preferred over vho at 3500(110 watt) lumens at 3" over the water surface.

Given that light from a point deliveres energy at the reciprocal of the distance squared, why would one prefer MH at the distances given above? DO I have wrong data?

This one is not an experience or opinion question folks. Just looking for what I must be missing in the physics department.

And I know I am talking about photometric measurements. Lets keep it at that for now, and worry about the radiometric measurements later.

b.

billsreef

02/18/2000, 08:06 PM

It's the difference between a point source (MH) and a non-point (difuse) source (FLO).

The diffuse flo. lighting looses intensity very rapidly with the increase in distance from the source, compared to the fall off in intensity from a concentrated point source of MH.

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Sorry Bill,

that doesn't fly.

b.

horge

02/18/2000, 08:19 PM

Won't comment on the photometric angle yet, I'm Mr. Natural Light, and so may not be qualified. Maybe you're misreading bill's comment?

Photometry aside, positioning your lights ala MH allows for pendant mounting --servicing your tank is a breeze compared to having a durn hood to flip up or lift off.

I'll shut up now :)

Psyduck

02/18/2000, 08:33 PM

Ok I dont know the answer to this question, but a VHO bulb is not a point source is it? I mean MH and VHO are 2 different designs the MH being a near point, and VHO being a length across the entire tank? I mean arent we comparing apples to oranges here? Just a thought, please dont yell at me :).

Larry M

02/18/2000, 08:54 PM

Hah! That subject line is pretty funny. :D Don't tell me we have another "I-don't-want to-spend-the-money-so-I-am-going-to-re-invent-the-(lighting)-wheel-again-and-go-a different-route reefer in denial????

;)

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MegaDeTH

02/18/2000, 09:01 PM

It's called the inverse squared law, double the distance, 1/2 the power, this is why I run my halides several inches from tank, as close as is possible. Vho spread this light out the lenght of the bulb, halides concentrate it in one point, I run a combination of both, vho simply dont cut it for the acro's that I keep (yes have kept many of my corals in different lighting) L8r mega

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Horge-

Getting you to shut up? Who has ever been able to accomplish that? :)

Psyduck,

I would never dream of yelling at you! :)

Bill,

I apologize, perhaps I replied too quickly.

Teacher in me, easy to forget I am the student here.

So--

No, we are not comparing apples an organges here. Or ease of one or the other. Just light delivery to the water suface.

I believe(and you can correct me here) what you all are referring to is the fact that MH lighting comes from a smaller surface area outside the enclcosure(bulb) than a tube.

Correct.

However, since light travels in a linear direction----

Intensity at a perpindicular angle for MH would be greater? But at greater and greater angles would, it in fact be less?

There is some serious math here, yes? :)

b.

NorthCoast

02/18/2000, 09:29 PM

BMW,

I'm not a lighting expert...someone will write a text book answer. I see your point. However, your linear assumption with the FO and VHO bulbs is wrong to a degree. A tube light is brighter in the center than on the ends. I don't have a formula in front of me to quote (and it depends and length and manufacture). It is that drop-off that you have to consider in the equation. It is not as easy as dividing the watts (or lumens, or whatever) by the length to get a constant value across the length of the tube.

A non-scientific answer FWIW, :)

NorthCoast

Whoa,

missed 2 replys while I was conversing with the wife.

No, Larry, money ain't the question here. No more than someone a few years back saying they don't need them wet-dry filters. :)

Megadeth,

Thanks, knew what it was called. :)

In fact, halides do not concentrate light on one point, and even given that the source is a single point, the envelope (glass housing) negates the "no diffusion" angle. You are trying to compare a single point placing all of its energy at a perpindicular angle to a source with infinite points radiating at perpindicular angles. MH does not radiate all of its energy at a perpindicular angle.

And thanks for the "experience" but not looking for that here.

Now if you are putting your MH 3" from the water surface--well that negates the whole question. There is no question of the fact that MH put out more light energy.

My question was--why do you think that putting MH 8" from the surface was more intense than putting VHO 3" from the surface.

b.

MegaDeTH

02/18/2000, 10:53 PM

well, your question is why would one prefer metal halide 8" away vs vho 3" away, simple, halides output MUCH MUCH more light. Back to the point source, halides are a 2" x 1" tube that produces light, encased in a UV shielding outer envlope. Vho (what vho produces 3500 lumens, 48"?)and put out that light over 48"x3". THe light radiates from the source in all directions in both cases. I dont fully understand the impacts of the inverse square law, however, your trying to compare apples vs oranges right now, halide at 4" vs halide at 8" would be a much better question. Seriously, that's 3500 lumens / 48" = 72 lumens per inch, vs 8000 lumens per inch of halide. Really are 2 different topics in your question that need to be discussed separatly.

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MegaDeTH

02/18/2000, 11:11 PM

Ok, using inversed square law, E=I/D^2, and not touching on Lamberts Cosine law (yet, I can do this, once I setup some senarios)

16,000 lumen halide 8" over water, with coral 12" under water will recieve 40 lumens

vho 3500 lumen 3" over water, with said coral 12" under water will recieve 15.55556 lumens. This is why we use halides for intense light.

I was bored and threw some bulb lumens into excel and played with some numbers, fwiw here's some I was able to find.

250w iwasaki 18200 lumens after 100 hours

96w pc 8250 lumens (per custom sea life) (update, freind just tested his and gets a MUCH lower number, 5220 lumens)

If you have the 96w pc & 96w blue pc 3" over the water, the lumins at a coral 12" under the water is 42.38. If you put the 250w iwasaki 8.75" above the water you will get the same, 42.27 lumens at the coral. is this the data you were after? Fwiw I'm trying to find measurements for vho and vho actinic lumens, and for some of the other halide bulbs. L8r mega

[This message has been edited by MegaDeTH (edited 02-19-2000).]

[This message has been edited by MegaDeTH (edited 02-19-2000).]

[This message has been edited by MegaDeTH (edited 02-19-2000).]

Wolverine

02/19/2000, 12:27 AM

I always thought that another of the big bonuses of point lighting is the shimmering effect. Now, this is all trying to remember from things I read a couple of years ago, but I think I have it right. When MH light goes through the waves at the surface, it gets focused into bands of more intense and less intense light which move across the corals (and everything else). I know that some feel this alternation of intensity is very important. And I know that the more intense bands end up delivering much more light to a given spot during that time than without them (I can't even begin to remember how much of an increase).

I also know, and can understand, why this happens with MH but not with tube lighting, but there's no way I could even begin to explain, especially mathematically, and especially of the top of my head. And visually this is far more striking (makes me wish we could have a MH with our current system). So, at least as far as this effect is concerned, it would be better to have the MH 8" away rather than the VHO 3" away from the water.

Buy anyway, I guess this is just a different angle that I wanted to throw into the conversation.

Dave

MegaDeTH

02/19/2000, 12:39 AM

ok, I've been looking at lamberts cosine law, trying to figure out what role it plays in our tanks, with reflectors it shouldnt factor too terrible much, at most 60 degree's is a 50 % loss (that was the easy one hehe) But considering it's a point source, but not a flashlight directed beam, directed in all directions, it doesnt factor in too much, or else my logic is wrong (trying to picture this in my head) However, the inverse squared law is different for a coral 12" down, and also 12" to the left of a halide, a senario that wont happin with a vho, as they emit light along a 48" source. Or am I way off on what your looking for? L8r mega

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horge

02/19/2000, 01:22 AM

Sigh. You were right Bruce. Can't shut up :p

Can I try to pick your mind?

With the math for your initial figures quickly put to sleep, it's easy to see your puzzlement.

On the nanolevel, apples and oranges are the same, as even 'non-pointsource' lighting is composed of conglomerations of pointsources. On the nanolevel, all radiated light (whether rays, waves or packets, all you Planckian activists..) is shooting out of the source at a 'perpendicular angle' anyway.

(Whether halide element or fluorescent phosphor, at least 50% percent of your lightsource is firing in the wrong direction.) Again on that 'nanolevel' you've pretty much erased any other answer than:

"It ISN'T better to have the specified MH at 8 than the specified Fluo at 3". Light bending would seem to be less an issue with fluorescents as the phosphor is right smack up against the glass, whereas an MH element can be less than centred within the bulb, and at the levels of output we're discussing, it's an insignificant factor.

So let's get to the real question I could think you're asking: "What structural and personal factors go into preferring MH over Fluo. despite --lookit!!-- my figures? "

If you suppress the macrolevel factors that differentiate apples from oranges, you're left with energy efficiency/spectrum rendition (one issue, really), footprint (in terms of installation space and energy broadcast), and cost-efficiency/lifespan as the bones of contention. And, oh yes, the facilitation of easy tank maintenance ;)

The importances of the first three vary for issues of relative personal wealth, a prevailing lack of knowledge about lighting requirements, and the variables inherent to different tanks and locations. The fourth is only slightly less so. There are bound to be others.

[This message has been edited by horge (edited 02-19-2000).]

NorthCoast,

One correction. One of the characteristics of a fluorescent lamp driven by an IceCap VHO ballast is that the light is more evenly distributed across the lamp. Conventional ballasts are up to 1/3 brighter in the center. That a major reason why some of our ballasts find themselves in indusrial application providing light for inspection machines.

Andy

Hi Megadeth,

thanks for all that--yes, that was the direction I was headed. couple of things--

You are treating water as air so I am not sure the calculations can apply(I know, we are treating air like a vacumn in the first place)

Second, yes, the characteristical differences between a 48" tube and and a 2" one does complicate comparison.

Third, calculate lumens hitting the water surface and the advantage seems to go to vho.Which I did comming home a couple of days ago --which prompted this whole thread.

Fourth, an accounting for the absorbsion of the longer wavelenghts in water would have to be accounted for(and I admit the original post said photometric measurements--this was why-to advoid this point for a while)

Fifth, I'm sick as a dog and going back to bed :)

Thanks again for putting in the time.I would like to carry this on awhile if you got any interest--but I am signing out for awhile.

And Horge, looking at your "plankian" comment--you know some quantum mechanics that might define things a bit?

b.

Gannet

02/19/2000, 09:51 AM

The reason this looks confusing is because of the numbers chosen.

For one thing, we don't care what the illumination at the surface is. We care what the illumination at the coral is.

Let's make some assumptions to simplify the math: 1) assume the MH is a point source, 2) assume the FL is a linear source with even distribution, 3) assume our reflectors are 100% efficient, 4) assume the area we are illuminating occupies the equivalent of a 4x4" square, 5) assume the area of interest is 8" below the surface and centered under the light, 6) assume 2 VHO tubes to make it a little fairer for the VHO, 7) assume other numbers you mentioned, although why you have the MH so high is beyond me. I run mine 2 3/4" off the water.

So:

If the MH is 8" above the water and coral is 8" down, we have the light going over a sphere with a 16" radius. Using the formula A = 4 * Pi * R^2, this will have a surface area of 3217 in^2. But since we assume the reflector is 100% efficient, the area is half that, or 1609. With 16000 lumens we have 9.94 lumens/in^2, or 159 lumens on our area of interest.

If the 2 VHOs are 3" over the water and coral is 8" down, we have a cylinder with an 11" radius that is 48" long. Using the formula A = L * Pi * R^2 this will have a surface area of 18246 in^2. Assume the 100% reflector again and the area is halved to 9123 in^2. With 7000 lumens we have 0.77 lumens/in^2, and our area of interest gets 12 lumens. Less than 10% as much as the MH.

Pretty big difference, no? :)

[This message has been edited by Gannet (edited 02-19-2000).]

npaden

02/19/2000, 12:58 PM

Okay, the math went right by me but one thing missing is # of MH bulbs for the length of the tank. There is no doubt that a coral smack dab under a MH is going to get way more light that one right under anywhere on a VHO. The key is length of tank and a combination of VHO & MH. I am not sure but don't think that 1 250W MH in the middle of a 4' tank will get near the light to the sides of the tank as 2 - 110W VHO's will. I think the old standard 1 MH for every 2' of tank With VHO supplementation is the way to go for almost everything. The math just shows how important placement of the high light loving corals is. (centered under the MH's) Oh well, thought I would throw in my thoughts on the subject. Nathan

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TheDonger

02/19/2000, 03:38 PM

My two cents.

You are trying to emulate the sun. Now it is hard to compare a light bulb to the sun, but Halides do it much better.

First, the fact of the matter is, halides are point source lighting. The filament produces the light inside the charged glass container. Light is coming from the mainly light filament which is a very small source. Flourescents spread light across the entire tube by the gas that is filled in the tube, spread out greatly. The sun is a point source light. Placement of halides over corals is not as an important concern as some may lead you to beleive. The fact is, the angle of the sun as the Earth rotates constantly changes, so corals are always experiencing different point sources of lighting and reflection, just like in your tank, sort of anyway.

Furthermore, the ripple effect, just like the sun creates in the ocean, is emulated much more accurately with halides than flourescents due to this point source lighting and intensity.

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Thank you all for your comments.

I do not think using quadratic equations are going to be a good model for this. I suspect a calculus equation involving some trigometric functions is probably the right direction.

I was kinda hoping someone that deals in this area, say Andy!, might give a reply.

Andy--how do your measure light output from various types of sources when you engineer your ballasts? What are the implications for aquarium use?

And does anyone know of readings being done underwater at the depths we are interested in? I know somewhere way back I saw some on depths in the ocean with sunlight. But how about using the artificial sources?

And finally, the lumen figures in the original post were just off the top of my head as a place to start. But the distance over the water figures are from what I have observed. Now I have seen two people say they are putting MHs much closer to the surface.

I had thought that putting MHs this close was asking for problems with water and fish comming into contact with a very hot piece of thin glass. I would like to here more from you on this.

Thanks alot folks.

b.

Canadian

02/20/2000, 10:03 AM

Someone is going to have to find out about the diffusion properties of water as well. If we want an accurate calculation we're going to have to apply the properties of water and not assume that the water behaves in a similar manner to air.

In addition, the turbidity of the water is key as well. FWIW, In the Baensch Marine Atlas Vol-1 Baensch has a half-assed calculation for "required wattage" and there is a table for "pollution factor". At 30cm 100% Clear water has a "polution factor" of 86 (Obviously not a particularily profound number).

Just trying to see that this question does get answered to bmw's standards, as I am quite interested in the physics involved here as well. I just want to see the raw calculations not "Metal halide is good." :rolleyes:

All in all, great question bmw!

Andrew

I'll add some less technical notes to this number crunching thread. Comparing a MH light source to the sun is interesting but not perfect. In the ocean if two objects are 6 inches apart they'll both be equally lit. In a tank where the point source is a foot away one object may be fully lit, the other only 1/2 lit. Point source in a tank = shadows, linear source = relatively no shadows.

Small correction TheDonger, when you say the "filament produces the light". Before you turn your MH on check it out. The arc exists without a filament. I also agree, the ripple effect is great and closest to what you observe snorkeling.

As far as measurements I'd look to someone like Richard Harker. He describes his methods in a linked article on IceCap's MH page. When you add water penetration to the mix you appreciate what started the 10K+ movement, as this is the spectral portion of light that goes deepest when full spectrum light is observed.

Andy

PhishHead

02/20/2000, 02:23 PM

canadian...my thoughts too

does the inverse square law apply at the same ratios for air/water

i still like my halide. law or no law, corals like it 10x more than the no/pc's i had before.

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BristolWorm

02/20/2000, 02:41 PM

I have to agree with Megadeaths analysis.

Water/Air ok fine the numbers might be different but both the MH and the VHO's light is going through the same medium (it has to for this to be a valid discussion).

Second I think that a Luxmeter is what is needed to end this debate.

Ok... to add to this mess, is lumens what should be measured to determine the most benifit light has on coral. And which types of coral? Does UV play a factor? What about alklinity in coloring and water that has more undissolved or dissolved organics than other water. All this will have an effect one way or another.

Thanks Megadeath for the math. I agree with your analysis. And I too run MH with supplimental lighting. IN MY EXPERIENCE it is the way to go.

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BW

'Semper Ubi Sub Ubi'

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

02/20/2000, 09:11 PM

"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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Tadashi

02/22/2000, 08:36 PM

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) :D Evan and I have been arguing about this for weeks someone please help us.

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MegaDeTH

02/22/2000, 08:43 PM

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

this url explains the inverse square law. L8r mega

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Cookoopod

02/22/2000, 09:59 PM

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

MegaDeTH

02/22/2000, 10:02 PM

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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Mega,

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

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MegaDeTH

02/22/2000, 10:45 PM

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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Mega,

lux = lumens per mÂ². As I said in the other thread, lumens don't make sense. Like measuring pressure in pounds, it doesn't work.

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

02/22/2000, 11:13 PM

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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MegaDeTH

02/22/2000, 11:24 PM

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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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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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

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MegaDeTH

02/23/2000, 07:26 AM

"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

02/23/2000, 08:24 AM

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

02/23/2000, 10:56 AM

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 (http://www.reefs.org)

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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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

02/23/2000, 03:16 PM

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

02/23/2000, 06:12 PM

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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fwiw, imo, ime, ymmv, etc.

when in doubt, change water

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.

<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>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

04/09/2000, 09:52 PM

WOW

Only just found this how come these sorts of discussion I have not come across.

just had to refresh printed it all

Martyn

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