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Thread: Another color balance question...

  1. #11
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    There are some furious PMs going back and forth in the background right now.

    I'm waiting on a ascii table of CIE standard observer numbers from a friend.

    We will solve this, but initial numbers look like 5 watts of 658 and closer to a watt to a watt and a half of 640 Stuka.

    However my buddy with a hungarian system using 640 and 473 tells me
    240 mW of red 260 mW of Blue and 130-150 of green for a "snowy white"
    He also says he can generate browns , golds ' and silvers with the system adjusted for those max powers. That would mean Stuka needs at least 800 mW of 640 if not more to balence his system, so now the question becomes how much does his 658 maxys contribute.

    This is getting really subjective and scientifically bullshit fast, so we're gonna suspend this thread till I have proven solid data andmath I can document.

    Or Stuka just buys 300 mW of red and the new dichro and lives within his means :-)

    Steve

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    Quote Originally Posted by mixedgas View Post
    .....
    Or Stuka just buys 300 mW of red and the new dichro and lives within his means :-)

    Steve
    Well......OK, I can live with that

    I'm probably going to regret this, but...

    Anybody got a good deal on 300mw of 635nm with good beam specs that will fit the mirrors on my ScanPro 40's?
    RR

    Metrologic HeNe 3.3mw Modulated laser, 2 Radio Shack motors, and a broken mirror.
    1979.
    Sweet.....

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    Cool stuff and really interesting.

    Thanks for your input Steve.

    Cheers

    Jem
    Quote: "There is a theory which states that if ever, for any reason, anyone discovers what exactly the Universe is for and why it is here it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another that states that this has already happened.”... Douglas Adams 1952 - 2001

  4. #14
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    By my estimates, if you're starting off with 400mW 532, 120mW 473 and 400mW 660, to match the green power you will need additional 690mW 635nm and 680 additional mW of 473.

    If you want to turn the green down to balance the blue power, you would turn it down to 60mW and at that point you should only need 6mW of 635 to balance, which is well within the margin of error here!!

    EDIT:

    If you ignore the common 2:1 for blue and green and go entirely by the white color balanced sensitivity chart, ignoring forward scattering effects, then:

    additional 431mW of 473 and 717 mW of 635nm to match green
    or if you turn down green to 87mW to match blue,
    additional 65mW 635nm to make white.
    Last edited by drlava; 09-23-2008 at 10:19.

  5. #15
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    SCORE:

    STEVE 440 mw first approximation, latter 800 mW of 633 total 1800 mW of red


    Estimed Dr Lava add 717 mw of 633 , total ~1100 mW of red

    Estimed Doctoral candidate who wishes not to be named but will post his results soon.

    ~1.5 watts of 633, ~5 watts of 658.

    DrLava was smart enough to add the scattering coefs. I ignored that at my peril, and assumed we were projecting on a screen, and that adds even more nonlinear effects.

    Comments from the peanut gallery ?
    Mecheng you want to chime in?

    This site was helpful:
    RadiantImaging.com

    had free downloadable RGB calculator (photometric, not radiometric)
    Use 3 led setting in program and reduce led bandwidths to 1 nm

    Why did we all have such different numbers ? Go look at this link and click on "math"and you will see why:

    http://www.brucelindbloom.com/index....alculator.html

    Lava and I agree, 800 mW total blue

    more to come,
    Steve Roberts
    Last edited by mixedgas; 09-23-2008 at 15:12.

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    Whoa!
    ~1.5 watts of 633, ~5 watts of 658

    I like!!

    If the "Estimed Doctoral candidate who wishes not to be named" has such beasts at his/her disposal, I would be glad to "field test" for as long as necessary to prove the theory...
    Last edited by Stuka; 09-24-2008 at 04:35.
    RR

    Metrologic HeNe 3.3mw Modulated laser, 2 Radio Shack motors, and a broken mirror.
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    I wish!

    I have revised my calculations using the 1931 CIE color matching function. My result for the "perfect" color balance is now:

    400mW 532nm
    400mW 660nm
    550mW 635nm
    500mW 473nm

    This gives a color temperature of roughly 6500K; the D65 standard illuminant, which is the white point in the sRGB space. It is a fairly blue white though and I personally like the white obtained by adding 400mW 473nm and 600mW 635nm better. It has a color temperature of about 5000K.

    I have not taken scattering effects into account though; this would increase the amount of red required while decreasing blue. Any tips on how to implement this are most welcome.

    My first estimate of 1500mW was based on a linear model, which supposedly works fine as long as colors are similar. Apparently it is not very valid for the full color space.

    I have written a matlab script to plot what color you get by adding an arbitrary number of lasers into a CIE diagram. I hope to be able to refine it sufficiently to release it as a standalone application. It is hardly perfect now, but I think it shows promise.

    Here is an example, it shows my two suggested balances:

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    This is interesting stuff..

    Can I ask a question... probably dumb knowing me.

    I often wonder when I see pictures of guys projectors that use 635 that a very high percentage of the red is "outside" the "mix"..
    Like if you have a green with a 3mm beam, a blue wih a 2.5mm beam and a red with a 6mm beam then like half the red is not in the "combined" beam hitting and leaving the scanner mirrors.
    Is this something that needs to be considered for color balance ??

    Is there some other factor I am not seeing here ?

    As I say just a dumb question..

    Thanks

    RAy
    NZ

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    I don't know if its possible to cheat here, but the way I would have approached it is look at the lasers used in a top class projector such as a Kvant or Jenlas and then just copied what they'd used in their equivolent powered model on the simple basis that they're so good at it! After all if they have a great white then with the same powers and wavelengths surely so should you.

    Sorry if this deflects from the real maths, just trying to think of a simple solution for the lazy or mathematically inept such as me.

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    another matlab user! I guess we know who the PhD candidate is. seems like our numbers agree relatively well.

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