A question on Source4 photometrics

A student of mine has asked this question from looking at the photometric data of the Source4 19 degree.

published numbers are: candelas = 243520, Field angle = 19 degrees, field lumens = 11,180

however, if you use the standard formula for calculating lumens,

F = Iv × 2π × (1 - cos(A/2))

you get F = 243520 x 2π x (1 -cos (19/2)) = 20,984 lumens.

Well, I couldn't give a good answer so I'm asking here....ANYONE??

 

  • Hello georgeg,

    The “standard formula for computing lumens” you have must have an assumption of what the beam profile is as this equation does not seem to take that into account.  Maybe that assumption doesn’t fit the Source 4 beam profile.

    Where did you get this formula from?

    Take a look at this info on the High End site by Mikhail Dubinovskiy (How to calculate lumens output is the last point)

    http://www.highend.com/support/training/lightingfaq.asp

    Mike Meskill

     

    European Service Manager, ETC

  • For some reason this question has fascinated me so I got one of our Optical Design Engineers involved and this is what he has to say about the matter:

     

    Okay, here is what is going on.  The equation used by the student is correct for a source that radiates uniformly in all directions, such as a bare HPL lamp.  In a 19-deg S4 light fixture, the center (or maximum) candela is 243,000, but decreases as you move off-axis.  You can see this in the candela plot on the data-sheet.

     

    A better approximate equation to use is this one from the High End website:

     

    P=2.1 R^2 (E1 + 0.5 E0).

     

    Using the data from the 19-degree data sheet, at 25 ft, the field diameter is 8 ft, and the center Illuminance is 392 fc (this is E0).  E1 is at the edge of the field, which is 1/10th the center value, by definition.  This yields P=7900 lumens.

     

    As you can see, this is not very accurate.  At ETC, we measure lumens using a sophisticated photometric camera made by Radiant Imaging.   The camera grabs a 512x512 image of the beam and computes the field lumens using geometric and photometric equations.

     

    I hope this covers it for you :-)

     

    - Mike

  • Thank you. I agree with you that the general equation the student was using assumes a completely uniform field. I can see that since the light distribution is (in a perfect world)  cosine, with the highest candela values at the centre, then one need to take that into account and integrate values working out from the centre to the 1/10th peak spot.

    Again thank you.

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