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Earl Kiosterud...
Posted: Fri May 09, 2008 12:42 pm
Guest
I have a friend who's constructing an internally lighted sign. He needs to ensure that the
light output will not be greater than 0.25 foot candles measured at 25' per city code. The
sign is to be illuminated with four 96" standard high-output fluorescent tubes, 8600 lumens
each, color temperature, 4100°K. We know there will be light lost internally and through the
translucent sign faces (both sides of the sign). How can we determine if such a light
source, worst case, would exceed this foot-candle limit? Thanks.
--
Regards from Virginia Beach,

Earl
...
Posted: Fri May 09, 2008 12:42 pm
Guest
On May 9, 11:42 am, "Earl Kiosterud" <some... at (no spam) nowhere.com> wrote:
Quote:
I have a friend who's constructing an internally lighted sign.  He needs to ensure that the
light output will not be greater than 0.25 foot candles measured at 25' per city code.  The
sign is to be illuminated with four 96" standard high-output fluorescent tubes,  8600 lumens
each, color temperature, 4100°K. We know there will be light lost internally and through the
translucent sign faces (both sides of the sign).  How can we determine if such a light
source, worst case, would exceed this foot-candle limit?  Thanks.
--
Regards from Virginia Beach,

Earl

Let's assume your sign is double sided so light goes out both sides,
and that it diffuses the light, so the light goes in all directions.

34,400 lumens into 4PI steradians gives an average of 2737 candela.

At 25 feet, 2737 candela will produce about 1.4 footcandles, which is
over 5 times the allowable limit...

Dean
Douglas G. Cummins...
Posted: Fri May 09, 2008 4:01 pm
Guest
Rather than try to calculate it - because you will most likely get it
wrong no matter how diligent you are - build a mockup and measure the
illuminance at the required distance. A decent illuminance meter
doesn't cost too much. Make your measurements at night or in a dark
room so that ambient light doesn't artificially inflate your measurement.

If you find your illuminance is too high, you can add reducing filters
between the light sources and the sign face.

If you're determined to do some calculations, the 0.25 fc at 25'
translates to ~150 Cd luminous intensity. The luminous flux of 8600
lumens is only an industry average. The actual flux will vary from tube
to tube and from ballast to ballast, but you can overestimate and say
that the flux will never exceed 9000 lumens per bulb and design around
that. You're already aware of the light loss due to the sign
transmittance, but if your sign box has a white or other non-black
interior, you may get internal reflections that increase the amount of
light being emitted from the sign. The relation between luminous
intensity (Cd) and luminous flux (lumens) is that 1 Cd is equal to 1
lumen per solid angle (lm/sr) for an isotropic source (a source that
puts and equal amount of light in all directions in a sphere). There
are 4pi sr in a sphere.



Earl Kiosterud wrote:
Quote:
I have a friend who's constructing an internally lighted sign. He needs to ensure that the
light output will not be greater than 0.25 foot candles measured at 25' per city code. The
sign is to be illuminated with four 96" standard high-output fluorescent tubes, 8600 lumens
each, color temperature, 4100°K. We know there will be light lost internally and through the
translucent sign faces (both sides of the sign). How can we determine if such a light
source, worst case, would exceed this foot-candle limit? Thanks.

--
Douglas Cummins
Calcoast - ITL
Earl Kiosterud...
Posted: Fri May 09, 2008 6:34 pm
Guest
<deanwil at (no spam) msn.com> wrote in message
news:4bd1ac3b-52ea-419e-a4e4-05c507a0c714 at (no spam) l17g2000pri.googlegroups.com...
On May 9, 11:42 am, "Earl Kiosterud" <some... at (no spam) nowhere.com> wrote:
Quote:
I have a friend who's constructing an internally lighted sign. He needs to ensure that the
light output will not be greater than 0.25 foot candles measured at 25' per city code. The
sign is to be illuminated with four 96" standard high-output fluorescent tubes, 8600
lumens
each, color temperature, 4100°K. We know there will be light lost internally and through
the
translucent sign faces (both sides of the sign). How can we determine if such a light
source, worst case, would exceed this foot-candle limit? Thanks.
--
Regards from Virginia Beach,

Earl

Let's assume your sign is double sided so light goes out both sides,
and that it diffuses the light, so the light goes in all directions.

34,400 lumens into 4PI steradians gives an average of 2737 candela.

At 25 feet, 2737 candela will produce about 1.4 footcandles, which is
over 5 times the allowable limit...

Dean

Thanks for your reply.

First of all, I don't know why my newsreader isn't putting the > characters in your reply.
I seems to only fail to do that with your reply.

This is all new to me. If we have a total of 34,400 lumens, then a 25 foot sphere, whose
surface area would be 4*PI*25^2, or 7854 sq feet, would be illuminated with those lumens.
So the lumens per square foot would be 34,400 / 7854, or 4.38 lumens/sq ft. Wouldn't that
be the same as 4.38 foot-candles? It's wrong, but I don't understand why.

--
Earl
RickR...
Posted: Sat May 10, 2008 11:23 am
Guest
On May 9, 2:01 pm, "Douglas G. Cummins"
<Douglas.Cumm... at (no spam) MAPSONcalcoast-itl.com> wrote:
Quote:
Rather than try to calculate it - because you will most likely get it
wrong no matter how diligent you are - build a mockup and measure the
illuminance at the required distance.  A decent illuminance meter
doesn't cost too much.  Make your measurements at night or in a dark
room so that ambient light doesn't artificially inflate your measurement.

If you find your illuminance is too high, you can add reducing filters
between the light sources and the sign face.

If you're determined to do some calculations, the 0.25 fc at 25'
translates to ~150 Cd luminous intensity.  The luminous flux of 8600
lumens is only an industry average.  The actual flux will vary from tube
to tube and from ballast to ballast, but you can overestimate and say
that the flux will never exceed 9000 lumens per bulb and design around
that.  You're already aware of the light loss due to the sign
transmittance, but if your sign box has a white or other non-black
interior, you may get internal reflections that increase the amount of
light being emitted from the sign.  The relation between luminous
intensity (Cd) and luminous flux (lumens) is that 1 Cd is equal to 1
lumen per solid angle (lm/sr) for an isotropic source (a source that
puts and equal amount of light in all directions in a sphere).  There
are 4pi sr in a sphere.

Earl Kiosterud wrote:
I have a friend who's constructing an internally lighted sign.  He needs to ensure that the
light output will not be greater than 0.25 foot candles measured at 25' per city code.  The
sign is to be illuminated with four 96" standard high-output fluorescent tubes,  8600 lumens
each, color temperature, 4100°K. We know there will be light lost internally and through the
translucent sign faces (both sides of the sign).  How can we determine if such a light
source, worst case, would exceed this foot-candle limit?  Thanks.

--
Douglas Cummins
Calcoast - ITL

A very useful number to have is the transmittance of the sign face
material. It should be pretty easily available from the supplier or
the manufacturer. The internal reflections are likely to be fully
symetric, what is lost on one side is gained on the other.

The big issue is the shape of your sign. A bit box with mostly
tranlucent sides and very little framing can be figured without the
internal losses. If you have a more complex or asymetric shape, then
it all gets really complex. Start with some basic calculations, then
see where you are likely off.

Sign makers and light fixture makers use 3D modeling and photometric
analysis software to do a virtual mock up. So, either hire someone or
do a real mock-up as Douglas suggests.

------
Richard Reid, LC
Ron Gibbs...
Posted: Mon May 12, 2008 5:04 am
Guest
"Douglas G. Cummins" <Douglas.Cummins at (no spam) MAPSONcalcoast-itl.com> top-posted in
message news:QfednXlrLqAsMLnVnZ2dnUVZ_uGdnZ2d at (no spam) earthlink.com...
Quote:
Rather than try to calculate it - because you will most likely get it
wrong no matter how diligent you are - build a mockup and measure the
illuminance at the required distance. A decent illuminance meter doesn't
cost too much. Make your measurements at night or in a dark room so that
ambient light doesn't artificially inflate your measurement.

As someone who uses raytracing software for a living, I agree this is good

advice. Before the days of non-sequential optical raytracing, I used to try
and calculate radiometric/photometric flux, but didn't expect to be accurate
within a factor af two or three. I would always build a mock-up if it was
critical.

Nowadays, good sofware can be quite accurate, depending on the time
spent on getting the model sufficiently detailed, which can be expensive in
time. Fluorescents are easy enough, but diffusing surfaces are difficult to
model accurately.

Ron Gibbs
--
Gibbs Associates
Optical Design Consultant
www.gibbsassociates.co.uk
 
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