I lifted this eight year old archived post from Andrei Broder
who says it still simply, but without those fuzzy edges

From: "Andrei Broder"
Date: Mon, 05 Feb 96 00:47:45 -0800
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Light flux is measured in lumens. Light sources are often labeled
with an output rating in lumens. //

Lumen is actually a "derived unit". The basic international unit,
measures luminous intensity and is called candela. Candles and
candlepower are the same thing but these are deprecated names. It
tells how much flux is flowing through a solid angle,
which is measured in steradians. A point source that has intensity
of one candle puts out a lumen per steradian. A unit area, all at
unit distance from a point covers exactly a steradian.

Illumination (illuminance) is the area density of incident luminous
flux: how many lumens per unit area. A lux is one lumen per one
square meter. Illumination from a point source falls off as the
square of the distance. So if you divide the intensity of a point
source in candles by the distance from it in meters squared, you
have the illumination in lux at that distance. (Remember this
assumes a single point source in a sphere - no reflectors, lenses,
etc. What we have are often multiple complex sources.)

Read the rec.photo FAQ for a good intro to all this. I cribbed from
it.
##########################################

Brian W

On Fri, 30 Jan 2004 19:53:38 +0800, Old Nick
wrote:

I am interested because I was looking at ultra-brite LEDS, and there
is all sorts of stuff quoted to give the usefulenss of these things,
mostly without the backing data to make it real.

OK. AFAICS. Please correct me or fine tune as necessary. If I am
pretty well right, then could somebody please simply say so G:

The CANDELA is the _unit_ of measure of emitted light from a source,
at the source.

The LUMEN is a measure of total light output. That is to say you
measure the CANDELAS, then assuming they are constant over the area of
light emitted (rare but for these purposes) then the LUMENs will be Cd
* area of emission. errr....no. There _is_ no area of emission from a
point source, at the source. OK. So it's the _angle_ over which the
light is emitted. How do you measure angle in three dimensions (bare
light globe), as distinct from a narrow beam or flat beam?

The LUX is a measure of light intensity, but measured at a distance
from a light source. Lumens/metre 2.

The CANDELLA would be the way to find out how bright the LED is at the
source. If the angle of emission is given also, then we trust we can
say that over that spread, we will se that brightness. A chart would
be nice. We could then approximate LUMENs (??)

BUT.

Then I get a site that gives the data for a LED-for-sale in LUX!!!
What does _that_ mean? No distance = no meaning AFAICS. The only thing
they did say was "Do not look into light source with remaining eye" or
words to that effect. This implies that this is collimated (sp?) to
almost lazer spread, and that therefore LUX could maybe be used, in
that LUX ~ Candela? WUP! No. It's a _wide angle_ LED!

The LED claims 1W , 3.6V 300mA 20 LUX, wide angle.

There is another site that has :
"LED-WHITE 5 mm SUPER-BRIGHT WHITE LED ~ 15 000 millicandela @ 20 mA,
+ or - 10 (20) degrees, VF @ 20 mA ~ 3.6 V, tolerance +/- 30%. Clear
Package"

I am having trouble getting more than a grunt out of the guy with the
15000 mCd LEd, either email or verbal.

BAH! I will try the other site.

So, I am assuming that one is the same intensity as the other (20
compared to 15) but for 1/20th the power. I have to assume that the
lower powered one is getting this from its narrower beam.

To add to that, Infrared LEDs are measured in Watts (or mW, and in one
case mW/sr.

So what is mW/sr?

Most spotlights are measured in Candlepower, which I gather is the
actual figure for the candelas emitted (as height is to metres).

So. If I have a LED with claimed Cd of 15000 mCd, or 15 Cd, then it is
a 15 CandlePower LED? Am I correct in comparing it with, say, a
spotlight with claimed output of 2,000,000 Candlepower? I realise
that:
- they are chalk and cheese (133,000 to one! G)
- I would have to look a the spread of the light in order to
compare, in that the luminous intensity at a given distance (LUX)
would be affected by that as well as Cd output.

I doubt this would (directly) affect the _visibility_ of the light
when looking at it from a distance in the dark. But it would certainly
affect its perceived brightness and ability as a source of
illumination.

Sorry for the ramble, but it is a bit messy.

************************************************* *** sorry
remove ns from my header address to reply via email

Spike....Spike? Hello?