This page notes that a *unifromly-emitting* 1 candela light source yields
12.5 lumens - the frst reference I'd found which mentioned this did not specify the "uniform" part.
ANyway, the age has a succinct chart that allows conversion of mcd into "real" lumens based upon the full-angle cone of the source's emission. THere is also a link to a nifty calculator.
THis helped me a lot, because before this, I'd not realized that bit about the *uniform* source. Now I know :)
Hopefully, this will be useful to others who have similar questions!
1 lumen is the unit of photometric 'power'--at the peak (540 nm), 1 W of optical power is 683 lumens (probably the largest prime number ever used for unit conversion).
1 lux is 1 lumen/m**2
1 candela is 1 lumen per steradian.
An isotropic source at the centre of a sphere illuminates the entire sphere surface uniformly--which is 4*pi steradians. Such an isotropic source might be a localized air plasma created by an induction source, for instance, or the Sun.
If the light shines only in one direction, it illuminates a hemisphere (2*pi steradians).
If instead of an isotropic illuminator, you pick a flat surface, then in addition there's obliquity to worry about. If you have a uniformly illuminated diffuse white card, it looks smaller if you see it obliquely, but the surface appears equally bright. Another way of putting this is that if you looked at it through a drinking straw, you couldn't tell by the surface brightness if it was tilted or not. Real surfaces aren't quite this diffuse, but things like white paper and packed MgO powder are close. A source that's perfectly uniform with angle (i.e. it passes the drinking-straw test) is said to be Lambertian.
Light from a Lambertian source has a cosine obliquity term, because a flat plate tilted by theta appears to be cos(theta) times its real length. When you do the integral over the hemisphere, you get
Effective solid angle = integral (0 to 2*pi)dPhi integral(0 to pi/2) sin(theta) cos(theta)dTheta = 2*pi*[cos(0)-cos(pi/2)] = pi steradians.
Lux is candela divided by square of distance in meters. The calculation for candela to lux lacks a 4pi.
Lumens is candela times 4pi if the light source is isotropic. For a uniform beam of a given solid angle, lumens within the beam is candela times the solid angle in steradians. A sphere covers 4pi steradians. A hemispherical shell covers 2pi steradians as seen from the point that is the center of the whole sphere that the hemisphere is half of.
For a uniform conical beam of a given angle from center to edge, lumens within the beam is candela times 2pi times (1-cos(theta)). The cone's whole included angle is 2theta. LED specifications often include "vieweing angle" noted as "two-theta-half", which is twice the angle from centerline of the beam to a point in the edge where the intensity is half the central value.
Make that 540 THz. The wavelength of this is approx. 555 nm.
For equivalence at other wavelengths, lumens in a watt is 683 times the photopic function of that wavelength. The official photopic function is now the one adopted by the CIE in 1988, and I think it is Vm(lambda), determined in 1978.
So far, so good.
I would have a quibble with that one - I would say a light source has to shine into half of all directions to illuminate a hemisphere. :)
A lambertian light source does indeed have lumens being pi times the axial intensity in candela. There are a fair number of LEDs that are supposed to approximate such a radiation pattern from a small source.
snipped-for-privacy@manx.misty.com (Don Klipstein) wrote in news: snipped-for-privacy@manx.misty.com:
OK - I get the concept that the angle-width of the LED's light influencing how much light is actually produced - but, if the Wikipedia page I mentioned in the original post is wrong, then how do I figure out whether an LED, or X combination of LEDs, will produce the illumination of a 15 Watt incandescent bulb (110 Lumens)?
If the lumen output of the LEDs is known, then simply add up the lumens.
If the lumen output is not known but candela and beam angle are known, then lumens are *ideally*:
Candela times 2 times pi times (1-cos(half the beam angle)).
Lumens will not be exactly this because the candela throughout the "specified beam angle" is unlikely to equal the rated candela, and there will also be some light outside the rated beam angle.
snipped-for-privacy@manx.misty.com (Don Klipstein) wrote in news: snipped-for-privacy@manx.misty.com:
Ah-ha! :) OK, I wrote that into my notebook.
It doesn't have to be precise, the above is great (I'll put, into the bottom of the units, either a curved mirror, or more economically, a "mirror pyramid" (i.e. buy an inexpnsive mirrir and take my glass-cutter to it)), to reflect light back up through the decorative glass.
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