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Illuminance Lux (lx=lm:m²)
PLEASE NOTE!!!:
Our German sites are much more detailed then our English sites. So please switch in our Germen sites and use the language translater (top right) with 60 languages.
1 lx = 1 lm/m2
The illuminance is a pure receiver size and serves the unit for brightness.
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Lighting conditions: Midday sun light in summer Covered skies in summer Rain weather with dark thunderclouds Office lighting Living room Stairway lighting Road lighting Twilight after sunset Midnight with full moon Moonless starry sky at night |
Illuminance: 100.000 Lux 10.000 Lux 1000 Lux 500 Lux 200 Lux 100 Lux 10 Lux 1 Lux 0,2 Lux 0,0005 Lux |
In the site link luminous intensity we describe the strength of a light source. If we now want to know with effect of a light source has, the illuminance gives the answer.
At home or in the job it interests less, how brightly the lamp or the fluorescent tube radiates, but if one wants to know, has I sufficient light around to read, is my working area sufficient illuminated , which mood I want to produce on the illuminated objects? The term "mood" was treated already in the chapter color temperature and one can easily imagine that the brightness of an object is related with the color temperature, but it is a big difference whether I use a 10W or 100W bulb.
Example:
We find the brightness of a candle as a very pleasant light and want to find out whether this brightness would be sufficient e.g.. to be able to read a book. Our cosy reading chair is 2m away from the candle.
The light-current of a candle is approx.. 12 lm. In the distance from 2 m the
Illuminance is („brightness“)= 12 lm/(4p · (2 m · 2 m)) = 0,24 lm/m² = 0,24 lx
A candle is an isotropic source of light, therefore we have here a spatial light propagation (Sphare surface = SR = 12.566)
The result:
Our white book sheet, illuminated by a candle in the distance from 2 m, appear approximately as bright as in the light of the full moon at midnight, thus for reading a little too weak.
A further example with an LED (anisotrope light source) with a luminous intensity of 6cd within a light cone of 30° and a distance r = 2m. 30° = SR of 0,2. (see table).
Luminus flux = 6 x 0,2 = 1,2 lm
The area, which a LED with 0,2 sr in a distance of r illuminates, amounts to:
A = 0,2sr x 4m² = 0,8m².
We know 1 lx = 1 lm/m2
The result:
Illuminanc = 1,2 lm : 0,8m² = 1,5 Lux.
Who has good eyes, could read there.
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© Markus Kottas, Heldenberg Ltd.2018