# $\rm Lux$ and $W/m^2$ relationship?

I am reading a bit about solar energy, and for my own curiosity, I would really like to know the insolation on my balcony. That could tell me how much a solar panel could produce.

Now, I don't have any equipment, but I do have a smartphone, and an app called Light Meter, which tells me the luminious flux per area in the unit lux.

Can I in some way calculate W/m2 from lux? E.g. the current value of 6000lux.

• lux are lumen/m². Lumen on the other hand are the power emitted in a steradian (i.e. a small portion of all possible directions your source could emit to) weighted by the luminosity function. The latter accounts for the variable sensitivity of the human eye to different wavelengths. So you'd have to 'de-weight' the lux in order to get the power/m². – user9886 Sep 15 '14 at 13:58
• If I might recommend a reference: the RCA E-O Handbook, a great compendium of definitions and formulas. It is freely available per the descendent company of RCA. One site is phy.davidson.edu/fachome/jny/optics/burle%20electro_optics.pdf – Carl Witthoft Sep 15 '14 at 14:07

I'm afraid that it is not easily possible to take the luminous flux and obtain the insolation (as radiant flux). Here's why:

The luminous flux $F$ is calculated from the radiant spectral power distribution $J(\lambda)$ by weighting each wavelength with a luminosity function $y(\lambda)$ as per

$$F = c \int J(\lambda)y(\lambda)\mathrm{d}\lambda$$

where $c$ is some unit conversion constant between lumen and watts. The total radiant flux $\Phi$ would be

$$\Phi = \int J(\lambda)\mathrm{d}\lambda$$

The problem is that the calculation of the luminous flux is not invertible - portions of $J(\lambda)$ lying outside the visible range are cut off by the luminosity function being zero there, and it is perfectly possible for two $J(\lambda)$ of different radiant flux $\Phi$ to have a similar luminous flux $F$.

However, in the case of solar radiation, there might be a way - we know the spectral composition of sunlight, and so we know the form of $J(\lambda)$ already quite well - you could try to run an algorithm that fits the scaling of the known spectrum $J(\lambda)$ to yield the value of $F$ you measure and then calculate $\Phi$ from that spectral function. I'm not sure how good this idea is though.

• Your approach is, as they say, "close enough for jazz." I wouldn't be surprised if solar panel mfrs or installers base their estimates on local luminous flux measurements plus a "simple" conversion formula. – Carl Witthoft Sep 15 '14 at 14:08
• I did an experiment measuring solar radiation with a pyranometer and a light meter that measured in lux. I implemented this algorithm using a black body distribution for the Sun with effective temperature 5780 K, and can confirm it does give results close to the W/m^2 pyranometer measurements! – binaryfunt Apr 13 '16 at 21:56

There is no simple conversion, it depends on the wavelength or color of the light.

However, for the sun there is an approximate conversion of $0.0079 \, \text{W/m}^2$ per Lux.

To plug in numbers as an example: if we read 75,000 Lux on a light sensor, we convert that reading to $\text{W/m}^2$ as follows: $$75,000 \times 0.0079 = 590 \, \text{W/m}^2 \, .$$

Source

Lux is a unit of "illuminance" and is based on the eye's response to light and each wavelength is weighted based on the percentage the eye is capable of perceiving. The curve is loosely a bell curve so at deep purple the eye may only "see" 5% of the W/m2 available while in the fat green part of the visible curve it may "see" 90%. So the ability to quantify lux, properly, means having a full spectral power distribution table of the incident light in W/m2/nm from about 390 to 810 nm, multiplying the value at each wavelength by the average human's eye response at that wavelength, and summing the results.

So lux as a measure of radiant energy only has value if you want to know how bright it will appear to a human, not any meaningful engineering measurement of spectral quality or heat energy.

This is my first post ever, sorry if I'm being Captain Obvious.