Very nice write up, Greg. May I cut and paste it into a Wiki article? This is not the first time this has come up.
-tk On Fri, Feb 8, 2019 at 7:04 AM Greg Troxel <[email protected]> wrote: > Colin Larsen <[email protected]> writes: > > > I believe it's a pretty complicated beast to do properly but a factor of > > Lux * 0.0079 gets you W/m2 (approx?) from what I've read ..... so just > the > > reverse? > > I'm looking at sensors at the moment that output Lux and converting it to > > Wm/2 which where I found that conversion. > > It simply does not make technical sense to convert a sensor value in > W/m^2 to lux (which is lm/m^2, lm being lumen). They are incompatible > units measuring different things. > > Sensors like the Davis solar radiation sensor measure "irradiance" in > W/m^2, which is the amount of power arriving in an area. By definition, > power at all wavelengths is treated equally. > > lux is a unit of illuminance, which measures the amount of light in an > area in terms of the response of the human eye, according to a > standardized response curve obtained from experiments. > > For a single wavelength, one can ask what the value of one measurement > is given the other, multiplying or dividing by the standarized response > curve. Using a spectroradiometer, one can measure the power at a > variety of wavelenghts, say at 10 nm intervals, and then sum the > converted powers. > > Another way to obtain lux is to use a device with a filter that matches > the standard curve. There are instruments for lighting, and for > photography that do this. There are also sky brightness meters (which > I have read about but do have not experience with): > http://unihedron.com/projects/darksky/ > > Typically people care about illuminance for management of artificial > lighting or light pollution, and the values are very low compared to > values that can be read on a solar radiation sensor. > > This stackexchange answer claims 6.8 mW/m^2 at full moon, and a full > moon is known to produce about 0.3 lux. Note that this ratio cannot be > applied to sunlight; moonlight has a very high color temperature, > meaning that it has more blue than red compared to sunlight. > > https://physics.stackexchange.com/questions/89181/how-is-the-earth-heated-by-a-full-moon#89197 > > So basically, you just cannot convert sensor output and weewx should not > try. If you want to measure lux you need a lux sensor, and to measure > W/m^w you need a radiation sensor. > > One could go out on a limb and start making assumptions about the > spectral distribution of different kinds of daylight that are likely for > various irradiance levels. For example, full sun with no clouds is well > characterized. But as soon as you get into dawn/dusk and clouds, there > is much more variability. Trying to guess spectral distribution based > on illuminance or irradiance and converting, and comparing that to a > sensor that measures the other, would be an interesting science > experiment. > > https://en.wikipedia.org/wiki/Photometry_(optics) > > -- > You received this message because you are subscribed to the Google Groups > "weewx-user" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "weewx-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
