# Illuminance Calculation in Scenes with Multiple Types of Light Sources

Currently, I am performing illuminance calculations for a scene with multiple types of light sources. Specifically, I am aiming to calculate the illuminance (lx) on the floor of an office where fluorescent lights are installed in the ceiling and sunlight is coming in through the windows. Additionally, I plan to use rtrace -I for these calculations.

The output from rtrace -I is in terms of irradiance (lm/㎡), and I have learned that to convert this to illuminance (lx), the following formula should be used:

`lx = 179 * (0.265 * R + 0.670 * G + 0.065 * B)`

My question is whether illuminance (lx) can be calculated using this formula even when there are multiple types of light sources. I am concerned that the luminous efficiency of fluorescent lights and sunlight might differ. Is it acceptable to use the “179” in the above formula as it is?

As a beginner who recently learned about Radiance, I might be mistaken, but I would appreciate your guidance.

Hi and welcome to the Radiance forum!

This is an intelligent question to ask, but using 179 in your conversion should be fine for both daylight and fluorescent light.

How are you calibrating the values for your sky? If you divide by the 179 factor in going from measure external illuminance, then this cancels out in the end. This is the usual procedure.

If you are just letting gensky or gendaylit estimate the solar and sky contributions, then there is no reference point for your calculation, anyway.

I hope this helps!
-Greg

Currently, I am using the files generated by the gensky command as they are, without any adjustments.

If you divide by the 179 factor in going from measure external illuminance, then this cancels out in the end. This is the usual procedure.

Does this mean that I should first measure the actual illuminance [lx] and then divide it by 179?

If you have access to global (sky+sun) and diffuse (sky w/o sun) measurements, then yes you can use these directly with gensky or gendaylit. In that case, divide illuminance by 179 to get irradiance values to the -R and -B options. The -B value will be the diffuse illuminance divided by 179. The -R value will be the global illuminance minus the diffuse, that quantity over 179.

I hope this is clear.
-Greg

This is an interesting post. Just a quick follow-up:

Let’s say I have a captured indoor HDR image after photometric calibration with luminance measurement.

Is it suitable to use this formula to get the illuminance map from the calibrated HDR image?
lx = 179 * (0.265 * R + 0.670 * G + 0.065 * B)

We can compute the scene luminance: L = 179 * (0.2126 * R + 0.7152 * G+ 0.0722 * B)/Exposure

My confusion is that a single image provides per-pixel information; calculating illuminance might require converting between scene radiance and irradiance. Is the formula above for lx related to this point?

Guanzhou

While an HDR image is not the most reliable way to get an absolute measurement of either luminance or illuminance, you can put a diffuse material of known reflectance R horizontally in your scene and use it to derive the global illuminance from its HDR-reported luminance L if Photosphere, hdrgen or raw2hdr is used in its creation.

Illuminance E = L*pi / R

The L luminance value uses the formula you quoted in your post, and the above relationship holds only for diffuse (Lambertian) gray surfaces.

If your HDR image was created by Photoshop or any other tools, then the output is uncalibrated and this method will not work at all.

Cheers,
-Greg

Hi Ji,

you may find this old paper interesting, since we demonstrated exactly such an application and also give some example data for what the “diffuse material” might be: https://www.research-collection.ethz.ch/handle/20.500.11850/618492 (this links to the full proceedings, for an extract try the copy at academia ).

Best, Lars.

Thanks for sharing, Lars,

I am a beginner in optics, just like with Radiance, and I realized I may lack some fundamental knowledge.

I reviewed the manuals and have understood the following. Could you please confirm if this is correct?

1. `gensky` generates a scene description based on the CIE standard sky distribution.
2. CIE standard sky distribution provides a relative luminance distribution, so a reference value must be specified to obtain an absolute luminance distribution.
3. In gensky, the reference value can be specified with `-b`, `-B`, or `-R`.

If you divide by the 179 factor in going from measure external illuminance, then this cancels out in the end. This is the usual procedure.

I thought that if no options were used, default reference values would be applied. Is that not the case?

I’m sorry for repeatedly asking questions and for any inconvenience this may cause.

Hi Ji,

I think you will find the Daylight Simulation chapter from the Radiance book useful. It is included in this PhD thesis as chapter 2 – Daylight simulation: validation, sky models and daylight coefficients . Although part of the book Rendering with Radiance, it should make sense as a standalone chapter. I’d also recommend much of the rest of the book also (if someone hasn’t already done so).

Best
John

Yes, do read up as John recommends.

Regarding “standard skies”, they are standard relative distributions, and do not address absolute quantities. This may be important for simulating the combination of daylight and electric light contributions in a space. Both gensky and gendaylit will choose a reasonable sky brightness, but there is no standard for these values and in fact they vary widely from place to place and day to day.

You really need to think about the question you are trying to answer to decide the best way to go about it. There are other resources as well, such as so-called “Typical Measurement Years” (TMYs), which attempt to capture the statistical character of different locations.

Daylight simulation is a vast and a mature field, which is not to say that we know all we need to know.

Best,
-Greg