Tough questions, Andy.

If you know the droplet size and density, you should be able to compute the likelihood of striking a particle in one unit's distance (e.g., in a world coordinate system of meters: 1 unit == 1 meter). To take a real example, let's say you assume spherical droplets of 100 microns in diameter at a density of 10^5 drops per cubic meter. That means that one droplet (on average) can be found in every 10^-5 of a cubic meter. More importantly, looking through one meter's distance in this volume, your chance of encountering a droplet equals the combined cross-sectional area of all the droplets in the volume divided by the volume. That is:

10^5 * pi*(50e-6 meter)^2 / 1 meter^3

or a 0.000785 (0.075%) probability. This is safely in the range of a single-scattering medium for an interior space, since a photon would need to travel nearly a kilometer to have a 50% chance of encountering a droplet.

Your extinction coefficient is approximately equal to the above probability, i.e., the average fraction of rays scattered per unit distance. The value needs to be computed more carefully as the density*size gets larger, since we really should be computing the probability of a ray *not* encountering a droplet after a unit's distance, and one minus that is the extinction coefficient. I was trying to work out the exact formula, but it gets into ODE's, which was one of my worst subjects in math, right after complex number analysis... It's only a problem when you get into the multiple scattering regime, where Radiance isn't really well-suited anymore.

The albedo for water droplets is quite high, somewhere around 0.999, and the H-G is about 0.84 (values taken from "Rendering with Radiance," p. 595). These values do not change with drop size or density.

I am not sure how visibility length relates to extinction. I did a quick Google search and found the following tidbit at <www.envirotechsensors.com>:

Conversion from extinction coefficient to visibility involves different algorithms, one for daytime and one for night. Daytime visibility is related to the viewing of dark objects against a light sky as previously mentioned. For measurement of visibility in the daytime, Koschmieder’s Law is used:

V = 3/σ

Where V is the visibility and σ is the extinction coefficient

Nighttime visibility is related to the distance at which a point source of light of known intensity can be seen. For measurement of visibility at night, Allard’s Law is used:

V = e-σ*V/0.00336

Where V is the visibility and σ is the extinction coefficient.

Most applications will use only the Koschmieder’s (V = 3/σ) formula. The aviation community typically uses a photometer to measure the day/night condition and applies both formulas depending on the ambient background light.

So, a daytime visibility of 0.8 meters corresponds to an extinction coefficient of 3.75 m^-1, which of course is nonsense because extinction must be < 1. I assume this approximation doesn't really apply below a certain visibility distance, >> 3 in whatever units you have. Again, this relates to the problem mentioned above with multiple scattering, and you need to make sure your in a single-scattering regime for any of this to make sense.

I hope this helps. It sure got me confused!

-Greg

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From: "Andy Stone" <[email protected]>

Date: December 6, 2006 5:24:01 AM PST

Hi all,

I am trying to use radiance to simulate a room filled with a fine water droplet mist. If I know a little about the composition of the mist (e.g. droplet size and density), is there a way to derive the extinction coefficient (-me), scattering albedo (-ma) and eccentricity parameter (-mg)?

Also, the beginning of chapter 14 of Rendering With Radiance suggests that the single scatter approximation used is only appropriate for thin mists – is there a guideline for how thin “thin” should be? I want to try several different thicknesses including some cases where visibility might be less than one metre…

Thanks,

Andy Stone