ExposureBlog

To someone not used to this sort of thing, depositions are not easy to read.  The question-and-answer format is similar to a script. But after awhile the words take on the character of a true discussion between two people–the lawyer and the person being deposed.  The trick is to figure out how to make sense of it.

I follow the advice of the first history professor I ever had—a guy named Walter H. Ryle IV.  He was a young guy back then, and the son of the college president, Walter Ryle III.  Unfortunately the only thing I took from his class was his comment that “a book that hasn’t been marked up, dog-eared, pages underlined and high-lighted—is a book that hasn’t been read.”

So, before I read a deposition I round up a new pack of highlighters, Post-It notes and a couple of red pens.  Then, I mark up everything.  Literally.  The deposition can probably gain several ounces during my perusal of it. And when I’m finished, it looks like a book with yellow feathers.  Or something. 

This is annoying to the opposing counsel.  To him or her, it’s obvious that each of the hundreds of notes (also called FLYTs—funny little yellow things) might involve information important to my conclusions.  Or not.  Regardless, there’s never time to read them all (and who would want to?).  So, the poor opposing counsel usually asks the court reporter to copy the depos *in color* and at a size that will include the Post It notes. 

In reading depositions, I’ve discovered it takes me about 2-1/2 hours per hundred pages.  That’s an average that hasn’t changed over ten years.  When a client asks me for a guestimate of the time involved, I usually respond by asking them how much paper they intend to send me. 

I’ve seen experts bring in depos they have claimed to have read—but there’s no notes attached. No marks of any kind. Pristine.  I don’t know how they do it.  But if were an attorney (and I’m not), I’d probably ask them, “is *this* document the one you read in preparation for this deposition?”

More on depositions in a later post.

I’m back writing on the Exposure Blog after spending the last month on a very interesting toxic exposure case. This one, like many, involved a person who was overcome by a toxic gas (in this case, sulfur dioxide) that was released from the top of a 36-foot tall tank. Even though he was literally at the side of the tank when the release occurred, it was not easy figuring out how much sulfur dioxide he inhaled.

The molecular weight of sulfur dioxide—at about 64 gram molecular weight (add the mass of the individual atoms–sulfur and two oxygens)—is heavier than air (generally considered 29 gmw) but that doesn’t necessarily mean that it will drop like a rock out of the sky. For reasons I’ll get into in a later post, wind generally has more of an effect on relatively “heavy” molecules like sulfur dioxide than gravity.

And that day, the wind was blowing at a relatively leisurely 9 miles an hour.

So, part of the case was in place—we knew that sulfur dioxide was involved, and we knew the wind speed. The next thing was tougher—finding out how much sulfur dioxide was actually involved. That was not easy, and in a later post I’ll tell you how we came up with a value for the amount of sulfur dioxide released. While our calculations showed that about a thousand pounds of sulfur dioxide was released, the other side came up with around 700 lbs.

Interestingly, an expert for a third party involved in the lawsuit suggested only 76 pounds of sulfur dioxide came out of that vent.

The bottom line was beginning to appear: no one really knew exactly how much sulfur dioxide came out. I suggested we analyze the release using a mathematical procedure called a Monte Carlo simulation. There are several good programs that will do this sort of thing, but the one I use is called @RISK.

I wanted to be sure of my answer at the 95% confidence interval—that is, I wanted to be sure that there was only a 1 in 20 chance that the amount (in parts per million) of sulfur dioxide that the individual was exposed to was less than the number I reported.

To do this, I used the most conservative figure presented by the experts—in this case, it was 76 lbs of sulfur dioxide. I then used it as the *maximum* amount and input that value into the computer program. Then, for a minimum amount, I arbitrarily chose a value equal to about 2/3 of that value—about 50 lbs of sulfur dioxide–and added that as a minimum value.

Then, I made an assumption: suppose all the sulfur dioxide was evenly distributed in a given volume of air—no areas of high concentration, no areas of low concentration. What would the concentration be? Well, the individual exposed was surrounded by tanks, and since tanks take up space, there wasn’t a lot of air available for dispersion. Still, I needed a maximum volume and a minimimum volume. So I looked at aerial photos of the site to follow the path that the individual followed to the control room after being exposed. I chose the block of air surrounding this path—up to 36 feet high—as a maximum volume. For a minimum volume (which would give a higher concentration) I chose a percentage of that.

I also knew that the exposed individual probably received his maximum exposure initially—where the volume was smaller and the concentration higher. After deciding on these values as endpoints of the range, I plugged them into the program.

Concentration in parts per million (ppm) can be calculated based on this general formula:

(ppm)(V)(mw) = (grams)(specific gravity)(24.45)

where ppm = parts per million
V = volume of the area in liters
mw = equals molecular weight of the gas (here, 64)
grams = grams of substance emitted—here, the minimum was 50 and the max was 76.
specific gravity = how much the liquid weighs per cc. Since sulfur dioxide is a gas anyway, it wasn’t used in this calculation.
24.45 = the volume that one mole (here, 64 grams) of a gas occupies at 70 degrees F—which, as it turns out, was almost exactly the temperature at the time of the event.

Now, to the Monte Carlo simulation. Here is what it did: it took all the data and produced a distribution of values–like a bell curve. It also returned values associated with various parts of that distribution. In other words, there was a value at the far right end (95%) that was much greater than the value associated with the far left side of the distribution (5%). I chose the value associated with the 5% end of the curve.

In other words, there was a 95 percent chance that the concentration of sulfur dioxide associated with the input parameters (which I believed reflected the actual event) were greater than that found at the 5 percent mark.

The value that resulted was around 600 parts per million. This is more than 6 times the Immediately Dangerous to Life and Health (IDLH) value.

More later.