I think the challenge was to name a figure without going over. If we count about 131/212k ounces, looks like lazyinvestor/tau are the clear winners. A bit of a sad figure, but good on you lazy & tau for your estimates. Saw your post lazy about the brookbank - any chance you'll post more numbers about KXL's deposit?

I must admit, I don't own KXL, but have a significant position in GRC. I came over here because things have been pretty quiet on that front while we await the 2010 drill program to begin. Does KXL have a table containing all of their drill results? If you can hunt that down, I can assist someone with the calculations, but it can be quite tedious manually typing the the data into excel, hole by hole.

Basically, how I do the calculation is by picturing a drill hole being in the center of a cylinder, with a radius of 25m. The volume of the cylinder will be pi * (25)^2 * true width of the intercept. In other words, 1963.49 * true width.

To calculate the true width, calculate cosine(drill angle) * core width. So as an example, KXL announces a drill hole that is 10g/t over 5 meters (drill angle of 45 degrees).

First, calculate true width:

cos(45) * 5m = 0.707 * 5 = 3.53m.

Now calculate the volume of our 'cylinder':

1963.49m^2 * 3.53m = 6941.99 m^3

We will assume that there are 2.7 tonnes of ore per m^3. This means that our cylinder has a mass of 6941.99 * 2.7 = 18743 tonnes.

We had 5 g/t, so 18743 * 5 = 93,716g.

There is about 31 g/Oz, so the number of ounces added by the example drill hole would be 93,716/31 = 3,023 Oz.