RE:RE:Area 52 programI would not be overly concerned in that BAR rarely if ever has specified VG in holes. In fact I have asked JF about this already at BAR and he would not comment on it since it never appeared in the press release. VG is an indicator but may not mean much about the actually gold content of the hole. Here is a method I ran across to evaluate drill holes which I posted at several sites but not here yet. So for thise that may be interested here it is using a drill hole from Wallbridge as an example and a link to the actual article.
For those that may be interested, here is a calculation you can use to estimate the oz’s of gold a drill hole might produce. I will take an example using WM hole FA-17-17 which was reported with VG when announced. This is not a good hole but it’s simple to use. I have a spreadsheet set up which makes the process much faster. Holes with long segments of gold will fare much better than those with a few short segments. First we need to understand a few terms – when you see a drill result and it states the length of the segment gold was found in you need to know if that length is the “true width” are just a length estimate. If it doesn’t say at the bottom of the PR “true widths” you must assume it’s a length estimate. WM has be producing “true width” between 66%-78% of estimated length so we will assume that the new holes run in the same range for this example and use the high 78% true width.
The process is fairly simple in that you take each segment gold was found in and perform the same set of calculations then add the oz’s of gold up. So let’s look at hole FA-17-17. Here are the segments.
.61/m – 2.81 g/t, 4.45-21.95, 5.22-6.13, 3.06-346.55
Another factor that is used is rock density which usually runs between a value of 10-15, so let’s use a midpoint value of 12.5. Usually, the mining company will provide you with a typical density value if you ask. So if we take the 1 segment .61-2.81 and perform the calculations you will get the idea.
Convert the sample length to feet then cube that value- .61x3.282=2.0 cubed = 8.02 Tons
Take the g/t value and multiple it by .032 to convert it to oz/t 2.81x.032 = .09 oz/t
Multiple the oz/t value by the true width in feet .09 x 2.0 = 1.2 and cube this value = 1.733
Divide this value by the ore body density 12.5 – 1.733/12.5 = .139 (final ore body)
Now multiply oz/t (.09) by the above value to get final oz’s of gold from this hole .09x.139 = .0125 Oz’s of gold for this hole segment if only the segment were mined. Now repeat for all the other segments and add them up.
I use this as a relative figure since it does not reflect the overall ore body but I can compare it against other drill holes to see which are stronger indicators.
Now before Mark has a chance to comment, don’t bother – here a link to the article you can argue with the guy the wrote the article.
Ralph Kettell is the President and founder of Nevada Fluorspar, Inc. He is also a Director of Piedmont Mining Company-OTCBB, SNS Silver Corp.-TSX/V, and Gold Summit Corp.-TSX/V.
Link =
https://news.goldseek.com/GoldSeek/1242751631.php