Re: Quotes with news release
posted on
Oct 02, 2007 10:08PM
NI 43-101 Update (September 2012): 11.1 Mt @ 1.68% Ni, 0.87% Cu, 0.89 gpt Pt and 3.09 gpt Pd and 0.18 gpt Au (Proven & Probable Reserves) / 8.9 Mt @ 1.10% Ni, 1.14% Cu, 1.16 gpt Pt and 3.49 gpt Pd and 0.30 gpt Au (Inferred Resource)
Based on some calculations I did the other day I ended up having to assume there was some drift in the drilling inorder to feel confortable with the results I got. If the drill head did follow a straight line path then I will have to wonder a little.
Got any calc's on these drillhole data. I wish I had access to some software to do the work prescicely but...
I will post a follow up to what I did on Sunday. Read through and tell me what you think.
Old Joe
On Sunday night I calculated #7 to show ~36.1m of mineralization in depth from ~51.6m almost directly below #5 (thus from ~51.6m to ~87.7m below #5's set up at ground level) and the horizontal distance NE to the end of mineralization from the ~87.7m point mentioned above of ~50m. The diagonal length of the mineralization in the core was stated as 51m (124m - 73m = 51m).
Thus, on a Rt angle triangle where 36.1m is the vertical and ~50m the horizontal and the 51m the hypothenuse, gives a Plane for drill hole #5-7.
Note: this an approximation as the deposit does appear to have an irregular boundry shape (given the holes that missed).
The same geometry can be done to form a plane for drill holes #1-2.
With this and using suitable reduction factoring to allow for the irregular shape and thus unmineralized rock, a reasonable though rough approximation of mineralization between these two planes can be made.
What would be a suitable reduction factor (to be on the conservative side is up to the person doing the calc's)?
By connecting the verticies of each triangle on each plane would be reasonable at this point in the game. Thus, without getting into the calculation of the distance between planes #1-2 and #5-7 (easily done if I had just completed the 2000 level university course in linear math -did do 30 years ago) for each line between verticies and thereby calculating a volume. I instead sketched a layout map from the given data and measured the space between to get the ~70m at the Rt angle corners and ~90m to 120m at the distal lower verticies (given the eventual shape this can be approximated at 90m). The top verticies distance I would approximate to be the same as the lower verticies (~70m). There you have it, what is the volume?
To sum: Plane #1-2 = ~50m X ~50m X ~75m and Plane #5-7 = ~36m X ~50m X ~50m. The distances between the two planes form corresponding vertex to vertex are ~70m, 70m, and 90m.
Doing a calculation on rounded measures (of a wedge shaped block) gives me a very approximate 86 000m cubed. ((~80m x 50m X ~43m) / 2) = 86 000m cubed. The 80m from difference between 70m and 90m - the top line and distal line connecting these verticies, and the 43m from difference between 36m and 50m - the depth along the #5 hole.
Plug in the Grades/rock densities info as needed and presto - What do you get?
This would be a good conservative estimate, given the info (data) to date. And I hope a little on the conservative side to be sure.
Old Joe