Bond BWi P100-P80 relationship
adapted from https://www.sagmilling.com/articles/28/view/PublicTestworkDatabase20201225.ods?s=1

Bond BWi P100-P80 relationship

TLDR version:

Two models work, take your pick from

  • P80, μm = 0.92 × (P100, μm)^0.96
  • P80, μm = 0.69 × (P100, μm) + 6.96

Data and discussion

I'm preparing the next release of the Public Grindability Database and wanted to share some interim results. The first one is useful when trying to interpret NI43-101 reports where an author has provided a ball mill work index text closing screen size, but not provided the P80. Digging into the interim database, we get a large number of potential fitted curves with similar R2 values.

Here are the first two candidates, a power model and a linear model:

P80, μm = 0.92 × (P100, μm)^0.96;  R2 = 0.96
P80, μm = 0.69×P100, μm + 6.96;   R2 = 0.95

(the grey points are published data that were excluded since the P80 was coarser than the P100).

These two models give almost identical predictions, so you can choose either and expect a reasonable prediction for the normal size ranges. There will be a problem with the linear model at very small P100 sizes (i.e. the Limit as P100 → 0 μm of P80 = 6.96 μm, which makes no sense). This problem won't manifest in the size range industry normally works with, so we can ignore this problematic boundary condition.

An alternative model is to use a linear model pinned at the origin, which looks like this

P80, μm = 0.73 × P100,μm

The R2 shown here is bogus because the pinned 0,0 point causes too much leverage. This model seems to work well in the middle size ranges, but can over-estimate the P80 when dealing with the coarsest P100 screens. I don't think this model is quite as robust at the previous candidates.

Conclusions

The power model and linear model both will give reasonable P80 estimates given the P100 of a Bond ball mill work index test product. The power model might be marginally more robust, but not in any meaningful way.

  • P80, μm = 0.92 × (P100, μm)^0.96
  • P80, μm = 0.69 × (P100, μm) + 6.96




Eduardo Osvaldo Rojas Ferreira

Vendedor Técnico para La Minería en AMTEX Chile

3 年

Sorry for the querry, but, Why can the P80 be bigger than the P100? How can such a thing exist?

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Marcos de Paiva Bueno

Founder & CEO | PhD in Mineral Processing | Process Optimization | Strategic Leadership

3 年

Alex Doll thanks for sharing another useful relationship. Do you see any trend/relationship to estimate a BWI at different closing sizes? e.g. if you have part of the testwork done at 150micron and a few other samples at 106, how to scale it to 150 so that all samples could be compared on the same basis

Alex Doll

Consultant at SAGMILLING.COM

3 年

I'll add that the average F80 of the database is about 2340 μm, so you can also use that as go-by if the NI43-101 didn't include it. The Bond equation isn't too sensitive to this value; whereas, it is quite sensitive to the P80.

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