# Finding mean and std for particles generated by crusher

I am trying to model particles generated by a mechanical crusher. We put a big chunk into crusher and we get small particles. These particles are categorised into 3 categories Size <2 mm, 2-50 mm and >50 mm. Assume these particles are spherical in shape.

Target is to find the particle size distribution such that.

Size <2 mm particles weight corresponds to 1.5% of input weight

Size 2-50 mm particles weight corresponds to 89% of input weight

Size >50 mm particles weight corresponds to 9.5% of input weight

I have assumed normal distribution for particle sizes and written a small code in python for varying mean and std, generate 1000 particles, categorise them and calculate their weights and check if the condition is satisfied or not.

I could not find a solution.

I did the same exercise for skewed normal distribution but no luck.

Note that normal distribution generates negative number, I wrote an if condition when negative number is observed take a number between 0.01 mm to 2 mm following uniform distribution

Is my approach correct or is their any simpler approach.

• There are infinitely many solutions -- but to be realistic, they should assume a distributional shape that has zero (or at least vanishlingly small) chance of being negative. A standard choice for particle size distributions is the family of lognormal distributions.
– whuber
Mar 2 at 15:58
• "I have assumed normal distribution for particle sizes" -- this doesn't seem likely to be realistic and I expect it will yield unsatisfactory results. Common choices for particle size distributions include lognormal and Weibull (sometimes called Rosin-Rammler in particle-size analysis), though a few other possibilities do occur. The Weibull seems a bit more common for analyses involving grinding/milling/ crushing of material but I expect the lognormal will likely do quite well. Mar 2 at 16:09
• @Glen FWIW, a nice property of the lognormal family is that when any of the diameters, volumes, or surface areas of the particles have a lognormal distribution, so do the other two properties.
– whuber
Mar 3 at 16:30
• Indeed. This handy property is related to the part of my recent answer discussing lognormality of $Y^p$ here: stats.stackexchange.com/a/641246/805 ... (nevertheless Weibull seems somwwhat more common in analyses with crushed media). However, I think this property is also true of the Weibull Mar 3 at 21:10
• Specifically, since Weibulls are powers of exponentials, they're also powers of each other. Indeed, that neat property should hold for the distribution of $Y= \exp(X)$ with $X$ being from a location-scale family (normal for lognormal, negative of a Gumbel for Weibull, etc) Mar 3 at 22:37

This is what I have done:

For lognormal distribution, $$\ln(x)\sim N(\mu,\sigma)$$

$$\int_0^2 x^3P(x)=0.015E[x^3]$$ After some math $$\operatorname{erf}\left(\frac{3\sigma^2+\mu-\ln 2}{\sqrt{2}\sigma}\right)=0.97$$

$$\int_2^{50} x^3P(x)=0.89E[x^3]$$ After some math $$\operatorname{erf}\left(\frac{3\sigma^2+\mu-\ln 2}{\sqrt{2}\sigma}\right)-\operatorname{erf}\left(\frac{3\sigma^2+\mu-ln50}{\sqrt{2}\sigma}\right) =1.78$$

For Weibull distribution $$f(x)=\frac{k}{\lambda}(\frac{x}{\lambda})^{k-1}\exp(-(\frac{x}{\lambda})^{k}),x\geqslant 0$$

Applying same conditions I got

$$1-\frac{\Gamma(1+\frac{3}{k},\frac{2^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})}=0.015$$

$$\frac{\Gamma(1+\frac{3}{k},\frac{2^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})}-\frac{\Gamma(1+\frac{3}{k},\frac{50^k}{\lambda^k})}{\Gamma(1+\frac{3}{k})} =0.89$$

Then calculated the variables.