School of Hard Rocks

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Some Rock Tumbling Math

Abstract

 

There is always a noticeable reduction in the volume of a batch of rocks when smoothing and polishing those rocks in a rotary tumbler. This paper explains why one should expect to grind away half of a typical batch of rocks.

 
Thumler Model B
Thumler Model A-R12
Thumler Model A-R2
 
Introduction

I enjoy tumbling rocks.  It is a task that takes a little bit of creativity, a fair amount of careful attention to detail and a whole lot of patience.  When the final tumbling run is completed and the stones are all silky smooth and shiny the weeks and months of process time are quickly forgotten.
 
In two years of tumbling rocks, Iíve noticed that the batch that comes out is much smaller than it was going in.  You can read many pages on the internet that advise adding beads to replace lost volume or that tell you to combine batches to keep the machines loaded at 1/2 to 5/8ths
 of the barrel.  So Iíve been doing that, but Iíve been experiencing reductions to half or less of the original volume.  That is pretty dramatic. 

Is it unusual to see half or more of the rocks being ground away?  You would think this large a reduction would be well known.  I do like smooth rocks, but have I been grinding them too long?  Now I pull out individual stones when they are smooth and hold them for the next step in the process.  The volume still drops to half.  So maybe it is not an overly aggressive process.

Is it something about the rocks I tumble? I usually tumble hammer broken rocks, as do most other rock tumblers.  I donít see a large fraction of the rocks breaking while tumbling, and have plenty of opportunities to remove the few that do fracture or scab off pieces, so this volume reduction is not driven by failure of the rock material.

The explanation turns out to be quite simple.  It is all in the geometry of the rocks.

Discussion
Hammer Broken Rock
Tumble Polished Stones
 

Here are before and after pictures from two batches of rock I tumbled.  My first assertion is that the hammer broken rock (on the left) is made up of pieces that closely resemble polyhedrons (solid figures with plane surfaces).  My second assertion is that the tumbled stones (on the right) closely resemble ellipsoids.  These are both approximations but they move us quickly to manageable arithmetic.


To keep things easy, letís assume that the broken stones can be modeled as some combination of the regular polyhedrons shown below:


Tetrahedron

Hexahedron

Octahedron

Regular Tetrahedron
Regular Hexahedron
Regular Octahedron
 

We will further assume that the polished stones can be modeled as general ellipsoids:


General Ellipsoid

The volume of a general ellipsoid is V = 4/3 π abc. If all axes are equal, the ellipsoid is a sphere and the volume is the familiar V = 4/3 π r3.  We will use a sphere in our consideration, again for easy math.


One final assumption is that the first two assumptions introduce compensating (actually correlated) errors.  The hammer broker rocks extend beyond the regular polyhedrons but those regular polyhedrons can be stretched to fit the observed rock surfaces.  Likewise, the sphere does not fill the finished ellipsoidal rocks but can be stretched to reasonably match the smoothed surface.  The defect introduced in each of the models used trends in the same direction from the observed state and appears to trend by similar amounts from the observed.


 

Development

A sphere is said to be inscribed in a regular polyhedron if there are points on the sphereís surface that are tangent to every surface of the polyhedron.  The radius  of the inscribed sphere can be determined by using the following ratio of the polyhedronís volume to its total surface area:


Eqn r = 3V/A

Letting a = the length of an edge of the polyhedron being considered, the volumes and total areas of the polyhedrons are calculated using the relations in the following table.

Our interest is the fraction of each polyhedronís volume contained in the inscribed sphere.  That fraction is shown in the third column. 


Solid Form

Equation

Fraction in Sphere

     

V tet = 0.1197 a


A tet = 1.7321 a2



Fraction of tetrahedron in sphere
Tetrahedron

   

V hex = 1.0000 a3 

A hex = 6.0000 a2

Fraction hexahedron in sphere
Hexahedron

   

V oct = 0.4714 a3

A oct = 3.4641 a2

Fraction of octahedron in sphere
Octahedron

   
Regular Dodedahedron

V dod = 7.6631 a3  

A dod = 20.6457 a2

Fraction of dodecahedron in sphere
Dodecahedron

   
Regular Icosahedron

V icos = 2.1817 a3 

A icos = 8.6603 a2

Fraction icosohedron in sphere
Icosahedron

   
Volume of general ellipsoid
Volume of sphere
General Ellipsoid

   
     

Following our assumption that most hammer broken rocks are similar to some combination of tetrahedrons, hexahedrons or octahedrons, we see that the average fraction of the rock contained in the spheres inscribed in these three solids is 0.476.  This is just below half, so we have a model that is consistent with our observations.

Can you do better than this, i.e. keep a larger fraction of rock?  Surely, if you settle for less smoothing.  Or if you start with nodules or river pebbles.  If you could find stones shaped like dodecahedrons or icosahedrons that would reduce the fraction lost but I have only seen rocks with those shapes when someone cut them to be that shape and the waste would dwarf any case considered here.

Can you end up with a smaller fraction than this?  You almost certainly will.  The fractions developed in the table will only be obtained if the grinding removes just the material that is outside our sphere.  The point of tangency on each face must remain for these fractions to hold.  Most likely, in tumble grinding the stones there will be some amount of material removed from those spots on each face, so the stone could be even smaller and the volume even less.  To try to quantify that pushes these simple models too far, so we will stop here.


Bottom line

If you are tumbling hammer broken rock you should always start with twice as much rock as you would like to have when you are finished – or more.

 

Addendum:

So you are not sure about these formulas relating the radius of an inscribed sphere to the volume and area of a polyhedron.  Let me use a cube Ė a hexahedron Ė as an example.


Sphere inscribed in a cube

We will use a cube that is 2 units long on each edge, that is a = 2 in the formulas of the table.  Recall that:


Now the sphere in the figure has a radius of 1 and diameter of 2 so it is clearly tangent to each of the faces of the cube.  The volume of the cube is 2 X 2 X 2 or 8 and the surface area is 6 X (2 X 2) or 24.  3V/A is then 3 X 8 / 24 or 1, which is indeed the radius of the inscribed sphere.

I leave the other cases as exercises for the reader.