# Proof of the formula for the angle of repose as a function of θ and φ

Four smooth identical spheres pack together in a regular tetrahedron. Suppose that the centres of three spheres are at A, B and C on a flat surface in the above diagram for a general φ' = 60° - φ , with the fourth sphere at D above the plane on which A, B and C lie. It is expedient to choose C to be at the origin (0, 0, 0) in an orthonormal \(\hat x \) - \(\hat y \) - \(\hat z \) space. Friction can be assumed to hold these particles together in a stable configuration.

When φ' = 0°, CB is parallel to the \(\hat x \)-axis, whereas when φ' = 60°, AB is parallel to the \(\hat x \)-axis.

If a rotation is now made about the \(\hat x \) axis by an angle θ, then A, B and D rotate to new positions in \(\hat x \) - \(\hat y \) - \(\hat z \) space. Projected onto the \(\hat x \) - \(\hat y \) plane, these positions will be A_{p}, B_{p} and D_{p}, respectively. The condition for stability of the tetrahedral arrangement (assumed still to be made stable by a sufficient level of friction) is that Dp must lie in the new projected triangle whose corners are A_{p}, B_{p} and C.

For simplicity, we can choose the sides of the regular tetrahedron to be of unit length.

B and A are initially at (cos φ', sin φ', 0) and (cos (φ' + 60°), sin (φ' + 60°), 0) respectively, while D is initially at

\[ \left( {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\sin (\varphi ' + 30^\circ ),\frac{{\sqrt 2 }}{{\sqrt 3 }}} \right){\rm{ }} \]

A rotation R of θ about the \(\hat x \)-axis rotates vectors R to new positions R' in \(\hat x \) - \(\hat y \) - \(\hat z \) space so that R' = R_{r} where

\[ R = \left( {\begin{array}{*{20}{c}} 1&0&0\\ 0&{\cos \theta }&{ - \sin \theta }\\ 0&{\sin \theta }&{\cos \theta } \end{array}} \right){\rm{ }} \]

Hence, for example, the ‘old’ vector [010] rotates to become the ‘new’ vector [0 , cos θ, sin θ]. Similarly, the vector CB, CA and CD rotate to the new positions

\[ \left( {\cos \varphi ',\cos \theta \sin \varphi ',\sin \theta \sin \varphi '} \right), \]

\[ \left( {\cos {\rm{ }}(\varphi ' + 60^\circ ),\cos \theta \sin {\rm{ }}(\varphi ' + 60^\circ ),\sin \theta \sin {\rm{ }}(\varphi ' + 60^\circ )} \right), \]

and

\[ \left( {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\cos \theta \sin (\varphi ' + 30^\circ ) - \frac{{\sqrt 2 }}{{\sqrt 3 }}\sin \theta ,\frac{1}{{\sqrt 3 }}\sin \theta \sin (\varphi ' + 30^\circ ) + \frac{{\sqrt 2 }}{{\sqrt 3 }}\cos \theta } \right){\rm{ }} \]

respectively.

The projection of D onto the \(\hat x \) - \(\hat y \) plane, D_{p}, will lie in the triangle, CB_{p}A_{p} until the vector CD_{p} is parallel to the vector CB_{p}, i.e., the limiting condition is that

\[ \left[ {\cos \varphi ',\cos \theta \sin \varphi ',0} \right] \]

is parallel to

\[ \left[ {\frac{1}{{\sqrt 3 }}\cos (\varphi ' + 30^\circ ),\frac{1}{{\sqrt 3 }}\cos \theta \sin (\varphi ' + 30^\circ ) - \frac{{\sqrt 2 }}{{\sqrt 3 }}\sin \theta ,0} \right]{\rm{ }} \]

i.e., a critical θ, θc, is defined by the equality

\[ \frac{{\cos \varphi '}}{{\cos {\theta _{\rm{c}}}\sin \varphi '}} = \frac{{\cos (\varphi ' + 30^\circ )}}{{\cos {\theta _{\rm{c}}}\sin (\varphi ' + 30^\circ ) - \sqrt 2 \sin {\theta _{\rm{c}}}}} \]

Now, φ' = 60° - φ. Substituting for φ' in this equation, the condition becomes

\[ \frac{{\cos (60^\circ - \varphi )}}{{\cos {\theta _{\rm{c}}}\sin (60^\circ - \varphi )}} = \frac{{\sin \varphi }}{{\cos {\theta _{\rm{c}}}\cos \varphi - \sqrt 2 \sin {\theta _{\rm{c}}}}} \]

i.e., the condition

\[ 2\sqrt 2 \sin {\theta _{\rm{c}}}\cos (60^\circ - \varphi ) = \cos {\theta _{\rm{c}}} \]

using some elementary trigonometry, or better still the condition

\[ {\theta _{\rm{c}}} = \arctan \frac{1}{{2\sqrt 2 \cos (60^\circ - \varphi )}}{\rm{ }} \]