θ2−α2s<0o
and
θ2+α2p>90o ??
And when μ2>μ1
θ2−α2s>θ1−α
α2s<θ2−θ1+α
and
θ2+α2p>θ1+α
α2p>α−(θ2−θ1)
If we define,
Δθ=θ2−θ1
α2s<α+Δθ
and
α2p>α−Δθ
and the split occurs on the two sides about α through the foot of the surface normal bounded by ±Δθ
When μ2<μ1
θ2−α2s<θ1−α
α2s>α+Δθ
and
θ2+α2p<θ1+α
α2p<α−Δθ