# Show that if C is a spherical helix then the projection of the helix on a plane orthogonal to its axis is an arc of an epicycloid.

This is what I have solved so far.

$\frac{d(\sigma{\rho\prime})}{ds}+(\rho/\sigma)=0$(Condition for a space curve lying on sphere) s$\rightarrow$arc length of the helix

For helix, $\rho$/$\sigma$=$\eta$(constant)

$\sigma\rho\prime=-\eta s+c$(integration constant)

$\rho\rho\prime/\eta=-\eta s+c$

$-dk/k^3=(-\eta^2 s+c)ds$

$\kappa^-2=\eta^2 s^2-pcs-2\beta$(integration constant)

$s_1\rightarrow$arc length of $C_1$

$\frac{ds_1}{ds}=sin\alpha$, $\alpha\rightarrow$angle between axis of the helix and its tangent

$s_1=(sin\alpha) s+\gamma$(integration constant)$~~Eq.(1)$

$\kappa=\kappa_1 sin^2\alpha, ~\kappa_1\rightarrow$curvature of the projected plane curve

$\kappa_1=\frac{cosec^2\alpha}{\sqrt{\eta^2 s^2-pcs-2\beta}}$

Use Eq.(1)

$\kappa_1=\frac{cosec^2\alpha}{\sqrt{\eta^2 cosec^2\alpha(s_1-\gamma)^2-pc(cosec\alpha)(s_1-\gamma)-2\beta}}$

I cannot proceed any further.