We have better and arrange to work with a double polar system,
- one ($R,\alpha$) on the $x-y$ plane describing the spiral,
- and one($r,\phi$) on the $n-z$ plane describing the helicoid
where $n$ is the (unit vector) normal to the spiral.
Both the spiral and the helicoid are supposedly linear.
Let's put
$$
\left\{ \matrix{
\alpha = 2\,\pi \,t \hfill \cr
\varphi = 2\,\pi \,m\,t \hfill \cr
{\bf k} = \left( {0,0,1} \right) \hfill \cr
{\bf p}(t) = \left( {x(t),\;y(t),\;z(t)} \right) \hfill \cr} \right.
$$
where the meaning of the various parameters is clear.
We have better and arrange to work with a double polar system,
- one ($R,\alpha$) on the $x-y$ plane describing the spiral,
- and one($r,\phi$) on the $n-z$ plane describing the helicoid
where $n$ is the (unit vector) normal to the spiral.
Both the spiral and the helicoid are supposedly linear.
Let's put $$ \left\{ \matrix{ \alpha = 2\,\pi \,t \hfill \cr \varphi = 2\,\pi \,m\,t \hfill \cr {\bf k} = \left( {0,0,1} \right) \hfill \cr {\bf p}(t) = \left( {x(t),\;y(t),\;z(t)} \right) \hfill \cr} \right. $$ where the meaning of the various parameters is clear.
Then we shall have $$ \left\{ \matrix{ {\bf R}(t) = \left( {R_{\,0} + c\,t} \right)\left( {\cos \alpha ,\;\sin \alpha ,\;0} \right) \hfill \cr {d \over {dt}}{\bf R}(t) = 2\,\pi \left( {R_{\,0} + c\,t} \right)\left( { - \sin \alpha ,\;\cos \alpha ,0} \right) + c\left( {\cos \alpha ,\;\sin \alpha ,\;0} \right) \hfill \cr {\bf N}(t) = {d \over {dt}}{\bf R}(t) \times {\bf k} = \left( {2\,\pi \left( {R_{\,0} + c\,t} \right)\cos \alpha + c\sin \alpha ,\;2\,\pi \left( {R_{\,0} + c\,t} \right)\sin \alpha - c\cos \alpha ,0} \right) \hfill \cr {\bf n}(t) = {\bf N}(t)/\left| {{\bf N}(t)} \right| \hfill \cr} \right. $$
And finally $$ {\bf p}(t) = {\bf R}(t) + r\cos \varphi (t)\;{\bf n}(t) + r\sin \varphi (t)\;{\bf k} $$