Building helixes with different torsion values

Question

I have following helix parametrization: $$ x(t)= r\cos(t) $$ $$ y(t)= r\sin(t) $$ $$ z(t)= h\cdot t $$ For example I have $ r=1 $ and $ h = 0.05 $

Then my torsion T will be: $$ T = \frac{h}{r^2+h^2} = \frac{0.05}{1+0.05^2} \approx 0.0476 $$ My question is how can I build helixes with $ 0.1T, 0.2T, -0.1T, -0.1T $

I dont get how changing torsion analytically change position of points of the helix.

Answer 1

Assuming $r$ is constant, you can find the $h$ corresponding to a given $T$ by solving for $h$ in the equation $$ T = \frac{h}{r^2 + h^2} $$ Rearranging, we get $Th^2 - h + Tr^2 =0$. This has real roots only when $r \le 1/(2T)$.

One possible solution is $r=h=1/(2T)$.

Answer 2

From $T=h/(r^2+h^2)$ we get $$r^2+\left(h-\frac{1}{2T}\right)^2=\frac{1}{4T^2},$$ that is, for given $T$ all possible values for $r$ and $h$ lie on a circle centered in $(0,1/2T)$ with radius $1/2T$.