<s>[INST] Relationship between constants so that the center of curvature of the helix is contained in the cylinder I am studying the topic of torsion and curvature and I ran into the following problem:
Consider the helix $\alpha(t)=(a\sin t,a\cos t,bt)$. Get the relationship between constants
so that the center of curvature of the helix is ​​contained in the cylinder
where the helix is ​​contained as well.
First I decided to parameterize the curve with the arc length to get the curvature:
$s(t)=\int_{0}^{t}|\alpha'(t)|dt=\int_{0}^t\sqrt{a^2(\sin^2t+\cos^2t)+b^2}dt=\int_0^t\sqrt{a^2+b^2}dt=\sqrt{a^2+b^2}t$
Hence, $t(s)=\cfrac{s}{\sqrt{a^2+b^2}}$, so
$\beta(s)=\alpha\left(t(s)\right)=\left(a\cos\cfrac{s}{\sqrt{a^2+b^2}},a\sin\cfrac{s}{\sqrt{a^2+b^2}},\cfrac{bs}{\sqrt{a^2+b^2}}\right)$
is a reparametrization of the curve by arc length. Differentiating, we see
that:
$\alpha'(s)=\left(\cfrac{-a\sin\cfrac{s}{\sqrt{a^2+b^2}}}{\sqrt{a^2+b^2}},\cfrac{a\cos\cfrac{s}{\sqrt{a^2+b^2}}}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}\right)$
Therefore,
$\alpha''(s)=\left(\cfrac{-a\cos\cfrac{s}{\sqrt{a^2+b^2}}}{a^2+b^2},\cfrac{-a\sin\cfrac{s}{\sqrt{a^2+b^2}}}{a^2+b^2},0\right)$
and so the curvature of the helix is given by:
$k(s)=|\alpha''(s)|=\sqrt{\cfrac{a^2\cos^2\left(\cfrac{s}{\sqrt{a^2+b^2}}\right)}{(a^2+b^2)^2}+\cfrac{a^2\sin^2\left(\cfrac{s}{\sqrt{a^2+b^2}}\right)}{(a^2+b^2)^2}}=\sqrt{\cfrac{a^2}{a^2+b^2}}=\cfrac{|a|}{a^2+b^2}$
So I checked that it has a curvature that depends on a and b and that it is always constant.
Does this help me to get the relationship between constants
so that the center of curvature of the propeller is contained in the cylinder
where in turn is said helix contained?
I appreciate your time and help.
 [/INST] Since the principal normal is horizontal, you will need $1/\kappa<2|a|$, so that seems to require $|b|<|a|$.
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