differential-geometrygeometry
i can understand well the graphical interpretation of the curvatura value of a curve (in R^3)
but… Could you help me to understand the graphical sense of the torsion of a curve?
I know that if torsion >0 the curve go trough osculating plane from below upwards.
if torsion <0 the curve go trough osculating plane from above downwards.
But…. how can i interprete a value os 2, 4, or -1.5 or -5 ?
Thank you.
The constant $k$ need not be $0$; that would be the case where $\alpha$ lies in a plane through the origin. You have $k=\alpha(0)\cdot v_0$, so for all $t$, $(\alpha(t)-\alpha(0))\cdot v_0=0$. This means that $\alpha(t)-\alpha(0)$ lies in the plane through the origin perpendicular to $v_0$, so $\alpha(t)$ lies in the plane through $\alpha(0)$ perpendicular to $v_0$. (If $0$ is not in the domain, then $0$ could be replaced with any point in the domain of $\alpha$.)
I had also came across the notion of "radius of torsion" in the context of a lecture on Frenet-Serret system of equations, and had great difficulties in trying to conceptualize it. The notion of "radius of torsion" is a little more abstract then the "radius of curvature", and cannot be given a direct geometric interpretation as in the case of the letter. One reason for this difference is that torsion cannot exist without curvature; if one tries to define a space curve with non-zero torsion but zero curvature, then the torsion degenerates into curvature.
However, great background and guidance is given to this subject simply by reading a little about the history of differential geometry of space curves. Much earlier than the publication of Frenet-Serret formulas, Alexis Clariault reffered to general space curves as "curves of double curvature". This simple renaming gives some clue about conceptualization of torsion and curvature of general space curves.
Interpretation:
Assume $\gamma$ is a space curve traced by a point particle. Let $TN$ be it's local osculating plane at some point $X$ (the plane spanned by the tangent vector $T$ and the normal vector $N$), and $BT$ also be a local plane perpendicular to it (it's the plane spanned by the tangent vector $T$ and the binormal vector $B$. Now, let us project the curve $\gamma$ into each of the planes $TN$ and $BT$. One can do this simply by illuminating on the particle from the (local) $B$ and $N$ directions, and then looking at the curves which the particle's shadow traces in each plane.
Obviously, the local curvature of the curve traced by the shadow in the $TN$ plane is $\kappa$. In the same way, the local curvature of of the curve traced in the $TB$ plane is the torsion $\tau$; this gives a straightforward interpretation of the "radius of torsion" - this is radius of the circle which has second order contact with the projection of $\gamma$ into the local $TB$ plane.
A partial illustation of this interpretation can be seen in this picture; it shows the osculating ($TN$) plane and the $BT$ plane.
Caution:
I have not proved yet this claim with full rigourosity; but but nevertheless i wrote it down to make some contribution to the resolution of this question. I think that proving this claim will be my final attempt before i give up the attempts to gain intuition about the "radius of torsion" (and i hope my claim is correct!).
Best Answer
Here's how my differential geometry professor explained it to me.
Now for some examples.