In my textbook (by do Carmo) and both in wikipedia.

There are descriptions of what a torsion is, and both of them says it is a measure of "how fast a curve twists out of the plane of curvature"

I am aware of the definition of torsion which is the magnitude of the derivative of the binomial vector, but I fail to see how this describes "how fast the curve is twisting out of the plane" or "pulling out of the plane".

If we are talking about how fast the curve is traveling outside of its osculating plane, then this makes absolutely no sense to me at all, since the tangent vector is on the osculating plane.

  • $\begingroup$ Not binomial, binormal. $\endgroup$ – Yves Daoust Jan 22 '18 at 8:47
  • $\begingroup$ "the tangent vector is on the osculating plane": this is true at the current point and stops being true at other points. $\endgroup$ – Yves Daoust Jan 22 '18 at 9:00

When a curve is planar, all osculating planes are equal.

When it is non planar, i.e. has some torsion, the osculating planes stop staying parallel when you move along the curve, and this change of direction is reflected by the binormal.

"Infinitesimal" insight:

Imagine the curve discretized with a fixed step.

Two successive points define a line, which is the tangent.

Three successive points define a plane, which is the osculating plane.

A fourth point can deviate from the plane and show the torsion. In other words, two triples of successive points will define two distinct planes and the angle between them corresponds to the torsion.

  • $\begingroup$ I tend to use infinitesimal to understand these concepts. So in terms of infinitesimal, can I think of a curve as many infinitesimal arc segment, and the osculating plane is the plane that is defined by these segments (so each osculating plane is the plane that contains an infinitesimal segment of a curve)? $\endgroup$ – Ecotistician Jan 22 '18 at 23:23
  • $\begingroup$ Actually my friend just gave a really good analogy of considering the curve as the trail of an airplane. Then I realize if we restricted the motion of the airplane to only 3 types of motion, foward translation, yaw, and roll. Then Roll is the torsion, yaw is the curvature. $\endgroup$ – Ecotistician Jan 22 '18 at 23:54
  • $\begingroup$ Another way I'm trying to picture this is, image the curves are wires with some dimensions and if we laid it out straight and slit a slot along its length, we now put the curve/wire back to where it was in space in such a way that the slot aligns with the tangent and place a toothpick sitting upright on the slot, One can image the toothpick/binormal only rotate(the only way it can change in a curve parametrized by arc length) about the tangent vector only if the wire starts to twists. And the torsion measures the magnitude of this rotation (how much the curve is bending its ocuslating plane). $\endgroup$ – Ecotistician Jan 22 '18 at 23:58
  • $\begingroup$ actually the slot will need to be concide with the tangent vector $\endgroup$ – Ecotistician Jan 23 '18 at 5:22
  • $\begingroup$ @Ecotistician: you should focus on the idea that torsion is what makes the curve non planar. $\endgroup$ – Yves Daoust Jan 23 '18 at 7:42

I'm confused by the relevance of the very last thing you said. The osculating plane is the plane spanned by the tangent vector and the normal vector. The binormal vector is the one pointing out of the plane, so it's derivative represents how fast you twist out of the (osculating) plane.

  • $\begingroup$ I guess I don't understand the meaning of the phrase "how fast a curve twists out". $\endgroup$ – Ecotistician Jan 22 '18 at 7:56
  • $\begingroup$ Ignore the word "twists". The plane we are considering is the osculating plane, the one formed by the tangent vector and normal vector. The vector pointing outside this plane is the binormal one. So how much we are pointing out of the plane is given by the binormal vector. How quickly it changes, i.e. its derivative, is how quickly we are leaving the plane. The word "twists" is used because in many examples, when the curve leaves the plane, it twists out of it. Think of the Helix, for example. $\endgroup$ – mathworker21 Jan 22 '18 at 8:00
  • $\begingroup$ when you say "how much (we) are pointing out of the plane" and how much "we" are leaving the plane what is (we) referring to in this context? $\endgroup$ – Ecotistician Jan 22 '18 at 8:03
  • $\begingroup$ Sorry. we = $\alpha(s)$.. $\endgroup$ – mathworker21 Jan 22 '18 at 8:46

The tangent vector is only the first-order approximation to the "motion" of the curve - if it told you everything about the curve then all curves would be straight lines! Thus, to understand what is being talked about here, you need to study the higher derivatives/Taylor expansions of the curve. On Wikipedia you can find the third-order expansion; the parts that are relevant to us are just the leading-order terms in the three principal directions: for a curve with $\gamma(0) = 0$ we have

  • $\gamma(s) \cdot T = s + o(s),$
  • $\gamma(s) \cdot N = \frac 12 \kappa(0)s^2 + o(s^2),$ and
  • $\gamma(s) \cdot B = \frac 16 \kappa(0)\tau(0)s^3+o(s^3).$

When we go to second order we get a term in the normal direction with magnitude determined by the curvature, so curvature tells you how the curve is "bending" out of its osculating line. Likewise, going to third order adds a term in the binormal direction determined by the torsion (and curvature), so torsion tells us how the curve "twists" out of its osculating plane.


To a curved curve $\gamma$ in $3$-space an orthonormal frame (in German: Dreibein) $F$ moving along the curve is associated. The first vector of this Frenet frame $F$ is the unit tangent vector ${\bf t}$. Since the curve is assumed curved this ${\bf t}$ is not constant, hence $\dot{\bf t}$ is a nonzero vector orthogonal to ${\bf t}$. The unit vector ${\bf n}$ in the direction of $\dot {\bf t}$ is the second vector of $F$. Together the two vectors ${\bf t}$ and ${\bf n}$ span the osculating plane of $\gamma$ at each point, and we in fact have $\dot{\bf t}=\kappa\,{\bf n}$, where $\kappa>0$ is the curvature.

The third vector ${\bf b}$ of $F$ is simply defined by ${\bf b}:={\bf t}\times{\bf n}$. This binormal ${\bf b}$ is automatically a unit vector orthogonal to the osculating plane. If the curve $\gamma$ happens to be planar the osculating plane, hence ${\bf b}$, is constant. If $\gamma$ is not planar the osculating plane, hence its normal vector, changes slightly from point to point. In fact one has $\dot{\bf b}=-\tau\,{\bf n}$. The angular velocity $\tau$ with which the binormal vector ${\bf b}$, resp. the osculating plane, turns, is called the torsion of $\gamma$ at the point under consideration.

  • $\begingroup$ Using a helix as an example (assumed parametrized by arc length). Which helixes would have a higher torsion? one with higher radius? Higher pitch? (distance between vertical points)? $\endgroup$ – Ecotistician Jan 22 '18 at 17:05
  • $\begingroup$ See here: mathworld.wolfram.com/Helix.html $\endgroup$ – Christian Blatter Jan 23 '18 at 8:59

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