The geometry of a parameterized curve in space can be understood by following the motion of its Frenet framing. We have previously discussed the way the framing $\{T, N, B\}$ is defined at teach point of a unit-speed curve $s\mapsto\alpha(s)$, and derived the basic equations:
\[ \begin{array}{ccccc} T'&=& &\kappa N & \\ N'&=&-\kappa T & & + \tau B \\ B'&=& & -\tau N & \end{array}\]These are the Frenet-Serret Equations. We would like to explore now what these equations mean and how they help us interpret the meaning of curvature, $\kappa$, and torsion, $\tau$.
This post has no picture in it. This is intentional. To get the most out of this description, you should follow along and embody the motion for yourself.
To get "oriented" (That's a joke, son), use your right hand to make a little moving frame for yourself: first make a fist. Then stick your index finger straight out and your thumb straight up. These two fingers will make a right angle. Then uncurl your middle finger so that it points out from your palm at a 90 degree angle. The three digits sticking out will be the vectors of your frame: the index finger is $T$; the middle finger is $N$; the thumb is $B$.
The important thing to remember is that your hand can move, turn, and spin, but it cannot change shape as it does so. The framing you are trying to represent is a rigid one: the fingers represent unit vectors that stay at right angles to one another.
First, note that the tangent vector $T$ tells us about the direction of travel of the curve. Way back from multivariable calculus, you know that this is the direction of the tangent line to the curve, which is the line that the curve is closest to at the given point. Tracking the motion of $T$ tells us how to how the curve changes direction, that is, how it curves. This is the job of the curvature $\kappa$ and the normal $N$. \[ T' = \kappa N \] The normal $N$ tells us the direction in which $T$ is changing, and $\kappa$ tells us how fast. For a stronger geometric interpretation, consider that $N$ points toward the center of the osculating circle, and $\kappa$ is the reciprocal of the radius of that circle. In a way, the curvature measures the how badly your curve fails to be a straight line.
If we think in terms of the physical model (put your fingers back in place!), as your hand travels along the curve, you should turn it so that your index finger turns toward your middle finger. The curvature tells you how fast to do that.
Now, the tangent vector $T$ and the normal vector $N$ span out the osculating plane. This is the plane which the curve is closest to lying in. (If your curve is a planar curve, $T$ and $N$ will be a basis for the plane containing the curve.) The next thing to do is to measure how badly the curve fails to lie in the osculating plane.
Now as we move along the curve, the osculating plane will change and move around, too. It is a bit tricky to measure this motion directly, but we can use its normal vector as a proxy. The binormal vector $B = T\times N$ is a normal vector to the osculating plane, and its rate of change $B'$ measures how the osculating plane moves around. Note that the equation \[ B'= -\tau N \] tells us that B moves in the direction $-N$ with speed $\tau$.
In terms of our physical model (fingers!), as your hand travels along the curve you should also rotate so your thumb turns away from your middle finger with speed $\tau$. If you do this correctly, positive values of $\tau$ will mean that you follow the curve upward in space (toward your head) as you curl around to the left, and negative values of $\tau$ will mean that you follow the curve downward in space (toward your feet) as you curl around to the left.
Okay, that last paragraph is only really true as long as your middle finger stays to the left side of your index finger. If you turn your hand over, things get funny. Once you are comfortable, you should be able to work out what happens if you meet a curve that is bending to the right when you meet it. (Hint: start with your thumb pointing down.)
Now, if you stare at it long enough, you can see why the last of the Frenet-Serret equations is forced on us. \[ N' = -\kappa T + \tau B \] If your hand is moving in the way indicated above (turning index finger toward middle finger while rotating thumb around axis of index finger away from middle finger), then your middle finger is doing a more complicated motion. But that complicated motion is just the combination of two motions forced on it: one, turning away from your index finger with rate $\kappa$ and rotating toward your thumb with rate $\tau$. This is the only way to keep the frame a rigid object while doing the motions described above.
