0

Consider the helix $h(t)=(a\cdot cos(t),a\cdot sin(t),bt)$ and the circle $c(t)=(a\cdot cos(t),a\cdot sin(t),0)$. I know that the curvature of them are:

  • $k(t)_h=\frac{a}{a^2+b^2}$
  • $k(t)_c=\frac{1}{a}$

respectively. The curvature measures the angular rate of change of the tangent unit vector per unit change in distance along the path. When we move in a circular or helicoidal movement, it happens that $T(0)=T(2\pi)=T(4\pi)=...$, that is when we loop the tangent unit vector repeats periodically. So in the case of the circle, the angular rate of change of the tangent unit vector... (or curvature) is $1/a$, so if we advance a whole loop, a distance of $2\pi a$ units, the rate of change becomes $2\pi a \cdot 1/a=2\pi$ rad. Thus, the rate of change of $T$ is $2\pi$ radian when we complete a loop.

Surprisingly, this does not happen for the helix! If we loop once along a helix we get that the distance "travelled" is $2\pi \sqrt{a^2+b^2}$, so multiplying by the curvature:

  • $2\pi \sqrt{a^2+b^2}\frac{a}{a^2+b^2} \neq 2\pi$ rad

Neverthelss $T(0)=T(2\pi)$! Why? Is my interpretation of curvature wrong?

Marius Alexandru Marinescu
  • 1

2 Answers2

1

Your reasoning only works in $2-$D, and this is why it works only for $b=0$. However, since you calculated $T$, you'll notice it traces out a circle, and the vector sweeps out a cone. Actually, the distance travelled $sk_h = 2\pi\frac{a}{\sqrt{a^2+b^2}}$ is precisely the length of that circle.

Btw, the fact that $T$ does not stay in a plane is characterized by torsion. Hope this helps.

LPZ
  • 2,788
0

You seem confused about curvature it is not "the global angular rate of change of of the tangent unit vector", but it is the differential change in the case of the circle it happens that it is also the global change.

trula
  • 1,472