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I know what curvature is; positive curvature, negative curvature, zero curvature, i understand. BUT THEN.. Signed curvature? I just can't seem to find something that distinguishes enough difference between what I already know and what I'm trying to find out...

From what I am seeing online signed curvature is equal to curvature which is the norm/modulus/magnitude of the differential of tangent... so what is with the name..

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  • $\begingroup$ Please give us the context. Are you talking about curves in the plane? Are you talking about curves on surfaces (in which case normal curvature is the signed curvature of a normal slice, the sign coming from whether the principal normal agrees with the surface normal or disagrees with it)? $\endgroup$ Apr 7 '17 at 20:25
  • $\begingroup$ I'm studying curvature in both $\endgroup$ Apr 7 '17 at 20:50
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If you understood what curvature is; positive curvature, negative curvature, zero curvature ....

THEN.. you innately understood what is signed curvature already.

By accepted conventions if curving above tangent is positive, then drooping below it is negative... the tangential rotation is positive (counter-clockwise) or negative (clockwise).

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Any sign is an indication of how much is its departure from zero. Zero curvature is a property of a straight line. So look at: downward bend 1 blue , no bend 2 black and upward bend 3 red as three progressively bent states..somewhat like a debt situation which had at first got even and then later looked up by means of strong growth action, that is positive curvature addition.

With respect to x-axis reference, Tangential rotation in blue is clockwise, black nothing no change and red anti-clockwise is positive.

enter image description here

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A "static" circle of radius $R>0$ in the plane or in ${\mathbb R}^n$ has (unsigned) curvature ${1\over R}>0$. If, however, a circle, or any curve for that matter, in the plane is traversed in increasing time in a certain direction, and if counterclockwise rotation is considered positive, then it makes sense to talk about signed curvature of this curve.

In this regard, let $$\gamma:\qquad t\mapsto\bigl(x(t),y(t)\bigr)$$ be a $C^2$ parametrization of $ \gamma$ with $\bigl(\dot x(t),\dot y(t)\bigr)\ne(0,0)$ for all $t$. Then
$$\theta(t):={\rm arg}\bigl(\dot x(t),\dot y(t)\bigr)\ ,$$ i.e., the polar angle mod $2\pi$ of the tangent vector of $\gamma$, is defined for all $t$. The quantity$$\kappa:= {d\theta\over ds}={\ddot y\dot x-\ddot x\dot y\over(\dot x^2+\dot y^2)^{3/2}}$$ is then the signed curvature of $\gamma$. This $\kappa$ is positive if the tangent vector turns counterclockwise with increasing $t$, and negative otherwise.

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