Is this Cornu spiral positively oriented or not?.

Cornu Spiral

I am learning about plane curves and I am told that if the curvature κ\kappa is a linear function of arc length ss. i.e κ=s\kappa = s we obtain the Cornu Spiral. I find this difficult to understand for the following reason:

If we take the xx axis in the above diagram to be arc length ss then, when ss is positive i.e the curvature is positive should the tangent vector to the curve be turning to the left?

It could be the case that if we start at the centre of the right hand spiral and then follow the curve then indeed the tangent vector is moving to the left and we have positive curvature. Is this the correct way to think of it? I think this is actually the correct way to think of it as when we get into the region of negative ss the tangent vectors to the curve move to the right. Is this correct?

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By accepted mathematical convention directions of x and y axes are taken as positive to the right and positive upwards from a reference origin respectively.

Slope ϕ=tan−1dy/dx \phi = \tan^{-1} dy/dx

So accordingly such a convention is made to apply to next derivative as well. Curvature is defined as the rate of slope / rotation angle change or rate of change of tangent direction to a curve with respect to arc. It means that the counterclockwise (CCW) rotation adds to curvature and clockwise rotation decreases curvature.

In Cornu’s spiral κ=dϕ/ds=as,ϕ=aϕ2/2 \kappa = d\phi/ds = a s, \phi = a \phi^2 /2 where aa is a constant. At origin curvature is zero.

If a>0a>0 and has zero slope tangent to x-axis it curls up CCW in quadrant 1 with increasing curvature. And if tangent to y-axis it curls up CCW in quadrant 2. This is as you have sketched.

Similarly if a<0,ϕ=aϕ2/2a<0, \, \phi = a \phi^2 /2 and has zero slope tangent to x-axis it curls down CW into quadrant 4 with high negative curvature. And if tangent to y-axis it curls to right CW in quadrant 1. Like the Cornu/Euler spiral here.