The roller mechanism follows the path indicated by the blue curve. A circle (closely matching the curve) can be drawn through the curve at this point. We are looking for the radius of this circle.

When x = 0.85, this is the situation. What is the radius?

`y=ln(sec\ x)`

so

`(dy)/(dx)=(sec\ x\ tan\ x)/(sec\ x)=tan\ x`

This is 1.138 when x = 0.85.

and

`(d^2y)/(dx^2)=sec^2x`

This is 2.2958 when x = 0.85.

So the radius of curvature is equal to:

`R=([1+((dy)/(dx))^2]^(3/2))/((d^2y)/(dx^2))`

`=([1+(1.138)^2]^(3/2))/2.2958`

`=3.4786/2.2958`

`=1.52\ "dm"`