Q: Transformation Shape - curvature values?

[DarkStar]

How does the popped-up curvature value (-68% in the screenshot) relate to the Y-value (about 79%)?

For a straight line it is 0%. If I drag it fully up / down the curvature value is +/-100% but it is not at the top/bottom of the graph.

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Supplementary question:
Give the points (0,0) and (100,100) and the curvature value of -68% how could I draw the same graph? And for a value of +68%?

[jmg8]

It might be me, being a bit thick, but I really dont understand the question. Could you re word it?

[DarkStar]

I'll try: Why is the popped-up value (of the draggable circle) -68% where the value for it on the Y-axis is 80%.

And, how would I draw that same graph myself? I've tried a Quadratic Bezier curve ( https://en.wikipedia.org/wiki/B%C3%A9zi ... ier_curves ) but need to do something with that -68% to get it to work

[jmg8]

What is a "pop-value"?
Also the -68% is just the curve amount and has no relationship to the X or Y values.
If you right click and set a different curve type, you could still have -68%, however it would be a completely different X/Y relationship.

[DarkStar]

it's a typo - it should be "popped-up value" (the tooltip). And, yes, I can see that a different Curvature type gives a different shape. I now see that my first question is moot.

As for the supplemental question, I guess I would need the formulas behind those curve types.

[jmg8]

Yes, indeed and unfortunately I am crap at maths, hehe.

[MeldaProduction]

This doesn't make any sense. The popup is the curvature, some factor of curving of the shape, it has no relation to the axis.

[DarkStar]

Yes, I understand that now, as jmg8 explained.