Curve Driving Generator

[Ether]

Originally Posted by theNerd

I'm sorry if its tedious but, could you show me how you got that formula for "Y = " ? thanks alot for your help!

If you set x1=0, y1=0, and m1=0 in these formulas:

a=-(-2*y2+2*y1+(m2+m1)*x2+(-m2-m1)*x1)/(-x2^3+3*x1*x2^2-3*x1^2*x2+x1^3);

b=(-3*x2*y2+x1*((m2-m1)*x2-3*y2)+(3*x2+3*x1)*y1+(m2+2*m1)*x2^2+(-2*m2-m1)*x1^2)/(-x2^3+3*x1*x2^2-3*x1^2*x2+x1^3);

c=-(x1*((2*m2+m1)*x2^2-6*x2*y2)+6*x1*x2*y1+m1*x2^3+(-m2-2*m1)*x1^2*x2-m2*x1^3)/(-x2^3+3*x1*x2^2-3*x1^2*x2+x1^3);

d=(x1^2*((m2-m1)*x2^2-3*x2*y2)+x1^3*(y2-m2*x2)+(3*x1*x2^2-x2^3)*y1+m1*x1*x2^3)/(-x2^3+3*x1*x2^2-3*x1^2*x2+x1^3);

... then you get:

a = (m₂X₂ - 2Y₂)/X₂³

b = (3Y₂ - m₂X₂)/X₂²

c = 0

d = 0

... and since c=0 and d=0, the cubic Y = aX³ + bX² + cX + d is reduced to:

Y = aX³ + bX²

I'm assuming that making the initial robot position always equal to the x axis simplified the problem?

Picking a coordinate system such that initial robot heading is aligned with +X axis (m1=0) and initial position is at origin (x1=0 and y1=0) simplified the problem.