Quote:
Originally Posted by Ether
You don't have to use matrix math if you don't want to. You can use plain ole algebra.
Attached PDF has the cookbook formulas for computing a, b, c, d.
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);
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Quote:
Originally Posted by Ether
Pick a coordinate system whose origin is the starting point and whose positive X axis aligns with the starting direction of motion.
Compute the ending point (X2,Y2) and slope m2 in that coordinate system.
Then the cubic becomes simply Y = aX3 + bX2
and the formulas for a and b are simply
a = (m2X2 - 2Y2)/X23
b = (3Y2 - m2X2)/X22
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As soon as I looked at the algebra method I wanted to run away from the problem. But I'm assuming that making the initial robot position always equal to the x axis simplified the problem? I'm sorry if its tedious but, could you show me how you got that formula for "Y = " ? thanks alot for your help!