Quote:
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!
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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 = (m
2X
2 - 2Y
2)/X
23
b = (3Y
2 - m
2X
2)/X
22
c = 0
d = 0
... and since c=0 and d=0, the cubic Y = aX
3 + bX
2 + cX + d is reduced to:
Y = aX
3 + bX
2
Quote:
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I'm assuming that making the initial robot position always equal to the x axis simplified the problem?
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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.