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Unread 31-03-2012, 00:56
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Ether Ether is offline
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Re: Curve Driving Generator

Quote:
Originally Posted by theNerd View Post
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


Quote:
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.


Last edited by Ether : 31-03-2012 at 01:08.