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View Full Version : solving multiple equations in matlab


samira_8511
22 Feb 2008, 09:03pm
hi all, i wanna do the inverse kinematic for T3 robot and now i have 6 equations with 5 variables but when i use solve i recieve no expilicit solution :mad:
thease are the equations :
v1 = 150*cos(t5)*sin(t1)+150*cos(t1)*sin(t2)*sin(t5)*sin(t4)*cos(t3)+150*cos(t1)*sin( t2)*sin(t4)*cos(t3)+150*cos(t1)*sin(t2)*cos(t4)*sin(t3)-150*cos(t1)*cos(t2)*cos(t4)*cos(t3)+1000*cos(t1)*sin(t2)*sin(t3)+150*cos(t1)*cos (t2)*sin(t4)*sin(t3)-150*cos(t1)*cos(t2)*sin(t5)*cos(t4)*cos(t3)-1000*cos(t1)*cos(t2)*cos(t3)-1000*cos(t1)*cos(t2)+150*cos(t1)*cos(t2)*sin(t5)*sin(t4)*sin(t3)+150*cos(t1)*sin (t2)*sin(t5)*cos(t4)*sin(t3)
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v2 = 1000*sin(t1)*sin(t2)*sin(t3)-150*sin(t1)*cos(t2)*sin(t5)*cos(t4)*cos(t3)-1000*sin(t1)*cos(t2)*cos(t3)+150*sin(t1)*sin(t2)*cos(t4)*sin(t3)+150*sin(t1)*cos (t2)*sin(t5)*sin(t4)*sin(t3)+150*sin(t1)*sin(t2)*sin(t5)*cos(t4)*sin (t3)+150*sin(t1)*sin(t2)*sin(t5)*sin(t4)*cos(t3)-1000*sin(t1)*cos(t2)+150*sin(t1)*sin(t2)*sin(t4)*cos(t3)+150*sin(t1)*cos(t2)*sin (t4)*sin(t3)-150*cos(t5)*cos(t1)-150*sin(t1)*cos(t2)*cos(t4)*cos(t3)
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v3 = -1300-150*sin(t2)*sin(t4)*sin(t3)+150*cos(t2)*sin(t4)*cos(t3)+150*cos(t2)*cos(t4)*sin( t3)+150*sin(t2)*cos(t4)*cos(t3)+1000*sin(t2)+150*sin(t2)*sin(t5)*cos(t4)*cos(t3)-150*sin(t2)*sin(t5)*sin(t4)*sin(t3)+150*cos(t2)*sin(t5)*sin(t4)*cos(t3)+150*cos( t2)*sin(t5)*cos(t4)*sin(t3)+1000*sin(t2)*cos(t3)+1000*cos(t2)*sin(t3)
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v4 = cos(t1)*cos(t2)*sin(t5)*cos(t4)*cos(t3)-cos(t1)*cos(t2)*sin(t5)*sin(t4)*sin(t3)-cos(t1)*sin(t2)*sin(t5)*cos(t4)*sin(t3)-cos(t1)*sin(t2)*sin(t5)*sin(t4)*cos(t3)-cos(t5)*sin(t1)
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v5 =sin(t1)*cos(t2)*sin(t5)*cos(t4)*cos(t3)-sin(t1)*cos(t2)*sin(t5)*sin(t4)*sin(t3)-sin(t1)*sin(t2)*sin(t5)*cos(t4)*sin(t3)-sin(t1)*sin(t2)*sin(t5)*sin(t4)*cos(t3)+cos(t5)*cos(t1)
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v6 = -sin(t5)*(sin(t2)*cos(t4)*cos(t3)-sin(t2)*sin(t4)*sin(t3)+cos(t2)*cos(t4)*sin(t3)+cos(t2)*sin(t4)*cos(t3))
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shwaip
23 Feb 2008, 02:23am
if you have six independent equations for five variables, it's possible that there is no solution that meets them all.

Those are quite complex equations too - there may be no nice symbolic solution.