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3 Degree of Freedom Manipulator
S G Krishnan and R Harilal
May 2, 2013
Position Analysis
Using (x,y ,) as inputwe first calculate the positions of A,B,C.
xCi = x lcicos(i)yCi = y lcisin(i)
1 = + (/6)2 = + (/2)3 = + (5/6)
D = (xC xA)2 + (yC yA)2
= cos1((lA)
2+D2(lB)2)
2lAD)
= tan1 yCyAxCxA
=
In the same way we can calculate the orientation of all the three legs.
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Velocity Analysis
For velocity analysis we have inputs as (x,y,, x, y, ). We obtain 1, 2, 3 by position analysis.
J X =
J is the jacobian. J is obtained by differentiating the contraints of the legs.
(xBi xCi)2
+ (yBi yCi)2
= (lbi)2
xC = x lccos1xB = xA lacos1yC = y lcsin1yB = yA lasin1
Differentiating with respect to time
2(xB xC)( xB xC) + 2(yB yC)( yB yC) = 0
On simplifying
1 la[(y yA)cos1 (x xA)sin1 + lccos(1 1)] =(x xA lccos1 lacos1) x + (y yA lcsin1 lasin1) y lc[(y yA)cos1 (x xA)sin1 + lasin(1 1)] 1
J can be represented as
p1/s1 q1/s1 r1/s1p2/s2 q2/s2 r2/s2p3/s3 q3/s3 r3/s3
pi = (x xAi lcicosi laicosi)qi = (y yAi lcisini laisini)ri = lci[(y yAi)cosi (x xAi)sini + laisin(i i)]si = lai[(y yAi)cosi (x xAi)sini + lcicos(i i)]
Thus given X, we can find
Acceleration Analysis
Accelerations are obtained by differentiating the velocity equation
J
X = on differentiation
JX+ JX = X = J1( JX)J can be represented as
P1 Q1 R1P2 Q2 R3P3 Q3 R3
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Pi =si pipisi
s2iQi =
siqiqisis2i
Ri =siririsi
s2i
pi = x + laisini + lbisiniqi = y laicosi lbicosiri = lci[(x xa)cosi + (y ya)sini] + lc[xsini ycosi] + lalc(i i)cos(i i)si = lai[(x xa)cosi + (y ya)sini] la[xsini ycosi] + lalc(i i)cos(i i)
Hence acceleration inversion is completed
Force and Torque Analysis
Force and torque analysis is done to determine the amount of torque needed by the motor tobring about the required motion.
Free Body Diagrams
Platform
Link
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Crank
Force equation for platform
Fx1 + Fx2 + Fx3 = MPlatformx
Fy1 + Fy2 + Fy3 = MPlatformy
(Fx1sin1 Fy1cos1)lc1+ (Fx2sin2 Fy2cos2)lc2+ (Fx3sin3 Fy3cos3)lc3 = Iplatform
Iplatform =Mplatforml
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ci
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xci = x lcicosixci = x + lcisinixci = x + lcisini + lcicosi
2
yci = y lcisiniyci = y lcicosi
yci = y lcicosi + lcisini2
xBi = laicosixBi = laisiniixBi = laisinii laicosii2
yBi = laisiniyBi = laicosii
yBi = laicosi
i laisini
i
2
= tan1[ yciybixcixbi
]
xcom =xBi+ xCi
2
ycom =yBi+ yCi
2
com =2( xCi xBi)sin+2( yci yBi)cos
lbi
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Force Equations of a link
Fx1 Fx4 = Mlinkxcom
Fy1 Fy4 = Mlinkycom
(Fy1cos Fx1sin+ Fy4cos Fx4sin) lbi2 = Ilinkcom
Ilink =Mlinkl
2
bi
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Solving for Fx1 to Fx6 , Fy1 to Fy6
Torque of the motor is obtained by
1 = (Fx4sin1 Fy4cos1)la1
Similarly we can calculate torque for the other two motors.
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