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Inverse Kinematics Tim Haines

Inverse Kinematics Tim Haines. Geometry based solution. Circle Intersection. Number of Arms = n Solution finds the intersection of 2*n-1 circles For each goal node there are between 4 and 0 solutions (using three links). clear all; clc; A = [0 0]; %# Origin

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Inverse Kinematics Tim Haines

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  1. Inverse Kinematics Tim Haines Geometry based solution

  2. Circle Intersection • Number of Arms = n • Solution finds the intersection of 2*n-1 circles • For each goal node there are between 4 and 0 solutions (using three links)

  3. clear all; clc; • A = [0 0]; %# Origin • B = [0 2.0]; %# center of goal node • l1 = 1.00; %# Length of base member • l2=1.00; %Length of second member • l3=0.5; %length of third member • r1 = 1.25; %# radius of the SECOND circle • %%Determines the location of first arm%% • c = norm(A-B) %# distance between circles • cosAlpha = (l1^2+c^2-r1^2)/(2*l1*c); • u_AB = (B - A)/c; %# unit vector from first to second center • pu_AB = [u_AB(2), -u_AB(1)]; %# perpendicular vector to unit vector • %# use the cosine of alpha to calculate the length of the • %# vector along and perpendicular to AB that leads to the • %# intersection point • inter1 = A + u_AB * (l1*cosAlpha) + pu_AB * (l1*sqrt(1-cosAlpha^2)) • inter2 = A + u_AB * (l1*cosAlpha) - pu_AB * (l1*sqrt(1-cosAlpha^2)) • %%Determines the location of the second and third arm Using inter1 %% • D=inter1; • d = norm(D-B) %# distance between circles • cosAlpha1 = (l2^2+d^2-l3^2)/(2*l2*d); • u_AB1 = (B - D)/d; %# unit vector from first to second center • pu_AB1 = [u_AB1(2), -u_AB1(1)]; %# perpendicular vector to unit vector • %# use the cosine of alpha to calculate the length of the • %# vector along and perpendicular to AB that leads to the • %# intersection point • inter3 = D + u_AB1 * (l1*cosAlpha1) + pu_AB1 * (l1*sqrt(1-cosAlpha1^2)) • inter4 = D + u_AB1 * (l1*cosAlpha1) - pu_AB1 * (l1*sqrt(1-cosAlpha1^2)) • %Plot Solution 1 • figure(1) • line([A(1) inter1(1)] ,[A(2) inter1(2)],[0 0],'Marker','.','LineStyle','-') • line([inter1(1) inter3(1)],[inter1(2) inter3(2)],[0 0],'Marker','.','LineStyle','-') • line([inter3(1) B(1)],[inter3(2) B(2)],[0 0],'Marker','.','LineStyle','-')

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