Created
November 12, 2018 01:05
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Kent Libby question 1 gist
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%% Kent Libby question 1 matlab script | |
L=.2; % in meters | |
end_point=[6 7]; % in meters | |
R=5; % in meters | |
D=sqrt(end_point(1)^2+end_point(2)^2) | |
%% This calculates angle of the arc the robot travels | |
arc=acos((-D^2+R^2+R^2)/(2*R*R)) % in radians | |
dt=20 % in seconds | |
omega=-arc/dt | |
%% This finds the wheel speeds based off radius and omega | |
Vr=omega*(R+L/2) | |
Vl=omega*(R-L/2) | |
%% calculate the new pose | |
init_x=0; % in meters | |
init_y=0; % in meters | |
tx_mat=[cos(omega*dt) -sin(omega*dt) 0;... | |
sin(omega*dt) cos(omega*dt) 0;... | |
0 0 1] | |
syms ICCx ICCy | |
%% Solve for ICCx and ICCy using the equation of the circle | |
x1=0; y1=0; x2=6; y2=7; | |
[solx,soly] = solve((0-ICCx)^2+(0-ICCy)^2==5^2, (6-ICCx)^2+(7-ICCy)^2==5^2); | |
clear ICCx ICCy | |
%% pick the first solution and find the ICC based off tha | |
ICCx=(double(solx(1))) | |
ICCy=(double(soly(1))) | |
theta=asin((ICCx-x1)/-R); % in radians | |
ICC=[ICCx; ICCy]; | |
%% display the new pose | |
newPose=tx_mat*[init_x-ICC(1); init_y-ICC(2); theta]+[ICC; omega*dt] |
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