"Index in position 1 exceeds array bounds (must not exceed 1)." error using spline.
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Ymbar Polanco
am 30 Okt. 2020
Beantwortet: Cris LaPierre
am 30 Okt. 2020
function [E,B,cmd,vx,vy] = trajplanner(a,x,N,T)
% trajplanner function creates a path and calculates the velocity of the
% end effector along path at various times
% inputs:
% a = column vector containing sections' lengths a1 & a2
% x = 2-by-N matrix containing target positions of end-effector;
% each column is one point (xi,yi)
% N = integer that equals # of intermediate configurations
% needing interpolation
% T = total duration of the trip from first to last point
% outputs:
% E = 2-by-N matrix containing path end-effector will follow
% B = 2-by-N matrix containing path joint B will follow
% cmd = 2-by-N matrix containing joint angle commands
% vx = array of length N containing x-component of velocity
% vy = array of length N containing y-component of velocity
eP=length(x); % # of effector points
xq=linspace(x(1,1),x(1,eP),N);
E=[xq;zeros(1,N)]; % create matrix for E
E(2,:)=spline(x(1,:),x(2,:),xq);
B=zeros(2,eP); % create matrix for B
cmd=zeros(2,eP);
% For loop to move through xqyq
for ii=1:N
cmd(:,ii)=ikFS(a,E(:,ii),[1;1]);
B(:,ii)=[a(1,1)*cos(cmd(1,ii));a(1,1)*sin(cmd(1,ii))];
end
% snapshots at intermediate positions
plot([zeros(1,N);B(1,:);E(1,:)],[zeros(1,N);B(2,:);E(2,:)],'k')
hold on
axis equal
plot(E(1,:),E(2,:),'g','linewidth',2) % trajectory at intermediate positions
pt_size = 1000; % mark user-defined points
scatter(x(1,:),x(2,:),pt_size,'.')
xpos=E(1,1:N);
ypos=E(2,1:N);
vx = (gradient(xpos)/steps)'
vy = (gradient(ypos)/steps)'
xE = cumtrapz(linspace(0,T,N),vx)+x(1,2);
yE = cumtrapz(linspace(0,T,N),vy)+x(1,2);
plot(xE,yE)
end
Running the program with a = [2,1], N = 50, T = 1, and x = [a(1) a(1)/2 0 -a(1) -a(1)-a(2) 0 a(1)/2 a(1) a(1) 0] returns:
"Index in position 1 exceeds array bounds (must not exceed 1).
Error in trajplanner (line 23)
E(2,:)=spline(x(1,:),x(2,:),xq);"
Any suggestions on how to fix this?
0 Kommentare
Akzeptierte Antwort
Cris LaPierre
am 30 Okt. 2020
I suspect the issue is your indexing of your variable x is the issue. I don't think it has a 2nd row, so your call to x(2,:) results in an error.
a = [2,1];
x = [a(1) a(1)/2 0 -a(1) -a(1)-a(2) 0 a(1)/2 a(1) a(1) 0]
x(2,:)
0 Kommentare
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!