You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
87 lines
2.6 KiB
87 lines
2.6 KiB
% Copyright (C) 1993-2013, by Peter I. Corke
|
|
%
|
|
% This file is part of The Robotics Toolbox for MATLAB (RTB).
|
|
%
|
|
% RTB is free software: you can redistribute it and/or modify
|
|
% it under the terms of the GNU Lesser General Public License as published by
|
|
% the Free Software Foundation, either version 3 of the License, or
|
|
% (at your option) any later version.
|
|
%
|
|
% RTB is distributed in the hope that it will be useful,
|
|
% but WITHOUT ANY WARRANTY; without even the implied warranty of
|
|
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
|
|
% GNU Lesser General Public License for more details.
|
|
%
|
|
% You should have received a copy of the GNU Leser General Public License
|
|
% along with RTB. If not, see <http://www.gnu.org/licenses/>.
|
|
%
|
|
% http://www.petercorke.com
|
|
|
|
%%begin
|
|
|
|
% Frequently we want to define a smooth sequence of positions (or poses) from
|
|
% one point to another. First consider the 1-dimensional case.
|
|
%
|
|
% We define the start and end position
|
|
|
|
p0 = -1;
|
|
p1 = 2;
|
|
|
|
% and a smooth path from p0 to p1 in 50 time steps is given by
|
|
|
|
p = tpoly(p0, p1, 50);
|
|
about p
|
|
% which we see has 50 rows. We can plot this
|
|
|
|
plot(p)
|
|
|
|
% and see that it does indeed move smoothly from p0 to p1 and that the initial
|
|
% and final derivative (and second derivative) is zero.
|
|
|
|
% We can also get the velocity and acceleration
|
|
|
|
[p,pd,pdd] = tpoly(p0, p1, 50);
|
|
subplot(3,1,1); plot(p); xlabel('Time'); ylabel('p');
|
|
subplot(3,1,2); plot(pd); xlabel('Time'); ylabel('pd');
|
|
subplot(3,1,3); plot(pdd); xlabel('Time'); ylabel('pdd');
|
|
|
|
% This path is a 5th order polynomial and it suffers from the disadvantage that
|
|
% the velocity is mostly below the maximum possible value. An alternative is
|
|
|
|
[p,pd,pdd] = lspb(p0, p1, 50);
|
|
subplot(3,1,1); plot(p); xlabel('Time'); ylabel('p');
|
|
subplot(3,1,2); plot(pd); xlabel('Time'); ylabel('pd');
|
|
subplot(3,1,3); plot(pdd); xlabel('Time'); ylabel('pdd');
|
|
% which we see has a trapezoidal velocity profile.
|
|
|
|
% Frequently the start and end values are vectors, not scalars, perhaps a 3D
|
|
% position or Euler angles. In this case we apply the scalar trajectory function
|
|
% to a vector with
|
|
|
|
p = mtraj(@tpoly, [0 1 2], [2 1 0], 50);
|
|
about p
|
|
% and p again has one row per time step, and one column per vector dimension
|
|
|
|
clf; plot(p)
|
|
|
|
%---
|
|
% Finally, we may wish to interpolate poses. We will define a start and end pose
|
|
|
|
T0 = transl(0.4, 0.2, 0) * trotx(pi);
|
|
T1 = transl(-0.4, -0.2, 0.3) * troty(pi/2) * trotz(-pi/2);
|
|
|
|
% and a smooth sequence between them in 50 steps is
|
|
|
|
T = ctraj(T0, T1, 50);
|
|
about T
|
|
% which is a 4x4x50 matrix. The first pose is
|
|
|
|
T(:,:,1)
|
|
|
|
% and the 10th pose is
|
|
|
|
T(:,:,10)
|
|
|
|
% We can plot the motion of this coordinate frame by
|
|
|
|
clf; tranimate(T)
|