How to plan path in Cartesian coordinate for robotic manipulator

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Kannan
Kannan 2025 年 3 月 15 日
コメント済み: Kannan 2025 年 3 月 15 日
I uploaded a 6Dof robotic arm rigid tree to matlab simulink , now i need to move the end effector in the shape ' M ' . How do i obtain the waypoints of the shape ' M '
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Sam Chak
Sam Chak 2025 年 3 月 15 日
Based on your description, if the manipulator starts from one end, you only need to specify the coordinates of the four waypoints. It is assumed that the manipulator can move the effector smoothly from Point A to Point B.
Since these are merely straight lines, is it necessary to accurately generate the time series data for the desired straight-line trajectories so that the effector reaches the data points according to the timestamps?
Kannan
Kannan 2025 年 3 月 15 日
Thank you Sam for your reply🫂💖 .
What if i need to move the end effector in a more complex path ( in 3D space ) . My doubt is how do i tranfsfer my robot to a xyz cartesian to find the waypoints . In many yt videos they are directly showing the waypoints of the path but how do they obtain that with respect to the robot

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Sam Chak
Sam Chak 2025 年 3 月 15 日
For a more complex path, such as the M-shaped Golden Arches, you will need to describe the path using parameterized equations to express the reference coordinates x, y, and z as functions of a parameter t. For example, you want the end-effector to trace the path on the x-y plane at a fixed height of 0.5 m, where x should vary from to 1.
If the path cannot be described mathematically, you will need to manually move the robotic arm to trace the desired path. Your robot algorithm should 'memorize' the movements of the robot's individual joints, specifying how each joint's angle changes over time to achieve the desired end-effector movement.
%% Define the time parameter, t
T = 10; % complete tracing the path at 10 sec
t = linspace(0, T, 1e3*T+1);
%% Parameterized equations in 3D
xA = -1; % lower bound of x
xB = 1; % upper bound of x
xR = xA + (xB - xA)*t/T; % for x-axis: Map t from [0, T] to x ∈ [−1, 1]
yR = 2 - sech(5*xR) - cosh(acosh(2)*xR); % for y-axis: M-shaped function, y = f(x(t))
zR = 0.5*ones(1, numel(t)); % for z-axis: Map t from [0, T] to z = 0.5
%% Plot the desired path
plot3(xR, yR, zR, 'color', '#FFBF00', 'LineWidth', 3), grid on
ylim([0, 0.8])
xlabel('x'), ylabel('y'), zlabel('z')
title('Desired Golden Arches Path in 3D space')
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Sam Chak
Sam Chak 2025 年 3 月 15 日
@Kannan Oh, I see; now I understand what you are looking for. The smooth curves between the waypoints can be generated using spline fitting. Please take a look at some examples produced by the spline() interpolation technique.
By the way, thank you for your acceptance.
Kannan
Kannan 2025 年 3 月 15 日
But how could i get those waypoints in that xyz scenario ? ( Thanks for giving your time to clear my doubt Sam ) :)

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