Kinematics of a Delta Robot in CGA

Hugo Hadfield

November 2019

hh409.user.srcf.net

Delta Robots

  • Very fast moving
  • Simple forward and inverse kinematics
  • Cheap and easy to build

History of the Delta Robot

  • Invented by Reymond Clavel at EPFL in 1985
  • A type of parallel manipulator

Model of a Delta Robot

Forward Kinematics

\begin{aligned} a_i &= (r_b - r_e + l\cos(\theta_i))s_i + l\sin(\theta_i)e_3\\ A_i &= \frac{1}{2}a_i^2n_\infty + a_i - n_0\\ \Sigma_i &= A_i - \frac{1}{2}\rho^2n_\infty\\ T &= I_5\bigwedge_{i=0}^{i=2}\Sigma_i\\ P &= 1 + \frac{T}{\sqrt{T^2}}\\ Y &= -\tilde{P}(T\cdot n_\infty)P\\ y &= \frac{-\sum_{j=1}^{j=3}(Y\cdot e_j)e_j}{Y\cdot n_\infty} \end{aligned}
r_e
Y
A_i
r_b
l
\rho
\theta

Reachable Volume

Inverse Kinematics

\begin{aligned} C_i &= \left( \text{up}(r_bs_i) - \frac{1}{2}l^2n_\infty \right)\wedge(I_3s_i\wedge e_3) \\ \Sigma_i &= \text{up}(y + r_es_i) - \frac{1}{2}\rho^2n_\infty \\ T_i &= (C_i\wedge\Sigma_i)I_5\\ P_i &= \frac{1}{2}(1 + \frac{T_i}{\sqrt{T_i^2}})\\ A_i &= -\tilde{P_i}(T_i\cdot n_\infty)P_i\\ a_i &= -\frac{\sum_{j=1}^{j=3}(A_i\cdot e_j )e_j}{A_i\cdot n_\infty} \\ z_i &= a_i - r_bs_i \\ \theta_i &= \arctan{\frac{z_i\cdot e_3}{z_i\cdot s_i}} \end{aligned}

If \(T_i^2 < 0\) then the point is unreachable

Forward Jacobian

- Move only one motor

 

- This fixes two of the spheres, they intersect in a circle

 

- We are therefore interested in the differential of the end point of the point pair made from intersection of the circle and the moving sphere

- First we need to find how the center of the sphere changes as we move theta

\Sigma_i = A_i - \frac{1}{2}\rho^2n_\infty \\

- How does point pair move with the sphere?

= \frac{\partial \Sigma_i}{\partial \theta_i} = \frac{\partial A_i}{\partial \theta_i} = -(r_b - r_e)l\sin(\theta_i)n_\infty - l\sin(\theta_i)s_i + l\cos(\theta_i)e_3
= \left( \frac{\partial a_i}{ \partial \alpha}\cdot a_i \right)n_\infty + \frac{\partial a_i}{\partial \alpha}
\frac{\partial \Sigma_i}{\partial \alpha} = \frac{\partial A_i}{\partial \alpha} = \frac{1}{2}\frac{\partial a_i^2}{\partial \alpha}n_\infty + \frac{\partial a_i}{\partial \alpha} \\

Forward Jacobian (2)

T = \Sigma_i\wedge C\\ \frac{\partial T }{\partial \theta_i} = \frac{\partial \Sigma }{\partial \theta_i}\wedge C\\

- Now we need to know how the conformal endpoint of the point pair changes

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } P = \frac{1}{2}\left(1 + \frac{T}{\sqrt{T^2}}\right)\\ Y = -\tilde{P}(T\cdot n_\infty)P\\ \pdiff{P}{\theta_i} = \frac{1}{2T^2}\left(\sqrt{T^2}\pdiff{T}{\theta_i} - T\frac{\pdiff{T}{\theta_i}\cdot T}{\sqrt{T^2}}\right)\\ \pdiff{Y}{\theta_i} = -\tilde{\pdiff{P}{\theta_i}}(T\cdot n_\infty)P -\tilde{P}(\pdiff{T}{\theta_i}\cdot n_\infty)P -\tilde{P}(T\cdot n_\infty)\pdiff{P}{\theta_i}

- How does the 3D endpoint change with the conformal endpoint change? (derivative of 'down' function)

Forward Jacobian (3)

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \pdiff{y}{\theta_i} = \frac{ -\sum_{j=1}^{j=3}\left(\pdiff{Y}{\theta_i}\cdot e_{j}\right)e_j (Y\cdot\ninf) + \sum_{j=1}^{j=3}\left(Y\cdot e_{j}\right)e_j (\pdiff{Y}{\theta_i}\cdot\ninf) }{ (Y\cdot\ninf)^2 }
\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } y = \frac{-\sum_{j=1}^{j=3}(Y\cdot e_j)e_j}{Y\cdot n_\infty}

- We now have a closed form expression for the derivative of the endpoint wrt. theta and we didn't have to use maple once!

- We can use this for control and optimisation. Consider a cost function:

Forward Jacobian In Use

g(y) = (y-t)^2 \\ \frac{\partial g}{\partial \theta_i} = 2\frac{\partial y}{\partial \theta_i}\cdot(y-t)

- We therefore have a closed form for

and so can do gradient descent

\nabla g

- Given the forward jacobian we can simply invert the 3x3 matrix at each point to get the inverse jacobian

OR

- We can differentiate our inverse kinematic equations..

Inverse Jacobian

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \begin{aligned} \Sigma_i &= X_i - \frac{1}{2}\rho^2n_\infty \\ \frac{\partial \Sigma_i}{\partial \alpha} &= \frac{\partial X_i}{\partial \alpha} = \left(\pdiff{x_i}{\alpha}\cdot x_i\right) \ninf + \pdiff{x_i}{\alpha}\\ x_i &= y + r_es_i\\ \frac{\partial x_i}{\partial \alpha} &= \frac{\partial y}{\partial \alpha} \end{aligned}

Starting with Sigma_i

Differentiate wrt. a scalar param \alpha

the sphere centred at the meeting of the 4 bar mechanism and the end platform

- Now we need to know how the elbow positions vary with \alpha

Inverse Jacobian (2)

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \begin{aligned} C_i &= \left( B_i - \frac{1}{2}l^2n_\infty \right)\wedge(I_3s_i\wedge e_3)\\ T_i &= (\Sigma_i\wedge C_i)^*\\ \pdiff{T_i}{\alpha} &= (\pdiff{\Sigma_i}{\alpha}\wedge C_i)^* \end{aligned}

- Here A_i is the conformal elbow position

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \begin{aligned} \pdiff{A_i}{\alpha} &= -\tilde{\pdiff{P_i}{\alpha}}(T_i\cdot n_\infty)P_i -\tilde{P_i}(\pdiff{T_i}{\alpha}\cdot n_\infty)P_i -\tilde{P_i}(T_i\cdot n_\infty)\pdiff{P_i}{\alpha}\\ \pdiff{a_i}{\alpha} &= \frac{ -\sum_{j=1}^{j=3}\left(\pdiff{A_i}{\alpha}\cdot e_{j}\right)e_j (A_i\cdot\ninf) + \sum_{j=1}^{j=3}\left(A_i\cdot e_{j}\right)e_j (\pdiff{A_i}{\alpha}\cdot\ninf) }{ (A_i\cdot\ninf)^2 } \end{aligned}

- Finally we need to know how the joints vary with elbow position

Inverse Jacobian (3)

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \begin{aligned} z_i &= a_i - r_bs_i \\ \theta_i &= \arctan{\frac{z_i\cdot e_3}{z_i\cdot s_i}}\\ \frac{\partial \theta_i}{\partial \alpha} &= \frac{z_i\cdot s_i \frac{\partial z_i}{\partial \alpha}\cdot e_3 - z_i\cdot e_3 \frac{\partial z_i}{\partial \alpha}\cdot s_i}{z_i^2} \end{aligned}

- We are done (Once again no Maple!)

- The inverse jacobian allows us to map velocity of the endpoint to velocity of joints

Inverse Jacobian in Use

g(y) = (y - t)^2 \\ \frac{\partial g}{\partial y} = 2(y-t)\\

- In other words, to minimise cost at maximum rate drive the end point linearly at our target point (obviously)

- With the inverse jacobian we can do exactly that

\newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\nn}{\nonumber} \newcommand{\ninf}{n_{\infty}} \newcommand{\einf}{e_{\infty}} \newcommand{\no}{n_{0}} \newcommand{\eo}{e_{0}} \newcommand{\wdg}{\wedge} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2} } \begin{aligned} \pdiff{y}{\alpha_j} \propto (2(t-y))\cdot e_j \end{aligned}

- CGA provides a very neat framework for solving the forward and inverse kinematics of parallel robots

 

- We can easily simulate the dynamics of the robot with clifford and ganja.js

Summary

- Trivectors in CGA represent lines and circles

 

- The direct interpolation of lines and circles results in trivectors interpretable as general screws

- Screw theory is widely used in robotics for analysis of kinematics and dynamics

 

- Lets bring screw theory and CGA (also PGA) together properly and analyse a whole load of systems

Future Work

Kinematics of a Delta Robot in CGA

By Hugo Hadfield

Kinematics of a Delta Robot in CGA

The delta robot is one of the most popular parallel robots that exist, known for its speed and precision. This presentation covers the forward and inverse kinematics of the delta robot using Conformal Geometric Algebra.

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