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So, I would like to do an Extra Credit problem for my Vector Analysis course (the last half of a traditional Calc III course) that involves robotics (my new toy). However, it's Week 05 already, I want the Extra Credit released in Week 08, and I haven't gotten as far in my robotics studies as I had hoped.

So, I'm told most robot arms have a 6-dim'l configuration space (c-space). Assuming this is the case and assuming one can do a work integral (e.g.) in the c-space and not in the workspace, would someone who knows more about robotics than me be able to suggest a robot arm with a 6-dim'l c-space $Q^6$ with $H^1_{dR}(Q) = 0$, a differential 1-form $M$ with $dM = 0$ on $Q$ (so that $M = df$ for a potential function $f$) representing a generalized force/torque field, and an oriented curve $[C]$ in $Q$ with $\displaystyle \int_{[C]} M = f(B) - f(A)$ representing the work done moving the robot arm "through" the c-space? I can change the form and the curve, of course; the robot arm, the c-space, and a coordinate patch for the c-space are the things about which I don't know yet. More than one robot arm, etc., would be nice.

See this post for doing something similar with a torque field on $SO(3)$.

Jeffrey Rolland
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  • Hint: If your robot arm is moving in the $d$-dimensional space with the initial joint at a fixed position, then the topology of $Q$ is that of the product of $k$ copies of the sphere $S^{d-1}$, where $k$ is the number of rods in the arm. However, I am surprised that this appears in Calc-III class; the right place for this material would be a differential topology class. – Moishe Kohan Apr 07 '21 at 15:51
  • (It's Extra Credit) – Jeffrey Rolland Apr 07 '21 at 16:30
  • (Actually, you don't need $H_{1,dR}(Q) = 0$; one could just start with a scalar function $f$ and take its exterior derivative. Starting for the students with a closed 1-form $\omega$, one could show all its periods are 0 if $H_{1,dR}(Q) \ne 0$ and conclude it has an anitderivative. I can edit the question statement to include this if it won't incur too much ire.) – Jeffrey Rolland Apr 07 '21 at 17:57

1 Answers1

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Actually, a gantry crane with a gimbal machine for the end-effector should work. https://app.adamcad.com/editor/2c8d19a6-5802-4376-ab95-7bcd34ed6bbd

So, a gantry crane has 3 actuators, one for each of the x-, y-, and z-coordinates of the end-effector (similar to some 3D printers or home CNC machines). One would then have the gimbal machine (similar to a multi-axis trainer) rotate to any direction in 3-space.

Per this post, this machine would have $$Q= [a_1, b_1] \times [a_2, b_2] \times [a_3, b_3] \times SO(3)$$ as its c-space, which strong deformation retracts onto $SO(3)$, and hence this c-space has the de Rham cohomology of $SO(3)$, which is a de Rham cohomology sphere. Thus, $H^{1}_{dR}(Q) = 0$, so any closed 1-form should have a potential function.

For any generalized force/torque 1-form $M$, by the above paragraph, if $M$ is closed, there is a potential function (0-form) $f$ with the property that $df = M$. By the Generalized Stokes's Theorem, we should then have $\displaystyle \int_{[C]} M = f(B) - f(A)$ (where $A$ and $B$ are the endpoints of the oriented curve $[C]$).

One could have $$M = 1dx_1 [N] + 1dx_2 [N] + 1dx_3 [N] + 2w_1dw_1 [Nm] - 2w_2dw_2 [Nm] + 2w_3dw_3 [Nm] - 2w_4dw_4 [Nm]$$ (where the units in brackets are the units of the formulas in front of the differentials [independent of the actual elementary formulas themselves; see, for instance, this post], where each $dx_i$ has units of m, and using the upper-half unit sphere quaternions, whose differentials are unitless, for the $SO(3)$ elements) and $[C]$ parametrized by $$\gamma: [0,1] \to[a_1, b_1] \times [a_2, b_2] \times [a_3, b_3] \times S^3$$ given by $$\gamma(t) = (t, t^2, t^3, \frac{\sqrt{2}}{2}t, \frac{\sqrt{2}}{2}\sqrt{1-t^2}, \frac{\sqrt{2}}{2}t, \frac{\sqrt{2}}{2}\sqrt{1-t^2})$$.

Then $$dM =0 + 0 + 0 + 2dw_1 \wedge dw_1 - 2dw_2 \wedge dw_2 + 2dw_3 \wedge dw_3 - 2dw_4 \wedge dw_4 = 0$$ so $M$ is closed. Hence, the 1-form would have a potential function, and a potential function would be $$f(x_1, x_2, x_3, w_1, w_2, w_3, w_4) = x_1 + x_2 + x_3 + w_1^2 - w_2^2 + w_3^2 - w_4^2$$ with $A = (0, 0, 0, 0, 0, \frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2})$ and $B = (1, 1, 1, \frac{\sqrt{2}}{2}, 0, \frac{\sqrt{2}}{2}, 0)$. So, $$f(B) - f(A) = (1+1+1+\frac{1}{2}-0+\frac{1}{2}-0) - (0+0+0+0-\frac{1}{2}+0-\frac{1}{2}) = 5\ J$$ would be the value of the generalized work integral.

This Mathematica script computes the line integral both ways using the The Fundamental Theorem of Calculus for Line Integrals (FTC for LI) using the DifferentialForms.m package (The HomotopyOperator function is not functioning properly with this script for some reason, so I had to compute the antiderivative by hand).

Jeffrey Rolland
  • 937