Car movement - differential geometry interpretation. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:36:47Z http://mathoverflow.net/feeds/question/66578 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66578/car-movement-differential-geometry-interpretation Car movement - differential geometry interpretation. Beni Bogosel 2011-05-31T17:58:43Z 2011-05-31T18:35:11Z <p>I've posted this on Math Stack Exchange and I didn't get any answer in a couple of days, so I'll try and post it here too.</p> <p>The <em>problem</em> presented below is from my differential geometry course. The initial reference is Nelson, <em>Tensor Analysis</em> 1967. The car is modelled as follows: </p> <p><img src="http://mathproblems123.files.wordpress.com/2011/05/car.png" alt="Image"></p> <p>Denote by $C(x,y)$ the center of the back wheel line, $\theta$ the angle of the direction of the car with the horizontal direction, $\phi$ the angle made by the front wheels with the direction of the car and $L$ the length of the car.</p> <p>The possible movements of the car are denoted as follows:</p> <ul> <li>steering: $S=\displaystyle\frac{\partial}{\partial \phi}$;</li> <li>drive: $D=\displaystyle\cos \theta \frac{\partial}{\partial x}+\sin\theta \frac{\partial}{\partial y}+\frac{\tan \phi}{L}\frac{\partial}{\partial \theta}$;</li> <li>rotation: $R=[S,D]=\displaystyle\frac{1}{L\cos^2 \phi}\frac{\partial }{\partial \theta}$;</li> <li>translation: $T=[R,D]=\displaystyle\frac{\cos \theta}{L\cos^2 \phi}\frac{\partial}{\partial y}-\frac{\sin\theta}{L\cos^2\phi}\frac{\partial}{\partial x}$</li> </ul> <p>Where $[X,Y]=XY-YX$ (I can't remember the English word now). All these transformations seem very logical. My question is:</p> <blockquote> <p>How can we justify the mathematical interpretation made above, especially the part with the rotations and translations?</p> </blockquote> <p>The interpretations are quite interesting: </p> <ul> <li>from the expression of $D$, when the car is shorter, you can change the orientation of the car very easily, but when it is longer, like a truck, you it is not that easy ( see the term with $\frac{\partial}{\partial \theta}$)</li> <li>the rotation is faster for smaller cars, and for greater steering angle</li> <li>translation is easier for smaller cars.</li> </ul> http://mathoverflow.net/questions/66578/car-movement-differential-geometry-interpretation/66583#66583 Answer by jvkersch for Car movement - differential geometry interpretation. jvkersch 2011-05-31T18:35:11Z 2011-05-31T18:35:11Z <p>I'm not sure what you mean by "justify", but one way of looking at this is through the Chow theorem. </p> <p>You've calculated that $R = [S, D]$. This is a statement about vector fields, which are infinitesimal flows. So if you steer by a small amount, followed by driving over a small amount, followed by turning the wheel back into the original position and driving backward, you end up performing a small rotation. You can interpret $T = [R, D]$ in the same way: if you want to parallel park the car (performing a translation), you turn ($R$), drive forward ($D$), turn back ($-R$) and drive backward ($-D$).</p> <p>A normal car has four degrees of freedom, the angles $\theta$ and $\phi$ on your figure and the coordinates $x, y$ of the center of mass. However, you as the driver can only be drive and steer the car, you can't just translate the car or rotate it around its axis. So you can only control $S$ and $D$. We say that this system is underactuated, meaning that you have less control inputs than there are degrees of freedom in the system.</p> <p>However, differential geometry comes to the rescue: if you consider the distribution spanned by $S$ and $D$, you'll find that it is totally nonholonomic, meaning that you can create any possible motion (any linear combination of $R$, $D$, $T$ and $S$) by a linear combination of $S$ and $D$ and their (iterated) commutators. Physically, this means that you can perform any motion with the car by driving and steering, although you will sometimes have to execute a nontrivial motion in $S$ and $D$ (look up "control by sinusoids") </p> <p>If now you have a metric on the distribution spanned by $S$ and $D$ (telling you, for instance, how much it costs to steer resp. drive the car) you have an instance of what is called a sub-Riemannian geometry. (I hope this will entice Richard Montgomery to comment on this question).</p> <p>I can polish/extend this answer a great deal, so please feel free to ask follow-up questions.</p>