It turns out that parallel parking is surprisingly simple.

In theory.

I first encountered this as a homework problem in general relativity. Graduate level.

Since I believe it is still assigned as a homework problem, I will merely highlight the key points, and not provide the solution.

Consider the infinite real two dimensional plane (or a Mall parking lot, after closing).

You are in a vehicle, which can be idealised as a rectangle, width a, length b, with four wheels. For simplicity, assume only the front wheels turn, on a rigid axle, and the rear wheels provide traction displacing the vehicle.

Then the entire degree of freedom of your vehicle is well described by two differential operators:

R, the infinitesimal rotation of the front wheels, and

D, the infinitesimal displacement of the wheels, translating the vehicle.

Pick your own sign convention.

Clearly, R and D do not commute: “RD” does not lead to the same effect as “DR”

Equally clearly, this is adequate to cover the entire plane, through repeated operation of R and D in some succession.

There are a number of interesting properties, notably the vehicle is not oriented on the plane, and by assumption R and D can have either sign.

Now consider the “parallel parking problem”:

b

———-

| |

| | a

———-

———— _______________

| |

| | W

——————

L

Through a succession of R, and D operations, the vehicle can in general be translated into the “space” limited by W >= a, and L >= b

Clearly there is some minimum L=L_{min} for which this is possible.

Calculating L_{min}, is left as an exercise for the reader.

For any L’ > L_{min}, parallel parking the vehicle is trivial, requiring merely some algorithmically determined sequence of R and D operations, depending on the initial conditions only.

QED.