Percy, DF

University of Salford Institutional Repository

Urban Maths: Car Park Mayhem∗Several years ago, my employer introduced an exit barrier on its main car park. In order to leave, driverswind down their windows and insert tokens or swipe passes. One cold wintry day, I burst out laughingon the way to retrieve my car, as some imaginative local youths stood by the barrier with a pyramid ofsnowballs. I was still laughing as I went through the ritual of winding down my window and receiving anicy wallop in the mush. Perhaps I would have received more facial wallops if I were to have suggested thata cannonball-like, face-centred cubic lattice would be more efficient for packing snowballs than would theirbody-centred cubic lattice. On the other hand, I might have escaped scot free if I were to have suggestedthat these happy teenagers charge some token protection money. But that’s all by the by. About the sametime, car park bays were re-painted in the usual rectangular pattern shown in Figure 1.         Figure 1: rectangular pattern.However, this traditional conformity set me thinking. First of all, the car park included some dead endswhereby the grid in Figure 1 continued to the boundary fence. How mad: cars need to reverse against theflow in order to escape the bedlam. One solution is to leave a free lane for traffic to move adjacent andparallel to the fence, rather like aisles along the ends of pews in churches. Another would be to move thislane slightly away from the fence so that a single row of parking bays could be placed along the fence. Thesepossibilities are illustrated in Figures 2(a) and 2(b) respectively, and the latter clearly packs more spacesinto the bounded region. That is not all though. There was no flow direction specified and vehicles cantravel in both directions. The problem with this is that the separation between columns of parking bays iswider than it would be for one-way flow, so this was a missed opportunity to fit more bays into this busycar park.      Figure 2: (a) boundary with lane; (b) boundary with bays.Now the challenge becomes more interesting. A nearby hospital has parking spaces that form a herring-bone pattern as illustrated in Figure 3(a). This pattern is attractive and appears in such things as woollen∗or Mayhematics as Professor Tony Mann’s letter in June’s issue might suggest1scarves, tweed jackets, parquet flooring and masonry. It can offer denser packing of parking spaces than therectangular pattern for small car parks. However, it is generally impractical for larger car parks becausetraffic cannot flow in opposite directions along adjacent aisles unless vehicles nose into some bays and reverseinto others, which is a recipe for disaster. A good solution for avoiding this problem is to consider a diagonalpattern as illustrated in Figure 3(b). The question of interest is whether a diagonal layout is more efficientthan a unidirectional rectangular layout.            Figure 3: (a) herringbone pattern; (b) diagonal pattern.The advantages of a diagonal layout over a rectangular layout are that the column width is less, because ofthe angled bays, and the gap between columns is less, because of the reduced turning circle arc. Conversely,the column length is substantially greater, so which arrangement is better? To answer this question, weformulate the problem using three parameters (bay length l, bay width w, turning circle radius r) and onevariable (bay angle θ), as displayed in Figure 4. In time-honoured fashion, we assume an infinite car park intwo dimensions. l wr θ Figure 4: bay angle optimisation.The total space required for each bay roughly corresponds to the area enclosed by a parallelogram, asillustrated in yellow within Figure 4. We specify that the car must be travelling parallel to the painted linesegments on entry to its bay, which allows it to cut the corner of the neighbouring bay. This is reasonable,as the bay width is already generous and the turning circle radius is considerably larger than this. If eithercondition were violated, we could instead require the car to be travelling parallel on reaching the start of itsneighbouring bay.Using simple trigonometry, the vertical distance (parallelogram base) occupied by each bay is w/ sin θ.Similarly, the horizontal distance (parallelogram height) occupied by each bay is found by summing threecomponents to give r(1−cos θ)+ l sin θ+(w cos θ)/2. The total area required for each bay is then the productof base times height, which isA =wsin θ{r(1− cos θ) + l sin θ + w cos θ2}= w{l + r csc θ +(w2− r)cot θ}. (1)2In order to minimise this area, we differentiate Equation (1) with respect to θ to givedAdθ= −wr csc θ cot θ − w(w2− r)csc2 θ= −w csc θ{r cot θ +(w2− r)csc θ}. (2)Setting the derivative in Equation (2) to zero for θ ∈ [arctan(w/2l), π/2] givesθ̂ = arccos(1− w2r). (3)This implies that the optimal angle is independent of the bay length, though the bay width and turningcircle are important. Although we should show that the second derivative is positive to deduce that thisturning point is a minimum, we rely upon a graph for the purposes of this article.Now let’s be specific. A Rolls-Royce Phantom (and why not?) is 5.84 metres long and 1.99 metres wide,with a turning circle of about 14 metres. For illustration, we round these parameters up to l = 6m, w = 3mand r = 8m to allow for pedestrian access, collision avoidance, door opening and axle track. Although theowner of such a fine vehicle is more likely to order home delivery than jostle for a supermarket parking space,this convenient template ensures that our bays accommodate almost all cars. Substituting these values of wand r into Equation (3) gives θ̂ ≈ 0.62 radians, which corresponds to about 36◦. Interestingly, Figure 5(a)displays a contour graph of θ̂ (degrees) against w and r, from which it is clear that the optimal angle is fairlyrobust against measurement errors for the bay width and turning circle radius. 30 35 40 45  10   9   8   7   6 r 2    2.5          3             3.5      4 w    45    40    35    30   20            40        60    80 θ A Figure 5: (a) contour graph of θ̂ against w and r; (b) graph of A as a function of θ.From Equation (1), this angle requires an area of about 32m2 per bay. For comparison, the equivalentarea for a rectangular pattern, corresponding to θ = π/2 radians, is 42m2. Hence, the optimal diagonalpattern compares favourably and represents an efficiency saving of about 24%. Figure 5(b) displays a graphof area as a function of angle (degrees), to illustrate this relationship in more detail. Ignoring our assumptionof infinite space, this analysis suggests that a car park accommodating 500 vehicles in a rectangular gridwould accommodate 619 vehicles in an optimal diagonal grid. The corresponding area for a 45◦ diagonalpattern as shown in Figure 3(b) is about 32 12m2, which is similar to the area for the optimal diagonal patternand would still accommodate 613 cars (23% efficiency saving). However, a grid using a 45◦ angle might beconsiderably easier to paint accurately than a grid using a 36◦ angle.Perhaps there is some analogy between this optimal parking bay problem and the recent use of beautifuldiagrids in architecture, which help to reduce the total mass of steel beams required. In any case, thistheory could also be adapted to address more important applications in finite space, such as the design ofterminal gates at airports and the allocation of yachting berths in marinas. Now to tackle the equivalentthree-dimensional problem ready for when future commuters travel in Chitty-Chitty-Bang-Bang style flyingcars . . .A. Townie3

Urban maths : car park mayhem [writing as A. Townie]

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Urban maths : car park mayhem [writing as A. Townie]

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