Question 4.2: Using the same data given for Example 4.1, let us compute th...

We first compute the parking width (PW). Recall that PW =\mathrm{SW} / sine \theta. That is, PW = 8.5 / sine 60=9.8^{\prime} . In reference to Figure 4.4, we next compute the value of y, which represents the distance lost due to parking at an angle.

By definition, y= SD \operatorname{cosine} \theta=16 \operatorname{cosine} 60=8^{\prime} . Since the lot width is equal to 180^{\prime}, we have

Module depth =y+( no. of stalls \times PW )=180,

which yields number of stalls =17.55 . That is, we can place 17 standard car stalls along the module depth, assuming a parking angle of 60^{\circ}. Using a 90^{\circ} parking angle in Example 4.1, we had obtained 180 / 8.5=21 cars. Hence, we reduced the module width by about 14^{\prime} but lost 4 \times 2=8 cars per module in the process. Also, parking at angles other than 90^{\circ} almost always requires the aisles to be one-way. The parking angle that will maximize the number of cars parked, in general, depends on the dimensions of the lot and how the individual modules are arranged within the lot. With increasing demand for parking spaces in virtually every major city and airport around the world—not to mention university campusesparking lot design and management continues to be a topic of interest.