The amount of freight that can be fit on a vehicle depends on the vehicle's weight capacity and volume capacity. In this paper mathematical equations are developed for evaluating the impact of weight capacity and volume capacity on total vehicle-miles. It is shown that the number of vehicle loads needed to carry a large amount of material is minimized when all vehicles are filled to the same capacity constraint. This is accomplished by mixing light items with heavy items in vehicle loads. Following this policy can reduce the number of vehicle loads and vehicle-miles. Under ideal circumstances, the reduction can be as large as 50 percent. Simple equations are provided for estimating the potential reduction in vehicle loads and vehicle-miles to be realized.

The cost of transporting a large quantity of items from one location to another depends on the number of vehicle loads required to carry the material and the distance traveled per vehicle load. Decreasing either the number of loads or the distance traveled per load reduces total vehicle-miles (the total distance traveled by all vehicles) and the cost of transporting the material. The number of vehicle loads depends on the quantity of items that can be fit on a vehicle. Typically, this quantity is determined by dividing the" capacity" of the vehicle by the" size" of each item. However, vehicle capacity and item size can be measured in more than one way. Most vehicles have both a weight capacity and a volume capacity. The vehicle is full when either capacity is reached. Depending on the type of items carried, some vehicles might be filled to the weight capacity, and others might be filled to the volume capacity (Figure 1)•