|
|||||||||||||||
|
|||||||||||||||
| |||||||||||||||
Hi Barry, The answer depends on the density of your aggregate rock. I looked on SImetric site and found that the density of crushed rock is 100 pounds per cubic foot. If that is a reasonable estimate for your aggregate rock then 14 tons is about $14 \times 2000 \times 100 = 2,800,000$ cubic feet. If the hole is a cube then each side would measure \[\sqrt[3]{2,800,000} = 141 \mbox{ feet.}\] For a rectangular hole it can be $L$ feet long, $W$ feet wide and $H$ feet deep as long as $L\times W \times H = 2,800,000$ cubic feet. I hope this helps, | |||||||||||||||
|
|||||||||||||||
Math Central is supported by the University of Regina and the Imperial Oil Foundation. |