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Question from Barry:

If I had 14 tons of inch and one half sized aggregate rock, how large of a square or rectangular shaped hole would I need to hold that amount?

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,
Penny

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