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Hi - I am trying to calculate the volume of dirt that would be removed if you dug a "pit" that is 380 acres across at the top, is square in shape, is 10 feet deep, and has sloping sides down at 1:4.  I also need to calculate volume for other configuations, and am trying to come up with a basic equation to stick all the info into to compare several different scenarios.  Basically, it's almost like an upside down pyramid, cut off after the first 10 feet, so it never comes to a point (is that clear as mud?)

Thanks SO much for your help!

Answer
Hi Jill,

I think I understand your description here.  However, I am a little confused by the 1 : 4 downward slope.  I'm assuming that you mean 1 unit down for every 4 units in.

Also, the "380 acres across" bothers me.  An acre is a measurement of area, equal to about 43,560 sq. ft.  It's about 208 ft on a side.  So, do you mean that the length of a side of the top square is 380 * 208 = 79040 ft?

If so, then the top base is a square, 79040 ft on a side.
If the slope is 1 ft down and 4 ft in, then in a 10 ft drop, it would go 40 ft in, and the sides of the bottom square base would be 79040 - 40 = 79000 ft.



First of all, I don't know of any generalized equation that would deal with different type of shapes, short of some applied calculus.  In any case, what is formed here is called a frustum of a pyramid.  The formula for the volume is:

   V = 1/3 * h * (A1 + A2 + sqrt(A1*A2))

where h = the vertical height
     A1 = area of bottom base
     A2 = area of top base

Okay, then, the calculations I made were as follows:

A1 = 79000^2 = 6,241,000,000 sq ft

A2 = 79040^2 = 6,247,321,600 sq ft

h = 10

So, using the formula,

 V = (10/3) * [6,241,000,000 + 6,247,321,600 +
               sqrt(6,241,000,000 * 6,247,321,600)]

   = (10/3) * [18,732,481,600]

   = 624,416,053,300  cu ft

   = 2,312652,049 cu yds.  ( a lot of mud!)

I sure hope this is something like what you wanted.  It's the best I could do with my understanding of the situation.

If it's not the correct set-up, let me know

Steve  

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Steve Holleran

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I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

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33 years teaching experience in NJ public schools

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B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

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