[Math] How to calculate the volume of a skip bin container knowing the height of the material inside

definite integralsfunctionsgeometryvolume

I need to know hot to calculate the volume of a skip bin (also known as a skip container or dumpster in some areas) with varying length and width.

It seems like a isosceles trapezoid when you look at it from the sides or the front. Like this:

A skip container

The width of the skip varies only slightly, but the length below and above vary by about a yard.

Or like this:

A skip container

The point is that in my work we put some semi-solids on it, and we use some aproximation to calculate the volume.
I'd like to know how to calculate the formula for exact value of the volume of liquid (i.e.) inside, but I keep thinking that it has something to do with integrals, infinitesimal sum, or something like it that exceeds my knowledge.

Anyone cares to help me, please?

Best Answer

You certainly could use integration, but that would be tediously unnecessary.

Thankfully, some simple geometry should suffice once you have some measurements:

Find the area of the trapezoid face (left side on your picture) through the following formula:

Then multiply that area by the "width" of the bin to find the total volume.

Conceptually, you could think of it as stacking up a bunch of those trapezoidal areas.

^{_{The picture of the formula is from a Google search for "area of trapezoid"}}

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[Math] Volume of n-dimensional solid w/ n-1 dimensional simplex as a base

I struggled with this one, but it finally became clear. I hope this answer helps someone else along the way.

The comment around the original MathForum question that suggested that one could calculate the volume of the kinds of solids described above by taking the hyperarea of the base and multiplying it by the average of the heights, while technically on the mark, its wording sent me down the wrong road.

To find the volume of these kinds of solids one needs to multiply the volume formula for a regular simplex (also shown in the original post):

enter image description here

times the average of the heights of the cap.

Calculating a solid with 3 heights to the cap will need a 2 dimensional regular simplex as a base (an equilateral triangle). In this case the formula for the volume of a regular simplex will give the triangle's area, hence the confusion about hyperarea.

When calculating for heights of n dimensions, the base simplex will have n-1 dimensions.

So, a little Mathematica function will do the trick:

volSimplexWithHyperCap[heights_List] := Module[{n, simplexVolume},
  n = Length[heights] - 1;
  simplexVolume = Sqrt[n + 1]/(n! * Sqrt[2^(n)]);
  N[simplexVolume * Mean[heights]]]

volSimplexWithHyperCap[3, {1, 1, 1}]

0.433013

This corresponds correctly to the prism volume in the question. It also, readily extends to higher dimensions:

volSimplexWithHyperCap[{1, 3, 2, 3, 5}]
0.0652186

This has the advantage over integration of being relatively simple and very fast to calculate. Not a Cholesky decomposition solution like I imagined at first, but it looks serviceable.

Maybe @MvG will think of a proof.

Thanks to everyone for their patients with all the edits.

[Math] How to calculate the height of a cone at particular volume

$$V=\frac{1}{3}\pi r^2h$$

$$3V=\pi r^2h$$

$$\frac{3V}{\pi r^2}=h$$

Best Answer

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[Math] Volume of n-dimensional solid w/ n-1 dimensional simplex as a base

[Math] How to calculate the height of a cone at particular volume

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