How to calculate the volume of a triangular prism where both triangles are irregular.

Question

I am trying to calculate the volume of a triangular prism where both triangles are a different irregular triangles, above shows the regular version. The shape I have will have varying lengths for a,b,c for the base and the top (so base and top will be different triangles) but h will be constant (ie no slope on the shape). I have so far been unsuccessful in finding an equation for this so any help is much appreciated. I am happy to clarify if my post is unclear.

I have just found the following equation online that uses a1,a2,a3, b1,b2,b3 a being the base and b being the top, as stated though im unfamiliar with this area of mathematics so if anyone could tell me whether the equation below looks correct it would help me a lot, thanks.

Answer 1

The volume is not uniquely determined by the sides of the bases and the distance between the basal planes. The twisting of one base relative to the other can change the volume. We prove this using a simple case.

Assume your bases are both equilateral triangles with edge length $1$ and separation $\sqrt{2/3}$. If the bases are aligned to form a proper triangular prism, then you can calculate the volume as $\sqrt2/4$. But if we twist one base by $60°$ and thus generate a regular octahedron with unit edges, the volume has grown to $\sqrt2/3$.

To see what happens geometrically, cut the figure with a plane parallel to and halfway between the two bases you designated. With the triangular prism the resulting cross-section is another equilateral triangle, whereas with the regular octahedron this cross-section is instead a regular hexagon. Both of these cross-sections have the same perimeter, but that makes the hexagon have a bigger area (closer to being a circle). Thus twisting the bases without changing their internal geometry or their separation still changes the volume by changing the intervening cross-sections.

Answer 2

EDIT 1:

A proof is given along similar lines of the previous post, but using known result.

It is assumed that the scalene triangles formed by an intersecting plane parallel to base, so can be scaled as similar triangles contained between same rays.

Assuming the solid to be frustum of a tetrahedron ( no more a prism !) frustum height $h~$ with each side reduced by factor $\mu<1$ in geometrical similarity seen from a point as vertex concurrent of slant heights. Base height from untruncated tetrahedron vertex $H$ and for cut plane $ H-h$ from vertex.

We should be given/specified, the similarity scaling ratio between sides and height.

$$ \mu= \frac{H-h}{H} \tag1 $$

As is known when a point is connected to corners of any polygon/ cone to vertex forming a tetrahedron /polyhedron/or cone a coefficient $ \frac13 $ that comes into play by a simple formula... that can be proved by integration (omitted).

Full tetrahedron volume $$ \frac13 \cdot A\cdot H $$

where given base area is as usual $$ A=\sqrt{s(s-a)(s-b)(s-c)};$$

Scaled down tetrahedron volume is $$ \frac13 \cdot A\cdot \mu^2 (H-h) $$

The difference is the truncated frustum volume of tetrahedron

$$ V=\frac13 \cdot A\cdot H -\frac13 \cdot A\cdot \mu^2 (H-h) \tag2 $$

which simplifies to $$ \frac13 \cdot A\cdot h\cdot(1+\mu+\mu^2 )\tag3$$

When $\mu=1,$ the above result reduces to the prism case:

$$ V= A\cdot h $$

as it should.

To check for case of oblique cone:

$$ \mu=r/R, A= \pi R^2, ~ V \to \pi h (R^2+r^2+r R)/3 ; $$ same as 3).