I am referring to the Polyhedron Volume formula here:

$$V = \frac{\sum_{i=1}^k \vec{x_{i}} \cdot \hat{n_i} A_i}{3}$$

My question is, does the above formula applicable for the Polyhedron below?

The given faces for the polyhedron are:

$$f_1: n_1,n_2,n_3, n_4$$ $$f_2:n_2, n_3,n_8,n_5$$ $$f_3:n_1, n_4,n_6,n_7$$ $$f_4:n_6, n_8,n_5,n_7$$ $$f_5:n_1, n_2,n_5,n_7$$ $$f_6:n_6, n_8,n_3,n_4$$

Note that for my application, the face $f_6$ may or may not intersect with the face $f_5$. Why is why I would prefer to apply the above formula mechanically, instead of having to determine the intersected line, and then compute the volume separately.

If it does, what would be the value of the above calculation?

  1. $|v_1-v_2|$
  2. $|v_1|+|v_2|$
  3. None of the above?

Is there a proof showing why the answer is it?

Note: Additional discussion on the validity of the above formula is shown here.

Edit: Update the formula according to the wiki page.

  • $\begingroup$ Your picture is misleading because even if the faces $f_5$ and $f_6$ intersect, the lines, say, $n_1n_7$ and $n_4n_6$ do not have to intersect. $\endgroup$ – timur Dec 28 '10 at 14:36
  • $\begingroup$ @timur, how if we can assume that if $f_5$ and $f_6$ intersects, then the lines, such as $n_{1}n_{7}$ and $n_{4}n_{6}$ must intersect? Would that change the outcome? $\endgroup$ – Graviton Dec 28 '10 at 14:45
  • $\begingroup$ Can you explain what $x,y,z$ are? I am having trouble with the derivation in the Wiki page you linked to. $\endgroup$ – timur Dec 28 '10 at 14:58
  • $\begingroup$ @timur, @Ngu: The explanation you saw on Wikipedia was very confused and, as far as I could tell, incorrect. I've gone ahead and edited the article. Just so you know. $\endgroup$ – Rahul Dec 28 '10 at 16:58
  • $\begingroup$ @Rahul: Still not clear to me because $\vec x_i\cdot \hat n _i$ is not constant on any face in general. I would believe if $\vec x_i$ was the mass center of the $i$-th face. $\endgroup$ – timur Dec 28 '10 at 22:10

It helps to observe what the formula is saying geometrically. Each term in the sum represents the signed volume of a pyramid; the cone whose apex is the origin and whose base is the face under consideration. The volume is positive if the normal to the face is pointing away from the origin, otherwise it is negative.

The formula assumes that the polyhedron has been equipped with an outward normal. Thus, for example, if the polyhedron is convex and contains the origin, then the volume is simply given by the sum of the pyramids that compose it, all of which have positive volume. If the origin is not contained in the polyhedron, then all the pyramids with base on the far side of the polyhedron will have positive volume, and the negative volumes corresponding to the faces on the near side will cancel out that portion of the volume that represents parts of the far-side pyramids that are outside of the polyhedron.

To see that it works, it is probably easiest if you play with it yourself. Consider the simpler case of a closed curve in the plane. The same formula works for computing the enclosed area, except now there'll be a 2 instead of a 3 in the denominator of the terms: we're adding the signed areas of triangles.

So, what happens if your closed curve is not simple? Consider a figure 8. The problem now is that your consistently oriented normals can't always point outside. They will point inside one of the loops and outside of the other. The area computed will be the area of the latter minus the area of the former.

So the answer to your multiple choice question is: 3) none of the above because you have put absolute value signs on the other answers. Your computed volume will be $\pm (v_1 - v_2)$, where the sign depends on how you chose to orient the normals. As moron mentioned, you need to ensure that the normals are oriented consistently between adjacent faces (so if you unfolded them onto a common plane their normals would agree).


I would say yes, it is applicable. The formula basically tells you that the volume is the sum of volumes of the two "component" polyhedrons, doesn't it?

  • $\begingroup$ that's my intuitive feeling also. And my intuitive feeling says that the result would be $|v_1-v_2|$, not $|v_1+v_2|$. But I can't prove it. $\endgroup$ – Graviton Jan 3 '11 at 8:55
  • $\begingroup$ @Ngu: You have to choose the normals to get a consistent orientation, which is possible in your case. The page you linked to in the question, mentions it (and provides a link). $\endgroup$ – Aryabhata Jan 3 '11 at 8:57
  • $\begingroup$ so which is the answer: $|v_1-v_2|$ or $|v_1+v_2|$? $\endgroup$ – Graviton Jan 3 '11 at 9:13
  • $\begingroup$ @Ngu: If you choose the normals consistently it should give you either v1+v2, or -(v1+v2). Note: The orientation should be consistent over all the 8 faces... $\endgroup$ – Aryabhata Jan 3 '11 at 16:56

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