Talk:Frustum

Is a frustrum a polyhedron?

Should a frustum be classified as a polyhedron?

If a truncated cone is a frustum, then not all frusta are polyhedra.

Am I correct? Please discuss. --Comment by 128.61.118.174

I don't know any specific policy on category usage. I'd think it can make sense to include a category even if it doesn't apply to all forms or definitions. Tom Ruen 21:04, 10 January 2007 (UTC)

Yes, in my opinion you're right. Or maybe the conical frustrum could be considered an infinite-faced polhyedron XD

A mathematician and linguist replies: A polyhedron is a geometric object with flat faces and straight edges (the definition at polyhedron), so yes, a frustum of a pyramid is definitely a polyhedron. The frustum of a cone is therefore definitely not a polyhedron, as its surfaces are not all planar. (Mathematically speaking, the limit as n tends to infinity of a polyhedron with n faces is not itself a polyhedron as it does not fit the definition of a polyhedron. Similarly, a circle can be considered to be the limit as n tends to infinity of a regular n-gon, but that does not make it a polygon.) — Paul G 12:10, 17 May 2007 (UTC)

The reply above is correct, but since some frustums are polyhedra, this article can be properly listed in the category "polyhedron".

Calculations missing

I think that some of the calculations on the page are missing. --Proficient 07:10, 8 April 2007 (UTC)

• Presumably this comment is moot now. --Jorge Stolfi (talk) 00:08, 28 December 2009 (UTC)

Frustum or frustrum?

My Maths teacher said it is a frustrum and not a frustum was this just misprnounciation on her part or can it actually be called a frustrum

A mathematician and linguist replies: The spelling frustrum is erroneous, according to the Oxford English Dictionary, and they know what they are talking about. This misspelling probably comes about because the other word beginning "frust..." that everyone knows is "frustrate". So, anonymous poster, I'm afraid your maths teacher was incorrect. Teachers can be wrong too, alas. — Paul G 12:10, 17 May 2007 (UTC)

I suppose that frusta must be commonly perceived to be very frustrating polyhedra. Double sharp (talk) 15:07, 8 November 2013 (UTC)

Incorrect formulas for areas

198.28.92.5 07:43, 24 May 2007 (UTC) David Ashby:

Warning: I think that the equation for surface area may be incorrectly given as:

A = π(R1S1 − R2S2)

Another website gives

A = pi * S * (R1+R2)

This makes more sense. S1 and S2 are not defined in the current wikepedia article, only S.

If I get time, I'll come back and definitively answer this, otherwise perhaps someone else can?

• WARNING: the equations for a circular frustum are wrong. Marcbark (talk) 22:42, 27 January 2009 (UTC) Marc
  • Hopefully these concerns have been fixed now. --Jorge Stolfi (talk) 00:08, 28 December 2009 (UTC)

Surface area

I see there is a formula for volume on the article. What about the surface area? 129.78.220.7 (talk) 05:33, 13 April 2009 (UTC)

• An anonymous user added this formula

The surface area is:

$SA=(B_{1}+B_{2})+{\sqrt {\frac {((P_{1}+P_{2})(r_{1}-r_{2}))^{2}+(h(P_{1}+P_{2}))^{2}}{4}}}$

where $B_{1}$ and $B_{2}$ are the base areas

where $P_{1}$ and $P_{2}$ are the base perimeters

where $h$ is the frustum height

where $r_{1}$ and $r_{2}$ are the inradius of the two bases

This formula cannot hold for arbitrary frustums. As a minimum the frustum must be right. Moreover, AFAIK, one can make two frustums with the same base perimeter, area, and inradius, but with different surface areas. Therefore the author probably was thinking of regular polygons. But for that case a separate and simpler formula was given, so this formula seems redundant. Is that so? --Jorge Stolfi (talk) 00:00, 28 December 2009 (UTC)

Text and image don't agree.

The article explicitly defines B sub 1 to be the smaller base and B sub 2 to be the larger in the section Formulae > Volume. However, in the image 'frustum with symbols.svg' has these two bases labeled in the opposite way. AnandaDaldal (talk) 21:02, 23 February 2010 (UTC)

Text and image don't agree

The triangular frustum in the image can't be reconciled with the caption, "A regular octahedron can be augmented on 3 faces to create a triangular frustum." There is no cut through a regular octahedron that yields anything like the image. If further explanation is required, then it isn't a clear, simple example of a frustum any longer and should be removed from the article.

pi=.....

Is it really so pathetic that we need to write the value of pi?(after linking it to the article 'pi') Just below the formula for volume of conical frustum.Gauravjuvekar (talk) 16:08, 6 January 2012 (UTC)

Quite. It is unlikely that someone reading this article and able to comprehend the formulae is unaware of the approximate value of π. I removed it. ehn (talk) 05:27, 26 May 2020 (UTC)

latin

Frustum means "piece cut off" in Latin. Please add the same.Gauravjuvekar (talk) 16:16, 6 January 2012 (UTC)

Is there and bounded volume that is NOT a frustum?

According to the article, "a frustum is the portion of a solid that lies between one or two parallel planes cutting it." Now, any bounded volume can be constructed this way, meaning that any bounded volume is a frustum? Is this correct, or have I missed something? So basically, frustra ares just the set of volumes that either are bounded, or for which there is some vector v such that the scalar product between the position of each point in the volume and v is bounded (this include all bounded volumes as well)? —Kri (talk) 05:32, 20 January 2022 (UTC)

It's definitely not true that every bounded volume, or even every finite polyhedron, has two parallel faces. —Tamfang (talk) 00:27, 30 October 2023 (UTC)

Potential error in the surface area of a right polygonal frustum?

Is there a source for the computation of the surface area of the right, polygonal frustum? I tried studying a limiting case, and it seems incompatible with the formula presented here. I am considering a regular hexagon with side length L, so its area is 3*sqrt(3)*L^2/2. For a frustum with side lengths L and 0 and perpendicular height 0, we expect its total surface area to be twice that of the regular hexagon, 3*sqrt(3)*L^2. This is not true with the formula as written. This can be recovered if the secant^2 is replaced with a cotangent^2 under the square root. Am I wrong in my thoughts on the limit? 145.101.40.28 (talk) 12:30, 23 October 2023 (UTC)

I have computed the lateral surface area myself and find that the cotangent appears in the final formula, not the secant. Here is a link to my proof. I will edit the formula accordingly. 145.101.40.28 (talk) 14:03, 23 October 2023 (UTC)

And I have a final argument. The side length and circumscribing radius are related via a = 2*r*sin(pi/n). The formula with the secant does not reduce to the circular case as n tends to infinity, but it does with the cotangent. 145.101.40.28 (talk) 14:42, 23 October 2023 (UTC)