A truncated cone (used in the strict mathematical sense), often used to describe a volume of space in computer graphics, particularly in terms of a view volume (i.e. the camera frustum) or a shadow volume.

The formula for the area of the curved part of a frustum is disproportionately hard to derive. But, here in this humble node, I shall attempt to demystify the process of the formula's derivation, for those two or three of you out there who actually care.

Those of you who recently had geometry will remember that the surface area of the curved part of a cone is πrl, where r is the radius of the base of the cone, and l is the slant height of the cone, or the length from the vertex of the cone to a point on the edge of the base. So, to find the surface area of the frustum, all you have to do is subtract the areas of the two cones that make up the top and bottom edges of the frustum, right? So, let's suppose the little and big radii of the frustum are r1 and r2, and the slant height of the frustum is l, and the slant height of the cone that you're subtracting is l1. Then, the surface area of the frustum would be

S = πr2(l1 + l) - πr1l1

S = π(r2l1 + r2l - r1l1)

Now, since the two cones that we're working with are similar, their respective slant heights and radii are proportionate; in other words,

(l1 + l)/r2 = l1/r1

From here, we isolate and get rid of l1.

r1(l1 + l)/r2 = l1

r1l1/r2 + r1l/r2 = l1

r1l/r2 = l1 - r1l1/r2

r1l/r2 = l1(1 - r1/r2)

l1 = (r1l/r2)/(1 - r1/r2) = l{(r1/r2)/[(r2 - r1)/r2]}

l1 = l[r1/(r2 - r1)]

Substituting this into our surface equation, we get

S = π({(r2 - r1)l[r1/(r2 - r1)]} + r2l)

S = π(r1l + r2l)

S = πl(r1 + r2)

And there it is: the formula for the curved surface area of a frustum. Note that the slant height of the subtracted cone is not in the final equation; this is a good thing. For calculus purposes, if you only have one radius to work with (i.e. the frustum is infinitessimally thin) the two radii can be combined as an average (the radius at the center of the frustum):

r = (r1 + r2)/2

S = 2πlr

Now, should you need to know the surface area of the entire frustum, you can just tack on the areas of the top and bottom circles:

S = πl(r1 + r2) + πr12 + πr22

Mommy, my fingers hurt from typing all these angle brackets!

Second in a series of MathRants by PMDBoi.

Node math, people!

NB: this is much easier to read in the printable version. Serifs are good.

The formula for the volume of a frustum is considerably easier to derive, and I'm surprised it wasn't here already. If it is, sorry, but I couldn't find it! It's actually more useful than you'd think; for example, I used it recently to determine the mass of a pile of marbles with only a ruler, two styrofoam cups, and some water. So if you're stranded in the desert with only two cups and some water, you can use this formula to accurately find the mass of small objects!

First of all, you need to know the formula for the volume of a cone.

V = 1/3 πr²h

Now, the easiest (and probably only) way to find the volume of a frustum is to subtract the volume of the missing cone from the volume of the original cone. Consider this horrid ASCII art of a frustum, seen from the side:

           ^               -
          /|\              |
         / | \             |
        /  |  \            |
       /   |   \           | missing cone; height = x
      /    |    \          |
     /     |     \         |
    /______.______\        -
   /       |  r2   \       |
  /        |        \      | frustum; height = h
 /         |         \     |
/__________.__________\    -

As you can see, I've put the missing cone back and drawn a line from the top of the full cone to the center of the base. Glancing quickly at these and applying the cone volume formula, the following are evident:

VB = 1/3 πr1²(h + x)
VS = 1/3 πr2²x

...where B is for the Big cone and S is for the Small cone. The volume of our frustum VF is equal to VB - VS, so we can subtract the bottom equation from the top equation and obtain:

VF = VB - VS = 1/3 πr1²(h + x) - 1/3 πr2²x
VF = 1/3 π * (r1²(h + x) - r2²x)

Now what? Obviously we need to do something with x, and this seems to be what throws most people. If you ignore the left part of the cone, you have two triangles that have two identical angles and are thus similar.

  |  \
x |   \
  |    \
  |_    \
- |_|____\
  |  r2   \
h |        \
  |_        \

The angle marked with a star and the two right angles are the identical ones. Now, finding x is a simple matter of proportions:

x / r2 = (x + h) / r1
x * r1 = (x + h) * r2
x r1 = x r2 + h r2
x r1 - x r2 = h r2
x = h r2 / (r1 - r2)

We now have an expression for x in terms of the other three variables. Now it's basic algebra:

VF = 1/3 π * (r1²(h + x) - r2²x)
VF = 1/3 π * (r1² h + r1² x - r2² x)
VF = 1/3 π * (r1² h + (r1² - r2²) * x)
VF = 1/3 π * (r1² h + (r1² - r2²) * h r2 / (r1 - r2))
VF = 1/3 πh * (r1² + (r1 - r2)(r1 + r2) * r2 / (r1 - r2))
VF = 1/3 πh * (r1² + (r1 + r2) * r2)
VF = 1/3 πh * (r1² + r1 r2 + r2²)

Which is, as not stated at the top of this writeup so you'd have to read the whole thing to find it, the formula for the volume of a frustum with height h and radii r1 and r2.


This has been another overly detailed formula derivation/explanation by your friendly neighborhood Eevee.

A frustrum is a truncated pyramid, one whose top has been cut off by a plane parallel to its base. This concept is often used in 3D graphics programming to describe the viewing volume for perspective projection. Perspective projection is characterized by foreshortening; that is the further an object is from the observer (or camera), the smaller it appears in the final image. Objects that fall within the viewing volume are projected towards the apex of the pyramid, which is the location of the observer. The viewing plane is the top of the frustrum, and the base of the frustrum is the furthest visible plane. All objects visible within the scene are within the frustrum. Objects closest to the viewpoint appear larger because they occupy a larger amount of the volume than those further away. Objects that overlap the boundaries of the frustrum are clipped, and only the parts that fall within the frustrum are rendered. A frustrum is defined by a pair of (x, y, z) coordinates for the front lower left and rear upper right corners, with distance from the observer on the z-axis.

A frustrum need not be symmetrical, nor aligned with the z-axis. This can be used to create false perspective.

Frus"tum (?), n.; pl. L. Frusta (#), E. Frustums (#). [L. fruslum piece, bit.]

1. Geom.

The part of a solid next the base, formed by cutting off the, top; or the part of any solid, as of a cone, pyramid, etc., between two planes, which may be either parallel or inclined to each other.

2. Arch.

One of the drums of the shaft of a column.


© Webster 1913.

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