BLR-310 · Lesson 4 of 12 · 20 min

Truncated and eccentric cones

The Tamarack Pulp dust collector hopper is the biggest piece of plate work in the shop this winter. It is a truncated cone of 10 mm (3/8 in) plate, 2,400 mm (96 in) across at the top where it bolts to the collector and 600 mm (24 in) across at the bottom where the rotary valve goes, 1,500 mm (60 in) high. On the next bench, the Riverbend fan inlet needs an eccentric reducer, a cone from a 1,200 mm (48 in) duct down to an 800 mm (32 in) duct, flat along the bottom so that ash does not collect in it. The first is a radial-line job; the second is the first triangulation job of the winter. This lesson covers both: the truncated right cone, which is a full cone with its top cut off, and the eccentric cone, which is not a right cone at all and needs a different method.

The truncated right cone

A truncated cone, or frustum, is a right cone with its top removed by a plane parallel to the base. Its pattern is part of a ring: the space between two arcs swung from the same centre, the apex of the full cone, and bounded by two straight lines along elements. The trick is that the apex is not on the part. It is the point where the elements would meet if the cone were extended, and the first step of the development is to find it.

From the large radius Rb, the small radius Rs and the height h between them, the height of the full cone to its imaginary apex comes from similar triangles: H equals h times Rb divided by Rb minus Rs. The large pattern radius is the slant height of the full cone, the square root of Rb squared plus H squared. The small pattern radius is the slant height to the small end, which is the large pattern radius times Rs divided by Rb. The sector angle is the same as for the full cone, 360 degrees times Rb divided by the large pattern radius.

\[H = \frac{h\,R_b}{R_b - R_s}, \qquad L_b = \sqrt{R_b^{2} + H^{2}}, \qquad L_s = L_b \frac{R_s}{R_b}\]
On the left, a truncated cone in elevation with its sides extended upward by dashed lines to meet at an imaginary apex, and labels for the large and small ends. On the right, the flat pattern: a large part ring swung about the apex, a little more than half a circle, divided by two dashed lines into three equal pieces.
A truncated cone: extend the sides to the apex, then develop the part ring.

The hopper pattern

Kiera works out the hopper on the mean diameters, 2,410 mm (96 3/8 in) and 610 mm (24 3/8 in). To keep the numbers simple for the first pass, Joanne has her run the shop code on the round mean figures of 2,400 mm (96 in) and 600 mm (24 in), then repeat it with the exact mean values for the cutting list.

# Truncated and eccentric cones (metric)
import math
from math import gcd
MODE = "mm"

def frac(x, den=16):
    sign = "-" if x < 0 else ""
    n = round(abs(x) * den); whole, rem = divmod(n, den)
    if rem == 0:
        return f"{sign}{whole} in"
    g = gcd(rem, den)
    return f"{sign}{whole} {rem // g}/{den // g} in" if whole else f"{sign}{rem // g}/{den // g} in"

def show(x):
    x = 0.0 if abs(x) < 1e-6 else x
    return f"{x:,.0f} mm" if MODE == "mm" else frac(x)
# 1. concentric truncated cone: hopper
Db, Ds, h = 2400, 600, 1500   # mean large and small diameters, height between them
Rb, Rs = Db / 2, Ds / 2
Hfull = h * Rb / (Rb - Rs)     # height to the imaginary apex
Lb = math.hypot(Rb, Hfull)     # large pattern radius
Ls = Lb * Rs / Rb              # small pattern radius
ang = 360 * Rb / Lb
print("apex height", show(Hfull), " pattern radii", show(Lb), "and", show(Ls))
print(f"sector angle {ang:.1f} deg; in 3 pieces of {ang / 3:.1f} deg")
print("slant length of the frustum", show(Lb - Ls))

# 2. eccentric reducer, flat on the bottom: triangulate between 12 points on each end
DB2, DS2, Lg = 1206, 806, 900  # large and small mean diameters, length along the axis
n = 12
def pt(R, k, x):               # point k on an end circle; both circles touch the bottom line
    a = 2 * math.pi * k / n
    return (x, R * math.sin(a), R - R * math.cos(a))   # (axial, sideways, height above bottom)
print("solid (k to k) and dashed (k to k+1) true lengths, half the reducer:")
for k in range(n // 2 + 1):
    a, b = pt(DB2 / 2, k, 0), pt(DS2 / 2, k, Lg)
    solid = math.dist(a, b)
    line = f"  {k:2} solid {show(solid):>10}"
    if k < n // 2:
        dashed = math.dist(a, pt(DS2 / 2, k + 1, Lg))
        line += f"   dashed {k}-{k + 1}' {show(dashed):>10}"
    print(line)
Output
apex height 2,000 mm  pattern radii 2,332 mm and 583 mm
sector angle 185.2 deg; in 3 pieces of 61.7 deg
slant length of the frustum 1,749 mm
solid (k to k) and dashed (k to k+1) true lengths, half the reducer:
   0 solid     900 mm   dashed 0-1'     924 mm
   1 solid     906 mm   dashed 1-2'     909 mm
   2 solid     922 mm   dashed 2-3'     914 mm
   3 solid     943 mm   dashed 3-4'     935 mm
   4 solid     964 mm   dashed 4-5'     968 mm
   5 solid     979 mm   dashed 5-6'   1,001 mm
   6 solid     985 mm

The apex is 2,000 mm (80 in) above the large end, well above the part. The large pattern radius is more than 2.3 m (93 in), too long for any trammel in the shop, and the sector is more than half a circle, so the pattern is far too big to cut from one plate. The answer is to cut it in three equal pieces of about 62 degrees each, with three seams spaced around the hopper. Each piece fits on a standard plate.

Laying out an arc with a radius of 2.3 m (93 in) without a trammel that long is a common problem with truncated cones. Joanne's method is to lay the pattern out by coordinates: calculate points on both arcs at equal angles, measure them off from a base line with a tape and a square, and bend a batten through them. The other method is to make one piece accurately on thin sheet metal, check it, and use it as a template for all three. For this job she does both: Kiera lays out one piece by coordinates and the other two are traced from it. In lesson 11, the same points go to the CNC table as a file.

Flat cones and the chord method

When a cone is very flat, like a tank roof, the apex can be tens of metres away and even coordinates become awkward. The pattern is then developed from the chord and the rise of its arcs: the chord across each end of a piece and the height of the arc above that chord, its sagitta, which Kiera met last winter for shell templates. The sagitta for an arc of radius L across a chord c is L minus the square root of L squared minus a quarter of c squared. With the chord and sagitta known for both the large and small arcs, the piece can be laid out from its straight edges without ever finding the apex.

The eccentric cone

An eccentric cone joins two circles whose centres do not lie on the same vertical line. The Riverbend reducer has its two ends tangent along the bottom, so the bottom of the duct stays straight and ash slides through. Because the elements do not all meet at one point and are not all the same length, there is no single apex to swing arcs from, and the radial-line method fails. The reducer is developed by triangulation.

Each end circle is divided into the same number of equal parts, twelve in this case, numbered from the bottom. Solid lines join each point on the large end to the matching point on the small end: 0 to 0, 1 to 1, and so on. Dashed lines join each point on the large end to the next point on the small end: 0 to 1, 1 to 2, and so on. Together, the solid and dashed lines divide the surface into narrow triangles. The true length of every line is found from its three distances: the length of the reducer along its axis, the sideways offset between its ends and the difference in their heights above the bottom line.

A side view of an eccentric reducer with a straight bottom edge, a tall large end on the left and a shorter small end on the right. Solid grey lines join matching points on the two ends, and dashed gold lines join each point on the large end to the next point on the small end. A legend explains the solid and dashed lines and notes that each true length comes from its run, rise and sideways offset, and the triangles laid in order give the pattern.
An eccentric reducer divided into triangles.

The pattern is built by laying out the triangles in order. Kiera starts with line 0, the bottom line, which on a flat-bottomed reducer is its true length along the bottom. From its ends, she swings the next pair of distances: from the large-end point 0, the dashed true length 0 to 1'; from the small-end point 0, the chord between small-end points. Where the arcs cross is point 1' on the small end. From that point and large-end point 0, she swings the solid true length 1 and the large-end chord to find large-end point 1. She continues, triangle by triangle, to line 6 at the top, then joins the points on each end with smooth curves. Because the reducer is symmetrical about its bottom line, half the pattern is laid out and the other half is a mirror image.

The lines used to triangulate the eccentric reducer.
LineJoinsFound from
Solid kLarge-end point k to small-end point kAxial length, sideways offset, height difference
Dashed k to k+1'Large-end point k to small-end point k+1Same three distances
Large-end chordTwo neighbouring large-end pointsLarge mean diameter and the number of divisions
Small-end chordTwo neighbouring small-end pointsSmall mean diameter and the number of divisions

Reading the reducer's true lengths

The second half of the code output gives the solid and dashed true lengths for half the reducer. Line 0, along the bottom, is exactly the reducer's length, because both ends touch the bottom line. Line 6, at the top, is the longest, because it rises the full difference between the two diameters. Kiera notices that the dashed lines are close to the solid lines near the bottom and grow apart toward the top; the triangles there are the most stretched, and errors there show most in the finished part.

Triangulation has one weakness: small errors build up as the triangles are laid out one after another. Joanne's defence is to check the pattern at the end. The curve on the large end must measure half the large mean circumference, and the curve on the small end half the small mean circumference. If either is off by more than a few millimetres, the triangles have drifted and the pattern is laid out again.

Worked problem

Checking the finished half pattern

Kiera measures her half pattern with a flexible rule. The large-end curve measures 1,892 mm and the small-end curve 1,267 mm. Are they right?

Show solution
  1. Half the large mean circumference = π × 1,206 ÷ 2 (48.25 ÷ 2) = 1,894 mm (75 13/16 in) (rounded to the nearest mm (1/16 in)).
  2. Half the small mean circumference = π × 806 ÷ 2 (32.25 ÷ 2) = 1,266 mm (50 11/16 in).

Answer: The large curve is 2 mm short and the small curve 1 mm long, both within Joanne's limit, so the pattern is accepted.

Eccentric reducers can also be made as truncated oblique cones and developed from the apex of the oblique cone, with elements of different lengths swung from that apex. The method is quicker than full triangulation when the reducer's sides, extended, meet at a point that is easy to reach on the layout floor. Joanne shows Kiera both and lets her use whichever she prefers; the triangulation method is the one that works on every shape.

Matching marks and seam positions

Before any piece of a split cone leaves the layout table, Kiera marks it. Each piece gets its job and part number, the hopper's 0 degree reference, the element lines for rolling, and matching marks at both ends of each seam edge, so that piece 1's right edge goes to piece 2's left edge and not to piece 3's. The marks are centre punched at the large and small ends, where they will still show after rolling, and written in paint marker beside them. On the eccentric reducer, she also marks the bottom line, because the reducer must be installed with its straight side down; a reducer turned over in the field would trap ash instead of shedding it.

The three seams of the hopper are placed so that none falls under the round side inlet or at the collector's support lugs, and they are staggered from the top course's long seam. Joanne's habit is to put one seam on the side that will face the platform, where it can be inspected easily in service.

Common mistakes

  • MistakeSwinging a truncated cone pattern from the small end as if it were the apex.

    InsteadFind the imaginary apex from similar triangles; both arcs are swung from it.

  • MistakeUsing the radial-line method on an eccentric cone.

    InsteadAn eccentric cone has no single apex with equal elements; triangulate it.

  • MistakeLaying out triangles without checking the result.

    InsteadCheck that each end curve equals half its mean circumference.

  • MistakeCutting a large pattern in pieces with seams across it.

    InsteadSplit along elements, so each piece is an equal part ring.

  • MistakeDrawing straight lines between the points on the end curves.

    InsteadJoin them with smooth curves through the points.

Summary and key terms

  • A truncated cone's pattern is a part ring swung from the imaginary apex of the full cone.
  • The apex height is h × Rb ÷ (Rb − Rs); the pattern radii are the slant heights to each end.
  • Large patterns are split into equal pieces and laid out by coordinates, by a master template, or by chord and sagitta.
  • An eccentric cone has no single apex and is developed by triangulation.
  • Solid lines join matching points and dashed lines join each point to the next; each true length comes from three distances.
  • Triangulated patterns are checked by measuring the end curves against the mean circumferences.
Frustum
A cone with its top cut off by a plane parallel to the base.
Imaginary apex
The point where a truncated cone's elements would meet if extended.
Eccentric cone
A cone joining two circles whose centres are offset.
Triangulation
Developing a surface by dividing it into triangles and laying out their true lengths in order.
Sagitta
The height of an arc above its chord.
Oblique cone
A cone whose apex is not directly above the centre of its base.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. A truncated cone has radii of 1,000 mm (40 in) and 500 mm (20 in) and a height of 800 mm (32 in). How high is its imaginary apex above the large end?

    Show answer

    H = 800 × 1,000 ÷ 500 = 1,600 mm.

  2. What is its large pattern radius?

    Show answer

    √(1,000² + 1,600² (40² + 64²)) = 1,887 mm.

  3. Why is the bottom line of the Riverbend reducer exactly the reducer's length?

    Show answer

    Both end circles touch the bottom line, so line 0 has no rise or sideways offset.

  4. How does Joanne check a triangulated pattern?

    Show answer

    By measuring each end curve and comparing it with half the mean circumference of that end.

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