How to lay out an eccentric reducer by hand

What you are making

The worked example is a flush reducer: a 300 mm large end, a 150 mm small end, 200 mm apart, with the small end offset by (300 − 150) ÷ 2 = 75 mm so that one side stands straight up — the usual reducer on a pump suction or a duct that must keep a flat side. For plate, work to the mid-thickness diameters; for thin sheet the difference is small.

Method 1 — from the apex

The classic method: the reducer is treated as a whole oblique cone with its tip cut off, laid out with lines radiating from the apex (Kittredge, Problems 163 and 167). Use it when the apex fits on your sheet or floor.

1. Find the apex

Extend the two sides of the side view until they meet. By similar triangles the apex sits D1 ÷ (D1 − D2) times the height above the large end — 300 ÷ 150 × 200 = 400.00 — and the same factor times the offset across from the large end's centre: 2 × 75 = 150.00, which on a flush reducer is straight above the short side.

2. Divide the large end

Divide the large end into equal parts — twelve here, 30° apart — and number them from the seam. With the seam on the short side, line 0 is the short side and line 6 the long side.

3. Find the true length of each line from the apex

Only lines 0 and 6 show their true length in the side view; the others lean towards or away from you. Find each one from a right triangle: one side is the apex height, the other the distance in plan from the point under the apex to the division point. Line 3, for example: in plan, from (150, 0) to (0, 150) is √(150² + 150²) = 212.13, so the true length is √(400² + 212.13²) = 452.77. Because the part is symmetric, lines 7 to 11 repeat lines 5 to 1.

The small end lies on the same lines at D2 ÷ D1 of the distance — exactly half here.

4. Lay the lines out

Mark the apex on the sheet and draw line 0, 400.00 long. From the apex swing an arc the length of line 1; from the end of line 0 swing the step to the next point; where the arcs cross is point 1. Carry on round to point 12, which comes back to line 0's length. Join the points with a smooth curve.

The one correction: step on the pattern, not the plan chord

The books step the spacing between divisions straight from the plan — here the chord 77.65. That builds the pattern from flat triangles inscribed in the cone, and — a chord being always shorter than its arc, as Kidder notes (p. 66) — it comes out short: twelve of those steps make 931.75, but the large edge has to be π × 300 = 942.48 long. The distance between neighbouring points on the real flat pattern is a little longer than the chord across the round part — 78.54 near the short side, 78.23 near the long side — and twelve of those make 940.85, with the smooth curve through them the full 942.48. The difference, about 9 mm on this part, is roughly 1 %: enough to leave a gap at the seam. The tool's table lists these pattern steps for every line.

5. Mark the small end

On each line measure in from the apex the apex-to-small length, or — the same point — one element length in from the large edge: 200.00 on line 0, 250.00 on line 6. Join those points with a smooth curve.

6. Check before you cut

Bend a flexible rule along the edges: the large edge should measure π × 300 = 942.48, the small π × 150 = 471.24, and both seam lines 200.00. The finished pattern for this part fits a blank about 681.93 × 395.43.

Method 2 — triangulation between the ends

When the two ends are close in size the apex is far away: a 300 to 290 reducer, 200 high, has its apex 6.00 m up. Atkins calls the radial method “somewhat inconvenient in practice” for exactly that reason (p. 117), and the part is triangulated between the two ends instead, quadrilateral by quadrilateral, without the apex (Kidder, Ch. VI; Coey, “Off-Centre Round Taper”).

1. Divide both ends alike

Divide both ends into the same number of equal parts and number them the same way from the seam. Join matching points with elements (large 1 to small 1, and so on) and each large point to the next small point with a diagonal.

2. Find the true lengths

Each element and diagonal is the hypotenuse of a right triangle: the height between the ends, and its length in plan. Element 6 runs in plan from (−150, 0) to (0, 0), so its true length is √(150² + 200²) = 250.00.

3. Build the strip

Draw element 0. From its small end swing the small-edge step; from its large end swing the diagonal; where they cross is small point 1. From small point 1 swing element 1, and from large point 0 the large-edge step: where they cross is large point 1. Repeat round to the seam.

The same correction applies. A textbook true-length diagram gives the straight diagonal across the part (line 0: 203.73) and the plan chords give the steps; together they build flat facets and the strip comes out about 1 % short round the edges. The diagonals and steps in the table are measured on the exact flat pattern, so every point lands where it should. Elements are straight lines on the part and need no correction.

Common mistakes

• Taking any line other than the two in the plane of the side view as true length from the side view.
• Measuring the offset anywhere but centre to centre in plan.
• Working to the outside diameters of heavy plate instead of the mid-thickness.
• Stepping plan chords and finding the seam does not close — see the correction above.

How this matches the tool

The eccentric reducer tool develops the cone exactly, by numerical integration around the apex, and a separate program checks it by triangulating between the ends with 8,192 divisions. Laid out by hand with the table's pattern steps, method 1 and method 2 put their points on that same exact outline. Every number on this page was produced by the tool's own code when the page was built.

Sources

1. G. W. Kittredge, The New Metal Worker Pattern Book (1901), Problem 167, pp. 311–312: round ends, parallel, placed eccentrically — the frustum of a scalene cone
2. Kittredge (1901), Problem 163, pp. 306–308: the envelope of a scalene cone, true lengths by right triangles, stepped triangle by triangle
3. J. S. Daugherty, Sheet-Metal Pattern Drafting and Shop Problems (1922), Ch. VII, Problem 75, pp. 96–99: scalene or oblique cone and its frustum
4. F. S. Kidder, Triangulation Applied to Sheet Metal Pattern Cutting (1917), Ch. VI, pp. 41–49: the frustum of an oblique cone triangulated between the ends, without the apex
5. E. A. Atkins, Practical Sheet and Plate Metal Work, 2nd ed. (Pitman, undated), p. 117: radial methods inconvenient when the taper is slight, because of the long radius
6. B. Coey, Pattern Development: Sheet Metal Level 1 (BCcampus, 2021, CC BY 4.0), “Off-Centre Round Taper”: an offset round-to-round triangulated directly
7. Kidder (1917), p. 66: a chord is always shorter than its arc, so spaces stepped off a drawn circle very likely come to something less than its circumference