Branch Pipe Development and the Hole in the Main
A ductwork firm is prefabricating a 90° branch for a ventilation system: a Ø60 branch pipe meets a Ø100 main pipe at 90°, both axes intersecting, both 1.2 mm galvanised sheet. The branch is 120 mm long, measured from the main pipe axis to its free end. The workshop cuts both the branch end profile and the hole in the main pipe from developments marked on flat sheet. The foreman has developed the branch as a plain rectangle, 188.5 mm × 120 mm, cut the end square, and relied on trimming on site. The result touches at two opposite points but leaves gaps of up to 10 mm elsewhere, and the hole in the main has been chain-drilled as a 60 mm circle, which leaves the joint open at the sides. Nobody can explain why a 'straight pipe' needs anything other than a straight cut.
1. Explain why the square-cut branch cannot fit and whether the intersection curve is a plane curve.
Read the worked answer
Why it cannot fit. A square end is a plane at right angles to the branch axis. The main pipe surface is curved, so the branch must be cut to a saddle that follows that curvature. A plane can only touch a cylinder along a line or at a point in this configuration, so the square end contacts the main at best at a point - in practice the two points where the branch stands farthest from the main axis - and gaps elsewhere. The observed 10 mm gaps are the depth of the saddle that was never cut.
Is it a plane curve? Only in the special case. When the two diameters are equal and the axes intersect, the intersection of two equal cylinders at 90° lies in two planes, so it is a plane curve and each pipe develops with a straight-line end (after the correct development). With unequal diameters - Ø60 into Ø100 here - the intersection is a space curve. It must be found point by point by projecting ordinates from the views, and it cannot be produced by any single straight cut, by an ellipse, or by a simple template.
Method consequence. Development is still by parallel lines, because both surfaces are cylinders and the elements are parallel to the axes; what changes is that the ordinate at each element varies. The developed branch is a rectangle in width, but its end is a curve plotted from those ordinates.
2. Compute the branch's developed profile: give the ordinates for 12 equal divisions and state the width of the development.
Read the worked answer
Development width. π × 60 = 188.50 mm, divided into 12 equal spaces of 188.50 / 12 = 15.71 mm.
Ordinates. Take the branch axis as z (up), the main axis as x, and let θ be measured round the branch from the side where the branch surface lies in the x–z plane. A point on the branch surface at angle θ has coordinates (30 cos θ, 30 sin θ, z). It lies on the main pipe surface when y² + z² = 50², so:
z = √(2500 − 900 sin²θ)
Measured from the branch's free end (120 mm from the main axis), the cut ordinate is 120 − z:
| θ | y (mm) | z (mm) | ordinate (mm) |
|---|---|---|---|
| 0° | 0.00 | 50.00 | 70.00 |
| 30° | 15.00 | 47.70 | 72.30 |
| 60° | 25.98 | 42.72 | 77.28 |
| 90° | 30.00 | 40.00 | 80.00 |
and mirrored for 120° to 360°.
Interpretation. The cut is deepest at the two sides (70 mm of pipe remains) and shortest at front and back (80 mm) - a 10 mm variation, exactly the gap seen on site. The shortest material is at θ = 90°, where the branch surface is nearest the main axis. Marking the ordinates on the wrapped development and joining them with a smooth curve gives the cutting line; the same ordinate table is transferred to the main pipe to mark the hole.
3. Show that the hole in the main is not a 60 mm circle, quantify its extent, and state the allowances that must be added to both developments.
Read the worked answer
The hole is not a circle. The opening in the main is the trace of the Ø60 branch cylinder on the Ø100 main cylinder. Parameterise the main surface by φ from the top: y = 50 sin φ, z = 50 cos φ. The branch surface requires x² + y² = 30², so:
x = √(900 − 2500 sin²φ)
This is only real while |50 sin φ| ≤ 30, i.e. |φ| ≤ 36.87°.
Extent.
- Along the main axis: at φ = 0, x = ±30, so the hole spans 60 mm along the main pipe - that part of the foreman's assumption was right.
- Around the main circumference: the hole wraps 36.87° each side of the top, an arc length of 50 × 0.6435 = 32.2 mm each way, 64.4 mm in total.
So a chain-drilled 60 mm circle is 60 mm along the axis but only 60 mm around a surface that needs 64.4 mm, and it is circular where the true boundary is not - hence the open sides. The hole must be marked from the same ordinate table, wrapped on the main pipe, or cut from a template taken off the branch's own development.
Allowances.
- Development is made on the neutral axis, not the outside surface; for 1.2 mm sheet the shift is about 0.6 mm - small, but it should be applied consistently rather than ignored on one part only.
- Add a lap/stitch-bolt allowance, typically 3 mm on one edge per joint, to one side of the seam only, not both.
- Cut the branch end on the curve with a small grinding allowance (1–2 mm) and finish to the line on assembly; do not rely on site trimming to correct a square cut, since the error is systematic, not random.