Development of an obliquely cut lamp shade
A metalwork shop produces a lamp shade as a truncated right cone in 0.8 mm steel sheet. The full cone has base radius 30 mm and height 40 mm. The top opening is cut by an oblique plane: it stands 20 mm above the base at the front generator and 10 mm above the base at the rear generator, the two generators being diametrically opposite and lying in the vertical plane of symmetry of the solid. A trainee develops the lateral surface as a 216° sector of radius 50 mm and draws the top edge as a concentric arc of radius 25 mm. The first article comes out 12.5 mm short at the rear, and the rim does not sit flat against the horizontal lamp holder because the cut surface is not a plane parallel to the base.
1. Verify the 50 mm and 216° values and state the development radius of a top edge produced by a perpendicular cut 20 mm above the base.
Read the worked answer
The generator length is L = √(30² + 40²) = 50 mm. When the lateral surface is unrolled, its arc must equal the base circumference 2π·30, so the sector angle is 360° × r / L = 360° × 30 / 50 = 216°.
For a cut perpendicular to the axis at 20 mm above the base, similar triangles give the section radius 30 × (1 − 20/40) = 15 mm, and the corresponding point on each generator lies 50 × (1 − 20/40) = 25 mm from the apex. So the top edge would be an arc of radius 25 mm, concentric with the 216° sector of radius 50 mm.
That part of the trainee's layout is correct - but only for a perpendicular cut. The 216° sector fixes the base circumference and is independent of how the top is cut; the top edge is where the error lies.
2. Explain why the oblique top edge cannot be an arc, and compute the distance from the apex at the front, at a quarter generator and at the rear.
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Every generator from the apex to the rim has its own length, and in the development each generator is a radial line, so the rim maps to a curve of varying radial distance - not to an arc of constant radius. Only a cut perpendicular to the axis produces a constant radius.
The cut height varies linearly along the line of maximum slope: 20 mm at the front, 10 mm at the rear, hence 15 mm at the two quarter generators. The radial distance is 50 × (1 − h/40): front 50 × 0.5 = 25 mm; quarter 50 × 0.625 = 31.25 mm; rear 50 × 0.75 = 37.5 mm.
The trainee drew 25 mm everywhere, so the rear edge is 37.5 − 25 = 12.5 mm short - exactly the defect found on the article. Correct practice is to plot at least eight or twelve generators, obtain each true length from the elevation by rotation into a plane parallel to the vertical plane (no generator except those in such a plane is seen in true length), transfer the distances along the corresponding radial lines of the sector, and join the points with a smooth curve.
3. Recommend where to place the seam and list the checks to carry out before cutting the sheet.
Read the worked answer
Place the seam on a generator, never across the middle of a panel, and preferably on the front generator: it is the shortest (25 mm) and the least visible in service, and splitting the sector there gives two mirror-image halves that are easy to nest on the sheet. Add the joint allowance (for example a 5 mm lap, or the folded allowance for a lock seam) on one side only - never on both - and cut the top edge as the plotted curve, not as an arc.
Checks before cutting: the development must be symmetric about the mid-generator, because the solid is symmetric about the vertical plane of symmetry; the developed base arc must measure (216/360) × 2π × 50 = 188.5 mm; the rim curve must reach exactly 25 mm at the front and 37.5 mm at the rear and must be tangent to those radial distances at the seam; and the sheet thickness means the neutral line, not the outer surface, is what should be developed on a 0.8 mm sheet with a tight rim.