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Cone and truncated cone flat pattern
Enter both diameters and the height. A diameter of 0 gives a full cone, an offset gives an eccentric reducer.
A truncated cone unrolls into a ring sector. With mean diameters of 500 and 250 mm and a height of 300 mm, the slant height is 325 mm, the outer radius 650 mm, the inner radius 325 mm and the sector angle 138.46°; the flat piece is 650.0 x 1,136.5 mm.
| Piece | Qty | Flat size, mm | Area, mm² | Cut length, mm |
|---|---|---|---|---|
| Shell | 1 | 650 × 1,136.5 | 382,881.6 | 3,006.2 |
Patterns are developed at mid-thickness. An inside dimension is increased by the thickness, an outside dimension is reduced by it.
CutMeter is software. Check every result before you cut or bend. The person who approves the part is responsible for its dimensions.
Formula
r1 = D1 / 2 and r2 = D2 / 2 (mean diameters, r1 > r2). Slant height s = sqrt(h² + (r1 - r2)²). Outer radius R1 = r1 x s / (r1 - r2); inner radius R2 = r2 x s / (r1 - r2). Sector angle = 360° x (r1 - r2) / s = 360° x r1 / R1, divided by the number of equal parts. Area of one piece = pi x (r1 + r2) x s / parts.
D1, D2 = diameters of the two ends; h = height between the end planes; s = slant height; R1, R2 = radii of the flat arcs. D2 = 0 gives a full cone (R2 = 0); equal diameters give a rectangle pi x D1 by h.
Concentric cones are drawn with exact arcs. An eccentric reducer (offset above 0) is developed along the straight lines of the oblique cone, 4 x divisions of them, as a polyline with its bend lines on layer BEND.
Worked example
Diameter 1 = 500 mm, diameter 2 = 250 mm, height 300 mm, thickness 3 mm, mid-thickness dimensions, 1 part.
s = sqrt(300² + 125²) = 325 mm. R1 = 250 x 325 / 125 = 650 mm. R2 = 125 x 325 / 125 = 325 mm. Angle = 360° x 125 / 325 = 138.46°.
One shell, 650.0 x 1,136.5 mm, area 382,881.6 mm², cut length 3,006.2 mm. Entered as inside diameters instead, the mean diameters become 503 and 253 mm and the radii 653.9 and 328.9 mm. In 2 equal parts, each piece has a 69.23° sector and measures 607.8 x 534.8 mm.
Cone flat patterns
| D1, mm | D2, mm | h, mm | Outer radius R1, mm | Inner radius R2, mm | Sector angle, ° |
|---|---|---|---|---|---|
| 300 | 150 | 200 | 427.2 | 213.6 | 126.4 |
| 500 | 250 | 300 | 650 | 325 | 138.46 |
| 600 | 200 | 400 | 670.82 | 223.61 | 161 |
| 800 | 400 | 300 | 721.11 | 360.56 | 199.69 |
| 1,000 | 500 | 600 | 1,300 | 650 | 138.46 |
| 400 | 0 | 300 | 360.56 | 0 | 199.69 |
Questions about cone flat patterns
How do I calculate the flat pattern of a truncated cone?
Find the slant height s = sqrt(h² + (r1 - r2)²), then the outer radius R1 = r1 x s / (r1 - r2), the inner radius R2 = r2 x s / (r1 - r2) and the sector angle 360° x (r1 - r2) / s. Draw two concentric arcs with that angle and join their ends.
Should I enter inside, outside or mean diameters?
Any of them; choose which under Dimensions given as. The pattern is developed at mid-thickness, so at 3 mm an inside diameter of 500 mm becomes 503 mm and an outside diameter 497 mm.
Can I split the cone into several pieces?
Yes, into 1 to 12 equal pieces for a concentric cone. The table shows the quantity, and the DXF holds one piece.
Can it unfold an eccentric reducer?
Yes. Enter the offset between the centres of the two ends. Equal diameters with an offset (an oblique cylinder) are not supported.
How exact is the end radius when I enter inside or outside diameters?
An inside or outside diameter is converted to mid-thickness by adding or subtracting the thickness. On a sloping cone wall the true shift is slightly larger, so each end radius is off by t/2 x (1 / cos φ - 1), where φ is the half-angle of the cone: 0.23 mm at t = 3 mm and φ = 30°. Mid-thickness diameters have no such error.
Sources
- Frustum: lateral area pi (r1 + r2) s en.wikipedia.org
- Cone: development of the lateral surface en.wikipedia.org