Quadric Family
Shape · (x, y, z) = f(u, v, t)
(sqrt(max(k, 0) +(3v/π − 1.5)²) · cos(u), sign(3v/π − 1.5) · sqrt((3v/π − 1.5)² − min(k, 0)) /(3v/π − 1.5 != 0 || k >= 0), sqrt(max(k, 0) +(3v/π − 1.5)²) · sin(u))
Open in the app The dials and keys named below are the app's.
What it draws
Every point of this surface satisfies x² + z² − y² = k, with k on its own slider. Wavelace draws y upward, so this is the textbook x² + y² − z² = k standing on its axis. The parameter u goes once around the axis, and w = 3v/π − 1.5 runs from −1.5 to 1.5 as v runs from 0 to π.
For k above zero, w is the height and the radius is √(k + w²). At the opening k = 1 the waist has radius 1 and the two rims have radius √3.25 ≈ 1.80. For k below zero the roles swap: w is the radius and the height is ±√(w² − k), so each sheet reaches its tip at w = 0.
The last factor of the middle expression is a cut. (w ≠ 0 || k ≥ 0) is 1 almost everywhere, and 0 on the one row where two sheets would otherwise be joined across the gap. Dividing by it leaves that row undefined, so nothing is drawn there.
One slider, three surfaces
At k = 1 the surface is a hyperboloid of one sheet, a single waisted tube. Lower k and the waist, of radius √k, narrows. At k = 0 it closes to a point, and the surface is the double cone x² + z² = y², its sides at 45°.
Below zero the cone tears apart into a hyperboloid of two sheets, two bowls facing away from each other. Their tips sit at heights ±√(−k), so at k = −1 they are 2 apart. The cone is the asymptote of every member of the family: far from the axis each surface hugs it, whatever k is.
The one-sheet hyperboloid is also doubly ruled. Through every point of it pass two straight lines that lie wholly in the surface, although it curves in every direction.
History
The ruling made the shape buildable. Vladimir Shukhov put up the first hyperboloid tower, a water tower for the All-Russian Exhibition at Nizhny Novgorod in 1896, as a lattice of straight steel bars. The cooling towers of power stations use the same shape for the same reason.
Try
- Drag k slowly from 1 down to −1. The waist closes at 0 and the tube splits into two sheets.
- Press Front in the deck. The outline is the hyperbola x² − y² = k, and at k = 0 it is a pair of crossing lines.
- Put a 2 in front of the first expression, 2·sqrt(max(k, 0) + (3v/π − 1.5)²)·cos u. The circles become ellipses twice as wide in x, the elliptic hyperboloid of the textbooks.
- Set Span of v (×π) to 0.5 and only the lower half is drawn, with w from −1.5 to 0. Below zero that is one sheet alone.