Torus
Shape · (x, y, z) = f(u, v, t)
((3 + cos(v)) · cos(u), sin(v), (3 + cos(v)) · sin(u))
Open in the app The dials and keys named below are the app's.
What it draws
The parameter v goes round the tube and u carries the tube round the ring. Read 3 + cos(v) first: it is the distance from the vertical axis, 3 at the top and bottom of the tube, 4 on the outside and 2 on the inside. The width and the depth are that distance turned through u as cos(u) and sin(u), so the ring lies flat on the ground plane.
The second expression is the height, sin(v), so the tube reaches ±1 above and below the ring. Nothing depends on t: the donut sits still.
The two radii
A torus is fixed by two numbers, and both are visible in the formula: the centre circle has radius 3 and the tube radius 1. Every point of the surface is exactly 1 from that centre circle. That is what makes it a surface of revolution: a circle of radius 1, drawn in a vertical plane with its centre 3 out from the axis, swept once round. Because 3 > 1 the hole stays open and this is a ring torus.
The deeper fact is in the parameters. Each point is named by a pair of angles, one round the ring and one round the tube, and neither can be combed away. That is what it means to say the torus is a product of two circles, and why the mesh lines close on themselves in two different ways.
Try
- Top in the deck: two concentric circles, at radius 2 and 4.
- Change both 3s to 1, giving (1 + cos(v))·cos(u) and (1 + cos(v))·sin(u). Now the two radii are equal and the hole shuts to a single point: a horn torus.
- Set Span of v (×π) to 1: only half of each tube section is drawn, leaving the top half as an open gutter.
- Set Turns of u to 0.5 for half a donut, with the circular cross-section on show at both cut ends.