I have read in a number of places --- though now I come to check up, I'm having difficulty tracking down a definitive source --- that the quartic surface known as the "cyclide of Dupin" has 4 nodes: conical singularities at which the gradient vector vanishes. Furthermore, it is averred that one pair of these may may be real or may be complex; the other pair is always complex. Like the trusting innocent I am, I have for years cheerfully accepted this scenario, without ever properly understanding why it might --- or might not --- obtain. The first pair can easily be seen in action in spindle or horn (real pair) versus ring (conjugate complex pair) cyclides --- at least, they could easily be seen before some brainstorm overtook the two pages at http://en.wikipedia.org/wiki/Dupin_cyclide http://mathworld.wolfram.com/Cyclide.html which I was sure used to show informative plots of the various forms! Pinning down the second pair is more problematic. In the special "parabolic" case --- an infinite cubic "full twist" surface --- there are indeed 4 complex nodes: for example differentiating y(x^2 + y^2 + z^2 - 1) - 2 x z = 0 wrt x,y,z and solving, we find solutions at [x, y, z] = [i, 1, i], [-i, 1, -i], [-i, -1, i], [i, -1, -i], where as usual i^2 = -1. However, in the general quartic case, Maple is quite insistent that there is only a single pair of solutions; and in consideration of the contradictions in which my earlier credulity was beginning to embroil me, I am beginning to think that Maple might just be right about this. In the case of the torus in standard postion with central and tube radii p and q, we (at least, Maple and me) find (x^2 + y^2 + z^2 - p^2 - q^2)^2 - 4 p^2(q^2 - z^2) = 0 has just the two nodes [x, y, z] = [0, 0, +sqrt(q^2-p^2)] , [0, 0, -sqrt(q^2-p^2)] . Inverting this with centre x = e say, we get a general shape with again just two nodes [x, y, z] = [e(e^2-1+q^2-p^2), 0, (+/-)sqrt(q^2-p^2)] / (e^2+q^2-p^2) unless e^2 = p^2 - q^2 (parabolic case). Can anyone cast light on my confusion? Fred Lunnon