By the way, if I recalled it correctly, if you know the speed [tex]\vec{v}_{A}[/tex] of a point A, and the angular speed [tex]\vec{\omega}[/tex], then you can find the speed of any point B with [tex]\vec{v}_{B}=\vec{v}_{A}+\vec{\omega}\times \vec{r}_{BA}[/tex], where [tex]\vec{r}_{BA}[/tex] is the vector from A to B. The condition on the centre of velocity is [tex]\vec{v}_{B} = \vec{v}_{c} = \vec{0}[/tex], so you can find its position.
Many thanks.
I found a site (eventually - it took me ages, even on google) that told me what I was supposed to do. I used Kennedy's theorem to find the IC's and worked things out from there.
It was faster the first way I did it though. I just resolved velocities at the link ends and worked my way through the mechanism. Instantaneous centres gave the same answer, but took longer