The middle parameter might not always correspond to the mid point of a curve. It's better to use the 'evaluate length' with the length as 0.5 and the boolean input set as true.
On May 6, 9:37 am, Rchitekt <andyopa...@gmail.com> wrote: > Frank, > The midpoint can be found using the Evaluate Curve component. Just > use 0.5 for the t-value and make sure you set your curve to > reparametrize. That should do it. > -Andy > > On May 5, 11:05 am, Marc Syp <marc....@gmail.com> wrote: > > > No need for all that. In Scalar components, there is a Min-Max node. > > Hook it up to your list and then extract the interval and voila, you > > have both min and max values to play with. > > > Marc > > > On May 5, 7:42 pm, Andrew Heumann <and...@heumann.com> wrote: > > > > @frankS > > > > I get around this without scripting by: > > > > 1. sorting the list, and taking the item at index=0 for the min > > > 2. reversing the sorted list and taking the item at index=0 for the > > > max > > > > On May 5, 12:37 pm, frankS <fffr...@googlemail.com> wrote: > > > > > 1.1.) find min/max of all list values. > > > > > i don't know if there is a smart way to find the lowest or highest > > > > value in a list. > > > > i use a vb-component (called "list mapper") friendly shared by someone > > > > on this forum. > > > > > frank
