I have a problem, i am having trouble getting a point or parameter (doesnt matter which one) of an arc by vector, i know which vector i have and it is on the line, but how do i get the corresponding point/ parameter of this vector on the line.

If it’s an arc just create a vector from the start and end points of the arc and normalise it. Or if it’s a segment, then your start and end point of the segment from the arc.

If you want the point on the arc segment there are many ways, one would be to get the mid point between the start point and end point and project it onto the arc.

I think you’d have to create an iterative function to loop through values until you converge on a match. Do you have a working tolerance or does it have to be exactly parallel?

In that case it’s fairly simple. Specify the number of points you want to check along your line and compare the angle between vectors. The closer to 0, the closer you are to the tangent. Then you can just return that parameter and get the point on the curve.

Ah ok, well thats really straightforward, just compute the cross product between your vector and the ‘normal’ vector of the arc (an arc is always on a plane, so I created one using the same inputs as the arc to obtain the ‘normal’, but you can easily do this from points on the arc etc) then project the centre point of the arc onto itself using the result of the cross product.

I have checked out the problem and i do now understand what you are doing. But the problem i am having is that the curve/ arc isnt always a perfect arc. So i cant use the centerpoint of the arc, because there isnt always one

As long as the curve doesn’t have multiple tangents along the same vector (can’t remember the mathematical term right now… degree of curvature?) my method should work on any curve. With a little modification you could even get it to show multiple points with similar tangents.

but could you explain the Point.Project node to me? I dont understand what is happening in that node and why you need an external point as input. @Thomas_Mahon