CurveLoop offset - CurveLoop normal vector?

[Anonymous]

Hi All, 

I'm struggling with this for a while. I need to offset a CurveLoop always to the "outside" of it. I'm using the CreateViaOffset(CurveLoop, Double, Vector) where the vector is normal to the curve.

If I have the original CurveLoop, How do I get the normal vector?

Thanks in advance!

[Anonymous]

Hi @jeremy_tammik, is this: https://forums.autodesk.com/t5/revit-api-forum/offset-a-face/m-p/3056230/highlight/true#M2049 still the case?

Cheers

[RPTHOMAS108]

Paraphrasing:

If you are walking along the direction of the curve then a normal in the direction from your feet to your head offsets right (for positive offset value), opposite normal from head to feet offsets left. Right hand rule is it?

Not sure what happens when some of your curves are reversed in the loop, I know for example that the curve loop isn't considered closed if all of the curve segments are not in the same direction. Knowing this could be a valid curveloop configuration leads me to wonder how that would offset i.e. which segment it would pick to define the global orientation of the offset?

[BardiaJahan]

If you would like to offset the curveloop to the 'outside':

if (curveLoop.IsCounterClockwise(normal))
   CurveLoop.CreateViaOffset(curveLoop, distance, normal));

else 
   CurveLoop.CreateViaOffset(curveLoop, - distance, normal));

and what I usually use as normal is the the normal of the same face I retrieve my CurveLoop from.

[Anonymous]

As far as I know, it's not possible for a line to have a normal in 3 space. It has an infinite number of "normals". If you're referring to the normal of the face you get the curve from, that certainly has a normal, and is probably the reference for which the method would be looking.

[BardiaJahan]

Yes, a line does not have a normal. And the normal of the face used to retrieve the CurveLoop is what needs to be passed to the CreateFromOffset method.