Vector field visualization

Hello everyone,
I was wondering have anyone tried to explore visualization of vector field like this below in dynamo inside 2d or 3d environment.

I have found in this site tried to explore such patterns in grasshopper environment.
I am unable to figure it out relevant node in dynamo.
Let me know if you have any idea.
Thank you.


What would be your initial input ?

i think like this.

points on curves having direction.
or it could be array of points and curves will be drawn later based on direction of each point.

So basically a list of vectors ?

yes exactly. a list of vectors and a list of correspondence points.
points with directions.

You could draw lines with the origin in the vector origin and set the length with a small unit

I have progressed whit this so far.
any ideas are welcome.

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@ROY See if this helps

vectorField.dyn (19.4 KB)


ita alot easier . Thanks!!!

Thanks @Vikram_Subbaiah but the problem is i need a output like this and its yet to figure it out. Rotation is partially doing the job for centre field but vectors far away from the influence point are suppose to be less rotated but if you see on left side the are rotated in to almost opposite direction. I believe there must be some function.


@ROY Seems like you’ll need equations and more.
Below are a few attempts at generating basic vector field illustrations
vectorField.dyn (33.0 KB)


@Vikram_Subbaiah thanks this is was i was looking for.

I’d suggest Structure Synth or Fugu (both no longer maintained) for procedural 3D graphics
or Nodebox for simpler 2D vector graphics

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thanks for these informations. Its completely new to me and it seems very interesting. I’ll check for more details.

i was searching further more for more operation on vector field and my findings are here.

vectorField.2.dyn (21.9 KB)
one more thing to ask, is there any way to draw curve along the vector direction. i know there is a node calls tangent at parameter but i think exact opposite node of this can do the job.

I don’t know the existence of such a node, but mathematically, if you want to do such thing, you’ll have to solve an differential equation, or at least do this tiny step by tiny step.