Find 4 points equidistant from a centroid in 3D space where all points lie on a predefined plane

Question

I have a plane defined as an xyz vector and a point that lies on the plane.

I would like to generate xyz coordinates for 4 points (N_points) on the plane surrounding the defined point (centroid) at a defined distance/radius (r).

My current solution only works in 2D. I would like to expand this to work in 3D but my knowledge of geometry is failing me. Any ideas would be much appreciated.

def circlePoints(r, N_points, plane=(1,1,1), centroid=(0,0,0), rotation=0):
    (plane_x, plane_y, plane_z) = plane
    (centroid_x, centroid_y, centroid_z) = centroid

    step = (np.pi*2) / N_points
    rot=rotation
    i=0
    points=[]
    for i in xrange(N_points):
        x = round(centroid_x + ((np.sin(rot)*r) / plane_x), 2)
        y = round(centroid_y + ((np.cos(rot)*r) / plane_y), 2)
        z=0 #?
        points.append((x,y,z))
        rot+=step
    return points

print circlePoints(1, 4, [1,2,0], [2,3,1])
print circlePoints(1, 4)

because I got a long float back for the default case `circlePoints(1,4)` — JoshuaBox, Jul 24 '17 at 18:38
Can you tell more about how you are defining your plane? You need 4 parameters to specify a plane in 3d, but you only have 3. — Imran, Jul 24 '17 at 19:08

score 1 · Accepted Answer · answered Jul 24 '17 at 20:28

We need to find two vectors perpendicular to plane (the normal). We can do so by the following procedure:

Normalize plane
Set a vector k = (1, 0, 0)
Calculate math.abs(np.dot(k, plane))
If > 0.9 then set k = (0, 1, 0)
Calculate a = np.cross(k, plane)) and b = np.cross(plane, a)
You now have two vectors in the plane. You can get any points in the plane by adding some number times these two vectors and adding to centeroid
If you want specific distances, you need to normalize a and b

Code:

import numpy as np
import math

def normalize(a):
    b = 1.0 / math.sqrt(np.sum(a ** 2))
    return a * b

def circlePoints(r, N_points, plane=(1,1,1), centroid=(0,0,0)):
    p = normalize(np.array(plane))
    k = (1, 0, 0)
    if math.fabs(np.dot(k, p)) > 0.9:
        k = (0, 1, 0)
    a = normalize(np.cross(k, p))
    b = normalize(np.cross(p, a))
    step = (np.pi * 2) / N_points
    ang = [step * i for i in xrange(N_points)]
    return [(np.array(centroid) + \
            r * (math.cos(rot) * a + math.sin(rot) * b)) \
            for rot in ang]

print circlePoints(10, 5, (1, 1, 1), (0, 0, 0))

This is fantastic. Thank you!! – JoshuaBox Jul 25 '17 at 11:06 — JoshuaBox, Jul 25 '17 at 11:06

Find 4 points equidistant from a centroid in 3D space where all points lie on a predefined plane

1 Answers1