math - Find point on a path in 3d space, given initial and final position and starting velocity vector -

let's have entity in 3d space , know position vector (x, y, z) , velocity vector (so orientation in space).

starting known point want reach known point b in 2 steps:

a) turn, following circular path known radius r, until point c

b) go straight point c final point b.

speed (scalar value) not important. velocity (vector) known, guess should define plane on circle resides, being tangent it, line between , b...

i want know how find coordinates (x, y, z) c , velocity vector of entity being there.

update: see working demo!

if understand correctly, situation (in plane containing a, b , v) shown in diagram below. points , b given, vector v , distance r. want find point c.

well, let vector w = (−v̂_y, v̂_x) unit vector perpendicular v. o = + r w.

now, |c − o| = r , (c − b)·(c − o) = 0 (where · dot product). combine these quadratic equation, can solve find 2 possible positions c. pick 1 right sign (c − b)×(c − o).

(there's second choice centre of circle, o = − r w, representing turning clockwise instead of anticlockwise. gives possibility c. guess you'll have use heuristic decide 1 prefer: maybe 1 smallest ∠aoc.)

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st0rm asks doing in 3d (see comments). that's easy! plane containing a, b, , v has normal vector n = (a − b) × v. let u = n × v vector perpendicular both n , v, , let w = û (the unit vector in direction of u).

you'll need take account constraint c lies in same plane a: c·n = a.n, , "the right sign (c − b)×(c − o)" becomes "the right sign (c − b)×(c − o)·n".

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having trouble solving system of equations?

well, if (c − b)·(c − o) = 0, (c − o + o − b)·(c − o) = 0, therefore (c − o)·(c − o) + (o − b)·(c − o) = 0, therefore c·(o − b) = o·(o − b) − r².

you'll note equation plane, , c·n = a.n. intersect these 2 planes (see wikipedia details — can use simpler solution since planes orthogonal , can made orthonormal) equation of line on c lies: c = h + λl, say, l = n×(b − o). use (c − o)·(c − o) = r² turn quadratic equation in λ. you'll find quadratic equation simplifies quite bit if rewrite equation of line c = h + λl + o occurrences of "− o" disappear.

here's implementation in python using numpy vector algebra. i'm sure can figure out how convert language of choice.

import math numpy import cross, dot numpy.linalg import norm  def unit(v):     """return unit vector in same direction v."""     return v / norm(v)  def turnpoints(a, b, v, r):     """generate possible turning instructions path b starts out in direction v, turns through part of circle of radius r until reaches point c (to determined), heads straight b. return each instruction in form (sense, c) sense -1 clockwise , +1 anticlockwise."""     n = unit(cross(a - b, v))   # unit normal plane containing a, b, v.     w = unit(cross(n, v))       # unit normal v lying in plane.     sense in (-1, +1):      # turn clockwise or anticlockwise?         o = + sense * r * w   # centre of turning circle.         bb = b - o         m = unit(bb)         # c lies on line h + l*l + o         h = dot(a, n) * n + (r**2 / norm(bb)) * m         l = cross(n, m)         # |c - o| = r**2 gives quadratic eqn in l coefficients a=1, b=0, c.         c = dot(h, h) - r**2         disc = - 4 * c          # discriminant of quadratic eqn.         if disc < 0:             continue            # no tangents (b inside circle).         elif disc == 0:         # 1 tangent (b on circle).             c = h + o             yield (sense, c)         else:                   # 2 tangents (b outside circle)             sign in (-1, +1):                 l = sign * math.sqrt(disc) / 2                 c = h + l * l + o                 # 1 choice c correct (the other involves                 # reversing direction).                 if dot(cross(c - b, c - o), n) * sense > 0:                     yield (sense, c)