Solving for the positions of vertices of 3 line segments

Question

I have 3 line segments of lengths p,q,r joined at their ends. Let's call the vertices A, B, C, and D. Suppose D is fixed at the origin. Suppose that A is constrained to move only in the Y direction. Finally, suppose that segments q and r are joined at point C and segments p and q are joined at point B. Thus, there are 3 line segments which are joined at their end points. One end is free to move and the other end is constrained. Please see the attached diagram:

(Large Version)

I would like to solve for the (x,y) positions of point B and C given a known displacement of point A in the y-dimension.

I have already done this in rectangular coordinates to get 4 equations in 4 variables (Bx,By,Cx,Cy). To solve the position requires solving the system of 4 nonlinear equations.

(1) Would this get simpler in polar coordinates? I'm rusty.

(2) Can anyone provide information on computer algorithms to solve this problem?

Answer 1

OK, so this is just about geometrical constraints, no physical complications like inertia.

We have the position vectors: \begin{align} r_A &= (0, y_A) \\ r_B &= (x_B, y_B) \\ r_C &= (x_C, y_C) \\ r_D &= (0, 0) \\ \end{align} and these constraint equations: \begin{align} p^2 &= \lVert r_B - r_A \rVert^2 = x_B^2 + (y_B - y_A)^2 \\ q^2 &= \lVert r_C - r_B \rVert^2 = (x_C-x_B)^2 + (y_C - y_B)^2 \quad (*) \\ s^2 &= \lVert r_D - r_C \rVert^2 = x_C^2 + y_C^2 \end{align} So we have these unknowns $$ x_B, y_B, x_C, y_C $$ as $y_A$ is given.

Here is a possible parametrization via $(y_A, \varphi)$:

$$ r_B = (x_B, y_B) = (p \sin\varphi, y_A - p \cos\varphi) $$

The feasible coordintes for $r_C$ are taken from the intersection of the two circles for $q$ and $s$ segments, see equations $(*)$.

Interactive Web GeoGebra Worksheet

(Large Version)

To the right, there are controls to choose the three segment lengths $p, q, s$:

• The $\color{red}{\text{red circle}}$ includes the area where the $p$ segment is allowed,
• similar a $\color{cyan}{\text{cyan coloured circle}}$ for the $q$ segment and
• the $\color{green}{\text{green circle}}$ for the $s$ segment.

These lengths go into the constraint equations $(*)$ given above.

The possible configuration is influenced by the two parameters:

• The $y$ coordinate of the point $A$, $y_a$ which is visualized as a $\color{red}{\text{purple segment}}$ on the $y$-axis.

• Then we have the $\color{olive}{\text{olive coloured angle } \varphi}$ which can be used to choose all positions for point $B$.

Further we can see the point $D$ which is fixed at the origin $(0,0)$.

So the only point whose positions is determined from the above is the position of the point $C$.

The above image shows a configuration where the $\color{cyan}{\text{cyan coloured circle}}$ for the middle $q$ segment does not intersect with the $\color{green}{\text{green coloured circle}}$ of the lower $s$ segment. It is not a feasble solution for the constraints.

(Large Version)

(Large Version)

This intersection is the set of feasible positions for the point $C$:

• It can be empty (no solution for $C$) or
• can contain one point (unique solution for $C$, see first image above),
• two points (two solutions for $C$, see second image above) and
• for a rare but existing configuration consisting of the points of the full circle, if both circles have same radius and center (infinite many solutions for $C$, see image below).

(Large Version)

I suggest to use the above link for the interactive GeoGebra worksheet and play a bit with the values to get a feeling.

Answer 2

Let $\angle BAD=\theta$, so that $x_B=p\sin\theta$ and $y_B=y_A-p\cos\theta$. Point $C$ can be found as the intersection of the circles having centers at $B$ and $D$ and radii $q$ and $r$ respectively: its coordinates (if they exist) are the solutions of the system $$ \left\{\eqalign {(x-x_B)^2+(y-y_B)^2&=q^2\\ x^2+y^2&=r^2 }\right. $$ In general, for a given value of $\theta$ you'll get two possible positions of $C$ (if $BD<q+r$), or none at all (if $BD>q+r$).