I have a relatively simple geometric system which is proving exceptionally difficult to solve. Any advice on how to proceed is much appreciated.
I know the coordinates of three points on a 2D plane A,B,C.
I need to find the the points D,E,F,G such that E and G lie on the lines AD and GD respectively and F is the midpoint of EG and is on the line CD. ED and DG are of equal length.
The lengths of AB, FC and EG are known. AB = 2950, FC = 300, EG = 130
The true objective of solving the problem is to simulate a hanging plotter where the pen is not anchored at the middle of the two strings, but is instead held at points E and G. AE and BG are steel cables, while the assembly made up of EFGC is made of wood. Most hanging plotters suffer from this issue, but it has historically been ignored in the designs because they are small enough that the error is not an issue and only operate well below the motors. Again, any advice at all is appreciated.
Thank you.
Let's use a left-handed coordinate system, where the origin is at the midway between points $$A = \left ( -\frac{L}{2}, 0 \right ) \; \text{ and } \; B = \left ( +\frac{L}{2}, 0 \right)$$ making their distance $L$, $x$ increasing right, and $y$ down; assume $y \ge 0$.
In the classic three-point assembly, we only have $A$, $B$, and $D$.
Let's use $r_A$ for the cable length from $A$ to $D$, and $r_B$ for the cable length from $B$ to $D$, and $D = (x,y)$, we have two equations, $$\begin{cases} \left(x + \frac{L}{2}\right)^2 + y^2 = r_A^2 \\ \left(x - \frac{L}{2}\right)^2 + y^2 = r_B^2 \end{cases}$$ Solving those for $x$ and $y$ yields $$\begin{cases} x = \frac{r_A^2 - r_B^2}{2 L} \\ y = \frac{\sqrt{-(L+r_A+r_B)(L+r_A-r_B)(L-r_A+r_B)(L-r_A-r_B)}}{2 L} \end{cases}$$ and solving for $r_A$ and $r_B$ yields $$\begin{cases} r_A = \sqrt{\frac{1}{4}L^2 + L x + x^2 + y^2 } \\ r_B = \sqrt{\frac{1}{4}L^2 - L x + x^2 + y^2 } \end{cases}$$
If we have cables from $A$ to $E$ (length $r_E$), and from $B$ to $G$ (length $r_G$), with $CDEFG$ being a rigid piece, we can no longer rely on the simple math.
If we use $E = (x_E, y_E)$ and $G = (x_G, y_G)$, with $d$ being the distance between $E$ and $G$, we have three equations $$\begin{cases} \left(x_E + \frac{L}{2}\right)^2 + y_E^2 = r_E^2 \\ \left(x_G - \frac{L}{2}\right)^2 + y_G^2 = r_G^2 \\ (x_G - x_E)^2 + (y_G - y_E)^2 = d^2 \end{cases}$$ but we have four unknowns: the remaining degree of freedom being the orientation of the $CDEFG$ part.
If we look at the situation from a physics point of view, gravity is no longer a simple force that pulls $D$ down like it is in the simple $ABD$ case (ignoring inertia and momentum; say, slow movement and friction and friction-like losses making them neglible).
In this case, we also have torque due to gravity. The location of the center of mass relative to $E$ and $G$ means gravity does not just pull the assembly down; it tends to turn it, too.
Furthermore, the steel cables have considerable weight compared to the had assembly $CDEFG$. Even while taut, they are not straight; there is some curvature. In practice, the steel cables' weight is also part of the assembly, affecting both the mass distribution, as well as the torque (pulling the assembly off from vertical).
If you know
the weight of the fixed assembly $CDEFG$
the steel cable weight per length unit
the distance between $E$ and $G$
the distance from $E$ and $G$ to the center of mass, with steel cables not attached (to $E$ and $G$)
(the center of mass can be found using a pin: if the pin is in the correct location, holding the assembly in the air, the assembly should not tip)
then it should be possible to physically model the situation. The math is hairy (just thinking about the integrals for the cables gives me the jeebies), and the model being quite nonlinear, will likely not allow a simple algebraic solution for $(x,y)$ or $(r_1, r_2)$.
Personally, I think it might be easier to model the physics directly in the microcontroller, with iterative controls. (At any time step, one can do one of nine things -- nothing, shorten or lengthen $AE$, or shorten or lengthen $BG$ --; calculating the effects of each, and picking the one that yields the results most desirable, should work.)
It might look funny, though. Similar control loops occasionally produce counterintuitive effects. Oscillation is a problem, so you'd probably need to model the movement of the head down to third derivatives (for both location and orientation).
Physically, the device can be trivially augmented to avoid the orientation problem.
For example, consider the $CDF$ part being replaced with a long rod, long enough to always intersect $AB$. At the midpoint of $AB$, a freely rotating bushing -- one that both rotates, but also allows the rod to move within linearly -- would fix the angle, making the math easy again.
I myself have played with a circular/rotary design. The frame is a circle, with a timing belt running on the outside. It is held in place by three rubber wheels. A stepper motor on the belt controls the rotation of the circular frame. The pen/head assembly moves in a linear slot in the center of the frame, also using a timing belt. The setup works both horizontally and vertically. The frame, two timing belts and pulleys, three outer wheels, and one freewheeling pulley are additional costs compared to the original minimal string-based hanging plotter, but not by that much: roughly 10-20 USD or EUR in late 2016.
Defining $E$ and $G$ using $F$ (at distance $d/2$ at angles $\varphi$ and $\varphi+180°$) yields two pairs of expressions for the coordinates of $F$ as a function of $\varphi$ using the three equations mentioned above, but they have a few dozen terms each.
If we assume the physics and geometry to be such that $\angle AEG = \angle EGB$, we get an additional equation, $$\frac{\vec{EA}\cdot\vec{EG}}{d r_E} = \frac{\vec{GE}\cdot\vec{GB}}{d r_G}$$ Solving the four equations for the coordinates of $E$ and $G$ yields a tangle of solutions. For example, there are three solutions for $r_E$, one of which is sane, but two have thousands of terms.