Page Comparison

The linkage lengths were approximated by measuring with a tape measure. The lengths are given below, measured in inches, as well as the vector equations in phasor form.

$$ R_0 = 10.5 e^{j\theta_0} $$
$$ R_2 = 2.5 e^{j\theta_2} $$
$$ R_3 = 7 e^{j\theta_3} $$
$$ R_4 = 6 e^{j\theta_4} $$
$$ R_5 = 6 \frac{5}{8} e^{j\theta_5} =  6 \frac{5}{8} e^{j(\theta_4-\pi)}$$
$$ R_6 = 7 e^{j\theta_6} $$
$$ R_7 = 12.5 e^{j\theta_7} + 1.25 e^{j(\theta_7-\pi/2)} $$

R7 has the perpendicular offset included as a separate vector, however the angle of that vector can always be known because the link is ternary.

A four-bar mechanism is formed between R0, R2, R3, and R4. The vector loop is provided by:

$$ R_2+R_3-R_4-R_0=0 $$

Theta for the ground link will be assumed 0 for all t, and theta for R2 will be considered as the input link. Measuring from the rocker.

r chair in real life, theta_2 varies from 165 to 15 degrees, with 165 as the closed position, and 15 as the open position.

The velocity loop equation is the time derivative of the vector loop, given by:

$$ \frac d{dt} (R_2+R_3-R_4-R_0) = V_2+V_3-V_4 = j\omega_2 R_2+j\omega_3R_3-j\omega_4R_4 = 0 $$

After solving the four bar composed of R0, R2, R3, and R4, R5 can be solved because the direction of R5 is always opposite R4 The remaining unknown vectors are R6 and R7, which means they can be solved using one more vector loop equation. The equation is given below:

$$ R_5 + R_6 + R_7 +R_3 - R_4 = 0 $$

The angular velocity of link 5 must also be the angular velocity of link 4. The velocity equation:

$$ \frac d{dt} (R_5 + R_6 + R_7 +R_3 - R_4) = V_5 + V_6 +V_7 +V_3 - V_4 = 0 $$
$$ j\omega_7 R_7 +j\omega_6 R_6 = j\omega_4 R_4 -j\omega_4 R_5 - j\omega_3 R_3 $$

The examples are shown in phasor notation, which means that a linear system of 2 equations can be developed from each equation. An example of developing and solving the linear equations is shown for the first vector loop equation, with numbers plugged in.

$$ R_2+R_3-R_4-R_0=0 $$
$$ R_3-R_4 = R_0 - R_2 $$
$$ 7(cos(\theta_3)+jsin(\theta_3)) - 6(cos(\theta_4)+jsin(\theta_4)) = 10.5 - 2.5(cos(\theta_2) + j sin(\theta_2)) $$

\begin{cases} 7cos(\theta_3) - 6cos(\theta_4) = 10.5 - 2.5cos(\theta_2) \\ 7jsin(\theta_3) - 6jsin(\theta_4) = - 2.5j sin(\theta_2) \end{cases}

This is a non-linear system of equations. There are an infinite number of solutions for theta_3 and theta_4. Even when we limit our range to [0,2π], there are 2 unique solutions, because the mechanism has a toggle position. To keep the solution consistent with the real world, all non-linear equations are solved numerically with initial conditions based on the measurements for the chair. In this case, the initial conditions and solutions are

$$ \theta_{3,guess} = 45 \hspace{1ex} degrees ; \theta_{4,guess} = 120 \hspace{1ex} degrees ; \theta_{2,t=0} = 165 \hspace{1ex} degrees $$
\begin{cases} \theta_{3,t=0} = 2.6 \hspace{1ex} degrees \\ \theta_{4,t=0} = 170.8 \hspace{1ex} degrees \end{cases}