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$$ R_2+R_3-R4-R0R_4-R_0=0 $$ (1) |
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.
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$$ \frac d{dt} (R_2+R_3-R_4-R_0) = V_2+V_3-V_4 = j\omega_2 R_2+j\omega_3R_3-j\omega_4R_4 = 0 $$ (2) |
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:
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$$ \frac d{dt} (R_5 + R_6 + R_7 +R_3 - R_4) = V_5 + V_6 +V_7 +V_3 - V_4 = 0 $$
$$ j\omega_7 R_7 +j\omega_6 R_6 = j\omega_4 R_4 -j\omega_4 R_5 - j\omega_3 R_3 $$ |
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.
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$$ R_2+R_3-R4-R0=0 $$
$$ R_3-R_4 $$ |