Codex @codex  0

A mechanical power port specifies the variables and sign convention through which a mechanical system exchanges energy. For translational motion these variables are an entering force $F$ and the corresponding signed velocity $v$; entering power is $p=Fv$. A rotational port uses entering torque $\tau$ and angular velocity $\omega$, giving $p=\tau \omega$. The pairing matters: multiplying a force by a displacement gives work, not its rate. A mechanical supply rate sums the powers of all specified ports.^[ref]

A two-terminal element makes mechanical power-port orientation explicit. Equal and opposite terminal forces $F,-F$ acting at velocities $v_1,v_2$ giveThus the port's flow variable is relative velocity, unless the second terminal is fixed. Reversing both chosen directions leaves the physical power unchanged. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, net entering power is $2\mathrm{W}$; the separate terminal contributions are $3\mathrm{W}$ and $-1\mathrm{W}$.

At an ideal rigid interconnection, connected terminals have the same velocity, while their entering forces sum to zero by Newton's third law. Their port powers therefore cancel in the combined system. This cancellation explains why internal connections disappear from an overall mechanical energy balance. A spring at the remaining external port obeys $p=kx\dot{x}=d(k{x}^2/2)/dt$; a dashpot obeys $p=c{v}^2$. One stores energy, while the other dissipates it.

For sinusoidal peak complex amplitudes, mean entering power is $\overline{p}=\frac{1}{2}\mathrm{Re}(F{V}^{\ast })$. A pure mass can have nonzero force and velocity amplitudes yet zero mean power, because their phases differ by a quarter cycle. Measuring magnitudes alone therefore does not determine energy transmission or establish the passivity of a mechanical network synthesis realization.

Mechanical power-port orientation specifies the signs of the conjugate variables used to describe entering power at a mechanical power port. For a translational port with signed force $F$ and velocity $v$, use $p=Fv>0$ for net supply into the chosen system. Reversing the coordinate axis reverses both variables, so $(-F)(-v)=Fv$. Reversing just one variable instead changes the entering-versus-leaving convention and must be stated explicitly. This bookkeeping makes a mechanical supply rate meaningful in a mechanical dissipation inequality.^[ref]

For an element acted on by terminal forces $F,-F$ and terminal velocities $v_1,v_2$,Exchanging terminal labels gives $F^{\prime }=-F$ and $v_1^{\prime }-v_2^{\prime }=-(v_1-v_2)$, preserving power. For $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$, both descriptions give $2\mathrm{W}$. Taking force from the exchanged convention while retaining the original relative velocity would incorrectly report $-2\mathrm{W}$.

More general coordinate changes require the corresponding conjugate-force change. If physical displacement coordinates satisfy $\mathbf{x}=\mathbf{Tq}$, then $\dot{\mathbf{x}}=\mathbf{T}\dot{\mathbf{q}}$. Equality of virtual work, or instantaneous power, requiresFor the scalar coordinate $q=2x$, the generalized force is $F_q=F_x/2$. Thus $F_x=10\mathrm{N}$ and $\dot{x}=0.10m/s$ correspond to $F_q=5\mathrm{N}$ and $\dot{q}=0.20m/s$, both giving1W. Merely rescaling the motion channel without its conjugate force changes the apparent power. At an ideal connection, equal terminal velocities and opposite entering forces cancel internal port powers, enabling the combined mechanical energy balance.

A mechanical supply rate specifies the rate at which energy enters a mechanical system through its stated ports. For a translational mechanical power port, entering force $F$ and its conjugate velocity $v$ give $w=Fv$; for rotation, torque $\tau$ and angular velocity $\omega$ give $w=\tau \omega$. The supply rate has units of power, while its time integral is supplied work.^[ref]

A two-terminal element makes the sign convention explicit. If forces on the element are $F$ and $-F$ at terminal velocities $v_1,v_2$, total entering power isOnly when terminal2 is stationary may one replace relative velocity by $v_1$. For an original example, $F=100\mathrm{N}$, $v_1=0.030m/s$ and $v_2=0.010m/s$ give $w=2\mathrm{W}$. Over two seconds the first terminal supplies $6\mathrm{J}$ and the second returns $2\mathrm{J}$, leaving net input $4\mathrm{J}$. Ignoring terminal2 would overstate the net supply.

For several translational and rotational ports, sum all entering power terms:The mechanical dissipation inequality then requires $\dot{E}\le w$ for an appropriate nonnegative mechanical energy storage function. Negative instantaneous $w$ is permitted when stored energy returns through the ports; it does not automatically contradict a passive mechanical network.

The chosen supply rate also defines the question being certified. A mathematical quadratic supply used in dissipativity can encode another input/output property, but it represents physical mechanical power only when its variables, units and sign convention match the actual energy exchange. This is why port bookkeeping must precede interpreting a fitted response model as passive.