Mechanical power port

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.