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Part VII: Acoustical circuits

3.4. Acoustical elements

Acoustical circuits are frequently more difficult to draw than mechanical ones because the elements are less easy to identify. As was the case for mechanical circuits, the more obvious forms of the elements will be useful as an intermediate step toward drawing the analogous circuit diagram. When the student is more familiar with acoustical circuits, he or she will be able to pass directly from the acoustic device to the final form of the equivalent circuit.

In acoustic devices, the quantity we are able to measure most easily without modification of the device is sound pressure. Such a measurement is made by inserting a small hollow probe tube into the sound field at the desired point. This probe tube leads to one side of a microphone diaphragm. The other side of the diaphragm is exposed to atmospheric pressure. A movement of the diaphragm takes place when there is a difference in pressure across it. This difference between atmospheric pressure and the incremental pressure created by the sound field is the sound pressure

\tilde{p}

$\tilde{p}$

Because we can measure sound pressure by such a probe tube arrangement without disturbing the device, it seems that sound pressure is analogous to voltage in electrical circuits. Such a choice requires us to consider current as being analogous to some quantity which is proportional to velocity. As we shall show shortly, a good choice is to make current analogous to volume velocity, the volume of gas displaced per second.

A strong argument can be made for this choice of analogy when one considers the relations governing the flow of air inside such acoustic devices as loudspeakers, microphones, and noise filters. Inside a certain type of microphone, for example, there is an air cavity that connects to the outside air through a small tube (see Fig. 3.22). Assume, now, that the outer end of this tube is placed in a sound wave. The wave will cause a movement of the air particles in the tube. Obviously, there is a junction between the tube and the cavity at the inner end of the tube at point A. Let us ask ourselves the question “What physical quantities are continuous at this junction point?ˮ.

First, the sound pressure just inside the tube at A is the same as that in the cavity just outside A. That is to say, we have continuity of sound pressure. Second, the quantity of air leaving the inner end of the small tube in a given interval of time is the quantity that enters the cavity in the same interval of time. That is, the mass per second of gas leaving the small tube equals the mass per second of gas entering the volume. Because the pressure is the same at both places, the density of the gas must also be the same, and it follows that there is continuity of volume velocity (cubic meters per second or m³/s) at this junction. Analogously, in the case of electricity, there is continuity of electric current at a junction. Continuity of volume velocity must exist even if there are several tubes or cavities joining near one point. A violation of the law of conservation of mass otherwise would occur.

Figure 3.22 Closed cavity connecting to the outside air through a tube of cross-sectional area S. The junction plane between the tube and the cavity occurs at A.

We conclude that the quantity that flows through our acoustical elements must be the volume velocity U in m³/s and the drop across our acoustical elements must be the pressure p in Pa. This conclusion indicates that the impedance type of analogy is the preferred analogy for acoustical circuits. The product of the effective sound pressure p times the in-phase component of the effective volume velocity U gives the acoustic power in W.

In this part, we shall discuss the more general aspects of acoustical circuits. In Chapter 4, we explain fully the approximations involved and the rules for using the concepts enunciated here in practical problems.

Acoustic mass M _A

Acoustic mass is a quantity proportional to mass but having the dimensions of kg/m⁴. It is associated with a mass of air accelerated by a net force which acts to displace the gas without appreciably compressing it. The concept of acceleration without compression is an important one to remember. It will assist you in distinguishing acoustic masses from other elements.

The acoustical element that is used to represent an acoustic mass is a tube filled with the gas as shown in Fig. 3.23.

The physical law governing the motion of a mass that is acted on by a force is Newton's second law, f(t) = M _M du(t)/dt. This law may be expressed in acoustical terms as follows:

\frac{f (t)}{S} = \frac{M_{M}}{S} \frac{d [u (t) S]}{d t S} = p (t) = \frac{M_{M}}{S^{2}} \frac{d U (t)}{d t}

$\frac{f (t)}{S} = \frac{M_{M}}{S} \frac{d [u (t) S]}{d t S} = p (t) = \frac{M_{M}}{S^{2}} \frac{d U (t)}{d t}$

p (t) = M_{A} \frac{d U (t)}{d t}

$p (t) = M_{A} \frac{d U (t)}{d t}$

(3.22)

where

p(t) is the instantaneous difference between pressures in Pa existing at each end of a mass of gas of M _M kg undergoing acceleration.
M _A = M _M/S ² is the acoustic mass in kg/m⁴ of the gas undergoing acceleration. This quantity is nearly equal to the mass of the gas inside the containing tube divided by the square of the cross-sectional area. To be more exact, we must note that the gas in the immediate vicinity of the ends of the tube also adds to the mass. Hence, there are “ end corrections” that must be considered. These corrections are discussed in Chapter 4 (page 121).
Figure 3.23 Tube of length l and cross-sectional area S.
Figure 3.24 (a) Impedance-type and (b) admittance-type symbols for an acoustic mass.
U(t) is the instantaneous volume velocity of the gas in m³/s across any cross-sectional plane in the tube. The volume velocity U(t) is equal to the linear velocity u(t) multiplied by the cross-sectional area S.

In the steady state, with an angular frequency ω, we have

\tilde{p} = j ω M_{A} \tilde{U}

$\tilde{p} = j ω M_{A} \tilde{U}$

(3.23)

where

\tilde{p}

$\tilde{p}$

and

\tilde{U}

$\tilde{U}$

are taken to be complex quantities.

The impedance-type analogous symbol for acoustic mass is shown in Fig. 3.24(a), and the admittance-type is given in Fig. 3.24(b). In the steady state, for either, we get Eq. (3.23). The arrows point in the direction of positive flow or positive drop.

Acoustic compliance C _A

Acoustic compliance is a constant quantity having the dimensions of m⁵/N. It is associated with a volume of air that is compressed by a net force without an appreciable average displacement of the center of gravity of air in the volume. In other words, compression without acceleration identifies an acoustic compliance.

The acoustical element that is used to represent an acoustic compliance is a volume of air drawn as shown in Fig. 3.25.

The physical law governing the compression of a volume of air being acted on by a net force was given as

f (t) = (1 / C_{M}) \int u (t) d t .

$f (t) = (1 / C_{M}) \int u (t) d t .$

Figure 3.25 Enclosed volume of air V with opening for entrance of pressure variations.

Converting from mechanical to acoustical terms,

\frac{f (t)}{S} = \frac{1}{C_{M} S} \int u (t) \frac{S}{S} d t or p (t) = \frac{1}{C_{M} S^{2}} \int U (t) d t

$\frac{f (t)}{S} = \frac{1}{C_{M} S} \int u (t) \frac{S}{S} d t or p (t) = \frac{1}{C_{M} S^{2}} \int U (t) d t$

p (t) = \frac{1}{C_{A}} \int U (t) d t .

$p (t) = \frac{1}{C_{A}} \int U (t) d t .$

(3.24)

where

p(t) is instantaneous pressure in Pa acting to compress the volume V of the air.
C _A = C _M S ² is acoustic compliance in m⁵/N of the volume of the air undergoing compression. The acoustic compliance is nearly equal to the volume of air divided by γP ₀, as we shall see in Chapter 4 (page 121 to 125).
U(t) is instantaneous volume velocity in m³/s of the air flowing into the volume that is undergoing compression. The volume velocity U(t) is equal to the linear velocity u(t) multiplied by the cross-sectional area S.

In the steady state with an angular frequency ω, we have

\tilde{p} = \frac{\tilde{U}}{j ω C_{A}},

$\tilde{p} = \frac{\tilde{U}}{j ω C_{A}},$

(3.25)

where

\tilde{p}

$\tilde{p}$

and

\tilde{U}

$\tilde{U}$

are taken to be complex quantities.

The impedance-type analogous element for acoustic compliance is shown in Fig. 3.26(a) and the admittance-type in Fig. 3.26(b). In the steady state for either, Eq. (3.25) applies.

Acoustic resistance R _A and acoustic conductance G _A

Acoustic resistance R _A is associated with the dissipative losses occurring when there is a viscous movement of a quantity of gas through a fine mesh screen or through a capillary tube. The unit is N·s/m⁵ or rayls/m².

Figure 3.26 (a) Impedance-type and (b) admittance-type symbols for an acoustic compliance.

Figure 3.27 Fine mesh screen which serves as an acoustical symbol for acoustic resistance.

The acoustic element used to represent an acoustic resistance is a fine mesh screen drawn as shown in Fig. 3.27.

The reciprocal of acoustic resistance is the acoustic conductance G _A. The unit is m⁵/N·s or acoustic siemens.

The physical law governing dissipative effects in a mechanical system was given by f(t) = R _M u(t) or, in terms of acoustical quantities,

p (t) = R_{A} U (t) = \frac{1}{G_{A}} U (t),

$p (t) = R_{A} U (t) = \frac{1}{G_{A}} U (t),$

(3.26)

where

p(t) is the difference between instantaneous pressures in Pa across the dissipative element.
R _A = R _M/S ² is acoustic resistance in N·s/m⁵.
G _A = G _M S ² is acoustic conductance in m⁵/N·s.
U(t) is instantaneous volume velocity in m³/s of the gas through the cross-sectional area of resistance.

The impedance-type analogous symbol for acoustic resistance is shown in Fig. 3.28(a) and the admittance-type in Fig. 3.28(b).

Acoustic generators

Acoustic generators can be of either the constant volume velocity or the constant pressure type. The prime movers in our acoustical circuits will be exactly like those shown in Fig. 3.7 and 3.9 except that

{\tilde{u}}_{2}

${\tilde{u}}_{2}$

often will be zero and

{\tilde{u}}_{1}

${\tilde{u}}_{1}$

will be the velocity of a small piston of area S. Remembering that

\tilde{u} = {\tilde{u}}_{1} - {\tilde{u}}_{2}

$\tilde{u} = {\tilde{u}}_{1} - {\tilde{u}}_{2}$

, we see that the generator of Fig. 3.7 has a constant volume velocity

\tilde{U} = \tilde{u} S

$\tilde{U} = \tilde{u} S$

and that of Fig. 3.9 a constant pressure of

\tilde{p} = \tilde{f} / S

$\tilde{p} = \tilde{f} / S$

Figure 3.28 (a) Impedance-type symbol for acoustic resistance and (b) admittance-type symbol for acoustic conductance.

Figure 3.29 (a) Impedance-type and (b) admittance-type symbols for a constant pressure generator.

Figure 3.30 (a) Impedance-type and (b) admittance-type symbols for a constant volume velocity generator.

The two types of analogous symbols for acoustic generators are given in Figs. 3.29 and 3.30. The arrows point in the direction of the positive terminal or the positive flow. As before, a wave inside the circle indicates zero impedance or admittance and an arrow inside the circle indicates infinite impedance or admittance.

Example 3.4. The acoustic device of Fig. 3.31 consists of three cavities V ₁, V ₂, and V ₃, two fine mesh screens R _A1 and R _A2, four short lengths of tube T ₁, T ₂, T ₃, and T ₄, and a constant pressure generator. Because the air in the tubes is not confined, it experiences negligible compression. Because the air in each of the cavities is confined, it experiences little average movement. Let the force of the generator be

f (t) = 10^{- 5} \cos ω t N

$f (t) = 10^{- 5} \cos ω t N$

so that

Figure 3.31 Acoustic device consisting of four tubes, three cavities, and two screens driven by a constant pressure generator.

Figure 3.32 Impedance-type analogous circuit for the acoustic device of Fig. 3.31.

\tilde{f} = 10^{- 5} e^{j ω t} N

$\tilde{f} = 10^{- 5} e^{j ω t} N$

| \tilde{f} | = 10^{- 5} N at ω = 1000 Hz .

$| \tilde{f} | = 10^{- 5} N at ω = 1000 Hz .$

Also, let the radius of the tube a = 0.5 cm; the length of each of the four tubes l = 5 cm; the volume of each of the three cavities V = 10 cm³; and the magnitude of the two acoustic resistances R _A = 10 N·s/m⁵. Neglecting end corrections, solve for the volume velocity

{\tilde{U}}_{0}

${\tilde{U}}_{0}$

at the end of the tube T ₄.

Solution. Remembering that there is continuity of volume velocity and pressure at the junctions, we can draw the impedance-type analogous circuit from inspection. It is shown in Fig. 3.32. The bottom line of the schematic diagram represents atmospheric pressure, which means that here the variational pressure

\tilde{p}

$\tilde{p}$

is equal to zero. At each of the junctions of the elements 1 to 4, a different variational pressure can be observed. The end of the fourth tube (T ₄) opens to the atmosphere, which requires that M _A4 be connected directly to the bottom line of Fig. 3.32.

Note that the volume velocity of the gas leaving the tube T ₁ is equal to the sum of the volume velocities of the gas entering V ₁ and T ₂. The volume velocity of the gas leaving T ₂ is the same as that flowing through the screen R _A1 and is equal to the sum of the volume velocities of the gas entering V ₂ and T ₃.

One test of the validity of an analogous circuit is its behavior for direct current. If one removes the piston and blows into the end of the tube T ₁ (Fig. 3.31), a steady flow of air from T ₄ is observed. Some resistance to this flow will be offered by the two screens R _A1 and R _A2. Similarly in the schematic diagram of Fig. 3.32, a steady pressure

\tilde{p}

$\tilde{p}$

will produce a steady flow

\tilde{U}

$\tilde{U}$

through M _A4, resisted only by R _A1 and R _A2.

As an aside, let us note that an acoustic compliance can occur in a circuit without one of the terminals being at ground potential only if it is produced by an elastic diaphragm. For example, if the resistance in Fig. 3.31 were replaced by an impervious but elastic diaphragm, the element R _A1 in Fig. 3.32 would be replaced by a compliance-type element with both terminals above ground potential. In this case, a steady flow of air could not be maintained through the device of Fig. 3.31 as can also be seen from the circuit of Fig. 3.32, with R _A1 replaced by a compliance.

Determine the element sizes of Fig. 3.32:

\tilde{p} = \frac{\tilde{f}}{S} = \frac{10^{- 5} e^{j 1000 t}}{π {(5 \times 10^{- 3})}^{2}} = 0.1273 e^{j 1000 t} Pa,

$\tilde{p} = \frac{\tilde{f}}{S} = \frac{10^{- 5} e^{j 1000 t}}{π {(5 \times 10^{- 3})}^{2}} = 0.1273 e^{j 1000 t} Pa,$

M_{A 1} = M_{A 2} = M_{A 3} = M_{A 4} = \frac{ρ_{0} l}{S} = \frac{1.18 \times 0.05}{7.85 \times 10^{- 5}} = 750 {kg/m}^{4},

$M_{A 1} = M_{A 2} = M_{A 3} = M_{A 4} = \frac{ρ_{0} l}{S} = \frac{1.18 \times 0.05}{7.85 \times 10^{- 5}} = 750 {kg/m}^{4},$

C_{A 1} = C_{A 2} = C_{A 3} = \frac{V}{γ P_{0}} = \frac{10^{- 5}}{1.4 \times 10^{5}} = 7.15 \times 10^{- 11} m^{5} /N,

$C_{A 1} = C_{A 2} = C_{A 3} = \frac{V}{γ P_{0}} = \frac{10^{- 5}}{1.4 \times 10^{5}} = 7.15 \times 10^{- 11} m^{5} /N,$

R_{A 1} = R_{A 2} = 10 N \cdot s / m^{5} .

$R_{A 1} = R_{A 2} = 10 N \cdot s / m^{5} .$

We now determine the ratio

\tilde{p} / {\tilde{U}}_{0}

$\tilde{p} / {\tilde{U}}_{0}$

{\tilde{p}}_{4} = j ω M_{A 4} {\tilde{U}}_{0} = j 7.5 \times 10^{5} \times {\tilde{U}}_{0} Pa

${\tilde{p}}_{4} = j ω M_{A 4} {\tilde{U}}_{0} = j 7.5 \times 10^{5} \times {\tilde{U}}_{0} Pa$

{\tilde{U}}_{5} = j ω C_{A 3} {\tilde{p}}_{4} = - 5.36 \times 10^{- 2} \times {\tilde{U}}_{0} m^{3} /s

${\tilde{U}}_{5} = j ω C_{A 3} {\tilde{p}}_{4} = - 5.36 \times 10^{- 2} \times {\tilde{U}}_{0} m^{3} /s$

{\tilde{U}}_{4} = {\tilde{U}}_{5} + {\tilde{U}}_{0} = 0.946 {\tilde{U}}_{0}

${\tilde{U}}_{4} = {\tilde{U}}_{5} + {\tilde{U}}_{0} = 0.946 {\tilde{U}}_{0}$

{\tilde{p}}_{3} = (R_{A 2} + j ω M_{A 3}) {\tilde{U}}_{4} + {\tilde{p}}_{4} = (9.46 + j 14.6 \times 10^{5}) {\tilde{U}}_{0}

${\tilde{p}}_{3} = (R_{A 2} + j ω M_{A 3}) {\tilde{U}}_{4} + {\tilde{p}}_{4} = (9.46 + j 14.6 \times 10^{5}) {\tilde{U}}_{0}$

{\tilde{U}}_{3} = j ω C_{A 2} {\tilde{p}}_{3} = (- 0.1043 + j 6.77 \times 10^{- 7}) {\tilde{U}}_{0}

${\tilde{U}}_{3} = j ω C_{A 2} {\tilde{p}}_{3} = (- 0.1043 + j 6.77 \times 10^{- 7}) {\tilde{U}}_{0}$

{\tilde{U}}_{2} = {\tilde{U}}_{3} + {\tilde{U}}_{4} = (0.842 + j 6.77 \times 10^{- 7}) {\tilde{U}}_{0}

${\tilde{U}}_{2} = {\tilde{U}}_{3} + {\tilde{U}}_{4} = (0.842 + j 6.77 \times 10^{- 7}) {\tilde{U}}_{0}$

{\tilde{p}}_{2} = (R_{A 1} + j ω M_{A 2}) {\tilde{U}}_{2} + {\tilde{p}}_{3} = (17.37 + j 2.091 \times 10^{6}) {\tilde{U}}_{0}

${\tilde{p}}_{2} = (R_{A 1} + j ω M_{A 2}) {\tilde{U}}_{2} + {\tilde{p}}_{3} = (17.37 + j 2.091 \times 10^{6}) {\tilde{U}}_{0}$

{\tilde{U}}_{1} = j ω C_{A 1} {\tilde{p}}_{2} = (- 0.1496 + j 1.242 \times 10^{- 6}) {\tilde{U}}_{0}

${\tilde{U}}_{1} = j ω C_{A 1} {\tilde{p}}_{2} = (- 0.1496 + j 1.242 \times 10^{- 6}) {\tilde{U}}_{0}$

\tilde{U} = {\tilde{U}}_{2} + {\tilde{U}}_{1} = (0.692 + j 1.919 \times 10^{- 6}) {\tilde{U}}_{0}

$\tilde{U} = {\tilde{U}}_{2} + {\tilde{U}}_{1} = (0.692 + j 1.919 \times 10^{- 6}) {\tilde{U}}_{0}$

\tilde{p} = j ω M_{A 1} \tilde{U} + {\tilde{p}}_{2} = (15.93 + j 2.61 \times 10^{6}) {\tilde{U}}_{0} .

$\tilde{p} = j ω M_{A 1} \tilde{U} + {\tilde{p}}_{2} = (15.93 + j 2.61 \times 10^{6}) {\tilde{U}}_{0} .$

The desired value of

{\tilde{U}}_{0}

${\tilde{U}}_{0}$

{\tilde{U}}_{0} = \frac{0.1273 e^{j 1000 t}}{15.93 + j 2.61 \times 10^{6}}

${\tilde{U}}_{0} = \frac{0.1273 e^{j 1000 t}}{15.93 + j 2.61 \times 10^{6}}$

\begin{matrix} U (t) \approx 4.88 \times 10^{- 8} \cos (1000 t - 90 °) \\ \approx 4.88 \times 10^{- 8} \sin 1000 t . \end{matrix}

$\begin{matrix} U (t) \approx 4.88 \times 10^{- 8} \cos (1000 t - 90 °) \\ \approx 4.88 \times 10^{- 8} \sin 1000 t . \end{matrix}$

In other words, the impedance is principally that of the four acoustic masses in series so that

{\tilde{U}}_{0}

${\tilde{U}}_{0}$

lags

\tilde{p}

$\tilde{p}$

by nearly 90 degrees.

Example 3.5. A Helmholtz resonator is frequently used as a means for eliminating an undesired frequency component from an acoustic system. An example is given in Fig. 3.33(a). A constant force generator G produces a series of tones, among which is one that is not wanted. These tones actuate a microphone M whose acoustic impedance is 500 N·s/m⁵. If the tube T has a cross-sectional area of 5 cm², l ₁ = l ₂ = 5 cm, l ₃ = 1 cm, V = 1000 cm³, and the cross-sectional area of l ₃ is 2 cm², what frequency is eliminated from the system?

Solution. By inspection we may draw the impedance-type analogous circuit of Fig. 3.33(b). The element sizes are

Figure 3.33 (a) Acoustic device consisting of a constant force generator G, piston P, tube T with length l ₁ + l ₂, microphone M, and Helmholtz resonator R with volume V and connecting tube as shown. (b) Impedance-type analogous circuit for the device of (a).

M_{A 1} = M_{A 2} = \frac{ρ_{0} l_{1}}{S_{T}} = \frac{1.18 \times 0.05}{5 \times 10^{- 4}} = 118 kg / m^{4},

$M_{A 1} = M_{A 2} = \frac{ρ_{0} l_{1}}{S_{T}} = \frac{1.18 \times 0.05}{5 \times 10^{- 4}} = 118 kg / m^{4},$

M_{A 3} = \frac{ρ_{0} l_{3}}{S_{R}} = \frac{1.18 \times 0.01}{2 \times 10^{- 4}} = 59 kg / m^{4},

$M_{A 3} = \frac{ρ_{0} l_{3}}{S_{R}} = \frac{1.18 \times 0.01}{2 \times 10^{- 4}} = 59 kg / m^{4},$

C_{A 3} = \frac{V}{γ P_{0}} = \frac{10^{- 3}}{1.4 \times 10^{5}} = 7.15 \times 10^{- 9} m^{5} / N,

$C_{A 3} = \frac{V}{γ P_{0}} = \frac{10^{- 3}}{1.4 \times 10^{5}} = 7.15 \times 10^{- 9} m^{5} / N,$

R_{A 1} = 500 N \cdot s / m^{5} .

$R_{A 1} = 500 N \cdot s / m^{5} .$

It is obvious that the volume velocity

{\tilde{U}}_{2}

${\tilde{U}}_{2}$

of the transducer M will be zero when the shunt branch is at resonance. Hence,

ω = \frac{1}{\sqrt{M_{A 3} C_{A 3}}} = \frac{10^{4}}{\sqrt{42.2}} = 1540 rad/s,

$ω = \frac{1}{\sqrt{M_{A 3} C_{A 3}}} = \frac{10^{4}}{\sqrt{42.2}} = 1540 rad/s,$

f = 245 Hz .

$f = 245 Hz .$

Mechanical rotational systems

Mechanical rotational systems are handled in the same manner as mechanical rectilineal systems. Table 3.4 shows quantities analogous in the two systems.

Table 3.4

Analogous quantities in rectilineal and rotational systems
Rectilineal systems			Rotational systems
Symbol	Quantity	Unit	Symbol	Quantity	Unit
$\tilde{f}$ $\tilde{f}$	Force	N	$\tilde{T}$ $\tilde{T}$	Torque	N·m
$\tilde{u}$ $\tilde{u}$	Velocity	m/s	$\tilde{θ}$ $\tilde{θ}$	Angular velocity	rad/s
$\tilde{ξ}$ $\tilde{ξ}$	Displacement	m	$\tilde{ϕ}$ $\tilde{ϕ}$	Angular displacement	rad
$Z_{M} = \tilde{f} / \tilde{u}$ $Z_{M} = \tilde{f} / \tilde{u}$	Mechanical impedance	N·s/m	$Z_{R} = \tilde{T} / \tilde{θ}$ $Z_{R} = \tilde{T} / \tilde{θ}$	Rotational impedance	N·m·s/rad
$Y_{M} = \tilde{u} / \tilde{f}$ $Y_{M} = \tilde{u} / \tilde{f}$	Mechanical admittance	m/N⁻¹·s⁻¹	$Y_{R} = \tilde{θ} / \tilde{T}$ $Y_{R} = \tilde{θ} / \tilde{T}$	Rotational admittance	rad/N·m·s
R _M	Mechanical resistance	N·s/m	R _R	Rotational resistance	N·m·s/rad
G _M	Mechanical conductance	m·N⁻¹/s⁻¹	G _R	Rotational conductance	rad/N·m·s
M _M	Mass	kg	I _R	Moment of inertia	kg·m²
C _M	Mechanical compliance	m/N	C _R	Rotational compliance	rad/N·m
W _M	Mechanical power	W	W _R	Rotational power	W