Theory of Repulsion Motors

The present article is an attempt to give a theory of the repulsion motor from the point of view of the transverse field produced by the rotor. Such a decomposition of the rotating field, analogous to the decompositions of Potier, Görges, and Sumer in the single-phase asynchronous motor, offers, in the author’s opinion, greater advantages than the decomposition of the stator winding, in the sense of greater clarity of the process and the convenience of the constants, and, as it seems to him, has more than merely mathematical significance.

Circular diagrams, which give such a vivid picture of the properties of a motor and predictions for the designer, have, since the well-known article by Ossan, become widely used; therefore the present article is devoted specifically to deriving and studying the circular diagram of the repulsion motor, at the same time covering all its properties, with the exception of commutation, to which the author hopes to devote a subsequent article. In deriving the circular diagram, the phase displacement between the magnetizing current and the magnetic flux is taken into account; the greater accuracy of the diagram and the possibility of allowing for losses in iron as a result of this, in the author’s opinion, compensate for the complexity of the formulas.

In what follows the following notation is adopted:

\(P_1\) — voltage at the stator terminals,

\(E_{cm} = E_1\) — electromotive force of pulsation in the stator winding.

\(E_2 = E_{nc_1}\) — electromotive force of pulsation in the longitudinal coil of the rotor due to the longitudinal flux,

\(E_{nc_2}\) — electromotive force of pulsation in the transverse coil of the rotor due to the transverse field,

\(E_{vr.1}\) — electromotive force of rotation of the longitudinal coil of the rotor in the transverse field,

\(E_{vr.2}\) — electromotive force of rotation of the transverse coil of the rotor in the longitudinal field,

\(\Phi_{nd}\) — longitudinal magnetic flux,

\(\Phi_{nr}\) — transverse magnetic flux,

$B_1, B_{nd}, B_{np}, B_2$ — the maximum inductions: of the total longitudinal, transverse, and resultant rotor fields,

$J_1, J_2, J_a$ — primary, secondary, and magnetizing currents,

$r_1, x_1$ and $z_1 = r_1 - jx_1$ — the ohmic, reactive, and apparent impedances of the stator winding,

$r_2, x_2$ and $z_2 = r_2 - jx_2$ — the ohmic, reactive, and apparent impedances of the rotor winding,

$g_a, b_a$ and $y_a = g_a + jb_a$ — the conductance of the watt component, the wattless component, and the apparent conductance of the exciting current,

$g_{np}, b_{np}$ and $y_{np} = g_{np} + jb_{np}$ — the conductance of the watt component, the wattless component, and the apparent conductance of the transverse coil of the rotor,

$x_{nd}, x_{np} = \dfrac{1}{b_{np}}$ — the reactance of the longitudinal and transverse rotor coils,

$x_3 = x_{np}\cos\gamma,\ x_4 = x_{nd}\cos\gamma,\ x_s = x_1 + x'_2,\ x_{\mathrm{II}} = x'_3 + x'_2,\ x_{\mathrm{III}} = x_4 + x'_2 + x'_3,$

$W_1 = P_1 J_1 \cos\varphi_1$ — the power received by the stator,

$W_a$ — the power received by the rotor,

$W_2$ — the power delivered by the rotor,

$V_1 = J_1^2 r_1$ — the ohmic losses of the stator,

$V_2 = J_2^2 r_2$ — the ohmic losses of the rotor,

$V_{a,1}, V_{a,2}$ — the iron losses from the longitudinal and transverse fields,

$\varphi_1$ — the phase-displacement angle between $P_1$ and $J_1$,

$\psi$ — the same between $E_1$ and $J_1$,

$\psi_2, v$ — the same between $E_2$ and $J_2$,

$\gamma$ — the same between $b_a$ and $y_a$, $b_{np}$ and $y_{np}$,

$\psi_1$ — the same between $r_1$ and $z_1$,

$\tau$ — pole pitch,

$\lambda$ — angle of displacement of the brushes from the longitudinal axis of the motor,

$\beta_1$ — ratio of the width of the stator winding to $\tau$,

$\beta_2$ — ratio of the width of the rotor winding lying within the limits of the angle $180 - 2\lambda$ to $\tau$,

$\beta_3 = 1 - \beta_2$ — ratio of the width of the rotor winding lying within the limits of the angle $2\lambda$ to $\tau$,

$\beta_4$ — ratio of the width of the rotor winding lying within the limits of the angle $180 - \lambda$ to $\tau$,

\[ c_1 = \frac{E_{vr.2\ \mathrm{synchr.}}}{E_{ps.1}} \]
— the ratio of the electromotive force of rotation $E_{vr.2}$ at synchronous speed to the electromotive force of pulsation $E_{ps.1}$,

\[ c_2 = \frac{E_{vr.1\ \mathrm{synchr.}}}{E_{ps.2}} \]
— the ratio of the electromotive force of rotation $E_{vr.1}$ at synchronous speed to the electromotive force of pulsation $E_{ps.2}$,

$\mathscr{N}_1$ — number of periods of the primary circuit,

$n_2$ — number of revolutions of the rotor per minute,

\[ \mathscr{N}_2=\frac{n_2}{60} \]
— number of periods of rotation,

\[ \nu=\frac{\mathscr{N}_1}{\mathscr{N}_2} \]
— ratio of the numbers of periods of pulsation and rotation,

\[ u=\frac{E_1}{E_2} \]
— coefficient of transformation,

$N_1, N_2, N_{nd}, N_{np}$ — number of bars of the stator, of the rotor, and of its longitudinal and transverse coils,

\[ \nu=\frac{N_2}{2\tau} \]
— specific number of bars,

\[ w_2=\frac{N_1}{2} \]
number of turns of the stator,

\[ w_2=\frac{N_2}{2} \]
— number of turns of the rotor,

$M_{nd}$ — moment of the longitudinal field with the transverse coil,

$M_{np}$ — moment of the transverse field with the longitudinal coil,

$M$ — resultant moment,

$D$ — diameter of the rotor,

$l$ — length of the rotor,

$\rho, \xi_0, \eta_0$ with the corresponding values — radius and coordinates of the center of the various circles,

$R, X, Z, Y$ with the corresponding values — active resistance, reactance, apparent resistance, and apparent conductance of the various circuits,

$m, n, p, q$ — scales of the circles $K_p'''$, $K_p''$, $K_p'$ and $K_p$.

A letter with a prime mark $'$ denotes the corresponding quantity referred to the primary circuit; for example, $J_2'$ is the rotor current referred to the primary circuit. A letter with a dot below denotes a quantity in symbolic notation.

All formulas refer to a two-pole motor, since converting them to a multipole motor presents no difficulty.

Here the assumption has been adopted of proportionality of the transverse flux $\Phi_{np}$ to the secondary current $J_2$, and the usual assumptions are also made: sinusoidality of the current, distribution of the winding in the form of a uniform layer, etc.

Let us consider the armature of a direct-current motor (Fig. 1-a), through which current from an external source flows via the brushes in the direction indicated in the drawing. This armature creates a magnetic field \(\Phi\) in the direction \(ab\). Let us mentally decompose its winding into two coils, \(ahf-egb\), which we shall call the longitudinal one, and \(ace-fdb\), which we shall call the transverse one. Accordingly, the triangular field of the rotor \(abc\) (Fig. 1-b) will be decomposed into two trapezoidal fields: the longitudinal \(\Phi_{nd}\) (defg, Fig. 1-b), with direction \(cd\), created by the longitudinal coil, and the transverse \(\Phi_{np}\) (hiklmn, Fig. 1-b), with direction \(hg\), created by the transverse coil. The field \(\Phi_{nd}\) with the transverse coil gives a torque \(M_{nd}=k\cdot B_{nd}\cdot N_{np}\cdot J_2\) (where \(k\) is a certain constant), directed counterclockwise, while the field \(\Phi_{np}\) with the longitudinal coil gives a torque \(M_{np}=k B_{np}\cdot N_{nd} J_2\), directed clockwise. It is not difficult to see that, so long as the magnetic permeability of the space surrounding the armature is the same in all directions, these torques mutually balance, since

\[ \frac{B_{nd}}{B_{np}}=\frac{N_{nd}}{N_{np}}, \]

and the armature is at rest; but as soon as iron masses are set opposite one of these coils, for example in the form of poles, as shown by the dotted line in Fig. 1-a, then \(B_{nd}\) increases, \(M_{nd}\) becomes

Fig. 1.

greater than \(M_{np}\), and the armature begins to rotate in the direction opposite to the motion of the hands of a clock (an unused reactive direct-current motor). If on these poles there is placed a winding creating a field directed opposite to \(\Phi_{nd}\), then the torques \(M_{nd}\) and \(M_{np}\) already act in the same direction, and the motor begins to rotate in the clockwise direction (an ordinary direct-current motor). Since in such motors against the transverse co—

…coil is absent in the vicinity of the iron masses, then the torque \(M_{np}\) is here very small in comparison with \(M_{nd}\), and thus these motors, and consequently all direct-current motors, belong to the class of motors with a longitudinal field.

Single-phase commutator motors likewise have representatives of this class in the form of the series motor alone. But here there also exists another class of motors, in which the dominant role is played not by the torque of the transverse coil in the longitudinal field, but, on the contrary, by that of the longitudinal coil in the transverse field; these are the so-called repulsion motors. To this class belong the Thomson motor (Fig. 2), the Deri motor (Fig. 3), and the Atkinson motor (Fig. 4).

These motors are usually constructed with a uniform inter-iron space and a distributed winding. The rotor winding is short-circuited by movable brushes (Thomson and Atkinson motors) or by a pair of fixed and a pair of movable brushes (Deri motor). The stator winding is connected directly to the network and represents the primary winding of a transformer, whose secondary winding is represented by the longitudinal coil of the rotor. In the Thomson motor it \((ahf—egb)\) is obtained as coaxial with the stator coil from the point of contact of the brushes (Fig. 1·a) and contains \(\beta_2 N_2\) conductors; in the Atkinson motor, in a similar way, with respect to the resultant field \(\Phi_{cm}\) (Fig. 4) from the fields of both stator windings. The transverse coil in these motors is an addition to the longitudinal one and contains \(\beta_3 N_2\) conductors. In the Deri motor the rotor winding is also resolved into longitudinal and transverse coils, but only in a more complicated manner. It is not difficult to see that in the parts \(np\) and \(qr\) (Fig. 5-a) there is no current; therefore we can…

the coil \(psr—qtn\) is resolved into a longitudinal one, occupying the whole circumference and having, over the spaces \(ps\) and \(tq\), a specific number of bars \(\nu=\dfrac{N_2}{2\tau}\), and over the spaces \(tnp\) and \(qrs\), a specific number of bars \(\nu=\dfrac{N_2}{2\tau}\); and into a transverse one, occupying the spaces \(tnp\) and \(qrs\) and having a specific number of bars \(\nu=\dfrac{N_2}{2\tau}\). Since on the parts \(np\) and \(qr\) the currents in the longitudinal and transverse coils are directed in opposite directions, when superposed they give the coil \(psr—qtn\) with the specific number of bars \(\nu=\dfrac{N_2}{2\tau}\). Accordingly, with this

Fig. 5.

the trapezoidal field of the rotor (Fig. 5-b), \(abcd\), is resolved into the longitudinal tent-shaped \(efbgh\) and the transverse trapezoidal \(iklmno\).

In all these motors the longitudinal and transverse coils constitute a closed circuit (Fig. 6).

Fig. 6.

2. Voltage Diagram.

Let us consider the electromotive forces induced in these motors. While the motor is at rest, only the electromotive force of pulsation of the longitudinal flux \(\Phi_{nd}\), \(—E_{nc1}\), is induced in the longitudinal coil of the rotor; it is balanced by the electromotive force of pulsation of the transverse flux \(\Phi_{np}\) in the transverse coil, \(E_{nc2}\), by the voltage drop due to reactance, \(—J_2x_2\), and by that due to ohmic resistance, \(—J_2r_2\), of the whole secondary circuit. When the rotor is in motion, to this are added the electromotive force of rotation of the transverse coil in the longitudinal field, \(—E_{bp2}\), and of the longitudinal coil in the transverse field, \(—E_{bp1}\).

Let \(O\Phi_{pd}\) (Fig. 7) represent the vector of the longitudinal magnetic flux; the electromotive force of pulsation \(E_{ps1}\) lags behind it by an angle of \(90^\circ\) and is represented by the vector \(Oa\); the electromotive force of rotation—\(E_{vr2}\), in phase with \(O\Phi_{pd}\)—by the vector \(ab\). Further, let \(O\Phi_{pr}\) be the vector of the transverse flux; then the vector of the secondary current \(OJ_2\) will lead it by the angle

\[ \gamma=\operatorname{arctg}\frac{g_{pr}}{b_{pr}}, \]

where \(g_{pr}\) is the watt conductance, and \(b_{pr}\) the wattless conductance, of the transverse coil, and the diagram of the secondary circuit is completed by the following four vectors: 1) \(bc\)—the electromotive force of pulsation \(E_{ps2}=J_2\cos\gamma\cdot x_{pr}=\dfrac{J_2\cos\gamma}{b_{pr}}\), at an angle of \(90^\circ\) to \(O\Phi_{pr}\); 2) \(Oe\)—the electromotive force of rotation \(E_{vr1}\), in phase with \(O\Phi_{pr}\); 3) \(cd\)—

Fig. 7.

the voltage drop from the leakage reaction, in quadrature with the current, and 4) \(ed\), the voltage drop from the ohmic resistance, in phase with the current.

The stator diagram differs in no way from the diagram of a transformer. The vector of the magnetizing current \(OJ_a\) leads the magneತ

... flux by the angle \(\gamma=\operatorname{arctg}\dfrac{ga}{ba}\) (where, of course, \(\dfrac{ga}{ba}=\dfrac{g_{pr}}{b_{pr}}\)); the vector of the primary current \(OJ\) is the geometric sum of \(OJ_2\) and \(OJa\). The vector of the voltage at the terminals \(OP_1\) is composed of: 1) the vector of the electromotive force of pulsation \(OE_1\), 2) the voltage drop from the reactance of primary scattering \(J_1 x_1 - fg\), and 3) the voltage drop from the ohmic resistance \(J_1 r_1 - gh\).

On starting, since \(Oe=0\), the vector \(OJ_2\) makes a small angle with the vector \(O\Phi_{nd}\); the moments \(M_{nd}\) and \(M_{pr}\) are added together and both have a considerable value, since \(OJ_2\) at starting is considerably greater than normal. Therefore repulsion motors start from rest with a large torque. At normal speed \(M_{nd}\) is very small, since \(OJ_2\) makes with \(O\Phi_{nd}\) an angle close to \(\dfrac{\pi}{2}\); at very high speeds \(OJ_2\) again makes small angles with \(O\Phi_{nd}\), but the moment \(M_{nd}\), having here a sign opposite to the sign of \(M_{pr}\), decreases the latter more and more as the speed increases. Thus, in repulsion motors the actual working torque is the torque of the longitudinal winding in the transverse field.

3. Transformation coefficient.

We shall regard all voltages, currents, and resistances of the rotor as reduced to the primary circuit, and shall distinguish them by a prime. Thus

\[ E'_2=E_2\cdot u;\quad J'_2=\frac{J_2}{u};\quad x'_2=x_2u^2;\quad r'_2=r_2u^2, \]

where \(u\) is the transformation coefficient, to the calculation of which we now turn.

The resultant longitudinal field, being the geometric sum of the fictitious fields of the stator and rotor, will have a rather complex form, changing moreover in dependence on the magnitude and phase of the primary and secondary currents; therefore an investigation with the exact form of the field is extremely difficult. At the same time, owing to the influence of iron saturation, it is by no means as exact as it may seem at first sight. Therefore, without making a large error, but gaining enormously in simplicity, we may assume that this field has a sinusoidal form.

In Fig. 8 the sinusoid \(abc\) represents the resultant longitudinal field pulsating in time; the stator winding is represented as developed...

of the form of the layer \(pq\), filled over the portion \(\beta_1\tau\); the layer \(rs\) in the Thomson and Atkinson motors is occupied over the portion \(gh=\beta_2\tau\) by the longitudinal coil, and over the portion \(fg=\beta_3\tau=(1-\beta_2)\tau\) by the transverse coil, with the same specific number of bars. For the Deri motor, the longitudinal coil occupies the whole layer \(rs\); over the portion \(gh=\beta_2\tau\) with the specific number of bars

\[ \nu=\frac{N_2}{2\tau}, \]

and over the portions \(ig\) and \(hk\), equal to

\[ \frac{\beta_3\tau}{2}=(1-\beta_2)\frac{\tau}{2}, \]

with the specific number of bars

\[ \nu=\frac{N_2}{4\tau}, \]

while the transverse coil occupies the portion \(fg=\beta_3\tau=(1-\beta_2)\tau\) with the specific number of bars

\[ \nu=\frac{N_2}{4\tau}. \]

The magnetic flux \(\Phi_x=B_x\,dx\,l\) is linked with \(2\nu x\) bars of the stator or rotor and induces in their windings the electromotive force

\[ dE_{nc}=4.44\,c_1\,2\nu x\,B_x\,l\,dx\,10^{-8}. \]

Fig. 8.

Taking into account the opposite side of the coil and denoting by \(w_{yd}\) the specific number of turns, we have

\[ E_{nc}=\int_{0}^{\tau/2}4.44\,c_1\,4w_{yd}\,x\,B_x\,l\,dx\,10^{-8}. \]

Let us apply this integral to the determination of the electromotive force in the longitudinal coil of the Deri motor. Here \(x=\dfrac{D}{2}a\) and \(B_x=B_{nd}\sin a\), where \(\dfrac{D}{2}\) is the radius of the armature, and \(a\) is the angle between the transverse axis of the motor and the point under consideration. Therefore

\[ \begin{aligned} E_{nd} &=10^{-8}\,4.44\,c_1\,4\,\frac{D^2}{4}\,B_1\,l \biggl\{ \frac{w_2}{\tau}\int_{0}^{\pi/2-\lambda} a\sin a\,da +\frac{w_2}{\tau}\left(\frac{\pi}{2}-\lambda\right) \int_{\pi/2-\lambda}^{\pi/2}\sin a\,da \\ &\qquad\qquad\qquad +\frac{w_2}{2\tau}\int_{\pi/2-\lambda}^{\pi/2} \left(a-\frac{\pi}{2}+\lambda\right)\sin a\,da \biggr\} =4.44\,c_1\,B_1D^2l\,\frac{w_2}{2\tau}(1+\cos\lambda)\,10^{-8}. \end{aligned} \]

where \(\lambda\) is the angle of displacement of the brushes from the longitudinal axis.

For the Thomson and Atkinson motors the third integral is equal to zero, and the electromotive force of pulsation of the longitudinal field is

\[ E_{n d}=4.44\,\omega_1 D^2 l\,\frac{w_2}{\tau}\,B_1\cos\lambda\cdot 10^{-8}. \]

In an exactly similar manner the electromotive force of pulsation of the longitudinal field in the stator winding for all motors is

\[ E_{c m}=4.44\,\omega_1 D^2 l\,\frac{w_1}{\beta_1\tau}\,B_1\sin\frac{\beta_1\pi}{2}\cdot 10^{-8}. \]

Thus the transformation coefficient for the Deri motor is

\[ u=\frac{E_{c m}}{E_{n d}} = \frac{2\sin\frac{\beta_1\pi}{2}}{\beta_1(1+\cos\lambda)}\cdot\frac{w_1}{w_2}, \tag{1·a} \]

and for the Thomson and Atkinson motors

\[ u= \frac{\sin\frac{\beta_1\pi}{2}}{\beta_1\cos\lambda}\cdot\frac{w_1}{w_2}. \tag{1·б} \]

4. Electromotive Forces and Reactions of the Motor Coils

As we saw above, the rotor field in the Deri motor is trapezoidal, and in the Thomson and Atkinson motors it is triangular. In both cases we shall replace it (Fig. 9) by a sinusoidal field, which we shall resolve into two components: 1) in the direction of the longitudinal axis of the motor, which together with the stator field gives the resultant field of the motor, and 2) in the direction of the transverse axis—the transverse field.

Fig. 9.

We shall make this substitution with such a calculation that the electromotive force induced by the sinusoidal field in the rotor winding shall be equal to the electromotive force from the trapezoidal or triangular field.

The electromotive force from the trapezoidal field (see Fig. 9) is determined as:

\[ E_{trp}=4.44\,\omega_1 l\cdot 4\cdot 10^{8} \left\{ \int_{0}^{\frac{\beta_4\tau}{2}} \frac{B_{trp}}{\frac{\beta_4\tau}{2}}\,x\,dx\cdot\frac{w_2}{\tau}\,x + \int_{\frac{\beta_4\tau}{2}}^{\frac{\tau}{2}} \frac{w_2}{\tau}\cdot\frac{\beta_4\tau}{2}\,B_{trp}\,dx \right\} = \]

\[ =4.44\,\omega_1 l\,w_2 B_{trp}\,\tau\,\beta_4 \left(1-\frac{2}{3}\beta_4\right)\cdot 10^{-8}. \]

Electromotive force from the triangular field

\[ E_{trp}=4.44\,\omega_1 l\cdot 4\cdot 10^{-8}\cdot \int_{0}^{\frac{\tau}{2}} \frac{B_{trp}}{\frac{\tau}{2}}\,x^2\,dx\,\frac{w_2}{\tau} = \frac{4.44}{3}\,\omega_1 l\,w_2 B_{trp}\,\tau\cdot 10^{8}. \]

Therefore, from the conditions \(E_{sin}=E_{trp}\) for the Deri motor, we have the greatest induction in the sinusoidal field

\[ B_{sin}= \frac{\pi^2 B_{trp}\left(1-\frac{2}{3}\beta_4\right)\beta_4} {4\cos\lambda}, \tag{2-a} \]

and for the Thomson and Atkinson motors

\[ B_{sin}=\frac{\pi^2}{12}\,B_{trp}. \tag{2-b} \]

The electromotive force of rotation of the transverse coil in the longitudinal field for the Deri motor is obtained (Fig. 8)

\[ E'_{vr}= \int_{\frac{\beta_2\tau}{2}}^{\frac{\tau}{2}} B_x v\,dl = 2\pi D\frac{n}{60}\cdot 2l\,10^{-8}\cdot\frac{1}{\sqrt{2}} \int_{\frac{\pi}{2}-\lambda}^{\frac{\pi}{2}} \frac{w_2}{2\tau}\cdot\frac{D}{2}\,d\alpha\,B_{nd}\sin\alpha = \]

\[ =\sqrt{2}\,\omega_2 l\,w_2 D\,B_{nd}\,10^{-8}\sin\lambda, \]

and in the Thomson and Atkinson motors

\[ \begin{aligned} E''_{vr} &= 2 \pi D \cdot \frac{n}{60} \cdot 2l \cdot 10^{-8} \cdot \frac{1}{\sqrt{2}} \int_{\frac{\pi}{2}-\lambda}^{\frac{\pi}{2}} \frac{w_2}{\tau}\cdot \frac{D}{2}\cdot d\alpha \cdot B_{nd}\sin\alpha \\ &= 2\sqrt{2}\,\mathfrak{s}_2\, l\, w_2\, D\, B_{nd}\cdot 10^{-8}\sin\lambda . \end{aligned} \]

The ratio of these rotational electromotive forces at synchronous speed to the corresponding electromotive forces of pulsation we shall denote by \(c_1\).

For the Dery motor it is equal to

\[ c_1= \frac{\sqrt{2}\,\mathfrak{s}_1\,l\,w_2\,D\,B_{nd}\,10^{-8}\sin\lambda} {4.44\,\mathfrak{s}_1\,B_{nd}\,D^2\,l\,\dfrac{w_2}{2\tau}\,(1+\cos\lambda)\,10^{-8}} = \frac{\sin\lambda}{1+\cos\lambda} = \tan\frac{\lambda}{2}, \tag{3-a} \]

and for the Thomson and Atkinson motors

\[ c_1=\frac{\sin\lambda}{\cos\lambda}=\tan\lambda . \tag{3-b} \]

The rotational electromotive force of the longitudinal coil in the transverse field (Fig. 10) for the Dery motor is

\[ \begin{aligned} E'_{vr} &= 2\pi D\,\frac{n}{60}\cdot 2l\cdot 10^{-8}\cdot \frac{1}{\sqrt{2}} \left\{ \int_0^{\frac{\pi}{2}-\lambda} \frac{w_2}{\tau}\cdot \frac{D}{2}\,d\alpha\,B_{np}\cos\alpha + \int_{\frac{\pi}{2}-\lambda}^{\frac{\pi}{2}} \frac{w_2}{2\tau}\cdot \frac{D}{2}\,d\alpha\,B_{np}\cos\alpha \right\} \\ &= \sqrt{2}\,\mathfrak{s}_2\,l\,w_2\,D\,B_{np}\,10^{-8}(1+\cos\lambda), \end{aligned} \]

and for the Thomson and Atkinson motors

\[ E'_{vr}=2\sqrt{2}\,\mathfrak{s}_2\,l\,w_2\,D\,B_{np}\,10^{-8}\cos\lambda . \]

The electromotive force of pulsation of the transverse field in the transverse coil for the Dery motor is

\[ \begin{aligned} E''_{ns} &= 4.44\,\mathfrak{s}_1\,D^2\,B_{np}\,l\,10^{-8}\,\frac{w_2}{2\tau} \left( \int_{\frac{\pi}{2}-\lambda}^{\frac{\pi}{2}} \left(\frac{\pi}{2}-\alpha\right)\cos\alpha\,d\alpha + \lambda \int_{\frac{\pi}{2}-\lambda}^{\frac{\pi}{2}} \cos\alpha\,d\alpha \right) \\ &= 4.44\,\mathfrak{s}_1\,B_{np}\,D^2\,l\,\frac{w_2}{2\tau}\cdot \sin\lambda \cdot 10^{-8}, \end{aligned} \]

and for the Thomson and Atkinson motors

\[ E''_{nc}=4.44\,\omega_1 B_{nr}D^2 l\,\frac{w_2}{\tau}\sin\lambda\cdot 10^{-8}. \]

Fig. 10.

The ratio of these electromotive forces of rotation at synchronous speed to the corresponding electromotive forces of pulsation we shall denote by \(c_2\). For the Deri motor,

\[ c_2= \frac{\sqrt{2}\,\omega_1 w_2 B_{nr}\cdot 10^{-8}} {4.44\,\omega_1 D^2 l\,\frac{w_2}{2\tau} B_{nr}\,10^{-8}} \cdot \frac{1+\cos\lambda}{\sin\lambda} = \operatorname{ctg}\frac{\lambda}{2}, \tag{4-a} \]

and for the Thomson and Atkinson motors

\[ c_2=\frac{\cos\lambda}{\sin\lambda}=\operatorname{ctg}\lambda. \tag{4-b} \]

The ratio \(\dfrac{c_2}{c_1}\) for the Deri motor is

\[ \frac{c_2}{c_1}=\operatorname{ctg}^2\frac{\lambda}{2}, \tag{5-a} \]

and for Thomson and Atkinson motors

\[ \frac{c_2}{c_1}=\operatorname{ctg}^2\lambda. \tag{5-b} \]

Thus, the assumption of sinusoidal fields is characterized by the fact that, under it, the product \(c_2 c_1 = 1\); for other forms of fields this condition is absent.

Reactions. By the reaction of the longitudinal and transverse coils of the rotor we shall understand the quotient obtained by dividing the electromotive force of pulsation of the corresponding sinusoidal component of the general rotor field by the secondary current. In this case (Fig. 10) it is evident that in the Déri motor

\[ B_{n d}=B_2\cos\frac{\lambda}{2} \quad\text{and}\quad B_{n p}=B_2\sin\frac{\lambda}{2}, \tag{6-a} \]

and in Thomson and Atkinson motors

\[ B_{n d}=B_2\cos\lambda \quad\text{and}\quad B_{n p}=B_2\sin\lambda. \tag{6-b} \]

Consequently, the reaction of the longitudinal coil in the Déri motor is

\[ x_{n d}= \frac{ 4{,}44\cdot c_1\, l\, D^2\cdot \dfrac{w_2}{2\tau}\, B_2 \cos\dfrac{\lambda}{2}(1+\cos\lambda)\cdot 10^{-8} }{ J_2 }; \]

substituting for \(B_2\) its value, equal to

\[ B_2=0{,}4\pi\cdot \frac{w_2\beta_4}{2\delta''}\, J_2\sqrt{2}, \]

(where \(\delta''\) is the air gap replacing the magnetic resistance of the chain of the flux tube, which we shall regard as the same for all tubes), we obtain

\[ x_{n d}= \frac{ 4\cdot c_1 l \beta_4 w_2^2 \tau \cdot 10^{-9}\cdot \cos\dfrac{\lambda}{2}(1+\sin\lambda) }{ \delta'' }, \tag{7-a} \]

and for Thomson and Atkinson motors

\[ x_{n d}= \frac{ 4{,}44\cdot c_1\, l\, D^2\, \dfrac{w_2}{\tau}\, B_2\cos^2\lambda\cdot 10^{-8} }{ J_2 }. \]

or, substituting the value

\[ \mathrm{B}_2 = 0.4 \cdot \pi \cdot \frac{w_2}{2\delta''}\cdot J_2\sqrt{2}, \]

we have

\[ x_{nd}=\frac{4 c_1 \cdot l w_2^2 \tau \cdot 10^{-9}\cos^2\lambda}{\delta''}. \tag{7b} \]

The reaction of the transverse coil for the Deri motor is

\[ x_{np}=\frac{4 \cdot c_1 l \cdot w_2^2 \cdot \beta_4 \cdot \tau\,10^{-9}}{\delta''}\sin\frac{\lambda}{2}\cdot \sin\lambda, \tag{8a} \]

and for the Thomson and Atkinson motors

\[ x_{np}=\frac{4 \cdot c_1 \cdot l \cdot w_2^2 \tau \cdot 10^{-9}}{\delta''}\sin^2\lambda. \tag{8b} \]

The ratio of the longitudinal reaction to the transverse reaction for the Thomson and Atkinson motors is

\[ \frac{x_{nd}}{x_{np}}=\frac{\cos^2\lambda}{\sin^2\lambda}=\operatorname{ctg}^2\lambda, \]

and for the Deri motor

\[ \frac{x_{nd}}{x_{np}}= \frac{\cos\frac{\lambda}{2}(1+\cos\lambda)} {\sin\frac{\lambda}{2}\sin\lambda} =\operatorname{ctg}^2\frac{\lambda}{2}. \]

Thus, for all three motors,

\[ \frac{x_{nd}}{x_{np}}=\frac{c_1}{c_2}. \tag{9} \]

It can be shown that a similar relation between the reactions \(x_{nd}\) and \(x_{np}\) also exists for certain other field forms; therefore, in what follows, for the sake of generality, \(\frac{c_2}{c_1}\) is not combined into a single constant.

5. Effective resistance, effective reactance of the rotor circuit, and the functional relation between them.

From triangle \(Obk\) (Fig. 7) we obtain the secondary current

\[ J_2= \frac{\sqrt{\mathrm{E}_2^2+\left(\mathrm{E}_2 c_1 v\right)^2}} {\sqrt{\left(x_{np}\cos\gamma\, v c_2+r_2\cos\gamma-x_2\sin\gamma\right)^2+ \left(x_{np}\cos\gamma+x_2\cos\gamma+r_2\sin\gamma\right)^2}}. \]

or, introducing the notation \(F\),

\[ \sqrt{(x_{np}\cos\gamma\, v c_2+r_2\cos\gamma-x_2\sin\gamma)^2+(x_{np}\cos\gamma+x_2\cos\gamma+r_2\sin\gamma)^2}=F, \]

\[ J_2=\frac{\sqrt{F^2+(E_2c_1v)^2}}{F}. \]

In what follows, for brevity, we shall denote \(x_{np}\cos\gamma\) by \(x_3\); then the effective power factor of the rotor circuit is

\[ \cos\psi_{2v}=\cos(\beta-\alpha-\gamma)= \]

\[ =\frac{(x_3vc_2+r_2\cos\gamma-x\sin\gamma)(b_{np}-cvg_{np})+(x_3+x_2\cos\gamma+r_2\sin\gamma)(c_1vb_{np}+g_{np})} {F\cdot\sqrt{1+(c_1v)^2}\cdot\sqrt{g^2_{np}+b^2_{np}}}, \]

where \(\angle\alpha=\angle aOb,\ \angle\beta=\angle kOb\) (see Fig. 7).

Thus the effective resistance is

\[ R_v=\frac{E_2\cos\varphi_2}{J_2}= \]

\[ =\frac{(c_2vx_3+r_2\cos\gamma-x_2\sin\gamma)(b_{np}-c_1vg_{np})+(x_3+x_2\cos\gamma+r_2\sin\gamma)(c_1vb_{np}+g_{np})} {(1+(c_1v)^2)\sqrt{g^2_{np}+b^2_{np}}}, \]

or, after expanding the brackets and collecting like terms,

\[ R_v=\frac{1}{1+(c_1v)} \{r_2+x_2c_1v+v\cos\gamma(c_1+c_2)x_3+\sin\gamma\,x_3-v^2c_1c_2\sin\gamma\,x_3\}. \tag{10} \]

In a similar manner,

\[ \sin\psi_{2,v}= \]

\[ =\frac{(x_3+x_2\cos\gamma+r_2\sin\gamma)(b_{np}-c_1vg_{np})-(c_2vx_3+r_2\cos\gamma-x_2\sin\gamma)(c_1vb_{np}+g_{np})} {F\cdot\sqrt{1+(c_1v)^2}\cdot\sqrt{b^2_{np}+g^2_{np}}} \]

and the effective reactance

\[ X_v=\frac{E_2\sin\varphi_2}{J_2}= \]

\[ =\frac{(x_3+x_2\cos\gamma+r_2\sin\gamma)(b_{np}-c_1vg_{np})-(c_2vx_3+r_2\cos\gamma-x_2\sin\gamma)(c_1vb_{np}+g_{np})} {[1+(c_1v)^2]\cdot\sqrt{b^2_{np}+g^2_{np}}}, \]

and after transformations,

\[ X_v=\frac{1}{1+(c_1v)^2} \{x_2-c_1vr_2-x_3\sin\gamma\,v(c_1+c_2)+x_3\cos\gamma-x_3\cos\gamma\,c_1c_2v^2\}. \tag{11} \]

The effective apparent resistance

\[ Z_v=\sqrt{R_v^2+X_v^2} =\sqrt{ \frac{1}{[1+(c_1 v)^2]^2} \left\{ (r_2^2+x_2^2)[1+(c_1 v)^2]+x_3^2+ \right. } \]

\[ \left. +v^2(c_1+c_2)^2x_3^2+(v^2c_1c_2x_3)^2 +2r_2x_3v\cos\gamma\,c_2-2x_2x_3v\sin\gamma\,c_2+ \right. \]

\[ \left. +2v^2r_2x_3\sin\gamma\,c_1^2 +2v^2x_2x_3\cos\gamma\,c_1^2 -2v^3x_2x_3\sin\gamma\,c_1^2c_2 -2v^3r_2x_3\cos\gamma\,c_1^2c_2+ \right. \]

\[ \left. +2r_2x_3\sin\gamma+2x_2x_3\cos\gamma -2v^2x_3^2c_1c_2 \right\}, \]

after collecting like terms and reducing by \(1+c_1^2v^2\),

\[ Z_v= \sqrt{ \frac{1}{1+c_1^2v^2} \left\{ r_2^2+x_2^2+x_3^2(1+c_2^2v^2) +2x_3\left[r_2(\sin\gamma+c_2v\cos\gamma)+ \right.\right. } \]

\[ \left.\left. +x_2(\cos\gamma-c_2v\sin\gamma)\right] \right\}. } \tag{12} \]

For \(v=0\),

\[ R_{v=0}=r_2+x_3\sin\gamma,\qquad X_{v=0}=x_2+x_3\cos\gamma \quad\text{and} \]

\[ Z_{v=0}=\sqrt{r_2^2+x_2^2+x_3^2+2x_3(r_2\sin\gamma+x_2\cos\gamma)}. \]

For \(v=\infty\),

\[ R_{v=\infty}=-\frac{c_1}{c_2}x_3\sin\gamma,\qquad X_{v=\infty}=-\frac{c_2}{c_1}x_3\cos\gamma \quad\text{and}\quad Z_{v=0}=-\frac{c_2}{c_1}x_3. \]

To obtain the functional relation between \(R_v\) and \(X_v\), we must eliminate \(v\) from both equations (10) and (11), for which we shall write them in the following form:

\[ (R_vc_1^2+c_1c_2\sin\gamma\,x_3)v^2 -\left[c_1x_2+(c_1+c_2)\cos\gamma\,x_3\right]v -(r_2+\sin\gamma\,x_3-R_v)=0 \]

and

\[ (X_vc_1^2+c_1c_2\cos\gamma\,x_3)v^2 +\left[c_1r_2+(c_1+c_2)\sin\gamma\,x_3\right]v -(x_2+\cos\gamma\,x_3-X_v)=0. \]

Denoting

\[ R_vc_1^2+c_1c_2\sin\gamma\,x_3=A_1,\qquad c_1x_2+(c_1+c_2)\cos\gamma\,x_3=B_1, \]

\[ r_2+\sin\gamma\,x_3-R_v=C_1,\qquad X_vc_1^2+c_1c_2\cos\gamma\,x_3=A_2, \]

\[ c_1r_2+(c_1+c_2)\sin\gamma\,x_3=B_2,\qquad x_2+\cos\gamma\,x_3-X_v=C_2, \]

and eliminating \(v^2\) and \(v\) from the equations thus obtained, we have

\[ v^2=\frac{C_1B_2+B_1C_2}{A_1B_2+A_2B_1}; \qquad v=\frac{A_1C_2-A_2C_1}{A_1B_2+A_2B_1}. \]

Equating,

\[ (A_1C_2-C_1A_2)^2=(A_1B_2+A_2B_1)(C_1B_2+B_1C_2); \tag{13} \]

whence, substituting the values \(A_1,\ A_2,\ B_1,\ B_2,\ C_1,\ C_2\),

\[ A_1C_2-C_1A_2 =c_1^2x_2R_v+c_1c_2x_3^2\sin\gamma\cos\gamma +R_vc_1^2x_3\cos\gamma-c_1^2X_vR_v+ \]

\[ \begin{aligned} &+c_1 c_2 x_3 x_2 \sin \gamma-c_1 c_2 \sin \gamma\, x_3 X_v-r_2 c_1^2 X_v-c_1^2 x_3 X^v \sin \gamma+c_1^2 X_v R_v\\ &-c_1 c_2 x_3^2 \sin \gamma \cos \gamma-r_2 c_1 c_2 x_3 \cos \gamma+c_1 c_2 \cos \gamma\, x_3 R_v-(c_1^2 x_2+c_1^2 x_3 \cos \gamma\\ &+c_1 c_2 \cos \gamma\, x_3)R_v-(c_1^2 r_2+c_1^2 x_3 \sin \gamma+c_1 c_2 \sin \gamma\, x_3)X_v+c_1 c_2 x_3(x_2 \sin \gamma\\ &-r_2 \cos \gamma); \end{aligned} \]

\[ \begin{aligned} A_1 B_2 + A_2 B_1 &=c_1^2[c_1 r_2+(c_1+c_2)\sin \gamma\,x_3]\,R_v+c_1^2[c_1 x_2\\ &\quad +(c_1+c_2)\cos \gamma\,x_3]\,X_v+c_1^2 c_2 x_3(\sin \gamma\,r_2+\cos \gamma\,x_2)+c_1 c_2(c_1+c_2)x_3^2; \end{aligned} \]

\[ \begin{aligned} C_1 B_2+C_1 B_2 &=c_1(x_2^2+r_2^2)+c_1 x_3(x_2\cos\gamma+r_2\sin\gamma)+(c_1+c_2)x_3^2\\ &\quad +(c_1+c_2)x_3(x_2\cos\gamma+r_2\sin\gamma)-[c_1 x_2+(c_1+c_2)\cos\gamma\,x_3]X_v\\ &\quad -(c_1 r_2+(c_1+c_2)\sin\gamma\,x_3)R_v. \end{aligned} \]

Denoting

\[ c_1 x_2+(c_1+c_2)\cos\gamma\,x_3=D;\qquad c_1 r_2+(c_1+c_2)\sin\gamma\,x_3=E; \]

\[ x_3(x_2\sin\gamma-r_2\cos\gamma)=H;\qquad x_3(x_2\cos\gamma+r_2\sin\gamma)=K;\qquad (c_2+c_1)x_3^2=L; \]

\[ c_1(x_2^2+r_2^2)=M; \]

and substituting these notations into eq. 5, after certain transformations we obtain the equation of the circle:

\[ \begin{aligned} &R_v^2(D^2+E^2)+X_v(D^2+E^2)+R_v\left[2c_2D\cdot H+c_2K E+\frac{c_2}{c_1}L\cdot E\right.\\ &\left.\quad -2(c_1+c_2)K\cdot E-ME-LE\right] +X_v\left[c_2K\cdot D+\frac{c_2}{c_1}L\cdot D-2c_2HE\right.\\ &\left.\quad -(2c_1+c_2)K\cdot D-M\cdot D-L\cdot D\right] +\left[c_2^2H^2-c_2(2c_1+c_2)K^2-c_2K\cdot M\right.\\ &\left.\quad -c_2LK-\frac{c_2}{c_1}(2c_1+c_2)K\cdot L-\frac{c_2}{c_1}L\cdot M-\frac{c_2}{c_1}L^2\right]. \end{aligned} \]

Substituting the values of \(D, E, H, K, L\), and \(M\) and carrying out the transformations, we have:

\[ D^2+E^2=c_1^2(r_2^2+x_2^2)+(c_1+c_2)^2x_3^2+2c_1(c_1+c_2)x_3(r_2\sin\gamma+x_2\cos\gamma) \]

\[ \begin{aligned} &2c_2DH-2c_1K\cdot E-ME-LE\left(1-\frac{c_2}{c_1}\right) =\left[-r_2-\left(1-\frac{c_2}{c_1}\right)x_3\sin\gamma\right]\cdot\\ &\quad \{c_1^2(r_2^2+x_2^2)+(c_1+c_2)^2x_3^2+2c_1(c_1+c_2)x_3(r_2\sin\gamma+x_2\cos\gamma)\}; \end{aligned} \]

\[ \begin{aligned} &-\left[2c_2H\cdot E+c_1K\cdot D+MD+LD\left(1-\frac{c_2}{c_1}\right)\right] =\left[-x_2-x_3\cos\gamma\left(1-\frac{c_2}{c_1}\right)\right]\cdot\\ &\quad \left(c_1^2(r_2^2+x_2^2)+(c_1+c_2)^2x_3^2+2c_1(c_1+c_2)x_3(r_2\sin\gamma+x_2\cos\gamma)\right); \end{aligned} \]

\[ c_2^2 H_2^2-c_2(2c_1+c_2)K^2-c_2KM-\frac{c_2}{c_1}(3c_1+c_2)KL-\frac{c_2}{c_1}LM-\frac{c_2}{c_1}L^2= \]

\[ =-\frac{c_2}{c_1}(x_3^2+x_3x_2\cos\gamma+x_3r_2\sin\gamma) \left[c_1^2(r_2^2+x_2^2)+(c_1+c_2)^2x_3^2+\right. \]

\[ \left.+2c_1(c_1+c_2)x_3(r_2\sin\gamma+x_2\cos\gamma)\right]. \]

Reducing by \(c_1^2(r_2^2+x_2^2)+(c_1+c_2)^2x_3^2+2c_1(c_1+c_2)x_3(r_2\sin\gamma+x_2\cos\gamma)\), we obtain the equation of the circle in the form

\[ X_v^2+R_v^2-X_v\left[x_2+x_3\cos\gamma\left(1-\frac{c_2}{c_1}\right)\right] -R_v\left[r_2+x_3\sin\gamma\left(1-\frac{c_2}{c_1}\right)\right] -\frac{c_2}{c_1}(x_3^2+ \]

\[ +x_3x_2\cos\gamma+x_3r_2\sin\gamma)=0, \]

which, after small transformations, can be represented in the following form:

\[ \left\{X_v-\frac{1}{2}\left[x_2+x_3\cos\gamma\left(1-\frac{c_2}{c_1}\right)\right]\right\}^2 +\left\{R_v-\frac{1}{2}\left[r_2+x_3\sin\gamma\left(1-\frac{c_2}{c_1}\right)\right]\right\}^2 \]

\[ =\frac{1}{4}\left\{r_2^2+x_2^2+x_3^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3\left(1+\frac{c_2}{c_1}\right)(x_2\cos\gamma+r_2\sin\gamma)\right\}. \tag{14} \]

6. Construction of the circular diagram by successive transformations.

Let us refer \(x_2\), \(r_2\), and \(x_3\) to the primary circuit by multiplying them by the square of the transformation coefficient, and denote their referred values by a prime:

\[ x_2'=u^2x_2,\qquad r_2'=u^2r_2,\qquad x_3'=u^2x_3, \]

then

\[ R_2'v=\frac{1}{1+(c_1v)^2}\left\{r_2'+x_2'c_1v+\right. \]

\[ \left.+x_3'\left[v\cos\gamma(c_1+c_2)+\sin\gamma-v^2\sin\gamma\,c_1c_2\right]\right\} \]

\[ X_2'v=\frac{1}{1+(c_1v)^2}\left\{x_2'-r_2'c_1v+x_3'\left[\cos\gamma-\right.\right. \]

\[ \left.\left.-v\sin\gamma(c_1+c_2)-\cos\gamma\,c_1c_2v^2\right]\right\}; \]

and the referred circuit takes the form of Fig. 11.

Fig. 11.

From equation 6, substituting there \(r_2'\), \(x_2'\), and \(x_3'\), and taking the scale \(m\) ohm per 1 cm, we obtain the radius \(\rho'''\) and the coordinates of the center \(\eta_0'''\), \(\xi_0'''\) of the circle \(K_p'''\):

\[ \rho'''=\frac{1}{2m}\sqrt{r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2+2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma)}; \tag{15} \]

\[ \eta_0'''=\frac{1}{2m}\left[x_2'+x_3'\cos\gamma\left(1-\frac{c_2}{c_1}\right)\right];\quad \xi_0'''=\frac{1}{2m}\left[r_2'+x_3'\sin\gamma\left(1-\frac{c_2}{c_1}\right)\right]. \tag{16} \]

The circle \(K\rho'''\) constructed from these data, with the origin of coordinates at the point \(O\), (Fig. 12*, Table I) represents the geometrical locus of the apparent resistances as \(v\) varies.

Let us mark on this circle the following points: 1) the point \(A''\), for which \(v=\infty\), with coordinates

\[ x_{2v=\infty}=-\frac{c_2}{c_1}\sin\gamma \quad \text{and} \quad y_{2v=\infty}=-\frac{c_2}{c_1}\cos\gamma; \]

it is obtained, therefore, simply by drawing a line at the angle \(\gamma\) to the axis of ordinates in the third quadrant.

2) The point \(K'''\), for which \(v=0\), with coordinates

\[ x_{2v=0}=r_2'+x_3'\sin\gamma \quad \text{and} \quad y_{2v=0}=x_2'+x_3'\cos\gamma, \]

to obtain which it is enough to draw, through the point \(A''\), a diameter; \(K'''\) lies at its other end. This is seen from the following: the abscissa of the point opposite \(A''\) will be

\[ \frac{c_2}{c_1}x_3'\sin\gamma+2\xi_0'''=r_2'+x_3'\sin\gamma, \]

and the ordinate

\[ \frac{c_2}{c_1}x_3'\cos\gamma+2\eta_0'''=x_2'+x_3'\cos\gamma, \]

which represents, as we have seen, the coordinates of the point \(K'''\).

3) The third point is the point \(C'''\), for which \(v=1\), with coordinates

\[ x_{2v=1}=\frac{1}{1+c_1^2}\{r_2'+c_1x_2'+(c_1+c_2)\cos\gamma x_3'+\sin\gamma x_3'-c_1c_2\sin\gamma x_3'\}; \]

\[ y_{2v=1}=\frac{1}{1+c_1^2}\{x_2'-c_1r_2'-(c_1+c_2)\sin\gamma x_3'+\cos\gamma x_3'-c_1c_2\cos\gamma x_3'\}. \]

Further, all the vectors \(OT'''\) for \(v=2, 3, -1, -2\), etc., go to points of the circle which, as is not difficult to prove, lie at the intersections of it with rays from the point \(A''\) to segments of a line perpendicular to \(A''O\), equal to the segment of this line between the rays \(A''K'''\) and \(A''C'''\).

As \(v\) changes from \(0\) to \(\infty\), the end of the vector \(OT'''\) moves along the upper half of the circle from \(K'''\) through \(C'''\) to \(A''\), and as \(v\) changes from \(0\) to \(-\infty\), it moves along the lower half.

* In Fig. 12 and in the following, the positive axis of abscissas is directed upward, and the positive axis of ordinates to the left, from the origin of coordinates.

Theory of Repulsion Motors.

Inverting the circle \(K_p'''\) with respect to the point \(O\), we obtain the circle \(K_p''\), which represents the geometric locus of the apparent conductances

\[ Y_{2v}=\frac{1}{Z_{2v}}; \]

in this case, the points \(K'''\), \(C'''\), \(A''\) of the circle \(K_p'''\) correspond to the points \(K''\), \(C''\), \(A'\) of the circle \(K_p''\).

Let us find the dependence between the radius and the coordinates of the center of the fundamental circle \(K_p'''\) and of the inverted circle \(K_p''\). Let (Fig. 13) \(K_{p_1}\) represent the fundamental circle, constructed to the scale \(m\ \Omega\) per 1 cm, and \(K_{p_2}\) the inverted one, to the scale \(n\ \Omega\) per 1 cm. Draw through the centers \(C_1\) and \(C_2\) and the origin of coordinates \(O\) the straight line \(a_1 a_2\). From the similarity of the inverted figures we have

\[ \frac{Od_1}{Od_2}=\frac{R_1}{R_2}=\frac{X_{01}}{X_{02}}=\frac{Y_{01}}{Y_{02}}, \]

where \(R_1\) and \(R_2\) are the radii, and \(X_{01}\), \(X_{02}\), \(Y_{01}\), and \(Y_{02}\) are the coordinates of the centers of the fundamental and inverted circles.

Fig. 13.

From the last equation,

\[ \frac{Od_1}{Oa_1}\cdot \frac{Oa_1}{Od_2}=\frac{R_1}{R_2}; \tag{17} \]

but

\[ Oa_1=R_1+\sqrt{X_{01}^{2}+Y_{01}^{2}}, \qquad Od_1=R_1-\sqrt{X_{01}^{2}+Y_{01}^{2}}, \]

\[ \frac{Oa_1}{Od_2} = m\cdot n\,(Oa_1)^2 = m\cdot n\left(R_1+\sqrt{X_{01}^{2}+Y_{01}^{2}}\right)^2. \]

Substituting these values into equation 17, we obtain

\[ R_2= \frac{R_1}{m\cdot n\left(R_1+\sqrt{X_{01}^{2}+Y_{01}^{2}}\right)^2}\, \frac{R_1+\sqrt{X_{01}^{2}+Y_{01}^{2}}}{R_1-\sqrt{X_{01}^{2}+Y_{01}^{2}}} = \frac{R_1}{m\cdot n}\cdot \frac{1}{R_1^{2}-X_{01}^{2}-Y_{01}^{2}}. \]

In the same way,

\[ X_{02}=\frac{X_{01}}{m\cdot n}\cdot \frac{1}{R_1^{2}-X_{01}^{2}-Y_{01}^{2}} \quad \text{and} \quad Y_{02}=\frac{Y_{01}}{m\cdot n}\cdot \frac{1}{R_1^{2}-X_{01}^{2}-Y_{01}^{2}}. \]

Applying these formulas to our inverted circle \(K_p''\) (Fig. 12), we obtain its radius

\[ \rho''= \frac{ \sqrt{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma) } }{ 2\cdot n.x_3'\frac{c_2}{c_1}(x_3'+x_2'\cos\gamma+r_2'\sin\gamma) } \]

and the coordinates of the center

\[ \xi_0''= \frac{ r_2'+x_3'\sin\gamma\left(1-\frac{c_2}{c_1}\right) }{ 2\cdot n.x_3'\frac{c_2}{c_1}(x_3'+x_2'\cos\gamma+r_2'\sin\gamma) } \]

and

\[ \eta_0''= \frac{ x_2'+x_3'\cos\gamma\left(1-\frac{c_2}{c_1}\right) }{ 2\cdot n.x_3'\frac{c_2}{c_1}(x_3'+x_2'\cos\gamma+r_2'\sin\gamma) }. \]

When \(v\) changes from \(0\) to \(+\infty\), the vector of apparent conductance \(OT''\) moves along the upper part of the circumference from the point \(K''\) through \(C''\) to \(A'\), and when \(v\) changes from \(0\) to \(-\infty\), along the lower part. Since, at constant rotor voltage \(E_{nc}\), its current is proportional to the vector of apparent conductance

\[ J_2'=E_{nc1}\cdot Y_{2v}, \]

and since \(E_{nc.1}\) coincides with the axis of abscissas, the abscissas of the points of the circumference represent, on a known scale, the power received by the rotor, and the axis of ordinates cuts the circumference at two points \(g''\) and \(h''\), where this power is equal to zero.

To obtain the resultant apparent conductance of the branches \(bc\) and \(ef\) (Fig. 11), we must add the apparent conductance \(y_a=g_a+jb_a\) of the excitation branch. It is not difficult to see that the vector

\[ A'O=\frac{1}{\frac{c_2}{c_1}x_3'}=\frac{c_1b_{nd}}{c_2\cos\gamma'}=\frac{b_a}{\cos\gamma'}, \]

so that the end of the vector \(y_a\) falls exactly at the point \(A'\) of the circumference. Thus the resultant conductance of the branches \(bc\) and \(ef\), \(Y_v'=Y_{2v}'+y_a\), is represented by the vector \(A'T'=OA'+OT'\), and the origin of coordinates moves to the point \(A'\).

Rotating the circle \(K_p''\) about the point \(A'\), we obtain the straight line \(K_p'\), perpendicular to \(A'P'\) and passing at the distance \(DA'\). If the scale is \(DA'\, p\, \Omega\) in \(1\) cm, then

\[ DA'=\frac{1}{2pnR''}= \]

\[ = \frac{\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right)} {p\sqrt{ r_2'^2+x_2'^2+x_3'^2\left(1+\dfrac{c_2}{c_1}\right)^2 +2x_3'\left(1+\dfrac{c_2}{c_1}\right)\left(x_2'\cos\gamma+r_2'\sin\gamma\right) }}. \tag{18} \]

The angle of inclination of \(A'P'\) to the axis of abscissas is equal to \(\operatorname{arctg} A'E' : P'E' = -\operatorname{arctg}\xi_0' : \eta_0'\),

but

\[ \xi_0' = OA'\sin\gamma-\xi_0'' = \frac{\sin\gamma}{n\dfrac{c_2}{c_1}x_3'} - \frac{ r_2'+x_3'\sin\gamma\left(1-\dfrac{c_2}{c_1}\right) }{ n\cdot 2\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) } = \]

\[ = \frac{ x_3'\sin\gamma\left(1+\dfrac{c_2}{c_1}\right)+x_2'\sin 2\gamma-r_2'\cos 2\gamma }{ 2n\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) }, \]

similarly,

\[ \eta_0' = OA'\cos\gamma-\eta_0'' = \frac{\cos\gamma}{n\dfrac{c_2}{c_1}x_3'} - \frac{ x_2'+x_3'\cos\gamma\left(1-\dfrac{c_2}{c_1}\right) }{ 2n\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) } = \]

\[ = \frac{ x_3'\cos\gamma\left(1+\dfrac{c_2}{c_1}\right)+x_2'\cos 2\gamma+r_2'\sin 2\gamma }{ 2n\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) }. \]

And the angle of inclination of \(P'A'\) to the axis of abscissas is

\[ \varepsilon = \operatorname{arctg} \frac{ x_3'\sin\gamma\left(1+\dfrac{c_2}{c_1}\right)+x_2'\sin 2\gamma-r_2'\cos 2\gamma }{ x_3'\cos\gamma\left(1+\dfrac{c_2}{c_1}\right)+x_2'\cos 2\gamma+r_2'\sin 2\gamma }. \tag{19} \]

Thus the equation of the straight line \(K_p'\) will be

\[ x\frac{ x_3'\left(1+\dfrac{c_2}{c_1}\right)\sin\gamma+x_2'\sin 2\gamma-r_2'\cos 2\gamma }{\rho'''} + \]

\[ +\,y\,\frac{x_3'\left(1+\dfrac{c_2}{c_1}\right)\cos\gamma+x_2'\cos2\gamma+r_2'\sin2\gamma}{p'''} - \]

\[ -\frac{\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right)}{p\cdot p'''}=0; \]

\[ x\,\frac{x_3'\left(1+\dfrac{c_2}{c_1}\right)\sin\gamma+x_2'\sin2\gamma-r_2'\cos2\gamma} {\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right)} + \]

\[ +\,y\,\frac{x_3'\left(1+\dfrac{c_2}{c_1}\right)\cos\gamma+x_2'\cos2\gamma+r_2'\sin2\gamma} {\dfrac{c_2}{c_1}x_3'\left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right)} -\frac{1}{p}=0, \]

or

\[ x\xi_0' + y\eta_0'=\frac{1}{2pn}. \tag{20} \]

The straight line \(K_p'\) represents the locus of the apparent resistances of the branches \(bc\) and \(ef\) as \(v\) varies. The vector \(A'T'\), representing this apparent resistance, is equal to

\[ A'T'=\frac{1}{Y_n'}=\frac{1}{Y'_{2v}+ya} =\frac{1}{\dfrac{1}{Z'_{2v}}+ya}. \]

To the points \(K''\), \(C''\), \(A'\) of the circle \(K_p''\) there correspond the points \(K'\), \(C'\), and \(\infty\) of the straight line \(K_p'\). As \(v\) varies from \(0\) to \(+\infty\), the vector \(A'T'\) moves along the upper part of the straight line \(K_p'\), from the point \(K'\) through \(C'\) to \(+\infty\); when \(v\) varies from \(0\) to \(-\infty\), it moves along the lower part from \(K'\) to \(-\infty\).

It is not difficult to prove that the vector \(A'T'\), for \(v=0,1,2,3\), etc., and also \(0,-1,-2,-3\), etc., passes through points lying on this straight line at equal distances from one another. Indeed, let us find the abscissa of an arbitrary point \(T'\) lying on this straight line. The corresponding points on the circles \(K_p'''\) and \(K_p''\) will be \(T'''\) and \(T''\). From the triangle \(T''OA'\) we have

\[ (T''A')^2=(T''O)^2+(OA')^2-2.T''O.OA'.\cos T''OA', \tag{21} \]

but

\[ \cos T''OA'=\cos(180-T''Oh''+\gamma)=-\cos(\delta_v-\gamma), \]

Theory of Repulsion Motors

where we shall denote \(\angle T''Oh''\) by \(\delta_v\). From the circle \(K_p''\)

\[ \cos(\delta_v-\gamma)= \frac{X'_{2v}\cos\gamma+R'_{2v}\sin\gamma}{Z'_{2v}}. \]

Substituting in Eq. 21 the values of the vectors

\[ T''O=\frac{1}{Z_{2v}},\qquad OA'=\frac{1}{\dfrac{c_2}{c_1}x_3'}, \]

and the value of \(\cos(\delta_v-\gamma)\), we obtain

\[ (T''A')^2= \frac{1}{Z_{2v}^{\,2}} + \frac{1}{\left(\dfrac{c_2}{c_1}x_3'\right)^2} + 2\frac{1}{Z_{2v}\dfrac{c_2}{c_1}x_3'} \frac{X'_{2v}\cos\gamma+R'_{2v}\sin\gamma}{Z'_{2v}}. \]

The abscissa of the point \(T'\).

\[ T'H_v' = \frac{1}{T''A'}\sin T'A'H' = \frac{1}{T''A'}\cdot \frac{T'H'+\dfrac{1}{\dfrac{c_2}{c_1}x_3'}\sin\gamma}{T'A'} = \]

\[ = \frac{ \dfrac{1}{Z'_{2v}}\sin\delta_v+ \dfrac{1}{\dfrac{c_2}{c_1}x_3'}\sin\gamma }{ (T''A')^2 } = \frac{ \dfrac{R'_{2v}}{Z_{2v}^{\prime 2}}+ \dfrac{1}{\dfrac{c_2}{c_1}x_3'}\sin\gamma }{ \dfrac{1}{Z_{2v}^{\prime 2}}+ \dfrac{1}{\left(\dfrac{c_2}{c_1}x_3'\right)^2} + 2 \dfrac{X'_{2v}\cos\gamma+R'_{2v}\sin\gamma} {Z_{2v}^{\prime 2}\dfrac{c_2}{c_1}x_3'} } = \]

\[ = \frac{ \left( R'_{2v}\dfrac{c_2}{c_1}x_3' + Z_{2v}^{\prime 2}\sin\gamma \right) \dfrac{c_2}{c_1}x_3' }{ \dfrac{c_2}{c_1}x_3' + Z_{2v}^{\prime 2} + 2\dfrac{c_2}{c_1}x_3' \left(X'_{2v}\cos\gamma+R'_{2v}\sin\gamma\right) }. \]

Substituting the values of \(R'_{2v}\), \(X'_{2v}\), and \(Z'_{2v}\), and, for the purpose of reducing like terms, we obtain

\[ T'H_v= \frac{ \left[ \left\{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right) +2x_3'\left(x_2'\cos\gamma+r_2'\sin\gamma\right) \right\}\sin\gamma +\frac{c_2}{c_1}x_3'r_2' \right. }{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right) \left(x_2'\cos\gamma+r_2'\sin\gamma\right) } \]

\[ \left. + \left( \frac{c_2}{c_1}x_3'\cos\gamma +x_3'\cos\gamma +x_2'\cos 2\gamma +r_2'\sin 2\gamma \right)c_2vx_3' \right] \frac{c_2}{c_1}x_3' \tag{22} \]

—a linear equation in the variable \(v\).

For \(v=0\)

\[ T'H'_{v=0}= \frac{ \left[ \left\{r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)+ 2x_3'(x_2'\cos\gamma+r_2'\sin\gamma)\right\}\sin\gamma +\frac{c_2}{c_1}x_3'r_2' \right]\frac{c_2}{c_1}\cdot x_3' }{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma) }. \]

And the difference of the abscissas of the points \(T'\) and \(K'\)

\[ T'H'_{v}-T'H'_{v=0}= \]

\[ = \frac{ \left[\left(1+\frac{c_2}{c_1}\right)x_3'\cos\gamma+x_2'\cos2\gamma+r_2'\sin2\gamma\right]c_2vx_3'\cdot\frac{c_2}{c_1}\cdot x_3' }{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma) }, \]

i.e. the elevation of the points \(T'_{v=1}\), \(T'_{v=2}\), etc., above the point \(T'_{v=0}\) is proportional to \(v\). The distance between the points \(T'_{v=0}\) and \(T'_{v=1}\), or between \(T'_{v=1}\) and \(T'_{v=2}\), etc., is expressed as:

\[ K'C=\frac{T'H'_{v=1}-T'H'_{v=0}}{\sin\varepsilon}= \]

\[ = \frac{ \left[\left(1+\frac{c_2}{c_1}\right)x_3'\cos\gamma+x_2'\cos2\gamma+r_2'\sin2\gamma\right]c_2vx_3'\cdot\frac{c_2}{c_1}\cdot x_3' }{ \left[x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right)+x_2'\cos2\gamma+r_2'\sin2\gamma\right] \sqrt{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2+ } } \]

\[ +2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma) \]

Adding to the point \(A'\) \(r_1-jx_1\), we obtain a new origin of coordinates at the point \(A\). Then the vector \(AT'\), representing the apparent resistance of our entire circuit, is equal to

\[ AT'=Z_v=z_1+\frac{1}{Y_{2v}+y_a}. \]

Rotating with respect to the point \(A\), we obtain a circle passing through the latter and representing the geometrical locus of the apparent conductivities, as \(v\) varies.

The diameter of this circle is

\[ 2\rho = \frac{1}{q \cdot p \cdot \left(A'D' + x_1 \sin \varepsilon + r_1 \cos \varepsilon\right)} = \]

\[ = \frac{ \sqrt{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right) \left(x_2'\cos\gamma+r_2'\sin\gamma\right) } }{ q\cdot \left\{ \frac{c_2}{c_1}x_3' \left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) +r_1\left[ x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\sin 2\gamma -r_2'\cos 2\gamma \right] +x_1\left[ x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\cos 2\gamma +r_2'\sin 2\gamma \right] \right\} }. \]

The angle of inclination of the radius from the origin remains the same; therefore the coordinates of the center will be:

\[ \xi_0=\rho\cos\varepsilon = + \frac{ x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\sin 2\gamma-r_2'\cos 2\gamma }{ 2q\left\{ \frac{c_2}{c_1}x_3' \left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) +r_1\left[ x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\sin 2\gamma-r_2'\cos 2\gamma \right] +x_1\left[ x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\cos 2\gamma+r_2'\sin 2\gamma \right] \right\} }. \]

\[ \eta_0=\rho\sin\varepsilon = + \frac{ x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\cos 2\gamma+r_2'\sin 2\gamma }{ 2q\left\{ \frac{c_2}{c_1}x_3' \left(x_3'+x_2'\cos\gamma+r_2'\sin\gamma\right) +r_1\left[ x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\sin 2\gamma-r_2'\cos 2\gamma \right] +x_1\left[ x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right) +x_2'\cos 2\gamma+r_2'\sin 2\gamma \right] \right\} }. \]

Usually in repulsion motors, owing to the weak saturation of the iron, the angle \(\gamma\) is very small, so that simplifications may be introduced into the formulae for the radius and the coordinates of the center of the latter circle.

In the common denominator we shall neglect the terms \(r_1x_3'\sin\gamma\), \(r_1x_2'\sin 2\gamma\), \(-r_1r_2'\cos 2\gamma\), and \(x_1r_2'\sin 2\gamma\), as being very small; and in the terms

\[ \frac{c_2}{c_1}x_3'x_2'\cos\gamma,\quad x_1x_3'\cos\gamma\left(1+\frac{c_2}{c_1}\right) \quad\text{and}\quad x_1x_2'\cos 2\gamma \]

we shall take \(\cos\gamma\) and \(\cos 2\gamma\) to be \(1\). In the numerator of the expression for the radius we set \(\gamma=0\), and, furthermore, in most cases we may also neglect \(r_2'^2\); then the following simple expressions are obtained:

\[ \rho= \frac{ \frac{c_2}{c_1}x_3' + x_3' + x_2 }{ q\cdot \left[ 2\frac{c_2}{c_1}x_3' \left\{ x_3' + (r_1+r_2')\sin\gamma + (x_1+x_2') \right\} +2x_1(x_3'+x_2') \right] }, \tag{23a} \]

\[ \xi_0 = \frac{x_3' \sin \gamma \left(1+\dfrac{c_2}{c_1}\right)-r_2'} {q \cdot \left[2 \dfrac{c_2}{c_1} x_3' \{x_3' + (r_1+r_2')\sin\gamma + (x_1+x_2')\} + 2x_1(x_3'+x_2')\right]}, \tag{23b} \]

\[ \eta_0 = \frac{x_3' \cos \gamma \left(1+\dfrac{c_2}{c_1}\right)+x_2'} {q \cdot \left[2 \dfrac{c_2}{c_1} x_3' \{x_3' + (r_1+r_2')\sin\gamma + (x_1+x_2')\} + 2x_1(x_3'+x_2')\right]} . \tag{23c} \]

An arbitrary vector \(AT\) represents the apparent conductance of the entire motor circuit,

\[ Y_v=\frac{1}{Z_v}=\frac{1}{z_1+\dfrac{1}{Y_{2v}+y_\mu}} . \]

To the points \(K'\), \(C'\), and \(\infty\) of the straight line \(K_\eta'\) there correspond the points \(K\), \(C\), and \(A\) of the circle \(K_\eta\).

With constant voltage at the terminals \(P_1\), the primary current is

\[ J_1=P_1 \cdot Y_v, \]

and consequently our diagram represents, on a known scale, the diagram of the primary current. Since the voltage at the terminals coincides with the positive direction of the axis of abscissas, each vector \(AT\) represents, in magnitude and direction, the primary current; its abscissa \(TH\) represents the watt current \(J_1\cos\varphi_1\), and the ordinate \(AH\) the wattless current \(J_1\sin\varphi_1\). The angle made by the vector \(AT\) with the axis of abscissas is the phase-displacement angle between \(E_1\) and \(J_1\). At starting, the primary current is represented by the vector \(AK\); at synchronism by \(AC\); and at an infinitely large number of revolutions by the point \(A\), i.e. it is equal to zero.

7. The magnetizing and secondary currents in the circle diagram.

At \(v=\infty\) the motor neither receives nor gives energy to the external circuit; \(J_1 z_1=0\), and the electromotive force induced in the stator, \(E_1\), is equal to the applied voltage at the terminals \(P_1\). In the rotor there circulates at this time a current \(J_{2v=\infty}\), equal to the magnetizing current

\[ J_{\mu v=\infty}=\frac{P_1}{\dfrac{c_2}{c_1}x'_{\eta\rho}}=\frac{P_1}{x'_{n\delta}}, \]

and consequently lagging behind \(P_1\) by the angle

\[ \frac{\pi}{2}-\gamma . \]

When \(v\) decreases, \(J_1 z_1\), and consequently \(E_1\) and \(J_\mu\)

move along circles, the position of which is determined by the following simple considerations.

Let (Fig. 14) \(K_{p_3}\) represent the circle of the primary currents known to us, and \(AP_1\) the voltage at the terminals. Let us take the greatest current \(AT_m\); from the point \(P_1\) construct \(P_1b=J_1x_1\) perpendicular to \(AT_m\), and \(bt_m=J_1r_1\) parallel to \(AT_m\). We obtain \(P_1t_m=J_1z_1\), at the angle \(\psi_1\) to \(AT_m\), where

\[ \psi_1=\operatorname{arctg}\frac{r_1}{x_1}; \]

then \(At_m\) represents \(E_1\), the electromotive force induced in the stator at the greatest current. Along the circumference having \(P_1t_m\) as its diameter, \(v-J_1z_1\) and \(E_1\) move as \(v\) varies. Laying off the magnetizing current \(J_a=\infty=Aa\) at the angle \(\frac{\pi}{2}-\gamma\) to \(AP_1\), drawing from the point \(a\) a straight line at the angle \(\varepsilon-\psi_1\) to \(Aa\), and laying off on it the segment

\[ at_{m_1}=\frac{P_1t_m\cdot Aa}{AP_1} =\frac{J_1z_1}{P_1}Aa =Y_{\sigma,\max}\cdot z_1Aa, \]

and describing a circle on \(at_{m_1}\), we obtain the geometrical locus along which the end of the vector of the magnetizing current moves as \(v\) varies. In this case, while the primary current moves along the upper part of the circumference, \(E_1\) moves along the left part and \(J_a\) along the lower part, and conversely.

Fig. 14.

The secondary current, representing the geometrical difference of the primary and magnetizing currents, moves with both its ends: one along the circle of the primary current \(K_{p_3}\), the other along the circle of the magnetizing current \(K_{p_1}\).

It is not difficult to show that such a vector can be replaced by a vector one end of which is fixed, while the other moves along a circle whose diameter represents the geometrical sum of the corresponding diameters \(K_{p_3}\) and \(K_{p_1}\). Indeed, let (Fig. 15) we have two such circles \(K_{p_3}\) and \(K_{p_1}\), with two mutually corresponding diameters \(AB\) and \(ab\). From the point \(B\) draw a line parallel to \(ab\) and lay off on it the segment \(BB_2=ab\); on the line \(AB_2\), as on a diameter, construct a circumference. Then every vector from the point \(a\) to the circumference of the circle \(K_{p_2}\) is equal in magnitude and direction to the vector between the corresponding points of the circles \(K_{p_1}\) and \(K_{p_3}\). Indeed, since each chord \(AE_2\) of the circle \(K_{p_2}\) is equal to the geometrical sum of the corresponding chords of the circles \(K_{p_1}\) and \(K_{p_3}\): \(AE\)

and \(ae\), then \(Ee\) is equal and parallel to \(ae\), and consequently \(aE_2\) is also equal and parallel to \(Ee\).

It is self-evident that it is much more convenient when the end of the vector of the secondary current moves along the same circle \(K_{p_3}\) as the vector of the primary current. This can be achieved by rotating the circle \(K_{p_2}\) through the angle \(B_2AB\) and reducing it by \(\dfrac{B_2A}{BA}\) times. To find the point \(a_0\), the fixed center of all vectors of the secondary current, one may proceed as follows: rotate, through the angle of rotation \(B_2AB\), the straight line \(Aa\), where \(D_4\) is an arbitrary point of the circle \(K_{p_4}\) described on \(B_4A\) as diameter; join it with \(a_1\), the new position of the point \(a\) after the rotation, and through \(D_3\) of the circle \(K_{p_3}\), corresponding to \(D_4\) of the circle \(K_{p_4}\), draw a parallel; the point \(a_0\) of its intersection with \(Aa_1\) is the center of the vectors of the secondary current. This is clear from the following: let \(D_3\) and \(D_4\) be corresponding points of the circles \(K_{p_3}\) and \(K_{p_4}\); from the similarity of the triangles \(AD_3C_3\) and \(AD_4C_4\) it follows that they lie on one straight line \(AD_3D_4\), and from the similarity of the triangles \(AD_3a_0\) and \(AD_4a_1\) it follows that \(a_0\) is precisely the required point.

Fig. 15.

The vector \(a_0D\) is, in magnitude, less than the true value of the vector of the secondary current by

\[ \frac{AB_2}{AB} = \frac{\sqrt{(AB)^2+(ab)^2+2AB\cdot ab\cdot \cos B_2BA}}{AB} \]

times, and in phase lags behind the latter by the angle \(BAB_2\). The angle

\[ B_2BB_4=\psi_1-\frac{\pi}{2}+\gamma \]

(see Fig. 14), and the angle

\[ ABB_2=\pi-\psi_1-\gamma+\frac{\pi}{2}, \]

whence

\[ \angle B_2AB_4 = \arcsin \frac{BB_2\sin\left(\gamma+\psi_1-\frac{\pi}{2}\right)}{B_2A}. \]

Since the angle of inclination between the corresponding diameters \(ab\) and \(AB\) is very small, and also since \(ab\) is usually small in comparison with \(AB\),

and \(Aa_1\), then in most cases the point \(a_0\) is found by dividing the line \(Aa\) into the parts \(aa_0\) and \(a_0A\), which are in the ratio

\[ aa_0 : Aa_0 = ab : AB. \]

8. Speed, phase displacement, and torque in the circle diagram.

In § 6 we saw that the segment \(K'T'\) of the line \(Kp'\) between the rays from the point \(A\), passing through \(K\) and an arbitrary point \(T'\), is proportional to the speed at this latter point. It follows from this that every line perpendicular to the diameter from \(A\) is a line of speed.

At low speeds, determining it from such a line presents no difficulty; the situation is otherwise when it is necessary to determine the speeds of points close to \(A\). In such cases one may proceed in the following manner. Joining (Fig. 16, Table II) \(d\) with \(A\) and \(b\), and extending \(Ad\) to its intersection with the line of speed at the point \(e\), from the similarity of the triangles \(Aef\) and \(Abd\) we have:

\[ \frac{fe}{db}=\frac{v\cdot ki+kf}{db}=\frac{fA}{dA} \]

whence

\[ v=\frac{fA\cdot db}{dA\cdot ki}-\frac{kf}{ki}. \]

By the same method it is also not difficult to find the point for a given speed.

The angle of phase displacement between the voltage at the terminals and the primary current is represented on the diagram directly by the angle between the vector \(AT'\) and the axis of abscissas. The power factor \(\cos \varphi_1\) is obtained in the form of the segment \(At_1\) on \(AT\), if along the axis of abscissas the segment \(Al=1\) is laid off and on it, as diameter, a semicircle is described.

The rotating torque, as we have seen, is composed of two torques: \(M_{np}\)—of the longitudinal coil in the transverse field, and \(M_{nd}\)—of the transverse coil in the longitudinal field, with the first playing the dominant role.

\[ M_{np}=k_1\Phi_{np}\cdot J_2\cos\gamma = kJ_2^2\cos\gamma, \quad M_{nd}=k_1\Phi_{nd}\cdot J_2\cos(\Phi_{nd},J_2) \]

\[ = k\cdot J_2J_a\cos(\Phi_{nd},J_2), \]

where \(k_1\) and \(k\) are constants determined below.

Thus the resulting torque of the motor is

\[ M=k\left[J_2^2\cos\gamma+J_2J_a\cos(\Phi_{nd},J_2)\right]. \]

Let us consider (Fig. 17) the circle diagram for a motor deprived of primary resistance and reactance. Here the vector \(Aa\), representing the vector of the magnetizing current, is fixed; \(AT\), representing the vector of the primary current, and \(aT\), the secondary, move along the circle; the vector of the longitudinal magnetic flux coincides with the axis of ordinates.

Fig. 17.

It is not difficult to see that

\[ M = k\{(X-X_a)^2\cos\gamma + (Y-Y_a)^2\cos\gamma + J_a(Y-Y_a)\}, \]

where \(X\) and \(Y\) are the coordinates of the point \(T\), and \(X_a\) and \(Y_a\) are the coordinates of the point \(a\).

From the equation of the circle we have

\[ X^2+Y^2=2X_0X+2Y_0Y-X_0^2-Y_0^2+R^2=2X_0X+2Y_0Y, \]

where \(X_0\) and \(Y_0\) are the coordinates of the center, and \(R=\sqrt{X^2+Y^2}\) is the radius of the circle. Substituting in the preceding equation, we obtain

\[ M=k[\cos\gamma\,X_a^2+\cos\gamma\,Y_a^2 +2\cos\gamma(X_0-X_a)X+2\cos\gamma(Y_0-Y_a)Y +J_a(Y-Y_a)], \]

but \(J_aY_a=J_a^2\cos\gamma\), consequently

\[ M=k\{2\cos\gamma(X_0-X_a)X+[2\cos\gamma(Y_0-Y_a)+J_a]Y\}. \]

This expression we can represent in the form:

\[ M=2\cdot\cos\gamma\cdot k\cdot \sqrt{(X_0-X_a)^2+\left[(Y_0-Y_a)+\frac{J_a}{2\cos\gamma}\right]^2} \times \]

\[ \times \left[ \frac{ (X_0-X_a)X+\left[(Y_0-Y_a)-\frac{J_a}{2\cos\gamma}\right]\cdot Y }{ \sqrt{(X_0-X_a)^2+\left[(Y_0-Y_a)+\frac{J_a}{2\cos\gamma}\right]^2} } \right] \tag{24} \]

The expression in square brackets represents, as is known, the distance of the current point with coordinates \(X,Y\) from a certain straight line with equation

\[ (X_0-X_a)X+\left[(Y_0-Y_a)-\frac{J_a}{2\cos\gamma}\right]Y=0 . \]

and which, obviously, passes through the origin of the coordinates. The angle of inclination of this straight line to the axis of abscissae is determined as

\[ \delta_m=\operatorname{arctg}\frac{2\cos\gamma\,(X_0-X_a)}{2\cos\gamma\,(Y_0-Y_a)-J_a}. \tag{25} \]

It is not difficult to see that this line intersects the circle at two points: \(A\) and at the point \(d\), where

\[ J_2\cos\gamma=J_a\cos(\Phi_{nd},J_2) \]

Hence an extremely simple construction of it follows. It is only necessary, from the point \(a\), with radius \(\dfrac{J_a}{\cos\gamma}\), to mark off the circle; the second point of the line of moments is obtained.

As for the expression

\[ \sqrt{(X_0-X_a)^2+\left[(Y_0-Y_a)+\frac{J_a}{2\cos\gamma}\right]^2}, \]

it is, obviously, represented by the line \(OD\), where the point \(D\) is obtained by the intersection of the perpendicular to the segment \(Aa\), from the midpoint of the latter, with the line parallel to the axis of ordinates from the point \(a\).

Thus the torque is represented by the product of \(2\cos\gamma\,k\) by the segment \(OD\) and by the distance \(TE\) of the corresponding point of the circle \(T\) from the straight line \(AE\).

\[ M=2\cos\gamma\,k\cdot OD\cdot TE \tag{26} \]

The angle of inclination of the line \(OD\) to the axis of ordinates is expressed as

\[ \angle\delta_{0D}=\operatorname{arctg}\frac{X_0-X_a}{Y_0-Y_a-\dfrac{J_a}{2\cos\gamma}}, \]

i.e. the same as that of the line of moments to the axis of abscissae. Thus, the segment \(OD\) is perpendicular to \(AE\). Hence one more simple construction of the line of moments follows. If \(OD\) is prolonged to its intersection with the circumference at the point \(b\), and \(bd=Ab\) is laid off, then we obtain the second point of the line of moments.

In the presence of primary reaction and resistance, the exciting current and the longitudinal magnetic flux, as we have seen, move along circles; therefore the line of moments no longer represents an exact straight line. But since, on the one hand, the oscillations of the exciting current and of the longitudinal flux are not very large, and, on the other hand, they affect only the second term in the sum of moments, which in itself is small in comparison with the first, we shall make an exceedingly small error if we take for the line of moments the straight line obtained-

…as follows. Let us suppose that, for a motor with primary reactive and ohmic resistances, the longitudinal magnetic flux coincides with the direction of the axis of ordinates, and the magnetizing current is stationary and coincides with the vector \(Oa_0\) (Fig. 16), where \(a_0\) is the center of the vectors of the secondary current; then the line of moments \(mm\) is obtained by the method just described.

9. Losses, power, and efficiency in the circle diagram.

Primary power. The abscissa of any point of the circle represents, as was indicated above, the watt current \(J_1 \cos \varphi_1\), and, consequently, at constant terminal voltage, on a known scale, the power supplied to the motor

\[ W_1 = E_1 J_1 \cos \varphi_1 . \]

The axis of ordinates thus serves as the line of primary power, and the distance of the point \(T\) from it is proportional to \(W_1\).

The line of primary power intersects the circle at two points, \(A\) and \(B\) (Fig. 16, Table II), for which \(W_1 = 0\); by these points it divides the circle into two parts: the upper—motor part, in which the power is positive, i.e. supplied, and the lower—generator part, in which the power is negative, i.e. delivered.

Losses in the stator winding. Following the example of Bragstad and La Cour, we represent the losses in the stator as proportional to the distance of the points of the circumference from the so-called semi-polar, i.e. a line parallel to the polar and passing through the middle of the tangent drawn from the fixed center of the vectors. Since in our case the center of the vectors of the primary current lies on the circumference itself, the semi-polar becomes the tangent to the point \(A\) of the circumference. The perpendicular from an arbitrary point of the circumference \(T\) to this tangent is

\[ TL = \frac{J_1^2}{2OA} = \frac{J_1^2}{J_{1\,max}}, \]

so that the primary losses in the stator winding are

\[ V_1 = J_1^2 r_1 = 2 \cdot q^2 \cdot \overline{OA} \cdot LT \cdot r_1 = q J_{1\,max} \cdot r_1 \cdot LT, \tag{27} \]

where \(q\) is the scale of the primary current.

Losses in the rotor winding. The secondary current has as its moving center the point \(a_0\), and its losses are again represented by \(TM\), the distance of the points of the circumference from the semi-polar to the point

$P$. The latter is the intersection of the tangents to the points $a_1$ and $a_2$, which are obtained as the intersection with the circle $K_p$ of the line $a_1a_2$, passing through the fixed center of the vectors $J_2$ and perpendicular to $Oa_0$. Therefore the losses in the rotor winding are

\[ V_2 = J_2^2 r_2 = 2s^2 Oa_0 r_2 \, TM, \tag{28} \]

where $s$ is the scale of the secondary current.

Iron losses due to the transverse field. We have taken the transverse field of the rotor as proportional to the secondary current and, consequently, the same line of losses in the rotor winding will serve as the line of iron losses due to the transverse field. Only the constant changes, being determined in the following way. As we had above,

\[ E_{nc}'' = J_2 x_{\eta p}\cos\gamma = J_2\cos\gamma\,\frac{x_{n0}}{\dfrac{c_2}{c_1}}. \]

Therefore the iron losses are

\[ V_{a_2} = g_{\eta p} E^2 = \frac{c_2}{c_1}\, g_a \left( \frac{x_{n0}}{\dfrac{c_2}{c_1}}\cos\gamma \right)^2 J_2^2 = \frac{g_a}{\dfrac{c_2}{c_1}(b_a^2+g_a^2)}\,J_2^2 = \frac{c_1}{c_2} r_a J_2^2 . \]

In the diagram they are represented by the segments $TM$ and are equal to

\[ V_{a_2}=2s^2\frac{c_1}{c_2}r_a Oa_0\,TM . \tag{26} \]

Iron losses due to the longitudinal field. In a similar manner we can represent the losses due to the longitudinal field by the segment $tq$ between the circle $k_p$, along which the end of the magnetizing current moves, and the polar line to it $qr$ from the fixed center of the latter $A$. These losses are equal to

\[ V_{a_1}=J_a^2\frac{g_a}{g_a^2+b_a^2}=J_a^2 r_a=2q^2 r_a\cdot oA\cdot tq . \tag{30} \]

But since, on the one hand, the circle $K_p$ is usually very small, and on the other, for the convenience of the further construction of the diagram it is necessary that all points be located on the circle $K_p$, we shall rotate the circle $k_p$ through the corresponding angle and enlarge it to the circle $K_p$. Then the point $A$ will take the position $A_1$ (see Fig. 18, tab—

$I$, representing Fig. 16 on a smaller scale, the half-poles $qr$, the position $QR$, and the losses from the longitudinal field

\[ V_a = 2q^2 r_a \cdot OA_1 \cdot TQ\left(\frac{ot}{OT}\right)^2 = 2q^2 r_a \, OA_1 \cdot TQ\left(\frac{z_1}{P_1} J_a{}_{v=\infty}\right)^2 \]
\[ = 2q^2 r_a \, OA_1 \cdot TQ \cdot z_1^2 y^2_{av=\infty} \tag{31} \]

The power delivered to the rotor is equal to the power received by the stator, minus the losses in the latter, $W_a = W_1 - J_1^2 r_1 - J_a^2 r_a$, and is likewise represented by the distance of the points of the circle from a straight line. Its position is determined by the points of its intersection with the circle for which $W_a = 0$.

Above (p. 22) we saw that on the circle $K_p''$ the points $g''$ and $h''$ represent points for which the power supplied to the rotor is zero. On the circle $K_p$ these points will correspond to the points $g$ and $h$; therefore the straight line $gh$ also represents the line of the power received by the rotor.

The total stator losses $V_c = J_1^2 r_1 + J_a^2 r_a$ are represented by the distance of the points of the circle from the line $S_1U$ (Fig. 16), passing through $S_1$, the point of intersection $LA$ and $RQ$ (Fig. 18), and through the point of intersection $U$ of the lines $hg$ and $BA$ (Fig. 16).

The power delivered by the rotor is equal to the power received by it minus the losses in it: $W_2' = W_a - J_2^2\left(r_2' + r_a \dfrac{c_1}{c_2}\right)$; and is represented by the distance of the points of the circle from the straight line passing through the points at which $W_2' = 0$. The position of these points is determined by the equation

\[ W_2' = W_a - V_2 - V_{a_2} = EJ_2 \cos \zeta_2 - J_2^2 r_2' - J_2^2 r_a \frac{c_1}{c_2} = J_2^2\left(R_r - r_2' - q_{m\varphi}x^2\right) \]
\[ J_2^2 \cdot \left[ \frac{1}{1+(c_1 v)^2} \cdot \left\{ r_2' + c_1 x_2'v + (c_1+c_2)\cos\gamma\, x_3' v + \sin\gamma\, x_3'(1-c_2 c_1 v^2) \right\} - r_2' - x_3'\sin\gamma \right] = \]
\[ J_2^2 \cdot \frac{1}{1+(c_1 v)^2} \left\{ c_1 v x_2' - r_2' c_1^2 v^2 + v x_3'\cos\gamma(c_1+c_2) - x_3'\sin\gamma\, c_1 v^2(c_1+c_2) \right\}. \]

From the last expression it is seen that $W_2'$ is equal to zero: 1) when $v=0$, i.e. at the point $K$, and 2) when

\[ v\left[x_3'\sin\gamma\, c_1(c_2+c_1)+c_1^2 r_2'\right] = c_1 x_2' + x_3'\cos\gamma(c_2+c_1), \]

i.e. when

\[ v = \frac{c_1 x_2' + x_3'\cos\gamma(c_2+c_1)} {c_1^2 r_2' + x_3'\sin\gamma(c_2+c_1)\cdot c_1} \tag{32} \]

Thus the second point is constructed according to the calculated speed.

Since the torque \(M = W'_2 : 2\pi \dfrac{n}{60}\), then for \(W'_2 = 0\) and a finite speed it is also equal to zero; consequently, the torque line intersects the circle at the same point as the line of mechanical power, which makes it possible to avoid calculating this point when it has already been obtained for the torque line.

The total losses in the rotor \(V_\rho = J_2'^2\left(r_2 + \dfrac{c_1}{c_2} r_u\right)\) are evidently represented by the distance of the points of the circle from the line \(MN\).

Since \(W'_2 = W_a - V_\rho\), the three lines representing \(W'_2\), \(W_a\), and \(V_\rho\)—\(JK\), \(gl\), and \(MN\)—must intersect at one point \(S_3\)*); this property serves as a check on the correctness of the construction.

The total losses of the entire motor \(V_e = V_c + V_\rho\) are determined by the distance of the points of the circle from the line \(FD\), which passes, first, through the point of intersection of the lines \(S_2 MN\) and \(S_1 U\) (Fig. 18, Plate I) and, second, through the point \(D\) of intersection of the lines \(KJ\) and \(AB\) (Fig. 16).

Dropping the perpendicular \(G_2 OG\) (Fig. 16) from the point \(O\) to the line \(KJ\), we obtain the greatest mechanical power of the motor in the form of the abscissa segment between the point \(G_1\) and the line \(KJ\), and for the generator, between \(G_2\) and \(JK\).

In the same manner, the abscissa segment through \(O\) between \(E_1\) and the axis of ordinates gives the greatest electrical power absorbed by the motor, and the similar segment between \(E_2\) and \(AB\) gives the greatest electrical power delivered by the generator.

The efficiency \(\eta = \dfrac{W'_2}{W_1} = \dfrac{W_1 - V_e}{W_1}\) is constructed as follows. Let us draw, at some distance below the axis of ordinates, a straight line parallel to it, and divide its segment between the lines \(KJ\) and \(FD\) into 100 parts; then the extension of the line connecting an arbitrary point of the circle with the point \(D\), the intersection of the lines \(KJ\) and \(FD\), cuts off on this parallel a segment representing, in percent, the efficiency of the motor.

When the point \(T\) coincides with \(K\) and \(A\), the ray \(TD\) coincides with \(KJ\), and the efficiency \(\eta = 0\); when the ray \(DT\) is tangent to the circle \(K_\rho\), the efficiency has its greatest value.

*) See Arnold, W. T., Vol. I.

10. Direct construction of the circular diagram.

In § 6 we saw that the radius and the coordinates of the center of the circle \(K_p\) are determined by formulas 24a, 24b, and 24c. Introducing into them the notations:

\[ x_4'=\frac{c_2}{c_1}\cdot x_3=\frac{c_2}{c_1}\cdot x'_{\mathrm{pr}}\cdot \cos\gamma=x'_{n0}\cdot \cos\gamma,\quad x_{\mathrm{II}}=x_3'+x_2',\quad x_{\mathrm{III}}=x_4'+x_3'+x_2', \]

\[ r_s=r_1+r_2'\quad \text{and}\quad x_s=x_1+x_2', \]
we obtain

\[ \rho=\frac{x_{\mathrm{III}}}{2x_4'(x_3'+r_s\sin\gamma+x_s)+2x_1x_{\mathrm{II}}}, \tag{33a} \]

\[ \xi_0=\frac{(x_3'+x_4')\sin\gamma-r_2'}{2x_4'(x_3'+r_s\sin\gamma+x_s)+2x_1x_{\mathrm{II}}}, \tag{33b} \]

\[ \eta_0=\frac{(x_3'+x_4')\cos\gamma+r_2'}{2x_4'(x_3'+r_s\sin\gamma+x_s)+2x_1x_{\mathrm{II}}}. \tag{33c} \]

By these formulas the circle \(K_p\) can also be obtained directly.

In the same paragraph we found that the abscissa of the point \(T'\) in the system of coordinates with origin at the point \(A'\) is expressed by formula 22; replacing in it \(\cos\gamma\) and \(\cos 2\gamma\) by unity and neglecting the terms

\[ r_2'^2\sin\gamma,\quad 2x_3'r_2'\sin^2\gamma,\quad c_2vx_3'r_2'\sin 2\gamma,\quad r_2'^2 \quad \text{and}\quad 2x_3'\left(1+\frac{c_2}{c_1}\right)r_2'\sin\gamma, \]

as small, after some transformations we obtain

\[ T'H'= \frac{ \left\{ \sin\gamma\left[(x_2'+x_3')^2+\frac{c_2}{c_1}x_3'^2\right] +\frac{c_2}{c_1}x_3'r_2' +c_2vx_3'\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] \right\}\frac{c_2}{c_1}x_3' }{ \left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]^2 }. \]

In a similar manner we shall also find the ordinate of the point \(T'\) (Fig. 12).

\[ A'H'=\frac{1}{T''A'}\cos T'A'H'=\frac{1}{T''A'}\cdot \frac{ OH''+\dfrac{\cos\gamma}{\dfrac{c_2}{c_1}x_3'} -\dfrac{\cos\delta_v}{Z_{2v}'} +\dfrac{\cos\gamma}{\dfrac{c_2}{c_1}x_3'} }{ T'A' } = \frac{1}{(T'A')^2} = \]

\[ = \frac{ \left(X_{2v}'\frac{c_2}{c_1}x_3'+Z_{2v}'^{\,2}\cos\gamma\right)\frac{c_2}{c_1}x_3' }{ \frac{c_2}{c_1}x_3'+Z_{2v}'^{\,2} +2\frac{c_2}{c_1}x_3'\left(X_{2v}'\cos\gamma+R_{2v}'\sin\gamma\right) }. \]

Whence, substituting the values \(X'_{2v}\), \(R'_{2v}\), and \(Z'_{2v}\), and carrying out the transformations, we obtain

\[ A'H' = \frac{ \left[ \cos\gamma \left\{ r_2'^2+x_2'^2+x_3'\left(1+\frac{c_2}{c_1}\right) +2x_3'(r_2'\sin\gamma+x_2'\cos\gamma) \right\} + \frac{c_2}{c_1}x_3'x_2' -vc_2x_3'\left(x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right)-r_2'\cos2\gamma+x_2'\sin2\gamma\right) \right]\frac{c_2}{c_1}x_3' }{ r_2'^2+x_2'^2+x_3'^2\left(1+\frac{c_2}{c_1}\right)^2 +2x_3'\left(1+\frac{c_2}{c_1}\right)(x_2'\cos\gamma+r_2'\sin\gamma) }. \]

Neglecting the terms \(2x_3'r_2'\sin\gamma\), \(r_2'^2\), \(2x_3'\left(1+\frac{c_2}{c_1}\right)r_2'\sin\gamma\), and replacing \(\cos\gamma\) by unity, after small transformations,

\[ A'H' = \frac{ \left\{ (x_2'+x_3')\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] - vc_2x_3'\left[ \sin\gamma\,x_3'\left(1+\frac{c_2}{c_1}\right)-r_2'\cos2\gamma+x_2'\sin2\gamma \right] \right\}\frac{c_2}{c_1}x_3' }{ \left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]^2 }. \]

The coordinates of the point \(T'\) in a system of coordinates with origin at the point \(A\) will be:

\[ AH'_1 = \frac{ \left\{ (x_2'+x_3')\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] - v\cdot c_2\cdot x_3'\left[ \sin\gamma\,x_3'\left(1+\frac{c_2}{c_1}\right)-r_2'\cos2\gamma+x_3'\sin2\gamma \right]\frac{c_2}{c_1}x_3' \right\} }{ \left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]^2 } +x_1, \tag{34} \]

\[ T'H'_1 = \frac{ \left\{ \sin\gamma\left[(x_2'+x_3')^2+\frac{c_2}{c_1}x_3'^2\right] + \frac{c_2}{c_1}x_3'r_2' +c_2v\cdot x_3'\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] \right\}\frac{c_2}{c_1}x_3' }{ \left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]^2 } +r_1, \tag{35} \]

and the angle which the vector \(AT'\) makes with the axis of abscissas ...

\[ \angle XAT=\angle \varphi_1=\operatorname{arctg} \frac{ \begin{gathered} \left\{(x_2'+x_3')\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] -vc_2x_3'\left[x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right)-r_2'\cos2\gamma+\right.\right.\\ \left.\left. +x_2'\sin2\gamma\right]\right\} \left\{\frac{c_2}{c_1}x_3'+x_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2 \end{gathered} }{ \begin{gathered} \left\{\sin\gamma\left[(x_2'+x_3')^2+\frac{c_2}{c_1}x_3'^2\right] +\frac{c_2}{c_1}x_3'r_2'\right\} \left\{\frac{c_2}{c_1}x_3'+x_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2\\ +c_2vx_3'\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] \end{gathered} }. \tag{36} \]

With the armature in a stationary position, it will be

\[ \angle XAK=\angle \varphi_{1k}= \]

\[ \operatorname{arctg}= \frac{ (x_2'+x_3')\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] \left\{\frac{c_2}{c_1}x_3'+x_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2 }{ \left\{\sin\gamma\left[(x_2'+x_3')^2+\frac{c_2}{c_1}x_3'^2\right] +\frac{c_2}{c_1}x_3'r_2'\right\} \left\{\frac{c_2}{c_1}x_3'+r_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2 } = \]

\[ =\operatorname{arctg} \frac{ x_{\mathrm{II}}\cdot x_{\mathrm{III}}\cdot x_4' + x_{\mathrm{I}}\cdot x_{\mathrm{II}}^{\,2} }{ \left\{\sin\gamma\left(x_{\mathrm{II}}^{\,2}+x_4'\cdot x_3'\right)+x_4'r_2'\right\}x_4'+r_1\cdot x_{\mathrm{III}}^{\,2} }, \tag{36a} \]

and at synchronism

\[ \begin{gathered} \left\{(x_2'+x_3')\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right] -c_2x_3'\left[x_3'\sin\gamma\left(1+\frac{c_2}{c_1}\right)-r_2'\cos2\gamma+\right.\right.\\ \left.\left. +x_2'\sin2\gamma\right]\right\} \left\{\frac{c_2}{c_1}x_3'+x_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2 \end{gathered} \]

\[ \angle \varphi_{1c}=\angle XAC=\operatorname{arctg} \frac{ \begin{gathered} \left\{\sin\gamma\left[(x_2'+x_3')^2+\frac{c_2}{c_1}x_3'^2\right]+\right.\\ \left.+\frac{c_2}{c_1}x_3'r_2' +c_2x_3'\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\} \left\{\frac{c_2}{c_1}x_3'+r_1\left[x_2'+x_3'\left(1+\frac{c_2}{c_1}\right)\right]\right\}^2 \end{gathered} } {} = \]

\[ =\operatorname{arctg} \frac{ x_{\mathrm{II}}\cdot x_{\mathrm{III}}\cdot x_4' + x_{\mathrm{I}}x_{\mathrm{III}} -c_2x_3'\left[(x_3'+x_4')\sin\gamma-r_2'\cos2\gamma+x_2'\sin2\gamma\right]x_4' }{ \left[\sin\gamma\left(x_{\mathrm{II}}^{\,2}+x_4'x_3'\right)+x_4'r_2'\right]x_4' +r_1\cdot x_{\mathrm{III}}^{\,2}+c_2x_3'x_{\mathrm{III}}x_4' }, \tag{36b} \]

Thus we obtain the points \(K\) and \(C\) of the circle \(Kp\). The secondary and magnetizing currents, the lines of losses, powers, torques, speeds, and efficiency are obtained as described above. If it is not desired to construct the lines of losses, then the efficiency is obtained with sufficient accuracy very simply by replacing the line of total motor losses with a line parallel to the line of speeds. This

possibly because the direction of the line of total motor losses, with the normal position of the brushes, usually differs very little from the parallel to the line of ohmic losses of the stator.

In a similar manner the diagrams in Figs. 19 (Table II), 20 and 21 (Table III) are constructed. Fig. 19 represents diagrams for the same data as in Fig. 16, but for different positions of the brushes. The circle \(K_p\) (Figs. 16 and 19) is constructed for \(\lambda=19^\circ\), the circle \(K_{p\mathrm{I}}\) (Fig. 19) for \(\lambda=11^\circ 30'\), and the circle \(K_{p\mathrm{II}}\) for \(\lambda=32^\circ\). With a decrease of \(\lambda\), as is seen from the figure, the diameter of the circle increases greatly, while the points \(K\) and \(C\) approach \(A\). In Fig. 20 circular diagrams are presented for the limiting positions of the brushes, \(K_{p90}\) for \(\lambda=90^\circ\) and \(K_{p0}\) for \(\lambda=0\), the scale of the circle \(K_{p90}\) being 5 times larger than that of \(K_{p0}\). In Fig. 21 three diagrams are presented for one and the same position of the brushes, but with different \(x_{nd}\). The circle \(K_p\) is the same as in Fig. 16; the circle \(K_{p\mathrm{II}}\) has \(x_{nd}\) equal to \(0.725\) of \(x_{nd}\) in the circle \(K_p\), and the circle \(K_{p\mathrm{I}}\) has \(x_{nd}\) 1.2 times greater than \(x_{nd}\) in the circle \(K_p\). With an increase of \(x_{nd}\) the diameter of the circle decreases, while the points \(K\) and \(C\) approach \(A\).

11. The primary, secondary, and magnetizing currents of the motor, and the influence of constants upon them.

As we saw above, the secondary current is

\[ \dot J_2'=\dot Y_{2v}'\dot E_2'=\dot Y_{2v}'\dot E_1=\frac{\dot E_1}{\dot Z_{2v}'}, \]

the magnetizing current is

\[ \dot J_a=\dot y_a\dot E_1 \]

and the primary current is

\[ \dot J_1=\dot J_2'+\dot J_a =\dot E_1\left(\dot Y_{2v}'+\dot y_a\right) =\dot E_1\left(\frac{1}{\dot Z_{2v}'}+\dot y_a\right) =\dot P_1\dot Y_\delta = \frac{\dot P_1}{\dot z_1+\dfrac{1}{\dot Y_{2v}'+\dot y_a}} . \tag{37} \]

Whence

\[ \dot E_1= \frac{\dot P_1}{1+\dfrac{\dot z_1}{\dot Z_{2v}'}+\dot z_1\dot y_a} = \frac{\dot P_1}{\dot z_1\left(\dfrac{1}{\dot z_1}+\dfrac{1}{\dot Z_{2v}'}+\dot y_a\right)} = \frac{\dot P_1}{\dot z_1\left(\dot y_1+\dot Y_{2v}'\right)}, \tag{38} \]

\[ \dot J_a= \frac{\dot y_a\dot P_1}{1+\dfrac{\dot z_1}{\dot Z_{2v}'}+\dot y_a\dot z_1} = \frac{\dot y_a}{\dot z_1}\cdot \frac{\dot P_1}{\dot y_1+\dot Y_{2v}'} . \tag{39} \]

and

\[ \dot J_2'= \frac{\dot P_1}{\dot Z_{2v}'(1+\dot z_1 y_a)+\dot z_1} = \frac{\dot P_1}{\dot z_1(\dot Z_{2v}'\cdot y_1+1)}, \tag{40} \]

where \(y_1=y_a+\dfrac{1}{\dot z_1}=y_a+y_1\).

In Fig. 22 (Table III) there are represented in rectangular coordinates, as functions of speed, for three different brush-shift angles, the primary and secondary currents obtained from the diagram of Fig. 19. The primary current \(J_1\) at \(v=\infty\) is equal to zero; as \(v\) decreases it slowly increases, at a certain \(v\) becomes equal to the secondary current, then greater than it; between \(v=1\) and \(v=0\) it increases rapidly, at a certain negative speed reaches a maximum, and then, decreasing as \(v\) increases, at \(v=-\infty\) becomes equal to zero. The secondary current \(J_2'\) at \(v=\infty\) is equal to the magnetizing current \(J_a\) with the rotor open; as the speed increases it increases, at a certain negative speed reaches a maximum and, decreasing with increasing \(v\), at \(v=-\infty\) is again equal to the magnetizing current. The electromotive force due to the pulsation of the longitudinal flux \(E_1\) (not represented in rectangular coordinates) at \(v=\infty\) is equal to the applied voltage \(P_1\); as \(v\) decreases it decreases, reaches a minimum at a certain negative speed at the point where \(J_1\) has a maximum, increases as \(-v\) increases, at a certain finite speed becomes equal to the applied voltage, then greater than it, passes through a maximum, and at \(v=-\infty\) is again equal to \(P_1\). The magnetizing current, in its variations, must of course follow the variations of \(E_1\).

With the brushes raised, \(\dot Z_{2v}'=\infty\), and from Eqs. 37, 39, and 40 we have the primary current

\[ \dot J_1= \frac{\dot P_1}{\dot z_1+\dfrac{1}{y_a}} = \frac{\dot P_1}{\dot z_1+\dot z_a} = \frac{\dot P_1}{\dot z_t} = J_a, \]

where \(\dot z_t=\dot z_1+\dot z_a\), and the secondary current \(\dot J_2'=0\).

For \(\lambda=90^\circ\) for the Thomson and Atkinson motors, and for \(\lambda=180^\circ\) for the Deri motor, \(c_1=0\), \(c_2=\infty\), \(\dot Z_{2v}'\), as is seen from formula 12, is, at all finite speeds, equal to infinity; \(J_1=J_a=\dfrac{P_1}{z_t}\). The vector of the primary current \(J_1\) moves along a circle (Fig. 20 \(K_{p90}\)) with radius:

$$ \rho = \frac{x_{3,0}+x_2}{2\left[x_4'(x_{3,0}+r_2\sin\gamma+x_2)+x_1(x_{3,0}+x_2)\right]} \sim \frac{1}{2(x_1+x_4')} $$

and the coordinates of the center

$$ \xi_0= \frac{x_{3,0}\sin\gamma-r_2}{2(x_1+x_4')(x_{3,0}+x_2)}, \qquad \eta_0= \frac{x_{3,0}\cos\gamma+x_2}{2(x_1+x_4')(x_{3,0}+x_2)}. $$

These formulas are obtained from 23a, 23b, and 23c by reducing to

$$ u^2\frac{c_2}{c_1}=0. $$

For \(v=\infty\) the primary current is equal to zero. The reduced secondary current \(J_2'\) is equal to zero at finite speeds and tends to infinity at \(v=\infty\), while the actual secondary current \(J_2=J_2'u\) at \(v=0\) is equal to zero, and at \(v=\infty\) is infinitely large (since at \(\frac{c_2}{c_1}=0\), \(u=\infty\)).

With a decrease in the brush-shift angle, the diameter of the circle increases, and the primary and secondary currents for all finite speeds increase; in Fig. 22 the current-variation curves as a function of speed are represented in rectangular coordinates for brush-shift angles of \(32^\circ\), \(19^\circ\), and \(11^\circ 30'\).

For \(\lambda=0^\circ\), \(x_3'=0\), and the radius of the circle is equal to

$$ \rho= \frac{x_4'+x_2}{2\{x_4'[(r_1+r_2')\sin\gamma+(x_1+x_2)]+x_1x_2'\}}, $$

and the coordinates of the center are

$$ \xi_0= \frac{x_4'\sin\gamma-r_2'}{2\{x_4'[(r_1+r_2')\sin\gamma+(x_1+x_2)]+x_1x_2'\}}, $$

$$ \eta_0= \frac{x_4'\cos\gamma+x_2'}{2\{x_4'[(r_1+r_2')\sin\gamma+(x_1+x_2)]+x_1x_2'\}}. $$

The primary current for \(\lambda=0^\circ\) and for \(v=0\), which may be called the short-circuit current, is obtained from Eqs. 34 and 35:

$$ J_{1km}= \frac{P_1}{\sqrt{(AH_1)^2+(TH_1)^2}} = $$

$$ = \frac{P_1(x_2'+x_4')^2} {\sqrt{[x_4'x_2'(x_2'+x_4')+x_1(x_2'+x_4')^2]^2+ [x_4'(x_2'^2\sin\gamma+x_4'r_2')+r_1(x_2'+x_4')^2]^2}}, $$

or, replacing \(x_2'^2\sin\gamma\) by \(x_2'r_2'\), we have

$$ J_{1km}\sim \frac{P_1(x_2'+x_4')} {\sqrt{[x_4'\cdot x_2'+x_1(x_2'+x_4')]^2+ [r_2'\cdot x_4'+r_1(x_2'+x_4')]^2}}. $$

It must be noted that this is not the greatest possible current of the motor. The magnetizing current of the primary circuit \(J_a\), at \(v=\infty\), has one and the same value for all positions of the brushes, and when the speed changes, at different angles of brush displacement, only the amplitude of the oscillations changes. The magnetizing current of the secondary circuit, equal to \(J_a \cdot u\), varies depending on the change of \(u\); at \(v=\infty\) and \(\dfrac{c_2}{c_1}=0\) it is infinitely large.

In Fig. 23 (Table III) are presented curves of the change of the primary and the actual secondary currents, as functions of the change of the angles of brush displacement, with the rotor at rest \(J_{1k}\) and \(J_{2k}\) (dotted line) and at synchronism \(J_{1c}\) and \(J_{2c}\), obtained from a series of circle diagrams. We see that at \(\lambda=90^\circ\) and \(\lambda=0^\circ\), \(J_{1c}=J_{2k}\), the explanation of which we shall find later. In addition, at large angles of brush displacement the current with the rotor at rest is smaller than at synchronism.

Equations 23a, 23b, and 23c show that the radius of the circle, and, consequently, the primary and secondary currents, at one and the same position of the brushes, increase with a decrease of \(x_3'\), \(x_1\), \(x_2'\), \(r_1\), \(r_2'\), and \(\gamma\). The resistances of the primary and secondary circuits and the angle \(\gamma\) do not have a large influence on the radius, but the last two quantities \(r_2'\) and \(\gamma\) have a great influence on the position of the center of the circle. As we have seen, in all diagrams the center of the circle lies above the axis of ordinates; if, however, we take \(\gamma=0^\circ\), then it will lie below it. This circumstance also compelled the author to take the angle \(\gamma\) into consideration, despite the fact that in repulsion motors, owing to the weak saturation of the iron, it is especially small.

Fig. 24.

Fig. 21 illustrates the change of the circle diagram as a result of a change in the reaction of the transverse coil with the other constants of the motor unchanged and with one and the same position of the brushes. Corresponding to the change of \(x_3'\), the magnetizing current also changes, of course, decreasing with an increase of \(x_3'\), and conversely.

12. The rotating moment of the motor and the influence of the constants on it.

Let there be, in the sinusoidal magnetic field \(abc\) (Fig. 24), a coil of width \(\beta\tau\), with number of turns \(\beta w_2\), through which

flows the current \(J_2\). The force of interaction between the field and the coil for a pair of poles is expressed in kg:

\[ F = 2 \cdot \frac{10^{-6}}{9.81}\cdot l \cdot 2 \int_{\frac{\tau-\beta}{2}}^{\frac{\tau}{2}} B_x \cdot J_2 \sqrt{2}\cdot \frac{w_2}{\tau}\, dx = \]

\[ = \frac{\sqrt{2}\cdot 4 \cdot 10^{-6}}{9.81} \cdot l \cdot B \cdot J_2 \frac{w_2}{\tau} \int_{x_0}^{\frac{\pi}{2}} \sin \alpha \cdot \frac{D}{2}\, dx . \]

Since \(J_2\) and \(B\) vary sinusoidally in time and are shifted by the angle \(\psi\), the mean force of interaction is

\[ F = \frac{4\sqrt{2}\cdot 10^{-6}}{9.81\cdot 2\cdot 2} \cdot l \cdot D \cdot B \cdot J_2 \frac{w_2}{\tau} \sin(90-\alpha_0)\cos\psi = \]

\[ = \frac{\sqrt{2}\cdot 10^{-6}\cdot 2}{9.81\pi} \cdot l \cdot B \cdot J_2 \cdot w_2 \cdot \cos\psi \cdot \sin(90-\alpha_0), \]

and the moment in kgm, if \(D\) is expressed in cm, is

\[ M = \frac{\sqrt{2}\cdot 10^{-8}}{9.81\cdot 2} \cdot D^2 \cdot l \cdot B \cdot J_2 \frac{w_2}{\tau} \cdot \sin(90-\alpha_0)\cos\psi = \]

\[ = \frac{\sqrt{2}\cdot 10^{-8}}{\pi\cdot 9.81} \cdot D \cdot l \cdot B \cdot J_2 \cdot w_2 \cdot \sin(90-\alpha_0)\cos\psi . \]

Thus, the moment of the longitudinal coil in the transverse field for the Thomson and Atkinson motors is equal to:

\[ M_{\mathrm{pr}} = \frac{\sqrt{2}\cdot 10^{-8}}{9.81\cdot \pi} \cdot D \cdot l \cdot B_{\mathrm{pr}} \cdot J_2 \cdot w_2 \cdot \cos\lambda \cdot \cos\gamma; \]

substituting from formulas 6b and 2b the value

\[ B_{\mathrm{pr}} = B_{m\mathrm{pr}}\cdot \frac{\pi^2}{12}\cdot \sin\lambda = \frac{0.4\cdot \pi \cdot w_2\sqrt{2}\, J_2\cos\gamma}{\delta''} \cdot \frac{\pi^2}{12}\sin\lambda, \]

we have

\[ M_{\mathrm{pr}} = \frac{10^{-9}\cdot 2\cdot \pi^2}{9.81\cdot 6} \cdot \frac{D\cdot l}{\delta''} J_2^2 \cdot w_2^2 \sin 2\lambda \cdot \cos^2\gamma, \]

and the torque of the transverse coil in the longitudinal field

\[ M_{nd}=\frac{2\cdot 10^{-8}}{\pi\cdot 9{,}81}\,D\cdot l\cdot B_{nd}\cdot J_2\cdot w_2\cdot \sin\lambda\cos(B_{nd},J_2). \]

We may assume that \(B_{nd}\), with sufficient approximation, is expressed by a formula (exact for \(v=\infty\)) analogous to that for \(B_{np}\):

\[ B_{nd}=\frac{0{,}4\cdot \pi\cdot w_2\sqrt{2}\cdot J_a\cdot \cos\gamma}{\delta''}\cdot \frac{\pi^2}{12}\cos\lambda, \]

Substituting this value of \(B_{nd}\), we obtain

\[ M_{nd}=\frac{2\cdot 10^{-8}\cdot \pi^2}{6\cdot 9{,}81}\cdot \frac{D\cdot l}{\delta''}\cdot w_2^2\cos\gamma\cdot J_2\cdot J_a\cos(B_{nd},J_2)\sin 2\lambda. \]

If the currents are expressed in reduced values, then

\[ M_{np}=\frac{2\cdot 10^{-9}\cdot \pi^2}{9{,}81\cdot 6}\cdot \frac{D\cdot l}{\delta''}\cdot J_2'^2\cdot u^2\cdot w_2^2\cos^2\gamma\sin 2\lambda \]

and

\[ M_{nd}=\frac{2\cdot 10^{-9}\cdot \pi}{9{,}81\cdot 6}\cdot \frac{D\cdot l}{\delta''}\cdot J_2'\cdot J_a u^2\cos(B_{nd}\cdot J_2')\,w_2^2\cos\gamma\sin 2\lambda. \]

For the Deri motor

\[ M_{np}=\frac{\sqrt{2}\cdot 10^{-8}}{\pi\cdot 9{,}81}\cdot D\cdot l\cdot B_{np}\cdot J_2\cdot w_2\cos\gamma\cdot \frac{1}{2}(1+\cos\lambda); \]

substituting the value

\[ B_{np}=\frac{0{,}4\cdot \pi\cdot w_2\sqrt{2}\cdot J_2\cos\gamma}{\delta''}\cdot \frac{\pi^2}{12}\sin\frac{\lambda}{2}, \]

we obtain

\[ M_{np}=\frac{10^{-9}\cdot 2\cdot \pi^2}{9{,}81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot J_2^2\cdot w_2^2\cdot \cos^2\gamma\cdot \sin\lambda\cos\frac{\lambda}{2}; \]

and the torque of the transverse coil in the longitudinal field

\[ M_{nd}=\frac{\sqrt{2}\cdot 10^{-8}}{\pi\cdot 9{,}81}\cdot D\cdot l\cdot B_{nd}\cdot J_2w_2\cos(B_{nd},J_2)\sin\lambda. \]

Substituting, as for the Thomson and Atkinson motors, the value

\[ B_{nd}=\frac{0{,}4\cdot \pi\cdot w_2\sqrt{2}J_a\cos}{\delta''}\cdot \frac{\pi^2}{12}\cos\frac{\lambda}{2}; \]

we have

\[ M_{n0}=\frac{2\cdot 10^{-9}\cdot \pi^{2}}{9.81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot J_{2}J_{a}\cos(B_{n0},J_{2})\,w_{2}^{2}\cos\gamma\sin\lambda\cos\frac{\lambda}{2}. \]

With the currents at the values given,

\[ M_{pr}=\frac{2\cdot 10^{-9}\cdot \pi^{2}}{9.81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot J_{2}^{2}\cdot u^{2}w_{2}^{2}\cos^{2}\gamma\sin\lambda\cos\frac{\lambda}{2} \tag{41} \]

and

\[ M_{n0}=\frac{2\cdot 10^{-9}\cdot \pi^{2}}{9.81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot J_{2}^{2}J_{a}\cdot u^{2}\cos(B_{n0},J_{2})\,w_{2}^{2}\cos\gamma\sin\lambda\cos\frac{\lambda}{2}. \tag{42} \]

Thus the constant mentioned in § 8, for the Thomson and Atkinson motor, is expressed as

\[ k=\frac{10^{-9}\cdot \pi^{2}}{9.81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot w_{2}^{2}\cdot u^{2}\cdot \cos\gamma\sin2\lambda, \tag{43a} \]

and for the Deri motor,

\[ k=\frac{2\cdot 10^{-9}\cdot \pi^{2}}{9.81\cdot 3}\cdot \frac{D\cdot l}{\delta''}\cdot w_{2}^{2}\cdot u^{2}\cos\gamma\sin\lambda\cos\frac{\lambda}{2}. \tag{43b} \]

In Fig. 25 (Plate IV) the curves of variation of the torque as a function of the speed are represented in rectangular coordinates, for \(\lambda=11^\circ30'\), \(\lambda=19^\circ\), and \(\lambda=32^\circ\), obtained from the diagram of Fig. 19. As is evident from these curves, at \(v=\infty\) the torque is zero; as the speed decreases it is at first negative, then passes through zero and slowly increases; shortly before \(v=1\) it begins to increase sharply; at \(v=0\) it has a value close to the maximum; at a small negative speed it passes through a maximum; then, decreasing at first sharply and then more slowly, it reaches zero at an infinitely large negative speed. Thus, in repulsion motors the greatest torque lies outside its working range.

When the angle of brush displacement is changed, the torque also changes greatly. At \(\lambda=90^\circ\) for the Thomson and Atkinson motor, and \(\lambda=180^\circ\) for the Deri motor, it is equal to zero, since at these angles the constant \(k\), \(J_{2}\), and \(J'_{2}\), at finite speeds, are equal to zero; as a result, the line of torques must pass through the point \(K\), at which, as we shall see below, all the other points at finite speeds merge. With a decrease in the angle of brush displacement, \(k\), \(J_{2}\), and

\(J_2'\) have finite values, the torque increases from zero, at first slowly, then rapidly, reaches a maximum at a very small brush-displacement angle, and then drops sharply to 0 at \(\lambda = 0\). In this position it is also equal to zero, because the currents \(J_2\) and \(J_2'\) have finite values, while the constant \(k\) is equal to zero.

For \(\lambda < 0^\circ\) and \(\lambda > 90^\circ\) for the Thomson and Atkinson motors, and for \(\lambda < 0^\circ\) for the Deri motor, the constant \(k\), as is not difficult to see, changes its sign, the torque becomes negative, and the motor changes its direction of rotation.

In Fig. 23 there are also shown curves of the variation of the rotating torque as a function of the brush-displacement angle, \(M_k\), for the stationary position of the motor, and \(M_c\)—for synchronism. From these curves it is seen that the maximum torque in the stationary position of the rotor is attained at a larger brush-displacement angle \((9^\circ)\) than the torque at synchronism \((4^\circ)\).

The angle which the line of torques makes with the axis of abscissas varies within narrow limits; therefore, for a given position of the brushes, the torque increases with an increase in the diameter of the circle \(K_p\), consequently (Fig. 21) with a decrease of \(x_3', x_1, x_2'\) and \(r_1\). As for \(r_2'\) and \(\gamma\), although they have little influence on the change of torque due to a change in the radius of the circle, they strongly influence the position of the line of torques through the change in the position of the center of the circle.

13. Primary and secondary powers of the motor and the influence of constants on them.

In Fig. 26 (Table IV) there are shown curves of the primary and secondary power as functions of the speed for three different brush-displacement angles, obtained from the diagram of Fig. 19. At \(v = \infty\) the primary power is equal to zero, and the secondary is negative; as \(v\) decreases, both slowly increase, with the curve of primary power lying above the curve of secondary power; at the speed determined by formula 32, the secondary power passes through zero; at speeds close to synchronism, both curves begin to increase more rapidly, reach a maximum, and, dropping sharply with further decrease of speed, pass through zero—the secondary at \(v = 0\), and the primary at a certain negative value. In the generator part, the roles of the curves change: the negative secondary power becomes greater than the primary. With an increase in negative speed, the powers increase sharply, reach a maximum, and with further increase of \(v\)

decrease; at \(v=-\infty\) the primary is equal to zero, and the secondary to a certain negative value.

With a change in the angle of brush displacement the powers change greatly. At \(\lambda=90^\circ\) for the Thomson and Atkinson motors, and \(180^\circ\) for the Deri motor, the secondary power is zero, since the secondary current is zero; the line of secondary power turns into a tangent (Fig. 20) at the point \(K\), owing to the fact that the speed at the second point, where \(W_2'=0\),

\[ v= \frac{ c_1 x_2 u^2+\dfrac{x_4' \cdot c_1^2}{c_2}\cos\gamma }{ c_1 r_2 u^2+\dfrac{x_4' \cdot c_1}{c_2}c_1^2\sin\gamma } =0 \]

(since here \(u=\infty\), \(c_2=0\), \(c_1=\infty\), and \(\dfrac{u^2 c_2}{c_1^2}=0\)). The primary power for such a position of the brushes is extremely small. With a decrease in the angle of brush displacement, both the primary and the secondary powers increase, reach a maximum at a certain position of the brushes close to the longitudinal axis, and then, with further decrease in the angle of brush displacement, fall sharply: the secondary to 0 at \(\lambda=0\), and the primary to a certain fairly considerable value, going only to cover the losses of the short-circuiting of the motor.

In Fig. 27 (Table IV) are shown the curves of the primary power \(W_1\)—I and of the secondary \(W_2'\)—II at synchronism as a function of the angle of brush displacement, obtained from a series of circular diagrams; moreover, the curve \(W_2'\) differs from the curve of the moments \(M_c\) in Fig. 23 evidently only in scale.

At \(\lambda=0\), \(c_2=\infty\), \(c_1=0\), and the speed at the second point, where \(W_2'=0\), from formula 32,

\[ v=\frac{x_3' c_2 \cos\gamma}{x_2' c_2 c_1 \sin\gamma}=\infty, \]

the line of secondary power is represented by the straight line \(KA\), but at the same time, as we shall see below, all points at finite speeds merge with the point \(K\); therefore the output power at all points with finite speed is equal to zero. The primary power at \(\lambda=0\) is represented by the segment \(KL\) (Fig. 20), the same for all finite speeds.

With a change in \(x_2'\), \(r_2'\), \(x_3'\), and \(\gamma\), the position of the line of secondary power, as is evident from formulas 36 and 32, changes very little (see Fig. 21); therefore, with a decrease in \(x_3'\), \(x_2'\), and \(x_1\), the primary and secondar...

power increase. As for \(r_2'\) and \(\gamma\), they, having very little influence on the secondary power, increase the primary power by their own increase, owing to the displacement of the axis of ordinates.

14. Power factor and efficiency of the motor, and the influence of the constants upon them.

In Fig. 28 (Table IV), curve I represents the variation of the power factor, and II the efficiency as a function of speed. We see that at \(v=0\), \(\cos \varphi_1\) has a finite value and, with increasing speed, continuously increases up to 1 at \(v=\infty\). In reality, in repulsion motors \(\cos \varphi_1\) has a maximum; this is explained by the strong influence upon it of the ampere-turns of the short-circuited coil. The efficiency is equal to zero at the speed determined by Eq. 32, slowly increases, reaches a maximum at \(v=1.75\), and then falls to zero at \(v=0\). The maximum efficiency is obtained, as is to be expected, because the iron losses in repulsion motors are not large at small currents and, consequently, at high speed.

For \(\lambda=90^\circ\), \(u=\infty\), \(c_2=0\), \(c_1=\infty\), the angles \(\varphi_{1k}\) and \(\varphi_{1c}\) are determined from Eq. 36:

\[ \tg \varphi_k=\tg \varphi_c = \frac{(x_2+x_{3,90})^2 x_4' + x_1 (x_2+x_{3,90})^2} {\sin \gamma (x_2+x_{3,90})^2 x_4' + r_1 (x_2+x_{3,90})^2} = \frac{x_4' + x_1}{\sin \gamma\, x_4' + r_1}, \]

and by this same expression the angle \(\varphi\) in general is determined at any finite speed; this follows from the fact that, in Eq. 36, the term depending on the speed is equal to zero. For this reason the efficiency at \(\lambda=90^\circ\), at all finite speeds, is equal to zero, since all points of the circle at finite speeds coincide with \(K\).

With a decrease in the brush displacement, the power factor and the efficiency increase, attain a maximum at brush-displacement angles fairly close to zero, and then both fall sharply to zero at \(\lambda=0\).

Curves III and IV of Fig. 27 represent the power factor and the efficiency at synchronous speed as a function of the brush position. From these curves it is seen that, in order for the motor to operate with the greatest possible power factor and efficiency, the brushes must be within \(10^\circ\)--\(25^\circ\) from the longitudinal axis.

Theory of Repulsion Motors

For \(\lambda=0,\ c_2=\infty,\ c_1=0,\ x_3'=0\), the angles \(\varphi_k,\ \varphi_c\), and in general for points with finite speed, are determined from Eq. 36,

\[ tg\,\varphi_k=tg\,\varphi_c= \frac{x'_{2,0}(x'_{2,0}+x_4')x_4' + x_1(x'_{2,0}+x_4')^2} {(x_2'^2\sin\gamma+x_4'r_2')x_4' + r_1(x_2'+x_4')^2}, \]

since here also the multiplier at the terms depending on \(v\) is equal to zero; i.e., here too all points of the circumference at finite speeds merge at \(K\), and the efficiency at finite speeds is equal to zero.

In Fig. 30 are shown the curves of the efficiency for \(v=2,\ v=1.5,\ v=1.0,\ v=1.75\), and \(v=0.5\) as functions of the change of the brush-shift angle. As may be seen from these curves, the maximum of the efficiency for a given speed, as the latter is reduced, moves toward larger brush-shift angles. In Fig. 31 (see p. 52), for the same five speeds, are shown the curves of the power factor as a function of the brush-shift angle. It must be noted, however, that the curves of the efficiency, and especially the power curves, are strongly affected by the short-circuited coil, so that the actual curves of the motor differ from those given in Figs. 30 and 31 all the more, the farther the motor speed is from synchronous.

Fig. 30. Efficiency for \(v=2,\ v=1.5,\ v=1.0,\ v=0.75\), and \(v=0.5\) as a function of the brush-shift angle.

With a decrease of \(x_3'\), \(r_1\), and \(\gamma\), and with an increase of \(x_1\), the phase-shift angle at synchronism \(\varphi_{1c}\), as can be seen from formula 36 and the diagram of Fig. 21, increases, so that the power factor decreases. The efficiency at synchronism decreases with decreasing \(x_3'\), owing to the fact that the point of synchronism moves higher along the circle, and with increasing \(r_1,\ r_2'\), and \(\gamma\), owing to the increase in losses. It should be noted here that the decrease of the efficiency due to a decrease of \(x_3'\) is associated with an increase in power.

In conclusion, in Fig. 29 (Table IV), for the normal position of the brushes, the curves of the primary current \(J_1\), the speed \(v\), the efficiency \(\eta\), and the power factor \(\cos \varphi_1\) are presented as functions of the delivered mechanical power.

Fig. 36. Power factor at \(v=2\), \(v=1.5\), \(v=1.0\), \(v=0.75\), and \(v=0.5\), as a function of the brush-shift angle.

In bringing the present article to an end, the author considers it his duty to point out that he has not taken into account the influence of the short-circuited coil, which is manifested chiefly in the abrupt change of the power-factor curve. The author hopes to devote his next article, in the near future, to the elucidation of this influence. For the same reason he does not touch upon the commutation of the motor.

Tomsk,
December 1911.

Vs. Khrushchov.

Noted Misprints

Page	Line	Printed	Should be
3	9 from top	$w_2=\dfrac{N_1}{2}$	$w_1=\dfrac{N_1}{2}$
17	13 from top	$Z_{\nu=0}=-\dfrac{c_2}{c_1}x_3$	$Z_{\nu=\infty}=-\dfrac{c_2}{c_1}x_3$
21	12 from top	$n\,\Omega$ in 1 cm	$n\,\overline{\omega}$ in 1 cm
40	8 from bottom	$+x_1 x_{\mathrm{III}}$	$+x_1 x_{\mathrm{III}}^2$

In Fig. 27, Table IV, $\cos\varphi_1$ for $\lambda=0$ should have the value $-0.3$, and for $\lambda=90^\circ$, $-0.09$.

Contents

No.	Section	Page
1.	Introduction	1
2.	Voltage diagram	6
3.	Transformation coefficient	8
4.	Electromotive forces and reactions of the motor windings	10
5.	Effective resistance, effective reactance of the rotor circuit, and the functional relation between them	15
6.	Construction of the circle diagram by means of successive transformations	19
7.	Magnetizing and secondary currents in the circle diagram	28
8.	Speed, phase displacement, and torque in the circle diagram	31
9.	Losses, power, and efficiency in the circle diagram	34
10.	Direct construction of the circle diagram	38
11.	Primary, secondary, and magnetizing currents of the motor and the influence of constants upon them	41
12.	Motor torque and the influence of constants upon it	44
13.	Primary and secondary powers of the motor and the influence of constants upon them	48
14.	Power factor and efficiency of the motor and the influence of constants upon them	50

To the article by V. I. [[unclear: author’s surname]]: “Theory of Repulsion Motors.”

Fig. 18.

Fig. 19.

Appendix II

To the supplement by V. M. Druzhinin, “Theory of Repulsion Motors.”

Line of speeds

Fig. 16.

According to the experiments of V. I. Dmitriev. Theory of the repulsion motor

Fig. 21. Change of the circular diagram with change in reactive conductance.

Fig. 22. Variable [[unclear: \(g\)-quantities]] and the rectified current, represented as currents as functions of the slip, for brush-displacement angles \(\lambda = 19^\circ\) and \(\lambda = 32^\circ\).

Table III.

Fig. 23. Variable and rectified currents \(g_1\), currents \(g_{2c}\), \(g_{2e}\), and maximum currents \(i_{1c}\) and \(i_{1e}\), as functions of the brush-displacement angle.

To the article by Prof. V. M. Khrushchev, “On the Theory of the Repulsion Motor.”

Table IV.

Fig. 26. Curves of \((M_k)\) and of the torque \((M)\), in dependence on the speed \(n\), for various brush-shift angles: \(\alpha = 11^\circ 30'\), \(\alpha = 17^\circ 30'\), [[unclear: further angles]].

Fig. 27. Primary diagram (I), secondary diagram (II), power factor (III), and efficiency (IV), as functions of the brush-shift angle.

Fig. 28. Efficiency coefficient (II) and power factor (I), as functions of the speed.

Fig. 29. Primary current (I), secondary current (II), efficiency coefficient (III), and power factor, as functions of the angle of brush lag (IV).

[[unclear: lower cropped figure caption fragment]] … abscissae … speed \(n\) and at angles \(\alpha = 11^\circ 30'\), \(\alpha = 19^\circ\), and \(\alpha = 52^\circ\).