Mechanical engineer A. V. Verkhovsky.

Kinematic and Dynamic Study of the Rake Mechanism of Self-Dumping Headers

The most widespread rake mechanism of self-dumping headers is a mechanism equipped with an “eccentric” mechanism with a “counter.” This mechanism provides the proper motion of the rakes by means of a special guide track—the “rake circle,” along which rolls a roller fastened to the elbow (bracket) of the rake.

To obtain the required motion of the rake, the profile of this track is given a rather complex shape. It should be noted that it is not entirely correct to call the mechanism under consideration “eccentric.” This mechanism is a spatial cam mechanism with a stationary cam (the “rake circle”) and with a swinging driven link (the “bracket”).

As is known, the purpose of the rakes is: 1) to bring uncut stalks up to the line of the knives, and 2) to dump from the machine platform the cut stalks when enough of them have accumulated for a sheaf.

Accordingly, the first of these functions is performed by all the rakes, while the second is performed by only some of them.

To achieve these two different motions of the rakes, the guide track (along which the roller rolls) has, over a certain section, a branching, with a special transfer device and an instrument—a counter—that regulates the passage of the roller along one track or another.

The smooth operation of the entire rake mechanism depends to a considerable extent on the rational construction of the profile of the guide track.

In this article we shall set ourselves the goal of carrying out a kinematic and dynamic study of the rake mechanism of self-dumping headers, outlining ways toward a critical evaluation of existing rake mechanisms (from the standpoint of their dynamics), and providing some material for the design of new mechanisms of analogous construction.

As an example, we shall present here a study of the rake mechanism of the header of the State Lyubertsy Plant. The law of motion of the rake was recorded by mounting on the rake head a sector marked with divisions in degrees. An arrow was attached to one of the elbows (brackets), which, during the motion of the elbow, indicated on the sector the angles of elevation of the rake in the vertical plane (Fig. 1).

When readings were taken on the sector scale, the vertical shaft of the rake head was driven by hand, and the reading on the sector was made when the vertical shaft was turned by half a tooth of its bevel gear. For the machine under consideration (of the Lyubertsy Plant), this rotation is \(5.3^\circ\).

Fig. 1.

From the readings thus obtained, one can construct diagram I (Fig. 2). Along the axis of abscissas are plotted the angles of rotation of the vertical shaft, and along the axis of ordinates—the angles of rotation of the rake in the vertical plane.

Scale of abscissas—\(m_\varphi = 3.3\ \text{deg/mm} = 0.058\ \text{rad/mm}\).

Scale of ordinates—\(m_d = 2\ \text{deg/mm} = 0.035\ \text{rad/mm}\).

Since the rake shaft rotates uniformly, it may also be assumed that time is laid off along the axis of abscissas; in that case the time scale will be

\[ m_t = \frac{T}{2\pi}, \]

where \(T\) is the time of one revolution of the vertical shaft of the rake apparatus, which will be equal to:

\[ T = \frac{2\pi}{\omega}, \]

where \(\omega\) is the angular velocity of rotation of this shaft. Then we shall have:

\[ m_t = \frac{2\pi}{2\pi\omega}\cdot m_\varphi = \frac{m_\varphi}{\omega}. \]

It will be found below that \(\omega = 1.34\ \frac{\text{rad}}{\text{sec}}\); then

\[ m_t = \frac{m_\varphi}{\omega} = \frac{0.058}{1.34} = 0.043\ \text{sec/mm}. \]

As the zero position in diagram I (Fig. 2) the position is taken in which the center of gravity of the rake is in the same horizontal plane with the axis of the elbow roller. For the non-discharging rake, the curves in the diagrams are plotted with a dotted line. Further, by graphical differentiation, diagrams of the angular velocities and accelerations of the rake lift were constructed (diagrams II and III, Fig. 2).

Fig. 2.

It is of primary interest to determine the moments acting on the rake in the vertical plane, since in some machines underestimation of the magnitude of the inertial moments acting in this plane leads to the fact that during operation this inertial moment, exceeding the other moments, separates the roller from the guide (at certain positions of the rake), after which the rake falls and the roller strikes the guide.

During the motion of the rake, the following moments act on it in the vertical plane (relative to the axis of the elbow roller):

a) \(M_1\)—inertial moment,

b) \(M_2\)—moment due to the rake’s own weight,

c) \(M_3\)—moment due to the centrifugal force arising in the rake as a result of its rotation about the vertical shaft.

Determination of the moment \(M_1\).

The inertial moment \(M_1\) will be equal to:

\[ M_1 = J \frac{d^2 \alpha}{dt^2}, \]

where \(J\) is the moment of inertia of the rake relative to the axis of the elbow shaft,
\(\dfrac{d^2 \alpha}{dt^2}\) is the angular acceleration of the rocking of the rake in the vertical plane.

The moment of inertia of the rake \((J)\) can be determined experimentally by rocking the rake in a suspended state about the axis of the elbow shaft (in this case the elbow shaft should be replaced by a specially made axle with prisms at its ends).

The time of one complete oscillation of the rake of the Lyubertsy Plant is

\[ T_0 = 2.03 \text{ sec.} \]

To determine the moment of inertia, there is the formula

\[ J = \frac{Q \cdot T_0^2 \cdot l}{4\pi^2} \ . . . . . . . . . . . . . . . \ (1) \]

where \(Q = 7\) kg is the weight of the rake,
\(l = 58\) cm is the distance from the center of gravity of the rake to the axis of oscillation.

Substituting the numerical values into formula (1), we obtain

\[ J = \frac{7 \cdot 2.03^2}{4 \cdot 3.14^2} \cdot 58 = 42 \text{ kg. cm. sec.}^2 \]

To determine the angular acceleration, as indicated above, graphical differentiation of the curve of the rake lift angles should be performed (diagram I). Diagram II represents the angular velocities \(\dfrac{d\alpha}{dt}\) of the rake lift on the scale:

\[ m'_{\alpha} = 0.14 \frac{\text{rad}}{\text{sec. mm}}. \]

Diagram III represents the angular acceleration of the rake lift on the scale:

\[ m''_{\alpha} = 1.11 \frac{\text{rad}}{\text{sec. mm}}. \]

Since

\[ M_1 = J \frac{d^2 \alpha}{dt^2}, \]

where \(J\) is a constant quantity, the diagram of angular accelerations may be regarded as the diagram of inertial moments \((M_1)\), changing only the scale.

This scale of moments will be:

\[ m_1 = m''_{\alpha} \cdot J = 1.11 \cdot 42 = 46 \text{ kg cm/mm.} \]

Determination of the moment \(M_2\).

The moment due to the rake’s own weight will be (Fig. 3)

\[ M_2 = Q \cdot l \cdot \cos \alpha \ . . . . . . . . . . . . . . . \ (2). \]

Using formula (2), diagram IV of the moments \(M_2\) is constructed to the scale \(m_1\).

Determination of the moment \(M_3\).

The moment from the centrifugal force arising as a result of the rotation of the rake about the vertical shaft of the rake mechanism will be (Fig. 3)

\[ M_3 = R \cdot l \cdot \sin \alpha \ldots \ldots \ldots (3) \]

where:

\[ R=\frac{Q}{g}\omega^2(a+l\cos\alpha) \]

is the centrifugal force,

\(\omega\)—the angular velocity of the rake during its rotation about the vertical shaft of the rake mechanism,

\(a=8.6\) cm—the distance from the axis of the roller to the axis of the vertical shaft of the rake mechanism.

Fig. 3.

Substituting the value of \(R\) into equation (3), we obtain

\[ M_3=\frac{Q}{g}l\omega^2(a\cdot\sin\alpha+l\sin\alpha\cos\alpha). \]

The angular velocity of rotation of the vertical shaft of the rake mechanism \((\omega)\) will be:

\[ \omega=\frac{2v\cdot i}{D}(1-\Psi), \]

where: \(D=82\) cm—the diameter of the running wheel,
\(v\)—the speed of motion of the header, which we take equal to \(1.3\) m/sec,
\(i=0.437\)—the transmission ratio,
\(\Psi=0.03\)—the slip coefficient.

Then

\[ \omega=\frac{2\cdot1.3\cdot0.437}{0.82}\cdot(1-0.03)=1.34\ \frac{\mathrm{rad}}{\mathrm{sec}}. \]

We shall show that the moment \(M_3\) is so small in comparison with the moments \(M_1\) and \(M_2\) that it may be neglected.

Let us find the maximum of the moment \(M_3\), for which we take the derivative and set it equal to zero.

\[ \frac{dM_3}{d\alpha}=\frac{Q}{g}\cdot\omega^2 l(a\cos\alpha+2l\cos^2\alpha-l)=0. \]

Whence

\[ \cos\alpha=-\frac{a}{4l}\pm\sqrt{\frac{a^2}{16l^2}+\frac{1}{2}}. \]

From this expression we find

\[ \cos\alpha_1=0.67;\qquad \alpha_1=47^\circ. \]

\[ \cos\alpha_2=0.74;\qquad \alpha_2=131^\circ. \]

The angle \(\alpha_2=131^\circ\) does not occur; at the angle \(\alpha_1=47^\circ\) we obtain

\[ M_3\max=\frac{7}{981}\cdot1.34^2\cdot58(8.6\cdot0.92+58\cdot0.93\cdot0.67)=33\ \text{kg·cm}. \]

With the previously obtained moment scale \(m_1 = 46\ \text{kg·cm/mm}\), this moment \((M_3)\) can change the diagram of moments \(M_2\) by no more than

\[ \frac{33}{46} = 0.7\ \text{mm}, \]

which may be neglected. To determine the resultant moment \((M)\) acting on the rake in the vertical plane, we perform a graphical summation of the moments \(M_1\) and \(M_2\), as a result of which we obtain diagram V. From this diagram it is seen that, at the adopted header speed \(v = 1.3\ \text{m/sec}\), the resultant moment will be negative in two sections (in the section between divisions 14 and 17 and in the section between divisions 31 and 33). This means that, under the action of this negative moment, the roller will separate from the guide track and then, in falling, will produce an impact. This phenomenon occurred because of a certain waviness of the guide track in these sections (see diagram I in Fig. 2), which may perhaps be accidental in character in the given specimen of the machine. At a header speed of \(1.1\ \text{m/sec}\), separation of the roller from the guide track will not occur, even in the presence of this waviness.

Checking the Rake for Strength.

The check was made for the machine of the Lyubertsy Plant. The greatest moment acting on the rake in the vertical plane will be, as is seen from diagram V of Fig. 2, at the 33rd division of this diagram. Its magnitude will be equal to the product of the corresponding ordinate by the scale:

\[ M \max = cd \cdot m_1 = 33 \cdot 46 = 1520\ \text{kg·cm}. \]

The section modulus of the rake bar will be

\[ W = \frac{b \cdot h^2}{6} = \frac{5 \cdot 5.3^2}{6} = 23.4\ \text{cm}^3, \]

where: \(b = 5\ \text{cm}\)—the width of the section of the rake bar,
\(h = 5.3\ \text{cm}\)—its height.

The bending stress in the vertical plane will be

\[ \sigma = \frac{M \max}{W} = \frac{1520}{23.4} = 65\ \text{kg/cm}^2, \]

which for wood may be considered allowable.

To check the rake bar for the forces acting parallel to the axis of oscillation of the rake and bending the rake bar in the plane of this axis, the magnitude of these forces must be determined.

Such forces will be:

1) the resistance of the mown grain when it is being thrown by the rake from the platform,
2) inertial forces caused by the Coriolis acceleration,
3) a component of the centrifugal force caused by the rotation of the rake about the axis of its oscillation.

The component of the last force arises because the body of the rake is not located perpendicular to the axis of oscillation, as is seen from the horizontal projection in Fig. (4).

This component will be so small that its influence may be neglected.

Indeed, the centrifugal force when the rake rotates about axis \(A\) (Fig. 4) will be:

\[ R_1=\left(\frac{d\alpha}{dt}\right)^2\cdot l\cdot \frac{Q}{g}. \]

Substitute into this formula the maximum value of \(\frac{d\alpha}{dt}\), which, according to diagram III (Fig. 2), will be \(2.95\ \frac{\text{rad}}{\text{sec}}\); then:

\[ R_1=2.95^2\cdot 58\cdot \frac{7}{981}=3.6\ \text{kg}. \]

The component of this force parallel to the axis of oscillation of the rake will be:

\[ N=R_1\cdot \operatorname{tg}\beta=3.6\cdot 0.13=0.47\ \text{kg}. \]

This force is so small that it may be neglected.

Turning now to the first and second forces mentioned above, it should first of all be noted that these forces act at different moments, since at the moment of discharge the rakes do not oscillate about axis \(A\), and Coriolis acceleration does not arise in them. Therefore the rake should be checked separately for the first and second forces. The first force is determined by the friction of the discharged grain against the platform. In addition, at the first moment the rake overcomes the inertia of the grain, setting it in motion along the platform. Taking the weight of the discharged grain as \(P=13\) kg and the coefficient of friction of the grain against the platform as \(f=0.23\), we obtain a friction force equal to:

\[ F=f\cdot P=0.23\cdot 13=3\ \text{kg}. \tag{4} \]

Fig. 4.

It may be assumed that this force is applied at the center of the tooth strip, i.e., at a distance of 110 cm from the oscillation axis \(A\).

The magnitude of the bending moment from the friction force will be:

\[ M_1=110\cdot 3=330\ \text{kg}\cdot\text{cm}. \]

The section modulus of the bar for the bending case under consideration will be:

\[ W_1=\frac{hb^2}{6}=\frac{5.3\cdot 5^2}{6}=22\ \text{cm}^3. \]

The stress obtained will be:

\[ \sigma'=\frac{M'_{\text{b}}}{W'}=\frac{330}{22}=15\ \text{kg}/\text{cm}^2. \]

In addition, as indicated above, the rake overcomes the inertia of the grain at the first moment of discharge. This inertial force will be:

\[ C=\frac{P}{g}\cdot j, \]

where \(P\) is the weight of the discharged grain,
\(g\) is the acceleration of gravity,
\(j\) is the acceleration of the discharged grain.

Let us determine this acceleration, assuming that the motion during the initial period of displacement of the grain over the platform will be uniformly accelerated, which may be assumed, since at the moment when the rake acts on the grain, the grain stalks will spring and will gradually acquire the velocity of the rake. In addition, let us specify the path over which the entire mass of grain reaches the velocity of the rake. Under these assumptions the force \(C\) will be constant, and for the calculation we can use the equation of live forces in the following form:

\[ \frac{m \cdot v_0^2}{2}=C \cdot S \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (5). \]

Whence

\[ C=\frac{m v_0^2}{2S}, \]

where \(S\) is the path over which the accelerated motion of the grain takes place; we take it equal to 5 cm;

\[ v_0 \]

is the velocity of the middle of the rake relative to the platform;

\[ v_0=\omega \cdot L_0 \quad (\text{Fig. } 9). \]

\[ v_0=1.34 \cdot 100=134\ \text{cm/sec}. \]

\(m\) is the mass of the dumped grain,

\[ \frac{P}{g}. \]

We shall have

\[ C=\frac{P}{g}\cdot \frac{v_0}{2S}=\frac{13}{981}\cdot \frac{134^2}{2 \cdot 5}=23\ \text{kg}. \ldots \ldots \ldots \ldots \ldots \ldots (6). \]

The bending moment due to this force will be:

\[ M'_{\text{b}}=C \cdot 110=23 \cdot 110=2530\ \text{kg}\cdot\text{cm}, \]

and the stress will be

\[ \sigma''=\frac{M''_{\text{b}}}{W_1}=\frac{2530}{22}=115\ \text{kg/cm}^2. \]

This stress may be considered permissible.

To find the force caused by the Coriolis acceleration, it is necessary to determine the value of this acceleration.

The Coriolis acceleration is determined as:

\[ j_k=2\omega v_2 \sin \alpha \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (7) \]

where: \(\omega\) is the angular velocity with which the transport motion is performed, in this case the angular velocity of the vertical shaft of the raking apparatus;

\(v_2\) is the velocity of the relative motion of the center of gravity of the rake (in Fig. 5, point \(B\)).

This velocity may be expressed as:

\[ v_2=\frac{d\alpha}{dt}\cdot l \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (8) \]

where \(\frac{d\alpha}{dt}\), as before, is the angular velocity of oscillation of the rake about the axis \(A\).

Substituting \(v_2\) from equation 8 into 7, we obtain

\[ j_k=2\omega \frac{d\alpha}{dt}\cdot l \cdot \sin \alpha \ldots \ldots \ldots \ldots \ldots \ldots (9) \]

In expressions 7 and 9 the angle \(\alpha\) is regarded as the angle between the direction of the velocity \(v_2\) and the axis about which the transport motion is performed (Fig. 5). In Fig. 5 this angle \(BCO\) is equal to the angle \(BAK\), i.e., to that angle \(\alpha\) for which diagram I in Fig. 2 was constructed.

Fig. 5.

To obtain the moment caused by the Coriolis acceleration, it is necessary to multiply the acceleration by the mass of the rake and by the arm \(l\); then we shall have:

\[ M_k = 2\omega \frac{d\alpha}{dt}\cdot l^2 \sin \alpha \cdot \frac{Q}{g}. \]

It should be noted that this moment has been found under the assumption that the entire mass of the rake is concentrated at its center of gravity.

The maximum value of this moment can be found by several trials, using diagrams I and II (Fig. 2), since this moment is a function of two variable quantities \(\frac{d\alpha}{dt}\) and \(\alpha\), which are shown on the indicated diagrams.

The maximum value of this moment will occur at the ninth and thirteenth division,

\[ M_{k\max}=35\ \text{kg·cm}. \]

The stress from the action of this moment in the rake bar will be:

\[ \sigma=\frac{M_{k\max}}{W_1}=\frac{35}{23.4}=1.5\ \text{kg/cm}^2. \]

Next we shall proceed to determine the moments acting on the vertical shaft of the rake apparatus.

These moments will be:

1) The moment \(M'_0\), obtained from the resistance experienced by the elbow roller from the guide track during its motion.

2) The moments obtained from the resistances of the dumping rake when removing the grain from the platform. These moments will be: \(M''_0\)—the moment obtained from the friction of the dumped grain against the platform, and \(M'''_0\)—the moment obtained from the inertial resistance of the dumped grain in the initial period of dumping.

3) The moment arising from the Coriolis acceleration, which we shall denote by \(M^{IV}_0\).

4) The moment arising as a result of air resistance during the motion of the rakes, and the moment of friction in the support of the vertical shaft. In view of their small magnitude, we shall neglect these moments.

Let us begin with the determination of the moment \(M'_0\).

In Fig. 6 an elbow is shown, the roller of which bears on the guide track; \(n m\) is the vertical shaft of the rake apparatus. The angle \(\xi\) is a certain constant angle between the axis of the roller and the axis of the rake. If, in the vertical plane, the rake is acted upon by

Fig. 6.

\[ y=\frac{\dfrac{d\alpha}{dt}}{m'_{\alpha}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots (17) \]

moment \(M\), determined by diagram \(V\) (Fig. 2), then the force \(H\) with which the roller will be loaded as a result of the action of this moment is determined as follows:

\[ H=\frac{M}{C} \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots (10) \]

Let us consider the conditions of motion of a roller loaded by the force \(H\) along the guide track. For this purpose we project the roller onto a plane perpendicular to its axis (Fig. 7).

Fig. 7.

Let us find the horizontal force \(P_0\) that must be applied to the roller so that the roller, being loaded by the force \(H\), begins to move along the guide track \(L\).

This force, as is known from the course “General Theory of Machines” (the case of lifting a body along an inclined plane), will be

\[ P_0=\frac{H(\operatorname{tg}\gamma+f')}{1-f'\operatorname{tg}\gamma} \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots (11) \]

Here \(\gamma\) is the angle of rise of the track,

\(f'\) is the total coefficient of friction of the roller.

The angle \(\gamma\) will be different for different positions of the roller. Let us determine its value.

For this purpose we shall trace the motion of the center of the roller. During the motion of the rake, the center of the roller moves parallel to the guide. Over a certain elementary interval of time \(dt\), the center of the roller will be displaced by a certain amount \(OD=dS\) (Fig. 8). This displacement can be resolved into the horizontal displacement \(dl\) and the displacement occurring in the vertical plane \(dh\).

Fig. 8.

The vertical shaft of the rake apparatus during the time \(dt\) will turn through a certain elementary angle \(d\varphi\). Then the center of the roller will be displaced in the horizontal direction by the amount

\[ dl=d\varphi\cdot \rho \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots (12) \]

where \(S\) is the distance of the center of the roller from the axis of the vertical shaft of the rake apparatus (Fig. 6).

At the same time, owing to the inclination of the guide track, the center of the roller in the vertical plane will be displaced by \(dh\), rotating about the axis \(A\). This displacement will be equal to

\[ dh=d\alpha\cdot c \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots (13) \]

where \(d\alpha\) is the corresponding angle of rotation of the rake (and together with it the roller in the vertical plane).

\(c\) is the distance from the center of the roller to the axis of the roller shaft \(A\).

From Fig. (8) it is seen that

\[ \operatorname{tg}\gamma=\frac{dh}{dl}=\frac{d\alpha\cdot c}{d\varphi\cdot \rho} \qquad \ldots\ldots\ldots\ldots\ldots\ldots\ldots (14) \]

Let us now determine the value of \(\rho\) for different positions of the rake.

From Fig. 6 it is seen:

\[ \rho = a + c \cdot \cos(\xi-\alpha) \ldots \ldots \ldots \ldots \ldots (15) \]

Now let us derive the final expression for the required moment \(M_0'\). From Figs. 6, 7, and 8 it is seen that

\[ M_0' = P_0 \cdot \rho \]

Using equations (11) and (14), we obtain

\[ M_0' = \frac{ H\left(\dfrac{d\alpha}{d\varphi}\cdot \dfrac{l}{\rho} + f'\right) }{ 1 - f'\dfrac{d\alpha}{d\varphi}\cdot \dfrac{c}{\rho} } \cdot \rho \]

Further, substituting \(H\) from expression (10), we obtain

\[ M_0' = \frac{M \cdot \rho}{c} \cdot \left[ \frac{ \dfrac{d\alpha}{d\varphi}\cdot \dfrac{c}{\rho} + f' }{ 1 - f'\dfrac{d\alpha}{d\varphi}\cdot \dfrac{c}{\rho} } \right] \ldots \ldots \ldots \ldots \ldots (16) \]

By this formula, and also using the expression for \(\rho\) (formula 15), it is not difficult to calculate the value of the moment \(M_0'\) for any position of the rake. To carry out this calculation it is also necessary to note that, for any positions of the rake, the value of the quantity \(\dfrac{d\alpha}{d\varphi}\) can be found from diagram II in Fig. 2. Indeed, this diagram (II) represents the angular velocities of rocking of the rake in the vertical plane \(\left(\dfrac{d\alpha}{dt}\right)\). If the abscissa axis of this diagram is regarded not as the time axis, but as the axis along which the angles of rotation of the vertical rake shaft are laid off, then the ordinates of this diagram, at a certain scale \((m_0)\), represent precisely the quantity \(\dfrac{d\alpha}{d\varphi}\). The scale \(m_0\) can be determined as follows.

Some arbitrary ordinate \(y\) of diagram II in Fig. 2 represents, at the scale \(m'_\alpha\), the angular velocity of rocking of the rake in the vertical plane, i.e. \(\dfrac{d\alpha}{dt}\); consequently one may write:

\[ y = \frac{ \dfrac{d\alpha}{dt} }{ m'_\alpha } \ldots \ldots \ldots \ldots \ldots (17) \]

\[ \left( m'_\alpha = 0.14 \frac{\mathrm{rad}}{\mathrm{sec.}\,\mathrm{mm}} \text{—see the beginning of the article} \right). \]

Further, the same ordinate \(y\) (according to what was said above) may represent \(\dfrac{d\alpha}{d\varphi}\) at some scale \(m_0\), i.e. one may write

\[ y = \frac{ \dfrac{d\alpha}{d\varphi} }{ m_0 } \ldots \ldots \ldots \ldots \ldots (18) \]

Comparing equations (17) and (18), we obtain

\[ \frac{\dfrac{d\alpha}{dt}}{m'_\alpha} = \frac{\dfrac{d\alpha}{d\varphi}}{m_0}. \]

Whence

\[ m_0=\frac{d\alpha}{d\varphi}\cdot\frac{dt}{d\alpha}\cdot m'_\alpha =\frac{dt}{d\varphi}\cdot m'_\alpha =\frac{m'_\alpha}{\dfrac{d\varphi}{dt}}. \]

Noting that \(\dfrac{d\varphi}{dt}=\omega\)—the angular velocity of the vertical shaft of the rake apparatus—we shall have

\[ m_0=\frac{m'_\alpha}{\omega}=\frac{0.14}{1.34}=0.104. \]

Using this scale, it is easy to find from diagram II in Fig. 2 the values of \(\dfrac{d\alpha}{d\varphi}\) for any position of the rake.

The value \(\rho\) should be determined by formula (15), substituting into it the corresponding values of the angle \(\alpha\) from diagram I in Fig. 2.

The magnitude of the moment \(M\), entering formula (16), is determined (as was indicated earlier) from diagram V in Fig. 2. The quantities \(c\) and \(f'\) are constant quantities; \(c\) is determined as a design dimension. As for the coefficient of friction \(f'\) of the roller against the guide, it can be found in the following way.

This coefficient of friction \(f'\) is composed of two parts: the coefficient of friction in the roller journal and the coefficient of rolling friction of the roller along the guide track. From the course “General Theory of Machines” it is known that, for the case of rolling of a roller rotating on a journal, the “general” coefficient of friction \(f'\) will be

\[ f'=f\cdot\left(\frac{r_1}{r_2}\right)+\frac{k}{r_2} \]

where \(r_1\) is the radius of the journal,
\(r_2\) is the radius of the roller,
\(f\) is the coefficient of friction in the journal,
\(k\) is the coefficient of rolling friction of the roller along the guide.

For the machine of the Lyubertsy plant we have \(r_1=13.5\) mm; \(r_2=22\) mm; and if we take \(f=0.2\) and \(k=0.05\) mm, then we obtain:

\[ f'=0.2\frac{13.5}{22}+\frac{0.05}{22}=0.121+0.0021=0.122. \]

For the machine under consideration \(C=78\) mm; \(a=100\) mm, \(\xi=60^\circ\).

We now have all the data in order, by formula (16), to calculate the magnitude of the moment \(M'_0\) for various positions.

Carrying out this calculation for 36 positions, i.e. at every \(10^\circ\) of rotation of the vertical shaft, one can construct from the obtained values the diagram of the moments \(M'_0\), both for the dumping and for the non-dumping rake.

Let us now proceed to the determination of the moments \(M''_0\) and \(M'''_0\). The moment \(M''_0\) will arise due to the friction of the dumped grain against the platform (as this

as was indicated above). The friction force was determined earlier (formula 4); it is equal to:

\[ F=f\cdot P=3\ \text{kg}. \]

The arm of this force relative to the vertical shaft of the rake mechanism will be \(L_0=100\ \text{cm}\) (Fig. 9), and the moment of this force will be:

\[ M_0''=F\cdot L_0= \]

\[ 3\cdot 100=300\ \text{kg}\cdot\text{cm}. \]

Fig. 9.

Further, the moment \(M_0'''\), arising as a result of the inertial resistance of the grain being dumped during the initial period of dumping, will be determined by the inertial force found from equation (6):

\[ C=23\ \text{kg}. \]

The moment of this force relative to the vertical shaft of the rake mechanism will be

\[ M_0'''=C\cdot L_0=23\cdot 100=2300\ \text{kg}\cdot\text{cm}. \]

It now remains to determine the moment \(M_0^{IV}\), arising due to the Coriolis acceleration. This moment is obtained if the Coriolis acceleration (formula 9) is multiplied by the mass of the rake and by the distance \(L\) from the center of gravity of the rake to the vertical shaft:

\[ M_0^{IV}=2\omega \frac{d\alpha}{dt}\cdot l\cdot \sin\alpha\cdot L\frac{Q}{g}\ . . . . . . . . . . (19) \]

It should be noted that, whereas for the dumping rake during the dumping period the distance \(L_0\) was constant, in formula (19) \(L\) will be variable, since this formula provides for different positions of the rake. Let us express \(L\) through the angle \(\alpha\). From Figure (3) it follows that:

\[ L=a+l\cos\alpha. \]

Substituting this expression into formula (19), we obtain

\[ M_0^{IV}=2\omega\frac{d\alpha}{dt}\cdot l\cdot \sin\alpha\,(a+l\cos\alpha)\frac{Q}{g}. \]

By this formula one can calculate \(M_0^{IV}\) for any position of the rake.

Now let us proceed to the construction of the resultant diagrams. In Fig. 10, for the Lyuberetskiy plant machine, the resultant diagram of the moments acting on the vertical shaft is constructed for the dumping rake, i.e. of the moments: \(M_0'\); \(M_0''\); \(M_0'''\); \(M_0^{IV}\).

Further, in Fig. 11 there is shown the diagram of moments acting on the vertical shaft for the non-dumping rake, i.e. of the moments \(M_0'\) and \(M_0^V\).

Fig. 10.

Since the rake head carries four rakes, which are simultaneously in motion and which are displaced relative to one another by an angle of 90°, then, in order to obtain the final resultant moment acting on the vertical shaft of the rake apparatus, it is necessary to add the diagrams shown in Figs. 10 and 11, shifting them relative to one another by 90°. In addition, it is necessary to take into account which rake, by number, will be the dumping rake. In Fig. 12 this summation has been carried out, with the fourth rake taken as the dumping rake, which is most often the case.

In the diagram of Fig. 12 this fourth rake is shown first, and after it come the three non-dumping rakes, the scale of the moments in all the diagrams being kept the same, namely \(m_1 = 46\) kg·cm/mm.

Fig. 11.

Fig. 12.

The mean ordinate \(y_0\) of this diagram makes it possible to calculate the power expended in setting the rake apparatus in motion. The mean moment acting on the vertical shaft of the rake apparatus will be:

\[ M_0 = y_0 \cdot m_1 = 9 \cdot 46 = 415 \text{ kg·cm.} = 4{,}15 \text{ kg·m.} \]

The power expended on this shaft will be

\[ A_0 = M_0 \cdot \omega = 4{,}15 \cdot 1{,}34 = 5{,}58 \text{ kg·m/sec.} \]

\[ N = \frac{5{,}58}{75} = 0{,}07 \ HP. \]

The power at the rim of the running wheel expended on rotating the rake apparatus will be

\[ N = \frac{N_0}{\eta} = \frac{0{,}07}{0{,}51} = 0{,}14 \ HP. \]

Here \(\eta\) is the efficiency of the transmission from the running wheel to the shaft of the rake apparatus, \(\eta = 0{,}8 \cdot 0{,}8 \cdot 0{,}8 = 0{,}51\).

To determine the efficiency of the toothed transmission, one must divide the power expended on dumping the grain by the total power expended on the vertical shaft of the rake apparatus.

The power expended on dumping will be found if the area of the rectangles (taken from the diagram of Fig. 12 and representing the dumping work) is multiplied by the scale and divided by the time of one revolution \(T\). This power will be

\[ A' = \frac{237}{T} = \frac{237 \cdot 46 \cdot 0{,}058}{4{,}69} = 135 \text{ kg·cm.} = 1{,}35 \text{ kg.} \]

Then the efficiency under consideration will be

\[ \eta'=\frac{A'}{A_0}=\frac{1.35}{5.58}=0.24. \]

The full efficiency of the rake apparatus, including losses in the transmission, will be

\[ \eta''=\eta\cdot\eta'=0.51\cdot0.24=0.12. \]

Figure 12 gives a diagram of the moments acting on the vertical shaft for a header of the Lyuberetsky Plant. For other headers the diagrams are obtained essentially the same.

Let us note some shortcomings of the existing rake apparatuses from the point of view of their dynamics.

In the diagram (Fig. 12) there are sections on which moments act that are directed in the direction of rotation of the rake apparatus. These sections in the diagram (Fig. 12) are marked by horizontal hatching, and the moments acting on them are denoted by a plus sign.

These positive moments must, of course, be recognized as undesirable, since during their action the vertical shaft of the rake apparatus becomes, as it were, the driving link, which will entail disengagement of the working profiles of the gear wheels transmitting rotation to it from the running wheel. Gaps will appear between the working profiles of the gear wheels. At the next instant the vertical shaft of the rake apparatus will again be loaded by the resistance moment, i.e., it will again become the driven link, and the gaps that had previously formed between the working profiles of the gear wheels will be traversed; moreover, at the instant the engagement of the profiles is restored, a certain impact will inevitably occur. Thus, in the presence of the above-mentioned positive torques in the diagram of Fig. 12, the rake apparatus will operate with jolts (not smoothly). This latter circumstance explains the fact that many plants have abandoned the use of chain transmission to the rake apparatus, since with chain transmission the noted phenomenon of jolts will manifest itself even more strongly.

Let us dwell on the causes by which the above-mentioned positive moments are produced, and on the question of eliminating them.

The positive moments in the diagram of Fig. 12 are produced by the work of the weight of the rake descending toward the line of the knives.

To eliminate the positive moments indicated above, it is necessary to make full use of the work of the weight of the rake descending toward the line of the knives for lifting the rake moving ahead. Then the positive moment developed by the descending rake will be used immediately at the head of the rake apparatus and, consequently, will not be transmitted to the vertical shaft of the rake apparatus and through it to the gear wheels.

To achieve this, the profile of the guide track should be slightly changed on the lifting section of the non-discharging rake, making it steeper over the interval on which positive values of the resultant moment are obtained.

Such elimination of positive moments can be achieved only in the case when grain is not being discharged from the platform. When the discharging rake, in performing its function, goes along the lower branch of the guide track, the direction of the track profile just proposed will not help. It should be emphasized, however, that the proposed correction of the track profile eliminates the positive moment when all non-discharging rakes pass, and leaves it unchanged only for the case of passage of the discharging rake.

Eliminating the positive moment throughout the entire operating cycle of the rake mechanism by any simple means apparently does not appear possible.

One further circumstance concerning the profile of the guide track should be emphasized. As already indicated above, in the diagrams of the resultant moments acting on the rake in the vertical plane (diagrams V in Fig. 2), for most machines the resultant moment is negative over certain sections. In these sections, at a somewhat increased machine speed, the roller will separate from the guide track and then fall back onto it with an impact. This phenomenon can easily be avoided by making only the slightest change in the profile of the guide track over the indicated sections.