Mechanical Engineer A. V. Verkhovsky.

THE WORKING PROCESS OF A HAY MOWER

Hay mowers, like many other agricultural machines, have not up to the present time had a sufficiently fully developed theory illuminating their working process.

The lack of theoretical elucidation of the phenomena occurring in a hay mower often makes it impossible to make sense of the contradictory material obtained from numerous field tests of these machines. The present work is an attempt to give a theoretical explanation of the working process taking place in a hay mower.

The tractive effort applied to a hay mower when it is operating on a horizontal plot of field is expended in overcoming the following resistances:

1. The resistances encountered by the knife when cutting the stems of plants, and the resistances associated with mechanical losses in the transmission mechanism,

2. The rolling resistance of the running wheels.

3. The resistance of the finger bar, arising as a result of its sliding over the surface of the field.

In accordance with this division of the resistances, the tractive effort, which we shall denote by \(P\), may also be divided correspondingly into three components \(P_1\), \(P_2\), and \(P_3\).

These components, like the force \(P\) itself, may be called the driving forces acting on the mower.

Let us dwell on the working process that will arise in the hay mower under the influence of the force \(P_1\).

The magnitude of the force \(P_1\) will depend on the magnitude of that resistance which this force overcomes. Since these resistances will be those indicated in the first item, it is clear that, in order to determine the magnitude of the force \(P_1\), it is necessary first of all to consider the working process taking place in the cutting apparatus of the machine.

For this purpose, Fig. 1 shows the operation of the knife apparatus. In view of the identity of the phenomena occurring in the intervals between the axes of any pair of fingers of the knife apparatus, it is sufficient to consider the cutting process taking place only in one of the indicated intervals.

In Fig. 1 the stroke of the knife is chosen equal to the distance between the axes of the fingers. The forward advance of the machine, during the time of movement of the knife from the right dead position to the left, may be determined as follows: during the motion of the knife under consideration, the crankshaft will turn through an angle

\[ \alpha_0 \cong \pi \qquad . . . . . . . . . . . . . . . . . . . . (1) \]

The running wheel will thereby turn through an angle

\[ \varphi=\frac{\alpha_0}{i}\ldots\ldots\ldots(2), \]

where \(i\) is the gear ratio.

Taking into account the slippage of the machine, the magnitude of the machine’s advance during one stroke of the knife will be

\[ l=\varphi\cdot R(1+\psi_0)\ldots\ldots\ldots(3) \]

where \(R\) is the radius of the running wheel, and \(\psi\) is the mean value of the slip coefficient. Using equation (2), from (3) we obtain:

\[ l=\frac{\alpha_0 R}{i}(1+\psi)\ldots\ldots\ldots(4) \]

If in formula (4) we take \(\alpha_0=\pi\), then the error will be about \(1.4^\circ\), or, relative to \(180^\circ\), the error will be less than \(1\%\), i.e. the error will be lost in the inaccuracy of the graphical construction.

To give concreteness to the further reasoning, let us substitute into equation (4) the particular values of the quantities taken in the present case as applied to the mower of the State Lyubertsy Plant “New Ideal,” namely—\(\alpha_0 \cong \pi,\ i=26.51,\ R=0.4,\ \psi=0.06\). Then we obtain

\[ l=\frac{\pi\cdot 0.4}{26.51}(1+0.06)=0.050\ \text{m}\ldots\ldots\ldots(5) \]

Neglecting the influence of the finite length of the connecting rod, the trajectory of the knife points may be regarded as a sinusoid. Then the magnitude \(l\) will determine half the pitch of the sinusoid. We shall not dwell on the construction of the sinusoid. (For the construction of the sinusoid see Weiss, Agricultural Machine Science, 1929 edition.)

Let us now consider the process of deflection and cutting of the stems. As is known, the principle of operation of the modern cutting apparatus of a hay mower consists in cutting the plant stems by the simultaneous action on the stem of the cutting edge of the knife segment and the edge of the guard ledger plate. It follows from this that the cut of the plant stems will occur at those points at which the blade of the segment, at some given moment, will intersect the edge of the guard ledger plate. The geometric locus of these points of intersection will be the line \(AB\) (Fig. 1), which we shall call the cutting line.

To clarify the method of constructing this line, let us trace the construction of one of its points, for example, point \(B\). Point \(B\) will lie on the sinusoid described by point \(D\), taken on the edge of the segment. To find the point of intersection of this sinusoid with the edge of the guard ledger plate (and thereby to find the point of the cutting line), point \(D\) must be transferred horizontally onto the edge of the guard ledger plate to point “\(b_1\).” Next, through point “\(b_1\)” draw a line parallel to

to the motion of the machine, which, at its intersection with the sinusoid under consideration, gives the desired point \(B\). Point \(B\) will be that point of the finger insert which, during the translational motion of the machine, will meet at point \(B\) with point \(D\) of the edge of the segment.

If the cutting edges of the segment and the insert are rectilinear, then the line \(AB\) will also be approximately rectilinear, so that for its construction it is sufficient, by the method described, to find only two of its points.

As the knife moves from the right dead position to the left and back, the knife will, over some part of its stroke, cut the stems of the plants; this part of the knife stroke is customarily called the “active stroke.”

Let us now make a very important reservation: all subsequent constructions will be made on the assumption of a uniform and homogeneous distribution of plant stems over the given mown area of the field. Although this assumption does not correspond to reality, it may be accepted because at any given moment not one segment will be working, but all the knife segments; and although each segment will be loaded differently (with a nonuniform distribution of vegetation), if one takes the average load of the segments, it will correspond precisely to that which would be obtained under the condition of a uniform and homogeneous distribution of vegetation.

Let us return to the further consideration of the cutting process. If the active stroke of the knife is divided into equal intervals, then (even under the assumption of a uniform distribution of stems) a different number of plant stems will be cut in each interval. To examine this phenomenon, let us divide the line of cuts \(AB\) into equal portions. In Fig. 1 this line is divided into 18 portions. Through all the division points let us draw sinusoids with such a calculation that these sinusoids will be the trajectories of points of the blade of the segment.

Let us make one more important observation.

If we proceed from the assumption of a uniform distribution of stems, then on each resulting portion of the straight line of cuts there will be cut a number of stems proportional to the area from which these stems will be brought to this portion. And, moreover, since the work expended on cutting (also taking into account the assumption of homogeneity of the stems) will be proportional to the number of stems, it follows that one may say that the cutting work expended on any portion of the line of cuts will be proportional to the magnitude of that area from which stems will be brought to this portion.

Thus, for example, on the portion \(bh\) of the line of cuts, stems will be cut from the area \(abcdeghi\).

The work expended on cutting on this portion will be proportional only to the indicated area. In exactly the same way, every other portion will have its own area, the stems from which will be brought to the given portion and cut on it. And ra-

the work expended on these sections will, according to the foregoing, be proportional to these areas.

Let us note that here, by the work expended by the knife on cutting the stems on some section of the cutting line, is meant all the work that the knife expends on this section, including also the work of friction on the back of the knife. This work of friction on the back of the knife, which arises as a consequence of the fact that the reactive force of the stems will act on the knife, pressing the back of the knife against the friction plates, will be the greater the more stems are cut on the given section of the cutting line.

Thus, this work of friction, being proportional to the number of stems cut, may be added to the work of cutting. Hence we shall assume that the work expended on cutting by the knife includes, as a certain part, the above-mentioned work of friction as well.

During the travel of the knife from the right dead position to the left, the stems located on the area shaded in Fig. 1 will be cut. Let us also note that the feeding of stems to the cutting line will be effected, on the one hand, by the cutting edge of the knife, and, on the other, by the deflecting action of the fingers. The deflection of the stems by the fingers is shown in Fig. 1 by inclined arrows. This inclination will result from friction between the stems and the edge of the finger insert.

Let us return again to considering the work expended on sections of the cutting line. To obtain a more visual picture of the relative magnitude of the work expended on each section separately, let us construct the diagram shown in Fig. 2.

In this diagram the work expended on each section of the cutting line is represented in the form of narrow rectangles constructed on the corresponding intervals of the knife travel. This diagram is constructed as follows. Along the horizontal line the knife travel \(S\) is laid off. From the right dead position, at a distance \(RK\), the point \(K\) is constructed, corresponding to the beginning of the active travel of the knife.

The segment \(RK\) is found as follows. From Fig. 1 it is seen that the active travel of the knife will begin when the point \(C\), taken on the edge of the segment in its right dead position, after describing part of a sinusoid, arrives at point \(A\). We project point \(A\) onto the line \(NM\). Then the segment \(CA_1\) will represent the path of the knife in its relative motion, which the knife will traverse from its right dead position before its active travel begins. This segment \(CA_1\) is laid off in Fig. 2 in the form of an equal segment \(RK\). To obtain the end of the active travel, one must in Figure 1 project points \(D\) and \(B\) onto the line \(NM\), and lay off the distance between the obtained projections in Figure 2 from point \(R\), in the form of segment \(RH\).

Thus, the active travel of the knife will be determined in Fig. 2 by the segment \(KH\).

For the further construction we divide the active travel of the knife into intervals corresponding to sections of the cutting line. There should be as many such intervals on the active travel of the knife as there are sections into which the cutting line is divided (in the case under consideration—18).

These intervals will turn out to be equal to one another for the following reasons: first, because the line of cuts is divided into equal sections, and, second, because the line of cuts, as well as the cutting edges of the segment and of the insert, are rectilinear. Under these conditions, as the cutting edge of the knife passes over equal sections of the line of cuts, the knife, in its relative motion, will also pass through intervals equal to one another. Thus, in order to mark off these intervals in Fig. 2 it is sufficient to divide the active stroke of the knife into 18 equal parts. On these intervals rectangles have been constructed, the areas of which are proportional to the areas from which stems are cut on the corresponding sections of the line of cuts (for clarity, one of these rectangles in Fig. 2 is hatched). Since the bases of these rectangles represent certain displacements of the knife, and the area of the rectangles, as indicated above, must be proportional to the work—that is, for the time being it represents, on an as yet unknown scale, the work of cutting—it may be concluded from this that the height of these rectangles will represent, on some scale, the cutting resistance force experienced by the knife as it passes through the corresponding intervals of its active stroke. This circumstance is extremely important for the further course of the reasoning.

Into this diagram it is also necessary to include the work expended in cutting the stems located on the area \(nAmpq\) of Fig. 1. The stems located on this area will be cut, after their deflection, at point \(A\). Therefore the work expended in cutting the indicated stems should, as it were, be added to the work represented by the first rectangle (counting from the right dead position in Fig. 2), by lengthening this rectangle accordingly. But since a rather large number of stems from the area \(nAmpq\) will be brought to point \(A\), these stems cannot, in practice, be cut at one point \(A\).

Under the action of the approaching cutting edge of the segment, these stems, before being cut, will be distributed over a certain section of the line of cuts. Depending on the density of the vegetation, they may occupy along the line of cuts a section \(1.5\)—\(2\) cm long, i.e. be distributed over several of the previously constructed sections on the line of cuts. In accordance with what has just been said, the work expended in cutting the stems from the area \(nAmpq\) has been added in Fig. 2 in the form of a horizontally hatched area to the previously constructed rectangles, taken over five intervals. The magnitude of this hatched area is determined on the basis of the proportionality described earlier. Its upper boundary is drawn as a straight line, in the absence of special experimental data on the actual character of this boundary.

The entire area thus obtained in Fig. 2 represents, on a certain (as yet unknown) scale, the work expended in cutting the stems during one stroke of the knife from the right dead position to the left dead position. Let us emphasize the circumstance that

in view of the identity of motion of all segments, it may be considered that this area of the diagram (Fig. 2) will represent, on a certain scale, the work expended on cutting by all segments over the course of the indicated stroke of the knife.

To determine the scale of the diagram it is necessary to know the magnitude of the work expended on cutting stalks from a unit area of the field (for example, from one square decimeter of the field) for each given vegetation. In view of the absence of direct data on the magnitude of this specific work (these data should be obtained by setting up special experiments, and some results of such experiments are given at the end of the article), we shall make use of certain indirect data available in the literature.

According to the data of tests of mowers by the machine-testing station of the Timiryazev Agricultural Academy, published in Vestnik Metallopromyshlennosti No. 3–4 for 1928 in an article by Prof. B. A. Krylov on p. 201, the cutting work for various machines amounts to from 1.1% to 45.9% of all the work expended. The causes of such large fluctuations can hardly be explained by the specific features of the cutting mechanisms of the machines tested. Apparently, these causes lie in a number of some other circumstances, unfortunately not indicated in the cited article.

Let us choose a certain average percentage of the work going to cutting, which we shall take as equal to 20%. We shall take the total power expended during operation of a two-horse mower as equal to 1.65 \(HP\), and the speed of its motion as equal to \(1.1 \dfrac{m}{sec}\). Then the power expended on cutting will be:

\[ N_1=\frac{N\,20}{100}=\frac{1.65\cdot 20}{100}=0.33HP. \]

The work expended on cutting during one stroke of the knife will be

\[ A_1=75\,N_1\,T, \]

where \(T\) is the time of one stroke of the knife.

This time will be:

\[ T=\frac{l}{v}, \]

where \(l\) is the advance of the machine during one stroke of the knife (see equation 5), \(v\) is the speed of motion of the machine

\[ T=\frac{l}{v}=\frac{0.05}{1.1}=0.046\ sec. \]

Then the work

\[ A_1=74\,N_1T=75\cdot 0.33\cdot 0.046=1.14\ kgm. \]

The area of the diagram (Fig. 2) represents precisely this work \(A_1\).

Consequently, the scale of this diagram can be found from the relation

\[ F \cdot m_p \cdot m_e = A_1 \qquad \ldots \ldots \ldots \ldots (6) \]

where \(F\) is the area of the diagram (Fig. 2), which can be found by planimetering the diagram; in our example \(F = 510\ mm^2\); \(m_e\) is the scale of the knife travel path, in our case

\[ m_e = 0.001\frac{m}{mm}; \]

\(m_p\) is the scale of the heights of the rectangles, or, more precisely, the scale of forces, since these heights represent the shearing resistance forces, as was described above.

We find the scale \(m_p\) from equation (6):

\[ m_p = \frac{1.14}{510 \cdot 0.001} = 2.24\ \frac{kg}{mm} \]

Thus, the diagram in Fig. 2 will represent the diagram of the working resistances acting on the knife, whose scale will be:

\[ m_p = 2.24\ \frac{kg}{mm}. \]

In order to obtain the complete picture of the forces acting on the knife, it is necessary to add the inertial forces to the working resistances just found.

The diagram of inertial forces is constructed over the knife travel path (Fig. 3) and is shown on it in the form of the inclined line \(EL\). In this construction, the effect of the finite length of the connecting rod is neglected.

To construct the straight line \(EL\), it is necessary to compute the magnitude of the inertial force in the dead positions of the knife. This inertial force will be:

\[ C = \frac{Q_1 + Q_2}{g} \cdot j \qquad \ldots \ldots \ldots \ldots (7) \]

where \(Q_1\) is the weight of the knife, \(Q_2\) is the weight of one third of the connecting rod, \(g\) is the acceleration of gravity, and \(j\) is the acceleration of the knife in the dead position.

The magnitude of the knife accelerations is determined by the formula:

\[ j = \omega^2 r \left( \Cos \alpha + \frac{a}{e}\Sin \alpha - \frac{r}{e}\Cos 2\alpha \right) \]

(see Vestnik Metal. Prom. No. 78 for 1927, article by Kovalenko).

In this formula, the angle \(\alpha\) is the angle of rotation of the crank, measured from the horizontal diameter; \(\omega\) is the angular velocity of the crankshaft; \(r\) is the crank radius; \(a\) is the elevation of the crankshaft.

For the right dead position:

\[ \Sin \alpha_1 = \frac{a}{e-r} = 0.353 \]

\[ \alpha_1 = 20.7^\circ. \]

For the left dead-center position:

\[ \operatorname{Sin}\alpha_2=\frac{a}{e+r}=0.326 \]

\[ \alpha_2=19.1^\circ. \]

Then the acceleration of the knife in the right dead-center position will be

\[ j_1=\omega^2 r(\operatorname{Cos}20.7^\circ+0.34\operatorname{Sin}20.7^\circ-0.04\operatorname{Cos}41.4^\circ)=\omega^2 r\cdot 1.025 \]

and in the left dead-center position it will be

\[ j_2=\omega^2 r(\operatorname{Cos}19.1^\circ+0.34\operatorname{Sin}19.1^\circ-0.04\operatorname{Cos}38.2^\circ)=\omega^2 r\cdot 1.021. \]

Thus it becomes clear that the difference between the accelerations of the knife in the left and right dead-center positions is very small; it amounts to 0.4%.

Allowing for a very small error, the acceleration in the right and left dead-center positions may be taken as equal to \(\omega^2\cdot r\cdot 1.023\), and on the diagram the acceleration curve may be drawn as a straight line.

To construct this straight line, let us determine the numerical value of the accelerations in the dead-center positions of the knife.

For this purpose, first find the angular velocity of the crank:

\[ \omega=\frac{a_0}{T}, \]

where \(T\) is the time of one stroke of the knife; \(a_0\) is the angle of rotation of the crank corresponding to one stroke of the knife.

The angle \(a_0\) may, as above, be taken equal to \(\pi\):

\[ a_0=\pi \]

\[ T=0.046\ sec \]

\[ \omega=\frac{a_0}{T}=\frac{\pi}{T}=\frac{3.14}{0.046}=68\frac{rad}{sac} \]

Substituting into equation (7) the value of \(\omega\) found and the values of all the other quantities found by direct measurement of the machine parts, we determine the inertial force:

\[ C=\frac{3.27+\frac{2.65}{3}}{981}\cdot 68^2\cdot 3.75\cdot 1.023=73\ kg. \]

In order to lay off this force on the drawing, it must be divided by the scale \(m_p=2.24\); we obtain

\[ \frac{73}{2.24}=32.6\ mm \]

This value is laid off in Fig. 3 along the verticals from the dead-center points, in the form of the segments \(DE\) and \(OL\), with the segment \(DE\) being laid off upward and \(OL\) downward, the laying off being done so that the ordinates of the dia-

grams of the working resistances were graphically summed with the ordinates of the diagram of inertial forces. For the summation in Fig. 3, the diagram of working resistances from Fig. 2 was transferred. The shaded area gives the law of variation of the resultant forces acting on the knife. The summed ordinates of the portions of the diagram marked with a minus sign depict resistance forces; those marked with a plus sign depict forces whose direction coincides with the direction of motion of the knife.

To obtain the summed ordinates of the forces, for the stroke of the knife from left to right, Fig. 3 also includes the diagram of working forces for this return stroke of the knife. It is plotted inverted downward so that its ordinates are summed with the ordinates of the previous diagram of inertial forces, thereby avoiding the need to construct a diagram of inertial forces for the return stroke of the knife. The diagram of working resistances for this stroke of the knife does not require, for its derivation, any special constructions, since this diagram must be completely identical to the diagram of the same name constructed for the forward stroke, because throughout the entire construction the influence of the finite length of the connecting rod was neglected. The summation of ordinates for the return stroke of the knife in Fig. 3 is not indicated by hatching.

For further analysis of the working process, let us construct a diagram of tangential forces. We shall find the magnitude of the tangential forces for various positions of the knife graphically. But in order to arrive at a graphical method for finding the indicated tangential forces, let us find an analytical expression for the tangential force.

Fig. 4 shows a certain arbitrary position of the crank-and-connecting-rod mechanism of the hay mower.

\(OB\)—crank.

\(AB\)—connecting rod.

For the position of the mechanism taken, the direction of the connecting rod makes an angle \(\beta\) with the horizontal. This angle \(\beta\) will be a variable quantity, but for a small value of the ratio

\[ \frac{OB}{AB}, \]

the changes in the angle will be insignificant. To neglect the influence of the finite length of the connecting rod is the same as neglecting the change in the magnitude of the angle \(\beta\), because with an infinitely long connecting rod the angle \(\beta\) will be constant. Since the entire construction was carried out without taking into account the influence of the finite length of the connecting rod, in finding the tangential forces we shall adhere to the same accuracy of construction, i.e. we shall consider the angle \(\beta\) constant; in other words, we shall take it as equal to a certain mean value, namely:

\[ \beta \simeq \beta_0 = \frac{\beta^{\max} + \beta^{\min}}{2} = \mathrm{Const}. \]

For the hay mower under consideration, \(\beta_0 = 18^\circ\).

Having made this reservation, we draw through point \(O\) a line at the angle \(\beta_0\) to the horizontal, and from this line we shall measure the angle of rotation of the crank—\(\alpha\).

At point \(A\) lay off the resistance force \(S\), which, for the given position of the mechanism, acts on the knife. Resolve this force along the connecting rod and along the normal to the finger bar. Then we obtain the force \(K\) acting along the connecting rod:

\[ K=\frac{S}{\operatorname{Cos}\beta}\simeq \frac{S}{\operatorname{Cos}\beta_0}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (8) \]

Next transfer the force \(K\) to point \(B\) and resolve it into the tangential force \(T\) and in the direction of the crank \(R\). The tangential force \(T\) is found from the force triangle.

\[ T=K.\operatorname{Sin}\alpha\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (9) \]

Substituting the value of \(K\) from equation (8) into equation (9), we obtain

\[ T\simeq \frac{S\operatorname{Sin}\alpha}{\operatorname{Cos}\beta_0} \]

Now, for the graphical determination of the tangential forces, let us apply the following graphical construction. Construct a semicircle of arbitrary radius (Fig. 5), and from a certain radius \(OD\) lay off the angle \(\alpha\), which in Fig. 4 determines the position of the crank. At this angle draw in Fig. 5 the radius \(OB\). Along this radius lay off the segment \(OL\), which, on the scale \(m_p\), will represent the force \(S\) acting, in the given position of the mechanism, on the knife. From point \(L\) drop the perpendicular \(LM\) to the line \(OD\). Next, through point \(L\), at the angle \(\beta_0\) to the line \(LM\), draw the straight line \(LE\).

Now we shall prove that the segment \(LE\) will represent the force \(T\) on the same scale on which the segment \(OL\) represents the force \(S\). From triangle \(ELM\) we write:

\[ LE=\frac{LM}{\operatorname{Cos}\beta_0}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (11) \]

Next, from triangle \(LOM\) we find:

\[ LM=OL\operatorname{Sin}\alpha\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (12) \]

Substituting \(LM\) from eq. (12) into eq. (11), we obtain

\[ LE=\frac{OL.\operatorname{Sin}\alpha}{\operatorname{Cos}\beta_0}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (13) \]

Multiply both sides of equality (13) by the force scale — \(m_p\):

\[ m_p.LE=\frac{OL.m_p.\operatorname{Sin}\alpha}{\operatorname{Cos}\beta_0}\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (14) \]

It was noted above that \(OL\), on the scale \(m_p\), represents the force \(S\). This means that

\[ OL.m_p=S\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ .\ (15) \]

Using equality (15), from (14) we obtain:

\[ LE \cdot m_p=\frac{S \sin \alpha}{\cos \beta_0}\ldots\ldots\ldots\ldots (16) \]

Comparing equality (16) with equality (10), we conclude:

\[ LE \cdot m_p \cong T, \]

and this means that the segment \(LE\) represents the force \(T\) on the same scale on which the segment \(OL\) represents the force \(S\).

In Fig. 6 the tangential forces are constructed in a similar manner for various positions of the crank.

Twelve such positions are taken on the semicircle. The resistance forces corresponding to these positions and acting on the knife were found from the diagram of total forces (Fig. 3). For this purpose, on the stroke of the knife (Fig. 3), as on a diameter, a semicircle was constructed, which was likewise divided into 12 equal parts. From the points of division, perpendiculars were dropped to the knife stroke, determining the corresponding positions of the knife. For these positions, according to the ordinates of the diagram, the total forces acting on the knife were found.

Determining the positions of the knife by means of a semicircle constructed on the knife stroke, as on a diameter, gives an error corresponding to neglecting the influence of the finite length of the connecting rod. The validity of what has just been said is seen in Fig. 7, which shows that the division of the knife stroke, with an infinitely large connecting-rod length, can be carried out for the crank-and-connecting-rod mechanism of the mower by means of a semicircle constructed on the knife stroke, as on a diameter.

Having developed the crank circle and divided it into 24 parts, let us lay off upward from the ordinates at each division the corresponding tangential forces found in Fig. 6. The diagram thus constructed (Fig. 8) will represent the diagram of tangential forces acting on the crank pin.

In this diagram, from point 0 to point 8 and from point 12 to 20, the ordinates are negative, since they represent the resistance forces reduced to the crank pin; from point 8 to point 12 and from point 20 to 24, the ordinates are positive, since the forces represented by them tend to turn the crank in the direction of its rotation. These latter forces will be driving forces; they have developed at the expense of the inertia forces arising in the knife.

In this diagram of tangential forces constructed for the crank pin, the losses in the crank-and-connecting-rod mechanism are not taken into account. To allow for these losses, the negative ordinates of this diagram, as representing resistance forces, should be multiplied by the coefficient of efficiency of the crank-and-connecting-rod mechanism; the positive ordinates, as representing driving forces, should be divided.

Taking the efficiency of the crank-and-connecting-rod mechanism as 0.9 and multiplying the negative ordinates by it, and dividing the positive ones by it, we obtain—

the diagram of corrected tangential forces shown in Fig. 9.

Further on, we shall reduce these tangential forces to the rim of the traveling wheel. Before carrying out this reduction, it is necessary to make some remarks concerning the diagram of tangential forces (Fig. 9).

Taking into account the circumstance that the torque from the traveling wheels to the crankshaft and, conversely, from the crankshaft to the traveling wheels can be transmitted only in one direction, owing to the presence of ratchet clutches in the mechanism, on the section from point 8 to point 12, due to the positive work represented by area \(B\), the speed of the crank disk will begin to increase and will exceed a certain mean speed of its rotation. In view of the clearance between the teeth of the bevel gear, and also because of the presence of the ratchet clutches, this increase in speed will not be communicated to the auxiliary and main shafts of the machine.

Thus, the crank disk, increasing its rotational speed on the section under consideration, will accumulate kinetic energy.

This accumulation of kinetic energy by the crank disk will continue up to point 12, after which this kinetic energy will be expended in performing the work of resistance lying beyond point 12. In order to determine the section on which the crank disk expends the accumulated excess of kinetic energy and again acquires the mean rotational speed, it is necessary to separate, from the area lying beyond point 8, an area \(B'\) equal to area \(B\). In Fig. 9 this area \(B'\) is hatched with horizontal lines, just as area \(B\) is. It should be noted that, in these arguments, the forward speed of the machine and the rotational speed of the traveling wheels have been assumed constant, which will not quite correspond to reality, since, while passing through the section under consideration, the gear wheel of the main shaft will not overcome any resistance, as a result of which the speed of motion of the machine and the rotational speed of the traveling wheels may increase somewhat. This increase in the rotational speed of the traveling wheels will lead to the teeth of the bevel gear of the auxiliary shaft catching up, by the amount of the clearance, with the teeth of the bevel gear of the crankshaft that have moved away from them, before the crank disk has completely expended the accumulated excess of kinetic energy. On the diagram (Fig. 9) this will be reflected in the fact that there will not be, between area \(B'\) and \(C\), so sharp a boundary as is given by the vertical ordinate \(ab\). This boundary will appear in the form of a certain inclined \(nm\), drawn (approximately) on the diagram with a dashed line.

Let us now proceed to the construction of the diagram of tangential forces reduced to the rims of the traveling wheels. For this it is first of all necessary to determine the length of the arc which the rim of the traveling wheel will pass through in one revolution of the crank. This arc length will be:

\[ S_0=\frac{2\pi R}{i}=\frac{0.3,14\cdot0.4}{26,51}=0,095\ m. \]

Choosing the previous length scale \(m_s = 0.001 \dfrac{m}{mm}\), we lay off this straightened arc in Fig. 10 in the form of the segment \(0—24\).

Next, having divided this segment into 24 parts, at each division we shall set up the values of the tangential forces reduced to the wheel rims. For this, the tangential forces acting on the crank pin (the diagram of Fig. 9) must be increased by the factor by which the peripheral speed of the rim of the traveling wheel is less than the peripheral speed of the crank pin, and divided by the efficiency of the two gear transmissions. This conversion factor, which we denote by \(q\), will be equal to

\[ q=\frac{V_0}{V_1 \eta_1 \eta_2}, \]

where \(V_0\) is the peripheral speed of the crank pin, \(V_1\) is the peripheral speed of the wheel rim, \(\eta_1\) is the efficiency of the bevel gear transmission, and \(\eta_2\) is the efficiency of the cylindrical gear transmission.

\[ V_1 = V(1-\psi_0)=1.1(1-0.06)=1.04\ \frac{m}{sec} \]

\[ \eta_1 = 0.9 \]

\[ \eta_2 = 0.8. \]

Then we obtain

\[ q=\frac{2.58}{1.04\cdot 0.9\cdot 0.8}=3.07. \]

Thus, the tangential force reduced to the rims of the traveling wheels will be

\[ H=T\cdot q, \]

where \(T\) is the tangential force at the crank pin.

In order not to obtain excessively large ordinates on the diagram, we choose a new force scale \(m'_p = 13.75 \dfrac{kg}{mm}\). Then, to obtain the ordinates of the diagram of tangential forces reduced to the wheel rims, the ordinates of the diagram in Fig. 9 must be multiplied by

\[ \frac{q\cdot m_p}{m_{1p}}=\frac{3.07\cdot 2.24}{13.75}=0.5, \]

i.e., the ordinates of the diagram (Fig. 9) must be reduced by half in order to obtain the ordinates of the diagram in Fig. 10. The scale of the latter diagram will be, as adopted above, \(m'_p = 13.7 \dfrac{kg}{mm}\).

In the diagram of Fig. 10 the areas denoted in the diagram (Fig. 6) by \(B\) and \(B'\) will disappear entirely, since the work represented by these areas, as indicated above, will go only toward changing the live force of the crank disk, and will not be transmitted to the wheel rims.

is transmitted. The character of the obtained diagram of tangential forces, reduced to the rims of the running wheels, indicates sharp fluctuations of these forces, and also their intermittency; these forces act over comparatively short sections, with two such sections occurring during one revolution of the crankshaft. The indicated nonuniformity of these forces will cause either corresponding fluctuations in the magnitude of the driving force \(P_1\) (\(P_1\) is part of the tractive force, as was noted at the beginning of the article), or fluctuations in the speed of motion of the machine.

If we assume that \(P = \mathrm{Const}\) (which is theoretically possible), then the magnitude \(P_1\) is determined as the mean force from the diagram in Fig. 10. This mean force is shown in the diagram in the form of a horizontal dashed line. The ordinates of all points of this line are equal to one another, and their value, as is known, is determined by the quotient of the division of the area of the diagram by the length of the developed arc, i.e.

\[ ab=\frac{\Omega_0}{S_0}, \]

where \(\Omega_0\) is the area shaded in the diagram of Fig. 10.

\[ ab=\frac{304}{35}=3.2\ mm \]

and the mean force

\[ P_{1\mathrm{cp}}=ab\cdot m'_p=3.2\cdot 13.75=44\ kg. \]

Changes in the live force of the machine, and consequently also fluctuations of speed (under the condition \(P_1=P_{1\mathrm{cp}}=\mathrm{Const}\)), can be calculated from the magnitude of the excess and deficiency of work, by means of the usual method applied in analogous cases.

The analysis of the working process of the hay mower given in this article may be applied, with slight changes, in the study of the cutting mechanisms of other harvesting machines, both horse-drawn and tractor-driven.

The theoretical elucidation of the question touched upon in the article, without claiming great accuracy, may prove (in the author’s opinion) useful in evaluating existing harvesting machines and in designing new ones.

Additional data on the work expended in cutting grass.

At the beginning of the article it was noted that, owing to the absence of direct data, in calculating the work expended in cutting grass (for one stroke of the knife), we were compelled to proceed from the power expended on the work of the knife, taken from experimental data of mower tests (see the calculation of the value \(A_1\) in equation—6).

Recently I have succeeded, using a specially made apparatus, in carrying out preliminary experiments to determine the specific work of cutting certain meadow grasses.

The indicated instrument is made, in general, in the same way as the pendulum instrument of engineer A. A. Vasilenko (see the journal S. Kh. Mashina for 1931, No. 9). The difference consists chiefly only in the fact that, instead of one cutting segment, the instrument made by me has two of them, each of which, in its arrangement, serves as the mirror image of the other. This achieves balancing of the transverse forces, which makes it possible to obtain very accurate proper contact of the segments (during cutting) with the two, correspondingly arranged, opposing finger inserts.

Of course, with such a design, the value of the work expended, obtained in the experiment, must be divided by two.

I give data, obtained by preliminary experiments, on the specific work of cutting two varieties of meadow grasses.

Experiment I. Red meadow clover. Density of the herbage: 15 stems per square decimeter. Mean diameter of the stems (at a height of 4 cm from the root): 2.7 mm.

The specific work (the work expended on cutting one square meter of the given vegetation) was determined to be equal to 15 kg m/m².

Experiment II. Grass of a marshy meadow, dense and bushy. Density of the herbage: 120 stems per square decimeter. Mean diameter of the stems: 1.5 mm.

The specific work was determined to be equal to 20 kg m/m².

In carrying out similar experiments in the future, it will be necessary to determine the influence on the specific work of the cutting speed, the sharpness of the segments, and similar factors.

Let us now try to compute that specific work which, in hidden form, appeared at the beginning of the article, proceeding from the value \(A_1 = 1.14\ kg\ m\) computed at the beginning of the article (see formula 6 and above).

This work \(A_1\), as was indicated, is equal to the work of cutting for one stroke of the knife. In one stroke of the knife, a horse mower cuts grass from an area:

\[ \Omega = Z \cdot e = 1.3 \cdot 0.05 = 0.065\ m^2, \]

where \(Z\) is the swath; \(l\) is the forward advance of the machine during one stroke of the knife.

Then the work expended on cutting grass from one square meter (the specific work) will be:

\[ A_0 = \frac{A_1 \cdot 1}{0.065} = 17.5\ kg\ m/m^2. \]

Thus, we see that the figures adopted at the beginning of the article were indeed average values and corresponded to average working conditions.

In all probability, it should be expected that, in more broadly organized experiments, grasses will be found that give considerably larger values of the specific work.

SUMMARY.

A method for determining the stresses acting in the cutting mechanism of a mowing machine, as well as a method for determining the power expended in accordance with the given characteristics of the grass cover, is offered in this article.

Fig. 1 shows the process of cutting accomplished in the knife mechanism of the mowing machine.

The line AB represents a geometrical locus of points where the points of the blade of the segment meet the corresponding points of the finger-bushing blade. The cutting of the plants according to the principle of shears will occur on this line. Let us term the line AB the line of shearing off.

Dividing this line into equal parts and drawing sinusoids of the paths of points of the segment blade through the points of division, it is possible to define the dimensions of the portions of the field, the grass of which will be cut off on one or another part of the shearing-off line. Assuming that the work spent on the shearing off of the stems is proportional to the areas of the field from which the stems will be shifted to the portions of the shearing-off line, it is possible to ascertain the amount of work performed on every part of the shearing-off line by measuring the mentioned areas.

In order to calculate this work it is necessary to know the value of the specific work of cutting the given grass cover, i.e. the amount of work which is to be spent for cutting the grass on one square unit of the field.

Accessory experiments gave the possibility to ascertain that, in order to cut the grass (it being dense) from 1 square meter of the field, 15 kg/met of work are spent.

Thus, multiplying the areas shown in Fig. 1 by this coefficient of specific work, we get the amount of work spent on each part of the shearing-off line.

In fig. 2 intervals, which during the movement of the knife with regard to the finger bar will correspond to the divisions drawn in fig. 1 on the line of shearing off, are plotted on the line of the course of the knife.

Constructing on each interval rectangles being proportional to the areas of cutting (fig. 1), we get a diagram the area of which will represent the work of shearing off performed by one segment during one course of the knife; and the heights of the rectangles will show the stresses of the shearing off developed upon the knife.

Multiplying these stresses by the number of segments in the mechanism of the knife (or altering respectively the scale of the diagram) we shall get the value of stresses as well as the amount of work evolved upon the whole knife.

Fig. 1.

Fig. 2.

Fig. 3.

Fig. 4.

Fig. 5.

Fig. 6.

Fig. 7.

Fig. 8.

Fig. 9.

Fig. 10.