Prof. S. K. Koniukhov.

Mine Hoisting Using the Koepe System

The hoisting arrangement according to the Koepe system may, in general terms, be described as follows: on the main shaft of the machine a main, or driving, grooved sheave is tightly fitted, and over this sheave is laid the metallic hoisting rope, which embraces the sheave for one half or more than one half of its circumference. The two free ends of the rope are led aside to a certain height and are passed over two guide sheaves. The guide sheaves are arranged either so that their axes lie in one horizontal plane, or so that their middle planes lie in one plane with the middle plane of the driving sheave. The latter arrangement of the guide sheaves is more favorable for the work of the rope, for in this case the angle of deviation of the rope is equal to zero, and thus the possibility of bending of the rope and its rubbing against the projecting edge of the groove is eliminated.

Hoisting cages are suspended from the free ends of the rope, and to the reinforced bottoms of the latter there is attached a balancing lower rope. This whole arrangement is shown schematically in Figure 1, where the middle planes of the sheaves are located in one plane.

A — the main, or driving, sheave
B and B₁ — two guide sheaves
C and C₁ — the loaded and empty cages
9-D-10 — the balancing lower tail rope
7-1-2-5-6-4-3 — the working upper head rope.

Following the generally accepted terminology, let us denote by \(L\) the dead load, or tare (the weight of the cage and the weight of the cars); by \(S\) the weight of the head rope; by \(S_1\) that of the tail rope; and by \(N\) the useful load being hoisted. All weights are taken in kilograms, and lengths and areas in meters and square meters, unless appropriate reservations are made. The hoisting of the load in the Koepe system is accomplished exclusively by means of the friction which arises, under the action of the moving loads, between the rope and the surface of the driving sheave embraced by it.

This dependence of the hoisting of the load on the coefficient of friction between the rope and the surface of the driving sheave is the principal cause of a certain uncertainty as to the reliable working capacity of the entire arrangement.

Hence there arises the first problem of a theoretical character, namely, to furnish a proper justification of this reliability of hoisting by determining the nature of the coefficient of friction and its numerical value.

An exact solution of the problem posed is of enormous interest from the theoretical point of view; but the solution of this problem is of still greater importance for the practice of mining. However, one must not lose sight of the fact that the accomplishment of this task is extremely difficult, and therefore it is necessary to confine oneself to an approximate solution. This is the aim pursued by the present article.

First of all, let us note that, in the operation of a hoist according to the Koepe system, it is necessary to lift only the useful load \(N\), since the weight of the rope and of the dead load in ...

...of the ascending cage are mutually balanced by the same weights at the descending cage. Thus, it is assumed that \(S = S_1\).

If \(R\) denotes the radius of the driving sheave, then the resisting moment during hoisting will remain constant at all times,

\[ M = NR\ \text{kg·m}. \]

In reality, the magnitude of this moment will constantly change during the entire hoist, and if one may speak of the constancy of the resisting moment, it is only within those limits when a constant speed of motion is established, the so-called maximum speed, at which the resisting moment reaches its maximum value; and it is this moment that must draw our attention, since with its aid the power of the machine is calculated. Suppose we have a stationary sheave (Fig. 2), to which, at point \(A\), a rope of weight \(Sa\) is attached. To this rope is attached a cage, the weight of which with the cars will be \(La\), and in the cars lies the useful load \(N\).

Under the described condition, at point \(A\) of the hoisting rope there arises a static stress uniformly distributed over the transverse section of the rope and equal to the sum of the acting weights \(= Sa + La + N\).

If we set the system under consideration in motion, causing the sheave to rotate in the direction of the arrow, then, assuming the acceleration equal to \(p\ \mathrm{m}/\mathrm{sec}^2\), and the acceleration due to gravity \(g = \mathrm{m}/\mathrm{sec}^2\), we obtain the additional load of the rope

\[ \frac{(Sa + La + N)p}{g}\ \text{kg}. \]

But the matter does not end there, since the transition from a state of rest to motion is accompanied by the appearance of resistance to hoisting, which grows with the increase in speed and comprises losses due to friction of the cage guides against the conductors, resistance due to the stiffness of the rope, etc.

The total magnitude of these resistances is expressed differently by different authors.

Professor von Hauer considers this resistance to hoisting constant:

\[ R = 0{,}04\,(N + La + Sa). \]

Professor Grebe believes:

\[ R = 0{,}05\,(N + La + Sa). \]

Von Reiche assumes that

\[ R = 0{,}04\,(N + La + Sa) + 0{,}06\,fV^2, \]

where \(f\) is the area of the cross-section of the cage in the direction perpendicular to the direction of motion, and \(V\) is the cage speed (maximum).

In the last years before the war, Havlíček and Rute were simultaneously engaged in determining the value of \(R\).

According to Havlíček,

\[ R = 0{,}012\,(N + La + Sa) + 4fv^{1{,}275}, \]

where \(f\) and \(v\) have the same meanings as in Reiche.

According to Rute,

\[ R = 0{,}015\,N + 0{,}60\,f(w^2 + v^2),\ \text{if } w < v, \]

\[ R = 0{,}015\,N + 1{,}20\,fwv,\ \text{if } w \ge v. \]

Here \(w\) is the velocity of the ventilation current in the hoisting shaft of the mine, and \(v\) is the speed of travel of the cages at the place where they meet.

It is not difficult to see even from this brief sample that the final conclusions will differ from one another, depending on which formula for \(R\) is adopted. However, the character of the general conclusions will nevertheless retain its direction. In order not to complicate the presentation too much, we shall adopt the formula given by von Hauer.

Then the tension of our rope will have the following form:

\[ S=N+S_a+L_a+(N+L_a+S_a)(0.1p+0.04)\ \mathrm{klg}. \]

Here \(g=10\) has been taken instead of \(9.81\).

Leaving, as before, the fastening of the rope at point \(A\) (Fig. 3), we shall pass the rope over the guide sheave \(D\); to the free end we shall suspend an empty cage, i.e., one without useful load, \(N\). Then the load of the rope at point \(B\) will be

\[ S_b+L_b. \]

If the sheave is made to rotate in the direction of the arrow, then additional resistances will again appear; but since the cage is now descending, these additional resistances will retard the motion, as it were reducing the weight of the descending cage.

In such a case the tension of the rope at point \(B\) will be

\[ S_1=S_b+L_b-(S_b+L_b)(0.1p+0.04). \]

At the moment of transition from uniformly retarded motion to motion established with constant speed, an unfavorable moment occurs for the motion of the descending cage.

In order that the tension of the descending end of the rope be maintained, it is necessary that \(S_1 \geq 0\), or

\[ S_b+L_b \geq (S_b+L_b)(0.1p+0.04). \]

Whence \(p=9.6\) (more exactly \(9.78\)).

This value \(p\) also represents the critical acceleration for the Koepe system.

Let us now combine both sheaves, discard the fastening at point \(A\), and suspend below the cages the balancing tail rope \(C\). Then the simplest scheme of Koepe hoisting will be obtained. The role of the fastening of the rope at point \(A\) will in this case be played by the friction arising between the rope and the rim of the sheave under the influence of the load on both ends of the rope.

We may assume that

\[ S=S_1+P. \]

On the other hand, Euler also showed that

\[ S=S_1 e^{\frac{fs}{r}}. \]

Where \(f\) is the coefficient of friction, \(s\) the length of the arc embraced by the rope on the sheave, and \(r\) the radius of the sheave. If the arc of contact is referred to the radius \(r=1\), then we obtain

\[ S=S_1 e^{fa}. \]

Then for the quantity \(P\) we obtain the following expression:

\[ P=S_1\left(e^{fa}-1\right). \]

If we put \(f=0.28\), then we obtain the following numerical values for \(e^{fa}\):

\[ a=0.4\pi \;—\; 0.6\pi \;—\; 0.8\pi \;—\; \pi \;—\; 1.2\pi \;—\; 1.4\pi \;—\; 1.6\pi \;—\; 1.8\pi \;—\; 2\pi \]

\[ e^{fa}=1.42 \;—\; 1.69 \;—\; 2.02 \;—\; 2.41 \;—\; 2.87 \;—\; 3.43 \;—\; 4.09 \;—\; 4.87. \]

We shall not enter into the subtleties of the question under consideration and take into account Renne’s correction in the tension of the approaching end of the rope (the tension is somewhat greater than is obtained by Euler’s formula, owing to the action of centrifugal force); we shall also leave aside the correction of the French engineer Kretz for the slipping of the rope over the sheave. To a certain extent these corrections are included in the assumption \(f \geq 0.3\). Then, for the quantity \(P\), with \(a=\pi\), we obtain the expression

\[ P=1.566\,S_1. \]

Thus, assuming \(L_a=L_b\) and \(S_a=S_b\), we obtain

\[ P=S-S_1=N+(0.1p+0.04)(N+2S_a+2L_a). \]

Hence we shall have

\[ p = 10 \left[ \frac{P - N}{N + 2S_a + 2L_a} - 0.04 \right]. \]

To come closer to the Koepe system, let us turn to Fig. 1. In this scheme we see two additional guide pulleys \(B\) and \(B_1\), the masses of which will receive acceleration from the rope, and whose resistance to motion must also be overcome by the rope. These additional terms will in turn cause a change in the tensions of the ropes, which may be taken into account by the method already known to us.

Let us suppose that the weight of each pulley, reduced to the middle of the rope, is \(Q\). Then the force for accelerating each pulley will be \(Q \frac{p}{g}\), and the resistance to motion will be \(0.04Q\).

Adding the action of these additional agents to the tensions of the ends of the rope that have been found, then for the ascending end of the rope we shall have

\[ S = N + S_a + L_a + (N + L_a + S_a + Q)(0.1p + 0.04). \]

For the descending end of the rope,

\[ S = S_a + L_a - (S_a + L_a + Q)(0.1p + 0.04). \]

The corresponding value of the acceleration \(p\) will receive the following expression:

\[ p = 10 \left[ \frac{P - N}{N + 2S_a + 2L_a + 2Q} - 0.04 \right]. \]

This formula is the simplest for calculating the acceleration for any hoist. In it the unknown is the quantity \(Q\) of the reduced weight of the pulley.

To determine this unknown, one may use the average values of the reduced weights observed in existing installations. The following table also gives the figures of the reduced weights that are of interest to us.

Turning to the last expression for \(p\), it is not difficult to see that the value \(p\), characterizing the lifting capacity of the Koepe pulley, will be greater the more \(P - N\) is developed; and since \(P = S_a + L_a\), this means the greater the weight of the rope and of the dead load in comparison with the useful load \(N\). Then the value \(p\) will be greater the smaller the reduced weight \(Q\).

The first conclusion is confirmed in practice by the fact that the Koepe system is preferably installed in deep shafts.

The second conclusion directly indicates that it is more advantageous to use pulleys of cast iron rather than cast-iron ones, although the difference in these weights is not so significant.

Apparently nothing more can be extracted from the formula for \(p\), but by indirect methods it is possible to obtain other interesting answers as well.

In practice, preference is given to working with sheaves of larger diameter, for here an increase in the working capacity of the sheave is observed. Meanwhile, the magnitude of the sheave diameter does not appear in the formulas for determining the acceleration.

Engineer M. Kaufhold in Essen poses this question in such an unacceptable form for theory that it is impossible to pass over his conclusion in silence.

Here is what he writes on this occasion: “Hier steht Theorie und Praxis in einem gewissen Widerspruch, denn letztere behauptet auf Grund Wahrnehmungen, dass das nicht Fall sei, vielmehr mit Vergrösserung der Scheibe auch eine Steigerung der Leistung verbunden sei...” etc.

In order to justify theory he resorts to the following argument: the grooves between the individual wires of the rope, although filled with lubricant, nevertheless on the working side of the rope one can note that almost every wire is positioned obliquely to the horizontal from the twist in the strand. These separate inclined wires, coming into contact with the soft lining of the sheave (usually wooden or leather), leave—under the influence of the heavy load on the rope—indentations in the lining. And in the end there is obtained something like a toothed engagement. The smaller the curvature of the sheave rim, the greater will be the degree of such small teeth. In order to make these teeth and hollows on the rim deeper, individual bars or strips of the lining are placed with their fibers transverse to the rope, and not along it.

All this is perfectly correct. But no contradiction between theory and practice is observed here.

Let us begin with the fact that there is a well-founded relation between the diameter of the individual wires and the diameter of the sheave. Thus, for ropes of iron wire, for which \(K_z = 6—8\ \mathrm{kg/mm^2}\), \(D = (1000—1200)\delta\), where \(D\) is the sheave diameter in mm, and \(\delta\) is the diameter of the wire. This ratio is minimal and is conditioned by the necessity of subjecting the wire to bending as little as possible.

For steel wires this ratio is taken still larger, \(D = (1200—1400)\delta\), and for wire of special steel \(D = (1400—1600)\delta\).

The deeper the shaft, the thicker the rope taken, and the more the wire will be subjected to bending, on which work is expended. Part of the load will be expended on this work, and, consequently, the force of friction between the sheave and the rope, equal to the coefficient of friction times the load, will be smaller. That is all.

With regard to the coefficient of friction, it must also be said that in new installations it is taken as equal to \(0.15—0.26\), but often it is taken as \(0.3—0.4\).

It is better to keep to the limit \(\mu \le 0.3\).

In this respect, practice could successfully come to the aid of theory by introducing an appropriate correction in the choice of the magnitude of the friction coefficients on the basis of the difference between the accelerations obtained by formula and those observed in reality.

The path to such a joint action of theory and practice may be as follows. Let \(p\) be the acceleration obtained by calculation according to the formula, and \(p_1\) the acceleration observed in an executed installation. Then the ratio \(p_1 : p = \eta\) will give the correction coefficient for the working capacity of the hoist.

Engineer M. Kaufhold gives the following expression for calculating \(p_1\):

\[ p_1 = \eta \cdot 10 \left[ \frac{1.566(S+L)-N}{3.566(S+L+Q)+N} - 0.04 \right] \]

If, on a number of installations, the value \(\eta\) were established, a simple and correct means would be obtained for determining the working capacity of the Koepe sheave in each individual case. At the same time, arbitrariness in the choice of the coefficient of friction, and hence also the assumed resistance to hoisting, would be eliminated.

For one installation the following numbers were found: \(N = 4400\), \(Q = 2500\), \(S = 6200\), and \(L = 7000\). It was established experimentally that \(p_1 = 2.04\), while calculation by the formula gave \(p = 2.28\). Thus,

\[ \eta = 2.04 : 2.28 = 0.875. \]

The calculations may be simplified in the following way.

If \(\mu = 0.3\) and \(g = 9.81 \simeq 10\ \mathrm{mtr}\), then one may put

\[ e^{\mu\alpha} - 1 = x \quad \text{and} \quad e^{\mu\alpha} + 1 = Z . \]

Then we obtain

\[ p = 9.81\left[\frac{x(S + L) - N}{Z(S + L + Q) + N}\right]. \]

Let us construct a diagram of the dependence (Fig. 5) between \(Z - 1\) and \(\mu\), plotting the value of \(\mu\) on the abscissa axis and \(e^{\mu\alpha}\) on the ordinate axis. Then it is possible, without computation, to take from this diagram the values

\[ x = e^{\mu\alpha} - 1 \quad \text{and} \quad Z = e^{\mu\alpha} + 1. \]

Example. Let \(S = 6200\), \(L = 7200\), \(N = 4000\), and \(Q = 2500\).

If we take \(\mu = 0.22\), then, according to the diagram, to this value of \(\mu\) there corresponds

\[ e^{\mu\alpha} = 2. \]

Consequently, \(x = 1\) and \(Z = 3\), while \(p = 1.37\ \mathrm{mtr}\).

Of considerable interest and importance is the calculation of the additional stresses to which the rope is subjected by the action of instantaneous forces. The calculation shows that these stresses also occupy an important place, and that they should, in all fairness, be called not additional but principal stresses.

The following hypotheses are made the basis of such a calculation:

1. The proportionality of the elongation of the rope to the magnitude of the acting force, i.e. Hooke’s law.
2. The conversion of all kinetic energy at the moment of greatest deformation into the potential energy of stretching.
3. The wave theory, etc.

The unsoundness of applying the first hypothesis to the study of rope deformation is proved by the fact that Hooke’s law is applicable to bodies in which small deformations arise, whereas in a rope these deformations are by no means small.

The inapplicability of the second hypothesis to the study of the phenomenon being analyzed may be justified by the fact that, under instantaneous forces or when the rope is stopped, elastic waves appear in it, whose energy is both potential and kinetic.

This is why the stresses calculated by this theory turn out to be larger than those given by the wave theory. The only exceptions are those sections where interference of waves is manifested. Then the wave theory gives excessive stresses.

If Hooke’s law is made the basis of the wave theory, then, as Professor Dinnik has shown, the problem of calculating rope stress can be solved very simply.

Let us apply this method to the calculation of the additional stresses in the rope when the cage is landed on the dogs. We shall consider the tail, or balancing, rope.

If the shaft is of considerable depth, then it may safely be assumed that the work of deformation is concentrated exclusively in the rope.

Let the upper cage land on the dogs with velocity \(V\). Then the upper end of the rope stops in place, while the lower moving end continues to move with the same velocity as that possessed by the moving system up to the ...

the cage onto the catches. As a result, tensile stresses arise in the rope, beginning at the top, and propagate along the rope at the speed of sound.

This is proved in the following manner.

Suppose that a compressive force \(P\) acts on the end of the bar \(AD\) (Fig. 6). The action of this force is transmitted along the face of the bar from layer to layer and, essentially, is nothing other than the motion of \(A\); if this is so, then the condition for applying the laws of dynamics is present. The simplest thing here is to apply the law of momentum, which is formulated as follows: the momentum is equal to the impulse of the force

\[ MV = Pt \]

If the speed of transmission of the compression is denoted by \(k\), then after the time \(t\) the compression is transmitted over the segment \(kt = AB\). From this moment on, the part of the bar \(ABCD\), compressed to the limit by the force \(P\), will move as an unchanging body, all points of which have one velocity \(V\), the determination of which presents no difficulty. In fact, during an infinitely small interval of time \(dt\), the compression is transmitted from the plane \(BC\) to the plane \(ET\). Calling the length \(BE\) \(ds\), we obtain

\[ V=\frac{ds}{dt}. \]

In the left-hand side of the law of momentum only the mass \(M\) remained undetermined. Let us find it.

If the cross-section of the bar is denoted by \(\omega\), then the volume \(ABCD\) will be

\[ \omega \cdot k \cdot t \]

Calling the density of the body of the bar \(\delta\), we obtain its weight

\[ \omega \cdot k \cdot t \delta, \]

and, dividing this weight by the acceleration of gravity \(g\), we find the mass

\[ M=\frac{\omega \cdot k t \delta}{g} \]

For an infinitely small mass contained in the volume \(BCEF\), one may write

\[ dM=\frac{\omega k dt \cdot \delta}{g} \]

By the well-known formula from the strength of materials, for an infinitely small compression \(ds\) one may write the following expression

\[ ds=Vdt=\frac{P \cdot kdt}{E\omega} \]

whence

\[ V=\frac{Pk}{E\omega}. \]

Now having all the data, let us write the law of momentum:

\[ Pdt=\frac{\omega \cdot kdt \delta}{g}\cdot \frac{Pk}{E\omega}; \qquad k=\sqrt{\frac{Eg}{\delta}} . \]

Thus, in due course, an expression was obtained for the velocity of propagation of elastic deformations in solid bodies by the famous Newton, whose imperishable glory is now being relinquished by physicists.

In acoustics, exactly the same expression is given for determining the speed of propagation of sound in rods or wires.

To the article by S. K. Konyukhov: “Mine Hoisting Using the Koepe System.”

Fig. 1.

Visible labels in the diagram: 1, 2, 3, 4, 5, 6, 7, 8, 10; \(A\), \(B\), \(B_1\), \(C\), \(C_1\), \(D\), \(E\), \(N\).

Fig. 2.

Visible label: \(N\).

Fig. 3.

Fig. 4.

Visible labels: \(A\), \(B\), \(N\).

Fig. 5.

Visible axis and scale labels: \(1\), \(1.5\), \(2\), \(2.5\), \(3\), \(3.5\); \(12\), \(16\), \(24\); \(0.1\), \(0.2\), \(0.3\), \(0.4\); \(\mu=\).

Fig. 6.

Visible labels: \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), \(P\).

If we assume that in time \(t\) the elastic deformation spreads over the segment \(AB = kt\), while the part of the bar beyond the section \(BC\) to the right is not yet being deformed, continuing to move with velocity \(V\), then the relative elongation on the segment \(AB\) will be

\[ Vt : kt = V : k . \]

To obtain the stress of the bar in the part \(AB\), it is necessary to multiply the relative elongation by Young’s modulus.

Thus, we shall have

\[ p = EV : k = V \sqrt{\frac{E\delta}{g}} . \]

The force \(F\) stretching the rope, or in general the deforming body, will be

\[ F = p\omega = V\omega \sqrt{\frac{E\delta}{g}} . \]

The formula for stresses gives grounds for the conclusion that the dynamic stress does not depend on the length of the deforming body. It is directly proportional to the velocity and to the square root of the product of Young’s modulus and the density of the body.

Let us apply these conclusions to our hoisting installation using the Koepe system.

Example: Mine depth \(H = 800\) mm, hoisted load \(Q = 1200\) kgv. Steel rope of uniform cross-section \((kz = 15\ \mathrm{kg/mm^2})\). Number of wires in the rope \(n = 36\). Wire diameter \(\delta = 2.5\) mm. Rope diameter \(d = 24\) mm. Weight of one running meter of rope \(= 1.76\) kg. The weight of a rope \(400 + 25\) meters long will be \(750\) kgv. Cross-section \(\omega = \sim 15\) sq. cm.

Young’s modulus for steel \(E = 2{,}000{,}000\ \mathrm{kg/cm^2}\).

The velocity of setting the cage onto the catches, according to data of Professor M. M. Fedorov, is

\[ V = 0.20\text{—}0.30\ \text{meter per second}. \]

Under these conditions, the velocity of the longitudinal wave will be

\[ k = \sqrt{\frac{1000 \cdot 9.81 \cdot 20000}{7 \cdot 7}} = 5048\ \mathrm{mtr}. \]

Stress

\[ p = EV : k = 2000.000 \times 0.25 : 5048 = \sim 60\ \mathrm{kg/cm^2}. \]

Force

\[ F = p \cdot \omega = 60 \times 15 = 900\ \mathrm{kgg}. \]

If, to the dynamic stress of the rope, we add also the static load \(= (1200 + 750) : 15 = 130\), then we obtain the total stress per sq. cm:

\[ p = 60 + 130 = 190\ \mathrm{kgg}. \]

From this calculation it is evident what a great role the velocity of setting the cage onto the catches plays in the stress of the rope. That is why the operator at the hoisting machine of a mine must have great experience in controlling the motion, so as not to cause, even against his will, a catastrophe, or, in the best case, distortion of the cage catches.