Prof. N. S. Lenz

GRAPHICAL CALCULATION

OF MINE

HOISTING MACHINES.

Prof. N. S. Pen.

Graphical Calculation of Mine Hoisting Machines.

PREFACE.

At every mine the hoisting installation is of great importance, and it would seem that there ought to exist precise and definite indications as to the conditions under which one system or another is advantageous.

Investigations in this direction were begun long ago, but until recently they did not go beyond the sphere of purely practical indications, and only in the last decade has mathematical analysis been applied. True, Reiche, in his work Der Dampfmaschinen—Constructeur, as early as 1880 applied mathematical analysis in comparing various types of hoisting installations; but these investigations did not acquire practical significance, on the one hand because the formulae he derived proved too complicated, and on the other because he did not divide the hoisting time into periods of acceleration, full speed, and retardation, which was necessary for determining the individual elements.

The period of grapho-analytical investigation of this question begins in 1902, when almost simultaneously (in May and July of that year) two articles appeared: Behr’s “Winding Plants for Great Depths” ¹)*) and Herrmann’s “Die dynamischen Verhältnisse der Schachtfördermaschinen” ²); moreover, Behr, in addition to mathematical analysis, gives a graphical construction of the operation of the hoisting machine, while Herrmann applies hyperbolic functions to determine the individual elements of hoisting. The weak side of Herrmann’s method consists in the fact that, owing to its complexity, it is applicable only to cylindrical drums. Behr’s method, however, may be called universal, since it is suitable for all possible types of hoisting installations.

In the following years the same question was studied by Habets, Hoffmann, Wells, and others.

Despite such attention to this question, no comparison of the various methods has been made, and the purpose of the present article is to fill this gap and, on the basis of the data of mathematical analysis—

*) An index of the articles is placed at the end.

to carry out a comparison of various systems of hoisting installations for mines with depths of 300 and 600 m. (the choice of this depth is indicated below);

In conclusion I consider it my duty to express my sincere gratitude to Prof. L. L. Tove, my mentor, who has always readily come to my aid with his valuable guidance, and also to Prof. S. Yu. Doborzhinsky, who kindly agreed to review the present work.

Tomsk. May 1, 1911.

Note. The article was delayed in printing owing to the author’s departure in 1913 on a foreign assignment, and until that same time owing to circumstances independent of the author.

CHAPTER I.

General Survey of the Various Types of Hoisting Devices.

§ 1. Cylindrical Drum.

Of all drum systems, the cylindrical drum is the most widely used, of course under the condition that the hoisting depth is not more than 500 m. Of the 70 examples described by Mellin,³) in 59 installations a cylindrical drum is used. Even in Westphalia, where the Koepe system, as well as spiral drums, have become widespread, we see that of the 70 examples given in “Die Entwickelung etc,” B. V,⁴)—25 belong to the Koepe system and 33 to cylindrical drums; moreover, in 6 cases the depth of the shaft exceeds 500 m. This applies all the more to English mines, where the use of cylindrical drums far exceeds the indicated limit. In part this has been reflected among us as well, in the Donets Basin. Here, at the two deepest mines in this district, the Novosmolyaninovka mines of the Novorossiisk Society (the depth of the Northern shaft is 744.9 m, and that of the Southern 745.5 m), a cylindrical drum is installed. I shall not discuss how rational this is, since a special chapter below is devoted to the comparative characterization of the types; I merely record the facts.

The arrangement of the device is shown in Fig. No. 1. In a direct-acting machine, one or two cylindrical drums are mounted on its shaft. The rope from the drum passes over guide pulleys into the shaft, where it is attached to the cage. Taking into account that the length of the rope changes over time, and that for testing its strength it is necessary from time to time to cut it off at the point of attachment to the cage, the advantage of the double drum is obvious.

§ 2. Conical Drum and Bobbin.

Since, with cylindrical drums without a counterweight rope, the fluctuations of the moments during hoisting are very large, at considerable shaft depth conical drums are used with different winding radii at the beginning and end of the hoist, choosing these radii so that the moments in either case are approximately equal. For complete equalization of the moments throughout the entire hoisting time it would be necessary to construct spiraloid drums; but, in view of the complexity of construction and their great weight, they are encountered very rarely, and in what follows we shall not mention them. The hoisting arrangement with conical drums is shown in Fig. 2.

The use of a flat rope makes it possible to replace the conical drum with a bobbin. This name is given to a spiral drum in which the individual turns of the rope are laid one upon another. The bobbin has found its greatest use in Belgian mines, where flat aloe ropes are widely employed. This method of balancing is simple, but it is hardly suitable for great depths.

§ 3. Systems in which the drums are replaced by pulleys.

The methods of equalizing moments considered in § 2 presuppose different winding radii; however, balancing may also be permissible with a constant drum radius in the case of 1) a counterweight rope, 2) replacement of the drum by a single pulley (the Koepe system), and 3) replacement of the drum by two pulleys (the Craven and Whiting systems).

The first case represents a very simple method of balancing, but, owing to the considerable increase in the weight of the moving masses, it is in many respects inferior to the Koepe system. The latter, as is known, consists in replacing the drum by the main pulley. The arrangement of the apparatus with a Koepe pulley is shown in Fig. 3, which indicates the installation at Mine No. 3 of the Ekaterininsky Society (Donets Basin). Let us mention that the arrangement of the pulleys one above the other is very advantageous, since the deviation (i.e., the angle of deflection of the rope in the extreme positions) is equal to 0, which helps to increase the service life of the rope. In this system the rope from one of the cages passes over a guide pulley to the main pulley, and from there over a second guide pulley to the other cage. The lower parts of the cages are connected to one another by a counterweight rope. A very important condition is the elimination of possible slipping of the rope along the rim. This condition will be discussed in more detail below; for now we shall only note that the frictional force depends on the coefficient of friction and on the arc of contact of the rope with the pulley. The former can be increased only slightly by choosing the best material for the segments; but if in that case the safety factor proves small, it will be necessary either to increase the dead load or the arc of wrap by using a second pulley. We pass to the Craven system.¹) The arrangement is shown in Fig. 4. The rope is fastened to one of the cages, passes over a guide pulley to the lower part of the main pulley \(a\), from there to the upper part of the auxiliary pulley \(b\), and, after going around the latter, returns to the main pulley and then, over a second guide pulley, into the shaft. The main (driving) pulley has two grooves, and the auxiliary pulley one groove. The pulleys on the headframe are set at an angle to the center line.

For deep mines in America and, in particular, in the gold mines of South Africa, the Whiting system is used. It is distinguished by the presence of several grooves on the driving pulleys, in this case connected by a common shaft. The arrangement is shown in Fig. 5. As in the Craven system, the rope from the cage, over a guide pulley, enters the lower part of the front driving pulley and, after making 3 or 4 turns on both pulleys, passes to the rear pulley mounted on carriage \(a\); from there, over pulley \(b\) and the second guide pulley, to the other cage. The movement of the carriage, allowing the length of the rope to be changed, makes it possible to hoist from different levels. It is carried out by means of a special machine \(c\) (Fig. 5), which tightens or releases rope \(d\), attached to the carriage. The Whiting system is not found in Europe.

§ 4. Systems that have not become widespread.

For completeness of the picture of the various hoisting systems, mention should be made of the Morgan system, which has no practical significance, but nevertheless

no less interesting, since it is based on an entirely new principle. A sketch of the installation is shown in Fig. 6. The engine and the drum are placed on a movable platform, which moves perpendicular to the direction of the rope. By means of a gear transmission the platform is displaced, for one revolution of the drum, by an amount equal to the thickness of the rope. The deviation of the rope is therefore always equal to 0. The Williams mine, at the Dalcoath mine in Cornwall, had a depth of 914.4 m, the weight of the cage and cars was 6750 kg, and the hoisting time was 1′42″. The drum diameter was 10′ (3.048 m) and its width 21′ (6.4 m). The engine was direct-acting, with a cylinder diameter of 24″ (0.609 m) and a piston stroke of 60″ (1.53 m), and a steam pressure of 140 lb/in². The valve gear was link-and-slide-valve. The steam pipe was made movable, with semimetallic packings, and in practice proved entirely adequate for its purpose.

One may also mention another arrangement for reducing the deviation of the rope, used exclusively for deep shafts—namely the Wallace system, the diagram of which is shown in Fig. 7. Here the hoisting engine is placed at the shaft itself, but the ropes do not go directly onto the drum; instead, from the guide sheaves they pass onto auxiliary sheaves aa, located behind the engine house, and from there are wound onto the drum. The ropes, both to the guide sheaves and to the drum, are supported by guide rollers. The advantages of this system include: the proximity of the engine room to the shaft, the possibility of seeing the cage, and also the small deviation of the rope; its disadvantages are: the great length of rope, the increased wear of it due to bending, and, in comparison with the Whiting system, the large weight of the drum.

Finally, let us point to the case of the use of pneumatic hoisting, invented by Blanchet, which took place at Épinac (France). The device was intended for the Hottinguer mine, for hoisting coal from a depth of 1000 m; however, workable seams were not found below a depth of 650 m, so that pneumatic hoisting could not be used on the full scale, although experiments showed the possibility of its practical realization. True, the cost of the installation is enormous, and the operating expenses, including amortization, do not bear comparison with ordinary types of installations. The diagram of this system is shown in Fig. 8, where one drawing shows the cage at the bottom of the shaft, and the other—at the surface.

The cage moves in a tube 1.6 m in diameter and with walls 8 mm thick. The piston, whose area = 2 m², consists of two parts, one above the cage and the other below it. The first part is divided into 2 compartments, arranged at such a distance from one another that, during the maneuvers of the cage, one of the compartments is always inseparably connected with the main tube; in this way constant pressure is ensured when the piston passes through the doors.

The lower part contains the parachute P. Let us indicate the scheme of operation: when the air above the piston is rarefied, the latter begins to rise. At this time the doors ff, serving for the unloading of the cars, are closed.

Upon reaching the upper pit bank, the motion is stopped as a result of the following conditions:

1) when the cage reaches point K, door C, connecting the main tube with the exhausting machine, closes automatically,

2) valve $e$ is raised and admits into the pipe air at atmospheric pressure, which is in chamber $S$; if, however, the hoisting still continues, valve $m$ is raised and the main pipe is connected with the atmosphere. To avoid the cage striking the rod $b$, which serves to raise valve $e$, the cage is provided with buffer $a$. The maneuvers are carried out by means of cocks $c$ and $i$, of which the former is installed on pipe $A$ and connects the main pipe with the atmosphere, while the latter, installed on pipe $B$, serves to connect the main pipe with the rarefying machine.

The cage has 3 double doors $ff$, so that all 9 cars can be unloaded in three movements of the cage. When the piston stops opposite $S_3$, cars 1, 4, and 7 are unloaded, and at $S$ cars 3, 6, and 9 can be rolled out.

At the bottom of the shaft there is a pipe $E$, running from the lower part of the main pipe parallel to the latter and ending in such a way that during all maneuvers the upper piston remains below the point at which pipe $E$ is attached to the main pipe. When cock $G$ in this pipe is closed, the air pressure on the piston holds the cage against the upper stops. When the cock is opened and both doors $h$ and $e$, leading to the workings, are simultaneously closed, the air under the piston is rarefied, and the cage descends.

For the main lowering of the cage there is a device analogous to that for hoisting, namely: when the cage reaches point $n$, located near the pit bottom, door $e$, which serves for the escape of air, closes automatically, and the cage gently comes to rest on the stops.

This apparatus also serves for the ventilation of the mine. During the lowering of the cage, door $h$ closes and $e$ opens, and all the air contained in the pipe passes into the workings; during hoisting, door $e$ is closed and $h$ is open, so that during this time an amount of air equal to the volume of the pipe is removed from the mine.

Below are given data taken from Habet’s course.^6) Dead load 4125 kg, useful load 5400 kg. Friction (5% of 9525 kg) 475 kg, total 10000 kg.

Since the cross-section of the pipe is 2 m², a degree of rarefaction of 0.5 atm is sufficient for hoisting; the hoisting speed is 6.38 m/sec.

When lowering the cage, the weight of the latter must be reduced by the friction; i.e., the force with which the lowering is accomplished is equal to $4125 - 475 = 3650$ kg, and since the pressure on the piston is 20000 kg, the required rarefaction is

\[ = 1 - \frac{20000 - 3650}{20000} \sim 0.2 \text{ atm.} \]

CHAPTER II.

Preliminary Calculations.

§ 5. General Conditions.

The operation of a hoisting machine is directly connected with the selected type of drum, since the hoisting time ultimately depends on the magnitude of the inertial forces of the moving masses. In fact, the dimensions of the machine are determined by the magnitude of the static moment; in other words, other conditions being equal,*) by the magnitude of the radius

*) For example, if one compares the Koepe system and a cylindrical drum with a counterbalancing rope.

drum; the system of the hoisting device selected, when determining the machine, plays no role. We observe something entirely different when the machine has already overcome the static moment, i.e., under dynamic conditions. Here the choice of drum system is of substantial importance, since, in addition to the work of hoisting, the machine must overcome the forces of inertia of the moving masses. The additional stresses in the rope, which likewise are of considerable importance when comparing hoists, must also be taken into account; but since they can be found only after determining the dimensions of the machine, they are not taken into account in the present system. The separation of the preliminary calculations into a special chapter is due to the circumstance that we do not wish to obscure, by superfluous calculations, the picture of the machine’s operating conditions which must be made the basis for comparing the various types of drums. Such a comparison can be made only under identical conditions, i.e., when the greatest output of the mines and the maximum hoisting speed are respectively equal. We have taken the former as \(=75\ t\) tons per hour. If, for example, we assume continuous hoisting operation for 12 hours*) and the number of working days per year as 250, we obtain an annual output of 22500 tons.

Wishing to indicate the influence of depth, in the subsequent calculations we selected depths of 300 and 600 m; the first figure approximates the average depth of the deeper mines of the Donets Basin,*) while the second figure is taken for comparison.

We shall adopt the following notation: \(H\)—depth of the mine (in meters), \(Q\)—weight of the useful load; \(P\)—weight of the cage and cars, \(G\)—weight of the rope; \(\gamma\)—weight of one running meter of rope (weights are given in kg); \(\delta\)—wire diameter, \(R\)—radius of a cylindrical drum, and also the greatest radius of winding in the case of a conical drum; \(r\)—smallest radius of winding; \(bd\)—width of the drum. \(J_1\)—moment of the inertial forces of one drum; \(J_2\)—the same for one sheave; \(J_3\)—the same for the loads; \(J\)—total moment of inertia.

Time of hoisting per day according to data for the Saarbrücken district.

Depth of mines	Time of hoisting per day	Number of mines
299–200 mt	from 20 to 15 hours	12
299–200 mt	„ 15–10 hours	4
299–200 mt	< 10 hours	5
666–400 mt	from 20 to 15 hours	16
666–400 mt	„ 15–10 hours	2
666–400 mt	< 10 hours	4
399–300 mt	from 20 to 15 hours	3
399–300 mt	„ 15–10 hours	2
399–300 mt	< 10 hours	7

Prof. Thime recommends taking the average hoisting time per day as from 12 to 15 hours.

A. Mine depth 300 m.

§ 6. Determination of the live and dead load; calculation of the rope.

To determine the amount of live load raised at one time, it is necessary to know the total hoisting time; it consists of the hoisting time in the strict sense of the word and the duration of the maneuvers.

The former, obviously, is equal to \(\frac{H}{V}\), where \(V\) is the average hoisting speed.

According to the data of Prof. Time ⁸).

For \(H = 100—200—300—400—500—1000\),

\[ V = \frac{H}{30} — \frac{H}{40} — \frac{H}{50} — \frac{H}{60} — \frac{H}{70} — \frac{H}{100}. \]

For our case the average hoisting speed \(= 6\) m/sec., and the time of the actual hoist is 50 sec. In determining the greatest hourly productivity of the mine, one must take the least time required for unloading and loading all decks of the cage. Prof. Skochinsky*)⁹ takes this time to be 30 sec. for each deck of the cage. With a two-deck cage the total hoisting time is \(50 + 2.30 = 110\) sec., and the quantity of load raised in one hoist is

\[ \frac{75000.110}{60.60} \sim 2300 \text{ kg.} \]

The capacity of a mine car on most mines is from 500 to 550 kg; taking the latter figure, with 4 mine cars we obtain \(Q = 2200\) kg.

The dead load consists of the weight of the cage and the mine cars. The former may be taken as equal to 3000 kg, on the basis of data from the mines of Germany, ** and the latter depends on the ratio of the dead load of the mine cars to their capacity. According to the data of Prof. Skochinsky ⁹)***, for mines of the Donets basin this ratio is from 0.4 to 0.6, so that the weight of a mine car may be taken as \(= 300\) kg; consequently,

\[ P = 3000 + 4.300 = 4200 \text{ kg.} \]

Weight of cages for 4 mine cars.

Mine	Weight	Mine	Weight	Mine	Weight	Mine	Weight	Mine	Weight
Dechen I.	3500	Reden IV	2830	Veniezsch.	2460	Pr. Regent	3250	Kaiserst. II.	3658
” II.	3500	Sälzer & .	2770	Prosper I	2412	Fridrich .	3200	D. Kaiser I.	3400
Eisenbahn	3400	Centrum I .	2700	Josephasch	2400	K. Ludwig I.	3020	Schürlank	3250
Ensdorfer.	3400	Gr. Moltke I	2700	Alstaden II.	2200	Hibernia I.	3000	Hugo III.	3150
Heinitz II.	3150	Mellin I.	2650	Constantin I	2100	Dahlbusch V.	3000	New-Coln	3100
” IV.	3150	” II.	2650	Schlägel &	2080	Zollern .	2980	Centrum II.	3000
Geisbeck.	3150	Itzcape I.	2600	Hannibal.	2000	Mathilde.	2900	Königsb. II.	3000
Hagenbeck	3060	Dorstfeld II.	2561	Nordstern I.	2000	Itzenpl III.	2600	Osterfeld.	2739
Germania .	2982	Reden II.	2556	Skalky .	1950	Annelung I.	2505	Erin . .	2602
Kaisert .	2875	Bommerb.	2550			Brefeld, . .	2500	Zollver III.	2320

The calculation of the rope must be based on two assumptions: 1) the temporary tensile strength and 2) the factor of safety. The first depends on the choice of material for the rope and, for mild steel, which withstands bending well, may be taken as equal to 12000 kg per cm.\(^2\)

The governmental instruction on supervision of private mining industry *10) requires that the load when iron ropes are used should not exceed 240 poods, and for steel ropes 520 poods, per 1 sq. dm., which corresponds to a stress of 610 kg per cm.\(^2\) for iron and 1320 kg/cm.\(^2\) for steel.

As for the factor of safety, it ranges from 6 to 8. In our calculations we take a factor of safety of 7, as Behr, Robeson, and Cook do; 11) although Hrabak *12) advises taking 7.5.

The weight of one running meter of rope is determined by Hrabak’s formula

\[ \gamma=\frac{Q+P}{\lambda-H}. \tag{1} \]

where \(\lambda\) is the fictitious length of a rope similar to the one sought, which is obtained by replacing the entire load by the weight of the rope, i.e. \(\lambda\gamma=qs\), where \(q\) is the cross-section of the rope, and \(s\) is the safe stress.

For ordinary ropes of 6 wires per strand, we have \(\lambda=0.97q\) and \(d=1.7\sqrt{\gamma}\) cm; for rather flexible ones (8 wires per strand) \(\lambda=0.93q\), \(d=1.77\sqrt{\gamma}\); for flexible ones (9 wires per strand) \(\lambda=0.99q\), \(d=1.81\sqrt{\gamma}\). Hence it is evident that \(\gamma\) is almost equal to \(q\), which is also true for flat ropes (wire ropes).

From the condition \(\gamma\lambda=qs\); with \(\gamma=q\), we have \(\lambda=s\); i.e. the fictitious length of the rope can be replaced by its safe stress.

Obviously, \(\gamma\lambda=Q+P+\gamma H\), whence formula (1) is obtained.

It is true that in an exact calculation one must also take into account the bending of the rope, by which the breaking stress is increased; but this circumstance has been omitted by the author for the sake of simplification.

Since the temporary tensile strength in our example is

\[ \lambda=s=\frac{12000}{7}=1715\ \text{m};\qquad \gamma=\frac{Q+P}{\lambda-H}=\frac{6400}{1715}=4.53\ \text{kg/m}, \]

then the diameter of the rope is

\[ d=1.715\sqrt{4.53}=3.72\ \text{cm}. \]

We take a rope by Felten & Guilleaume with a diameter of 38 mm, wire diameter 2.2 mm, weight per running meter 4.8 kg, and breaking load 60660 kg.

The weight of the working rope is about 1500 kg.

Determination of Twisting Moments and Inertial Forces.

§ 7. Cylindrical Drum.

\[ D=4.5\ \text{m}. \]

As is known, the diameter of the drum depends on the thickness of the wire, and an increase in the power of the machine with a larger diameter is compensated in part by an increase in the service life of the rope. Hrabak **12) says: “the …”

the larger the diameter, the better.” For a flat rope he proposes using the formula

\[ R > 700\,\delta . \tag{2} \]

In all other cases,

\[ R > 800\text{--}900\,\delta . \tag{3} \]

We take the drum diameter \(D = 4.5\) m.

Let us consider the change in loads during hoisting, and for this use Deschamps’s method\({}^{13}\), namely: on an arbitrary straight line (Fig. 9) we lay off, to the chosen scale, the depth of the mine, with point \(D\) corresponding to the level of the lower mine yard, and \(D_1\) to the upper. On the perpendicular \(DA\) we lay off the loads: \(CD\)—the useful load, \(CB\)—the dead load, and \(AB\)—the weight of the rope.

The load diagram for the descending rope is obtained by rotating the diagram \(AB_1C_1C\) through \(180^\circ\) and referring it to the origin \(D_1D\), so that \(AC = D_1L_1\) and \(B_1C_1 = DL\).

The load diagram for the ascending rope is represented by the area \(ABD_1D\), and for the descending rope by \(L_1LDD\); the resultant diagram is evidently equal to \(ABL_1L\), or, when referred to the origin \(DD_1\), is represented in the form \(FF_1D_1D\).

For calculating the machine it is necessary to know the moments: 1) at the beginning of hoisting; 2) at the end of hoisting; 3) from half the weight of the cage; 4) from friction in the shaft.

These moments are most simply determined graphically as follows: on the line \(F_2H_2\), drawn from \(D_1\) at a distance of \(1\) m (in the chosen scale), we lay off the magnitudes of the loads and draw rays to point \(D_1\), extending them to their intersection with the straight line \(F_3H_3\), passing from \(D_1B_1\) at a distance equal to the radius of the drum; then the segments of the line \(F_3H_3\) will give the required moments.

In the following Table 1 their numerical values are given.

TABLE 1.

	Loads, kgf	Moments, kgf·m
At the beginning of hoisting . . . . . .	\(DF = 3700\) kgf	\(H_3F_3 = 8350\)
At the end of hoisting . . . . . .	\(F_1D_1 = 700\) kgf	\(H_3K_2 = 1570\)
Half the weight of the cage . . . . . .	\(\dfrac{BC}{2} = 700\) kgf	\(H_3G_3 = 4730\)
Friction in the shaft . . . . . . . .	\(0.05(Q + 2P + G) = 600\)	\(H_3E_3 = 1350\)

To determine the forces of inertia, we note that

\[ J = 2J_1 + 2J_2 \frac{R^2}{R_1^2} + J_3 . \]

An exact calculation of the moments of the inertia forces of the drums and sheaves can be made in each particular case; here, however, we shall limit ourselves—

we shall confine ourselves to approximate calculations, assuming that the arm of the inertia forces is \(= R \sqrt{\dfrac{\rho}{2}}\); a similar assumption was made by Guters⁶.

Hughes[^15] advises, for determining the masses, taking \(2/3\) of the weight of the drums and sheaves. The weight of one drum will probably be \(\sim 12000\) kg, and the weight of a sheave of diameter 5 m \(=4000\) kg;** then

\[ J_1=\frac{12000\cdot 2{,}25^2}{2\cdot 9{,}81}=3030;\quad J_2=\frac{4000\cdot 2{,}25^2}{2\cdot 9{,}81}=1280; \]

\[ J_3=\frac{(Q+2P+2q)R^2}{9{,}81} =\frac{13600\cdot 2{,}25^2}{9{,}81}=6870. \]

The total moment of inertia is

\[ J=2J_1+2J_2\frac{R^2}{R_1^2}+J_3=14960. \tag{4} \]

Such is the moment of the inertia forces during the period of acceleration; the moment of inertia of the masses moving with retardation differs somewhat from the above-indicated moment of inertia, namely, the moment of inertia of the sheaves \(J_2\) is here multiplied by the ratio \(\dfrac{R}{R_1}\), whereas in the increment of live forces the multiplier is the square of the velocity.

Thus, during retardation,

\[ J_0=2J_1+2J_2\frac{R}{R_1}+J_3=15230. \tag{5} \]

§ 8. Koepe sheave of 6 m diameter.

With the same output and depth of the mine, we took a larger diameter, being guided by the following considerations:

1. A larger drum diameter ensures less wear of the wooden segments, which is of no small importance.

Cook⁹ says: Spengel, in his article on electric hoists, indicates the smallest diameter of a Koepe sheave as 16 ft (4.88 m). The use of such sheaves would lead to a considerable improvement of the Koepe hoist. Apparently, the reason for using a large diameter of the Koepe sheave is that, with small sheaves, the wooden segments are subjected to such rapid wear that proper operation is not sufficiently assured.

In objecting to him, Spengel[^16] points out that in Germany large diameters are usually used for drums as well, and the reason for using large sheaves is hardly to be found in this.

1. The second consideration, expressed in “Die Entwickelung etc.”[^4], seems to us more substantial. As investigations have shown, the best results with Koepe sheaves, with respect to the load on the rope and its cost per ton-km, were given by ropes twisted according to Albert’s method, so that cross-lay is now no longer used with the Koepe sheave in Westphalia.

The table below gives examples of installations existing in Germany.

Name of mines	Year of installation	Depth, m	Number of cars	Diameter: Koepe sheave	Diameter: guide sheave	Deviation	Rope: diam.	Rope: wire thickness	Lay*
Hibernia II . . .	1855	610	4	5,4	3,8	1°45′	40	2,3	K
” I . . .	1855	520	4	5,4	3,8	1°45′	40	2,3	K
Prosper I . . .	1862	382	4	6,0	3,0	0°	45	2,4	A
Dahlbusch . . .	1869—97	882	2	5,0	3,84	1°24′	33	2,6	—
Westhausen . . .	1872	413	2	7,0	4,7	?	40	2,5	A
Consolid I . . .	1873	434	6	7,0	4,75	1°52′	50	2,3	K
” .	—	—	—	—	—	—	—	2,9	A
Concordia II . . .	1874	300	4	7,0	5,0	0°	42	2,0	—
”	—	—	—	—	—	--	—	2,8	—
Prosper II . . .	1874 · 88	300	8	6,0	5,0	34′	50	2,9	A
Centrum II . . .	1874—95	316	4	6,5	3,76	—	40	2,1–2,4	A
Consolid II . . .	1885	540	8	8,17	4,75	0°	50	2,9	A
Zollver I . . .	1885	274	4	8,06	3,00	1°57′	44	2.2	A
Hansa II . . .	1889	660	8	8,0	5,0	50′	55	2,5	A
Ewald II . . .	1891	500	6	7,5	5,0	1°30′	50	2,9	A
Mont—Cenis II .	1896	300	6	7,0	6 0	42′	44	2,5	A
Christ. Lev . . .	1897	430**)	8	8,0	5,0	0°	50	3,0	A
Neu Essen II . . .	1897	435	8	8,0	5,0	0°	50	3,0	A
Schl & Ess II . .	1897	490	4	8,0	5,0	33′	44	2,5	—

The use of this method of lay leads to an increase in the thickness of the wire, and, with it, of the diameter of the rope; this circumstance prompted the author to take the diameter of the Koepe sheave as 6 m.

*) K—cross lay; A—lay according to Albert’s method.
**) According to Makovsky’s data, “An Outline of Ore Deposits, etc.,” the depth of the mine is now 800 m.

The determination of the twisting moments is carried out analogously to the preceding case. Since the rope is counterbalanced and, consequently, it is not necessary to construct the load diagram, the construction is simplified. In Fig. 10 the line \(A H_1\) is drawn at a distance of 1 m (to scale), and \(H_2 A_2\) at a distance of the radius (3 m).

Table 2 gives the numerical values of the loads and twisting moments that we need to know in determining the dimensions of the machine,

TABLE 2.

	Loads, kg	Moments, kg·m
Hoisting of the useful load . . . . . .	\(H_1 A_1 = 2200\)	\(H_2 A_2 = 6600\)
Half the weight of the cage . . . . . .	\(H_1 G_1 = 2100\)	\(H_2 D_2 = 6300\)
Friction in the shaft*) . . . . . .	\(H_1 K_1 = 680\)	\(H_2 K_2 = 2040\)

According to Ehrlich’s data \(^{17}\), a Koepe pulley of this diameter weighs 12000; consequently

\[ J_1 = \frac{12000 \cdot 9}{2 \cdot 9.81} = 5500. \]

With the former pulleys of 5 m diameter, \(J_2 = 1280\); and

\[ 2 J_2 \frac{R}{R_1} = 3700; \]

\[ J_3 = \frac{(Q + 2P + 2G)}{9.81} = \frac{13600 \cdot 9}{9.81} = 12500. \]

Consequently,

\[ J = 5500 + 3700 + 12500 = 21700. \]

The moment of the inertia forces during retardation is

\[ J_0 = J_1 + 2J_2 \frac{R}{R_1} + J_3 = 21070. \]

§ 9. Cylindrical drum of 6 m diameter.

To illustrate the influence of the drum diameter on the dimensions of the machine, we have analyzed a case analogous to the first, but the drum diameter has been taken equal to the diameter of the Koepe pulley.

In this case the service life is increased and the deviation is decreased, since the latter is determined by the width of the drum. The magnitude of the deviation affects the greater or lesser rubbing of the separate coils against one another. Without dwelling further on this question, since it is developed in detail in Milkovsky’s work \(^{18}\), we shall only point out that the harmful influence of friction can be avoided to a certain extent by sheathing the stays on the surface of the drum.

With a cylindrical drum of 4.5 m diameter, the number of turns of the active rope is

\[ \frac{300}{\pi D} = 21.2. \]

*) It is \(= 0.05(Q + 2P + G) = 680\); the increase is due to the weight of the other rope.

With a rope diameter of 38 mm, the width of the working part of the drum is \(21.2 \cdot 0.038 = 0.805\) m; and, taking the distance between the centers of the sheaves and the drum as 40 m, we find that the fleet angle is \(1^\circ 12'\).

With five reserve turns of rope on the drum, we obtain its width (not counting the rim of the brake sheave—\(26.2 \cdot 0.038\)) \(\sim 1.0\) m.

With the same drum diameter of 6 m, the width of the latter is \(= 0.8\) m, and the fleet angle is \(0^\circ 51'\).

In each particular case the question must be decided as to the extent to which the reduction in expenditures on ropes is compensated by the increase in depreciation charges for the larger machine, as well as by the cost of the entire installation.

The determination of the twisting moments, carried out graphically in Fig. 11, in essence differs in no way from the first example.

Table 3 gives the corresponding loads and moments.

TABLE 3.

	Loads, kg	Moments, kg·m
At the beginning of hoisting	\(H_1F_1 = 3700\)	\(H_2F_2 = 11100\)
At the end of hoisting	\(H_1K_1 = 700\)	\(H_2K_2 = 2100\)
Half the weight of the cage	\(H_1G_1 = 2100\)	\(H_2G_2 = 6300\)
Friction in the shaft	\(H_1E_1 = 600\)	\(H_2E_2 = 1800\)

The weight of one drum will probably be about 15,000 kg.

\[ J_1 = \frac{15000 \cdot 9}{2 \cdot 9.81} = 6900. \]

The sheaves remain the same as in the first two cases, i.e.

\[ 2J_2 \frac{R^2}{R_1^2} = 3700. \]

Consequently,

\[ J = 13800 + 3700 + 12500 = 30000. \]

The moment of inertia in deceleration is

\[ J_0 = 2J_1 + 2J_2 \frac{R}{R_1} + J_3 = 29370. \]

Before proceeding to further calculations, let us point out that conical drums and bobbins will be considered for shafts of great depth.

Shaft depth 600 m.

§ 10. Determination of useful and dead loads; calculation of the rope.

For ease of comparison we shall retain the former capacity. Although the depth of the shaft has doubled, we can confine ourselves to a 3-deck cage, since the average hoisting speed will be greater. Taking it as 10 m/sec, we have a hoisting time of 60 sec, for

maneuvering and unloading, 90 sec. will be spent, so that the total time between two hoists is 150 sec.

\[ \text{Amount of useful load per hoist } \frac{75000 \cdot 150}{60 \cdot 50}=\sim 3200 \text{ kg.} \]

We take, as before, the capacity of the cars to be 550 kg; then

\[ Q=6 \cdot 550=3300 \text{ kg.} \]

The weight of the cage may be taken as 3800 kg.^81)

Weight of 3-deck cages.

Mine	Weight	Mine	Weight	Mine	Weight	Mine	Weight
Hannover II	4800	Eisenbahn II	4000	Camphausen I	3500	Hannover I	5200
Eintracht	4534	Frohl. Morgensonne	4000	Camphausen II	3500	Victor I	4900
Centrum III	4500	Marie	3770	Brefeld II	3500	Hannover III	4600
Ko’nigsgrube	4350	Frieda	3770	W. Victoria I	3500	Recklinghausen	3300
Gr. Beust.	4200	Wilhelm III	3500	Hermine	3355	V. d. Heysdt.	3220
Eisenbahn I	4000			Skalley III	3200

The average weight of the cages listed below is somewhat higher than that taken by us; this depends both on the fact that cages are for the most part built without exact calculation, and on the material and construction of the cage. With good material it is entirely possible to adopt the weight = 3800 kg, dead load.

\[ P=3800+6.300=5600 \text{ kg.} \]

We take the rope of the same material as for \(H=300\) m, i.e. with a temporary tensile strength of 12000 kg/cm. With a factor of safety of 7, the weight per running meter is

\[ \gamma=\frac{Q+P}{\lambda-H}=\frac{8900}{1115}=7.98. \]

Rope diameter

\[ d_2=1.75\sqrt{\gamma}=4.95 \text{ cm.} \]

We select a rope from the Felten & Guilleaume works, 50 mm in diameter, with 133 wires, wire diameter 3.1 mm, and weight per running meter 8.2 kg.

\[ \text{Weight of the entire rope}=8.2 \cdot 600=\sim 5000 \text{ kg.} \]

Determination of twisting moments and forces of inertia.

§ 11. Cylindrical drum 6.5 m in diameter, with rope of constant cross-section.

Guided by the considerations set forth in § 7, we chose

\[ R=3.25 \text{ m.} \]

In Fig. 12 a load diagram for this case is drawn by the method of Deschamps, and the magnitudes of the twisting moments are also determined, enabling us to compile the following Table 4.

TABLE 4.

	Loads, kg	Moments, kg·m
At the beginning of hoisting . . . . . . . . .	$H_1F^1 = 8330$	$H_2F^2 = 27000$
At the end of hoisting . . . . . . . . .	$H_1K^1 = 1700$	$H_2K^2 = 5440$
Half the weight of the cage . . . . . . . . .	$H_1G^1 = 2800$	$H_2G^2 = 9100$
Friction in the shaft . . . . . . . . .	$H_1E^1 = 980$	$H_2E^2 = 3100$

To determine the weight of the drum it is necessary to know its width; the latter, excluding the brake rim, is equal to $\left(\dfrac{600}{\pi D} + 5\right)dr = 1.67$ m.

Its weight may be taken as equal to 25000 kg,^{32} and the moments of the inertia forces

\[ J_1 = \frac{25000\,3.25^2}{2 \cdot 9.81} = 13500. \]

The sheaves, 6 m in diameter, will probably weigh about 6000 kg; then

\[ J = 2J_1 + 2J_2 \frac{R^2}{R_1^2} + \frac{(Q + 2P + 2G)R^2}{9.81} = 60000. \]

The moment of inertia of these forces during retardation

\[ J_0 = 2J_1 + 2J_2 \frac{R}{R_1} + J_3 = 59500. \]

§ 12. Cylindrical drum 6.5 m in diameter, with a rope of variable cross-section.

Let us suppose that the cross-section of the rope changes every 200 m; obviously, the weights of these separate parts of the rope will not be identical, but will be proportional to the weight of one running meter. The latter is determined by Hraba’k’s formula

\[ \gamma = \frac{Q + P}{\gamma - H}. \]

For the section 200 m from the shaft mouth, $Q + P = 8900$, $\gamma - H = 1515$; therefore $\gamma' = 5.87$ kg; the diameter of this part is $dr' = 1.75\sqrt{\gamma} = 42.5$ mm, and the weight $G' = 5.87 \cdot 200 = 1174$ kg. In the same way we find that $\gamma'' = 6.65$ kg, $dr'' = 45.2$ mm, $G'' = 1330$ kg; $\gamma''' = 7.48$ kg, $dr''' = 47.3$ mm, $G''' = 1496$ kg, so that the total weight $G = 4000$ kg, i.e. 1000 kg less than in the preceding example.

^{32} Hoffmann. Z. V. J. 1904, 1348, § 15.

The load diagram and the determination of the torques are presented in Fig. 13*).
The results obtained are presented in Table 5.

TABLE 5.

	Loads, kg	Moments, kg·m
At the beginning of hoisting	$H_2F_2 = 7300$	$H_1F_1 = 23800$
At the end of hoisting	$H_2K_2 = 700$	$H_1K_1 = 2300$
Half the weight of the cage	$H_2G_2 = 2800$	$H_1G_1 = 9100$
Friction in the shaft	$H_2E_2 = 1100$	$H_1E_1 = 3600$

With the same drums and sheaves we find that $I = 57800$.
And the moment of the inertia forces during retardation is $I_0 = 57300$.

§ 13. Cylindrical drum 6.5 m in diameter with a counterbalancing rope.

Owing to the equality of the loads throughout the entire hoisting period, the construction of the load diagram is unnecessary, so that in Fig. 14 only the determination of the moments is useful, making it possible to compile the following Table 6.

TABLE 6.

	Loads, kg	Moments, kg·m
Hoisting of the useful load	$H_1A_1 = 3300$	$H_2A_2 = 10700$
Half the weight of the cage	$H_1G_1 = 2800$	$H_2G_2 = 9100$
Friction in the shaft	$H_1E_1 = 1200$	$H_2E_2 = 3900$

The moments of the inertia forces will be greater than in both examples considered, namely:
$I = 27000 + 6500 + 31800 = 65300$; during retardation,
$I_0 = 27000 + 6000 + 31800 = 64800$.

It should be noted that the integral line of the rope weight in the present case will be broken, though differing very little from a straight line, and therefore, without detriment to accuracy, in constructing the diagram of Fig. 13 we assume that the change in weight takes place according to the law of a straight line.

§ 14. Koepe Pulley of 8 m Diameter

For the reasons set forth in § 8, we selected the diameter of the Koepe pulley as 8 m.

The twisting moments for this case are determined graphically in Fig. 15; the results of the determination are given in the following Table 7.

TABLE 7.

	Loads, kg	Moments, kg · m
Hoisting of useful load	$H_1 A^0 = 3300$	$H_2 A_2 = 13200$
Half the weight of the cage	$H_1 G_1 = 2800$	$H_2 G_2 = 11200$
Friction in the shaft	$H_1 E_1 = 1200$	$H_2 F_2 = 4800$

The weight of the main pulley of diameter 8 m will probably be 20,000 kg,¹⁷) so that $I = 66050$, and the moment of the inertia forces during retardation $I_0 = 63650$.

Let us note, incidentally, that the pulley diameter we have selected is in full agreement with existing installations. The table of installations given in § 8 convinces us that a diameter of 8 m is common practice for ropes of diameter 45–50 mm.*)

§ 15. Conical Drum

For conical drums and reels, Hrabák¹⁸) advises taking

$r \geq 700 \delta$.

Leaving the same rope, with wire diameter 3.1 mm, we have $r \geq 2.17$ m. We take $r = 2.25$ m.

The greatest winding radius is determined from the condition of equality of the moments at the beginning and at the end of the hoist. These moments will be, respectively,

\[ M_1 = (Q + P + G)r - PR; \]

\[ M_2 = (Q + P)R - (P + G)r, \]

whence, for $M_1 = M_2$, we obtain

\[ R = \frac{r(Q + 2P + 2G)}{Q + 2P}. \tag{6} \]

For the given case,

\[ R = \frac{2.25 \cdot 24500}{14500} \sim 3.8\ \text{m}. \]

We take $R = 3.75$ m.

*) The mines of the Donets Basin are not an exception (the Schmidt shafts and Kapitalnaya Ekat. O-va).

In view of the fact that the moments of the loads during hoisting change not only in proportion to the change in the loads themselves (as in the cases considered previously), but also in proportion to the radii of the drum, the graphical determination of these moments leads to more complicated constructions.

The most accurate is the method of Dechamps⁴, which makes it possible to carry out the construction on a very large scale.

For all its merits, it has two serious shortcomings which compel one to prefer Behr’s method⁸ (all the more since accuracy here does not play such a role). These shortcomings consist: 1) in the mass of auxiliary constructions; 2) in the impossibility of combining the curves of the static moments with the curves of the inertia forces, since the drawing becomes too confused and it is difficult to make out the individual curves.

Proof of this is furnished by the diagrams obtained by Habets and placed on Plate I, Figs. 1 and 4.⁷ The work is also slowed by the transfer of the curves for the descending rope, which in Behr’s method is eliminated by the use of a second pole.

For comparison of these methods we have constructed in Fig. 16 a curve of the moments on the drum by Dechamps’ method, and in Fig. 17 the same curve by Behr’s method. Without dwelling on the first method, especially since there is a translation of this article in the Mining Journal for 1903, we shall set forth Behr’s method with the explanations given by Wells.[^21]

With a constant winding radius we considered the change of moments as a function of distance. Here, however, we shall have to consider their change as a function of the number of revolutions. The mean diameter of the drum in the present case is \(3.75 + 2.25 = 6\) m, and the circumference is 18.85 m; consequently, the number of revolutions made during the hoist is

\[ \frac{600}{18.85} = 31.8. \]

In Fig. 17 the line \(a_1 a_2\) represents this number of revolutions on the scale \(1\) revolution \(= 1\) mm; from \(a_1\) we lay off the length \(a_1 c_2 = r\), and from \(a_2\) the length \(a_2 b_1 = R\), on an arbitrary scale, and continue the line until it intersects the base at the point \(O_1\); then \(O_1 C_2\) will represent the generator of the additional cone, the presence of which makes it possible to shorten the constructions very considerably. The vertex of the cone is distant from \(a_1\) by 47.7 revolutions. The length of rope wound on the additional cone is obviously \(\pi D \cdot 47.7 = 337\) m, and its weight is \(337 \cdot 8.2 = 2760\) kg. Behr and Wells call the rope wound on the full cone “ideal.” In the present case its weight is \(5000 + 2760 = 7760\) kg, which we shall denote by \(G_1\).

From the point \(a_2\), corresponding to the end of hoisting, lay off the moment

\[ a_2 f = R(Q + P + G_1), \]

in the present case \(= 62500\) kg·m, and draw the straight line \(o_1 f\).

If the load remained constant, and only the radius of the drum changed, then the moments would be determined by the segments of straight lines between \(O_1 a_2\) and \(o_1 f\); in reality, however, these moments must be diminished by the magnitude of the moment of the rope wound on the cone. The weight of the latter is proportional to \(\rho^2\); hence the moment will be proportional to \(\rho^3\), so that the ends of the ordinates will lie on a cubic parabola. To construct the latter, lay off from \(a_2\) the moment of the ideal rope

\[ = R G_1 = 29100 \]

kg·m.

We connect the point \(d\) thus obtained with the pole \(O_1\), and project the intersections of the line \(de\) with the selected divisions onto \(a_1 f\); in this way the points \(e_0', e_1', e_2', e_3', e_4'\) are obtained. On \(a_3 d\), as on a diameter, we construct a semicircle and, from the center \(a_3\), cut off arcs of radius \(a_3 e_0'\), \(a_3 e_1'\), up to their intersection with the circle at the points \(E_0, E_1, E_2, E_3, E_4\), which we then project onto the diameter, thereby obtaining the points \(E_0', E_1', E_2', E_3', E_4'\). Now drawing rays to the pole \(O_1\), we obtain the drum points \(d_1, d_2, d_3\ldots\). The ordinates between this parabola and the line \(O_1 f\) give the moments for the loaded cage, while the moments of the rope on the drum are measured by the ordinates between the parabola and the line \(O_1 d\).

Instead of transferring the ordinates \(d_0 h_0, d_1 h_1\), etc., to the base \(fR\), the curve for the loaded cage can be found independently of the parabola, namely: from the points \(E_0', E_1', E_2'\ldots\) we draw vectors to the pole \(O_3\); the intersections of these vectors with the verticals passing through the points \(k_1, 1, 2\ldots\) will give the required curve \(\beta_0 \beta_1 \beta_2\ldots\).

The diagram of moments for the empty cage is found in a similar manner, by drawing vectors from the points \(g, I, II\ldots\). The point \(g\) lies below the point \(n\), corresponding to \(E_0\), on the other drum, by the amount of the moment from the useful load \(Q_1 R = 12400\); the points \(I, II, III, IV\) and \(\gamma\) are determined by parallels from the points \(E_1', E_2'\ldots\). The intersections of the vectors with the corresponding verticals give the required curve of moments \(\gamma_0, \gamma_1, \gamma_2\ldots\) for the empty cage. The ordinates between these curves, transferred to the origin \(kf\), give the resultant curve \(AB\), corresponding to the curve \(AB\), Fig. 16.

We now proceed to the determination of the moments of the inertia forces.

Let us note that the moment of the inertia forces of the drum is constant throughout the whole time of hoisting, while the moment of these forces from the load and the rope is found by multiplying the ordinates obtained from the preceding drawing by the quantity

\[ \frac{\rho}{9.81}, \]

where \(\rho\) is the radius of an arbitrary section.

Graphically these moments are determined in Fig. 18. From the point \(k\) we lay off the segment

\[ kB_1=\frac{r^2(Q+P+G)}{9.81}=7200, \]

and from the point \(f\) the segment

\[ fB=\frac{R^2(P+Q)}{9.81}=18700. \]

The intermediate points are found, by means of the pole \(O_5\), by a construction clear from the drawing. The curve \(BB_1\) thus found gives the moments of the inertia forces for the loaded cage. In an analogous manner we find the curve \(AA_1\), using the pole \(O_6\), for the empty cage, where

\[ Af=\frac{r^2(P+G)}{9.81}=5500, \qquad A_1 f=\frac{R^2 P}{9.81}=8000. \]

The resultant curve \(CC_1\) is found by summing the ordinates, so that \(BC_1=5500\) and \(CB_1=8000\).

To the moments of inertia of the loads it is necessary to add the magnitude of the inertia forces of the ropes remaining on the drum. In the presence of an additional cone it is easy to see that the length of the ideal rope for an arbitrary radius \(\rho\) is, obviously, proportional to the surface of the cone, i.e. to \(\rho^2\), while the moment of inertia of the cone surface is proportional to \(\rho^4\).

The construction of the parabola of the fourth degree is facilitated by the presence of the parabola of the third degree \(d_0 d_1 d_2\ldots\). For this purpose we lay off

\[ a_1 m_1=\frac{R_2 G_1}{2\cdot 9.81}=5500 \]

and determine the curve \(m_1 m_1'\), using the pole \(O_1\), and also draw ...

the symmetrical curve \(m_3 m_4'\) for the other rope. The straight line \(m_1' m_2'\) will be offset from the line \(a_1 a_2\) by the amount

\[ \frac{(G_1-G)\, r^2}{2 \cdot 9.81} = 700. \]

The resulting curve \(m_1 m_0 m_2\) gives the value of the moments of the inertia forces of both ropes. Adding the results obtained to the curve \(CC_1\), we find the line \(DD_1\), to which there remains to be added the moment of inertia of the pulleys.

With the same pulleys as in § 11, i.e., with \(D_1 = 2750\), we have: for the large drum radius,

\[ J_1' = J_1 \frac{R_1^2}{R^2} = 4280. \]

For the small radius,

\[ J_2' = J_2 \frac{r^2}{R_1^2} = 1550. \]

Thus the resultant moment of the inertia forces of both pulleys, which we may regard as constant throughout the whole hoisting period, will be \(= 5830\).

The weight of one drum will probably be 3000 kg,^[42] and the arm of inertia 2.5 m; consequently,

\[ J_1 = \frac{30000 \cdot 2.5^2}{9.81} = 20000, \]

and for both drums,

\[ J = 40000. \]

Returning to Fig. 18, we obtain that the magnitude of the moments of the inertia forces at the beginning of the hoist is expressed by the ordinate \(FE\), and at the end of the hoist by the ordinate \(F_1E_1\); the trapezoid \(FEE_1F_1\) makes it possible to determine the moments of inertia at any time, which will be necessary in calculating the machine.

§ 16. Bobbin.

With a flat rope the factor of safety must be taken greater than with a round rope, since the load is not distributed uniformly over its cross-section.

If any strand is subjected to an excessively large load, reaching the elastic limit of the material, then the wires begin to break, so that the stress on the other wires increases. In Westphalia, over 7 years the rupture of flat ropes accounts for 10.75%, and of round ropes 5.2%,^[22] of the total number of ropes. In Silesia, over 8 years, of 78 flat ropes, 6 broke (7.69%), and of 671 round ropes only 9 (1.34%). The service life of a flat rope is 13–15 months, whereas that of a round rope is 18–20 months.

Hrabák^[12] advises taking the factor of safety for a flat rope as \(= 9\).

Taking a rope of 6 strands (as is usual), we obtain the load on 1 strand as

\[ \frac{8900}{6} = 1480 \text{ kg}. \]

The effective length

\[ \lambda = \frac{12000}{9} = 1333 \text{ m}. \]

Let us assume that the rope section changes every 200 m; then for the lower part of the rope we have

\[ \gamma' = \frac{1480}{1133} = 1.31;\quad J' = 1.31 \cdot 6 = 7.86 \text{ kg}. \]

^[42] On the basis of data from various English mines.

The weight of this part of the rope is \(G'=1580\) kg.

For the middle part of the rope,

\[ \gamma'=\frac{1750}{1133}=1.54;\quad J'=9.25\ \text{kg};\quad G'=1860\ \text{kg}. \]

For the upper part:

\[ \gamma''=\frac{2060}{1133}=1.82;\quad J''=11\ \text{kg};\quad G''=2200\ \text{kg}. \]

The total weight of the rope is

\[ G=G'+G''+G'''=5630\ \text{kg}. \]

On the basis of Hrabák’s tables one may take the wire diameter \(\delta=2.5\) mm, so that on the basis of formula (2), § 7, the smallest winding radius is \(=1.75\) m.

From formula (6), § 15, we find

\[ R=\frac{r\left(Q+2P+2G\right)}{Q+2P}=3.1\ \text{m}. \]

The mean diameter of the reel is \(4.85\) m, and the next number of its revolutions during the hoist is

\[ n_1=\frac{600}{\pi\cdot 4.85}=39.4. \]

In Fig. 19, by a construction analogous to that of the preceding section, the vertex of the additional cone \(O_1\) is found, with \(O_1a_1=47.5\) revolutions.

From the point \(O_1\) we set off, in an arbitrary scale, \(O_1d_0\)—the length of the rope wound on the additional cone⁵, and \(d_0d_1\), \(d_1d_2\), and \(d_2d_3\), of which each is equal to the length of rope of constant section, i.e. 200 m. We project \(d_0\) onto the straight line \(a_1c_1\), and \(d_3\) onto the straight line \(a_2b_1\), thus obtaining the points \(d'_0\) and \(d'_3\). Finally, we carry the vector \(O_1d'_0\) over until it meets the straight line \(a_2b_1\) at the point \(d''_0\), and divide the segment \(d''_0d''_3\) into the same number of equal parts as the straight line \(a_1a_2\) (in the present case, into 4 parts).

The intersection of the vectors \(x_1o_1\), \(x_2o_1\), and \(x_3o_1\) with the verticals through the points 1, 2, and 3 will give the sought points \(a_1\), \(a_2\), and \(a_3\), belonging to the parabola. The points where this parabola meets the horizontals through \(d_1\) and \(d_2\) determine the position of the moments at the end of each section and their radii \(\rho_1\) and \(\rho_2\). From the drawing we have that \(\rho_1=2.3\) m and \(\rho_2=2.75\) m.

The construction of the moments for the loaded and empty cages is carried out in Fig. 20, where

\[ \begin{aligned} k\beta_0&=r(Q+P+G'+G''+G''')=25400,\\ l_1\beta_1&=\rho_1(Q+P+G'+G'')=28400,\\ z_2\beta_2&=\rho_2(Q+P+G')=28800,\\ f'\beta_3&=R(Q+P)=27600, \end{aligned} \qquad \begin{aligned} k_0&=RP=17400,\\ l_1\gamma_1&=\rho_1(P+G')=19700,\\ b_2\gamma&=\rho_1(P+G'+G)=20800,\\ f'\gamma_3&=r(P+G'+G''+G''')=19700. \end{aligned} \]

The magnitude of the ordinates between these curves gives the value of the resulting moment. Referring the latter to the origin \(kf\), we obtain the curve \(AB\), with \(kA=8000\) kgm, and \(fB=7900\) kgm.

The moments of the inertia forces are determined in Fig. 21, where

\[ kA=\frac{R^2P}{9.81}=5500,\qquad BC=\frac{R^2P}{9.81}=5500, \]

\[ fA_1=\frac{r^2(P+G)}{9.81}=3500,\qquad BC_1=\frac{r^2(P+G)}{9.81}=3500, \]

\[ kB_1=\frac{r^2(Q+P+G)}{9.81}=4550,\qquad CD-C_1D_1=\frac{R^2G_1}{2\cdot 9.81}-\frac{r^2(G_1-G)}{2\cdot 9.81}=3500, \]

\[ fB=\frac{R^2(P+Q)}{9.81}=8700,\qquad ED=E_1D_1=5830,\quad kF=fF_1=2J_1=15000. \]

Thus, the variation of the moments of the forces of inertia is represented by the trapezoid \(F E E_1 E_1\), to the consideration of which we shall return when determining the work of the machine.

§ 17. Cylindro-conical drums

The transition from cylindrical to conical drums is constituted by cylindro-conical drums. We shall carry out only the calculations necessary for comparing these drums with others. Leaving the smallest winding radius as before, we may take the largest radius equal to \(3.5\) m.

Let us consider 3 cases:

a) the number of rope turns on the conical part is 12 (which corresponds to the period of acceleration);

b) the number of these turns is 18, and finally,

c) the number of turns is 24.

The schematic drawing of the drum for case a) is shown in Fig. 22; for b)—24, and for c)—26.

The analogous drawing for the conical drum considered in § 15 is shown in Fig. 28.

The diagrams of twisting moments, shown in Figs. 23, 25, and 27, are drawn similarly to Fig. 17, and therefore we shall not dwell on the constructions, but shall compile Table 8, on the basis of which this construction has been made.

We shall adopt the following notation:

\(n_1 = a_1 a_1' = a_2 a_2'\)—the number of revolutions of the conical part of the drum; \(n_2 = a_1 a_2 = a_3' a_1\)—the same for the cylindrical part; \(n_3 = O_1 a\)—the number of revolutions of the additional cone; \(S_1 = c_1 b_1' = c_2' b_2'\)—the length of rope on the conical part of the drum, and \(G_1\)—its weight; \(S_2 = b_1 b_1' = b_2' b_2\)—the length of rope on the cylindrical part of the drum, and \(G_2\)—its weight; \(S_3 = n_2 n_3 = O_1 c_1 = O_2 c_2'\)—the length of rope on the additional part of the cone, and \(G_3\)—its weight; \(Ak = m \beta_0\)—the resultant moment at the beginning of hoisting; \(Bf = b_1\)—the same at the end of hoisting.

TABLE 8.

Case	\(n_1\)	\(n_2\)	\(n_3\)	\(S_1\) m	\(G_1\) kg	\(S_2\) m	\(G_2\) kg	\(S_3\) m
a	12	17.4	21.6	217	1780	383	3220	153
b	18	12.4	32.4	326	2680	274	2320	230
c	24	7.55	43.2	434	3560	66	1440	308

Case	\(G_3\), kg	\(a_1 d = R(G_1 + G_3)\), kg	\(d_2 = R(Q + P + G_2)\), kg	\(A k\)	\(B f\)	\(M \max\)	\(M \min\)
a	1260	10600	42800	12000	7000	17000	5000
b	1890	16000	39400	12000	7000	12000	7000
c	2520	21200	36200	12000	7000	12500	7000

The constant quantities, except \(Ak\) and \(Bf\), will be \(R P = mk = 1980\) kgm, \(R Q = mg = 11500\) kgm, \(R(P+Q)=lf=31100\) kgm.

We shall not determine the work of the machine for each case, and therefore the determination of the moments of the forces of inertia has also been omitted.

CHAPTER II.

Operating Conditions and Methods for Calculating Hoisting Machines

§ 18. General remarks.

There is hardly another branch of machine construction in which the methods of calculation would be so varied as in the calculation of mine hoists, excluding electric hoisting machines, where it is easy to take into account all operating conditions, the influence of the inertia forces of the moving parts, and to carry out the calculation with mathematical precision.

Steam engines are far from being in such favorable conditions. The presence of the connecting-rod and crank mechanism forces one to modify the course of the calculation somewhat. The point is that, in calculating machines of this type, a condition must be imposed that one cylinder be capable of overcoming the greatest static moment obtained at the circumference of the drum. But this is not enough: when one of the cranks is in a position corresponding to cut-off, the other may be not far from a dead position (which, however, depends entirely on the degree of cut-off), and consequently the tangential force obtained from the driving crank will be insignificant.

Below the influence of cut-off on the magnitude of this tangential force will be indicated; for the present we shall only note this fact in order to point out that it must be reckoned with and that this condition must be included in the calculation.

Usually, in order to impart motion to a machine in analogous cases (with coupled engines having cranks arranged at \(90^\circ\)), it is sufficient to give a reverse stroke and thereby bring the other crank into operation. With mine hoists the matter is not so simple.

The large diameter of the drum (with the Koepe sheave, even where the depth is not very great, one can often encounter a diameter of 8 m.*) does not always allow a considerable reverse stroke, since even with a small rotation of the drum the level of the cage changes considerably.

It is therefore reasonable to take into account the most unfavorable position of the cranks and, proceeding from this, to determine the dimensions of the cylinders. Rehr¹), Norton²³), Wells²¹), Fruters¹⁴), Dam²⁴), and others arrive at such a conclusion.

Moreover, when calculating the machine for the greatest twisting moment at the beginning, middle, and end of the hoisting operation, it is necessary to take into account also the case when one of the cages is at rest on the keeps, and consequently the weight of the other cage is unbalanced.

In order that the machine may also satisfy this condition, Behr, Wells, and others propose adding to the initial moment the moment due to half the weight of the cage; in other words, the product of half the weight of the cage by the radius of the drum.

* Capital mine ( \(H\) up to 225 m, \(D = 8\) m); Schmidt 3 ( \(H = 165\) m, \(D = 8\) m); Zollverein I ( \(H = 274\) m, \(D = 8.06\) m); Kaiser Friedrich ( \(H = 296\) m, \(D = 8\) m).

§ 19. Calculations proposed by various authors.

Before proceeding to the calculations, let us determine the magnitudes of the twisting moments in the various periods of hoisting.

To the notation adopted in § 5 we shall add the following: \(M_1\)—the moment at the beginning of the hoist; \(M_2\)—the same in the middle of the hoist, \(M_3\)—the same at the end of the hoist, \(M_4\)—the moment of hoisting the upper loaded cage above the mouth of the shaft, when the lower one rests on the keps, \(M_5\)—the moment of hoisting the lower cage when the upper one rests on the keps, \(B\)—the moment due to friction in the shaft.

The twisting moments are expressed by the following formulas of Table 9:

TABLE 9.

Values of moments	Magnitude of moments	Magnitude of moments
Values of moments	Conical drum or bobbin.	Cylindrical drum.
\(M_1\)	\((Q + P + G)r - PR + B\)	\((Q + G)R + B\)
\(M_2\)	\(\left(Q + P + \dfrac{G}{2}\right)r - \left(P + \dfrac{G}{2}\right)R + B\)	\(QR + B\)
\(M_3\)	\((Q + P)R - (P + G)r + B\)	\((Q - G)R + B.\)
\(M_4\)	\((Q + P)R - Gr + B\)	\((Q + P - G)R + B\)
\(M_5\)	\((P + G)r + B\)	\((P + G)R + B.\)

The value \(B\) will be given in the separate calculations presented below.

Habets*) for determining the twisting moments considers 3 cases:

a) with keps requiring preliminary raising of the cage, one must take the greatest of the following moments \(M_1, M_3, M_4\).

He believes that in this case, for a steam engine, it is useless to take into account the dynamic moment during the period of acceleration, since an engine calculated for static conditions will prove quite sufficient to overcome the inertia forces by the work of two cylinders. Such an assertion, however, is not always correct. With large cylindrical drums without a counterbalancing rope, which are widely used in England and are also employed in the mines of continental Europe, the influence of the inertia forces on the hoisting time is very considerable and must be taken into account.

b) with keps not requiring raising of the cage, \(M_{\max}\) is chosen from \(M_1, M_3\).

c) With balanced ropes (Koepe pulley) the static moment is equal to \(M_2\), and maneuvers are often carried out without keps.

In Habets’s formula the moment of friction in the shaft is not included; true,

in his article,²⁴) Ilabeck takes the frictional force to be equal to 15% of the useful load (with the Koepe pulley).

As regards the dimensions of the machine, they are determined, in cases a and b, by the formula

\[ M_{\max}=K(p_0-p_1)\frac{\pi D^2}{4}\, r \ \mathrm{m}, \]

and in case c,

\[ (M_1+C)\,2\tau=2K(p_0-p_1)\frac{\pi D^2}{4}\,4\,r\ \mathrm{m}, \]

where \(r\ \mathrm{m}\) (crank radius) \(=0.75D-D\), so that the piston speed does not exceed \(1.5\text{--}2\ \mathrm{m/sec}\). \(K\) (coefficient of useful action) for large machines is \(0.68\text{--}0.7\), for medium ones \(0.6\text{--}0.65\), for winches \(0.5\text{--}0.55\). \(p_0\) and \(p_1\) are the admission pressure and the back pressure; \(C\) is the dynamic moment.

Hügel²⁵) proposes, when

\[ G>\frac{P}{2}, \]

to carry out the calculation according to the twisting moment \(M_1\), and when

\[ G<\frac{P}{2}, \]

according to \(M_4\).

At the same time he takes the friction in the shaft to be equal to \(0.04(Q+2P+G)\). *)

The dimensions of the machine are determined by the calculation formulae proposed by Hrabák. The latter²⁶) calculates the machine so that it should overcome

a) the initial moment \(M_1\), under the condition that one of the cranks is in the dead position and the other in the most favorable position, **) i.e.

\[ M_1=\bigl[Q+G+0.05(Q+2P+G)\bigr]R \le 10000\,r_n\,0.05\,L, \]

or

\[ M_1 \le 5000\,r_n\,O\,L \]

(filling \(0.8\)).

b) the mean moment

\[ M_2=\bigl[Q+0.05(Q+2P+G)\bigr]R \]

with the most advantageous filling by means of both cylinders, i.e.

\[ \frac{1}{2}M_1 \le 0.5\,L\left(\frac{2.10000}{\pi}\,r_n\,O\right) \]

or

\[ \frac{1}{2}M_2 \le 3183\,r_n\,O\,L \]

(the most advantageous filling).

c) the moment at the end of the hoist (when the loaded cage is raised above the catches and the empty one rests on the catches),

\[ M_4=\bigl[Q+P-G+0.05(Q+P+G)\bigr]R \]

by both cylinders, even at the greatest filling, i.e.

\[ \frac{1}{2} \le 3183\,r_n\,O\,L \]

(filling \(0.8\)).

Hrabák’s point of view is shared by Riche²⁷) and Durham²⁸).

Prof. Time quite rightly proposes taking into account the moment \(M_5\) (see above) and calculating the machine by the larger of the moments \(M_4\) and \(M_5\).

The same requirement for hoisting machines is put forward by Makovsky, pointing out that the machine of the Central Mine of the Novorossiisk Society cannot overcome the moment \(M_5\) with the force of one cylinder. ***

) Prof. Kondrat’ev (Course of Hoisting Machines*, p. 29), carrying out the calculation analogously to Pavésy, proposes for \(M_1\) to increase the friction by the amount

\[ \frac{Av^2}{8}, \]

where \(A\) is the area of the cage, \(v\) its speed.

**) Italics are ours.

***) Incidentally, let us note that Makovsky’s fears are hardly well founded, since the efficiency coefficient \(0.86\) obtained by him (at \(Mk=13725\)) corresponds exactly to the coefficient proposed by Hrabák (Hilfobuch f. D—m T., J. 221), namely \(0.865\) for \(Mk=1000\) and \(0.872\) for \(Mk=20000\).

One cannot pass over in silence the very detailed calculations of Futers’s⁶. It has already been indicated above that he takes into account the most unfavorable position of the cranks, as a result of which the machines obtained are of somewhat larger dimensions than under the usual calculations.

Sometimes a machine calculated on the basis of static moments may prove insufficient for overcoming dynamic forces; in this case it must be calculated in a somewhat different way. By specifying the acceleration time \(t_1\), it is easy to determine the work of the machine during this period, and consequently also its dimensions. Knowing the greatest permissible speed, \(v_1\), we determine the distance \(S_1\) traversed during the acceleration time by the formula \(S_1 = \dfrac{v_1 t_1}{2}\), taking the acceleration to be constant, and also the number of revolutions of the drum during this time

\[ n_1 = \frac{S_1}{2 \pi D_a}. \]

For the most frequently occurring ratio of piston stroke to cylinder diameter \(= 2\), the value of the latter is determined by the formula

\[ d = \sqrt[3]{\frac{F + E\,[Q + G - \gamma S_1 + \mu (Q + 2P + G)]\,S_1}{2 \pi \cdot 10000\, p i \cdot n_1 \cdot \xi}}. \]

Here:

\(F\)—work to overcome the forces of inertia of the drum and sheaves, \(E\)—work to overcome the forces of inertia of the loads, \(\mu\)—coefficient of friction; for rope guides it may be taken as \(=0.02\); for wooden and rail guides Futers proposes taking \(0.025\), \(p i\)—mean indicated pressure, \(\xi\)—coefficient of efficiency.

The introduction of the calculation of a steam hoisting machine on the basis of dynamic forces is also noticeable among German authors. One may note the detailed analytical study of cage motion published by Prof. Herrmann.⁷ Herrmann’s method has a purely theoretical significance and is not applicable to all those cases that may arise in calculations with bobbins or conical drums.

The most convenient and simple is Behr’s method,⁸ which we shall use for comparing various types of drums. However, before setting forth the method itself, it is necessary to derive the basic formula which Behr gives on p. 12 of the cited article:

\[ M_1 = \frac{8 F f}{2 \pi c_1} \tag{7} \]

Here:

\(M_1\)—the mean moment under the action of both cylinders, \(F\)—the initial moment when one of the cranks is in the cut-off position and the other in the most unfavorable position, \(f\)—coefficient of fullness of the indicator diagram, \(c\)—sine of the angle between the crank and the connecting rod.

To derive this formula and determine the numerical values of the expressions entering into it, we must consider the relation between the degree of filling and the tangential forces developed on the circumference of the crank.

§ 20. Relation between the degree of filling and the tangential forces for static conditions.

It is necessary to clarify this relation because the value of the tangential forces developed along the circumference of the crank depends—

reduce the dimensions of the machine. Above, § 18, the considerations were indicated according to which the most unfavorable position of the cranks should be taken into account. This position is not included in the calculation by the authors of continental Europe. It is true that Hauer, ^25) taking tangential forces into account, says that for ordinary machines without compensation the ratio of the piston areas for static and dynamic conditions is less than unity, and therefore it is sufficient to carry out the calculation on the basis of static moments; but

1) Hauer does not take into consideration the work of overcoming the forces of inertia;

2) he takes the admission pressure as equal to 1.2 of the receiver pressure when both cranks are working, which is true only for established motion at the greatest speed. Numerous diagrams obtained by Wallichs and Huffmann over the period of the entire hoist show a gradual fall of pressure, so that during the period of acceleration the mean value of this pressure scarcely corresponds to the figure given by Hauer.

For determining the tangential forces for a single stroke we have adopted the graphical method, as being more visual. As may be seen from Fig. 29, the admission pressure has been taken by us as 6 atm.

For the admission conditions we may assume that the back pressure is equal to 0, all the more so since the pressure drop from the boilers will be smaller at the established speed.

In Fig. 30 the tangential forces are determined by the method of Prof. Ugarov, ^29) the auxiliary semicircle being divided into 20 equal parts. These tangential forces make it possible to construct a force diagram both for one and for the other crank. This construction is made in Fig. 31, where the solid line refers to the leading crank, and the dotted line to the following one.

Let us consider the mutual position of the cranks at the moment of cutoff, which we shall take as equal to 0.8.

The determination of the angles of the cranks with the line of dead centers is made in Fig. 30. Setting off from point \(A\) along the line of dead centers the value \(AA_1 = 0.8\,AB\), and projecting point \(A_1\) onto the main semicircle, we obtain point \(C_2\). The angle of inclination of the crank to the line of dead centers is equal to \(\angle AOC_2\); from the drawing, \(\angle AOC_2 = 122^\circ\).

Obviously, the angle of the second crank is \(122^\circ - 90^\circ = 32^\circ\). In Fig. 33 (case 1) this relation is shown. Here the steam-distribution organs cut off steam into cylinder \(A\), and the motion is performed only by crank \(B\) up to the moment of lead admission of steam into cylinder \(A\). Let us find the tangential force developed by crank \(B\). For this it is sufficient to draw a perpendicular to the line \(OC_2\) and determine its intersection with the auxiliary semicircle. The point of intersection falls midway between 3 and 4. However, one may dispense with this construction, recalling that an arc of \(90^\circ\) (the angular distance between the cranks) corresponds to 10 divisions; and since \(C_2O\) intersects the auxiliary semicircle midway between 13 and 14, it is obvious that the position of crank \(B = 3\frac{1}{2}\) divisions. Taking the corresponding ordinate from the diagram for crank \(B\), Fig. 31, we find that the tangential force is \(T_1 = 3.75\) atm. (kg/cm\(^2\)).

The second critical position will be when crank \(B\) takes the place of crank \(A\) (in the preceding case).

The angle of crank \(A\) with the zero position will then be \(129^\circ + 90^\circ = 212^\circ\), and the corresponding tangential force \(T_2 = 2{,}7\) atm.

To determine the 3rd and 4th critical positions it is necessary to know the angles of the cranks at the moment of cut-off. They are found by a method analogous to the preceding one, with the difference that the quantity \(0{,}8\,AB\) is laid off from point \(B\). Since \(\Gamma C_1O = 131^\circ\), crank \(A\) will obviously be at an angle \(131 + 180 = 311^\circ\), and crank \(B\) at an angle \(311 - 90 = 221^\circ\). The steam in cylinder \(A\) is shut off, and only \(B\) is the driving crank.

In consequence of the fact that \(OC_1\) intersects the auxiliary semicircle between points 34 and 35 (nearer to 35), to determine the tangential force acting on crank \(B\), the ordinate must be taken between 24 and 25; its numerical value is \(T_3 = 3{,}4\) atm.

Finally, for the 4th critical position, Fig. 33, IV, the angle of crank \(A\) will be \(41^\circ\), and crank \(B\)—\(311^\circ\); the tangential force acting on crank \(A\), from Fig. 31, is determined as \(T_8 = 4{,}55\) atm.

From the foregoing it is clear that the most unfavorable position is obtained in the second case, when \(T = 2{,}7\) atm.

The dimensions of a machine calculated to overcome the greatest moment by means of this force would prove very large for ordinary operation; in view of this, the majority of English works equip hoisting machines with a special valve-link steam distribution, making it possible to give a cut-off of up to 0.95. Examples of such steam distribution may be found in the works of Fluters’a, ^14), Hughes’a, ^15) Hurst’a, ^30).

Leaving this question aside as being beyond the scope of the present article, we proceed to the determination of tangential forces when working with both cranks.

§ 21. Determination of tangential forces for dynamic conditions.

We shall apply the same method as in the preceding paragraph, but the forces acting along the connecting rod are taken on the basis of indicator diagrams. In Fig. 29*) diagrams are constructed for the forward and return strokes, and in Fig. 30 the tangential forces for separate positions are determined.

The resultant curve when working with both cranks is constructed in Fig. 32 and makes it possible to determine the mean tangential force for dynamic conditions.

In the following table, 10, are given the numerical values both of the individual ordinates and of their arithmetic mean for each quadrant.

TABLE 10.

First	Second	Third	Fourth
5,75	5,75	5,75	5,75
6,55	6,2	6,75	7,1
7,2	6,5	7,3	8,1
7,75	6,7	7,75	8,65
8,05	6,95	8,0	9,1
8,3	7,0	8,25	9,4
8,0	6,9	8,1	9,2
7,5	6,6	7,8	8,7
7,0	6,25	7,35	8,0
6,45	6,0	6,6	7,15
7,255	6,485	7,365	8,115

*) Figs. 29, 31 and 32 were intended to be printed as clichés and are now omitted.

Therefore, the mean force on both cranks will be

\[ \frac{7.255+6.485+7.365+8.115}{4}=7.305 \text{ atm.} \]

§ 22. Derivation of the fundamental formula.

The least tangential force for static conditions, as is seen from Fig. 31, is equal to 4.8 atm.; and consequently the ratio between the dynamic and static forces, which is obviously equal to the ratio of the corresponding moments, will be

\[ \frac{7.305}{4.8}\simeq 1.5. \]

This ratio may be derived analytically as follows: let \(P\) be the total steam pressure on the piston; \(F\)—the static moment; \(M_1\)—the dynamic moment; \(r\)—the crank radius; \(f\)—the coefficient of fullness of the indicator diagram; \(c\)—a coefficient depending on the degree of cut-off (for \(\frac{S_1}{S}=0.95,\ c=0.8\)).

The work of both cylinders in 1 revolution is equal to \(2P\cdot 4fr=8(Pr)f=8Ff\). The mean turning moment on both cranks \(M_1\) is equal to the mean tangential force multiplied by \(r\), and the work in 1 revolution is \(2\pi M_1\). From the equality of the works we obtain \(8Ff=2\pi M_1 c\), whence

\[ M_1=\frac{8Ff}{2\pi c}. \tag{8} \]

This fundamental formula is also used by Behr in calculations. Substituting the numerical values (\(f=0.95\) and \(c=0.8\)), we have \(M_1=1.51F\). In calculations we take \(M_1=1.5F\).
\[ \tag{9} \]

Derivation of formulas for determining the distance traveled and the time of hoisting.

§ 23. The rope is balanced.

In this case the work of hoisting is constant and is equal to \(QS\), where \(S\) is the distance traveled in the given time. The work to overcome friction in the shaft is also taken by us as constant for all cases, and here it will be \(=0.05(Q+2P+2G)S\).

The sum of these works will give the effective work of the machine, if the work to overcome the forces of inertia is not taken into account. The latter will be expended in imparting kinetic energy to the moving masses, which, as is known, for bodies rotating about a fixed axis is

\[ C=J\frac{\omega^2}{2}, \tag{10} \]

where \(J\)—the polar moment of inertia, \(\omega\)—the angular velocity.

Finally, the work of retardation in the absence of braking (the ideal case) is equal to the work of acceleration. It remains, for completeness, to add also the work of machine friction, and then we have all the data for a graphical representation of the work of the hoisting machine in the various periods of motion.

Let us lay off distances along the axis of abscissae, and moments along the axis of ordinates.

In Fig. 34 \(aa_1\)—the depth of the shaft, \(ob\)—the moment from friction in the shaft \(=0.05(Q+2P+2G)R\), \(bc\)—the moment from hoisting the useful load \(QR\).

\(cd\)—the moment from half the weight of the cage, \(\dfrac{P}{2}R\), \(cf\)—the moment accepted by us for starting, equal to \(\left(\dfrac{P}{2}+\xi\right)R\), where \(\xi\) depends on the static conditions; \(ae\)—the friction moment of the machine (a fictitious quantity) \(=25\%\) of \(af\), so that \(e=80\%\) of \(ef\).

Then the static moment \(F=ef\).

We have derived that the dynamic moment, when both cranks are working, is

\[ M_1=1.5\,F. \]

Laying off the numerical value of \(M_1\) from point \(e\), we obtain point \(h\).

Approximately it may be assumed that the moment \(m_1=ch\) is spent in overcoming the forces of inertia.

To be sure, this is not entirely correct, since the driving forces, and not the moments, must be taken into account; but, since in the diagram we everywhere take moments, the ratio of the works for the different periods of hoisting remains equal to the actual one.

Denoting the moment at the end of hoisting by \(M_2\), we obtain that the additional moment remaining from the work of acceleration at its end is

\[ m_2=M_2-A_1. \]

The moment \(M_2\) is chosen in accordance with the operation of the machine and the required speed. Having \(m_1\) and \(m_2\), we can graphically find the work of the machine during the period of acceleration.

In this there may be three cases: a)—the acceleration decreases gradually (the case of a steam engine, when the cut-off during the period of acceleration is reduced), b)—the acceleration gradually increases in the case of an electric motor, c)—the acceleration is constant.

Case a).

Assuming that the acceleration varies according to the law of a straight line, we observe that the trapezoid \(chkl\), whose area is \(\dfrac{(m_1+m_2)S_1}{2}\), is proportional to the work.

On the other hand, the same area represents the product of the drum radius by the work of acceleration, since in Fig. 34 the ordinates are moments, and not simple forces.

On the basis of (10) we have

\[ \frac{(m_1+m_2)S_1}{2}=CR=RJ\frac{\omega_1^2}{2} \tag{10′} \]

where \(\omega_1\)—the greatest angular velocity.

If \(v_1\)—the greatest hoisting speed, then

\[ \omega_1=\frac{v_1}{R}. \]

Consequently,

\[ \frac{(m_1+m_2)S_1}{2}=\frac{Jv_1^2}{2R}. \tag{11} \]

Whence

\[ S_1=\frac{Jv_1^2}{R(m_1+m_2)}. \tag{12} \]

Let us investigate the dependence between the variables \(m\), \(s\), \(v\), and \(t\). For this purpose we shall replace \(v_1\), \(m_2\), and \(s_1\) by the variable quantities \(v\), \(m\), and \(s\), whereby the depen-

...dependence retain their force.

\[ \frac{(m_1+m)S}{2}=\frac{Jv^2}{2R}. \tag{13} \]

From equation 13) we have

\[ v^2=(m_1+m)S\cdot\frac{R}{J}, \]

or

\[ v=\pm\sqrt{\frac{R}{J}}\sqrt{(m_1+m)S}. \]

Since velocity is the derivative of distance with respect to time, then

\[ \frac{ds}{dt}=\pm\sqrt{\frac{R}{J}}\sqrt{(m_1+m)S}. \tag{14} \]

Let us denote \(\dfrac{m_1-m}{S}\) by \(a\), i.e.,

\[ m=m_1-as. \tag{15} \]

Substituting the value \(m\) from (15) into expression (14), we obtain

\[ \frac{ds}{dt}=\pm\sqrt{\frac{R}{J}}\sqrt{2m_1s-as^2}, \]

whence

\[ dt=\pm\sqrt{\frac{J}{R}}\,\frac{ds}{\sqrt{2m_1s-as^2}}. \]

Integrating, we have

\[ t=\pm\sqrt{\frac{J}{R}}\int_0^s \frac{ds}{\sqrt{2m_1s-as^2}}. \]

The integral

\[ x=\int \frac{ds}{\sqrt{2m_1s-as^2}} \]

is reduced to the form

\[ x_1=-\frac{1}{\sqrt{a}}\arcsin\left(\frac{m_1-as}{m_1}\right)+C_0. \]

For our case, integrating within the limits, we obtain

\[ t=\pm\sqrt{\frac{J}{R}}\left(\int_{s_1}\frac{ds}{\sqrt{2m_1s-as^2}}-\int_0\frac{ds}{\sqrt{2m_1s-as^2}}\right)= \]

\[ =\pm\sqrt{\frac{J}{R}}\left\{-\frac{1}{\sqrt{a}}\arcsin\left(\frac{m_1-as_1}{m_1}\right)+\frac{\pi}{2}\right\}. \]

Taking into account that, by equation (15), \(m_1-as=m_2\), we have

\[ t=\pm\sqrt{\frac{J}{R}}\left(\frac{\pi}{2}-\arcsin\frac{m_2}{m_1}\right). \]

Since \(t_1\) is essentially positive, the plus sign must be taken before the radical quantity; then

\[ t_1=\sqrt{\frac{J}{Ra}}\left(\frac{\pi}{2}-\arcsin\frac{m_2}{m_1}\right). \]

Substituting for \(a\) its value from (15), we obtain

\[ t_1=\sqrt{\frac{J S_1}{R(m_1-m_2)}}\left(\frac{\pi}{2}-\arc \sin \frac{m_2}{m_1}\right). \tag{16} \]

Formula (16) makes it possible to determine the time for given \(R, J, m_1\) and \(m_2\); the path, however, traversed during the time of acceleration is determined from formula (12).

Case b.

Let us use formula (14)

\[ \frac{ds}{dt}=\pm \sqrt{\frac{R}{J}}\sqrt{(m_1+m)s} \tag{17} \]

with the difference that here

\[ m=m_1+as. \]

Substituting in formula (14) the quantity obtained from (17), i.e.

\[ \frac{ds}{dt}=\pm \sqrt{\frac{R}{J}}\sqrt{2m_1s+as^2}, \]

or

\[ dt=\pm \sqrt{\frac{J}{R}}\frac{ds}{\sqrt{2m_1s+as^2}}. \]

Integrating, we obtain

\[ t_1=\pm \sqrt{\frac{J}{R}}\int_0^{S_1}\frac{ds}{\sqrt{2m_1s+as^2}}. \]

Since the coefficient of \(S^2>0\), the integral is reduced to the form

\[ \int \frac{dx}{\sqrt{a+2bx+cx^2}} = \frac{1}{\sqrt{c}}\lg e\left[b+cx+\sqrt{c}\sqrt{a+2bx+cx^2}\right]+C. \]

For our case

\[ \begin{aligned} t_1 &=\pm \sqrt{\frac{J}{R}}\int_0^{S_1}\frac{ds}{\sqrt{2m_1s+as^2}} \\ &=\sqrt{\frac{J}{Ra}}\lg e\left(m_1+as_1+\sqrt{a}\sqrt{2m_1s_1+as_1^2}\right)-f_0. \end{aligned} \tag{18} \]

The integral \(f_0\) is determined from the condition that when \(S_1=0\) also \(t_1=0\), i.e.

\[ f_0=\pm \sqrt{\frac{J}{Ra}}\lg e\, m_1. \]

Substituting the value of \(f_0\) into formula (18), we have

\[ t_1=\pm \sqrt{\frac{J}{Ra}} \left\{ \lg e\left(m_1+as_1+\sqrt{a}\sqrt{2m_1s_1+as_1^2}\right)-\lg e\,m_1 \right\}, \]

or

\[ t_1=\pm \sqrt{\frac{J}{Ra}}\lg e \left( \frac{m_1+as_1+\sqrt{a}\sqrt{2m_1s_1+as_1^2}}{m_1} \right); \tag{18 bis} \]

\[ m_2=m_1+as_1 \tag{17 bis} \]

but consequently

\[ t_1=\pm \sqrt{\frac{J}{Ra}}\lg e \left\{ \frac{m_2+\sqrt{as_1(2m_1+as_1)}}{m_1} \right\}. \]

The expression \(\sqrt{aS_1(2m_1+aS_1)}\) is transformed as follows:

\[ \sqrt{aS_1(m_1+Sa_1+m_1)}=\sqrt{aS_1(m_2+m_1)} =\sqrt{(m_2-m_1)(m_2+m_1)}=\sqrt{m_2^2-m_1^2}. \]

Finally we have

\[ t=\sqrt{\frac{J S_1}{R(m_2-m_1)}}\lg e\left(\frac{m_2}{m_1}+\sqrt{\left(\frac{m_2}{m_1}\right)^2-1}\right). \tag{19} \]

The distance traversed during the time of acceleration is determined from the same formula (12).

Case C.

Finally, with constant acceleration, which occurs in the case when the cutoff remains at the maximum throughout the entire acceleration period, the preceding formulas are considerably simplified.

Here \(m_2=m_1\), therefore

\[ S_1=\frac{Jv_1^2}{R(m_1+m_2)}=\frac{Jv_1^2}{2Rm_1}. \tag{20} \]

On the other hand,

\[ S_1=\frac{t_1 v_1}{2}, \]

whence

\[ t_1=\frac{2S_1}{v_1}. \tag{21} \]

The preceding formulas make it possible to determine the time and distance for the acceleration period.

We now pass to the period of retardation. Returning to Fig. 34, we see that the area \(chkl\) is equal to the area \(c_2e_1e_1l_2=OR=\frac{Jw_1^2}{2}R\).

Obviously,

\[ A_1S_3=RJ_0\frac{w_1^2}{2}=\frac{J_0v_1^2}{2R}, \]

whence

\[ S_3=\frac{J_0v_1^2}{2RA_1}. \tag{22} \]

The time of retardation is

\[ t_3=\frac{2S_3}{v_1}. \tag{23} \]

The distance \(S_2\), traversed at full speed, will be

\[ S_2=H-(S_1+S_3). \tag{24} \]

The time at full speed is

\[ t_2=\frac{S_2}{v_1}. \tag{25} \]

The total hoisting time is

\[ T=t_1+t_2+t_3. \tag{26} \]

§ 24. The rope is not balanced

The work diagram of the machine, constructed analogously to the preceding section, is presented in Fig. 35.

The work of hoisting here is a variable quantity and is determined from the analytical expression

\[ E=\int_0^{S_1}\{Q+\gamma H-2\gamma s\}\,ds, \]

where obviously \(\gamma H=G\).

Solving the integral, we obtain

\[ E = Q S_1 + \gamma S_1 (H - S_1) + C_0 . \]

The constant \(C_0\) is determined from the condition that, when \(S_1 = 0\), \(E_0 = 0\), i.e.

\[ E = Q S_1 + \gamma S_1 (H - S_1) \tag{27} \]

The work of friction in the shaft is equal to

\[ (Q + 2P + G) S_1 . \tag{28} \]

The formulas of the preceding section for determining the distance and time of acceleration remain valid here also, but the expression for the static moment at the end of the acceleration period will be different. Let \(M_2 - m_2 = A_2\), and the moment at the end of the hoist be \(A_4\); then

\[ A_2 = A_4 + (H - S_1)\tg\alpha . \]

Here \(\alpha\) is the angle of inclination between the segments \(cc\) and \(ee\); in this case \(\alpha\) is a fictitious quantity, since the ordinates are expressed in kgm and the abscissas in m; it has been introduced only to facilitate the calculations.

The additional moment at the end of the acceleration period is

\[ m_2 = M_2 - A_2 = M_2 - \{ A_4 + (H - S_1)\tg\alpha \}. \tag{29} \]

On the basis of (10′) we have

\[ S_1 = \frac{2CR}{m_1 + m_2} \tag{30} \]

Solving equations (29) and (30) with respect to \(m_2\), and taking into account that \(A_1 - A_4 = H\tg\alpha\) and \(A_1 + m_1 = M_1\), we obtain

\[ m_2 = -\frac{M_1 - M_2}{2} \pm \sqrt{ \left(\frac{M_1 - M_2}{2}\right)^2 + (M_2 + A_1)m_1 + 2CR\tg\alpha }. \]

Here it is necessary to take the positive value of the root, since the quantity \(m_2\) is in essence \(> 0\), and the line of dynamic moments does not intersect the line of static moments. Consequently, finally we have

\[ m_2 = -\frac{M_1 - M_2}{2} + \sqrt{ \left(\frac{M_1 - M_2}{2}\right)^2 + (M_2 - A_1)m_1 + 2CR\tg\alpha } \tag{31} \]

Equation (31) is very complicated and hardly applicable for calculations; it is much more convenient to use equation (12), substituting various values of \(m_2\) and obtaining the value of \(S_1\). Determining from the drawing the \(A_2\) corresponding to the found \(S_1\), and adding \(m_2\), we obtain the dynamic moment at the end of the acceleration period.

If \(M_2\) proves insufficient for the given conditions, then a new \(m_2\) is specified. Behr says that even the first substitution gives satisfactory results.

The formulas for the distance and time of retardation likewise acquire a new form, since the moment curve here is represented by an inclined line.

Obviously, the area of the trapezoid \(c_2 c_1 e_1 e_2\), Fig. 35, is equal to

\[ \frac{(A_3 + A_4)}{2} = CR = \frac{J_0 v_1^2}{2R}, \]

whence

\[ S_3 = \frac{J_0 v_1^2}{R(A_3 + A_4)} \tag{32} \]

On the other hand, the relationship between \(S_3\) and \(A_3\) and \(A_4\) can be expressed through the tangent of the angle of the lines \(cc_1\) and \(ee_1\), namely,

\[ \tg\alpha = \frac{A_3 - A_4}{S_3} = \frac{ce - c_1e_1}{H} = \beta . \]

\[ A_3 = A_4 + \beta s_3. \tag{33} \]

Substituting the value of \(A_3\) into equation (32), we obtain the equation

\[ \beta s_3^2 + 2 A_4 s_3 - \frac{J_0 v_1^2}{R} = 0; \]

solving it with respect to \(s_3\), we find

\[ s_3 = \frac{ -A_4 \pm \sqrt{A_4^2 + \dfrac{J v_1^2 \beta}{R}} }{\beta}. \]

Here the \(+\) sign must be taken, i.e.

\[ S_3 = \frac{ \sqrt{A_4^2 + J v_1^2 \dfrac{\beta}{R}} - A_4 }{\beta}. \tag{34} \]

To determine the retardation time, let us note that expression (33) is analogous to expression (17 bis) \(m_2 = m_1 + a s_1\), and therefore we can use the corresponding formulas already determined in case b. Only here it will not be possible to use the final formula (19), since the moment \(A_3\) at the beginning of acceleration is unknown to us. The most suitable is formula (18 bis)

\[ t_1 = \sqrt{\frac{J}{R a}}\, \lg e \left( \frac{ m_1 + a s_1 + \sqrt{a}\sqrt{2 m_1 s_1 + a s_1^2} }{m_1} \right), \]

where it is necessary to replace:

\(J\) by \(J_0\), \(a\) by \(\beta\), \(s_1\) by \(s_3\), \(m_1\) by \(A_4\).

Thus, for the retardation time we have the following formula

\[ t_3 = \sqrt{\frac{J_0}{R\beta}}\, \lg e \left\{ \frac{ A_4 + \beta s_3 + \sqrt{2 A_4 \beta s_3 - (\beta s_3)^2} }{A_4} \right\}. \tag{35} \]

§ 25. Conical drum.

The formulas of the preceding two sections are unsuitable for the present case, since the distances traveled here must be expressed as a function of the number of revolutions; moreover, the changes in the static moments are represented by a curved line, both for conical drums and for bobbins, and only with spiraloid drums are the moments equal throughout the entire hoisting time. True, introducing this change of moments into the calculation would make the formulas excessively complicated, and therefore we may take as the basis of the calculations a certain mean moment. The polar moments of inertia during acceleration and retardation are likewise taken as mean values—for the given number of revolutions—and are found directly from the drawing.

The distribution of the work of the hoisting machine is shown in Fig. 36, analogous to Fig. 34.

The work of acceleration here is also proportional to the area of the trapezoid, which is equal to

\[ \frac{(m_1 + m_2)\cdot S_1}{2} = c\rho = \frac{J w_1^2}{2}\rho. \]

The distance traveled during the acceleration time, where \(\rho\) is the mean radius during acceleration.

\[ S_1 = 2 \pi \rho n_1,\quad \text{whence } (m_1 + m_3)\,2\pi \rho n_1 = Jw_1^2 \rho,\quad \text{or } Jw_1^2 = 2\pi n_1(m_1 + m_1). \tag{36} \]

We determine the dependence between \(J_1\), \(n_1\), \(w\), and \(t\) analogously to case \(a\), § 23, with the difference that instead of \(v\) and \(s\) we take \(w\) and \(n\).

We shall write expression (36) in the general form

\[ Jw^2 = 2\pi n(m_1 + m). \tag{37} \]

Similarly to expression (15), we may write

\[ m = m_1 - an. \tag{38} \]

Substituting the value of \(m\) from (38) into expression (37), we have

\[ Jw^2 = 2\pi n(2m_1 - an^2). \]

Whence

\[ w = \pm \sqrt{\frac{2\pi}{J}}\sqrt{(2m_1 n - an^2)}. \tag{39} \]

On the other hand, \(v = \rho w = \dfrac{ds}{dt}\), but \(S = 2\pi \rho n\), therefore

\[ \rho w = \frac{2\pi \rho\, dn}{dt}, \]

whence

\[ w = \frac{2\pi\, dn}{dt}. \tag{40} \]

From expressions (39) and (40), we have

\[ \frac{2\pi\, dn}{dt} = \pm \sqrt{\frac{2\pi}{J}}\sqrt{2m_1 n - an^2}. \tag{41} \]

From equation (41)

\[ dt = \pm \sqrt{2\pi J}\, \frac{dn}{\sqrt{2m_1 n - an^2}}. \tag{42} \]

Integrating equation (42) within the limits from \(0\) to \(n_1\), we obtain

\[ t_1 = \pm \sqrt{2\pi J}\int_0^{n_1} \frac{dn}{\sqrt{2m_1 n - an^2}}. \tag{43} \]

The definite integral

\[ \int_{n_1} \frac{dn}{\sqrt{2m_1 n - an^2}} = -\frac{1}{\sqrt{a}}\arcsin \frac{m_1 - an}{m_1}. \]

For the other limit we have

\[ \int_0 \frac{dn}{\sqrt{2m_1 n - an^2}} = -\frac{1}{\sqrt{a}}\arcsin \frac{m_1}{m_1} = -\frac{1}{\sqrt{a}}\cdot\frac{\pi}{2}. \]

Consequently,

\[ \int_0^{n_1} \frac{dn}{\sqrt{2m_1 n - an^2}} = \int_{n_1} \frac{dn}{\sqrt{2m_1 n - an^2}} - \int_0 \frac{dn}{\sqrt{2m_1 n - an^2}}, \]

\[ \int_0^{n_1} \frac{dn}{\sqrt{2m_1 n - an^2}} = \frac{1}{\sqrt{a}} \left( \frac{\pi}{2} - \arcsin \frac{m_1 - an}{m_1} \right). \tag{44} \]

Substituting the value from equation (44) into equation (43), we have

\[ t_1 = \pm \sqrt{\frac{2\pi J}{a}} \left\{ \frac{\pi}{2} - \arcsin\left(\frac{m_1 - an}{m_1}\right) \right\}. \]

But from equation (38) \(m_1 - an = m_2\), therefore

\[ t_1 = \pm \sqrt{\frac{2\pi J}{a}} \left( \frac{\pi}{2} - \arcsin \frac{m_2}{m_1} \right). \]

Taking the positive sign of the root, we have

\[ t_1 = \sqrt{\frac{2\pi J}{a}} \left( \frac{\pi}{2} - \arcsin \frac{m_2}{m_1} \right). \]

Replacing \(a\) by its value obtained from expression (38), we finally have

\[ t_1=\sqrt{\frac{2\pi J n'_1}{m_1-m_2}}\left(\frac{\pi}{2}-\arcsin\frac{m_3}{m_1}\right), \tag{45} \]

where \(n'_1\) is determined from formula (36)

\[ n'_1=\frac{J w_1^2}{2\pi(m_1+m_2)}. \tag{46} \]

When \(m_2=0\), equation (45) is considerably simplified, namely,

\[ t_1=\frac{\pi}{2}\sqrt{\frac{2\pi J n'_1}{m_1}} =\pi\sqrt{\frac{\pi J n'_1}{2m_1}}; \]

but from expression (36)

\[ m_1=\frac{J w_1^2}{2\pi n'_1}, \]

therefore

\[ t_1=\pi\sqrt{\frac{\pi J n'_1\cdot 2\pi n'_1}{2Jw_1^2}} =\frac{\pi^2 n'_1}{w_1}, \tag{45 bis} \]

where

\[ n'_1=\frac{J w_1^2}{2\pi m_1}. \tag{46 bis} \]

Without the use of the brake, the work of retardation is equal to the work of acceleration, and the area of the rectangle \(c_2 c_1 e_1 e_2\) is equal to the area of the trapezoid \(c h k l\), Fig. 36, i.e.

\[ 2\pi R n'_3 A=\frac{J_0 w_1^2}{2}R, \]

whence

\[ n'_3=\frac{J_0 w_1^2}{4\pi A}. \tag{47} \]

With constant retardation

\[ S_3=\frac{v_1 t_3}{2}=\frac{w_1 p t_3}{2}. \]

On the other hand, \(s_3=2\pi n'_3 p\); consequently,

\[ t_3=\frac{4\pi n'_3}{w_1}. \tag{48} \]

The number of revolutions passed at full speed is

\[ n'_2=N-(n'_1+n'_3), \tag{49} \]

and the time at full speed, where \(N=n_1+n_2\) (see § 17), the number of revolutions during the hoist, is

\[ t_2=\frac{2\pi n'_2}{w_1}. \tag{50} \]

CHAPTER IV.

Operation of the machine with drums of various systems.

A. Mine depth 300 m.

§ 26. Cylindrical drum 4.5 m in diameter.

In view of the fact that the rope here is not counterbalanced, we shall use the formulas of § 24.

First of all, let us determine the static moment \(F\).

As is seen from Fig. 35, it consists of the following moments:
\(ab\)—the friction moment in the shaft; \(bc\)—the moment of the loads; \(cf\)—the additional moment for starting; \(ae\)—the friction moment of the machine.

From Table 1 we have \(ab = 1350;\ bc = 8350\).

The moment of half the cage weight (ibid.) \(d = 4730\), therefore we may take the additional moment \(cf = 5000;\ af = ab + bc + cf = 14700\).

The friction moment of the machine is equal to \(0.25\,af = 3700\). The static moment \(F = ae + af = 18400\). The dynamic moment \(M_1 = 1.5F = 27800\). The moment remaining for acceleration,

\[ m_1 = 27800 - (3700 + 1350 + 8350) = 14400. \]

The moment at the end of acceleration, \(m_2\), may be taken as \(2500\). The polar moment of inertia during acceleration is \(J = 14960\), expression (4). Taking the greatest hoisting speed \(v_1 = 8\ \text{m}\), we determine, by formula (11), the distance traveled during the period of acceleration:

\[ s_1 = \frac{J v_1^2}{R(m_1 + m_2)} = \frac{14960 \cdot 8^2}{2.25(14400 + 2500)} = 25.1\ \text{m}. \]

The acceleration time is determined by formula (16)

\[ t_1 = \sqrt{ \frac{J s_1}{R(m_1 - m_2)} \left( \frac{\pi}{2} - \arcsin \frac{m_2}{m_1} \right) } \]

\[ = \sqrt{ \frac{14960 \cdot 25.1}{2.25(14400 - 2500)} \left( \frac{\pi}{2} - \arcsin \frac{2500}{14400} \right) } = 5.28\ \text{sec}. \]

For the period of retardation, according to eq. (34),

\[ s_3 = \frac{ \sqrt{ A_4^2 + J_0 v_1^2 \frac{\beta}{R} - A_4 } }{\beta}, \]

Here, from Table 1,

\[ A_4 = 3700 + 1350 + 1570 = 6620; \]

\[ \beta = \frac{ce - c_1 c_1}{H} = 22.6 \]

\[ J_0 = 15230. \]

Consequently,

\[ s_3 = \sqrt{ \frac{ 6620^2 + 15230 \cdot 8^2 \frac{22.6}{2.25} - 6620 }{22.6} } = 31\ \text{m}. \]

The retardation time will be found from eq. (35).

\[ t_3 = \sqrt{\frac{J_0}{R\beta}}\, \lg e \left\{ \frac{ A_4 + \beta s_3 + \sqrt{2A_4 \beta s_3 - (\beta s_3)^2} }{A_4} \right\}. \]

\[ t_3 = \sqrt{\frac{15230}{2.25 \cdot 22.6}}\, \lg e \left\{ \frac{ 6620 + 22.6 \cdot 31 + \sqrt{2 \cdot 6620 \cdot 22.6 \cdot 31 - (22.6 \cdot 31)^2} }{6620} \right\} \]

\[ t_3 = 7.6\ \text{sec}. \]

Expression (24) gives the distance traveled at full speed,

\[ s_2 = H - (s_1 + s_3) = 243.9\ \text{m}, \]

in the time

\[ t_2 = \frac{s_2}{v_1} = 30.5\ \text{sec}. \]

The total hoisting time

\[ T = t_1 + t_2 + t_3 = 43.38\ \text{sec}. \]

The operating diagram is shown in Fig. 37.

We have taken here, as in the following examples, the friction of the machine to be constant. In reality, however, it varies with the load of the machine, and this circumstance must be taken into account in detailed calculations. The line of moments \(h k l a c_2 e_2 e_1\) passes into the position \(h k l' c_2^{1} e_2^{1} e_1\).

Considering this friction during starting as 20% of the static moment, Behr says that during the entire hoist it will be, probably, no more than 8%. This question has been the subject of numerous debates, as is evident from the discussions of Sederholm’s, Webb’s, Thurston’s.

Without going into the details of this question, we shall note the following propositions: 1) at the moment of starting, the force of the machine must be sufficient to overcome also the friction; 2) the work of friction decreases with increasing speed, at first rapidly, and at high speeds more slowly; 3) the magnitude of the work of friction for hoisting machines fluctuates within considerable limits (from 10 to 25%), and the first figure still requires careful verification; the most unfavorable conditions will be in the case when two parallel drums are connected by a drawbar (spiral drums or Whiting sheaves). Behr*) gives a comparison with locomotives, which, with coupled drawbars, have a significantly lower efficiency than with direct transmission.

As has already been indicated, in the following examples we take the friction of the machine as constant, especially since it has very little effect on the time of the complete hoist.

Let us now determine the work of the machine and the hoisting time at constant acceleration.

The distance and time of acceleration here are determined by formulas (20) and (21), namely,

\[ s_1=\frac{Jv_1^2}{2Rm_1}=\frac{14960\cdot 8^2}{2\cdot 2\cdot 25\cdot 14400}=14.8\ \text{m}, \]

\[ t_1=\frac{2s_1}{v_1}=\frac{2\cdot 14.8}{8}=3.7\ \text{sec.} \]

Since \(s_3\) and \(t_3\) remain as before, i.e. \(s_3=31.0\) m, and \(t_3=7.6\) sec., then

\[ s_2=H-(s_1+s_3)=254.2\ \text{m}, \]

\[ t_2=\frac{254.2}{8}=31.9\ \text{sec.} \]

The total hoisting time is

\[ T=t_1+t_2+t_3=43.2\ \text{sec.} \]

Consequently, the gain in time will be only 0.18 sec., or \(<0.5\%\) of the total hoisting time. Therefore it is more advantageous to reduce the cutoff during the acceleration period, which is what we have adopted in the subsequent calculations.

The diagram of moments at constant acceleration is shown in Fig. 38.

§ 27. Koepe sheave, diameter 6 m.

From Table 2 we have \(ab=2040\), \(bc=6600\), \(cd=6300\). We take the additional moment as 6500, \(af=ab+cb+cf=15140\). The friction moment of the machine is \(0.25\,af=3800=ac\). The static moment is \(F=af+ac=18940\). The dynamic moment is \(M_1=1.5F=28400\). Hence \(m_1=15960\); we take \(m_2=3000\). The polar moment of inertia during acceleration is \(J=21700\). At the same greatest speed (\(v_1=8\) m.) we have, according to formulas (12) and (16),

\[ s_1=\frac{Jv_1^2}{R(m_1+m_2)} =\frac{21700\cdot 8^2}{3\cdot 18900} =24.4\ \text{m}, \]

\[ t_1=\sqrt{\frac{21700\cdot 24.4}{3\cdot 12960} \left(\frac{\pi}{2}-\arcsin\frac{3000}{15960}\right)} =5.15\ \text{sec.} \]

*) T. J. M. & E. M., p. 421.

For the period of retardation \(J_0 = 21070\)

according to eq. (22)

\[ s_3=\frac{J_0 v_1^2}{2 R A}=\frac{21070\cdot 8^2}{2\cdot 3\cdot 12440}=18\ \text{m}. \]

\[ t_3=\frac{2s_3}{v_1}=4.5\ \text{sec.} \]

In the period of full speed,

\[ s_2=H-(s_1+s_3)=257.6\ \text{m}. \]

\[ t_2=\frac{257.6}{8}=32.2\ \text{sec.} \]

The total hoisting time is

\[ T=t_1+t_2+t_3=41.85\ \text{sec.} \]

From this we see that the hoisting time, in comparison with the cylindrical drum, has been reduced. With a cylindrical drum 6 m in diameter, the advantages of the Koepe pulley are even greater.

§ 28. Cylindrical drum, 6 m in diameter.

Here the moments at the beginning of hoisting are the following (Table 3): \(ab=1800\), \(bc=11100\), \(cd=6300\), \(cf=6500\), \(af=19400\).

The friction moment of the machine is

\[ 0.25\cdot 19400=4800=ac. \]

The static moment is

\[ F=af+ac=24200. \]

The dynamic moment is

\[ M_1=1.5F=36300,\qquad m_1=18600,\qquad m_2=3000. \]

For the period of acceleration \(J=30000\)

\[ s_1=\frac{Jv_1^2}{R(m_1+m_2)} =\frac{30000\cdot 8^2}{3\cdot 21600}=29.6\ \text{m}, \]

\[ t_1=\sqrt{\frac{30000\cdot 29.6}{3\cdot 15600} \left(\frac{\pi}{2}-\arcsin\frac{3000}{18600}\right)} =6.15\ \text{sec.} \]

The moment at the end of hoisting, counting the friction of the machine, is

\[ A_1=1800+2100+4800=8700=c_1e_1. \]

Consequently,

\[ \beta=\frac{ce-c_1e_1}{H} =\frac{17700-8700}{300}=30.\quad (\text{p. 75}). \]

In the period of retardation, \(J_0=29370\) (p. 29)

\[ s_3= \sqrt{ \frac{A_1^2-J_0v_1^2\frac{\beta}{R}}{\beta} } -A_1 = \sqrt{ \frac{8700^2-29370\cdot 8^2\frac{30}{3}}{30} } -8700 =35\ \text{m}. \]

\[ t_3= \sqrt{\frac{J_0}{R\beta}}\, \lg e \left\{ \frac{ A_1+\beta s_3+\sqrt{2A_1\beta s_3+(\beta s_3)^2} }{A_1} \right\}, \]

\[ t_3= \sqrt{\frac{29370}{3\cdot 30}}\, \lg e \left\{ \frac{ 8700+1050+\sqrt{2\cdot 8700\cdot 1050-1050^2} }{8700} \right\} =8.5\ \text{sec.} \]

For the period of full speed,

\[ s_2=235.4\ \text{m},\qquad t_2=29.5\ \text{sec.} \]

The total hoisting time is

\[ T=t_1+t_2+t_3=44.15\ \text{sec.} \]

It is more convenient to make a comparison of the different types of hoisting devices by determining the influence of the depth on the operation of the machine.

For comparison, we take a deeper mine, of 600 m, and see how this affects the operation of the machine.

At a mine depth of 600 m.

§ 29. Cylindrical drum, 6.5 m in diameter (rope of constant cross-section).

The moments at the beginning of hoisting (Table 4, p. 31): \(ab=3100\),

\(bc = 27000,\ cd = 9100;\) we take \(cf = 10000,\ af = 40100.\) Machine friction \(0.25\,af = 10000 = ac.\) Static moment \(F = +af + ca = 50100.\) Dynamic moment \(M_1 = 1.5F = 75150.\) Hence \(m_1 = he - ce = 35050,\) and the moment remaining at the end of acceleration, \(m_2,\) may be taken as \(5000.\)

We take the greatest speed \(v_1,\) in consequence of the increase in depth, as \(12\ \text{m/sec};\) then, in the period of acceleration, \(J = 6000\) (p. 32).

\[ s_1=\frac{Jv_1^2}{R(m_1+m_2)} =\frac{60000\cdot 12^2}{3.25\cdot 40050} =66.3\ \text{m}. \]

\[ t_1= \sqrt{\frac{Js_1}{R(m_1-m_2)}}\cdot \left(\frac{\pi}{2}-\arcsin\frac{m_2}{m_1}\right) \]

\[ = \sqrt{\frac{60000\cdot 66.3}{3.25\cdot 30050}} \left(\frac{\pi}{2}-\arcsin\frac{5000}{35050}\right) =9.1\ \text{sec}. \]

For the period of retardation,

\[ s_2= \sqrt{\frac{A_4^2+J_0v_1^2\frac{\beta}{R}}{\beta}-A_4} \qquad \text{(formula 34, p. 35)}. \]

Here: \(J_0 = 59500\) (p. 32) from table 4, \(A_4=b_1e_1+b_1c_1=\)

\[ 13100+(-5440)=7660. \]

\[ \beta=\frac{40100-7660}{600}=54; \]

\[ s_2= \sqrt{ \frac{ 7660^2+\frac{54\cdot 59500\cdot 12^2}{3.25}-7660 }{54} } =121\ \text{m}. \]

\[ t_2= \sqrt{\frac{J_0}{R\beta}}\, \lg e \left\{ \frac{ A_4+\beta s_2+\sqrt{2A_4\beta s_2-(\beta s_2)^2} }{A_4} \right\} \]

\[ t_2= \sqrt{\frac{59500}{3.25\cdot 54}}\, \lg e \left( \frac{ 7660+6520+\sqrt{2\cdot 7660\cdot 6520-6520^2} }{7660} \right) =19.18\ \text{sec}. \]

For full speed,

\[ s_3=H-(s_1+s_2)=412.7\ \text{m}. \]

\[ t_3=\frac{412.7}{12}=34.38\ \text{sec}. \]

Thus the total hoisting time

\[ T=t_1+t_2+t_3=62.66\ \text{sec}. \]

It differs only slightly from the value adopted by us at the beginning (see p. 29), \(T_0=60\ \text{sec}.\)

The dimensions of the machine are somewhat reduced when a rope of variable cross-section is used. Let us consider this case.

§ 30. Cylindrical drum of diameter 6.5 m (rope of variable cross-section).

Here the moments at the beginning of hoisting, table 5, p. 33, are: \(ab = 3600;\ bc = 23800,\ cd = 9100,\ cf = 10000,\ af = 37400.\) Machine friction \(=0.25\,af=9300=ac.\)

The static moment at the beginning of hoisting \(F = ae + af = 46700\).

Dynamic moment

\[ M_1 = 1{,}5\,F = 70000 = eh,\qquad m_1 = eh - ec = 33300. \]

Similarly to § 29, we take \(m_2 = 5000\).

For the period of acceleration we have \(J = 57800\),

\[ s_1 = \frac{57800 \cdot 12^2}{3{,}25 \cdot 38300} = 66{,}8\ \text{m}. \]

\[ t_1 = \sqrt{ \frac{57800 \cdot 66{,}8}{3{,}25 \cdot 28300} \left(\frac{\pi}{2} - \arcsin \frac{5000}{33300}\right) } = 9{,}28\ \text{sec}. \]

For the period of retardation \(J_0 = 57300\) (p. 33).

From Table 5

\[ A_4 = 9300 + 3600 - 2300 = 10600, \]

\[ \beta = \frac{36700 - 10600}{600} = 43{,}5. \]

\[ s_3 = \frac{ \sqrt{ A_4^2 + \dfrac{J_0 v_1^2 \beta}{R} } - A_4 }{\beta} = 104{,}5\ \text{m}. \]

\[ t_3 = \sqrt{\frac{J_0}{R\beta}}\, \lg e \left\{ \frac{ A_4 + \beta s_3 + \sqrt{2A_4\beta s_3 - \beta^2 s_3^2} }{A_4} \right\} = 17{,}7\ \text{sec}. \]

For the period of full speed \(s_2 = 428{,}7\ \text{m}\), \(t_2 = 35{,}7\ \text{sec}\). The total hoisting time is \(T = 62{,}68\).

It is evident from this that, with a rope of variable cross-section, the hoisting time, in comparison with a rope of constant cross-section, remains the same, while the dimensions of the machine are reduced (in the present case by \(6{,}8\%\)).

Let us consider the operating conditions of the machine with a cylindrical drum and balanced ropes.

§ 31. Cylindrical drum of diameter 6.5 m (the rope is balanced).

The operating conditions are shown in Fig. 43; here, according to Table 6, p. 34,

\[ ab = 3900,\quad bc = 10700,\quad cd = 9100,\quad cf = 10000,\quad af = 24600. \]

The friction moment of the machine

\[ ae = 0{,}25\,af = 6200. \]

The static moment

\[ F = af + ae = 30800 = ef. \]

The dynamic moment

\[ M_1 = 1{,}5\,F = 46200 = eh. \]

The moment remaining for acceleration,

\[ m_1 = eh - ec = 25400. \]

We take the additional moment at the end of acceleration \(m_2 = 4000\).

Then for the period of acceleration we have \(J = 65300\) (p. 34)

\[ s_1 = \frac{65300 \cdot 12^2}{3{,}25 \cdot 29400} = 100\ \text{m}. \]

\[ t_1 = \sqrt{ \frac{65300 \cdot 100}{3{,}25 \cdot 29400} } \left( \frac{\pi}{2} - \arcsin \frac{4000}{25000} \right) = 13{,}4\ \text{sec}. \]

To determine the elements of the retardation period, we shall use equations (22) and (23) (p. 72),

\[ s_3 = \frac{J_0 v_1^2}{2RA_2}, \]

where

\[ J_0 = 64800\ \text{(p. 34)},\qquad A_1 = ec = 20800, \]

\[ s_3 = \frac{64800 \cdot 12^2}{2 \cdot 3{,}25 \cdot 20800} = 69{,}2\ \text{m}. \]

\[ t_3 = \frac{2s_3}{v_1} = 11{,}52\ \text{sec}. \]

For the period of full speed \(s_2 = 430.8\) m, \(t_2 = 35.9\) sec.
The total hoisting time \(T = 60.82\) sec.
In comparison with the preceding case, the time has been reduced by only 1.84 sec., but the dimensions of the machine will be less by \(1 \tfrac{1}{2}\) times.

§ 32. Koepe pulley of diameter 8.0 m.

For this case, according to Table 7, p. 35, \(ab = 4800\), \(bc = 13200\), \(cd = 11200\), \(cf = 12000\), \(af = 30000\).

The friction of the machine \(ae = 0.25\,af = 7500\).

The static moment at the beginning of hoisting \(F = 37500 = ef\).

The dynamic moment \(M_1 = 1.5\,F = 56400\);

Hence \(m_1 = 30900\), and we take \(m_2 = 5000\); then for the acceleration period \(j = 66050\) (p. 35)

\[ s_1 = \frac{66050 \cdot 12^2}{4 \cdot 35900} = 66.3\ \text{m}. \]

\[ t_1 = \sqrt{\frac{66050 \cdot 66.3}{4 \cdot 25900}} \left( \frac{\pi}{2} - \arcsin \frac{5000}{30900} \right) = 9.16\ \text{sec.} \]

For the retardation period: \(J_0 = 63650\) (p. 35), \(s_3 = 45\) m, \(t_3 = 7.5\) sec.

For full speed

\[ s_2 = H - (s_1 + s_3) = 488.7\ \text{m}, \]

\[ t_2 = \frac{s_2}{v_1} = 40.7\ \text{sec.} \]

The total hoisting time

\[ T = t_1 + t_2 + t_3 = 57.36\ \text{sec.} \]

The moment diagram is presented in Fig. 44. We proceed to the determination of the work with a conical drum.

§ 33. Conical drum \(D = 7.5\) m and \(d = 4.5\) m.

Here the resultant moment is variable; its numerical value has been obtained graphically in Fig. 17 and is equal to 10200 kgm at the beginning of hoisting and 9500 kgm at the end.

The friction in the shaft is equal to 5% of the twisting moments; i.e.,

\[ 0.05\,(k\beta_0 + f\gamma_5) = 2850\ \text{kgm}. \]

The additional moment at starting along \(cf\) is

\[ 2800 \cdot 3.75 + x = 10500 + x \simeq 11000. \]

The friction moment of the machine

\[ ae = 0.25\,(11000 + 10200 + 2850) \simeq 6000. \]

The static moment at the beginning of hoisting

\[ F = ef = 30050\ \text{kgm}. \]

The dynamic moment

\[ M_1 = eb = 45000\ \text{kgm}. \]

The moment for overcoming the forces of inertia

\[ m_1 = 25950, \]

since

\[ A_1 = 6000 + 2850 + 10200 = 19050. \]

The remaining moment at the end of acceleration, \(m_2\), may be taken equal to 1000, and the polar moment of inertia of the moving parts is found from Fig. 18 after an approximate determination of the number of revolutions in the acceleration period; for the given case \(j = 66000\).

The elements of the acceleration period are determined by formulas (46) and (45), namely,

\[ n_1^{1} = \frac{j\,w_1^2}{2\pi(m_1 + m_2)}. \]

The determination of the linear velocity at different radii has been made in Fig. 45 (lower drawing), whence it is seen that \(rw = 12\), or \(w_1 =\)

\[ \frac{12}{2{,}25}=5{,}35;\quad \text{then}\quad n'_1=\frac{66000\cdot 5{,}35^2}{2\cdot 3{,}14\cdot 26950}=10{,}8\ \text{rev.} \]

\[ t_1=\sqrt{\frac{2\pi J n'_1}{m_1-m_2}\left(\frac{\pi}{2}-\arcsin\frac{m_2}{m_1}\right)}=29{,}5\ \text{sec.} \]

To determine the time and the number of revolutions in the period of retardation, we shall use equations (47) and (48), namely:

\[ n'_3=\frac{J_0 w_1^2}{4\pi A_4}. \]

Here \(J_0=68500\) (see Fig. 18), \(w_1=5{,}35\), \(A_4=6000+2850+9500=18350\).

\[ n'_3=\frac{68500\cdot 5{,}35^2}{4\cdot 3{,}14\cdot 18350}=8{,}5\ \text{rev.} \]

\[ t_3=\frac{4\pi n'_3}{w_1}=\frac{4\cdot 3{,}14\cdot 8{,}5}{5{,}35}=20\ \text{sec.} \]

The number of revolutions passed at full speed is

\[ n'_2=N-(n'_1+n'_3)=12{,}5\ \text{rev.} \]

during the time

\[ t_2=\frac{2\pi n'_2}{w_1}=14{,}8\ \text{sec.} \]

Thus the total hoisting time is \(T=55{,}3\) sec.

The velocity and acceleration diagram is shown in Fig. 45; there the distribution of the work of the machine is also indicated, with the paths laid off along the abscissa axis analogously to Figs. 37–44.

The mean winding diameter in the acceleration period is \(d_1=5{,}01\), and \(s_1=170\) m; for the retardation period \(d_3=7{,}10\) and \(s_3=190\) m; whence \(s_2=240\) m.

The use of a bobbin permits the dimensions of the machine to be reduced somewhat.

§ 34. Bobbin \(D=6{,}2;\ d=3{,}5\) m.

Here: \(ab=0{,}05(28000+20000)=2400\) (Fig. 20), \(be=8000\) (Fig. 20), \(cd=8700\), \(cf=9000\), \(ea=0{,}25\,af=5000\), \(F=ef=24400\), \(M_1=1{,}5F=36600\), \(A_1=A_4=15400\), \(m_1=M_1-A_1=21200\).

Moment at the end of acceleration: \(m_2=500\).

In the acceleration period, \(J=35000\) (Fig. 21).

\[ n'_1=\frac{Jw_1^2}{2\pi(m_1+m_2)} =\frac{35000\cdot 6{,}85^2}{2\cdot 3{,}14\cdot 21700} =12{,}3\ \text{rev.} \]

\[ t_1=\sqrt{\frac{2\pi J n'_1}{m_1-m_2}\left(\frac{\pi}{2}-\arcsin\frac{m_2}{m_1}\right)}=17{,}7\ \text{sec.} \]

\(d_1=3{,}922\), \(s_1=152\) m.

For retardation, \(J_0=36500\) (Fig. 21).

\[ n'_3=\frac{36500\cdot 6{,}85^2}{4\cdot 3{,}14\cdot 15400}=9{,}05\ \text{rev.},\qquad t_3=\frac{4\cdot 3{,}14\cdot 9{,}05}{6{,}85}=16{,}6\ \text{sec.} \]

\(d_3=5{,}89\), \(s_3=167\) m.

In the period of full speed,

\[ n'_2=39{,}4-21{,}35=18{,}05. \]

\[ t_2=\frac{6{,}28\cdot 18{,}05}{6{,}85}=16{,}5\ \text{sec.} \]

\[ s_2=281\ \text{m.} \]

The time at full speed is \(T=50{,}8\) sec.

Let us consider the magnitude of the various stresses arising in the ropes for the cases adopted by us.

The replacement of drums by sheaves has an especially sensitive effect on the magnitude of the dynamic stress; for example, with Koepe sheaves it is very important to take the influence of this stress into account, since its magnitude is inversely proportional to the polar moment of inertia of the moving parts, which with Koepe sheaves is considerably smaller than with cylindrical drums. When Whiting’s system is used, a differential stress arises because the diameters of the grooves on the driving sheaves are not entirely identical. According to Robeson’s experiments⁹, the magnitude of this differential stress is very great, so that it must be taken into consideration.

In the following Table 11 we give the results obtained from the calculations of the examples we have considered; moreover, for the reasons indicated in § 8, the wire diameter is not specified, and consequently the bending stresses in the rope cannot be derived.

TABLE 11.

Drum type	No.	Diameter, m	Shaft depth, m	Weight of useful load, kg	Weight of cage and cars, kg	Rope: diameter, mm	Rope: total weight, kg	Rope: weight per running meter, kg	Weight of drums, kg	Weight of sheaves, kg	Sheave diameter, m	Polar moment of inertia at acceleration, \(J_1\)	Polar moment of inertia at retardation, \(J_0\)	Twisting moments: static load, kg·m	Twisting moments: at acceleration, kg·m	Twisting moments: at retardation, kg·m	Maximum speed, m/sec
Cylindrical		4.5	300	2200	4200	38	21500	4.8	21200	24000	5	14960	15280	18400	27800	14400	8
Koepe sheave		6.0	300	2200	4200	—	—	—	12000	24000	5	21700	21070	18940	43400	16960	8
Cylindrical		6.0	300	2200	4200	38	21500	4.8	21500	24000	5	30000	29370	24200	36300	18600	8
Cylindrical		5.5	600	3300	5600	50	21500	8.2	22500	26000	6	60000	59500	50000	75150	35050	12
Cylindrical		6.5	600	3300	5600	50	24000 variable	—	22500	26000	6	57800	57300	46700	70000	33300	12
Cylindrical		8.0	600	3300	5600	50	35000	8.2	22500	26000	6	63550	64800	39600	46200	25400	12
Koepe sheave		8.0	600	3300	5600	—	—	—	20000	26000	6	69050	68650	37500	56400	30900	12
Conical		4.5–7.5	600	3300	5600	50	25000	8.2	23000	26000	6	66000	68500	30050	45000	25950	18
Bobbin		3.5–6.2	600	3300	5600	—	25630 variable	—	[[unclear: drum weight]]	26000	6	30000	36500	24100	36600	21200	20

TABLE 12.

Drum	Radius, m	Acceleration distance \(s_1\), m	Full-speed distance \(s_2\), m	Deceleration distance \(s_3\), m	Acceleration time \(t_1\), sec	Full-speed time \(t_2\), sec	Deceleration time \(t_3\), sec	Total hoisting time \(T\), sec	Wire diameter \(\delta\)	\(R/\delta\)	Bending stress, kgf/mm²	Tensile stress, kgf/mm²	Dynamic stress, kgf/mm²	Total stress, kgf/mm²	Static safety factor	Dynamic safety factor	Application
T and P. Cylindrical	2.25	25.1	273.9	31.0	5.28	30.38	7.6	43.38	1000	4.4	16.5	1.0	21.9	5.45	7.3	5.45	Application
T and P. Cages in the cage compartment	3.0	24.4	257.6	18.0	5.15	32.2	4.5	41.85	—	—	—	—	—	—	—	—	Application
T and P. Cylindrical	3.0	29.6	235.4	35.0	6.15	28.5	8.5	41.5	1350	3.25	16.5	0.62	20.37	5.9	7.3	5.9	Application
T and P. Cylindrical	3.25	66.3	412.7	121.0	9.1	44.38	19.18	62.66	1060	4.2	16.9	0.52	21.92	5.48	7.1	5.48	Application
T and P. Cylindrical	3.25	66.8	427.7	104.5	9.28	35.7	17.7	62.68	1060	4.2	17.2	0.8	22.5	5.35	7.0	5.35	Cage hoist, northern section
T and P. Cylindrical	3.25	100	430.8	69.2	13.4	39.9	11.52	60.82	1060	4.2	16.9	0.55	21.65	5.31	7.1	5.31	Cage hoist
T and P. Cages in the cage compartment	4.0	66.3	188.7	45	9.16	30.7	7.5	57.36	—	—	—	—	—	—	—	—	Application
T and P. Conical, \(2.25–3.75\)	\(2.25–3.75\)	170	240	190	20.5	[[unclear: value obscured]]	20.0	55.3	—	—	16.2	0.5	22	5.45	7.1	5.45	Application
T and P. Bobbin	\(1.75–3.1\)	152	281	167.0	17.7	16.8	16.6	50.8	—	—	—	—	—	—	—	—	Application

Literature

1) Behr, “Winding Plants for Great Depths.” T. J. m. & m. Vol. XI.

2) Herrmann, “Die dynamischen Verhältnisse der Schachtfördermaschinen.” Dingl. Pol. journ. 1902, B. 317, S. 469,485.

3) Mellin, “Der Steinkohle bergbau des Preussishen Staates etc.” III Teil, s. 20—21.

4) “Die Entwickelung des Niederrheinisch—Westfälischen Steinkohlen.” B. V, Förderung 1902.

5) Coll. Guard. 1882, June 9.

6) Habets, “Cours d’ Exploitation des Mines,” vol. I, p. 623.

7) Makovsky, “Essay on Mine Hoists, etc.” 1909.

8) Time, “Handbook for Mining Engineers,” p. 52.

9) Skochinsky, “Essay on the Mines of the Donets Basin.” Mining Journal, 1903.

10) Appendix to § 30 of the Instructions on Supervision of Private Mining Industry. 1892.

11) Robeson & Co k.—Tran. J. m. & m. Vol. XI, p. 1.

12) Hrabák, “Die Drahtseile.”

13) Dechamps. Revue U. d. M. V. LVIII.

14) Euters, “The mechanical Engineering of Collieries,” vol. II.

15) Hughes, “Text-book of Coal Mining,” p. 298.

16) Spengel. Transactions J. m. & m. vol. XI, p. 324.

17) Ehrlich. Zeitschrift d. V. d. J.

18) Milkovsky, “Wire Rope, etc.” Part II.

19) Hoffmann. Z. d. V. d. J. 1904. S. 151.

20) Habets, “Etude dynamique de la machine d’ Extraction.” R. U. d. M. 4 S, XI.

21) Wells, “Winding Engine Design.” Coll Guard. 1. 1909, p. 1289.

22) Mines and minerals, March 1905, p. 379.

23) Norton. Discussion. T. J. m. & m. v. XI.

24) Dam, “Winding Engines.” Min. & Min. Vol. XXV, p. 426.

25) Hauer, “Die Fördermaschinen der Bergwerke,” s. 270.

26) Hrabák, “Hilfsbuch der Dampfmaschinen—Techniker,” B. II, s. 214.

27) Reiche, “Der Dampfmaschinen—Constructeur,” B. II, s. 79.

28) Durham, “Determination the Size of Hoisting Plants.” T. A. J. m. E., vol. XXXIII, p. 156.

29) Ugarov, “A New Method for Determining Tangential Forces.” Izv. t. t. I., 1910, No. 2.

30) Hurst, “Valves and Valve—Gearing.”

Contents

	Page
Preface	1

CHAPTER I

General survey of the various types of hoisting devices.

	Page
§ 1. Cylindrical drum	3
§ 2. Conical drum	3
§ 3. Systems in which drums are replaced by pulleys	4
§ 4. Systems that have not come into general use	4

CHAPTER II

Preliminary calculations.

	Page
§ 5. General conditions	6
A. Mine depth 300 m	8–14
§ 6. Determination of the useful and dead loads; calculation of the rope	8
§ 7. Determination of twisting moments and forces of inertia for a cylindrical drum \(D = 4.5\) m	9
§ 8. The same for a Koepe pulley \(D = 6.0\) m	11
§ 9. The same for a cylindrical drum \(D = 6.0\) m	13
B. Mine depth 600 m	14–23
§ 10. Determination of the useful and dead loads	14
§ 11. Determination of twisting moments and forces of inertia for a cylindrical drum \(D = 6.5\) m	15
§ 12. The same with a rope of variable cross-section	16
§ 13. The same with a balanced rope	17
§ 14. The same for a Koepe pulley \(D = 8.0\) m	18
§ 15. The same for a conical drum	18
§ 16. The same for a bobbin	21
§ 17. The same for cylindrical-conical drums	23

CHAPTER III

Working conditions and methods for calculating hoisting machines.

	Page
§ 18. General remarks	24
§ 19. Calculations proposed by various authors	25
§ 20. Relationship between the magnitude of filling and tangential forces for static conditions	27
§ 21. Determination of tangential forces for dynamic conditions	29
§ 22. Derivation of the basic formula	30
§ 23. Determination of the travel and time of hoisting for a balanced rope	30
§ 24. The same for an unbalanced rope	34
§ 25. The same for a conical drum and a bobbin	36

CHAPTER IV

Operation of the machine with drums of various systems.

	Page
A. Mine depth 300 m
§ 26. Cylindrical drum \(D = 4.5\) m	38
§ 27. Koepe pulley \(D = 6.0\) m	40
§ 28. Cylindrical drum \(D = 6.0\) m	41
B. Mine depth 600 m
§ 29. Cylindrical drum \(D = 6.5\) m	41
§ 30. The same with a rope of variable cross-section	42
§ 31. The same with a balanced rope	43
§ 32. Koepe pulley \(D = 8.0\) m	44
§ 33. Conical drum	44
§ 34. Bobbin	45
Bibliography	49

1. See § 15. ↩

2. See works ⁶, [^19], ⁸. ↩

3. See Hoffmann’s article.[^10] ↩

4. Dechamps, “Application de la méthode graphique etc.” R. U. d. M. ↩

5. This length is evidently equal to \(47.5\cdot \pi\cdot 1.75=261\) m. ↩

9. Robeson’s. ↩↩