Prof. S. K. Konyukhov.

Theory of the Bobbin

In mine hoisting installations here in Russia, three types of drums are most often encountered: cylindrical, bobbin, and conical. More complicated types of drums are not encountered here at all.

Some attention is also given to the Köpe system, in which the frictional force of the rope on the sheave is utilized for hoisting. Because of the originality of its idea, this system occupies a special position, and therefore we shall not discuss it. Likewise we shall not discuss the system of hoisting with a cylindrical drum when a tail balancing rope is used, making it possible to carry out so-called harmonic hoisting.

Within these limitations, the cylindrical drum, as though exclusively widespread in the Siberian mines, is the most imperfect with respect to the expenditure of power energy.

After it comes the bobbin, and after that the conical drum.

Thus, the bobbin occupies the middle place.

Strictly speaking, the cylindrical drum and the bobbin are only special cases of the conical drum.

Indeed, if we take as the basic elements the radii of the bases of a truncated conical drum, \(R\) and \(r\), and the height of the cone \(H\), then, by setting \(R = r\), we obtain a cylindrical drum, while by taking \(H = 0\), we obtain a bobbin.

From the conical drum one may also obtain the Köpe system, if one sets \(R = r\) and \(H = 0\). A simple example of a cylindrical drum with a rope wound or coiled on it is a spool of thread, a bobbin in the shuttle of a sewing machine, or the drum on which telegraph wire is usually wound before being strung.

An equally simple example of a bobbin with a flat-rope tape wound on it is a spool with the tape wound on it in a typewriter, a reel of telegraph tape, or a wound clock spring. An example of a conical drum is a reel on which wire is wound during rolling and drawing.

The theory of the bobbin presents a good example illustrating the power and vitality of mathematical analysis in application to engineering.

Beginning the exposition of this theory, we introduce into our discussion the following notations:

\(N\) — useful weight of the load being lifted, in kilograms,

\(L\) — dead load (weight of the cage and empty mine cars), in kilograms,

\(S\) — weight of the hoisting rope of length \(= H\) meters, i.e. the depth of the mine,

\(g\) — weight of one running meter of rope of equal cross-section, in kilograms. It is equal to \(0.0077\, n \delta^{2}\), where \(n\) is the number of wires in the rope, and \(\delta\) is the diameter of the wire in millimeters,

\(d\) — thickness of the flat-rope tape in millimeters,

\(r\) — inner radius of the bobbin in millimeters,

\(R\) — greatest radius of the bobbin when the portion of the rope equal to the depth of the mine \(= H\) plus five reserve turns is wound on it,

\(H\) — depth of the mine in meters,

\(\theta\) — angle of winding of the rope over the whole period of hoisting.

When winding a rope onto a bobbin, the axis of the wound strip bends along an Archimedean spiral, in which the radius vector increases with each complete revolution of the bobbin by the thickness of the rope strip \(= d\).

Let us suppose that the working stroke of the bobbin begins with the winding of the strip onto a bobbin of radius \(r\).

The equation of the Archimedean spiral will then be written for the entire working stroke in the following form:

\[ R = r + \frac{\theta}{2\pi}\cdot d \]

Any intermediate variable value of the radius vector can then be expressed as:

\[ x = r + \frac{\varphi}{2\pi}\cdot d \]

Let us express the length of an element of the arc of the Archimedean spiral in polar coordinates (see drawing), for which we introduce the following notation:

\[ \begin{aligned} OB &\ldots\ldots\ldots\ldots r \\ \angle BOE &\ldots\ldots\ldots\ldots \varphi \\ \angle AOB &\ldots\ldots\ldots\ldots d\varphi \\ AD &\ldots\ldots\ldots\ldots dr \\ AB &\ldots\ldots\ldots\ldots ds \end{aligned} \]

The infinitely small element of the arc \(BD\), normal to the radius vector \(OD\), will be equal to \(r\,d\varphi\). Taking \(AB\) as a straight line, one may write from the right triangle \(ABD\)

\[ ds^2 = dr^2 + r^2 d\varphi^2 \]

\[ ds = \sqrt{dr^2 + r^2 d\varphi^2} \]

Since \(r\) and \(\varphi\) are variable coordinates, the expression for a finite arc may be written in the following form:

\[ S = \int_0^\theta \sqrt{dx^2 + x^2 d\varphi^2} \]

In practice, the first term of the radical expression is very small in comparison with the second term. Indeed, the expression for the radius vector has the form

\[ x = r + \frac{\varphi \cdot d}{2\pi} \]

Since the initial radius in hoisting devices is a constant quantity, \(r = \mathrm{Const}\), differentiation gives

\[ dx = \frac{d}{2\pi}\cdot d\varphi \]

On the other hand, multiplying all terms of the expression for the variable radius vector by \(d\varphi\), we have

\[ x d\varphi = r d\varphi + \frac{d}{2\pi}\varphi d\varphi \]

Whence we find that

\[ \frac{dx}{x d\varphi} = \frac{\dfrac{d}{2\pi}}{r + \varphi \dfrac{d}{2\pi}} \]

The left-hand side attains its greatest value in the case when in the right-hand side we put \(\varphi = 0\), i.e.

\[ \frac{d}{2\pi r}. \]

The maximum value of \(d\): \(2r\), i.e. the ratio of the diameter of the rope to the diameter of the drum onto which the rope is wound, is obtained when there is a manual winch and the rope is organic. In this case the ratio \(d : 2r\) reaches up to

\(\frac{1}{10}—\frac{1}{15}\). With driving winches and a thicker rope this ratio is close to the value \(\frac{1}{25}—\frac{1}{30}\). Finally, with drums of hoisting machines this ratio is in no case less than \(\frac{1}{40}\), and with metal ropes, especially steel ones, this ratio is \(\frac{1}{50}—\frac{1}{60}—\frac{1}{100}\). If we take \(d : 2r = 0.1\), we obtain

\[ d : 2\pi r=\frac{10}{314}\simeq \frac{1}{31}. \]

The smaller this ratio, the more favorable the conditions under which the rope works, since less work is spent on bending the rope, and, moreover, the strength of the rope is not reduced to such a great extent.

Thus, the length of the rope wound onto the drum depends essentially only on one term in the radical expression, while the influence of the other term is practically insignificant.

This conclusion is of essential importance in the subsequent calculations, since it makes it possible to introduce simplifications into the computations and thereby facilitate the solution of the problem.

This consideration is most easily illustrated by an example.

The simplest form of the equation of the Archimedean spiral is represented as follows:

\[ r=a\varphi, \]

where \(r\) is the variable radius vector, \(a\) is a constant quantity, and \(\varphi\) is the variable argument.

The expression for the element of arc in this case will be written as

\[ ds=\sqrt{a^2d\varphi^2+a^2\varphi^2d\varphi^2}=a\,d\varphi\sqrt{1+\varphi^2}. \]

The magnitude of the finite arc is found by integration

\[ S=\int_0^\theta a\,d\varphi\sqrt{1+\varphi^2}. \]

This integral can be found in general form by using the rule of integration by parts.

It is known that

\[ \int x^{n-1}\,dx\,\sqrt{a^2+x^2} = \frac{x^n}{n+1}\sqrt{a^2+x^2} + \frac{a^2}{n+1} \int \frac{x^{n-1}\,dx}{\sqrt{a^2+x^2}}. \]

Putting \(n=1\) in this equation, we obtain

\[ \int dx\cdot\sqrt{a^2+x^2} = \frac{x}{2}\sqrt{a^2+x^2} + \frac{a^2}{2}\operatorname{Lg}\left(x+\sqrt{a^2+x^2}\right)+C. \]

Carrying out the integration within the limits from \(0\) to \(\theta\), we obtain

\[ S=\frac{a}{2}\left[\theta\sqrt{1+\theta^2}+\operatorname{Lg}\left(\theta+\sqrt{1+\theta^2}\right)\right]. \]

As can be seen, even for the simplest case the integral has a rather complicated form. The same picture of complexity remains when finding the radius of curvature of the spiral.

Indeed, the general expression for the radius of curvature in polar coordinates has the following form

\[ \rho= \frac{ \left[ r^2+\left(\frac{dr}{d\varphi}\right)^2 \right]^{3/2} }{ r^2+2\frac{dr^2}{d\varphi^2}-r\frac{d^2r}{d\varphi^2} }. \]

For the Archimedean spiral

\[ \frac{dr}{d\varphi}=a,\qquad \frac{d^2r}{d\varphi^2}=0. \]

Therefore

\[ \rho=\frac{(a^2\varphi^2+a^2)^{3/2}}{a^2\varphi^2+2a^2} =\frac{a(1+\varphi^2)^{3/2}}{2+\varphi^2}. \]

If one takes the equation of the Archimedean spiral in the form

\[ x=r+\frac{d}{2\pi}\,\varphi, \]

then the solution of the problem of finding the length of the arc and the magnitude of the radius of curvature becomes considerably more complicated, which will make it difficult to use the formulas in practice. It is therefore important to introduce simplifications into the calculations. Such a simplification in our case will be obtained if, in the expression for the element of arc, we put \(dx^2=0\).

For this simplification we shall have

\[ dS=x\,d\varphi \quad \text{and} \quad S=\int_0^\theta \left(r+\frac{d}{2\pi}\varphi\right)d\varphi =r\theta+\frac{d}{2\pi}\theta^2. \]

The meaning of this simplification amounts to the fact that, instead of the straightened turn of the Archimedean spiral, one takes the length of a circumference whose radius is, in magnitude, the arithmetic mean between the extreme radius vectors of the turn.

If we denote by \(\theta_1\) the angle described by the radius of the shaft when lifting the cage from the bottom of the shaft to the surface, then the final radius vector of the bobbin spiral \(R\) and the depth of the shaft \(H\) are expressed through \(\theta_1\) and \(r\) as follows:

\[ R=r+\frac{\theta_1}{2\pi}d \]

\[ H=r\theta_1+\frac{d}{4\pi}\theta_1^2. \]

In these expressions the quantity \(r\) is known and prescribed in advance, since it can easily be determined from the condition

\[ 2r=D_{\min}=\frac{E\delta}{S-k_z}, \]

where \(E\) is the modulus of elasticity of the material from which the rope is made, \(\delta\) is the diameter of the wire, \(k_z\) is the allowable tensile stress in the wire, and \(S\) is the total stress of the wire (in tension and bending). In practice \(D_{\min}\) has the following values:

1. Iron wire with breaking strength
   \(60\ \mathrm{kg/mm^2}\) . . . . . . . . . . . . . . . . \(1000\delta\)—\(1200\delta\)

2. Crucible-steel wire \((120\ \mathrm{kg/mm^2})\) . . . . . . . \(1200\delta\)—\(1400\delta\)

3. Patented steel wire \((180\ \mathrm{kg/mm^2})\) . . . . . . . \(1400\delta\)—\(1600\delta\)

For these grades of wire the allowable tensile load is taken, in the first case, as \(7\)—\(8\ \mathrm{kg/mm^2}\), in the second \(15\)—\(16\ \mathrm{kg/mm^2}\), and in the third \(22\)—\(24\ \mathrm{kg/mm^2}\).

As for the modulus of elasticity, for the same cases it is taken equal to \(18000\ \mathrm{kg/mm^2}\), \(19000\ \mathrm{kg/mm^2}\), and \(21000\ \mathrm{kg/mm^2}\). Finally, the diameter of the wire in millimeters is calculated by the formula

\[ \delta=\sqrt{\frac{1.27\,Q}{n\,(k_z-0.01H)}}, \]

where \(Q\) is the load at the lower end of the rope in kilograms, \(n\) is the number of wires in the rope, \(k_z\) is, as before, the allowable tensile load of the rope in kilograms, and \(H\) is the vertical height or depth of the shaft in meters. To these considerations it is necessary to add certain practical instructions, given by factories engaged in the manufacture of wire ropes. Factories issue—

must prepare wire of specified numbers, depending on the machines available at these plants. In a rope, the individual wires are twisted into separate strands or cords, and the strands are then interwoven with one another, enclosing within them a metal or hemp core. It is not recommended to twist a rope from five strands, since such a rope comes out angular and wears out quickly.

The number of wires in a rope is limited to definite ranges, different for different plants.

Thus, for example, the plants in Pishbram prepare ropes with the number of wires

\[ n = 30—36—42—48—60—72—84—96—108—126—144—216 \]

The Felten–Guilleaume plant in Mülhausen manufactures ropes for the market with the following number of wires

\[ n = 36—42—49—56—84—96—114—133—180 \]

The following German millimeter gauge is very convenient for selecting a suitable wire diameter.

Wire No.	Diameter \(\delta\) in millimeters	Wire No.	Diameter \(\delta\) in millimeters
4	0.4	18	1.8
4.5	0.45	20	2
5	0.5	22	2.2
5.5	0.55	25	2.5
6	0.6	28	2.8
7	0.7
8	0.8
9	0.9
10	1
11	1.1
12	1.2
13	1.3
14	1.4	31	3.1
16	1.6	34	3.4
		38	3.8
		42	4.2
		46	4.6
		50	5
		55	5.5
		60	6

The thickness of the rope is found fairly well from the formula \(d = a \cdot \delta\). The quantity \(a\) is determined as follows:

\[ \begin{aligned} n &= 36—42—48—60—72—84—96—108—126—144—216,\\ a &= 9.5—10—11—12—13—14—15—16—17.5—19—30. \end{aligned} \]

The most suitable wire Nos. for mine ropes are from 14 to 28.

When the wire diameter \(\delta\) has been calculated and replaced by a suitable number of the millimeter gauge, then the minimum drum diameter, or in our case the initial diameter of the core of the bobbin, \((2r)\), is also found without difficulty.

The result obtained may be checked by the following table.

Wire No.	\(2r = D_{\min}\) in millimeters	Wire No.	\(2r = D_{\min}\) in millimeters
4	300	14	1700—1800
5	400	16	2000—2200
6	500	18	2400—2600
7	600	20	2800—3000
8	700	22	3200—3400
9	800—900	25	3600—3800
10	1000—1100	28	4000
11	1100—1200	31	4500
12	1300—1400	34	5000
13	1500—1600

As can be seen, practice protects the interests of engineering very well, providing ready results of calculations reduced to tables. These tables can be used without difficulty even by a non-engineer.

And now the question involuntarily arises: what, then, is theory needed for? This question has great meaning, and therefore it must be treated seriously.

A false view of engineering as a craft has become so deeply rooted that it has long been sung, like a song, by both learned and unlearned alike.

The separation of engineering from science, observed from the most ancient times, occurred because engineers were not sufficiently skilled in fulfilling scientific requirements. Now this separation is gradually being smoothed out, and engineering is placing into the hands of science such pearls of its labor as previously one could not even dream of. Engineering gives flesh to the most refined thoughts of scholars. Without science it would have been impossible to realize aviation and aeronautics; and, on the other hand, without engineering, scientific researches would have remained simply dead capital. Only interaction between science and engineering helps to solve the most difficult tasks simply, confidently, and economically.

Practical data indicate how a problem was solved in one case or another, but how to solve a new problem under different conditions—practice alone does not provide this; theory and the engineer’s creative work provide it. It is not without reason that Boltzmann writes: “a good theory gives the best practice.” Theory is created not for isolated cases, but for an entire series, and the art of the engineer consists in making full use of theory for the successful execution of the work.

For an engineer, as the highest representative of engineering, it is simply improper, by the very nature of things and of his title, to be a practitioner in the everyday sense. Such an engineer will be a craftsman among engineers, and such a craftsman cannot be entrusted with responsible work: he will not be able to cope with it. The foreign words “technician” and “engineer,” in Old Russian, sound far more richly significant: “master craftsman” and “one of reflection.” A good engineer must be able to design, i.e. to mark out his best path toward attaining the goal, and not simply to imitate, i.e. to live by another’s mind, without duly comprehending his own work.

Design, as the intermediate stage from conception to the realization of an idea in actuality, always requires a thorough weighing of all the conditions under which the idea must be carried out.

This intellectual work resembles the casting pattern according to which the mold for the casting will be prepared, and then the casting itself.

What should one be guided by in such work? Here “rules” are often brought to the fore. But did the Americans not know these everyday rules when they were working on the installation of large turbines at Niagara Falls, and yet the turbines were still swept away by the water? But then Europeans arrived and installed their turbines in such a way that the flow of water was not able to tear them from their place. The matter is not in rules, but in good theory, which the practical Americans ignored.

The Quebec Bridge collapsed several times, despite all rules; but theory made it possible to do without a catastrophe.

The best rule for an engineer is to be original and independent, and then his work will not be marking time in place, but creative and progressive work.

From the most ancient times to our own day, rules and practice have given mankind only poor imitation of the original.

The Pyramid of Cheops, the Colosseum, the church of Hagia Sophia in Constantinople, St. Peter’s Cathedral in the Vatican, the Grand Opera in Paris, the Pantheon, Pagodas, the Eiffel Tower, the Diesel engine, the Titanic, tanks, the Junkers airplane, and

i.e., all this has been created not according to rules, but according to a fresh conception that does not recognize rules.

The designer or planner must solve the problem while taking into account a number of conditions and assumptions that do not lower the dignity of science and technology at the present moment, not only in the eyes of specialists, but also in the eyes of the broad mass of uninformed people.

Thus, the laws of mechanics must be taken into account, the selected coefficients strictly weighed, the suitability of the hypotheses of the given science clarified, and successes in solving similar questions—if not as a whole, then in part—identified; all this must serve as material for the designer, from which he must compose the image of his original thought in the form of a design.

Then the result will be not routine work according to rules and practice, but a work of creative “conception.”

The most celebrated designer of our time, Professor Riedler, writes the following on this subject: “In the technical world, mathematical calculations alone are not enough. The difficult mental work consists not in the calculation, but in the establishment of the data for the calculation. This establishment of data for calculation must correspond to the many-sided conditions of reality.”

A good theory best reveals this many-sidedness.

Now we can proceed to the exposition of the theory of the bobbin, without fearing that the complexity of the theory may frighten an interested person away from studying it.

The work of winding a flat rope onto a bobbin does not take place independently, but under the action of the work of the engine. Therefore, in clarifying the working conditions of the bobbin, one must also keep in mind the possibility of easy performance of this work by the motor, with an expedient expenditure of energy. Among the conditions that facilitate the machine’s task is the condition of the smallest possible fluctuation in the magnitude of the resisting moment on the shaft during the entire working stroke.

For the bobbin this condition is fulfilled only in part, when equality of the resisting moments at the beginning and at the end of the lift is attained.

Analytically, this requirement, with the notation adopted by us, is reduced to the following equation:

\[ r(N+L+gH)-RL=R(N+L)-r(L+gH) \]

Whence we have

\[ \frac{R}{r}=\frac{N+2L+2gH}{N+2L}=1+\frac{2gH}{N+2L}=1+\beta \]

Consequently,

\[ R=r+r\beta=r+\frac{2gHr}{N+2L}. \]

On the other hand,

\[ R=r+\frac{d}{2\pi}\cdot\theta_1, \]

Comparing the right-hand sides, we obtain

\[ \theta_1=\frac{4\pi rgH}{(N+2L)d} \]

With the aid of this angle \(\theta_1\) one can determine the depth of the shaft \(H\).

\[ H=r\theta_1+\frac{d}{4\pi}\theta_1^2 =\frac{4\pi r^2 gH}{(N+2L)d} +\frac{4\pi r^2 g^2H^2}{(N+2L)^2d}. \]

After insignificant transformations of an algebraic nature, we shall have

\[ H=\frac{(N+2L)^2(1-4\pi r^2 g)d}{4\pi r^2 g^2}. \]

In turn, for \(r\) and \(R\) the following expressions are obtained:

\[ r=\frac{N+2L}{2}\sqrt{\frac{d}{\pi g\,(N+2L+gH)}} \]

\[ R=\frac{N+2L+2gH}{2}\sqrt{\frac{d}{\pi g\,(N+2L+gH)}} ; \]

Further,

\[ \frac{\theta_1}{2\pi}=H\sqrt{\frac{g}{\pi d\,(N+2L+gH)}} . \]

It can be shown that, when the condition of equality of the moments of resistance to motion at the beginning and at the end of the ascent of the cage is fulfilled, we simultaneously attain the same moment of resistance also when the cages meet, i.e.

\[ M_{\mathrm{n}}=M_{\mathrm{v}}=M_{\mathrm{k}} . \]

It is easiest of all to carry out this problem if, during the raising and lowering of the cages, one follows the change in the sum of the moments of resistance applied to the shaft of the bobbins.

The simple mechanics of a mine hoisting installation reduces to the following: when the loaded cage begins to rise from the lower horizon, the upper empty cage (with empty cars) begins to descend, and by the end of the ascent the cages have exchanged places.

For both cages the moment of resistance to displacement changes all the time, but when the cages are at the same height, the moments of resistance become equal, and this instant corresponds precisely to half the number of revolutions necessary for raising the cage from the bottom of the shaft to the surface level; in other words, when the angle of rotation of the bobbin shaft is equal to \(\dfrac{\theta_1}{2}\).

Indeed, suppose that, before the cages meet, the loaded cage has made the bobbin turn through \(n_1\) revolutions; during the same time the empty cage has made its bobbin turn \(n_2\) times.

Let us transfer the load from one cage to the other and make the machine work in the reverse direction. Quite naturally, the first cage will descend to the bottom of the shaft after \(n_1\) revolutions, while the second cage will rise to the surface level after \(n_2\) revolutions.

Thus the first bobbin has made altogether \(2n_1\) revolutions, and the second \(2n_2\). The cages meeting in the shaft hang on ropes of one and the same length, and therefore, from a mechanical point of view, it is entirely immaterial on which of them the raising of the load is to be completed.

If the load were not transferred from the first cage to the second, then, in order to complete the raising of the cage, the working shaft of the bobbin would have had to make \(n_2\) revolutions.

It follows, therefore, that in order to raise the load from the bottom of the shaft to the surface level in the first cage, the bobbin shaft must be made to make a number of revolutions \(n_1+n_2\). But the very same effect is attained when the load is transferred from the first cage to the second, in which case the working shaft of the bobbin makes \(2n_2\) revolutions. Hence it is clear that \(n_1+n_2=2n_2\), or \(n_1=n_2\).

When the bobbins have a common shaft, it is clear that \(n_1=n_2\).

It is very interesting to calculate exactly where the meeting of the cages in the shaft will take place.

To solve this problem, we shall reason as follows.

When the bobbin shaft is turned through some angle \(\varphi\), the initial radius of the bobbin will become \(r_1\), so that one may write

\[ r_1=r+\frac{d}{2\pi}\varphi . \]

From the other bobbin the rope will be unwound, so that the initial radius \(R\) will decrease and become \(R_1\); in this case

\[ R_1=R-\frac{d}{2\pi}\,\varphi . \]

Adding these equalities, we obtain \(r_1+R_1=R+r=\mathrm{Const}=2\rho\), where \(\rho\) represents the mean radius of the winding.

Let us write the last equation in the following form

\[ R_1-\rho=\rho-r_1=x \]

We shall show that at the moment when the cages meet, \(R_1=r_1=\rho\).

If we look at the rope ribbon wound in a spiral on the bobbin, then, with ribbon thickness \(d\) and its length \(H\), the area covered by it will be approximately equal to \(Hd\). On the other hand, this area will be approximately equal to \(\pi(R^2-r^2)\).

For any two arbitrary depths \(H_1\) and \(H_2\), or lengths of the rope ribbon, one may write analogously

\[ H_1d=\pi(R^2-y_1^2), \]
\[ H_2d=\pi(R^2-y_2^2), \]

where \(y_1\) and \(y_2\) will be the variable radii of the winding.

At the moment when the cages meet, \(H_1=H_2\), and this condition leads to \(y_1=y_2\), or \(R_1=r_1\). But \(R_1+r_1=2\rho\). Consequently,

\[ R_1=r_1=\rho \]

Thus, at the moment when the cages meet, the outer radius of unwinding and winding of the ribbon on the bobbins will be one and the same, namely

\[ \frac{R+r}{2}=\rho \]

Now one may write the following relation

\[ \frac{x\cdot d}{H\cdot d}=\frac{\pi(R^2-\rho^2)}{\pi(R^2-r^2)} \]

Whence one may write

\[ X=H\frac{R^2-\rho^2}{R^2-r^2} =H\frac{R^2-\rho^2}{(R+r)(R-r)} \]

But \(R+r=2\rho\); \(R-r=R-(2\rho-R)=2(R-\rho)\).

Substituting the found values for the sum and difference of the radii, we shall have

\[ X=H\frac{R^2-\rho^2}{4\rho(R-\rho)} =\frac{H}{4}\left(1+\frac{R}{\rho}\right) \]

By this formula the place where the cages meet is also determined. Since \(R>\rho\), the place where the cages meet will be below half the depth of the shaft.

Let us now trace how the magnitude of the moment of resistance on the bobbin shaft changes during the whole working stroke.

The condition of equality of the moments of resistance at the beginning and at the end of the lift was expressed analytically by the equation

\[ (N+2L)R=(N+2L+2gH)r \]

Dividing the right- and left-hand sides of this equality by 2, we obtain

\[ \left(\frac{N}{2}+L\right)R=\left(\frac{N}{2}+L+gH\right)r \]

These equalities will not be violated if to the right- and left-hand sides we add

\[ \frac{rN}{2}-LR \]

After collecting like terms and reducing, we obtain

\[ r(N+L+gH)-LR=\frac{N(R+r)}{2}=N\rho \]

The right-hand side of this equality expresses the resisting moment during lifting at the point where the cages meet. Thus, assuming the condition of equality of the moments at the beginning and at the end of the lift to be fulfilled, we at the same time obtain that at the moment when the cages meet the same resisting moment is obtained.

\[ M_{\text{beginning}} = M_{\text{meeting}} = M_{\text{end}} \]

The same conclusion may be reached in a somewhat different way.

Let us write the equations of the resisting moments at the beginning and at the end of the lift, and add them:

\[ \begin{aligned} M_{\text{beginning}} &= (N+L+gH)r - LR,\\ M_{\text{end}} &= (N+L)R - (L+gH)r,\\ \hline M_{\text{beginning}} + M_{\text{end}} &= N(R+r)=2N\rho . \end{aligned} \]

If \(M_{\text{beg.}} = M_{\text{end}}\), then each of them is equal to \(N\rho\), i.e. is equal to the twisting moment at the instant when the cages meet.

The nature of the change in the magnitude of the twisting moment during the entire lift can best be studied either analytically or graphically.

Let us dwell on the first method.

Let us write an expression for the magnitude of the twisting moment at some instant, assuming that on one bobbin this instant corresponds to the winding radius \(Y_1\), and on the other to the radius \(Y_2\). Let the ends of the ropes on which the cages hang have weights \(gH_1\) and \(gH_2\). In that case the variable magnitude of the twisting moment will be

\[ M=(N+L+gH_1)Y_1-(L+gH_2)Y_2 . \]

In the second right-hand part of this equality, or more precisely equation, the variables are \(Y_1\), \(Y_2\), \(H_1\), and \(H_2\). But we shall immediately express them through a single quantity, and then \(M\) will be a function of one independent variable.

Indeed, we have already had the following two equations:

\[ \begin{aligned} H_1 d &= \pi(R^2-Y_1^2),\\ H_2 d &= \pi(R^2-Y_2^2). \end{aligned} \]

Let us add to these equations also the equality

\[ Hd=\pi(R^2-r^2). \]

From these relations we obtain without difficulty

\[ \frac{H_1}{H}=\frac{R^2-Y_1^2}{R^2-r^2}; \qquad \frac{H_2}{H}=\frac{R^2-Y_2^2}{R^2-r^2}; \]

or

\[ gH_1=gH\frac{R^2-Y_1^2}{R^2-r^2}; \qquad gH_2=gH\frac{R^2-Y_2^2}{R^2-r^2}. \]

If these values of the weights of the rope ends are substituted into the basic equation for the twisting moment, we obtain the dependence of \(M\) only on the two quantities \(Y_1\) and \(Y_2\):

\[ M= \left[ N+L+gH\frac{(R^2-Y_1^2)}{R^2-r^2} \right]Y_1 - \left[ L+gH\frac{(R^2-Y_2^2)}{R^2-r^2} \right]Y_2 . \]

But

\[ Y_1+Y_2=R+r=2\rho. \]

Or otherwise,

\[ Y_2-\rho=\rho-Y_1=X. \]

Whence we obtain

\[ Y_2=\rho+X \quad \text{and} \quad Y_1=\rho-X . \]

Substituting these values of \(Y_2\) and \(Y_1\) into the equation for the magnitude of the twisting moment,

\[ M=\{N+L+m[R^2-(\rho-x)^2]\}(\rho-x) -\{L+m[R^2-(\rho+x)^2]\}(\rho+x), \]

where

\[ m=\frac{gH}{R^2-r^2}. \]

Thus, the magnitude of the twisting moment is a function of only one variable, \(x\), and therefore is easily investigated with respect to its magnitude. Expanding the brackets and collecting like terms, we obtain

\[ M=\frac{2gH}{R^2-r^2}x^3-\left(N+2L+\frac{2gHR^2}{R^2-r^2}-\frac{6gH}{R^2-r^2}\rho^2\right)x+N\rho \]

Thus, the law of variation of the magnitude of the twisting moment is expressed by a curve of the third order, in whose equation there is no term with \(x^2\).

By the general rules of differential calculus we shall at once find the conditions for maxima and minima of the function.

\[ \frac{dM}{dx}=\frac{6gH}{R^2-r^2}x^2-\left(N+2L+\frac{2gHR^2}{R^2-r^2}-\frac{6gH\rho^2}{R^2-r^2}\right)=0 \]

\[ \frac{d^2M}{dx^2}=\frac{12gHx}{R^2-r^2} \]

The values \(X\) which make our function a maximum and a minimum have the following form:

\[ x=\pm\sqrt{\frac{(N+2L)(R^2-r^2)}{6gH}+\frac{R^2}{3}-\rho^2} \]

It is not difficult to construct the curve of variation of the magnitude of the twisting moment, taking the values \(X\) as ordinates or abscissae, and the values \(M\) as abscissae or ordinates; moreover, three points are determined at once from the equation for the magnitude of the moment by putting \(X=0\) in this equation.

Indeed, carrying out this operation, we obtain \(M=N\rho\), and \(N\rho\) is the magnitude of the resisting moment at the meeting of the cages. But we have already seen that this moment is equal to the moment at the beginning and at the end of the hoist.

Thus, three points of the curve are determined, and they are situated on a horizontal line parallel to the axis of \(X\)’s, at a distance from the axis of \(X\)’s equal to \(N\rho\).

The abscissae of these points will be

\[ x_1=0;\quad x_2=\frac{H}{4}\left(\left(1+\frac{R}{\rho}\right)\right)\quad \text{and}\quad x_3=H. \]

If the vertical passing through the end of the abscissa \(X_2\) is taken as the dividing line, then the branches of the curve will be unequal, and hence the mean speeds of travel of the descending cage and the ascending cage will also be different. The ascending cage will move with a lower mean speed than the descending one.

Example. \(N=600\) kg, \(L=700\) kg, \(gH=550\) kg, \(H=400\) meters, \(r=0.90\) meter, \(R=1.52\) meters, \(\delta=1.1\) millimeters, \(d=10.5\) millimeters, and \(n=36\) wires.

Substituting these numerical values into the equation for the moment \(M\), we obtain

\[ M=733.33x^3-473.86x+726 \]

This result is obtained after the following calculations:

\[ 2gH=1100;\quad R^2-r^2=1.5;\quad 1100:1.5=733.33 \]

\[ N+2L+\frac{2gN}{R^2-r^2}\cdot R^2=600+1400+733.33\times2.31=2000+1693.99= \]

\[ =3694;\quad \frac{6gH}{R^2-r^2}\rho^2=2199.99\cdot1.464=\sim 3220.8 \]

By the general method of differential calculus we find the maximum and minimum of the function \(M\), regarding \(x\) as the independent variable.

\[ \frac{dM}{dx}=2200x^2-473.86=0 \]

Whence we obtain for \(x\) two values

\[ x=\pm 0.464 \]

The second derivative is written as

\[ \frac{d^{2}M}{dx^{2}} = 4400x \]

Substituting into the second derivative the value of \(x\) from the first derivative, we obtain for max

\[ \frac{d^{2}M}{dx^{2}} = -2041.6 \]

and for min

\[ \frac{d^{2}M}{dx^{2}} = +2041.6 \]

The value \(M\) itself for the one and the other case will have the following values

\[ M_{\max} = \sim 871.7\ \mathrm{kg \cdot mtr.} \]

\[ M_{\min} = \sim 580.28\ \mathrm{kg \cdot mtr.} \]

The fluctuations of the magnitude of the moment of resistance from the magnitude of the same moment at the beginning and at the end of the hoisting amount to

\[ +20\% \text{ and } -20\% \]

The place where the cages meet will be at a depth from the surface of the earth

\[ x_{0} = \frac{H}{4}\left(1+\frac{R}{\rho}\right) = 225.5\ \mathrm{mtr.} \]

Let us carry out the solution of this same problem again in general form, by a somewhat different method. Let \(Y_{1}\) be the variable radius of winding on one bobbin at the moment when the ascending cage is at a height \(H_{1}\) from the bottom of the shaft. For the descending cage we shall denote the same quantities by \(Y_{2}\) and \(H_{2}\). Then one may write

\[ Y_{1} = r + d\frac{\varphi}{2\pi} \quad \text{and} \quad H_{1} = r\varphi + \frac{d\varphi^{2}}{4\pi} \]

\[ Y_{2} = R - d\frac{\varphi}{2\pi} \quad \text{and} \quad H_{2} = R\varphi - \frac{d\varphi^{2}}{4\pi} \]

Whence it follows that

\[ Y_{1} + Y_{2} = r + R \quad \text{and} \quad H_{1} + H_{2} = (r+R)\varphi . \]

The distance between the cages will be equal to

\[ H - (H_{1}+H_{2}) = H - (r+R)\varphi \]

At the moment when the cages meet this distance will be equal to zero, and therefore

\[ \varphi = \frac{H}{r+R} \]

This angle will also correspond to the meeting of the cages when \(H_{1}=H_{2}\). But if so, and

\[ r\varphi + d\frac{\varphi^{2}}{4\pi} = R\varphi - \frac{d\varphi^{2}}{4\pi}, \]

or \(Y_{1}=Y_{2}\). From this equality

\[ Y_{1}=Y_{2} \]

it follows that

\[ r + \frac{d\varphi}{2\pi} = R - \frac{d\varphi}{2\pi}; \quad \varphi = \frac{\pi(R-r)}{d}. \]

Comparing the angles \(\varphi\) by the one and by the other formula, we obtain

\[ \frac{H}{r+R} = \frac{\pi(R-r)}{d} \quad \text{or} \quad \frac{dH}{\pi} = R^{2}-r^{2} \]

Let us denote the radius vector of the Archimedean spiral at the moment of meeting by \(\rho\). Then one may write

\[ \rho = \frac{r+R}{2} \]

Let us now express the variable magnitude of the radius vectors \(Y_{1}\) and \(Y_{2}\) in terms of \(\rho\). At the moment when the cages meet, both radii \(r\) and \(R\) turn into \(\rho\), and therefore one may write

\[ Y_{1} = \rho + \frac{d}{2\pi}\omega \quad \text{and} \quad Y_{2} = \rho - \frac{d\omega}{2\pi} \]

The magnitude of the new angle \(\omega\) may vary within the limits from \(-\dfrac{H}{2\rho}\) to \(\dfrac{H}{2\rho}\).

The total angle \(\Theta\) through which the bobbin shaft turns during the whole lifting period can now be expressed as

\[ \Theta=\omega+\frac{H}{2\rho}. \]

The length of the part of the rope wound onto the bobbin can now be expressed by means of the following general formulae

\[ H_1=\rho\omega+\frac{d}{4\pi}\omega^2+C_1 \]

\[ H_2=\rho\omega-\frac{d}{4\pi}\omega^2+C_2 \]

The constant quantities \(C_1\) and \(C_2\) are determined from the condition that, for

\[ \omega=-\frac{H}{2\rho}, \]

the quantities \(H_1\) and \(H_2\) are equal to zero.

Then one may write

\[ H_1=\rho\omega+\frac{d}{4\pi}\omega^2+\frac{H}{2}-\frac{dH^2}{16\pi\rho^2} \]

\[ H_2=\rho\omega-\frac{d}{4\pi}\omega^2+\frac{H}{2}+\frac{dH^2}{16\pi\rho^2} \]

At any instant of lifting, the variable magnitude of the twisting moment can be expressed in general form as:

\[ M=[N+L+g(H-H_1)]Y_1-[L+gH_2]Y_2. \]

Substituting into this equation the values of \(H_1\) and \(H_2\), \(Y_1\) and \(Y_2\), as functions of \(\rho\), we then obtain the following complicated expression

\[ M=\left[N+L+g\left(H-\rho\omega-\frac{d\omega^2}{4\pi}-\frac{H}{2}+\frac{dH^2}{16\pi^2\rho^2}\right)\right]\left(\rho+\frac{d\omega}{2\pi}\right)+ \]

\[ -\left[L+g\left(\rho\omega-\frac{d\omega^2}{4\pi}+\frac{H}{2}+\frac{dH^2}{16\pi^2\rho^2}\right)\right]\left(\rho-\frac{d\omega}{2\pi}\right) \]

Opening the brackets and combining like terms, we obtain a considerably simplified equation

\[ M-\rho N=\omega\left[\left(N+2L+gH\right)\frac{d}{2\pi}-2g\rho^2+\frac{gd^2H^2}{16\pi^2\rho^2}-\frac{gd^2}{4\pi^2}\omega^2\right]. \]

The right-hand side of this equation becomes zero, on the one hand, when \(\omega=0\), and on the other hand, for two values of \(\omega\) equal in magnitude and opposite in sign, which are obtained after equating to zero the polynomial enclosed in square brackets.

The first condition \(\omega=0\) gives us the magnitude of the moment when the cages meet:

\[ M=N\rho. \]

The second condition gives us the deviations from the mean moment in the direction of increase and in the direction of decrease.

Turning to the expression standing in square brackets, it is not difficult to observe that the choice of the numerical magnitudes of the separate terms and their components, though it allows freedom, does so only to a limited degree. In practice, \(N\), \(L\), \(g\), \(H\), and \(d\) may be regarded as sufficiently constant in magnitude, and then there remain at our disposal two quantities with a more considerable amplitude of variation: \(\rho\) and \(\omega\).

It is necessary to choose \(\rho\) by calculation in such a way that the two values of \(\omega\) which make the polynomial in square brackets equal to zero correspond to real conditions. In this way all imaginary solutions are rejected. The second still bo—

where the restriction of the choice of \(\rho\) by the condition consisting in the fact that, for all values of \(\omega\) within the limits

\[ \text{from }-\frac{H}{2\rho}\text{ to }+\frac{H}{2\rho}, \]

the quantity

\[ M-N\rho \]

should be as small as possible. If it is desired that \(M-N\rho\) at the beginning and at the end of the ascent be zero, then one must choose for \(\rho\) such a value that would make the polynomial in the square brackets vanish at

\[ \omega=+\frac{H}{2\rho}\quad \text{or}\quad \omega^2=\frac{H^2}{4\rho^2}. \]

Using this consideration, we write

\[ \left(N+2L+gH\right)\frac{d}{2\pi}-2g\rho^2+\frac{gd^2}{4\pi^2}-\frac{gd^2}{4\pi^2}\cdot\frac{H^2}{4\rho^2}=0. \]

Hence we obtain

\[ \rho=\frac{1}{2}\sqrt{\frac{\left(N+2L+gH\right)d}{\pi g}}. \]

The value of the left-hand side of the equation will then be

\[ M-N\rho=\frac{gd^2}{4\pi^2}\left(\frac{H^2}{4\rho^2}-\omega^2\right)\omega. \]

In the final result there was obtained the same cubic equation with the variable quantity \(\omega\).

Let us analyze when the values of \(M-N\rho\) assume the greatest and the least magnitude as functions of \(\omega\).

The rules for finding max and min give us the right to write

\[ \frac{d(M-N\rho)}{d\omega}=\frac{H^2}{4\rho^2}-3\omega^2=0. \]

Hence we obtain

\[ \omega=\pm\frac{H}{2\rho}\sqrt{\frac{1}{3}}. \]

Under this condition the values of \(M-N\rho\) will be equal in magnitude but opposite in sign:

\[ (M-N\rho)_{\max}=\frac{gd^2\sqrt{3}}{144\pi^2}\frac{H^3}{\rho^3}, \]

\[ (M-N\rho)_{\min}=-\frac{gd^2\sqrt{3}}{144\pi^2}\frac{H^3}{\rho^3}. \]

These expressions represent the measure of the greatest deviation of the resisting moment in the direction of increase and in the direction of decrease from \(N\rho\).

We see that \(M-N\rho\) becomes zero at

\[ \omega=-\frac{H}{2\rho}\quad \text{or}\quad \theta=0, \]

and remains negative for all negative values of \(\omega\).

This same difference \(M-N\rho\) reaches its greatest value at

\[ \omega=-\frac{H}{2\rho}\sqrt{\frac{1}{3}}\quad \text{or when}\quad \theta=\frac{H}{2\rho}\left(1-\sqrt{\frac{1}{3}}\right). \]

The decrease of the difference \(M-N\rho\) continues up to the moment corresponding to

\[ \omega=0\quad \text{or}\quad \theta=\frac{H}{2\rho}. \]

when \(M-N\rho\) again becomes zero. This instant corresponds to the meeting of the coils.

After this, \(\omega\) acquires a positive value and \(M\) becomes greater than \(N\rho\), continuing to increase up to the value corresponding to

\[ \omega=\frac{H}{2\rho}\sqrt{\frac{1}{3}} \quad \text{or} \quad \theta=\frac{H}{2\rho}\left(1+\sqrt{\frac{1}{3}}\right). \]

Having reached its greatest magnitude, the difference \(M-N\rho\) begins to decrease and becomes zero when

\[ \omega=\frac{H}{2\rho} \quad \text{or} \quad \theta=\frac{H}{\rho}. \]

Imposing the condition that \(M=N\rho\), it can be shown that the value of \(\omega\) lies between the following limits

\[ \omega_1=-\frac{H}{2m\rho} \quad \text{and} \quad \omega_2=\frac{H}{2m\rho}, \]

where \(m\) is a positive number greater than unity.

Let us write the general equation for the moment of resistance (it is also the twisting moment of the main shaft)

\[ \left[(N+2L+gH)\frac{d}{2\pi}-2g\rho^2+\frac{g d^2 H^2}{16\pi^2\rho^2}-\frac{g d^2}{4\pi^2}\omega^2\right]\omega=M-N\rho. \]

Having the condition that \(M-N\rho=0\), let us substitute into the left-hand side of the general equation the value

\[ \omega^2=\frac{H^2}{4m^2\rho^2}. \]

Then we obtain

\[ (N+2L+gH)\frac{d}{2\pi}-2g\rho^2+\frac{g d^2(m^2-1)}{4\pi^2}\cdot\frac{H^2}{4m^2\rho^2}=0. \]

This equation is biquadratic with respect to \(\rho\). The solution of this equation gives four roots for \(\rho\):

\[ \rho=\pm\sqrt{\frac{(N+2L+gH)d}{8\pi g}\pm \sqrt{\frac{(N+2L+gH)^2d^2}{64\pi^2g^2} +\frac{d^2H^2(m^2-1)}{32\pi^2m^2}}}. \]

Since \(m>1\), the two imaginary roots immediately fall away. Likewise the real root with the negative sign also falls away.

Thus only one root remains:

\[ \rho= \sqrt{\frac{(N+2L+gH)d}{8\pi g}+ \sqrt{\frac{(N+2L+gH)^2d^2}{64\pi^2g^2} +\frac{d^2H^2(m^2-1)}{32\pi^2m^2}}}. \]

This expression is to be used in determining the mean radius of the winding of the rope on the bobbin.

Knowing its magnitude, one can then determine \(R\) and \(r\) without difficulty.

Indeed,

\[ R+r=2\rho; \quad \text{and} \quad \pi(R^2-r^2)=dH. \]

From this we obtain

\[ R=\frac{4\pi\rho^2+dH}{4\pi\rho} \quad \text{and} \quad r=\frac{4\pi\rho^2-dH}{4\pi\rho}. \]

To facilitate finding \(\rho\) by means of the complicated formula, let us proceed as follows. Let us write the biquadratic equation without denominators:

\[ 128\pi^2g\rho^4-32\pi d(N+2L+gH)\rho^2-gd^2H^2=0. \]

Further,

\[ R=\rho+\frac{dH}{4\pi\rho} \quad \text{and} \quad r=\rho-\frac{dH}{4\pi\rho}. \]

Let us introduce a new unknown

\[ x=\frac{4\pi\rho^{2}}{dH}\quad\text{and}\quad \rho=\sqrt{\frac{dHx}{4\pi}}. \]

If this value of \(\rho\) is substituted into the biquadratic equation, it will assume the very simple form

\[ x^{2}-\frac{(N+2L+gH)}{gH}x-\frac{1}{8}=0. \]

The values of the largest and smallest radii of winding will be

\[ R=\left(1+\frac{1}{x}\right)\rho \quad\text{and}\quad r=\left(1-\frac{1}{x}\right)\rho . \]

In the simplified equation the thickness of the rope \(d\) is not present in explicit form, but this thickness enters into the expression for determining \(\rho\), depending on whose magnitude the values of \(R\) and \(r\) vary proportionally to the square root of \(d\).

It is now not difficult to see that \(R\) and \(r\) will come out larger in calculations for round ropes, with which flat ropes were formerly replaced, by choosing narrower bobbins. This sort of replacement of a flat rope or belt is also justified in our time by the consideration that round ropes take up the stress more uniformly and therefore are more durable in operation.

The equivalent round section of the rope will, of course, have a larger diameter than the thickness of the belt.

From the quadratic equation it is further evident that the positive root for \(x\) increases as the coefficient

\[ \frac{N+2L}{gH}+1 \]

increases.

In other words, it grows proportionally to the useful load plus double the dead load and inversely proportionally to the weight of a linear meter of rope \(g\). We do not yet take into account the depth of the shaft \(H\).

Let us now trace analytically the course of variation of \(r\) as a function of \(x\). The small radius of winding \(r\) is determined as follows:

\[ r=\sqrt{\frac{dH}{4\pi}}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right). \]

Hence it is quite obvious that the magnitude \(r\) grows together with the growth of \(x\).

In order to reveal the influence of the shaft depth \(H\) on the magnitude of \(x\), and therefore also on the radii \(R\) and \(r\), we write the quadratic equation

\[ x^{2}-\frac{(N+2L+gH)}{gH}x-\frac{1}{8}=0, \]

in the form

\[ H=\frac{N+2L}{g}\cdot\frac{x}{x^{2}-x-\frac{1}{8}} =\frac{N+2L}{g}\cdot\frac{1}{x-1-\frac{1}{8x}}. \]

In this case one may write that

\[ \rho=\sqrt{\frac{(N+2L)d}{4\pi g}}\cdot \frac{x}{\sqrt{x^{2}-x-\frac{1}{8}}}, \]

\[ r=\sqrt{\frac{(N+2L)d}{4\pi g}}\cdot \frac{x-1}{\sqrt{x^{2}-x-\frac{1}{8}}}. \]

In order that \(r\) be real, it is necessary that the condition be satisfied

\[ x>\frac{1}{2}\left(1+\sqrt{\frac{3}{2}}\right), \quad \text{i.e. } x>1.1123 \]

From the expression for \(H\) it is seen that, for \(x=\infty\), \(H=0\), while as \(x\) decreases the quantity \(H\) will increase.

By differentiating \(r\) with respect to \(x\) we obtain

\[ \frac{dr}{dx} = \sqrt{\frac{(N+2L)d}{4\pi g}} \cdot \frac{x-\frac{5}{4}}{2\left(x^{2}-x-\frac{1}{8}\right)^{3/2}} . \]

It follows from this that, so long as \(x\) is greater than \(\frac{5}{4}\), the derivative remains positive. And this circumstance suggests the conclusion that \(r\) changes in the same way as \(x\), but inversely to the change of \(H\).

For values of \(x\) in the interval \(1.25—1.11\), the conclusion will be the opposite.

If \(x=\infty\), then \(H=0\) and

\[ r=\sqrt{\frac{(N+2L)d}{4\pi g}} . \]

And hence also the converse conclusion: with a decrease of the radius \(r\), the depth \(H\) will increase.

This radius \(r_{11}\) reaches its minimum value at \(x_{11}=\frac{5}{4}\), for a depth

\[ H_{11}=\frac{20}{3}\frac{(N+2L)}{g}; \qquad r_{11}=\sqrt{\frac{(N+2L)d}{12\pi g}} . \]

The ratio

\[ r_{11}:r=1:\sqrt{3}=0.5773 . \]

From this moment the radius \(r\) begins to increase together with the depth, and at \(x_{111}=1.11\) the quantities \(H_{111}=\infty\) and \(r_{111}=\infty\).

The old French rule, expressed by Cabany’s formula, gives for determining \(H_{11}\), corresponding to the minimum radius \(r_{11}\), the following dependence

\[ N+L+gH=gh . \]

Hence we obtain

\[ g=\frac{N+L}{h-H} \quad \text{and} \quad H_{11}=\frac{20}{3}\frac{N+2L}{N+L}(h-H). \]

If it is desired to know the condition under which \(r_{11}\) does not reach its minimum, then one must turn to the inequality

\[ \frac{20}{3}\frac{N+2L}{N+L}(h-H)\geq H, \quad \text{or} \]

\[ \frac{H}{h}\leq \frac{20}{3} + \frac{60}{989+529\frac{N}{L}} . \]

The right-hand side in this inequality reaches its minimum at \(L:N=0\). In this case one may practically put

\[ \frac{H}{h}=\frac{20}{23}. \]

Substituting this ratio into Cabany’s formula, we obtain

\[ N+L=g(h-H)=\frac{3}{20}gH . \]

In other words, the weight of the rope \(gH\) will be \(\frac{20}{3}\) times greater than \(N+L\), which is not allowed.

Hence it is evident that \(r_{11}\) does not in practice reach its \(\min\), and the initial winding radius \(r\) will always decrease with the depth \(H\), or rather with increasing \(H\).

For moderate depths the magnitude of the change becomes scarcely noticeable even for considerable values of \(H_{11}\).

This is clear from the following table.

\(\dfrac{H}{H_{11}}\)	\(\dfrac{r}{r_{11}}\)	\(x\)
0	1	\(\infty\)
0.16	0.73	2
0.36	0.63	\(3/2\)
0.53	0.60	\(11/8\)
1.00	0.58	\(5/4\)

Let us now trace the variation of \(R\).

For this purpose we write the value of \(R\)

\[ R=\sqrt{\frac{dH}{4\pi}}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) \]

If one begins the change of \(x\) from infinity, then the value of \(R\) will decrease in magnitude; as \(x\) decreases, the value of \(R\) reaches a minimum:

\[ \frac{dR}{dx}=\frac{1}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}=0;\quad x=1. \]

But \(x\) cannot be less than \(1.11\). Consequently, however we change \(R\) in the direction of decreasing \(x\), it should never reach unity.

Hence it follows that \(R\) changes in one direction with \(x\), i.e. its magnitude will be proportional to \(N\) and inversely proportional to \(g\).

Let us now determine the influence of the depth \(H\) on the change in the magnitude of \(R\).

\[ R=\sqrt{\frac{(N+2L)d}{4\pi g}}\cdot \frac{x+1}{\sqrt{x^{2}-x-\frac18}} \]

\[ \frac{dR}{dx}= \frac{3\left(\frac14-x\right)} {2\left(x^{2}-x-\frac18\right)^{3/2}} \sqrt{\frac{(N+2L)d}{4\pi g}} \]

Since the minimum value \(X=1.11\), the obtained value of the derivative will be negative.

Thus, the magnitude of the radius \(R\) changes oppositely to the change in \(X\), but in the same way as \(H\). Hence it follows that the magnitude of the shaft depth affects the magnitudes of the radii \(r\) and \(R\) in opposite directions.

Let us now investigate the nature of the component \(m\), by means of which the limits of variation for \(\omega\) are established.

Let us write the general expression for the resisting moment

\[ M-N\rho=\frac{gd^{2}}{4\pi^{2}}\left(\frac{H^{2}}{4\rho^{2}}-\omega^{2}\right)\omega. \]

In this expression the quantity \(\omega\) oscillates within the limits

\[ -\frac{H}{2\rho}\ \text{to}\ +\frac{H}{2\rho}. \]

Under the new conditions, when the quantity \(\omega\) oscillates within the limits

\[ -\frac{H}{2m\rho}\ \text{to}\ +\frac{H}{2m\rho}, \]

the expression for the moment of resistance will be written as follows

\[ M-N\rho=\frac{g d^{2}}{4\pi^{2}}\left(\frac{H^{2}}{4m^{2}\rho^{2}}-\omega^{2}\right)\omega . \]

The quantity \(M-N\rho\) attains a max or min under the condition

\[ \omega=\pm \frac{H}{2m\rho}\sqrt{\frac{1}{3}} . \]

Substituting this value of \(\omega\) into the equation of the moment, we obtain

\[ M-N\rho=\mp \frac{g d^{2}\sqrt{3}\,H^{3}}{144\pi^{2}m^{3}\rho^{3}} . \]

From this it becomes clear that the value of \(\rho\) increases with increasing \(m\). As for the change of the moment \(M\) itself for values of \(\omega\) within the limits

\[ \text{from }-\frac{H}{2m\rho}\text{ to }+\frac{H}{2m\rho}, \]

the magnitude of this moment decreases as \(m\) increases; but more rapidly than the cube of the fraction \(\frac{1}{m}\).

Let us now determine how \(M-N\rho\) will vary when \(\omega\) varies within the limits

\[ -\frac{H}{2\rho}\text{ to }-\frac{H}{2\rho m}, \]

on the one hand, and, on the other, when \(\omega\) varies within the limits

\[ +\frac{H}{2\rho m}\text{ to }+\frac{H}{2\rho}. \]

For solving this problem it is simplest to turn to the equation

\[ M-N\rho=\pm \frac{g d^{2}}{4\pi^{2}}\left(\frac{1}{8m^{2}}-\frac{1}{8}\right)\frac{H^{3}}{\rho^{3}} . \]

Let us determine \(m\) in such a calculation that the deviation of \(M\) from its mean value should not exceed the maximum deviation of \(M\) for values of \(\omega\) lying between

\[ -\frac{H}{2\rho m}\text{ and }+\frac{H}{2\rho m}. \]

For this it is sufficient to satisfy the condition

\[ \left(\frac{1}{8}-\frac{1}{8m^{2}}\right)\leq \frac{\sqrt{3}}{18}\frac{1}{m^{3}} . \]

This inequality makes it possible to find the greatest value of \(m\) from the equation

\[ m^{3}-m-0.3849=0 . \]

With a sufficient degree of accuracy one may put \(m=1.155\sim 1.16\) (see S. K. Konyukhov, Finding the Real Root of an Equation of Any Degree by a Graphical Method).

Then we obtain

\[ (m-1.155)\left(m^{2}+1.155m+0.334\right)=0 . \]

The other two roots are discarded, as being less than positive unity and therefore not satisfying the condition of the problem.

And so,

\[ m_{\max}=1.155 . \]

In the theory analyzed here, harmful resistances have deliberately not been taken into account, even in the simplified form proposed by Tauroj. The form of the formulas then becomes somewhat more complicated. But the question will become still more complicated if the forces of inertia are taken into account, using d’Alembert’s principle.

This should form the subject of a subsequent article on the same topic.

For the article by S. K. Konyukhov, “Theory of the Bobbin”

[Figure: geometric diagram with labeled points \(O\), \(A\), \(B\), \(M\), \(M_1\); angles \(\varphi\) and \(d\varphi\) are marked.]

It remains only to add that the theory of the bobbin is very simply applied to the theory of the conical drum.

Indeed, if we denote by \(\beta\) the angle made by the generatrix of the truncated conical drum with the axis of the drum, then the equation of the Archimedean spiral will take the following form:

\[ x = r + \frac{d \operatorname{Sm}\beta}{2\pi}\,\varphi \]

instead of

\[ x = r + \frac{d}{2\pi}\,\varphi . \]

Example. \(H = 550\) mm, \(N = 1600\), \(L = 2670\), \(S = 2200\), \(g = 4\) kg; carding ribbon \(= 80 \times 18\) mm; \(d = 18\) mm. Determine \(\rho\).

\[ x^2 - \frac{1600 + 5340 + 2200}{2200}\,x - 0.125 = 0 \]

\[ x^2 - 4.16x - 0.125 = 0;\quad x = 4.39 \]

\[ \rho = \sqrt{\frac{dHx}{4\pi}} = \sqrt{\frac{550 \cdot 0.018 \cdot 4.39}{12.56}} = \sim 1.89 . \]

Determine \(r\) and \(R\).

\[ r = \sqrt{\frac{dHx}{4\pi}} = \rho\left(1 - \frac{1}{x}\right) = 1.89\left(1 - \frac{1}{4.39}\right) \]

\[ = 1.89 \cdot 0.77 = 1.455 \sim 1.46\ \text{mt.} \]

\[ R = \rho\left(1 + \frac{1}{x}\right) = 1.89(1 + 0.228) = 2.31\ \text{mt.} \]

This result of the calculations may be considered quite satisfactory.

Indeed, the maximum and minimum deviation of \(M\) will be

\[ \frac{g d^2}{4\pi^2}\sqrt{\frac{3H^3}{36 \cdot \rho^3}} = \frac{4 \cdot 0.018^2 \cdot 1.73 \cdot 550^3}{4 \cdot 9.86 \cdot 36 \cdot 1.89^3} = \sim 464\ \text{kg·mt} \]

The equation for determining the moment, after the corresponding substitutions and reductions, will take the following form:

\[ 1375x^3 - 31.25x + 3024 = M \]

The value \(x\), determining the max. and min. of this function, will be

\[ \frac{dM}{dx} = 4125x^2 - 31.25 = 0 \]

\[ x = \pm 0.76 . \]

Substituting this value of \(x\) into the expression for \(M\), we obtain

\[ M_{\max} = 3603.88\ \text{kg mt} \]

\[ M_{\min} = 2444.75\ \text{kg mt} \]

The deviation in the first case is \(13\%\), and in the second \(19\%\).

But this result can be further improved if the coefficient \(m = 1.16\) is introduced into the calculation.

Then the deviation in the first case will amount to only about \(8.3\%\), and in the second \(12.2\%\). The value of \(\rho\) for \(m = 1.16\) will be equal to

\[ \rho = 2.19\ \text{mtr};\quad r = 1.69\quad \text{and}\quad R = 2.68\ \text{mm}. \]

Such is the simplified static theory of the bobbin.

In writing this work I tried to make use of all the technical literature on this question, but chiefly I used the works of Combe, Tagan de la Gutilière, Kondrat’ev, etc., introducing corrections for accidental errors in the works of these talented people.