Engineer-technician S. F. Lebedev.

The Standard Safety Catch of Siberian Cages from the Standpoint of Hoisting Safety

(With 2 pages of drawings.)

A safety catch is an extremely widespread accessory of hoisting cages used in mining. Its purpose is to stop the hoisting cage in the event of a break in the hoisting rope and thereby to prevent the death or maiming of a whole number of persons who at that moment might be in the cage (the lowering and raising of workers and employees); but, in view of the fact that almost always one and the same cage serves for both passenger and freight trips, it is now customary to install a safety catch on all cages. Nevertheless, the safety catch has been very little studied theoretically, and especially in Russian literature. Therefore, in practice one happens to see safety catches which will never operate under the required circumstances and, by their weight, constitute only an unnecessary load on the rope. Here I shall attempt to give a more or less exact characterization of the safety catch with the direct action of a spring, which is the most widespread in the Siberian mines.

I. The Idea of the Safety Catch

Schematically, the safety catch is shown in Fig. 1. The hoisting rope is attached to bolt A, which, by means of levers D, transmits rotation to two shafts arranged across the cage. The ends of these shafts carry stops K—the latter, during operation, will be pressed against the shaft guides. The same bolt A serves as the axis of a helical spring, which, with one plane of its cut, bears against the upper fastening of the cage, and with the other against the washer of the bolt.

In operation the hoisting rope acts on bolt A and compresses the spring. At the same time the rods B rotate the shafts and move the stops away from the shaft guides. In the event of a rope break, bolt A is drawn downward by the spring; rods B rotate the shafts of the stops and press the latter against the guides. The friction that develops stops the cage. Such, in idea, is also the safety catch in Fig. 2, with one structural difference: the spring has, as its axis, the same shaft on which the stop is mounted, whereby the need is eliminated for an articulated connection of this shaft with bolt A; it is connected by chain B. In that and the other safety catch the hoisting rope acts on the spring only until the supporting chains (at the corners of the cage) are tightened.

II. Excitation of the Force of the Safety Catch

From this description it is clear that the active force of the safety catch is the spring, and, of course, it is also clear that the force of gravity (the weight of the load) cannot, and must not, be used here. Let us imagine that a load had been taken instead of the spring. Then, when the rope broke, the cage and the safety catch connected with it, together with the load, would be in a state of free fall with acceleration “g” (roughly speaking); the relative position of the parts, in particular of the load as well, would not change at all, and, consequently, the stops of the safety catch would hang freely in the air in complete inactivity. I dwell on this question because up to now safety catches are encountered in which the role of the exciting force is assigned to the load, and the designers of such safety catches, evidently, are not yet free from error.

III. Details of the safety catch.

According to diagram 1, the safety catch consists of the following parts: 1. the rod (or bolt) of the safety catch “A,” 2. two rods “B” (connecting bolt “A” with the shafts of the stops), 3. two levers “D,” 4. two shafts, on both ends of which the stops are mounted (with the necessary supports), 5. these stops “K,” 6. springs, 7. two wooden guides fastened to the shaft of the mine.

According to diagram 2, the safety catch consists of the same parts with the following modifications: 1) rods “B” are replaced by chains, 2) the springs are placed on the same shafts as the stops and, consequently, have a horizontal axis.

The whole meaning of the first modification (replacing rods by chains) is perfectly clear, since the springs here directly turn the shafts of the stops (and not through bolt “A” and rods “B”)—and there is no need whatever for a rigid connection.

IV. Factor of safety in calculating the safety catch and its parts.

According to the Mining Regulations (see Rules for the Conduct of Mining Operations with Respect to Their Safety, ed. M. T. P., 1911, §§ 116, 117, 118 and 128), metal ropes for hoisting must possess no less than a sixfold strength in relation to the greatest load being lifted, including also the weight of the rope itself. It is also stated there that the ultimate (temporary) tensile resistance of the wires must be not less than 18,000 kg/cm². Thus, for hoisting purposes it is prescribed to use only material of high quality and under the indispensable condition of a sixfold factor of safety (not less!!).

Proceeding from considerations of equal strength of all the details of the structure, equally responsible in operation, I consider it necessary to adopt the factor of safety in calculating the safety catch also as 6.

Thus, for metals necessarily of high quality, the following should be adopted:

1. for iron, not more than 700 kg/cm².
2. for steel, not more than 1000 kg/cm².

V. Forces acting on the safety catch.

a. Hoisting (lowering) on the rope.

During hoisting or lowering on the rope, forces act on the cage which are schematically shown in Fig. 3.

The resultant of all forces “P” is equal to

\[ 1.\quad P = G + G_1 + G_0 \pm m \frac{d^2 x}{dt^2} + \Sigma T; \]

here \(G\) is the weight of the cage,
\(G_1\) is the weight of the useful load with the empty mine car,
\(G_0\) is the weight of the tail rope,
\(m\) is the mass of the moving parts, equal to

\[ m = \frac{G + G_1 + G_0}{g} \]

\[ \frac{d^2 x}{dt^2} \]

is the acceleration of motion (the sign \(+\) for hoisting, \(-\) for lowering), \(\Sigma T\) is the friction force in the guides, always directed opposite to the motion.

With complete balancing, \(G_0\) must be taken equal to 0 (in the expression for the mass \(G_0\) remains even then).

The friction force \(\sum T\) is equal to \(4\%\) of the sum of all moving weights, including also the weight of the hoisting rope¹.

\[ \text{2. }\sum T = 0.04\,(G+G_1+G_0+G_k), \]
where \(G_k\) is the weight of the hoisting rope.

The acceleration \(\left(\dfrac{d^2x}{dt^2}\right)\) may be determined as follows.

An approximate diagram of the speed of the cage for a steam hoist is presented in Fig. 4 and is entirely typical.

The dotted curve corresponds to those operating conditions under which the live force of the system is wholly converted into work against harmful resistances—a condition possible only in a mine of great depth². The solid curve corresponds to ordinary conditions, when the forces of inertia of the system are opposed, together with the harmful resistances, by the braking force of the steam or hand brake and, sometimes, by counter-steam. The periods of acceleration \((t_1)\) are usually shorter than the run-up period \((t_3)\), and especially so, as experiments show, in cylindrical drums without a tail rope³. In absolute magnitude these periods are, for a steam hoist: acceleration period 3–10 sec., retardation period 12–40 sec. without a tail rope, and, with the latter present, 6–20 sec.⁴ The curve \(OA\) may with sufficient accuracy be taken as a parabola. Then the acceleration (during the acceleration period \(t_1\)) will be found as follows.

The area of the parabola \(OAK\) is equal to the distance traveled, i.e.

\[ \text{3. } St = \frac{2}{3} V_m t_1, \]
and for any arbitrary instant “\(t\)” in this period, analogously,

\[ \text{4. } St = \frac{2}{3} Vt, \]
where “\(V\)” will be the velocity at that instant.

Taking the derivative, we obtain

\[ \frac{ds}{dt}=\frac{2}{3}\left(V+t\frac{dv}{dt}\right), \quad \text{but } \frac{ds}{dt}=V, \]
whence

\[ V-\frac{2}{3}V=\frac{2}{3}t\frac{dv}{dt}, \]
and further,

\[ \text{5. } w=\frac{dv}{dt}=\frac{V}{2t}. \]
Here “\(w\)” is already the acceleration for an arbitrary instant of time (in the acceleration period).

For the beginning of this period \((t=0)\), \(w\to\infty\) has the greatest value; for the end,

\[ \text{6. } w_1 = t_1 = \frac{V_m}{2t_1}. \]

The mean acceleration

\[ \text{7. } W_m=\frac{V_m}{t_1}. \]

The fact that the motion begins with an impact is undoubtedly extremely harmful for all parts of the mechanism. As the design acceleration we shall nevertheless take the mean value and, consequently, must determine for this purpose the velocity of the period of uniform motion. This velocity of motion in the mine is established rather definitely by the rules of mining operations and constitutes a certain fraction of the depth of the mine.

Namely: for a mine 75 m deep, the speed is \(1/25\) of the depth, i.e. 3 m/s

105 m > > \(1/30\) > > 3.5 >

150 m > > \(1/35\) > > 4.3 >

225 m > > \(1/40\) > > 5.6 >

300 m > > \(1/45\) > > 6.6 >

450 m > > \(1/55\) > > 8.2 >

900 m > > \(1/85\) > > 10.6 >

more > > \(1/100\) > > up to 16 >

Hence the mean acceleration (during the period of acceleration) for steam hoisting is determined as from 0.3 to 1.2 m/s\(^2\), which is in full agreement also with the experimental data, both my own personally and the extensive experiments of Futers¹.

In the case of electric hoisting, the diagram of cage velocity has the typical form of Fig. 5.

The periods of acceleration \((t_1)\) and retardation \((t_3)\) are equal here, and the change in velocity follows the law of a straight line.

The duration of both periods ranges from 3 to 20 seconds, depending on the depth of the mine.

The acceleration is constant, obviously, and is also equal to from 0.3 to 1.2 m/s\(^2\).

In view of the infinitely large values of acceleration at the beginning of motion in steam hoisting, in calculations the greatest possible value, i.e. 1 m/s\(^2\), should be taken as the mean acceleration; for electric hoisting, however, the acceleration may always be individualized with variations within the limits indicated above.

During lowering or hoisting on the rope, the parachute is kept in complete inactivity by a part of this force \(P\) (see the reverse below); thus this is the period of the passive state of the parachute.

B. Hoisting (lowering) when the rope breaks.

I. The rope breaks during lowering.

At this moment the moving masses of the cage possess considerable kinetic energy, which the parachute must absorb.

It is clear that the expended potential energy of the parachute spring must be just equal to the kinetic energy of the moving masses. The very phenomenon of stopping the cage may be represented schematically as follows (see Nos. 6, 7, and 8).

From the figures it is evident that the angle of rotation of the stop during the braking process is equal to the difference between the full angle of rotation and the angle of free travel \((a_2+\varphi_1)\). In practice it is always possible to attain equality \(a_2=\varphi_1\), and therefore \(a_0=2\varphi_1\). The full angle of rotation is usually different, depending on the inertia of the parts of the parachute. If this angle is small, the parachute comes into action even when the direction of travel of the cage is changed (at the beginning of hoisting and lowering when the cage is removed from the cams). Therefore it should be made not less than

\[ \frac{\pi}{2}\left(\frac{\pi}{2}\right) \]

But in all cases the braking angle \((\varphi_2)\) varies within about

\[ \frac{\pi}{4} \]

In what follows it may be taken equal to

\[ 8 \cdot \varphi_2=\frac{\pi}{4} \]

The pressure of the stops produces a frictional force between the stop and the guide, which, acting over a known distance \((s)\), stops the cage (see Fig. 8). Accordingly—

relationship between the acting forces, etc., obtained on the basis of the law of living forces:

\[ 9.\quad mgS+\frac{mv^2}{2}=\int_0^S F\,ds=\int_0^S N\mu\,ds, \]

where

\(m\)—mass of the moving parts (see above)
\(N\)—normal force
\(T\)—frictional force
\(\mu\)—coefficient of sliding friction

The condition of the spring during braking will be as follows:

a) Spring with a vertical axis (see Fig. I)

The spring, released from the action of the hoisting rope, straightens, and instead of its former height “\(h\)” acquires the height “\(h_1\).” Between the force acting on the spring and its deformation there is a linear dependence, namely¹):

round sections—for a cylindrical spring

\[ 10.\quad f=\frac{64PnR^3}{G\delta^4}, \]

for a conical spring

\[ 11.\quad f=\frac{16PnR^3}{G\delta^4} \]

\(P\)—acting force
\(n\)—number of turns
\(R\)—mean radius of the spring (for a conical spring, in the most remote part)
\(G\)—modulus of shear
\(\delta\)—diameter of the wire

rectangular sections
cylindrical

\[ 12.\quad f=\frac{7.2\Pi nR^3(b^2+h^2)}{b^3h^3}\,\frac{P}{G}; \]

conical

\[ 13.\quad f=1.8\Pi R^3 n\,\frac{b^2+h^2}{b_c+h^3}\,\frac{P}{S}; \]

\(f\)—deformation in the direction of the force
“\(bh\)”—sides of the rectangular section (\(b\)—smaller, \(h\)—larger)

All the formulas may be given one and the same form:

\[ 14.\quad P=Af. \]

where “\(A\)” denotes a constant quantity.

Graphically this dependence is expressed by a straight line passing through the origin of coordinates (with axes \(P\) and \(f\); see Fig. 9). In Fig. 9 the origin of coordinates corresponds to a completely free spring, point “M” to the maximum compression of the spring under the action of the rope (\(l_{\max}\)), point “A” to the compression at the beginning of braking, and “D” to the end of braking.

¹) See Hütte 1902, 419.

If the reduced transmission ratio from the spring to the stop is denoted by “\(k\),” the force “\(N\)” (see Fig. 8) will be expressed for each instant of braking as

\[ 15.\quad N = k P_x \quad \text{where } P_x \text{ is the spring force} \]

And the integral

\[ \int_0^S N \mu\, ds = \int_0^S \mu k P_x\, ds = \int_0^{fs} \mu k P_x \frac{df}{k_1} \]

and

\[ 16.\quad \int_0^S N \mu\, ds = \mu \frac{k}{k_1} S_{ABCD} \left(k_1 \text{ is the ratio } \frac{fs}{S}\right), \]

i.e. it will be proportional to the area of the trapezoid \(S_{ABCD}\), which expresses the difference between the values of the potential energy at the beginning and at the end of braking. \(\left(\frac{1}{K_1}\right.\)—the ratio of the path to the deformation \(fs\)). The magnitude of the force “\(P_e\)” at the end of braking is quite evident. It must be equal (allowing for transmissions) to the total weight of the cage, i.e.

\[ 17.\quad N_1 \mu = P_e k = mg,\quad \text{whence} \]

\[ 18.\quad P_e = \frac{mg}{\mu k} \]

The area of the trapezoid \(S_{ABCD}\) will be equal to

\[ 19.\quad S_{ABCD} = \frac{(P_e + P_b)fs}{2} = \frac{mg}{2\mu k}fs + \frac{P_b fs}{2} \]

The equation of living forces, after substitution, becomes the following:

\[ mg\frac{fs}{k_1} + \frac{m v^2}{2} = \frac{1}{2}\frac{mg\,fs}{k_1} + \frac{P_b fs\,\mu k}{2k_1} \]

Whence

\[ 20.\quad P_b = \frac{G+G_1+G_0}{\mu} \left( \frac{v^2}{q\,fs}\frac{k}{k_1} + \frac{1}{k} \right) \]

The greatest force acting on the spring exceeds “\(P_b\)” by no more than 10%, since the free travel of the spring is taken, for design considerations, as \(0.1\,fs\)

\[ 21.\quad P_{\max} = 1.1\,P_b = 1.1\,\frac{G+G_1+G_0}{\mu} \left( \frac{v^2}{b\,fs}\frac{k}{k_1} + \frac{1}{k} \right) \]

The transmission ratios “\(k\) and \(k_1\)” may be determined on the basis of the following considerations.

In Fig. 10 the transmission rods are shown in two positions, with the spring passive and at the end of braking. On the basis of the principle of possible displacements:

\[ 22.\quad P\,\delta fs = N l_1 \frac{1}{\cos \alpha_1}\,\delta \alpha_1 \quad (\text{for any angle of rotation}). \]

For an infinitely small rotation, the motion of an arc of radius “\(l_2\)” may be taken as equal to “\(fs\)”—the vertical displacement of the point of application of the force \(P\), i.e.

\[ 23.\quad \delta fs = l_2 \delta \alpha_1 \]

after substitution

\[ P l_2 = N l_1 \frac{1}{\cos \alpha_1} \]

and

\[ 24.\quad K=\frac{N}{P}\frac{l_2\cos\alpha_1}{l_1}. \]

In accordance with the previous reasoning, the angle “\(\alpha\)” takes values from “\(\varphi_1\)” to “\(\varphi_2\),” i.e., from \(0\) to \(\alpha \frac{\Pi}{4}\). Consequently, “\(K\)” varies from

\[ K=\frac{l_2}{l_1}\cdot 1=\frac{l_2}{l_1} \quad \text{to} \quad K=\frac{l_2}{l_1}\frac{\sqrt{2}}{2}=0.7\frac{l_2}{l_1}. \]

The ratio \(\frac{l_1}{l_2}\) is usually equal to \(1:2\).

Then

\[ \begin{aligned} K &= 1\cdot 2 = 2 && \text{at the beginning of braking,}\\ K &= 0.7\cdot 2 = 1.5 && \text{at the end } \quad \text{, i.e.} \end{aligned} \]

toward the end of braking “\(K_1\)” becomes less favorable; we find it in an analogous way. Motion along the path “\(S\)” may be considered uniformly retarded, the acceleration of which will obviously be \(\frac{V}{T}\), where \(V\) is the velocity of fall, \(T\) the time of retardation to 0. Therefore

\[ 25.\quad S=\frac{VT}{2}. \]

The time of retardation at the speeds in use must not be large; otherwise the stop will occur too late to save people from the impact. According to the experiments of the Australian Mining Inspection¹, the stopping time with the aid of parachutes is so insignificant that the stopping may be considered instantaneous. We shall take the time as equal to 1. Then

\[ 26.\quad S=\frac{V}{2}. \]

For economic as well as structural reasons, the travel of the spring is not made more than 50 mm².

Hence

\[ 27.\quad K_1=\frac{k}{S}=\frac{2\cdot 50\cdot 10^{-3}}{V}=\frac{1}{10V}. \]

The expression for the force \(P_{\max}\) will be

\[ 28.\quad P_{\max}=l_1 l\cdot \left( \frac{G+G_1+G_0}{2\mu} \left( \frac{2v}{g}+1 \right) \right). \]

The coefficient of friction “\(\mu\)” may be taken equal to 0.5, as the mean between the coefficient of friction with water and dry friction, since, although the guides are almost never dry, the grips have sharp teeth, which undoubtedly is very favorable for braking. Substituting this value, we finally obtain

\[ 29.\quad P_{\max}=l_1 l\ldots \left( \frac{2v}{g}+1 \right)(G+G_1+G_0). \]

B) Spring with a horizontal axis (see Fig. 2)

Here the springs act directly on the stops; their number is not less than 2, more often 4. As is seen from Fig. 12, the spring force and the normal force are in the following relation:

\[ 30.\quad i P l_2 = N l_1, \]

where \(i\) is the number of springs on the shaft.

The number of shafts is usually 2; therefore

\[ 2 P i l_2 = 2 N l_1, \]

and consequently the transmission ratio is

\[ k=\frac{N}{g}=\frac{i l_2}{l_1}=i \quad \left(\text{here } \frac{l_2}{l_1}=1\right). \]

The force \(P\) is determined as follows: the angle of deformation under bending is equal to

\[ d\Delta\varphi=\frac{M\,ds}{EJ} = \frac{P(l_2-y)}{EJ}\,l_2\,d\varphi \quad\text{and} \]

\[ l_2-y=l_2(1-\cos\varphi). \]

Then

\[ \Delta\varphi = \int_0^{2\pi n} \frac{P l_2^2(1-\cos\varphi)\,d\varphi}{EJ}, \]

where \(n\) is the number of turns,

\[ \Delta\varphi = \frac{P l_2^2\,2\pi n}{EJ}, \]

or

\[ 31.\quad P=A\Delta\varphi \left( A=\frac{EJ}{2\pi n l_2^2} \right), \]

i.e., the dependence is likewise linear (see Fig. 13).

Point \(M\) corresponds to the maximum tension of the spring during hoisting on the rope; point A is the beginning of braking, D is the end; the difference between the angles of deformation for A and D, as was indicated above, is

\[ \frac{\pi}{4}. \]

Then the integral

\[ 32.\quad \int_0^S N\mu\,ds = 2 i \mu A \int_0^S \Delta\varphi\,ds \]

Over the braking path \(S\), the change in the angle of deformation corresponds to

\[ \frac{\pi}{4}. \]

If the paths and angles of rotation are assumed proportional, then

\[ \int_0^S N\mu\,ds = \frac{(P_b+P_e)}{2}\,S\,2 i\mu = (P_b+P_e)S i\mu. \]

But

\[ S=\frac{U}{2} \]

\[ 33.\quad \int_0^S N\mu\,ds = \frac{(p_b+p_l)\,v\,i\mu}{2}. \]

The equation of live forces takes the following form:

\[ 34.\quad mg\frac{V}{2} + \frac{m v^2}{2} = \frac{(p_b+p_l)\,v\,i\mu}{2}. \]

The force “P” at the end of braking of the safety catch is equal to

\[ 35.\quad \rho l \ge \frac{(G+G_1+G_0)}{2\mu i} \]

Finally,

\[ 36.\quad \rho b=\frac{G+G_1+G_0}{i}\left(\frac{2v}{g}+1\right) \]

\[ 37.\quad P_{\max}=1{,}1\,\rho b =\frac{1{,}1(G+G_1+G_0)}{i}\left(\frac{2v}{g}+2\right) \]

(per spring).

With one spring on the shaft (\(i=1\)), \(P_{\max}\) is obviously twice as great with a vertical spring as with a horizontal one. It follows from this that the vertical spring is in more favorable working conditions and is therefore always preferable to the horizontal one.

2. The rope breaks during hoisting.

In this case the cage at first continues its motion by inertia; the braking force of the safety catch retards this motion and, finally, eliminates the inertial motion of the cage, bringing it to rest.

It is clear that the force of the safety catch is, at the beginning of braking,

\[ 38.\quad T_{\text{res}}\cdot \rho N=\rho kP = m\frac{d^2x}{dx^2}-\Sigma T-G-G_1-G_0 . \]

\[ P=\frac{1}{\rho k}\left[m\frac{d^2x}{dx^2}-\Sigma T-G-G_1-G_0\right]. \]

At the end of braking,

\[ 39.\quad P \sim \frac{1}{\rho k}(G+G_1+G_0), \text{ i.e.} \]

The greatest value of the force of the safety catch in the case considered is less than when the rope breaks during lowering.

VI. Calculation of Details.

A. Safety-catch bolt.

The dimension “\(\delta\)” is determined by the diameter of the shank of the safety hook (“\(\delta_0\)”) and is equal to

\[ \delta=1{,}05\,\delta_0 . \]

The dimension “\(k\)” is determined by the free travel of the safety hook and depends on the magnitude of the load. These dimensions may conveniently be taken from the table of M. Ketchum¹. Their average value for loads:

	up to \(4^t\)	\(12^t\)
\(\delta_0=\)	\(1''2''\)	\(3\frac{7}{16}''\)
\(k=\)	\(7''\)	\(17\frac{1}{4}''\)

The dimensions “\(a\)” and “\(S\)” are equal to

\[ a = 1.5\delta^{*}) \]

\[ \delta = 1.2d \]

“\(b\)” is found from the formula:

\[ (b-\delta) S k_2 = P \]

\[ 40.\quad b = \frac{P}{k_2\,1.2\,d}+\delta. \]

The length “\(L\)” is determined from the following considerations. 1. It is convenient to attach the safety hook to it; 2. the rollers for the rods and chains are successfully distributed in it. These conditions are satisfied by

\[ L \approx 20d \]

The diameter “\(d\)” is established as a function of the force acting on the parachute:

\[ 41.\quad \frac{\pi d^2}{4}\,k_2 = 2P_{\max}^{1}) \]

The length of the bolt “\(L_1\)” is determined by the free length of the springs (see below). The diameters of the holes for the rods “\(\delta\)” and for the chains are equal to

\[ 42.\quad \delta_1 = \frac{d}{2} \]

B. Rods (see Fig. 1).

The rod is under the action of the spring force. The greatest force acting upon it is, obviously, equal to

\[ 43.\quad P_1 = P_{\max}\cos\alpha \]

Usually the angle “\(a\)” is equal to \(30^\circ\) and is always, of course, determined graphically for the given cage dimensions and arrangement of the parachute parts. The length of the rod is also most conveniently determined graphically, but for preliminary calculations it may be taken as

\[ 44.\quad l = (10—15)d, \]

where \(d\) is the diameter of the parachute bolt.

The diameter of the rod is determined for compression by the formula:

\[ 45.\quad \frac{\pi d_1^2}{4}\,k_2 = P_1 \]

It is checked for longitudinal bending, as a bar with free ends, with a factor of safety of 6

\[ (\text{of course, with } l_1 = 35d_1) \]

C. Levers D (see Fig. 1.)

The levers “D” are subjected to bending under the action of a force equal to \(2N\)

\[ \left(\text{with } \frac{l_1}{l_2}=1\right); \]

moreover, obviously,

\[ 46.\quad 2N = P_{\max} \]

\[ ^{*})\ \text{See D. M. Bobarykov.} \]

A force acts on one lever:

$$ 47.\quad P_1=\frac{P_{\max}}{2} $$

With lever length \(l_2=(4-6)d\), the strength equation for a rectangular cross-section of the lever is

$$ 48.\quad \frac{b h^2}{6}k_b=P_{\max}(3-2)d. $$

The ratio \(\frac{b}{h}\) is usually equal to \(\frac{1}{2}\); then the dimensions will be known.

D. Stops \(K\) (see Fig. I)

The stops are also subjected to bending and, consequently, must be calculated similarly to the levers “D.” In view of the fact that the impact falls directly upon them, the dimensions of the stops should be determined at an allowable stress not exceeding one half of that for the levers, i.e.

$$ K_1=K_2=300\ \mathrm{kg/cm^2} $$

E. Shaft.

The shaft (more precisely, the axle) is subjected to bending by the concentrated force

$$ P_1=\frac{P_{\max}}{2} $$

The distance between its supports is equal to the cage width (usually \(a=1.2-1.5\ \mathrm{m}\)).

Its diameter is equal to

$$ 0.1d^3=\frac{P_{\max}l}{24k_b}. $$

For \(k_b=700\ \mathrm{kg/cm^2}\) and \(l=1.2\ \mathrm{m}\),

$$ 49.\quad d\ \mathrm{cm}=0.5\sqrt[3]{P_{\max}} $$

I. Spring.

1. With a vertical axis.

The transverse section is most often a rectangle, rarely a circle.

The dimensions of the transverse section are determined from the conditions of strength in torsion by the formula:

$$ 50.\quad \text{for a rectangle:}\quad P_{\max}R=\frac{2}{9}b^2h k_s, $$

$$ 51.\quad \text{for a circle}\quad P_{\max}R=\frac{1}{5}d^3k_s $$

Here “\(R\)” is the mean radius of the spring and may be taken equal (approximately) to the radius of the parachute bolt for a cylindrical spring; for a conical one, “\(R\)” will denote the greatest radius, equal to “\(d\)” (the bolt diameter).

The stress, in view of the excellent material for springs, may be taken as

$$ P_s=4500-8000\ \mathrm{kg/cm^2}. $$

The first value is for light structures, the second for heavy ones. The number of coils \((n)\) of the spring is determined by the formula:

$$ 52\quad n=\frac{Gd^3h^3f}{7.2\,nR^3(d^2+h^2)P}. $$

for a rectangular conical spring, \(R_0\) is the radius of the upper base; the remaining designations are as before.

\[ 53\quad n=\frac{G b^3 h^3 f}{1.8\pi P(d^2+h^2)(R^2+R_0^2)(R+R_0)}. \]

for a round cylindrical spring

\[ n=\frac{g d^4 f}{64 R^3 P} \]

for a round cylindrical spring

\[ n=\frac{G d^4 f}{16 R^3 P}. \]

The deflection “\(f\),” as indicated above, is equal to

\[ f=50\text{ m/m}. \]

This is the usual travel of the spring.

In the process of design, the deflection is specified by the travel of the stops and other parts of the parachute, and is conveniently determined from the drawing, so that 50 m/m should be considered a preliminary and greatest dimension.

2. With a horizontal axis.

The form of the cross-section is most often a circle.

The dimensions are determined by the formula:

\[ 54.\quad 2 P_{\max} R=\frac{1}{10}\,\varepsilon^3 k_0 \]

where “\(R\)” is equal to the diameter of the shaft on which the spring is located, \(k_v=6500\ \text{kg/cm}^2\) *).

The number of turns is determined by the formula

\[ 55.\quad n=\frac{\Lambda\varphi E\pi \delta^4}{P\cdot 2\pi R^2\cdot 64}\quad (\text{not }P_{\max}!). \]

\(\Lambda\varphi\) is the total angle of rotation corresponding to the force “\(P\).” Obviously, for \(P_{\max}\), “\(\Lambda\varphi\)” will be greater. Usually, as stated above, when the cage is being hoisted, the stop rotates through an angle

\[ \frac{3\pi}{4} \]

Whence, i.e., the number of turns is established.

Indeed,

\[ \Lambda\varphi_{\max}=\frac{P_{\max}\,2\pi R^2 n}{E I}. \]

\[ (\Lambda\varphi)_P=\frac{P\,2\pi R^2 n}{E I}, \]

hence

\[ \Lambda\varphi_{\max}-(\Lambda\varphi)_P=\frac{3\pi}{4} =\frac{(P_{\max}-P)\,2\pi R^2 n}{E I} \quad\text{and} \]

\[ 56.\quad n=3.75\,\frac{EI}{PR^2} \]

Thus

\[ 57.\quad n=\frac{3.75\cdot 1.1\cdot EI}{P_{\max}R^2} \]

* For stresses see Hütte.

At the same time, the angle through which the spring must be twisted during installation will be

\[ 58.\quad A\varphi=\frac{P\cdot 2\pi R^{2}n}{EI}. \]

From the additional angle of twist under the action of the rope there arises a considerable additional stress. It can be estimated in advance and, depending on its magnitude, the number of turns may be chosen.

Let us imagine that we have twisted the spring through an angle \(\varphi_o\) by a moment \(M\). Then, by the law of conservation of energy, we have

\[ \frac{M\varphi_o}{2}=\frac{M^{2}\,2\pi R n}{2EI}, \]

and hence

\[ 59.\quad M=\frac{EI\varphi_o}{2\pi R n}. \]

From the strength equation

\[ kb=\frac{M}{I}\cdot\frac{\delta}{2} \]

and the preceding, we obtain

\[ 60.\quad kb=\frac{E\varphi_o\delta}{4\pi R n}. \]

Usually \(\varphi_o=\dfrac{3\pi}{4}\); in this case the additional stress is

\[ 61.\quad kb=\frac{3Ed}{16Rn}. \]

For \(d=25\) mm, \(n=30\), \(R=70\) mm, \(E=2.0\cdot 10^{6}\ \mathrm{kg/cm^{2}}\), \(kb \simeq 4000\ \mathrm{kg/cm^{2}}\).

It follows from this that the additional angle of rotation is very large and should, as far as possible, be reduced. Otherwise it is necessary to increase the number of turns. In any case, taking a definite percentage of the stress for the additional angle, from the preceding formula we shall obtain the number of turns.

It is best to take the additional \(kb_{\text{add.}}=(30\text{–}35)\%\ Kd\) of the permissible value, namely \(1500\ \mathrm{kg/cm^{2}}\).

VII. Characteristics of a standard parachute.

Let us consider a parachute with a vertical spring for a two-deck cage at a mine depth of 100 sazhens (200 m) (one mine car on each deck).

I. The weight of a two-deck cage varies from 1.2 t for a steel cage to 2.0 t for an iron cage¹.

We are in Siberian conditions, where there are as yet no steel cages. Hence,

\[ G=2000\ \text{kg}. \]

Let us further assume, as is usual, that the parachute must operate during passenger trips; during freight trips it will be disengaged. Then, with a simultaneous descent of no more than 4 persons,

\[ G_{1}=80\cdot 4=320\ \text{kg}. \]

We shall take

\[ G_{1}=350\ \text{kg}. \]

For the tail rope, however, this is not arranged, hence \(= 0\).

Thus

\[ G + G_1 + G_0 = 2350 \ \mathrm{kgr}. \]

The average speed of motion of the cage (according to the preceding table)

\[ U = \frac{5}{2} = 2.5 \ \mathrm{m/s}, \]

since the speed of passenger trains is half that of freight trains.

The force acting on the spring is equal to

\[ P_{\max} = 1.1 \left( \frac{2 \cdot 2.5}{9.81} + 1 \right) 2350 = 3900 \ \mathrm{kg}. \quad P_{\max} = 3900 \ \mathrm{per}. \]

Parachute bolt.

\[ \frac{\pi d^2}{4} = \frac{7755}{700} = 12.3 \]

\[ d = 35 \ \mathrm{m/m} \]

Tie rods.

\[ \frac{\pi d_1^2}{4} = \frac{0.8 \, 3900}{700} = 4.5 \]

\[ d_1 = 30 \ \mathrm{m/m}. \]

Check for longitudinal bending at \(l = 10d = 350 \ \mathrm{m/m}\).

\[ f_{\min} = \frac{6 \cdot 0.8 \, 3900 \, 35^2}{10 \cdot 2 \cdot 10} = 2.56 \ \mathrm{cm}^4 \]

For \(d = 30 \ \mathrm{m/m}\), \(f_{\min} = 3.97 \ \mathrm{cm}^4\).

Levers.

\[ \frac{h^3}{12} = \frac{3900 \, 20}{2 \cdot 700} \quad (l_2 = 4d = 200 \ \mathrm{m/m}). \]

\[ h = 150 \ \mathrm{m/m}. \]

\[ b = 75 \ \mathrm{m/m}. \]

Stops.

Since the transmission ratio from the levers to the stops is 2, the dimensions of the former will change only in relation to the stresses, increased twofold.

\[ h = 150 \sqrt{2} = 200 \ \mathrm{m/m}. \]

\[ b = \ldots = 100 \ \mathrm{m/m}. \]

Shaft.

\[ d = 0.5 \sqrt[3]{3900} = 75 \ \mathrm{m/m}. \]

\[ d = 75 \ \mathrm{m/m}. \]

Spring.

Let the radius of the spring be, as usual,

\[ R = d = 35.0\ \text{m/m}. \]

Then, from the strength equation at \(kb = 8000\ \text{kg}/\text{cm}^2\),

\[ {}^{1}/_{9}\, b^2 h = \frac{3900 \cdot 3.5}{8000} = 1.75 \]

with \(b = 5\ \text{m/m} = 0.5\ \text{cm}\).

\[ h = \frac{1.75 \cdot 9}{2 \cdot 0.25} = 31.5\ \text{m/m}. \]

The height of the springs in use varies

\[ \text{from } h = 120\ \text{m/m},\quad \text{and } b = 2\ \text{m/m}, \]

\[ \text{to } h = 200\ \text{m/m},\quad \text{and } b = 5\ \text{m/m}. \]

Thus it will be necessary to install not fewer than two springs.

The number of coils at \(f = 25\ \text{m/m}\) (the amount of springing) is

\[ n = 8. \]

From this it is evident what the dimensions of the spring must be in order for braking to be reliable during descent even under these easy conditions. During hoisting the situation will be considerably more favorable.

During hoisting:

\[ P_{\max} = \frac{1.1\,(g + g_1 + g_0)}{2\mu} \]

“\(\mu\)” may be taken equal to 1 (friction at rest).

\[ P_{\max} = \frac{1.2 \cdot 2250}{2} = 1200\ \text{kg}. \]

Further,

\[ {}^{2}/_{9^2}\, b h = \frac{1200 \cdot 3.5}{8000} = 0.2 \]

\[ h = \frac{0.2 \cdot 9}{2 \cdot 0.25} = 4.5 = 5\ \text{m/m}. \]

Correspondingly, with the same “\(n\),” the amount of springing is

\[ f = \frac{2.5}{3} = 0.8 = 8\ \text{m/m}. \]

VIII. Conclusions.

From the foregoing it follows with obviousness that the action of the standard parachute, with ordinary dimensions, is reliable only during hoisting. Increasing the dimensions, in view of the probable buckling of the spring as its height increases, is structurally impossible; increasing the number of springs is difficult, since this greatly complicates the apparatus and makes it heavy, as a result of which the overall coefficient of efficiency is considerably reduced.

Therefore it is necessary to instruct the operators servicing machines with such safety catches to observe the utmost caution when lowering people into the mine and, of course, to prohibit them decisively from taking “a breeze ride.”

In any case, the present study shows that the problem of reliable braking by a safety catch with the direct action of a spring is not solved. It is necessary to redesign the safety catch so that the spring plays only the role of the initial exciter of the force, while during the braking process all the force and importance must be transferred to reliable mechanical brakes and—best of all—without any participation of the spring.

The significance of the safety catch (its spring) is very great, although indirect, from another standpoint as well. It consists in the fact that the shocks inevitable in the operation of a hoisting machine (reversal of motion, sudden stops, etc.) are weakened by the action of the spring (especially with a large number of coils), and the effect of the shocks on the condition of the rope is very slight—they almost do not increase the stress in the rope, provided the spring has a sufficient number of coils and the necessary dimensions.

To the article by Eng. S. F. Lebedev: “The Standard Safety Catch of the Siberian Mines from the Standpoint of Safety in Hoisting.”

No. 1

No. 2

No. 3

No. 4

No. 5

No. 6

No. 7

No. 8

Mutual position of the guide and the stops
during a break
at the end of braking.

For the article by Eng. S. F. Lebedev: “The Standard Safety Catch of the Siberian Mines from the Standpoint of Safety in Hoisting.”

No. 9

Visible labels: \(P\), \(O\), \(L_1\), \(L\), \(C\), \(P_c\), \(D\), \(df\), \(f_s\), \(B\), \(A\), \(P_b\), \(P_{\max}\), \(M\).

No. 10

Visible labels: \(A\), \(P\), \(D\), \(t_s\), \(l\), \(l_1\), \(l_4\), \(e_1\).

No. 11

Visible labels: \(P_b\), \(P_c\),

\[ S=\frac{1}{k_i} f_s \]

No. 12

Visible labels: \(P\), \(l_2\), \(l_1\), \(N\).

No. 13

Visible labels: \(P\), \(O\), \(\Delta \varphi_{\max}\), \(\Delta \varphi_1\), \(\Delta \varphi_i\), \(C\), \(P_c\), \(D\), \(B\), \(A\), \(P_b\), \(P_{\max}\), \(M\), \(z\).

No. 14

Visible labels: \(s\), \(b\), \(a\), \(k\), \(\delta\), \(\delta_1\), \(L\), \(L_1\).

No. 15

Visible labels: \(L\), \(P_1\), \(\alpha\), \(P_{\max}\).

1. M. ketchum M. S. p. 30. ↩↩↩↩↩

2. See the work “Hoisting at the Sudzhenka Mines.” ↩↩

3. See Enters, [[unclear: repeated title marks]]. ↩

4. ibidem p. 127. ↩