On the Design of Jig Ramps
Abstract Generated abstract
This article develops a design theory for mine jig runs, curved track connections that allow cars to move between inclined and horizontal tracks without turntables. It addresses the lack of established methods for steeper bremsbergs and slopes, especially where dip angles exceed those commonly considered practicable, by deriving geometric and dynamic equilibrium conditions for the main track sections. The analysis relates seam dip, turnout angle, curve radius, car speed, gauge, center of gravity height, and available pillar width to determine safe curve angles, rail elevation, transition-curve form, and layout parameters. A numerical example illustrates the calculation procedure and suggests that jig runs can be designed with reliable and economical parameters under the stated conditions.
Full Text
ON THE DESIGN OF JIG RUNS
N. G. KAPUSTIN
(Presented by Prof. P. M. Alabuzhev, Doctor of Technical Sciences)
A jig run is a curvilinear connection at the point where inclined and horizontal tracks intersect, carried out by means of a switch turnout and a subsequent curve.
Sometimes the connection of inclined and horizontal tracks is effected by inserting a turntable, on which workers manually turn each mine car in the required direction.
In contrast to this, a jig device makes it possible, by means of the mechanical hoisting installation serving the inclined tracks, to bring cars from the inclined track onto the horizontal tracks and back. With jig devices, the staff of turntable workers is almost completely eliminated; their labor, with large-capacity cars, becomes the heaviest in the mine.
Thus, in order to increase the labor productivity of the main workers by reducing the number of auxiliary workers, and to create safer conditions for miners, it is necessary to introduce jig runs more widely instead of turntables, which are an obsolete solution in mine transport.
On gently inclined tracks \((<12^\circ)\), experience already exists in the design and construction of jig runs. At large angles of inclination of hoists, there is neither construction experience nor a theory for designing jigs. Moreover, erroneous opinions are encountered that jig runs on bremsbergs and slopes with a dip angle greater than \(12—15^\circ\) are supposedly structurally unfeasible, and, if built, will prove unsuitable in operation.
The task of the present article is to develop a theory for the design of jig runs in mines working coal seams of inclined dip, where questions of the economical operation of bremsbergs and slopes are especially important.
In the layout of the jig tracks, four characteristic sections should be distinguished (Fig. 1), namely:
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Section \(ab\)—curved tracks of constant rounding radius \(R\) and with the outer rail raised (by lining). The principal parameters of this section are the radius \(R\) and the central angle \(\lambda\) of the arc \(ab\). The value of \(R\) is established according to the rigid base of the rolling stock. The angle \(\lambda\) is the subject of the calculations made below.
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Section \(bc\)—curved tracks of variable radius \(\rho\), which changes from infinity at point \(c\) to \(R\) at point \(b\). This transition curve ensures the elimination of the impact at point \(b\) from centrifugal force by the gradual increase of the pressure of the wheel flange on the outer rail.
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Section \(cd\)—curved tracks, also of constant radius \(R\), laid in the plane of the switch turnout, i.e. the floor of the coal seam. Therefore the outer rail at point \(c\) is lower than the inner rail. Point \(c\) is the most—
…more dangerous with respect to self-tipping of the car under the action of two forces: the centrifugal force and a certain component of the car’s weight, arising in connection with the car entering the curve \(cd\), which lies in the plane of the switch.
Fig. 1
The latter condition follows from the very idea of the jig. The curved section \(cd\) has a central angle of the arc \(\varphi\), which cannot exceed a certain critical value ensuring the dynamic equilibrium of the car at point \(c\).
- The section \(de\) is a switch lying with all its points in one plane (usually the plane of the bed).
When designing the track layout of a jig, it is necessary to take into account the need to place all the curves of the jig entry within the limits of the Bremsberg pillar of width \(M\).
Fig. 2
From the description of the jig layout it was evident that, in order to determine its unknown parameters, the two conditions specified above may be used: the prescribed value \(M\), within which the jig is placed, and the equilibrium position of the moving car at point \(c\). On the basis of these two conditions, all the parameters necessary for laying out the jig tracks will subsequently be determined.
Let us consider some relations between these parameters. On the basis of the geometric considerations in Fig. 2, we may write:
\[ h = z \sin \alpha,\quad y = z \operatorname{ctg}(\gamma+\varphi),\quad z^2+y^2=u^2,\quad u=h:\sin\omega. \]
After the corresponding substitutions we obtain:
\[ \sin\omega=\sin\alpha\sin\gamma\cos\varphi+\sin\alpha\cos\gamma\sin\varphi, \tag{1} \]
where \(\alpha\) is the dip angle of the bed, and \(\gamma\) is the angle of the switch frog at point \(e\).
Let us consider the condition of equilibrium of the car at point \(c\). The center of gravity of the car lies at a height \(h\) above the rail head. The centrifugal force on the curved track will be \(Qv^2:9.8R\), where \(Q\) is the weight of the car and \(v\) is its speed.
The component of the weight of the mine car \((T)\) of the same sign as the centrifugal force, \(T=Q\sin\omega\). It is opposed by the other component of the weight, \(N=Q\cos\omega\). Then the equilibrium condition will be:
\[ h\,\frac{Qv^2}{9.8R}+hT=0.5aN \quad \text{or} \quad \frac{hv^2}{9.8R}+h\sin\omega=0.5a\cos\omega, \tag{2} \]
where \(a\) is the track gauge.
Solving equation (2) with respect to \(\sin\omega\), we obtain
\[ \sin\omega=\sqrt{B^2+D}-B, \tag{3} \]
where
\[ B=v^2:\left[1+\left(\frac{a}{2h}\right)^2\right]\,9.8R, \tag{3a} \]
\[ D=\left(\frac{a}{2h}+\frac{v^2}{9.8R}\right) \left(\frac{a}{2h}-\frac{v^2}{9.8R}\right): \left[1+\left(\frac{a}{2h}\right)^2\right]. \tag{3b} \]
Equating expressions (1,3), we obtain an equation with one unknown \(\varphi\):
\[ \sqrt{B^2+D}-B=\sin\alpha\sin\gamma\cos\varphi+\sin\alpha\cos\gamma\sin\varphi. \tag{4} \]
Solving equation (4), we find the angle \(\varphi\).
The section of track \(ab\) is safe against overturning of the car under the action of the force \(T\), since at point \(b\) the necessary elevation of the outer rail relative to the inner rail by the height \(\Delta=av^2:9.8R\) will already have been attained, which follows from the equilibrium condition of the car on the curve \(ab\). Similarly, for any point of the arc \(bc\) we have
\[ \Delta=\frac{av^2}{9.8\rho}. \tag{5} \]
The rail superelevation \(\Delta\) should be created by a uniform rise of the outer rail on the section \(bc\) of length \(S\). This can be achieved by the relative grade \(i\) of the outer rail.
\[ i=\Delta:S,\quad \Delta=iS \quad \text{or} \quad \Delta=iS=av^2:9.8\rho; \]
\[ S\rho=\frac{av^2}{9.8i}=\text{constant } C. \tag{6} \]
As is known, the curve for which the condition \(S\rho=\) constant is satisfied is a “radial spiral” with characteristic (Fig. 1):
\[ y=x^3:6C,\quad C \text{— in our case — } \frac{av^2}{9.8i}; \tag{7} \]
\[ x=\frac{C}{\rho}-\frac{C^3}{40\rho^5}; \quad y=\frac{C^2}{6\rho^3}-\frac{C^4}{336\rho^7}. \]
The subtracted terms in these formulas are negligibly small, and therefore we obtain
\[ x=C:\rho;\quad y=C^2:6\rho^3; \tag{8} \]
\[ \frac{dy}{dx}=\operatorname{tg}\beta=\frac{x^2}{2C}=\frac{C}{2\rho^2}. \tag{8a} \]
Here we replace the value of \(C\) by expression (6):
\[ x=\frac{av^2}{9.8\rho i}; \quad y=\frac{a^2v^4}{6\rho^3 9.8^2 i^2}. \tag{9} \]
The angle \(\beta\) (Fig. 1) is determined by formula (8a):
\[ \tg \beta=\frac{a v^2}{19.6\,\rho^2 i}. \tag{10} \]
Next we find the projection of the broken line “\(a—1—b—4—c—2—d—e—3\)” onto the line of extent—“\(a3\)”:
\[ R\sin\lambda+y\cos(\gamma+\varphi)+x\sin(\gamma+\varphi)+R\cos\gamma+x\sin\gamma = \]
\[ = M+R\cos(\gamma+\varphi). \tag{11} \]
For point \(b\) we have:
\[ x=\frac{a v^2}{9.8Ri};\quad y=\frac{a^2 v^4}{576\,R^3 i^2}. \tag{12} \]
The polygon (Fig. 1) with vertices “\(a—1—b—5—c—2—d—e—3\)” has the sum of its interior angles \(180^\circ(9-2)=1260^\circ\). This sum is made up as follows:
\[ 1260=90+\lambda+270+180+\beta+270+\varphi+270+\gamma+90; \]
\[ \lambda+\beta+\varphi+\gamma=90. \tag{13} \]
Here the angles are: \(\gamma\), specified by the construction of the turnout; \(\varphi\), by formula (4); \(\beta\), by formula (10). Solving equations (13, 12, 11, 10) simultaneously, we obtain the values of \(\lambda\) and \(i\). Having the slope \(i\), by formula (12) we find \(x, y\), which serve for staking out the curve \(bc\), whose formula in the coordinate system shown in Fig. 1 is
\[ y=x^3:6C,\quad \text{where } C=\frac{a v^2}{9.8i}. \]
The other sections of the track curves are staked out in the field in the usual manner. Using the method set forth, let us solve a concrete example, taking the following initial data: \(\alpha=25^\circ,\ \gamma=15^\circ36',\ a=0.6\ \text{m},\ h=0.67\ \text{m},\ M=18\ \text{m},\ v=3\ \text{m/sec},\ R=12\ \text{m}\).
\[ B=\frac{3^2}{9.8\cdot12[1+(0.66:2\cdot0.67)^2]}=0.066\quad \text{or}\quad B^2=0.0043. \tag{3a} \]
\[ D=\left(\frac{0.6}{2\cdot0.67}+\frac{3^2}{9.8\cdot12}\right)\cdot\left(\frac{0.6}{2\cdot0.67}- \]
\[ -\frac{3^2}{9.8\cdot12}\right):[1+(0.6:2\cdot0.67)^2]=0.16; \tag{3b} \]
\[ \sin\omega=\sqrt{0.0043+0.16-0.066}=0.34; \tag{3} \]
\[ \sin\omega=0.34=0.422\cdot0.27\cos\varphi+0.422\cdot0.96\sin\varphi, \tag{4} \]
where \(\varphi=38^\circ\) transforms expression (4) into an identity.
\[ 12\sin\lambda+\frac{0.6^2\cdot3^4}{576\cdot12^3 i^2}\cos(15^\circ36'+38^\circ)+\frac{0.6\cdot3^2}{9.8\cdot12\,i}\sin(15^\circ36'+38^\circ)+ \]
\[ +12\cos15^\circ36'+2.94\sin15^\circ36'=25+12\cos(15^\circ36'+38^\circ) \tag{11} \]
\[ \lambda=\arc\sin [F(i)]. \tag{11a} \]
\[ \tg\beta=\frac{0.6\cdot3^2}{2\cdot9.8\cdot12^2 i}=\frac{0.002}{i}. \]
\[ \lambda + \operatorname{arctg}\beta + 38 + 15^\circ 36' = 90^\circ. \]
\[ \lambda + \operatorname{arctg}\beta = 90^\circ - 53^\circ 36' = 36^\circ 24'. \]
\[ \lambda = 36^\circ 24' - \operatorname{arctg}\frac{0.002}{i}. \tag{14} \]
We solve equations (11a, 14) jointly by a graphical method and obtain the values \(\lambda, i\).
After calculation we have:
\[ \lambda = \arcsin\left(1.06 - \frac{0.145}{10^5 i^2} - \frac{0.002}{i}\right), \tag{11a} \]
\[ \lambda = 36^\circ 24' - \operatorname{arctg}\frac{0.002}{i}. \tag{14} \]
The desired roots of these equations will be: \(\lambda = 8^\circ 36'\), \(i = 0.0038\). Then \(x = 12\) m, \(y = 2\) m, \(\beta = 27^\circ 42'\).
The calculations performed make it possible to design a jig run with reliable and economical parameters.