CN103033720B - A kind of long distance discrimination method of polycore cable - Google Patents

Abstract

The invention discloses a kind of long distance discrimination method of polycore cable, " resistance-cable connector " that make with test resistance m=10K, can reuse after producing once, be applicable to common all wireline tests.This method utilizes the relevant knowledge of electric aspect and mathematical knowledge to combine, and by popularization and the use of the method, the real practical problems solved in engineering, efficiently, fast completes identification and the wiring of polycore cable.

Description

A long-distance discrimination method for multi-core cables

technical field

The invention belongs to the technical field of construction and relates to a long-distance discrimination method for multi-core cables.

Background technique

Under coal mines or in construction roadways, long-distance multi-core cables are often laid for communication or control. Sometimes it is unreliable to only rely on the color identification of the cable core, and it may even be misjudged or misconnected. For example, when connecting communication cables in a coal mine roadway, there is such a problem: a certain signal line at the equipment side is blue, but it is found to be brown through repeated inspections at the control room side. In practice, there will be joints in the middle of hundreds of meters of cables, and it is possible that the two ends of the same core wire have different colors. Sometimes the distance between the two ends of the cable is relatively long, and there are spacers in the middle. For example, one end of a communication cable is on the ground and the other end is underground in a coal mine. It is not an easy task to find out each wire and then connect it correctly.

The multi-core cable discrimination test method in the prior art is shown in Figure 1: a group of personnel short-circuit any two core wires at one end, and short-circuit "1" and "2" on the left side of the cable, and another group of personnel At the other end, on the left side of the cable in Figure 1, use tools such as a multimeter to find the two ends of the short circuit. It is time-consuming to find it blindly, and both ends require personnel to communicate with each other by shouting or using the phone. Communication is very inconvenient. The traditional method is cumbersome. If the distance is long, you have to contact each other by phone, which is easy to make mistakes and takes a long time. When wiring the cable in the roadway without communication signal, because the construction personnel at both ends of the cable cannot get in touch, the wiring becomes inconvenient and difficult.

In view of this, it is particularly necessary to find a convenient, practical, simple, fast, efficient and accurate multi-core cable identification method.

Contents of the invention

In order to solve the above technical problems, the present invention provides a multi-core cable long-distance discrimination method, which uses the combination of electrical knowledge and mathematical knowledge, and through the promotion and use of this method, it can truly solve practical problems in engineering, which is efficient and fast Complete the identification and wiring of multi-core cables.

Its technical scheme is as follows:

A multi-core cable long-distance discrimination method, comprising the following steps:

In the first step, according to the length L (unit: meter) of the cable, the resistivity ρ (unit: ohm m) of the material selected for the cable core wire and the cross-sectional area S (unit: square meter) of the cable core wire, use the formula r ₀ =ρ×L÷S estimated resistance r ₀ (unit: ohm);

The second step is to select the configured test resistance m, as long as m>>r ₀ is satisfied, the value of N is determined according to the total number of cable core wires Z, N=Z-1, and the corresponding resistance R ₀ , R ₁ , The values of R ₂ , R ₃ , ..., R _N are respectively 0×m, 1×m, 2×m, 3×m, ......, N×m;

The third step is to make a resistor-cable connector, and number and connect the multi-core cables according to the order of 0, 1, 2, 3, ..., N at one end of the cable;

The fourth step is to randomly select three core wires at the other end of the cable, set the numbers as a _i , a _j , a _k , measure the resistance values M _ij , M _jk , M _ki between the two with a multimeter, list the equations in turn and Solve N _ij , N _jk , N _ki , and deduce the conclusion formula.

(1) Core wire numbers a _i and a _j

① Resistance composition between core wires: (i×m+r ₀ )+(j×m+r ₀ )

②Measurement data: M _ij

③ Calculated data [rounded]: N _ij = [M _ij /m]

(2) Core wire numbers a _j and a _k

① Resistance composition between core wires: (j×m+r ₀ )+(k×m+r ₀ )

②Measurement data: M _jk

③ Calculated data [rounded]: N _jk = [M _jk /m]

(3) Core wire numbers a _k and a _i

① Resistance composition between core wires: (k×m+r ₀ )+(i×m+r ₀ )

②Measurement data: M _ki

③ Calculated data [rounded]: N _ki = [M _ki /m]

According to the above data, the following three equations are obtained:

$\left\{\begin{array}{c}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; ij</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; jk</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; the\; ki</span>}}\end{array}\right.$

Considering "m>>r0", a new equation is obtained:

After that, the solution is: (remember this conclusion, just apply it directly)

(conclusion formula) $\left\{\begin{array}{c}i=(N_{\mathrm{ij}}-N_{\mathrm{jk}}+N_{\mathrm{the\; ki}})\&divide;2\\ j=(N_{\mathrm{ij}}+N_{\mathrm{jk}}-N_{\mathrm{the\; ki}})\&divide;2\\ k=(-N_{\mathrm{ij}}+N_{\mathrm{jk}}+N_{\mathrm{the\; ki}})\&divide;2\end{array}\right.$ (Launch conclusion) $\left\{\begin{array}{c}{a}_i=i\\ a_j=j\\ a_k=k\end{array}\right.$

The fifth step, using the previous conclusion formula, the values of i, j, and k obtained are the numbers of the corresponding cable core wires a _i , a _j , a _k ;

The sixth step, on this basis, use any one of a _i , a _j , a _k (such as a _i ) to combine with any other a _n , measure the resistance value M _in between them, and calculate N after rounding _in = [M _in /m], get n=N _in -i, n is the number of the corresponding cable core wire a _n ; and so on, put a ₀ , a ₁ , a ₂ , a ₃ ,.... .., the numbers of a and _n are all found out, so that the wiring is correct.

Compared with prior art, the beneficial effect of the present invention:

The scope of application of the method of the present invention: underground coal mines, roadways or tunnels during road and bridge construction, and the method shows its superiority in the occasions where the communication signal is poor.

Since the resistance value of the metal core wire of the cable is generally very small (for example, when the length is about 1000 meters, when the cross-sectional area of the core wire is 1 square millimeter, the resistance r ₀ is on the order of ohms), as long as m>>r ₀ is satisfied, so Generally, the selected test resistance m=10K is sufficient.

The "resistance-cable connector" made by the present invention with the test resistance m=10K can be used repeatedly after being made once, and is suitable for all common cable tests.

The invention saves manpower, has low manufacturing cost of the resistance-cable connector, greatly facilitates the connection and maintenance of on-site engineers and technicians, and has high speed and high efficiency.

Description of drawings

Figure 1 is a traditional multi-core cable discrimination test diagram;

Figure 2 is a schematic diagram of a multi-core cable;

Figure 3 is a connection diagram for the multi-core cable discrimination test.

Detailed ways

The technical solutions of the present invention will be further described in detail below in conjunction with the accompanying drawings and specific embodiments.

Referring to Figure 2, it is a schematic diagram of a multi-core cable.

Table 1 Resistivity of common metal core wires (take L=1000m, S=1×10 ^-6 m ² as an example)

As shown in Figure 3, the multi-core cable long-distance discrimination method of the present invention comprises the following steps:

In the first step, according to the length L (unit: meter) of the cable, the resistivity ρ (unit: ohm m) of the material selected for the cable core wire and the cross-sectional area S (unit: square meter) of the cable core wire, use the formula r ₀ =ρ×L÷S estimated resistance r ₀ (unit: ohm);

The second step is to select the configured test resistance m (as long as m>>r ₀ is satisfied), determine the value of N (N=Z-1) according to the total number of cable core wires Z, and determine the corresponding resistance R ₀ , R ₁ , R ₂ , R ₃ , ..., R _N are respectively 0×m, 1×m, 2×m, 3×m, ......, N×m;

The third step is to make a resistor-cable connector, and number and connect the multi-core cables according to 0, 1, 2, 3, ..., N at one end of the cable;

The fourth step is to randomly select three core wires at the other end of the cable, set the numbers as a _i , a _j , a _k , measure the resistance values M _ij , M _jk , M _ki between the two with a multimeter, list the equations in turn and Solve N _ij , N _jk , N _ki , and deduce the conclusion formula.

(1) Core wire numbers a _i and a _j

① Resistance composition between core wires: (i×m+r ₀ )+(j×m+r ₀ )

②Measurement data: M _ij

③ Calculated data [rounded]: N _ij = [M _ij /m]

(2) Core wire numbers a _j and a _k

① Resistance composition between core wires: (j×m+r ₀ )+(k×m+r ₀ )

②Measurement data: M _jk

③ Calculated data [rounded]: N _jk = [M _jk /m]

(3) Core wire numbers a _k and a _i

① Resistance composition between core wires: (k×m+r ₀ )+(i×m+r ₀ )

②Measurement data: M _ki

③ Calculated data [rounded]: N _ki = [M _ki /m]

According to the above data, the following three equations are obtained:

$\left\{\begin{array}{c}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; ij</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; jk</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; the\; ki</span>}}\end{array}\right.$

Considering "m>>r ₀ ", a new equation is obtained:

After that, the solution is: (remember this conclusion, just apply it directly)

(conclusion formula) $\left\{\begin{array}{c}i=(N_{\mathrm{ij}}-N_{\mathrm{jk}}+N_{\mathrm{the\; ki}})\&divide;2\\ j=(N_{\mathrm{ij}}+N_{\mathrm{jk}}-N_{\mathrm{the\; ki}})\&divide;2\\ \end{array}\right.$ (Launch conclusion) $\left\{\begin{array}{c}{a}_i=i\\ a_j=j\\ \end{array}\right.$

k＝(-N _ij +N _jk +N _ki )÷2a _k ＝k

The fifth step, using the previous conclusion formula, the values of i, j, and k obtained are the numbers of the corresponding cable core wires a _i , a _j , a _k ;

The sixth step, on this basis, use any one of a _i , a _j , a _k (such as a _i ) to combine with any other a _n , measure the resistance value M _in between them, and calculate N after rounding _in = [M _in /m], get n=N _in -i, n is the number of the corresponding cable core wire a _n ; and so on, put a ₀ , a ₁ , a ₂ , a ₃ ,.... .., the numbers of a and _n are all found out, so that the wiring is correct.

[Application example] Take a 200-meter-long 10-core cable (each core cable is 2 square millimeters, and the resistivity of the copper conductor ρ=1.75×10 ^-8 Ωm) is introduced as an example.

Known conditions: L=200 meters, S=2×10 ^-6 square meters, ρ=1.75×10 ^-8 ohm·meter, N=9

The first step is to estimate r ₀ =ρ×L÷S=(1.75×10 ^-8 )×200÷(2×10 ^-6 )=1.75Ω;

The second step is to select the configured test resistance m (as long as m>>r ₀ is satisfied): select m=10K, and the corresponding resistances a ₀ , a ₁ , a ₂ , a ₃ ,..., a The values of ₉ are 0, 10K, 20K, 30K, ..., 90K;

The third step is to make a "resistance-cable connector" and connect it to one end of the cable; the numbers corresponding to the core wires connected to the resistance 0, 10K, 20K, 30K, ..., 90K are 0, 1, 2, 3, ..., 9;

The fourth step is to randomly find three core wires at the other end of the cable, and test their resistance values in pairs;

Table 2 Resistance value between three core wires (example)

The fifth step, using the previous conclusion formula, can calculate i=5, j=8, k=3; these three lines correspond to No. 5, No. 8 and No. 3 lines on the left;

The sixth step is to find the numbers of the remaining wire ends in turn, so as to connect them correctly.

The above is only the best implementation mode of the present invention, any simple changes or equivalent replacements of the technical solutions that can be clearly obtained by any person skilled in the art within the technical scope disclosed in the present invention all fall into the scope of the present invention within the scope of protection.

Claims (1)

1. a multi-core cable long-distance discrimination method is characterized in that, comprising the following steps: In the first step, according to the length L of the cable, the resistivity ρ of the material selected for the cable core wire and the cross-sectional area S of the cable core wire, use the formula r ₀ =ρ×L÷S to estimate the resistance r ₀ ; The second step is to select the configured test resistance m, m>>r ₀ , determine the value of N according to the total number of cable core wires Z, N=Z-1, and determine the corresponding resistance R ₀ , R ₁ , R ₂ , R ₃ , ......, R _N values are 0×m, 1×m, 2×m, 3×m, ......, N×m; The third step is to make a resistor-cable connector, and number and connect the multi-core cables according to 0, 1, 2, 3, ..., N at one end of the cable; The fourth step is to randomly select three core wires at the other end of the cable, set the numbers as a _i , a _i , a _k , measure the resistance values M _ij , M _jk , M _ki between the two with a multimeter, list the equations in turn and Solve N _ij , N _jk , N _ki , and derive the conclusion formula: (1) Core wire numbers a _i and a _j ① Resistance composition between core wires: (i×m+r ₀ )+(j×m+r ₀ ) ②Measurement data: M _ij ③Calculation data, rounding up: N _ij ＝[M _ij /m] (2) Core wire numbers a _j and a _k ① Resistance composition between core wires: (j×m+r ₀ )+(k×m+r ₀ ) ②Measurement data: M _jk ③Calculation data, rounding up: N _jk ＝[M _jk /m] (3) Core wire numbers a _k and a _i ① Resistance composition between core wires: (k×m+r ₀ )+(i×m+r ₀ ) ②Measurement data: M _ki ③Calculation data, rounded: N _ki = [M _ki /m] According to the above data, the following three equations are obtained: $\left\{\begin{array}{c}<span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; ij</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; j</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; jk</span>}}\\ <span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; k</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; +</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; i</span>}<span\; class="notranslate">\; \&times;</span>\mathrm{<span\; class="notranslate">\; m</span>}<span\; class="notranslate">\; +</span>{\mathrm{<span\; class="notranslate">\; r</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; =</span>{\mathrm{<span\; class="notranslate">\; m</span>}}_{\mathrm{<span\; class="notranslate">\; the\; ki</span>}}\end{array}\right.$ Considering "m>>r0", a new system of equations is obtained: After that, the solution is: conclusion formula $\left\{\begin{array}{c}i=(N_{\mathrm{ij}}-N_{\mathrm{jk}}+N_{\mathrm{the\; ki}})\&divide;2\\ j=(N_{\mathrm{ij}}+N_{\mathrm{jk}}-N_{\mathrm{the\; ki}})\&divide;2\\ k=(-N_{\mathrm{ij}}+N_{\mathrm{jk}}+N_{\mathrm{the\; ki}})\&divide;2\end{array}\right.$ conclusion $\left\{\begin{array}{c}{a}_i=i\\ a_j=j\\ a_k=k\end{array}\right.$ In the fifth step, using the conclusion formula in the fourth step, the values of i, j, and k obtained are the numbers of the corresponding cable core wires a _i , a _j , and a _k ; The sixth step is to combine any one of a _i , a _j , a _k with any other a _n , measure the resistance value M _in between them, and calculate N _in = [M _in /m] after rounding to get n=N _in -i, n is the number of the corresponding cable core wire a _n ; by analogy, find out the numbers of a ₀ , a ₁ , a ₂ , a ₃ ,..., a _n , so that the wiring is correct.