CN106123848A - A kind of measuring method of conducting wire sag - Google Patents

Abstract

The invention discloses a method for measuring the sag of a wire. The two ends of the wire are respectively marked as A ₁ and B ₁ , and the projections of A ₁ and B ₁ on the horizontal plane where O ₁ and O ₂ are located are respectively A ₀ , B ₀ , mark the lowest point of the wire inside A ₁ and B ₁ as C ₁ , and the projection of C ₁ on the projection of O ₁ and O ₂ on the horizontal plane is C ₀ ; measure the distance L between O ₁ and O ₂ , according to Calculate the distance h ₂ of C ₁ C ₀ by the law of cosines; extend A ₁ A ₀ to point D so that B ₁ , C ₀ , and D are on a straight line, and obtain the distance h ₀ from A ₁ B ₁ to C ₀ according to the principle of similar triangles , h ₀ ‑h ₂ is the wire sag. The method for measuring wire sag of the present invention can effectively solve the problems of visual error in the direct measurement method of sag plate and the problem of being unable to measure due to the limitation of eyesight; it can also solve the problem that the distance between poles and towers needs to be known in the method of theodolite measurement.

Description

A method for measuring wire sag

technical field

The invention relates to a measuring method, in particular to a measuring method for wire sag.

Background technique

The vertical distance between any point on the wire and the line connecting the suspension point is called the sag of the wire at that point. The current measurement method of wire sag generally adopts manual direct measurement with the help of sag plate, and also uses theodolite measurement. Although the manual measurement method using the sag plate is relatively simple and convenient, when measuring the wire sag between two towers, because of the long distance, the vision is limited, and the visual error will lead to deviations in the measurement results. When the deviation is large, It will lead to insufficient wire-to-ground distance or insufficient crossing distance, which will lead to grid safety accidents. To measure with a theodolite, it must be measured at the bottom of a pole tower or a place very close to the pole tower, and the horizontal distance between the two pole towers must be known. It is difficult or even impossible to measure the horizontal distance. Therefore, in order to accurately measure the sag of the grid conductor and ensure the safe operation of the grid, the existing measurement methods must be improved.

Contents of the invention

The purpose of the present invention is to provide a method for measuring wire sag, which solves the problems of manual direct measurement with sag plate, or measurement with theodolite, which is difficult to measure, and the measurement results deviate and even cause power grid safety accidents.

The technical solution adopted in the present invention is a method for measuring wire sag, and the specific steps are as follows:

Step 1, select any two points M and N on the horizontal ground, support the theodolite at M and N respectively, the distance from M point to O ₁ and the distance from N point to O ₂ are the height of the theodolite, at this time the theodolite The measurement points are O ₁ and O ₂ respectively, and O ₁ and O ₂ are on the same horizontal plane, and the two ends of the wire are respectively marked as A ₁ and B ₁ , and the projections of A ₁ and B ₁ on the horizontal plane where O ₁ and O ₂ are located are respectively are A ₀ and B ₀ , mark the lowest point of the wires in A ₁ and B ₁ as C ₁ , and the projection of C ₁ on the horizontal plane where O ₁ and O ₂ are located is C ₀ ;

Step 2, measure the distance L between O ₁ and O ₂ , and calculate the distance h ₂ of C ₁ C ₀ according to the law of cosines;

Step 3: Extend A ₁ A ₀ to point D so that B ₁ , C ₀ , and D are on a straight line. According to the similar triangle principle, the distance h ₀ from A ₁ B ₁ to C ₀ is obtained, and h ₀ -h ₂ is the wire Sag.

The present invention is also characterized in that,

The specific calculation process of step 2 is as follows:

Measure the distance L between O ₁ and O ₂ , adjust the observation angle of theodolite at O ₁ , and measure ∠A ₀ O ₁ A ₁ = α ₁ , ∠A ₀ O ₁ O ₂ = β ₁ , ∠O ₂ O ₁ B ₀ ＝ω ₃ , ∠C ₀ O ₁ C ₁ ＝α ₂ , ∠C ₀ O ₁ O ₂ ＝β ₂ ; at O ₂ , adjust the observation angle of theodolite, and measure ∠O ₁ O ₂ A ₀ ＝ω ₁ , ∠B ₀ O ₂ B ₁ ＝α ₃ , ∠B ₀ O ₂ O ₁ ＝β ₃ , ∠O ₁ O ₂ C _O ＝ω ₂ ;

Make the perpendicular segment O ₁ O ₃ from O ₁ to A ₀ O ₂ , and record the length as d1, then in the triangle O ₁ O ₂ O ₃ :

d1＝L×sinω ₁ (1)

Because in triangle A ₀ O ₁ O ₃ :

A ₀ O ₁ =d1/[sin(180-β ₁ -ω ₁ )] (2)

Put (1) into (2) to get:

A ₀ O ₁ =(L×sinω ₁ )/[sin(180-β ₁ -ω ₁ )] (3)

Denote A ₁ A ₀ as h ₁ in triangle A ₁ A ₀ O ₁ because:

h ₁ =A ₀ O ₁ ×tanα ₁ (4)

Substitute (3) into (4) to get:

h ₁ =(L×sinω ₁ ×tanα ₁ )/[sin(180-β ₁ -ω ₁ )] (5)

Make the perpendicular segment O ₂ O ₄ from point O ₂ to B ₀ O ₁ , and record the length as d2, then in the triangle O ₁ O ₂ O ₄ :

d2＝L×sinω ₃ (6)

Because in triangle B ₀ O ₂ O ₄ :

B ₀ O ₂ =d2/[sin(180-β ₃ -ω ₃ )] (7)

Substitute (6) into (7) to get

B ₀ O ₂ =(L×sinω ₃ )/[sin(180-β ₃ -ω ₃ )] (8)

Denote B ₁ B ₀ as h ₃ , because in triangle B ₁ B ₀ O ₂ :

h ₃ =B ₀ O ₂ ×tanα ₃ (9)

Substitute (8) into (9) to get:

h ₃ =(L×sinω ₃ ×tanα ₃ )/[sin(180-β ₃ -ω ₃ )] (10)

According to the law of sines, in triangle C ₀ O ₁ O ₂ ,

C ₀ O ₁ =(L×sinω ₂ )/sin(180-β ₂ -ω ₂ ) (11)

C ₀ O ₂ =(C ₀ O ₁ ×sinβ ₂ )/sinω ₂ (12)

Denote C ₁ C ₀ as h ₂ , then in triangle C ₁ C ₀ O ₁ :

h ₂ ＝C ₀ O ₁ ×tanα ₂ (13)

Substitute (11) into (13) to get

h ₂ =(L×sinω ₂ ×tanα ₂ )/sin(180-β ₂ -ω ₂ ) (14)

The specific calculation process of step 3 is:

Note ∠C ₀ O ₁ A ₀ ＝θ ₁ , A ₀ C ₀ ＝L ₁ , then

θ ₁ =β ₁ -β ₂ (15)

According to the law of cosines, in triangle A ₀ C ₀ O ₁

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; +</span>{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>{\mathrm{<span\; class="notranslate">\; COS\&theta;</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 16</span>}<span\; class="notranslate">\; )</span>$

Substituting (2)(11)(15) into (16), we can calculate L ₁ ,

Note ∠C ₀ O ₂ B ₀ ＝θ ₂ , B ₀ C ₀ ＝L ₂ , then

θ ₂ =β ₃ -ω ₂ (17)

According to the law of cosines, in triangle C ₀ B ₀ O ₂

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; B</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; +</span>{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; B</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>{\mathrm{<span\; class="notranslate">\; COS\&theta;</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 18</span>}<span\; class="notranslate">\; )</span>$

Substitute (7)(12)(17) into (18) to calculate L ₂ ;

The distance between the marks A ₀ D is h ₄ , since the triangles B ₁ B ₀ C ₀ and A ₀ C ₀ D are similar, we get:

h ₄ =(L ₁ ×h ₃ )/L ₂ (19)

Substitute (10)(16)(18) into (19) to calculate h ₄ ;

In triangle A ₁ B ₁ D, because:

h ₀ =[L ₂ (h ₁ +h ₄ )]/(L ₂ +L ₁ ) (20)

Substitute (19) into (20) to get

h ₀ =(h ₁ L ₂ +h ₃ L ₁ )/(L ₂ +L ₁ )

Finally, the sag=h ₀ -h ₂ is obtained by calculation.

The beneficial effect of the present invention is that, the method for measuring the wire sag of the present invention can accurately obtain the wire sag according to the cosine theorem and the principle of similar triangles by measuring the distance between any two observation points on the horizontal ground. There is no visual error, which can effectively solve the visual error of the direct measurement method of the sag plate and the problem that it cannot be measured due to the limitation of vision. This method does not need to be close to the bottom of the tower, and does not need to know the distance between the two towers, and can also solve the problem that the distance between the two towers and the proximity to the bottom of the tower need to be known when measuring the theodolite.

Description of drawings

Fig. 1 is the measuring method structural representation of wire sag of the present invention;

Fig. 2 is a schematic diagram of the measurement structure of the distance h ₀ from A ₁ B ₁ to C ₀ in the method for measuring wire sag of the present invention.

detailed description

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

The measuring method of wire sag of the present invention, concrete steps are as follows:

Step 1, as shown in Figure 1, select any two points M and N on the horizontal ground (M, N is on the same side of the line connecting A ₁ and B ₁ , and M is on the right side of A ₁ , and N is on the side of B ₁ On the left side, the distance between M and N is smaller than the distance between A ₁ and B ₁ ), the theodolite is respectively supported at M and N, the distance from M point to O ₁ and the distance from N point to O ₂ are theodolite the height of. At this time, the measurement points of the theodolite are O ₁ and O ₂ respectively. O ₁ and O ₂ are on the same horizontal plane. Mark the two ends of the wire as A ₁ and B ₁ respectively. A ₁ and B ₁ are on the horizontal plane where O ₁ and O ₂ are located. The projections on are respectively A ₀ and B ₀ , and the lowest point of the inner wire marked A ₁ and B ₁ is C ₁ , and the projection of C ₁ on the horizontal plane where O ₁ and O ₂ are located is C ₀ ;

Step 2, measure the distance L between O ₁ and O ₂ , and calculate the distance h ₂ of C ₁ C ₀ according to the law of cosines;

Specifically: measure the distance L between O ₁ and O ₂ , the measuring point of the theodolite is at O ₁ , adjust the observation angle of the theodolite, and measure ∠A ₀ O ₁ A ₁ =α ₁ , ∠A ₀ O ₁ O ₂ =β1 , ∠O ₂ O ₁ B ₀ ＝ω ₃ , ∠C ₀ O ₁ C ₁ ＝α ₂ , ∠C ₀ O ₁ O ₂ ＝β ₂ ; move the theodolite measurement point to O ₂ , adjust the theodolite observation angle, measure ∠O ₁ O ₂ A _{0 =} ω ₁ , ∠B ₀ O ₂ B ₁ = α ₃ , ∠B ₀ O ₂ O ₁ = β ₃ , ∠O ₁ O ₂ C _O = ω ₂ ;

Make the perpendicular segment O ₁ O ₃ from O ₁ to A ₀ O ₂ , and record the length as d1, then in the triangle O ₁ O ₂ O ₃ :

d1＝L×sinω ₁ (1)

Because in triangle A ₀ O ₁ O ₃ :

A ₀ O ₁ =d1/[sin(180-β ₁ -ω ₁ )] (2)

Put (1) into (2) to get:

A ₀ O ₁ =(L×sinω ₁ )/[sin(180-β ₁ -ω ₁ )] (3)

Denote A ₁ A ₀ as h ₁ in triangle A ₁ A ₀ O ₁ because:

h ₁ =A ₀ O ₁ ×tanα ₁ (4)

Substitute (3) into (4) to get:

h ₁ =(L×sinω ₁ ×tanα ₁ )/[sin(180-β ₁ -ω ₁ )] (5)

Make the perpendicular segment O ₂ O ₄ from point O ₂ to B ₀ O ₁ , and record the length as d2, then in the triangle O ₁ O ₂ O ₄ :

d2＝L×sinω ₃ (6)

Because in triangle B ₀ O ₂ O ₄ :

B ₀ O ₂ =d2/[sin(180-β ₃ -ω ₃ )] (7)

Substitute (6) into (7) to get

B ₀ O ₂ =(L×sinω ₃ )/[sin(180-β ₃ -ω ₃ )] (8)

Denote B ₁ B ₀ as h ₃ , because in triangle B ₁ B ₀ O ₂ :

h ₃ =B ₀ O ₂ ×tanα ₃ (9)

Substitute (8) into (9) to get:

h ₃ =(L×sinω ₃ ×tanα ₃ )/[sin(180-β ₃ -ω ₃ )] (10)

According to the law of sines, in triangle C ₀ O ₁ O ₂ ,

C ₀ O ₁ =(L×sinω ₂ )/sin(180-β ₂ -ω ₂ ) (11)

C ₀ O ₂ =(C ₀ O ₁ ×sinβ ₂ )/sinω ₂ (12)

Denote C ₁ C ₀ as h ₂ , then in triangle C ₁ C ₀ O ₁ :

h ₂ ＝C ₀ O ₁ ×tanα ₂ (13)

Substitute (11) into (13) to get

h ₂ =(L×sinω ₂ ×tanα ₂ )/sin(180-β ₂ -ω ₂ ) (14)

Step 3, as shown in Figure 2, extend A ₁ A ₀ to point D, so that B ₁ , C ₀ , and D are on a straight line, and obtain the distance h ₀ and h ₀ from A ₁ B ₁ to C ₀ according to the principle of similar triangles -h ₂ is the wire sag.

The specific calculation process is:

Note ∠C ₀ O ₁ A ₀ ＝θ ₁ , A ₀ C ₀ ＝L ₁ , then

θ ₁ =β ₁ -β ₂ (15)

According to the law of cosines, in triangle A ₀ C ₀ O ₁

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; =</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; +</span>{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; A</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}<span\; class="notranslate">\; )</span>{\mathrm{<span\; class="notranslate">\; COS\&theta;</span>}}_{\mathrm{<span\; class="notranslate">\; 1</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 16</span>}<span\; class="notranslate">\; )</span>$

Substituting (2)(11)(15) into (16), we can calculate L ₁ ,

Note ∠C ₀ O ₂ B ₀ ＝θ ₂ , B ₀ C ₀ ＝L ₂ , then

θ ₂ =β ₃ -ω ₂ (17)

According to the law of cosines, in triangle C ₀ B ₀ O ₂

${\mathrm{<span\; class="notranslate">\; L</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; =</span>\sqrt{{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; B</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; +</span>{<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>}^{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; -</span>\mathrm{<span\; class="notranslate">\; 2</span>}<span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; B</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span><span\; class="notranslate">\; (</span>{\mathrm{<span\; class="notranslate">\; C</span>}}_{\mathrm{<span\; class="notranslate">\; 0</span>}}{\mathrm{<span\; class="notranslate">\; o</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}<span\; class="notranslate">\; )</span>{\mathrm{<span\; class="notranslate">\; COS\&theta;</span>}}_{\mathrm{<span\; class="notranslate">\; 2</span>}}}<span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; -</span><span\; class="notranslate">\; (</span>\mathrm{<span\; class="notranslate">\; 18</span>}<span\; class="notranslate">\; )</span>$

Substitute (7)(12)(17) into (18) to calculate L ₂ ;

The distance between the marks A ₀ D is h ₄ , since the triangles B ₁ B ₀ C ₀ and A ₀ C ₀ D are similar, we get:

h ₄ =(L ₁ ×h ₃ )/L ₂ (19)

Substitute (10)(16)(18) into (19) to calculate h ₄ ;

In triangle A ₁ B ₁ D, because:

h ₀ =[L ₂ (h ₁ +h ₄ )]/(L ₂ +L ₁ ) (20)

Substitute (19) into (20) to get

h ₀ =(h ₁ L ₂ +h ₃ L ₁ )/(L ₂ +L ₁ )

Finally, the sag=h ₀ -h ₂ is obtained by calculation.

The method of the invention is not limited by eyesight, and has no visual error, and can effectively solve the problems of visual error and inability to measure due to the limitation of eyesight in the direct measurement method of the sag plate. The general method of measuring and calculating sag with theodolite must be measured at the bottom of a pole tower or very close to the pole tower, and the horizontal distance between the two pole towers must be known. When the two towers are on the banks of the ditch or on two hills, it is difficult or even impossible to measure the horizontal distance between the two towers; or because there are trees and weeds at the bottom of the towers, it is difficult to approach the towers. This method does not need to be close to the bottom of the tower, and does not need to know the distance between the two towers. It can effectively solve the problem that it is difficult to approach the tower when measuring and calculating the sag with theodolite and the horizontal distance between the two towers needs to be known. This method The required data is easy to measure, and the sag of the wire can be obtained through mathematical calculation of the measured data.

Claims (3)

1. the measuring method of a conducting wire sag, it is characterised in that specifically comprise the following steps that

Step 1, any two points M, N on selected level ground, theodolite is supported in M, at N, the now measurement point of theodolite It is respectively O₁、O₂, O₁、O₂Being in same level, labelling wire two ends are respectively A respectively₁、B₁, A₁、B₁At O₁、O₂Place water Projection in plane is respectively A₀、B₀, labelling A₁、B₁Inside conductor minimum point is C₁, C₁At O₁、O₂Being projected as in the horizontal plane C₀；

Step 2, measures O₁、O₂Between distance L, according to the cosine law calculate C₁C₀Distance h₂；

Step 3, extends A₁A₀To D point, make B₁、C₀, D point-blank, obtain A according to similar triangle theory₁B₁To C₀Away from From h₀, h₀-h₂It is conducting wire sag.

The measuring method of conducting wire sag the most according to claim 1, it is characterised in that it is as follows that step 2 specifically calculates process:

Measure O₁、O₂Between distance L, Theodolite Measuring Point is at O₁Place, adjusts theodolite observation angle, records ∠ A₀O₁A₁=α₁, ∠ A₀O₁O₂=β₁, ∠ O₂O₁B₀=ω₃, ∠ C₀O₁C₁=α₂, ∠ C₀O₁O₂=β₂；Theodolite Measuring Point is moved to O₂Place, adjusts longitude and latitude Instrument view angle, records ∠ O₁O₂A₀₌ω₁, ∠ B₀O₂B₁=α₃, ∠ B₀O₂O₁=β₃, ∠ O₁O₂C_O=ω₂；

It is O₁To A₀O₂Vertical line section O₁O₃, length is designated as d1, then at triangle O₁O₂O₃In:

D1=L × sin ω₁ (1)

Because at triangle A₀O₁O₃In:

A₀O₁=d1/ [sin (180-β₁-ω₁)] (2)

(1) is brought into (2) obtain:

A₀O₁=(L × sin ω₁)/[sin(180-β₁-ω₁)] (3)

Note A₁A₀For h₁, at triangle A₁A₀O₁In, because:

h₁=A₀O₁×tanα₁ (4)

(3) are substituted into (4) obtain:

h₁=(L × sin ω₁×tanα₁)/[sin(180-β₁-ω₁)] (5)

It is O₂Point arrives B₀O₁Vertical line section O₂O₄, length is designated as d2, then at triangle O₁O₂O₄In:

D2=L × sin ω₃ (6)

Because at triangle B₀O₂O₄In:

B₀O₂=d2/ [sin (180-β₃-ω₃)] (7)

(6) are substituted into (7) obtain

B₀O₂=(L × sin ω₃)/[sin(180-β₃-ω₃)] (8)

Note B₁B₀For h₃, because at triangle B₁B₀O₂In:

h₃=B₀O₂×tanα₃ (9)

(8) are substituted into (9) obtain:

h₃=(L × sin ω₃×tanα₃)/[sin(180-β₃-ω₃)] (10)

According to sine, at triangle C₀O₁O₂In,

C₀O₁=(L × sin ω₂)/sin(180-β₂-ω₂) (11)

C₀O₂=(C₀O₁×sinβ₂)/sinω₂ (12)

Note C₁C₀For h₂, then at triangle C₁C₀O₁In:

h₂=C₀O₁×tanα₂ (13)

(11) are substituted into (13) obtain

h₂=(L × sin ω₂×tanα₂)/sin(180-β₂-ω₂) (14)。

The measuring method of conducting wire sag the most according to claim 2, it is characterised in that the concrete calculating process of step 3 is:

Note ∠ C₀O₁A₀=θ₁, A₀C₀=L₁, then

θ₁=β₁-β₂ (15)

According to the cosine law, at triangle A₀C₀O₁In

$L_1=\sqrt{{\left(A_0{O}_1\right)}^2+{\left(C_0{O}_1\right)}^2-2\left(A_0{O}_1\right)\left(C_0{O}_1\right){\mathrm{COS\&theta;}}_1}---\left(16\right)$

(2) (11) (15) are substituted into (16), L can be calculated₁,

Note ∠ C₀O₂B₀=θ₂, B₀C₀=L₂, then

θ₂=β₃-ω₂ (17)

According to the cosine law, at triangle C₀B₀O₂In

$L_2=\sqrt{{\left(B_0{O}_2\right)}^2+{\left(C_0{O}_2\right)}^2-2\left(B_0{O}_2\right)\left(C_0{O}_2\right){\mathrm{COS\&theta;}}_2}---\left(18\right)$

(7) (12) (17) are substituted into (18), calculates L₂；

Labelling A₀Distance between D is h₄, due to triangle B₁B₀C₀And A₀C₀D is similar:

h₄=(L₁×h₃)/L₂ (19)

(10) (16) (18) are substituted into (19), calculates h₄；

At triangle A₁B₁In D, because:

h₀=[L₂(h₁+h₄)]/(L₂+L₁) (20)

(19) are substituted into (20) obtain

h₀=(h₁L₂+h₃L₁)/(L₂+L₁)

Finally by being calculated sag=h₀-h₂。