CN109835200A

CN109835200A - A kind of paths planning method of the uniline wireless charging sensor network of bridge

Info

Publication number: CN109835200A
Application number: CN201910001921.2A
Authority: CN
Inventors: 刘贵云; 彭百豪; 蒋文俊; 向建化; 张杰钊; 唐冬
Original assignee: Guangzhou University
Current assignee: Guangzhou University
Priority date: 2019-01-02
Filing date: 2019-01-02
Publication date: 2019-06-04
Anticipated expiration: 2039-01-02
Also published as: CN109835200B

Abstract

The invention discloses a kind of paths planning method of the uniline wireless charging sensor network of bridge, specific steps include: the Bridge Joints being arranged needed for (1) is obtained based on linear topology structure；(2) the case where being 2 and minimum line number according to coverage, solution obtains optimal interstitial content；(3) in the case of being uniformly distributed according to two node linears it is optimal spiral a little and optimal path distribution；(4) in the case of being uniformly distributed according to three node linears it is optimal spiral a little and optimal path distribution；(5) according to two nodes and three node optimal disk reconnaissance situations, optimal in the case of n node spiral a little and the optimal path of unmanned plane charging is obtained.The present invention provides the planing methods of the optimal charge path of unmanned plane, and the sensor network in bridge parameter monitoring system can be made to run for a long time.

Description

A kind of paths planning method of the uniline wireless charging sensor network of bridge

Technical field

The present invention relates to wireless energy transmission technology field more particularly to a kind of uniline wireless charging sensor networks of bridge The paths planning method of network.

Background technique

With the high speed development of domestic economy, the construction level of domestic bridge is constantly promoted, various across river bridge spanning the seas, public affairs The dual-purpose bridge of iron is announced to the world splendidly, and as bridge construction must be longer, volume is bigger, and structure is more complicated, becomes to the maintenance of bridge construction It obtains particularly important.And domestic many bridges expose to the sun and rain by long-term, the damage that internal structure generates is difficult to discover, once this A little structural damages cannot be discovered and be repaired in time, it will cause irremediable heavy losses.

Therefore, carrying out regularly monitoring to bridge is highly desirable.But traditional manpower maintenance, not only human cost Height, and the accuracy of measurement may be caused inadequate due to subjective reason.But it is universal with wireless sensor network, this Problem has obtained very good solution.But continuous with wireless network scale increases and the continuous growth of message transmission rate, Consumption explosive growth of the wireless terminal device to energy in wireless network, and the battery capacity of the wireless device in wireless network The performance of wireless network is significantly limited with battery life, it may be said that limited battery life is design and development wireless network Bottleneck where.And in recent years, wireless charging technology starts prevailing, which can well solve that extend wireless network raw The problem of ordering the period.

Summary of the invention

It is an object of the invention to be based on common bridge structure, for wireless chargeable sensor network, take optimal Nodal Arranging Scheme extends the life cycle of wireless chargeable sensor network, provides a kind of uniline wireless charging fax of bridge The paths planning method of sensor network.

The purpose of the present invention can be achieved through the following technical solutions:

A kind of paths planning method of the uniline wireless charging sensor network of bridge, specific steps include:

(1) Bridge Joints being arranged needed for being obtained based on linear topology structure；

(2) the case where being 2 and minimum line number according to coverage, solution obtains optimal interstitial content；

(3) in the case of being uniformly distributed according to two node linears it is optimal spiral a little and optimal path distribution；

(4) in the case of being uniformly distributed according to three node linears it is optimal spiral a little and optimal path distribution；

(5) according to two nodes and three node optimal disk reconnaissance situations, obtain in the case of n node it is optimal spiral a little with And the optimal path of unmanned plane charging.

Specifically, in step (3), for two maximized situations of node energy, lemma 1, lemma 2 and lemma are proposed 3, it is based on lemma 1, lemma 2 and lemma 3, theorem 1 is proposed to the maximized unmanned plane charge path of energy in the case of two nodes With theorem 2.

Further, lemma 1, lemma 2 are identical with the scene that lemma 3 is based on, it is assumed that scene are as follows: fix two nodes and exist On bridge, it is parallel to the side of bridge, using cartesian space rectangular coordinate system as reference, two nodes are symmetrically placed on zero Point both ends, it is assumed that the linear distance between two nodes is D, node is placed in x-axis, the coordinate of left node isAnd the coordinate of right side node is thenBase station is located at the same side of bridge with two nodes, and distance is right Side gusset certain distance, it is assumed that the coordinate of base station is (ξ, 0,0), whereinAt the beginning of charge cycle, unmanned plane is sailed Start to fly from base station；The prestissimo of unmanned plane during flying is V, and unmanned plane flies with the line of node in parallel；Unmanned plane distance x The distance perseverance of axis is h, i.e. the horizontal flying height of unmanned plane is h, and unmanned plane during flying point coordinate is (ξ-Vt, 0, h)；Unmanned plane from from It opens base station and rises and start to charge to node, including flight course and the process spiraled；

The lemma 1 specifically:

One and only one optimal spirals a little when the following conditions are met:

Condition 1: the corresponding Lagrange multiplier of node is equal；

Condition 2: the ratio between euclidean distance between node pair D and the flying height H of unmanned plane is equal to or less than

Then optimal spiral is a little (0,0,0)；

The lemma 2 specifically:

When the following conditions are met, there can be two optimal to spiral a little:

Condition 1: the corresponding Lagrange multiplier of node is equal；

Condition 2: the ratio between euclidean distance between node pair D and the flying height H of unmanned plane is greater than

The then optimal coordinate to spiral a little are as follows: (X1,0,0) and (X2,0,0), wherein X1 and X2 meets:

The lemma 3 specifically:

When the following conditions are met, it optimal spirals a little there is only one:

Condition 1: the corresponding Lagrange multiplier of node is unequal；

The optimal coordinate that spirals corresponding at this time is (x, 0,0).Wherein x will not be determined depending on specific numerical value Quantic.

Further, the theorem 1 specifically:

Assuming that the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station For (ξ, 0,0)；The full-flight velocity of unmanned plane is V；The charge cycle of unmanned plane is T, and the numerical value of T is greater than to spiral a little back and forth Time；Assuming that the flying height of unmanned plane is h；Corresponding coordinate in the flight path of two side gussets, base station and unmanned plane Middle y is 0；

In the case where meeting lemma 1 or lemma 3, that is, contain only one it is optimal when spiral, unmanned plane it is optimal The description in path are as follows:

Unmanned plane sails out of base station since 0 moment,Time in, unmanned plane need to be flown to straight most with speed V Flash disk revolves point, and spirals the time of a little spiraling optimalFinally it is in remaining timeUnmanned plane needs It is maked a return voyage immediately with speed V, returns to base station.

During unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two nodes The gross energy of capture is E2, then it is based on lemma 1 or lemma 3, the ceiling capacity of unmanned plane capture are as follows: E=E1+E2.

The theorem 2 specifically:

Consider the unmanned plane charge path under two node energies maximize, assume initially that: the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station is (ξ, 0,0)；The most fast flight speed of unmanned plane Degree is V；Unmanned plane charge cycle is T, and the numerical value of T is greater than a little required time of spiraling back and forth；The flying height of unmanned plane is h；The airborne period of unmanned plane is T, and T has to be larger thanIt is right in the flight path of two side gussets, base station and unmanned plane Y is 0 in the coordinate answered；

In the case where meeting lemma 2, i.e., containing there are two when optimal spiral, the description of optimal path are as follows:

Unmanned plane sails out of base station from 0 moment, in the timeIt is interior, unmanned plane need to be flown to speed V right side first It is optimal to spiral a little, it spirals the time of a little spiraling at thisIt then needs to make a return voyage immediately with speed V, returns to base station.

During unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two nodes The gross energy of capture is E2, is based on lemma 2, the ceiling capacity of unmanned plane capture are as follows: E=E1+E2.

Specifically, in step (4), for three maximized situations of node energy, lemma 4 and lemma 5 is proposed, is based on Lemma 4 and lemma 5 propose theorem 3 and theorem 4 to the maximized unmanned plane charge path of energy in the case of three nodes.

Further, lemma 4 is consistent with scene corresponding to lemma 5, it is assumed that scene are as follows: by node symmetry distribution, divides For 3 points of left, middle and right, linear topology structure, base station is located at right side node side, and the coordinate of left node is (- d, 0,0), in The coordinate of intermediate node is then (0,0,0), and corresponding to the position of origin, the coordinate of right side node is then (d, 0,0).Base station x-axis away from From for ξ, corresponding coordinate is (ξ, 0,0).Unmanned plane starting point is located at base station；Unmanned plane full-flight velocity is v；

The lemma 4 specifically:

When meeting following three conditions:

Condition 1:Wherein

Condition 2:Wherein Δ₁=720h⁸+2304h⁴d⁴+576h²d⁶,

Δ₂=10368h¹²+82944h⁸d⁴+165888h⁴d⁸；

Condition 3:When；

When being unsatisfactory for the following conditions 4, node capture energy, which maximizes corresponding optimal spiral a little, to be had and only one A, coordinate is (0,0, h):

Condition 4:

When meeting above-mentioned condition, but being unsatisfactory for the following conditions 5:

Condition 5:

The energy of node capture at this time maximizes corresponding optimal spiral and has two, coordinate respectively (X1,0, h), (X2,0, h), wherein X1, X2 are as follows:

When node meets condition 4, meets condition 5, but is unsatisfactory for condition 6:

Condition 6:

Node capture energy maximizes corresponding spiral and has 3, respective coordinates respectively (X1,0, h), (X2,0, H) and (0,0, h), wherein corresponding spiral in the left and right sides a little is spiraled a little for maximum, corresponding spirals of origin is a little maximum value It spirals a little, wherein the coordinate of X1 and X2 corresponds to:

When node meets 4 condition 5 of condition and condition 6, node capture energy maximizes corresponding spiral and a little deposits At 4, the corresponding coordinate that spirals is (X1,0, h), (X2,0, h), (X3,0, h) and (X4,0, h), X1, X2, X3 and X4 Corresponding coordinate is respectively as follows:

The lemma 5 specifically:

When being unsatisfactory for 4 conditional 1 of lemma, when 3 either condition of condition 2 or condition, one and only one is optimal to spiral a little:

Only having the optimal a little corresponding coordinate that spirals at this time is (0,0, h).

, will be left according to unmanned plane to a formula for charging power consumption, after the coordinate of neutralization right side node substitutes into respectively, respectively To Q₁, Q₂And Q₃.It is correspondingAnd it enables

ψ=Q₁+Q₂+Q₃ (4.28)。

Further, the theorem 3 specifically:

The charge path of unmanned plane in three maximized situations of node energy, it is assumed that three node linear distributions；Left side The coordinate of node is (- d, 0,0)；The coordinate of intermediate node is then (0,0,0), the position corresponding to origin；The seat of right side node Mark is then (d, 0,0)；The coordinate of base station is (ξ, 0,0)；The flying height of unmanned plane is h；The charge cycle of unmanned plane is T；Nothing Man-machine full-flight velocity is V；The corresponding y value of the flight path coordinate of three nodes, base station and unmanned plane is 0.

According to the difference of 4 conditional of lemma of satisfaction, following 4 kinds of situations can be divided into:

Situation 1: when the condition 1 in the condition for meeting lemma 4, under the premise of condition 2 and condition 3, being unsatisfactory for condition 4, The charge cycle that must satisfy unmanned plane simultaneously is greater thanConsider that the maximized movement routine of unmanned plane rechargeable energy is retouched at this time It states and answers are as follows:

Unmanned plane in zero moment from base station, with speed V fly to straight it is optimal corresponding to origin spiral a little, when spiraling BetweenAfterwards, V flies back base station straight at a same speed.

Situation 2: when the condition 1 in the condition for meeting lemma 4, condition 2 under the premise of condition 3 and condition 4, is unsatisfactory for Condition 5, the charge cycle that must satisfy unmanned plane at this time are greater thanConsider the maximum mobile road of unmanned plane rechargeable energy Diameter description is answered are as follows:

Unmanned plane in zero moment from base station, with speed V fly to straight it is optimal corresponding to X1 spiral a little, spiral the timeAfterwards, it is flown back straight base station with speed V.

Situation 3: when the condition 1 in the condition for meeting lemma 4, condition 2, condition 3, condition 4 and condition 5, but It is unsatisfactory for condition 6, guarantees that the charge cycle of unmanned plane is greater than at this timeConsider that unmanned plane rechargeable energy maximumlly moves road Diameter description is answered are as follows:

Situation 4: when meeting conditions all in lemma 4, there are four maximum to spiral a little, if X1 corresponds to maximum Maximum corresponding greater than X2 guarantees that the charge cycle of unmanned plane is greater thanConsider that unmanned plane rechargeable energy is maximized Movement routine description is answered are as follows:

If the corresponding maximum of X1 is less than the corresponding maximum of X2, guarantee that the charge cycle of unmanned plane is greater than Consider that the maximized movement routine description of unmanned plane rechargeable energy is answered are as follows:

Unmanned plane sails out of base station from zero moment, with speed V fly to straight it is optimal corresponding to X2 spiral a little, spiral the timeAfterwards, then with speed V it flies back straight base station.

Assuming that the gross energy of three nodes capture is E1, and when unmanned plane spirals, three nodes are caught when unmanned plane shuttle flight The gross energy obtained is E2, then, the correlated condition at this time based on lemma 4, the ceiling capacity of node capture, i.e. unmanned plane charge Ceiling capacity is E=E1+E2.

The theorem 4 specifically:

The charge path of unmanned plane in three maximized situations of node energy, it is assumed that three node linear distributions；Left side The coordinate of node is (- d, 0,0)；The coordinate of intermediate node is then (0,0,0)；Corresponding to the position of origin, the seat of right side node Mark is then (d, 0,0)；The coordinate of base station is (ξ, 0,0)；The flying height of unmanned plane is h；The flying speed of unmanned plane is V；Nothing Man-machine airborne period is T, and wherein T has to be larger thanThe coordinate pair of the flight path of three nodes, base station and unmanned plane The y value answered is 0；

When meeting the condition of lemma 5, optimal spiral a little there is only 1.At this point, to make node capture energy maximum, The description of unmanned plane charge path is answered are as follows:

Since 0 moment, after unmanned plane sails out of base station, it is necessary to prestissimo V drive towards it is optimal spiral a little, in the optimal disk Rotation point spirals the timeBase station is returned to prestissimo V at once afterwards.

Specifically, it in step (5), according to two nodes and three node optimal disk reconnaissance situations, obtains theorem 5, determine Reason 6 and theorem 7, so as to obtain optimal in the case of n node spiral a little and the optimal path that charges of unmanned plane.

Further, the theorem 5 specifically:

In the case that n node is uniformly distributed about origin symmetry, the derivative of the gross energy of node capture contains only surprise It is several.

The theorem 6 specifically:

In the case that n node is uniformly distributed about origin symmetry, when an optimal number of spiraling is even number, origin must be Minimum point.

The theorem 7 specifically:

In the case that n node is uniformly distributed about origin symmetry, when an optimal number of spiraling is odd number, origin It must be maximum of points.

The present invention compared to the prior art, have it is below the utility model has the advantages that

The present invention provides the optimal Nodal Arranging Scheme of linear topology based on bridge, i.e. guarantee number of nodes is least same When, coverage is 2 interstitial content, then solve to obtain interstitial content be 2 and interstitial content be 3 in the case of the optimal of unmanned plane fill Then power path summarizes certain law when n node, to obtain optimal charge path.

Detailed description of the invention

Fig. 1 is specific steps flow chart of the invention.

Specific embodiment

Present invention will now be described in further detail with reference to the embodiments and the accompanying drawings, but embodiments of the present invention are unlimited In this.

Embodiment

Specifically, in the step (2), according to the symmetry on the direction x and the symmetry on the direction y, by bridge Three regions are divided into, based on the number for reducing node as far as possible, only consider the case where coverage is 2.

Bridge width is W=24m, and the search radius of sensor node is R=10m, sensor node apart from bridge one end, Here what is defined is that the length of hithermost one end is D, the half of two sensor node circle center line connectings on the direction y, definition For H.

As sensor node position is constantly close to bridge center, i.e. the euclidean distance between node pair of bridge two sides constantly reduces, Here in order to which operation is convenient, the same round numbers of the value of D.

Situation one: D=3m, H=9m.In the case that coverage is 2, the distance of node will obviously reduce on the direction x, also It is that the circle on right side and the intersection point of upper end bridge wanted the intersection point in the left node center of circle and upper end bridge vertical range on the direction x, feelings Condition is as follows:

Two sections can be found out by the right angled triangle that the intersection point of two nodes and right circles and bridge upper end forms The distance of point, the distance of two nodes are as follows:Numerical value is 9.5m, at this time it can be found that being 1 than above-mentioned coverage In the case of the distances of corresponding two nodes also want small, condition is unsatisfactory for verification condition when coverage is 2 at this time.

Situation two: D=4m, H=8m.In the case that coverage is 2, with the method for situation one, find out between two nodes away from From for 9.2m, the region for coverage occur and being 1 only possible at this time, for the direction x, a certain node, the node of the left and right sides The node is substantially completely covered, unique region is exactly the node by a fan section of two coverages of left and right Domain, if circle can cover this region on the direction y, region of the node on bridge can be completely covered.By verifying, Circle on the direction y, which can be crossed, covers the fan-shaped region.At this point, the number of nodes needed altogether is 43.

Situation three: D=5m, H=7m.In the case that coverage is 2, distance of center circle 7.1m, verifying is found, on the direction y The fan-shaped region that coverage is 1 can be completely covered in node circle, and the node number needed at this time is 47.

Situation four: D=6m, H=6m.In the case that coverage is 2, distance between two nodes is 8, and when verifying finds, the side y Upward circle can just cover fan-shaped region, the node number needed at this time 50.

Situation five: D=8m, H=4m.The node center of circle is vertically met on the intersection point and node circle and bridge of bridge upper end at this time The distance of the intersection point at end is greater than the distance in the two node centers of circle, it is meant that the intersection point of two nodes circle is located at the downside of bridge upper end, deposits In the region being not covered with.So at this time in order to cover the region, right side node must be as left, until just covering this Region, in the case that coverage is 2, it is 66 that two nodal distances, which are interstitial content required for 12m, at this time.

In conclusion with node more mobile, the case where coverage is 2 required interstitial content meeting toward the middle line of bridge It is more and more.In the case that coverage is 2, the interstitial content at least needed is 43.

Specifically, in step (3), lemma 1, lemma 2 is identical with the scene that lemma 3 is based on, it is assumed that scene are as follows: fixed Two nodes are parallel to the side of bridge on bridge, using cartesian space rectangular coordinate system as reference, by two nodes Symmetrically it is placed on zero point both ends, it is assumed that the linear distance between two nodes is D, node is placed in x-axis, the coordinate of left node ForAnd the coordinate of right side node is thenBase station is located at the same side of bridge with two nodes, distance Right side node certain distance, it is assumed that the coordinate of base station is (ξ, 0,0), whereinAt the beginning of charge cycle, unmanned plane Base station is sailed out of to start to fly；The prestissimo of unmanned plane during flying is V, and unmanned plane flies with the line of node in parallel；Unmanned plane away from Perseverance is h with a distance from x-axis, i.e. the horizontal flying height of unmanned plane is h, and unmanned plane during flying point coordinate is (ξ-Vt, 0, h)；Unmanned plane It charges since leaving base station to node, including flight course and the process spiraled；

Assuming that unmanned plane possesses the output energy consumption P of continuous-stable, then each node energy work rate obtained can by with Lower formula is described:

Wherein, η indicates the energy conversion efficiency of each node, and the value of η is greater than 0 less than 1, usually in the ideal case, false If the value perseverance of η is 1.β indicates to obtain channel power gain when distance is by 1 meter in a reference distance.H indicates unmanned plane When flight path is parallel to two node lines, the vertical range of two parallel lines.x_k, y_kX corresponding to node is respectively indicated, y-axis is sat Mark.X, y then indicate x and y coordinates corresponding to the track of unmanned plane during flying.

Further, the lemma 1 specifically:

Condition 1: the corresponding Lagrange multiplier of node is equal；

Then optimal spiral is a little (0,0,0).

Further, the lemma 2 specifically:

Condition 1: the corresponding Lagrange multiplier of node is equal；

Further, the lemma 3 specifically:

Condition 1: the corresponding Lagrange multiplier of node is unequal；

Specifically, when two node capture ceiling capacities, it is necessary to meet the following conditions:

Condition 1: establishing a constraint equation, constrains two nodes, the sum of energy of node is expressed as E, left side Energy percentage shared by node is a₁, the percentage of energy shared by the node of right side is a₂, therefore have a₁+a₂=1.

Condition 2: carrying out constraint to left node hasSimilarly, right side node energy values obtained It is greater than and is equal to a₂E.It can go to solve MaxE by way of solving Lagrangian.

According to condition 1 and 2, maximum ENERGY E is obtained, it is necessary to meet following three formula:

In order to obtain Max L, problem is analyzed using Lagrange duality method, according to the original of Lagrange duality method Then, it is necessary to first L be assumed, it is assumed that Max L bounded can get maximum value, i.e. E can get maximum value.

On the basis of this assumption, it can be deduced that formula below:

Because when E is intended to infinite, the multinomial of E will beIfSo E will Conclusive influence can be generated on L, that is, L will also tend to infinite, so the presence in order to guarantee Max L, it has to be assumed that Formula 4.5 is set up.

In primal problem, MaxE is objective function, and constraint condition is formula 4.3 and formula 4.4, wherein α₁+α₂=1, It is first fixed Lagrange multiplier using Lagrange duality method, corresponding herein is λ₁And λ₂Wherein λ₁And λ₂It is all larger than Equal to 0, then solve optimal unknown number, herein it is corresponding be unmanned plane during flying track x coordinate, finally solve again Optimal λ₁And λ₂, so establishing new Equation f (λ using Lagrange duality method₁,λ₂), it is as follows:

It, can be by solving Minf (λ according to paired method₁,λ₂) method acquire MaxE.

When solving Max L, L function can be split into two parts, two parts take maximum, then L is affirmed Maximum value can be obtained.L is split into the content of following two brackets:

In first bracket due toSo its numerical value perseverance is 0, so Max L is in this case, it can only It is determined by the content of second bracket, that is, in second bracket, can first assume unmanned plane in all the period of time all optimal It spirals a little, first finds outAfter finding out optimal spiral a little again, after unmanned plane flies to optimal spiral a little with prestissimo, Guarantee that the last period can return to base station just with prestissimo, also ensure that has the time as much as possible most within the period On flash disk rotation point.

It enablesIn order to seek the maximum value of ψ, the symbol of this field interior derivative is verified Whether change judges whether the point is extreme point.

Because left and right sides node is respectively positioned in x-axis, correspond to its y-axis, horizontal coordinate is 0, so unmanned plane pair The formula of node charge power can be with indentily transformation are as follows:

Corresponding to left node,And for right side nodeIt is updated in ψ and obtains following formula:

Then it differentiates to ψ, variable x, that is, ask

It needs exist for that λ is discussed₁λ₂Two kinds of situations: λ₁=λ₂With λ₁≠λ₂；

The proof of lemma 1 and lemma 2 is as follows:

Work as λ₁=λ₂When establishment, the constant term occurred after the abbreviation of molecule will be cancelled completely, remaining Xiang Quan Portion is that the multinomial comprising x forms, as follows:

It enablesMolecule turn to it is most simple after be the biquadratic equation containing x, enable a=x²Afterwards, after molecule abbreviation are as follows:

Using the radical formula of quadratic equation, substitutes into and solves and can obtain:

Because of a=x²If then x, so needing to carry out Taxonomic discussion to the content of a, can enable here a without solution so a < 0 =0, solution obtainsSo working asWhen, the root or repeated root that there are two x is not mutually equal, whenOne and only one null solution at this time.WhenWhen, it is false in the solution of a because according to formula 4.13 If sign takes negative sign, then a will be without solution, so sign can only take positive sign, and guarantee that a is greater than zero, so non trivial solution Are as follows:

Null solution is also to work asWhen one solution, verifying can obtain, in section (- ∞, x₁) function monotonic increase, Section (x₁, 0) when function monotone decreasing, in section (0, x₂) when corresponding function monotonic increase, in section (x₂,+∞) when it is corresponding Function monotone decreasing, so ψ is corresponding in x1, the position of origin and x2 obtain extreme value.Extreme value is maximum value at this time, by pole ψ (x is found after being worth numerical value comparison₁)=ψ (x₂) > ψ (0), it is minimum corresponding to zero point, so working asWhen, have Two optimal to spiral a little.So far the proof of lemma 2 finishes.

WhenWhen, one and only one null solution, verifying obtains, when x < 0, function monotonic increase, Function monotone decreasing when working as x > 0, so that is, at this time x=0 is optimal spirals a little at this point, obtaining maximum value at x=0.Extremely This lemma 1 proof finishes.

The proof of lemma 3 is as follows:

Work as λ₁≠λ₂When establishment, molecule will become 5 power formulas, and the multinomial one including x from 0 to 5 time shares 24 , due to not corresponding Method of Seeking Derivative, at this time only one it is optimal spiral a little, it is as a result as follows:

1, whenAnd λ₁> λ₂When, the section of optimal solution is

2, whenAnd λ₁< λ₂When, optimal solution section is

3, whenAnd λ₁> λ₂When, optimal solution section is

4, whenAnd λ₁< λ₂When, optimal solution section is

According to λ₁=λ₂Comparative analysis in, when base station is closer to right side node, it is clear that λ₁> λ₂It is uncomfortable It closes, because optimal at this time spiral a little closer to left node, and unmanned plane leaves to sail to optimal from right side node and spiral a little In the process, total reception energy of sensor node can undergo the process of a reduction, guarantee that only one optimal spirals a little Under the premise of, the solution section closer to right side node is obviously optimal solution, so when base station is closer to right side node.Select λ₁< λ₂The case where, and when base station is closer to left node, select λ₁> λ₂The case where be then optimal.So far lemma 3 proves It finishes.

Pass through above-mentioned discussion proof procedure, it can be deduced that such conclusion:

In the case where two nodes, when λ is equal, the ratio of the flying height h of euclidean distance between node pair D and unmanned plane is big InWhen, there are two it is optimal spiral a little, when the ratio is less thanWhen, only one optimal to spiral a little.And work as λ not When equal, all only one optimal spirals a little anyway.

Further, the theorem 1 specifically:

Consider the unmanned plane charge path under two node energies maximize, assume initially that: the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station is (ξ, 0,0)；The most fast flight speed of unmanned plane Degree is V；The charge cycle of unmanned plane is T, and the numerical value of T is greater than the time spiraled a little back and forth；Assuming that the flying height of unmanned plane For h；Y is 0 in corresponding coordinate in the flight path of two side gussets, base station and unmanned plane；

During remembering unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two sections The gross energy of point capture is E2, then it is based on lemma 1 or lemma 3, the ceiling capacity of unmanned plane capture are as follows: E=E1+E2.

Further, the theorem 2 specifically: consider the unmanned plane charge path under two node energies maximize, Assume initially that: the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station is (ξ,0,0)；The full-flight velocity of unmanned plane is V；Unmanned plane charge cycle is T, and the numerical value of T is greater than required for round-trip spiral a little Time；The flying height of unmanned plane is h；The airborne period of unmanned plane is T, and T has to be larger thanTwo side gussets, base station And y is 0 in corresponding coordinate in the flight path of unmanned plane；

During remembering unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two sections The gross energy of point capture is E2, then it is based on lemma 2, the ceiling capacity of unmanned plane capture are as follows: E=E1+E2.

The proof procedure of theorem 1 is as follows:

Consider only one optimal the case where spiraling, the distance of base station distance origin is ξ, works as λ₁=λ₂And When, it is optimal to spiral a little just in origin.

WhenWhen, unmanned plane sails out of base station, flies to that origin is corresponding optimal to spiral a little straight with speed V. Left side and right side node are respectively as follows: in period energy obtained

AndPeriod in, unmanned plane is just above origin, and two nodes of left and right obtain at this time Energy is same are as follows:

In last remaining time,Interior, unmanned plane finishes two node chargings, returns to base station and carries out It rests and reorganizes, left node energy obtained will are as follows:

The energy that right side node obtains is then are as follows:

The sum of the gross energy that unmanned plane charges to two nodes in whole cycle, i.e. two nodes are based on lemma 1 or draw The ceiling capacity captured under the conditions of reason 3 are as follows:

Work as λ₁≠λ₂When, situation is analyzed as above-mentioned proof procedure, is not just repeated excessively here.So far theorem 1 Proof finishes.

The proof procedure of theorem 2 is as follows:

When containing there are two when optimal spiral, i.e.,Also assume that unmanned plane full-flight velocity is V, Other assume situations with only one it is optimal spiral when it is the same.Proof line at this time is consistent with the proof line of theorem 1, letter For it, i.e., in some cycles T, the flight time is most short, time longest of spiraling.

At this time because the coordinate of base station be (ξ, 0,0), two optimal coordinates to spiral a little are (X1,0,0), (X2,0, 0) two optimal solutions that, X1, X2 respectively correspond lemma 2 only need at this time wherein one spiral and a little spiral, do not consider energy Balanced problem is measured, only considers energy maximization problems, hence it is evident that a little spirals in nearest the spiraling apart from base station, can node be caught Obtain most energy.

?In time, unmanned plane sails out of base station, with speed V fly to straight right side it is optimal spiral a little, at this time To the energy of left node supplement are as follows:

The received energy of right side node is then are as follows:

When the time of timeUnmanned plane is located at that right side is optimal to spiral a little.At this time two nodes by It is different in distance, so received energy is also different, have at this point for left node:

And for right side node, have at this time:

Right side spiral point X1 spiraled one section when after, need immediately with speed V return base station.

The received energy of return course interior joint with reach the received energy values of point process node of spiraling from base station and be The same, have at this time:

So unmanned plane equally only needs to undergo 3 periods when there are two node.Firstly, from base station to arrival The optimal point X1 that spirals in first, right side.Secondly, spiraling a little to spiral and charging to two nodes at this.Finally, when guaranteeing remaining Between in enough situations, from the optimal base station of a little directly flying back of spiraling in first, right side, complete a wheel charging, two nodes are received The ceiling capacity that the sum of gross energy, i.e. two nodes are captured based on lemma 2 are as follows:

So far the proof of theorem 2 finishes.

To sum up, it can be deduced that such as draw a conclusion:

1, when unmanned plane only have one it is optimal spiral when, it is assumed that condition is based on theorem 1, after unmanned plane sails out of base station, needs It directly flies to and spirals a little, when return also needs to fly back straight, and the ceiling capacity of two nodes capture at this time is as shown in formula 4.20.

2, when there are two when spiraling for unmanned plane, it is assumed that condition is based on theorem 2, and when unmanned plane during flying all must be with most Fast speed flight, including in the optimal a little flight between base station of spiraling a little and spiral, but only needs nearest apart from base station Spiral and a little spiral, it is not necessary to spiral at two and a little all spiral.The ceiling capacity such as formula 4.27 that two nodes are captured at this time It is shown.

Further, the lemma 4 specifically:

When meeting following three conditions:

Condition 1:Wherein

Condition 2:Wherein Δ₁=720h⁸+2304h⁴d⁴+576h²d⁶,

Δ₂=10368h¹²+82944h⁸d⁴+165888h⁴d⁸；

Condition 3:When；

Condition 4:

Condition 5:

Condition 6:

Further, the lemma 5 specifically:

, will be left according to unmanned plane to a formula 4.1 for charging power consumption, after the coordinate of neutralization right side node substitutes into respectively, point Q is not obtained₁, Q₂And Q₃.It is corresponding And it enables

ψ=Q₁+Q₂+Q₃ (4.28)

Lemma 4 and lemma 5 are proved with the method for derivation.

In conclusion the case where three nodes, is available to draw a conclusion:

1, when the formula that is unsatisfactory for 4.46, formula 4.47 and formula 4.48, to catch in the case where three node linear arrangements Capacitation amount is maximum, and optimal spiral of unmanned plane puts one and only one.

2, when meeting formula 4.46, when formula 4.47 and formula 4.48, to capture in the case where three node linears arrangements Energy is maximum, according to condition difference, it is understood that there may be 1,2,3 and 4 optimal the case where spiraling.

Further, the theorem 3 specifically:

When the condition 1 in the condition for meeting lemma 4, under the premise of condition 2 and condition 3, it is unsatisfactory for condition 4, while must The charge cycle that unmanned plane must be met is greater thanConsider that the maximized movement routine description of unmanned plane rechargeable energy is answered at this time are as follows:

Assuming that the gross energy of three nodes capture is E1, and when unmanned plane spirals, three nodes are caught when unmanned plane shuttle flight The gross energy obtained is E2, then, the correlated condition at this time based on lemma 4, the ceiling capacity of node capture, i.e. unmanned plane charge Ceiling capacity is E=E1+E2；

Further, the theorem 4 specifically:

Consider three maximized problems of node energy, assume initially that: three node linear distributions；The coordinate of left node For (- d, 0,0)；The coordinate of intermediate node is then (0,0,0), the position corresponding to origin；The coordinate of right side node be then (d, 0, 0)；Assuming that the flying height of unmanned plane is h；The corresponding y value of the coordinate of three nodes, base station and unmanned plane is 0；

The proof procedure of theorem 4 is as follows:

Unmanned plane need to fly to straight first after taking off from base station with speed V and spiral a little, in this process, a corresponding left side The received energy of side gusset are as follows:

The received energy of intermediate node is then are as follows:

And the reception energy of right side node is then are as follows:

Equally work as the timeWhen, unmanned plane is located at right above origin, spirals a little in optimal, at this point, left Right two side gussets are due to symmetrical, so corresponding received energy are as follows:

Intermediate node energy obtained is maximum, because the line distance apart from intermediate node is recently, right at this time for unmanned plane The reception energy answered is

After guaranteeing sufficient remaining time,Time in, unmanned plane must be with most fast flight Speed flies back base station, at this time a corresponding left side, neutralizes the right side received energy of node and is respectively

Then unmanned plane is captured in the case of i.e. three nodes based on lemma 5 at this time to the energy summation of three node-node transmissions Ceiling capacity are as follows:

So far, the proof of theorem 4 finishes.

The proof procedure of theorem 3 is as follows:

Unmanned plane from base station, with speed V drive towards straight this it is optimal spiral a little, i.e., origin is corresponding spirals a little, mobile In the process, the received energy of left node are as follows:

The received energy of intermediate node is then are as follows:

And the reception energy of right side node is then are as follows:

So far 3 situation 1 of theorem proof finishes.

To sum up two kinds of situations, in the case where three nodes, it can be deduced that with next conclusion:

1, when only one spirals, in order to which the energy for capturing node is maximum, unmanned plane should sail out of base When standing erectly to optimal spiral, is flown with prestissimo always, also to be flown during returning to base station with most fast speed, The gross energy of node capture at this time is as shown in formula 4.57；

2, when not only one spiral that a little namely there are three unmanned planes when spiraling in addition to when spiraling, Other when it is mobile with prestissimo when being moved between node.Meet condition, Yi Gongke according to meeting a variety of of lemma 4 at this time To be divided into 7 kinds of situations.

Specifically, it in step (5), according to two nodes and three node optimal disk reconnaissance situations, obtains theorem 5 and determines Reason 6, so as to obtain optimal in the case of n node spiral a little and the optimal path that charges of unmanned plane.

Further, the theorem 5 specifically:

From two, the proof procedure of three and four nodes, it can be seen that at most there is two between adjacent node two-by-two Maximum is spiraled a little, and relative to adjacent node, on line, peripheral node number, which spirals to maximum, can only a little make maximum spiral Point position is moved, and its number cannot be made to change, so corresponding to n node, theoretically, according to it is unknown for several times The relationships of number and solution, that is, the number solved with unknown number number be it is equal, according to the derivation formula of above-mentioned 2,3 and 4 nodes From the point of view of, highest number meets following formula:

Highest time several=1+4 (n-1) (4.123)

So when contain n node when, should theoretically contain up to 4n-3 maximum and spiral a little, and origin no matter It how to be all polynomial solution, because being found from the operation proof procedure of 2,3 and 4 nodes, even item (including number is 0) it can be canceled each other out because of the symmetrical reason of node linear.

Because node linear is symmetrical, even order terms all will by positive and negative counteracting, if that is, there are also even number node, node It is uniformly distributed about origin symmetry, node number 2n+2, linear homogeneous distribution, in the derivative for receiving gross energy, even item Understand and controls and balance out because node is uniformly distributed about origin symmetry.

It is proved with mathematics formula, when node number is even number, the derivative of the corresponding node in the left and right sides about divides it Afterwards, denominator will be the same, and molecule only have two it is different, left and right node derivative reduction of a fraction after molecule be shown below respectively:

(x+d)[x⁴-4x³d+6x²d²+2x²h²-4xd³-4xdh²+d⁴+2d²h²+h⁴]......

(4.124)

(x-d)[x⁴+4x³d+6x²d²+2x²h²+4xd³+4xdh²+d⁴+2d²h²+h⁴]......

(4.125)

Wherein d is the position of left and right sides nodal distance origin, corresponding node derivative reduction of a fraction at left and right sides of ellipsis part Identical part after integration, different parts also arrange to have write and come out.It is found after formula 4.124 and formula 4.125 are unfolded, even item Numerical value is identical but symbol is on the contrary, odd term numerical value is identical and symbol is identical.Therefore because ellipsis part is identical, multinomial Number has odd number also to have even number, and after multiplication, symbol is constant, i.e., after the even item of corresponding formula 4.124 is multiplied by ellipsis part Numerical value it is equal multiplied by the numerical values recited behind ellipsis part with 4.125 even item of formula, symbol is on the contrary, eventually cancelled out each other Fall, the odd term of remaining corresponding formula 4.124 is multiplied by the numerical value behind ellipsis part with 4.125 odd term of formula multiplied by ellipsis Numerical values recited behind part is equal, and symbol is identical, and final two formula is added, and becomes original twice.Other symmetrical nodes two-by-two Analysis and above formula are the same, just do not repeat excessively herein.

When node number is odd number, corresponding node analysis in the left and right sides is the same with above formula, unique different such as figure Shown in 4-21, more nodes positioned at origin.Denominator and other nodes after the derivative reduction of a fraction integration of origin node energy Corresponding multinomial is identical, and molecule is shown below:

x[(x-d)²+h2]²[(x+d)²+h2]²....[(x-(2n+1)d)²+h²]²[(x+(2n+1)d)²+h²]² (4.126)

4.126 clipped of formula is identical as the right and left multinomial, and above formula being understood to, part in addition to x be interpreted as pair Claim the product of square product of node energy capture formula denominator.By one pair of them multinomial, it is assumed that be [(x-d)²+h²]²[(x+ d)²+h²]²It is shown below after expansion:

x⁸-4x⁶d²+4x⁶h²-4x⁴h²d²+6x⁴d⁴+6x⁴h⁴-4x²d⁴h²-4x²d⁶+4x²d²h⁴

4x²h⁶+6h⁴d⁴+4h²d⁶+d⁸+h⁸ (4.127)

It is found from formula 4.127, odd term is all about fallen, and remaining is all even item.Other than this pair, remain Under symmetrical nodes polynomial multiplication and formula 4.127 it is similar, only corresponding coefficient is different.So two/n is to multiplication Afterwards, i.e., countless even items are multiplied, and as a result also must be even item, and the x being multiplied by formula 4.126 all becomes odd term.

So far, n node situation lower derivate contains only odd term proof and finishes.

It it is easy to show that minterm number must be 1 simultaneously, i.e., there was only 0 solution at this time.

Accordingly, because solution must include Min-max, i.e., maximum number adds minimum number that must be 1+4 (n-1), by Spiral a little in reaching first maximum from infinity, the received gross energy of node must monotonic increase, and spiral from the last one Point drives towards infinity, and the received gross energy of node must monotone decreasing.The left and right sides must tend to negative infinity, and maximum point necessarily compares Minimum point is one few.The proof of theorem 5 finishes.

It is former when an optimal number of spiraling is even number in the case that 6:n node of theorem is uniformly distributed about origin symmetry Point must be minimum point.

It proves: in 2,3 and 4 nodes, the case where which centainly sets up, expand to multiple nodes.It is minimum Value point means that field is negative on the left of the derivative, and right side field is positive.

It is identical with the proof of lemma 6, all symmetrically occur two-by-two because the left and right sides is spiraled a little, it is impossible to only occur one Side, and the case where another side is not present, so maximum, which is spiraled, a little to be only possible in the presence of in origin when a number of spiraling is even number The left and right sides, origin are only minimum point.The proof of theorem 6 finishes.

In the case that 7:n node of theorem is uniformly distributed about origin symmetry, when an optimal number of spiraling is odd number It waits, origin must be maximum of points.

Prove: method of proof is consistent with 7 analysis method of lemma when four nodes, and this method is named as symmetry analysis Method.Symmetry analysis is directed to two kinds of situations, and constantly superposition is compared, final to determine that origin is maximum of points.

Firstly, because origin left and right sides maximum is spiraled a little and all occurred two-by-two, the corresponding maximum of origin is had only The number spiraled a little could be become odd number from even number by the appearance spiraled a little, and existence proof finishes.

The first situation using symmetry analysis is node two sides the case where spiraling there are two maximum, at this time with Any node is vertical line of the intersection point work perpendicular to x-axis, it is assumed that the node right end still has x node, the first maximum of vertical line right end It spirals a little, certainly exists a point a in vertical line left end, spiral in first maximum of vertical line right end a little and in a for unmanned plane Point spirals, and to centered on the node, x, right side node, x, the left side received energy of node is necessarily equal.In comparison, Node on the left of origin, due to a point than the node closer to origin, the received energy of the node on the left of origin must at this time So it is more before than, so the received energy ratio of a point first maximum of vertical line right end spiral a little received energy want it is more, At this point, the maximum for being located at vertical line left end spirals and is a little only possible to be overlapped with a point or in the left side of a point.So node more The maximum a little received energy that spirals close to the side of origin necessarily a little connects than spiraling in node far from origin side maximum The energy of receipts is much bigger.

It is that there are two maximum to spiral a little between adjacent node that second situation, which obtains application scenarios, obtains midpoint in adjacent node Make vertical straight line, it is assumed that contain x node on the right side of vertical line, obtain maximum positioned at vertical line right end and spiral a little, certainly exist one Point b and vertical line right end the maximum range vertical that spirals are equidistant, but the b point is located on the left of vertical line.Point b or this hang down On the right side of line maximum is spiraled a little, on the right side of vertical line on the left of x node and vertical line x node receive energy is equally certainly , and it is left to obtain the n-2x node in left side, hence it is evident that since b point is closer to origin, so remaining n-2x node is received Energy certainly than on the right side of vertical line first maximum spiral a little spiral when received energy more than it is more.So in adjacent node Between maximum spiral and a little spiral, when the maximum closer to origin is spiraled, the energy of node capture can be more.

N node linear, which is uniformly distributed, to be made of the first above-mentioned situation with second situation, passes through above-mentioned two The continuous comparison superposition of kind of situation, because node is about origin symmetry, maximum is spiraled a meeting identical appearance two-by-two, so only It needs to analyze wherein one side, the meeting maximum disk more all than the left and right sides so the maximum of final at the origin is spiraled Rotation point will be much bigger, i.e., it is a little that maximum value is spiraled a little that origin, which spirals to deserved maximum, at this time, i.e., optimal to spiral a little.

So the corresponding maximum point of origin must be maximum of points when node number is odd number.Theorem 7 has proved Finish.

According to above-mentioned theorem, using exhaustive search method, obtain in the present embodiment based on the optimal unmanned plane charging spiraled a little Optimal path: distance between one side gusset of bridge is 9.2 meters, and the spacing of bridge two rows node is 16 meters；The upside of node is sat Mark is (Xi, 8,0), and wherein Xi is the abscissa of corresponding node, and the downside coordinate of node is (Xi, -8,0), and wherein Xi is The abscissa of corresponding node；The coordinate of base station is (ξ, -12,0), and wherein ξ is greater than any one Xi；Node is symmetrical with x-axis and y Axis distribution；The full-flight velocity of unmanned plane is V, and the lower limit of drone flying height h is 5 meters；The airborne period of unmanned plane is T, T has to be larger than at this time

When meeting the flying height h of unmanned plane and being higher than 5 meters lower than 11 meters, there are two two it is optimal spiral a little, Corresponding coordinate is [0, Y, h] and [0 ,-Y, h], and wherein the value range of Y is between 0 to 8.So to make bridge two sides Node captures ceiling capacity, and the description of unmanned plane charge path is answered are as follows:

Unmanned plane sails out of base station from zero moment, is flown to speed V and is spiraled a little apart from nearest optimal in base station straight, at this time away from The spiral coordinate nearest from base station is [0 ,-Y, h], so spiraling the time of a little spiraling at this Afterwards, then with speed V it makes a return voyage straight base station.The energy of node capture at this time is maximum.

When the flying height for meeting unmanned plane is greater than 11 meters, only one optimal to spiral a little, and corresponding coordinate is [0,0, h], then two side gusset of bridge to be made to capture ceiling capacity, the description of unmanned plane charge path is answered are as follows:

Unmanned plane since zero moment sails out of base station, with maximum speed V drive towards this it is optimal spiral a little, spiralIt is maked a return voyage immediately with speed V after time.The energy of node capture at this time is maximum.

The above embodiment is a preferred embodiment of the present invention, but embodiments of the present invention are not by above-described embodiment Limitation, other any changes, modifications, substitutions, combinations, simplifications made without departing from the spirit and principles of the present invention, It should be equivalent substitute mode, be included within the scope of the present invention.

Claims

1. a kind of paths planning method of the uniline wireless charging sensor network of bridge, which is characterized in that specific steps include:

(5) according to two nodes and three node optimal disk reconnaissance situations, optimal in the case of n node spiral a little and nothing is obtained The optimal path of man-machine charging.

2. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 1, special Sign is, in step (3), for two maximized situations of node energy, proposes lemma 1, lemma 2 and lemma 3, based on drawing Reason 1, lemma 2 and lemma 3 propose theorem 1 and theorem 2 to the maximized unmanned plane charge path of energy in the case of two nodes.

3. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 2, special Sign is, lemma 1, and lemma 2 is identical with the scene that lemma 3 is based on, it is assumed that scene are as follows: fixes two nodes on bridge, puts down Two nodes, using cartesian space rectangular coordinate system as reference, are symmetrically placed on zero point both ends in the side of bridge by row, false If the linear distance between two nodes is D, node is placed in x-axis, the coordinate of left node isAnd right side The coordinate of node is thenBase station is located at the same side of bridge with two nodes, false apart from right side node certain distance If the coordinate of base station is (ξ, 0,0), whereinAt the beginning of charge cycle, unmanned plane sails out of base station and starts to fly；Nothing The prestissimo of man-machine flight is V, and unmanned plane flies with the line of node in parallel；Distance perseverance of the unmanned plane apart from x-axis is h, i.e., The horizontal flying height of unmanned plane is h, and unmanned plane during flying point coordinate is (ξ-Vt, 0, h)；Unmanned plane since leaving base station to section Point charges, including flight course and the process spiraled；

The lemma 1 specifically:

Condition 1: the corresponding Lagrange multiplier of node is equal；

Then optimal spiral is a little (0,0,0)；

The lemma 2 specifically:

Condition 1: the corresponding Lagrange multiplier of node is equal；

The lemma 3 specifically:

Condition 1: the corresponding Lagrange multiplier of node is unequal；

The optimal coordinate that spirals corresponding at this time is (x, 0,0)；Wherein x will be depending on specific numerical value, generation for not determining Number form formula.

4. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 3, special Sign is, the theorem 1 specifically:

Assuming that the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station is (ξ,0,0)；The full-flight velocity of unmanned plane is V；The charge cycle of unmanned plane is T, and the numerical value of T is greater than to spiral a little back and forth Time；Assuming that the flying height of unmanned plane is h；Y in corresponding coordinate in the flight path of two side gussets, base station and unmanned plane It is 0；

In the case where meeting lemma 1 or lemma 3, that is, contain only optimal when spiral a, optimal path for unmanned plane Description are as follows:

Unmanned plane sails out of base station since 0 moment,Time in, unmanned plane need to fly to straight optimal spiral with speed V Point, and spiral the time of a little spiraling optimalFinally it is in remaining timeUnmanned plane need immediately with Speed V makes a return voyage, and returns to base station；

During unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two node captures Gross energy be E2, then be based on lemma 1 or lemma 3, unmanned plane capture ceiling capacity are as follows: E=E1+E2；

The theorem 2 specifically:

Consider the unmanned plane charge path under two node energies maximize, assume initially that: the coordinate of left node isAnd the coordinate of right side node is thenThe coordinate of base station is (ξ, 0,0)；The most fast flight of unmanned plane Speed is V；Unmanned plane charge cycle is T, and the numerical value of T is greater than a little required time of spiraling back and forth；The flying height of unmanned plane For h；The airborne period of unmanned plane is T, and T has to be larger thanIn the flight path of two side gussets, base station and unmanned plane Y is 0 in corresponding coordinate；

Unmanned plane sails out of base station from 0 moment, in the timeIt is interior, unmanned plane need to be flown to speed V right side first it is optimal It spirals a little, spirals the time of a little spiraling at thisIt then needs to make a return voyage immediately with speed V, returns to base station；

During unmanned plane during flying, the gross energy of two nodes capture is E1, and unmanned plane is during spiraling, two node captures Gross energy be E2, be based on lemma 2, unmanned plane capture ceiling capacity are as follows: E=E1+E2.

5. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 1, special Sign is, in step (4), for three maximized situations of node energy, proposes lemma 4 and lemma 5, is based on 4 He of lemma Lemma 5 proposes theorem 3 and theorem 4 to the maximized unmanned plane charge path of energy in the case of three nodes.

6. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 5, special Sign is that lemma 4 is consistent with scene corresponding to lemma 5, it is assumed that scene are as follows: by node symmetry distribution, is divided into left, middle and right 3 points, linear topology structure, base station is located at right side node side, and the coordinate of left node is (- d, 0,0), the seat of intermediate node Mark is then (0,0,0), and corresponding to the position of origin, the coordinate of right side node is then (d, 0,0)；Base station x-axis distance is ξ, corresponding Coordinate be (ξ, 0,0)；Unmanned plane starting point is located at base station；Unmanned plane full-flight velocity is v；

The lemma 4 specifically:

When meeting following three conditions:

Condition 1:Wherein

Condition 2:Wherein Δ₁=720h⁸+2304h⁴d⁴+576h²d⁶,

Δ₂=10368h¹²+82944h⁸d⁴+165888h⁴d⁸；

Condition 3:When；

When being unsatisfactory for the following conditions 4, node capture energy maximizes corresponding optimal spiral and puts one and only one, sits It is designated as (0,0, h):

Condition 4:

Condition 5:

Condition 6:

Node capture energy maximizes corresponding spiral and has 3, respective coordinates respectively (X1,0, h), (X2,0, h) with And (0,0, h), wherein corresponding spiral in the left and right sides is a little spiraled a little for maximum, corresponding spiral of origin is a little spiraled for maximum value Point, wherein the coordinate of X1 and X2 corresponds to:

When node meets 4 condition 5 of condition and condition 6, node capture energy, which maximizes, corresponding spirals that there are 4 A, the corresponding coordinate that spirals is (X1,0, h), (X2,0, h), (X3,0, h) and (X4,0, h), X1, X2, X3 and X4 pairs The coordinate answered is respectively as follows:

The lemma 5 specifically:

Only having the optimal a little corresponding coordinate that spirals at this time is (0,0, h)；

, will be left according to unmanned plane to a formula for charging power consumption, after the coordinate of neutralization right side node substitutes into respectively, respectively obtain Q₁, Q₂And Q₃；It is correspondingAnd it enables

ψ=Q₁+Q₂+Q₃ (4.28)。

7. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 5, special Sign is, the theorem 3 specifically:

The charge path of unmanned plane in three maximized situations of node energy, it is assumed that three node linear distributions；Left node Coordinate be (- d, 0,0)；The coordinate of intermediate node is then (0,0,0), the position corresponding to origin；The coordinate of right side node is then For (d, 0,0)；The coordinate of base station is (ξ, 0,0)；The flying height of unmanned plane is h；The charge cycle of unmanned plane is T；Unmanned plane Full-flight velocity be V；The corresponding y value of the flight path coordinate of three nodes, base station and unmanned plane is 0；

Situation 1: when the condition 1 in the condition for meeting lemma 4, under the premise of condition 2 and condition 3, it is unsatisfactory for condition 4, simultaneously The charge cycle that must satisfy unmanned plane is greater thanConsider that the maximized movement routine description of unmanned plane rechargeable energy is answered at this time Are as follows:

Unmanned plane in zero moment from base station, with speed V fly to straight it is optimal corresponding to origin spiral a little, spiral the timeAfterwards, V flies back base station straight at a same speed；

Situation 2: when the condition 1 in the condition for meeting lemma 4, condition 2 under the premise of condition 3 and condition 4, is unsatisfactory for condition 5, the charge cycle that must satisfy unmanned plane at this time is greater thanConsider that the maximum movement routine of unmanned plane rechargeable energy is retouched It states and answers are as follows:

Unmanned plane in zero moment from base station, with speed V fly to straight it is optimal corresponding to X1 spiral a little, spiral the timeAfterwards, it is flown back straight base station with speed V；

Situation 3: when the condition 1 in the condition for meeting lemma 4, condition 2, condition 3, condition 4 and condition 5, but it is discontented Sufficient condition 6 guarantees that the charge cycle of unmanned plane is greater than at this timeConsider that the maximized movement routine of unmanned plane rechargeable energy is retouched It states and answers are as follows:

Situation 4: when meeting conditions all in lemma 4, there are four maximum to spiral a little, if X1 corresponds to maximum and is greater than The corresponding maximum of X2 guarantees that the charge cycle of unmanned plane is greater thanConsider that unmanned plane rechargeable energy maximumlly moves Path description is answered are as follows:

If the corresponding maximum of X1 is less than the corresponding maximum of X2, guarantee that the charge cycle of unmanned plane is greater thanConsider The maximized movement routine description of unmanned plane rechargeable energy is answered are as follows:

Unmanned plane sails out of base station from zero moment, with speed V fly to straight it is optimal corresponding to X2 spiral a little, spiral the timeAfterwards, then with speed V it flies back straight base station；

Assuming that the gross energy of three nodes capture is E1 when unmanned plane shuttle flight, when unmanned plane spirals, what three nodes captured Gross energy is E2, then, the correlated condition at this time based on lemma 4, the ceiling capacity of node capture, the i.e. maximum of unmanned plane charging Energy is E=E1+E2；

The theorem 4 specifically:

The charge path of unmanned plane in three maximized situations of node energy, it is assumed that three node linear distributions；Left node Coordinate be (- d, 0,0)；The coordinate of intermediate node is then (0,0,0)；Corresponding to the position of origin, the coordinate of right side node is then For (d, 0,0)；The coordinate of base station is (ξ, 0,0)；The flying height of unmanned plane is h；The flying speed of unmanned plane is V；Unmanned plane Airborne period be T, wherein T has to be larger thanThe coordinate of the flight path of three nodes, base station and unmanned plane is corresponding Y value is 0；

When meeting the condition of lemma 5, optimal spiral a little there is only 1；At this point, to make node capture energy maximum, nobody The description of machine charge path is answered are as follows:

Since 0 moment, after unmanned plane sails out of base station, it is necessary to prestissimo V drive towards it is optimal spiral a little, optimal spiral a little at this It spirals the timeBase station is returned to prestissimo V at once afterwards；

Assuming that the gross energy of three nodes capture is E1 when unmanned plane shuttle flight, when unmanned plane spirals, what three nodes captured Gross energy is E2, then, the correlated condition at this time based on lemma 4, the ceiling capacity of node capture, the i.e. maximum of unmanned plane charging Energy is E=E1+E2.

8. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 1, special Sign is, in step (5), according to two nodes and three node optimal disk reconnaissance situations, obtains theorem 5, theorem 6 and theorem 7, so as to obtain optimal in the case of n node spiral a little and the optimal path of unmanned plane charging.

9. a kind of paths planning method of the uniline wireless charging sensor network of bridge according to claim 8, special Sign is, the theorem 5 specifically:

In the case that n node is uniformly distributed about origin symmetry, the derivative of the gross energy of node capture contains only odd term；

The theorem 6 specifically:

In the case that n node is uniformly distributed about origin symmetry, when an optimal number of spiraling is even number, origin must be minimum Value point；

The theorem 7 specifically:

In the case that n node is uniformly distributed about origin symmetry, when an optimal number of spiraling is odd number, origin must be Maximum of points.