Simulation Model of Target Exposure

2017 2nd International Conference on Computational Modeling, Simulation and Applied Mathematics (CMSAM 2017) ISBN: 978-1-60595-499-8

Simulation Model of Target Exposure

Ji-rong XUE, Yong SUN

, Yuan FENG and Tie-zheng LIU

China Research and Development Academy of Machinery Equipment, Beijing, China

*Corresponding author

Keywords: Target exposure, Simulation, Model.

Abstract. This paper proposes a simulation model of target exposure, including simulation model of target exposure height, simulation model of the target exposure height relative to the observation line, simulation model of target exposure. The simulation model of target exposure height is used to calculate height of target exposure part. The simulation model of the target exposure height relative to the observation line is used to calculate height of the target exposed part relative to the observation line during the search. The simulation model of the target exposure is used to calculate the parameters and geometric center of the exposed part of the target of the vehicle or airplane relative to the observation line or the incident line. The model is simple and feasible.

Simulation Model of Target Exposure Height

Input Parameters

For the vehicle class object, the input parameters include Z coordinate Z1 at the bottom of the target under the vehicle body coordinate system, Z coordinate Z2 at the top of the vehicle body, Z coordinate Z3 at the top of the turret (for a target without a turret, Z3 = Z2), the above parameters are obtained from the combat unit integrated protection performance data file, the protection state of the target (obtained from the received entity status information).

Similar definition of the aircraft class target.

Output Parameters

Target exposure part height hh2.

Simulation Model

For vehicles or aircraft class target, if the target is not within the fortifications, hh2Z3Z1 . For the vehicle class target, if the target in the fortifications, hh2Z3Z2.

For combat personnel, if the protection status is standing, the combatant height hh2 = 1.7m, if the protection status is not standing, the combatant height hh2 = 0.3m.

Simulation Model of the Target Exposure Height Relative to the Observation Line

Calculate the target exposure height relative to the observation line during the search.

Input Parameters

For the vehicle and aircraft class object, the input parameters include Z coordinate Z1 at the bottom of the target, Z coordinate Z2 at the top of the vehicle body, Z coordinate Z3 at the top of the turret (for a target without a turret, Z3 = Z2), the above parameters are obtained from the combat unit integrated protection performance data file, the target coordinates （X2, Y2, Z2） (obtained from the received

entity status information), the observation point coordinates ( , , 1/) / 1 /

1 Y Z

X _{, the protection state of the}

For the combatant class object, the input parameters include the coordinates of the target (X2, Y2,

Z2), the target equivalent height, the observation point coordinates ( , , 1/) / 1 /

1 Y Z

X _{, and the terrain data.}

Output Parameters

The output parameter is the target exposure height hh at the searching time.

Simulation Model A B C D E O Fi

For vehicles that are not within fortifications, or for aircraft targets, make

3 2 1 2 2 max 2 min Z Z Z Z Z Z    

If the target is within the fortifications, make

3 2 2 2 2 max 2 min Z Z Z Z Z Z    

For combat personnel, make

2 2 2 2 max 2 min hh Z Z Z Z   

Z2min is the lowest point of the target observable part, Z2max is the highest point of the target observable part, hh2 is the combatant equivalent height.

Known AB is the observation equipment, B is the view point, CE is the goal, if the length OD is find from the terrain, the target exposure height DE could be find. Each point A, B, C, E geographical

coordinates are (lA,bA,hA), (lB,bB,hB), (lC,bC,hC)_, (lE,bE,hE).

Known latitude and longitude coordinates of the lower left corner of the map is (l0,b0,h0)_{, the}

latitude and longitude coordinates of the top right is (l1,b1,h1)_{, the map length and width are L and W}

respectively, each point A, B, C, E latitude and longitude coordinates are:

The each point coordinates of the latitude and longitude into the geocentric coordinates of the geocentric system, each point coordinates marked (XA,YA,ZA), (XB,YB,ZB),(XC,YC,ZC),(XE,YE,ZE), A,

C correspond to the geodetic coordinates are (xA,yA,zA)_, (xC,yC,zC)_.

Insert N equal points Fk(xk,yk,zk)(k0,,N1) _{between AC:}

  

   

 



   

   

) , (

1 1

k k k

A B A k

A B A k

y x f z

k N

y y y y

k N

x x x x

Into geocentric coordinates for Fk(Xk,Yk,Zk).

The topographic data (l, b) are arranged in a grid, the sides of the grid are Δl and Δb, and Δd = max (△ l, Δb), N is obtained as follows:









max(      

 l l d b b d

N _{B A B A}

hk according to the following method:

An adjacent four points in the terrain data form a rectangle (as shown in the figure below).

(i,j) (i,j+1)

(i+1,j+1) (i+1,j)

First, according to (lk,bk) _{determine the value of i, j, and obtain the latitude and longitude} coordinates:





i ( _k  ₀)/





j ( _k  ₀)/

l j lj  

b i b_i  

According to the terrain file, read the elevation values at the four vertices:

) , ( 1 f lj bi

h 

) ,

( ₁

2 f lj bi

h  _

) ,

( ₁

3  f lj bi

h

) ,

( _{1 1}

4  f lj bi

h

According to the elevation values h1, h2, h3, h4 at the four vertices, the elevation hk corresponding to the inner （lk,bk） of the rectangle is obtained by linear interpolation method.

l l l h h h

h k j

    

 ₁ ( _{2 1}) '

l l l h h h

h k j

    

 3 ( 4 3) ''

b b b h h h

h k i

k _

   

For k  0,,N 1, B, Fk _{The point where the extension of the two-point line intersects the} OE is D, calculate the lengthOD of OD respectively:

) , , (X_B Y_B Z_B OB

) , , (X_E Y_E Z_E OE 

) , , ( ) ,

,

( k B k B k B k k k

k X X Y Y Z Z X Y Z

BF        

2 2 2 2 2 2 1 cos

E E E B B B

E B E B E B

Z Y X Z Y X

Z Z Y Y X X BOE

  



  

 

2 2 2 2 2 2

1 ( )

cos

k k k B B B

k B k B k B

Z Y X Z Y X

Z Z Y Y X X OBD

      

      

 

OBD BOE

ODB  

 

ODB OBD OB

OD

  

sin sin

Thus corresponding to each Fk_{, D point elevation} r

OD h_kD  

Where

2 2 2

C C

C Y Z

X

r  

is earth radius at point C, set maxhmax{hkD}_{, and if} maxhhE_{, then}

the target is not visible, hh0, and if maxhhC _{, the target is completely transparent,}

min

max 2

2 Z

Z hh 

, and if hC maxhhE_{, and if} hC maxhhE_{, the target is partially visible and}

the exposure height is hhhE maxh_.

Target Exposure Body Simulation Model

The model is used to calculate the vehicle type or aircraft type target relative to the observation line or the exposure line of the exposed surface parameters and geometric center.

Input Parameters

Enter the target exposure height hh (relative to the observation line or incident line), the target outline size (obtained from the combat unit integrated protection performance data file), the coordinates of the target in the geodetic coordinate system (X2 , Y2 , Z2) and the attitude (obtained from the received entity status information), the top of the target Z coordinate Z3.

Output Parameters

Output parameters of the outer surface of the exposed body in the geodetic system(vertex, center coordinates), geometric center of the exposed body.

Calculate the Vertex and Center Coordinates of the Exposed Body

Let the target have n outer surfaces p1…p n, and there are rectangular, The center of each surface is E1…En, let a outer surface center Ei coordinates is （XiE，YiE，ZiE）under the vehicle system, the

four vertices are respectively Ai,Bi,Ci,Di_{, the vertices coordinates are}(XiA,YiA,ZiA)，(XiB,YiB,ZiB)， )

, ,

(X_iC Y_iC Z_{iC ，}(X_iD,Y_iD,Z_iD)

under car body coordinate system

The equivalent shading height in the body (target) coordinate system is:

hh Z

If ZiA,ZiB,ZiC,ZiD _{are all greater than or equal to h, the outer surface is unobstructed; if}

iD iC iB iA Z Z Z

Z , , , _{are all less than h, the outer surface is completely obscured, eliminating the surface; if}

there are two greater than h in ZiA,ZiB,ZiC,ZiD_{, the outer surface is partly obscured,}

Suppose there is k1 partly obscured outer surface, set ZiA,ZiD _{is greater than h, then the area} covered by the total area of the ratio is:

iB iA iB i Z Z Z h K   

Calculate the vertices and center coordinates of the unobstructed rectangles in the body coordinate system: ) , , ( ) , ,

(x_iA y_iA z_iA  X_iA Y_iA Z_iA

) , , ( ) , ,

(x_iD y_iD z_iD  X_iD Y_iD Z_iD

) , , ( ) , , ( ) , ,

(xiB yiB ziB  XiB YiB ZiB Ki XiA XiB YiAYiB ZiA ZiB

) , , ( ) , , ( ) , ,

(x_iC y_iC z_iC  X_iC Y_iC Z_iC K_i X_iDX_iC Y_iDY_iC Z_iDZ_iC

4

/

)

,

,

(

)

,

,

(

x



y



z





x





x





x





x



y





y





y





y



z





z





z





z



Suppose there is k2 unobstructed outer surface, calculate the rectangular vertex and center coordinates under the car body coordinate system:

) , , ( ) , ,

(x_iA y_iA z_iA  X_iA Y_iA Z_iA (x_iD ,y_iD ,z_iD)(X_iD,Y_iD,Z_iD)

) , , ( ) , ,

(x_iB y_iB z_iB  X_iB Y_iB Z_iB

) , , ( ) , ,

(x_iC y_iC z_iC  X_iC Y_iC Z_iC

4 / ) , , ( ) , ,

(x_iE y_iE z_iE  x_iA x_iB x_iC x_iD y_iA y_iB y_iC y_iD z_iA z_iB z_iC z_iD

The rectangular vertex and center coordinates of the unobstructed rectangles under the geodetic coordinate system: ) , , ( ) , , ( ) , ,

( 2 2 2 iA iA iA

n b iA

iA

iA y z X Y Z C x y z

x     

) , , ( ) , , ( ) , ,

( 2 2 2 iD iD iD

n b iD

iD

iD y z X Y Z C x y z

x     

) , , ( ) , , ( ) , ,

( 2 2 2 iB iB iB

n b iB

iB

iB y z X Y Z C x y z

x     

) , , ( ) , , ( ) , ,

( 2 2 2 iC iC iC

n b iC

iC

iC y z X Y Z C x y z

x     

) , , ( ) , , ( ) , ,

( 2 2 2 iE iE iE

n b iE

iE

iE y z X Y Z C x y z

x     

Calculate the Geometric Center of the Exposed Body

Target exposure part height hh2.

) , ,

(xiE yiE ziE _{is the exposed body outer surface coordinates in the geodetic system, and the}

geometric center of the exposed body is ( , , )

' 2 ' 2 '

2 Y Z

X _{, according to the exposed body outer surface}

p1… pk ,(k=k1+k2), we can get:

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