PATENT DOCUMENT

Abstract:
A method and apparatus for guiding a vehicle to intercept a target is described. The method iteratively estimates a time-to-go until target intercept and modifies an acceleration command based upon the revised time-to-go estimate. The time-to-go estimate depends upon the position, the velocity, and the actual or real time acceleration of both the vehicle and the target. By more accurately estimating the time-to-go, the method is especially useful for applications employing a warhead designed to detonate in close proximity to the target. The method may also be used in vehicle accident avoidance and vehicle guidance applications.

Full Description:
FIELD OF THE INVENTION 
     The present invention relates to a method of and apparatus for guiding a missile. In particular, the present invention provides for a method of guiding a missile based upon the time of flight until the missile intercepts the target, i.e., the time-to-go. 
     BACKGROUND OF THE INVENTION 
     There is a need to estimate the time it will take a missile to intercept a target or to arrive at the point of closest approach. The time of flight to intercept or to the point of closest approach is known as the time-to-go τ. The time-to-go is very important if the missile carries a warhead that should detonate when the missile is close to the target. Accurate detonation time is critical for a successful kill. Proportional navigation guidance does not explicitly require time-to-go, but the performance of the advanced guidance law depends explicitly on the time-to-go. The time-to-go can also be used to estimate the zero effort miss distance. 
     One method to estimate the flight time is to use a three degree of freedom missile flight simulation, but this is very time consuming. Another method is to iteratively estimate the time-to-go by assuming piece-wise constant positive acceleration for thrusting and piece-wise constant negative acceleration for coasting. Yet another method is to iteratively estimate the time-to-go based upon minimum-time trajectories. 
     Tom L. Riggs, Jr. proposed an optimal guidance method in his seminal paper “Linear Optimal Guidance for Short Range Air-to-Air Missiles” by (Proceedings of NAECON, Vol. II, Oakland, Mich., May 1979, pp. 757-764). Riggs&#39; method used position, velocity, and a piece-wise constant acceleration to estimate the anticipated locations of a vehicle and a target/obstacle and then generated a guidance command for the vehicle based upon these anticipated locations. To ensure the guidance command was correct, Riggs&#39; method repeatedly determined the positions, velocities, and piece-wise constant accelerations of both the vehicle and the target/obstacle and revised the guidance command as needed. Because Riggs&#39; method did not consider actual, or real time acceleration in calculating the guidance command, a rapidly accelerating target/obstacle required Riggs&#39; method to dramatically change the guidance command. As the magnitude of the guidance command is limited, (for example, a fin of a missile can only be turned so far) Riggs&#39; method may miss a target that it was intended to hit, or hit an obstacle that it was intended to miss. Additionally, many vehicles and targets/obstacles can change direction due to changes in acceleration. Riggs&#39; method, which provided for only piece-wise constant acceleration, may miss a target or hit an obstacle with constantly changing acceleration. 
     Computationally, the fastest methods use only missile-to-target range and range rate or velocity information. This method provides a reasonable estimate if the missile and target have constant velocities. When the missile and/or target have changing velocities, this simple method provides time-to-go estimates that are too inaccurate for warheads intended to detonate when the missile is close to the target. 
       FIG. 1  illustrates two different prior art methods for determining time-to-go.  FIG. 1  shows a missile  100  with a net velocity v relative to the target at a missile-to-target angle relative to the LOS between the missile  100  and a target  104 . The net velocity v is a function of both the missile  100  and the target  104  velocities. The missile-to-target range is shown as r. As such a target intercept scheme occurs in three-dimensional space, vectors will be shown in bold, while the magnitudes of such vectors will be shown as standard text. 
     Assuming the missile and target velocities are constant, the distance between the missile  100  and target  104  at time t is:
 
 z=r+vt.   Eq. 1
 
The miss distance is minimized when
 
                       ∂     (     z   ·   z     )         ∂   t       =   0.           Eq   .           ⁢   2               
Substituting Eq. 1 into Eq. 2 yields:
   r·v+v·vt= 0.  Eq. 3 
Solving Eq. 3, the time-to-go τ is:
 
                   τ   =     -         v   ·   r       v   ·   v       .               Eq   .           ⁢   4               
Eq. 4 yields the exact time-to-go if the missile  100  and target  104  have constant velocities.
 
     The minimum missile-to-target position vector z can be obtained by substituting Eq. 4 into Eq. 1 resulting in: 
                   z   =             (     v   ·   v     )     ⁢   r     -       (     v   ·   r     )     ⁢   v         v   ·   v       =           (     v   ×   r     )     ×   v       v   ·   v       .               Eq   .           ⁢   5               
The zero-effort-miss distance, corresponding to the magnitude of the minimum missile-to-target position vector z, illustrated as point P in  FIG. 1 , is:
 
     
       
         
           
             
               
                 
                   
                      
                     z 
                      
                   
                   = 
                   
                     
                        
                       
                         
                           
                             ( 
                             
                               v 
                               × 
                               r 
                             
                             ) 
                           
                           × 
                           v 
                         
                         
                           v 
                           · 
                           v 
                         
                       
                        
                     
                     = 
                     
                       
                         
                           
                             v 
                             2 
                           
                           ⁢ 
                           r 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           sin 
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           α 
                         
                         
                           v 
                           2 
                         
                       
                       = 
                       
                         r 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         sin 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           α 
                           . 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   6 
                 
               
             
           
         
       
     
     The prior art time-to-go formulation is simply: 
                     τ   =     r     r   .         ,           Eq   .           ⁢   7               
where {dot over (r)} is the range rate. The difference between Eq. 4 and Eq. 7 is apparent in  FIG. 1 . Eq. 4 estimates the flight time for the missile  100  to reach the point of closest approach, P. Eq. 7, however, estimates the flight time for the missile  100  to reach point Q. If the missile  100  and target  104  have no acceleration, then Eq. 4 is exact. However, if a missile guidance system is trying to align the relative velocity with the LOS, the missile  100  is likely to travel the range r. In this case, Eq. 7 is more appropriate for estimating the time-to-go. On the other hand, if zero-effort-miss distance is needed by the missile guidance system, Eq. 4 is more appropriate. It must be emphasized that Eqs. 4 and 7 are only accurate when both the target  104  and the missile  100  have constant velocities.
 
     A simple technique that includes the effect of acceleration by the missile  100  and/or the target  104  uses the piece-wise average acceleration along the LOS. The time-to-go τ using this technique by Riggs is calculated according to: 
                     τ   =       2   ⁢   r         v     c   ⁢               +         v   c   2     +     4   ⁢     a   m     ⁢   r               ,           Eq   .           ⁢   8               
where v c =−{dot over (r)}, the closing velocity, and a m  is the piece-wise average acceleration along the LOS. When a m =0, then Eqs. 7 and 8 are the same. If a m  is known, then the time-to-go can be obtained directly from Eq. 8. If a m  is not known, the piece-wise constant acceleration is approximated as:
 
                       a   m     =           a   max     ⁡     (       t   e     -     t   0       )       +       a   min     ⁡     (       t   f     -     t   e       )         τ       ,           Eq   .           ⁢   9               
where t 0  is the initial time, t f  is the terminal time, t e  is the thrust-off time, a max  is the average acceleration when the thrust is on from t 0  to t e , and a min  is the average acceleration (actually deceleration) primarily due to drag when the thrust is off from t e  to t f . Since the time-to-go estimate is a function of a m  and a m  is a function of time-to-go, an iterative solution is required.
 
     OBJECT OF THE INVENTION 
     A first object of the invention is to provide a highly accurate method of estimating the time-to-go, which is not computationally time consuming. A further object of the invention is to provide a method of estimating the time-to-go that remains highly accurate even when the vehicle and/or target velocities change or at large vehicle-to-target angles. 
     Yet another object of the invention is to provide a highly accurate method of guiding a vehicle to intercept a target based on the time-to-go. Such a guidance method will not be computationally time consuming. The guidance method will also remain highly accurate in spite of changes in vehicle and/or target velocities and large vehicle-to-target angles. 
     These objects are implemented by the present invention, which takes actual, or real time acceleration into account when estimating the anticipated locations of a vehicle and a target/obstacle. By using actual acceleration information, the present invention can generate guidance commands that need only small adjustments, rather than requiring dramatic changes that may be difficult to accomplish. Furthermore, because the present invention more accurately anticipates the locations of the vehicle and the target/obstacle, the present invention provides more time for carrying out the guidance commands. This is especially useful as the small adjustments may be made at lower altitudes where aerodynamic surfaces, such as fins, are more responsive. In the thin air at higher altitudes, aerodynamic surfaces are less responsive, making dramatic changes more difficult. 
     Each of these methods can be incorporated in a vehicle and used for guiding or arming the vehicle. The method finds applicability in air vehicles such as missiles and water vehicles such as torpedoes. Vehicles using the invention may be operated either autonomously, or be provided additional and/or updated information during flight to improve accuracy. 
     While the invention finds application when a vehicle is intended to intercept a target, it also finds application when a vehicle is not intended to intercept a target. In particular, a further object of the invention is to guide a vehicle during accident avoidance situations. In like manner, another object of the invention is to guide a first vehicle relative to one or more other vehicles and/or obstacles. Such objects of the invention may readily be implemented by notifying a vehicle operator of potential accidents and/or the location of other vehicles and/or obstacles. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       The present invention is described in reference to the following Detailed Description and the drawings in which: 
         FIG. 1  shows a geometry of a vehicle-target engagement, 
         FIG. 2  shows a geometric relationship between a fixed reference frame and a LOS reference frame, 
         FIG. 3  is a plot of a guidance scaling factor as a function of initial angle α 0  and proportional navigation gain N, 
         FIG. 4  is a plot of the estimated time-to-go τ for different time-to-go equations using a first set of initial conditions, 
         FIG. 5  is a plot of the estimated time-to-go τ for different time-to-go equations using a second set of initial conditions, 
         FIG. 6  illustrates the trajectories of missiles using three different guidance methods to intercept a target, 
         FIG. 7  illustrates the magnitude of the acceleration command using three different guidance methods, 
         FIG. 8  illustrates the cumulative amount of energy required to implement the acceleration commands of three different guidance methods, 
         FIG. 9  illustrates the miss distance for one embodiment of the present invention as a function of target acceleration error, 
         FIG. 10  illustrates the cumulative amount of energy required to implement the acceleration commands of two different guidance methods as a function of target acceleration error, 
         FIG. 11  illustrates a first missile system according to the present invention, and 
         FIG. 12  illustrates a second missile system according to the present invention. 
     
    
    
     DETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS 
     The following Detailed Description provides disclosure regarding two target interception embodiments. These embodiments provide two methods for estimating the time-to-go τ with differing degrees of accuracy, and corresponding different magnitudes of computational requirements. 
     First Embodiment 
     Deriving a more accurate time-to-go estimate that accounts for the actual or real time acceleration in the first embodiment begins by modifying the zero-effort-miss distance to include acceleration: 
                     z   =     r   +   vt   +       1   2     ⁢     at   2           ,           Eq   .           ⁢   10               
where a is the missile-to-target acceleration. As with the velocity v, the missile-to-target acceleration a is a net acceleration and is a function of both the missile and target accelerations. Substituting Eq. 10 into Eq. 2 yields:
 
     
       
         
           
             
               
                 
                   
                     
                       
                         1 
                         2 
                       
                       ⁢ 
                       
                         a 
                         · 
                         
                           at 
                           3 
                         
                       
                     
                     + 
                     
                       
                         3 
                         2 
                       
                       ⁢ 
                       
                         a 
                         · 
                         
                           vt 
                           2 
                         
                       
                     
                     + 
                     
                       
                         ( 
                         
                           
                             a 
                             · 
                             r 
                           
                           + 
                           
                             v 
                             · 
                             v 
                           
                         
                         ) 
                       
                       ⁢ 
                       t 
                     
                     + 
                     
                       v 
                       · 
                       r 
                     
                   
                   = 
                   0. 
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   11 
                 
               
             
           
         
       
     
     The following equations (Eqs. 12-14) simplify the remainder of the analysis.
 
 v·r=vr  cos α  Eq. 12
 
 a·r=ar  cos β  Eq. 13
 
 a·v=av  cos γ  Eq. 14
 
When a≠0, the following additional equations (Eqs. 15, 16) further simplify the analysis.
 
     
       
         
           
               
             
               
                 
                   
                     
                       v 
                       _ 
                     
                     = 
                     
                       v 
                       a 
                     
                   
                 
                 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     15 
                   
                 
               
               
                 
                   
                     
                       r 
                       _ 
                     
                     = 
                     
                       r 
                       a 
                     
                   
                 
                 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     16 
                   
                 
               
             
           
         
       
     
     Substituting Eqs. 12-16 into Eq. 11 yields:
 
 t   3 +3   v   cos γ t   2 +2(   r   cos β+   v     2 ) t+ 2   v   r   cos α=0.  Eq. 17
 
Defining τ as the time-to-go solution, Eq. 17 becomes:
 
( t −τ)( t   2   +bt+c )=0.  Eq. 18
 
     Eq. 18 has only one real solution, when b 2 −4c&lt;0. Expanding Eq. 18 yields:
 
 t   3 +( b −τ) t   2 +( c−b τ) t−cτ= 0.  Eq. 19
 
Equating Eqs. 17 and 19 yields:
 
 b−τ= 3   v   cos γ,  Eq. 20
 
 c−bτ= 2(    r    cos β+   v     2 ), and  Eq. 21
 
− cτ= 2   v   r  cos α.   Eq. 22
 
     Rewriting Eq. 20 as:
 
 b= 3   v   cos γ+τ,  Eq. 23
 
and substituting Eq. 23 into Eq. 21 yields:
 
 c= 2(    r    cos β+   v     2 )+3   v   cos γτ+τ 2 .  Eq. 24
 
Assuming
 
               -     π   2       ≤   β   ≤     π   2           
and
 
                 -     π   2       ≤   γ   ≤     π   2       ,         
then c&gt;0. Returning to Eq. 22, a real positive time-to-go τ for c&gt;0 occurs when:
     v   r  cos α&lt; 0.  Eq. 25 
     Rewriting Eq. 24 as 
                     c   =       2   ⁢     r   _     ⁢           ⁢   cos   ⁢           ⁢   β     +       (     τ   +       3   ⁢     v   _     ⁢           ⁢   cos   ⁢           ⁢   γ     2       )     2     +       (       8   -     9   ⁢           ⁢     cos   2     ⁢   γ       4     )     ⁢       v   _     2           ,           Eq   .           ⁢   26               
c will be positive if:
 
     
       
         
           
               
             
               
                 
                   
                     
                       - 
                       
                         π 
                         2 
                       
                     
                     ≤ 
                     β 
                     ≤ 
                     
                       
                         π 
                         2 
                       
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       and 
                     
                   
                 
                 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     27 
                   
                 
               
               
                 
                   
                     
                       
                         8 
                         9 
                       
                     
                     &gt; 
                     
                       cos 
                       ⁢ 
                       
                           
                       
                       ⁢ 
                       
                         γ 
                         . 
                       
                     
                   
                 
                 
                   
                     Eq 
                     . 
                     
                         
                     
                     ⁢ 
                     28 
                   
                 
               
             
           
         
       
     
     Combining Eqs. 23 and 24 yields:
 
 b   2 −4 c =−(8−9 cos 2 γ)   v     2 −8   r   cos β−6   v   cos γ−3τ 2 .  Eq. 29
 
Satisfying Eqs. 27 and 28 also ensures that b 2 −4c is negative. In this case, only one real solution to the time-to-go τ can be obtained from Eq. 17:
 
                     τ   =         (       -     e   2       +           e   2     4     +       d   3     27           )       1   3       +       (       -     e   2       -           e   2     4     +       d   3     27           )       1   3       -       v   _     ⁢           ⁢   cos   ⁢           ⁢   y         ,           Eq   .           ⁢   30               
where
   d= 2(    r    cos β+   v   2)−3    v     2  cos 2 γ, and  Eq. 31   e= 2   v     3  cos 3 γ−2   v   cos γ(    r    cos β+   v     2 )+2   v   r   cos α.  Eq. 32 
     For 
                     e   2     4     +       d   3     27       ≤   0     ,         
there are three possible solutions for the time-to-go τ:
 
                     τ   =       2   ⁢         -   d     3       ⁢   cos   ⁢     {       1   3     ⁢       cos     -   1       ⁡     (         -   e       2   ⁢         -     d   3       /   27           +   φ     )         }       -       v   _     ⁢           ⁢   cos   ⁢           ⁢   γ         ,           Eq   .           ⁢   33               
where φ=0, 2π/3, and 4π/3. For the initial estimated value of the time-to-go, the angle φ is used that yields the solution closest to that predicted by Eq. 7. For all subsequent iterations, the time-to-go solution that is closest to the previously estimated time-to-go is used.
 
     The result leads to zero-effort-miss with acceleration compensation guidance (ZEMACG). The corresponding acceleration command for the ZEMACG system is the equation: 
                     A   =       r     τ   2       +     v   τ     +       1   2     ⁢   a         ,           Eq   .           ⁢   34               
in which the estimated time-to-go τ found in Eqs. 30 or 33 is then inserted. The numerical examples below show that ZEMACG is an improvement over proportional navigation guidance (PNG).
 
     The advantage of Eq. 30 over Eq. 8 is the actual or real time acceleration direction is accounted for more properly. For true proportional navigation acceleration, the acceleration is perpendicular to the LOS. In this case a m =0, and therefore Eq. 8 is the same as Eq. 7. Although β=0 when the acceleration is perpendicular to the LOS, the contribution of acceleration in Eq. 30 to the time-to-go is through the term containing γ. The difference between Eqs. 8 and 30 will be illustrated by an example below. 
     The zero-effort-miss position vector z using Eq. 34 is: 
                   z   =     r   +     v   ⁢           ⁢   τ     +       1   2     ⁢   a   ⁢           ⁢       τ   2     .                 Eq   .           ⁢   35               
The zero-effort-miss position vector z yields a zero-effort-miss distance of:
 
     
       
         
           
             
               
                 
                   
                       
                   
                   ⁢ 
                   
                     
                       
                         
                            
                           z 
                            
                         
                       
                       
                         
                           = 
                           
                             
                               
                                 ( 
                                 
                                   r 
                                   + 
                                   
                                     v 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     τ 
                                   
                                   + 
                                   
                                     
                                       1 
                                       2 
                                     
                                     ⁢ 
                                     a 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       τ 
                                       2 
                                     
                                   
                                 
                                 ) 
                               
                               · 
                               
                                 ( 
                                 
                                   r 
                                   + 
                                   
                                     v 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     τ 
                                   
                                   + 
                                   
                                     
                                       1 
                                       2 
                                     
                                     ⁢ 
                                     a 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     
                                       τ 
                                       2 
                                     
                                   
                                 
                                 ) 
                               
                             
                           
                         
                       
                       
                         
                             
                         
                       
                     
                   
                   ⁢ 
                   
                       
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   36 
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         
                           
                             
                               
                                 r 
                                 2 
                               
                               + 
                               
                                 
                                   ( 
                                   
                                     2 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     vr 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     cos 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     α 
                                   
                                   ) 
                                 
                                 ⁢ 
                                 τ 
                               
                               + 
                               
                                 ( 
                                 
                                   
                                     ar 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     cos 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     β 
                                   
                                   + 
                                   
                                     v 
                                     2 
                                   
                                 
                                 ) 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               
                                 τ 
                                 2 
                               
                               + 
                               
                                 
                                   ( 
                                   
                                     av 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     cos 
                                     ⁢ 
                                     
                                         
                                     
                                     ⁢ 
                                     γ 
                                   
                                   ) 
                                 
                                 ⁢ 
                                 
                                   τ 
                                   3 
                                 
                               
                               + 
                               
                                 
                                   
                                     a 
                                     2 
                                   
                                   ⁢ 
                                   
                                     τ 
                                     4 
                                   
                                 
                                 4 
                               
                             
                           
                         
                       
                     
                     . 
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   37 
                 
               
             
           
         
       
     
     Second Embodiment 
     In the second embodiment, equations based upon three-dimensional relative motion will be developed leading to an analytical solution for true proportional navigation (TPN). The analytical solution to the TPN is then used to derive the time-to-go estimate that accounts for TPN acceleration. 
     Let [ E   1   , E   2   , E   3 ] be the basis vectors of the fixed reference frame. Two additional reference frames will also be employed: the LOS frame and the angular momentum frame. Let [e 1   L , e 2   L , e 3 L] be the basis vectors of the LOS frame, with unit vector e 1   L  aligned with the LOS. Let [e 1   h , eh 2   h , e 3   h ] be the basis vectors of the angular momentum frame, with unit vector e 3   h  aligned with the angular momentum vector. As will be shown below, the unit vector e 1   h  is aligned with unit vector e 1   L . Further, the missile-to-target acceleration components expressed in the angular momentum frame can be solved analytically. 
     Let λ 2  and λ 3  be the LOS elevation and azimuth angles, respectively, with respect to the fixed reference frame. These LOS elevation and azimuth angles are illustrated in  FIG. 2 . The transformation between the LOS frame and the fixed reference frame is the matrix: 
     
       
         
           
             
               
                 
                   
                     [ 
                     
                       
                         
                           
                             e 
                             1 
                             L 
                           
                         
                       
                       
                         
                           
                             e 
                             2 
                             L 
                           
                         
                       
                       
                         
                           
                             e 
                             3 
                             L 
                           
                         
                       
                     
                     ] 
                   
                   = 
                   
                     
                       
                         [ 
                         
                             
                         
                         ⁢ 
                         
                           
                             
                               
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                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   λ 
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                                 ⁢ 
                                 
                                     
                                 
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                                 ⁢ 
                                 
                                     
                                 
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                                   3 
                                 
                               
                             
                             
                               0 
                             
                           
                           
                             
                               
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                                 ⁢ 
                                 
                                     
                                 
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                                 ⁢ 
                                 
                                     
                                 
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                                   λ 
                                   3 
                                 
                               
                             
                             
                               
                                 sin 
                                 ⁢ 
                                 
                                     
                                 
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                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
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                                   3 
                                 
                               
                             
                             
                               
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                                 ⁢ 
                                 
                                     
                                 
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                         ] 
                       
                       ⁢ 
                       
                           
                       
                       [ 
                       
                         
                           
                             
                               E 
                               1 
                             
                           
                         
                         
                           
                             
                               E 
                               2 
                             
                           
                         
                         
                           
                             
                               E 
                               3 
                             
                           
                         
                       
                       ] 
                     
                     . 
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   38 
                 
               
             
           
         
       
     
     The angular velocity ω and angular acceleration {dot over (ω)} associated with the LOS frame are: 
     
       
         
           
             
               
                 
                   ω 
                   = 
                   
                     
                       
                         ω 
                         1 
                       
                       ⁢ 
                       
                         e 
                         1 
                         L 
                       
                     
                     + 
                     
                       
                         ω 
                         2 
                       
                       ⁢ 
                       
                         e 
                         2 
                         L 
                       
                     
                     + 
                     
                       
                         ω 
                         3 
                       
                       ⁢ 
                       
                         e 
                         3 
                         L 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   39 
                 
               
             
             
               
                 
                   
                     = 
                     
                       
                         
                           - 
                           
                             
                               λ 
                               . 
                             
                             3 
                           
                         
                         ⁢ 
                         sin 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           λ 
                           2 
                         
                         ⁢ 
                         
                           e 
                           1 
                           L 
                         
                       
                       + 
                       
                         
                           
                             λ 
                             . 
                           
                           2 
                         
                         ⁢ 
                         
                           e 
                           2 
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                       + 
                       
                         
                           
                             λ 
                             . 
                           
                           3 
                         
                         ⁢ 
                         cos 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           λ 
                           2 
                         
                         ⁢ 
                         
                           e 
                           3 
                           L 
                         
                       
                     
                   
                   , 
                   and 
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   40 
                 
               
             
             
               
                 
                   
                     ω 
                     . 
                   
                   = 
                   
                     
                       
                         
                           ω 
                           . 
                         
                         1 
                       
                       ⁢ 
                       
                         e 
                         1 
                         L 
                       
                     
                     + 
                     
                       
                         
                           ω 
                           . 
                         
                         2 
                       
                       ⁢ 
                       
                         e 
                         2 
                         L 
                       
                     
                     + 
                     
                       
                         
                           ω 
                           . 
                         
                         3 
                       
                       ⁢ 
                       
                         e 
                         3 
                         L 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   41 
                 
               
             
             
               
                 
                   
                     
                       
                         = 
                           
                         ⁢ 
                         
                           
                             
                               { 
                               
                                 
                                   
                                     - 
                                     
                                       
                                         λ 
                                         ¨ 
                                       
                                       3 
                                     
                                   
                                   ⁢ 
                                   sin 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     λ 
                                     2 
                                   
                                 
                                 - 
                                 
                                   
                                     
                                       λ 
                                       . 
                                     
                                     2 
                                   
                                   ⁢ 
                                   
                                     
                                       λ 
                                       . 
                                     
                                     3 
                                   
                                   ⁢ 
                                   cos 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     λ 
                                     2 
                                   
                                 
                               
                               } 
                             
                             ⁢ 
                             
                               e 
                               1 
                               L 
                             
                           
                           + 
                           
                             
                               { 
                               
                                 
                                   λ 
                                   ¨ 
                                 
                                 2 
                               
                               } 
                             
                             ⁢ 
                             
                               e 
                               2 
                               L 
                             
                           
                           + 
                         
                       
                     
                   
                   
                     
                       
                           
                         ⁢ 
                         
                           
                             { 
                             
                               
                                 
                                   
                                     λ 
                                     ¨ 
                                   
                                   3 
                                 
                                 ⁢ 
                                 cos 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   λ 
                                   2 
                                 
                               
                               - 
                               
                                 
                                   
                                     λ 
                                     . 
                                   
                                   2 
                                 
                                 ⁢ 
                                 
                                   
                                     λ 
                                     . 
                                   
                                   3 
                                 
                                 ⁢ 
                                 sin 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   λ 
                                   2 
                                 
                               
                             
                             } 
                           
                           ⁢ 
                           
                             
                               e 
                               3 
                               L 
                             
                             . 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   42 
                 
               
             
           
         
       
     
     It follows that:
 
 ė   1   L   =ω×e   1   L =ω 3   e   2   L −ω 2   e   3   L ,  Eq. 43
 
 ė   2   L   =ω×e   2   L =ω 3   e   1   L −ω 1   e   3   L ,  Eq. 44
 
 ė   3   L   =ω×e   3   L =ω 3   e   1   L −ω 1   e   2   L .  Eq. 45
 
     The missile-to-target position r, velocity v, and acceleration a, respectively, are: 
     
       
         
           
             
               
                 
                   
                     r 
                     = 
                     
                       re 
                       1 
                       L 
                     
                   
                   , 
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   46 
                 
               
             
             
               
                 
                   
                     v 
                     = 
                     
                       
                         r 
                         . 
                       
                       = 
                       
                         
                           
                             
                               r 
                               . 
                             
                             ⁢ 
                             
                               e 
                               1 
                               L 
                             
                           
                           + 
                           
                             r 
                             ⁢ 
                             
                               
                                 e 
                                 . 
                               
                               1 
                               L 
                             
                           
                         
                         = 
                         
                           
                             
                               r 
                               . 
                             
                             ⁢ 
                             
                               e 
                               1 
                               L 
                             
                           
                           + 
                           
                             r 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               ω 
                               3 
                             
                             ⁢ 
                             
                               e 
                               2 
                               L 
                             
                           
                           - 
                           
                             r 
                             ⁢ 
                             
                                 
                             
                             ⁢ 
                             
                               ω 
                               2 
                             
                             ⁢ 
                             
                               e 
                               3 
                               L 
                             
                           
                         
                       
                     
                   
                   , 
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   47 
                 
               
             
             
               
                 
                   a 
                   = 
                   
                     
                       v 
                       . 
                     
                     = 
                     
                       
                         
                           r 
                           ¨ 
                         
                         ⁢ 
                         
                           e 
                           1 
                           L 
                         
                       
                       + 
                       
                         2 
                         ⁢ 
                         
                           r 
                           . 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         ω 
                         ⁢ 
                         
                             
                         
                         × 
                         
                           e 
                           1 
                           L 
                         
                       
                       + 
                       
                         r 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           ω 
                           . 
                         
                         ⁢ 
                         
                             
                         
                         × 
                         
                             
                         
                         ⁢ 
                         
                           e 
                           1 
                           L 
                         
                       
                       + 
                       
                         r 
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         ω 
                         ⁢ 
                         
                             
                         
                         × 
                         
                           ( 
                           
                             ω 
                             ⁢ 
                             
                                 
                             
                             × 
                             
                                 
                             
                             ⁢ 
                             
                               e 
                               1 
                               L 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   48 
                 
               
             
             
               
                 
                   
                     
                       
                         = 
                           
                         ⁢ 
                         
                           
                             
                               { 
                               
                                 
                                   r 
                                   ¨ 
                                 
                                 - 
                                 
                                   r 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         ω 
                                         2 
                                         2 
                                       
                                       + 
                                       
                                         ω 
                                         3 
                                         2 
                                       
                                     
                                     ) 
                                   
                                 
                               
                               } 
                             
                             ⁢ 
                             
                               e 
                               1 
                               L 
                             
                           
                           + 
                           
                             
                               { 
                               
                                 
                                   2 
                                   ⁢ 
                                   
                                     r 
                                     . 
                                   
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     ω 
                                     3 
                                   
                                 
                                 + 
                                 
                                   r 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     
                                       ω 
                                       . 
                                     
                                     3 
                                   
                                 
                                 + 
                                 
                                   r 
                                   ⁢ 
                                   
                                       
                                   
                                   ⁢ 
                                   
                                     ω 
                                     1 
                                   
                                   ⁢ 
                                   
                                     ω 
                                     2 
                                   
                                 
                               
                               } 
                             
                             ⁢ 
                             
                               e 
                               2 
                               L 
                             
                           
                           - 
                         
                       
                     
                   
                   
                     
                       
                           
                         ⁢ 
                         
                           
                             { 
                             
                               
                                 2 
                                 ⁢ 
                                 
                                   r 
                                   . 
                                 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   ω 
                                   2 
                                 
                               
                               + 
                               
                                 r 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   
                                     ω 
                                     . 
                                   
                                   2 
                                 
                               
                               - 
                               
                                 r 
                                 ⁢ 
                                 
                                     
                                 
                                 ⁢ 
                                 
                                   ω 
                                   1 
                                 
                                 ⁢ 
                                 
                                   ω 
                                   3 
                                 
                               
                             
                             } 
                           
                           ⁢ 
                           
                             
                               e 
                               3 
                               L 
                             
                             . 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   49 
                 
               
             
           
         
       
     
     The angular momentum h, using Eqs 46 and 47, is defined as:
 
 h=r×{dot over (r)}=r   2 {ω 2   e   2   L +ω 3   e   3   L }.  Eq. 50
 
Rewriting Eq. 50 yields:
 
 h=he   3   h ,  Eq. 51
 
where:
 
 h=r   2 √{square root over (ω 2   2 +ω 3   2 )}=r 2   ω , and Eq. 52
 
                       e   3   h     =             ω   2     ⁢     e   2   L       +       ω   3     ⁢     e   3   L               ω   2   2     +     ω   3   2           =           ω   _     2     ⁢     e   2   L       +         ω   _     3     ⁢     e   3   L             ,           Eq   .           ⁢   53               
based upon:
 
                         ω   _     2     =       ω   2       ω   _         ,           Eq   .           ⁢   54                     ω   _     3     =       ω   3       ω   _         ,   and           Eq   .           ⁢   55                 ω =√{square root over (ω 2   2 +ω 3   2 )}.  Eq. 56 
     From Eq. 53, it is clear that e 3   h  is perpendicular to e 1   L . By aligning e 1   h  with e 1   L , i.e.:
 
 e   1   h   =e   1   L ,  Eq. 57
 
then:
 
     
       
         
           
             
               
                 
                   
                     e 
                     2 
                     h 
                   
                   = 
                   
                     
                       
                         e 
                         3 
                         h 
                       
                       × 
                       
                         e 
                         1 
                         h 
                       
                     
                     = 
                     
                       
                         
                           
                             
                               ω 
                               3 
                             
                             ⁢ 
                             
                               e 
                               2 
                               L 
                             
                           
                           - 
                           
                             
                               ω 
                               2 
                             
                             ⁢ 
                             
                               e 
                               3 
                               L 
                             
                           
                         
                         
                           
                             
                               ω 
                               2 
                               2 
                             
                             + 
                             
                               ω 
                               3 
                               2 
                             
                           
                         
                       
                       = 
                       
                         
                           
                             
                               ω 
                               _ 
                             
                             3 
                           
                           ⁢ 
                           
                             e 
                             2 
                             L 
                           
                         
                         - 
                         
                           
                             
                               ω 
                               _ 
                             
                             2 
                           
                           ⁢ 
                           
                             
                               e 
                               3 
                               L 
                             
                             . 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   58 
                 
               
             
           
         
       
     
     The transformation matrices between the LOS frame [e 1   L , e 2   L , e 3   L ] and the angular momentum frame [e 1   h , e 2   h , e 3   h ] are: 
                       [           e   1   h               e   2   h               e   3   h           ]     =       [         1       0       0           0           ω   _     3           -       ω   _     2               0           ω   _     2             ω   _     3           ]     ⁡     [           e   1   L               e   2   L               e   3   L           ]         ,   and           Eq   .           ⁢   59                 [           e   1   L               e   2   L               e   3   L           ]     =         [         1       0       0           0           ω   _     3             ω   _     2             0         -       ω   _     2               ω   _     3           ]     ⁡     [           e   1   h               e   2   h               e   3   h           ]       .             Eq   .           ⁢   60               
These transformation matrices are orthogonal if ω 2   2 +ω 3   2 ≠0.
 
     The missile-to-target acceleration a can be expressed as:
 
 a=a   1   L   e   1   L   +a   2   L   e   2   L   +a   3   L   e   3   L   =a   1   h   e   1   h   +a   2   h   e   2   h   +a   3   h   e   3   h .  Eq. 61
 
     By comparing Eqs. 49 and 61 and substituting with Eqs. 52, 53, 59, and 60, the missile-to-target acceleration components are: 
                       a   1   L     =       {       r   ¨     -     r   ⁡     (       ω   2   2     +     ω   3   2       )         }     =     {       r   ¨     -       h   2       r   3         }         ,           Eq   .           ⁢   62                 a   2   L =2 {dot over (r)}ω   3   +r{dot over (ω)}   3   +rω   1 ω 2 ,  Eq. 63   a   3   L =−2 {dot over (r)}ω   2   −r{dot over (ω)}   2   +rω   1 ω 3 ,  Eq. 64 
                       a   1   h     =       a   1   L     =     {       r   ¨     -       h   2       r   3         }         ,           Eq   .           ⁢   65                       a   2   h     =       ⁢           ω   _     3     ⁢     a   2   L       -         ω   _     2     ⁢     a   3   L                       =       ⁢       2   ⁢       r   .     ⁡     (           ω   _     2     ⁢     ω   2       +         ω   _     3     ⁢     ω   3         )         +     r   ⁡     (           ω   _     2     ⁢       ω   .     2       +         ω   _     3     ⁢       ω   .     3         )           ,   and                 Eq   .           ⁢   66                 a   3   h =  ω   2   a   2   L +  ω   3   a   2   L   =r{  ω     1 (ω 2   2 +ω 3   2 )+(  ω   2 {dot over (ω)} 3 −  ω   3 {dot over (ω)} 2 )}.  Eq. 67 
     The resulting angular momentum rate {dot over (h)} is obtained by differentiating Eqs. 50 or 51: 
     
       
         
           
             
               
                 
                   
                     h 
                     . 
                   
                   = 
                   
                     
                       
                         
                           h 
                           . 
                         
                         ⁢ 
                         
                           e 
                           3 
                           h 
                         
                       
                       + 
                       
                         h 
                         ⁢ 
                         
                           
                             e 
                             . 
                           
                           3 
                           h 
                         
                       
                     
                     = 
                     
                       r 
                       × 
                       
                         r 
                         ¨ 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   68 
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         - 
                         
                           ra 
                           3 
                           L 
                         
                       
                       ⁢ 
                       
                         e 
                         2 
                         L 
                       
                     
                     + 
                     
                       
                         ra 
                         2 
                         L 
                       
                       ⁢ 
                       
                         
                           e 
                           3 
                           L 
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   69 
                 
               
             
           
         
       
     
     With the help of transformation matrix Eq. 60, Eq. 69 becomes: 
     
       
         
           
             
               
                 
                   
                     h 
                     . 
                   
                   = 
                   
                     
                       - 
                       
                         
                           ra 
                           3 
                           L 
                         
                         ⁡ 
                         
                           ( 
                           
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 3 
                               
                               ⁢ 
                               
                                 e 
                                 2 
                                 h 
                               
                             
                             + 
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 2 
                               
                               ⁢ 
                               
                                 e 
                                 3 
                                 h 
                               
                             
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         ra 
                         2 
                         L 
                       
                       ⁡ 
                       
                         ( 
                         
                           
                             
                               - 
                               
                                 
                                   ω 
                                   _ 
                                 
                                 2 
                               
                             
                             ⁢ 
                             
                               e 
                               2 
                               h 
                             
                           
                           + 
                           
                             
                               
                                 ω 
                                 _ 
                               
                               3 
                             
                             ⁢ 
                             
                               e 
                               3 
                               h 
                             
                           
                         
                         ) 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   70 
                 
               
             
             
               
                 
                   = 
                   
                     
                       
                         - 
                         
                           r 
                           ⁡ 
                           
                             ( 
                             
                               
                                 
                                   
                                     ω 
                                     _ 
                                   
                                   2 
                                 
                                 ⁢ 
                                 
                                   a 
                                   2 
                                   L 
                                 
                               
                               + 
                               
                                 
                                   
                                     ω 
                                     _ 
                                   
                                   3 
                                 
                                 ⁢ 
                                 
                                   a 
                                   3 
                                   L 
                                 
                               
                             
                             ) 
                           
                         
                       
                       ⁢ 
                       
                         e 
                         2 
                         h 
                       
                     
                     + 
                     
                       
                         r 
                         ⁡ 
                         
                           ( 
                           
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 3 
                               
                               ⁢ 
                               
                                 a 
                                 2 
                                 L 
                               
                             
                             - 
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 2 
                               
                               ⁢ 
                               
                                 a 
                                 3 
                                 L 
                               
                             
                           
                           ) 
                         
                       
                       ⁢ 
                       
                         
                           e 
                           3 
                           h 
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   71 
                 
               
             
           
         
       
     
     By comparing Eqs. 68 and 71, and using Eqs. 63, 64, and 67, the following equations are obtained: 
     
       
         
           
             
               
                 
                   
                     
                       
                         
                           h 
                           . 
                         
                         = 
                           
                         ⁢ 
                         
                           r 
                           ⁡ 
                           
                             ( 
                             
                               
                                 
                                   
                                     ω 
                                     _ 
                                   
                                   3 
                                 
                                 ⁢ 
                                 
                                   a 
                                   2 
                                   L 
                                 
                               
                               - 
                               
                                 
                                   
                                     ω 
                                     _ 
                                   
                                   2 
                                 
                                 ⁢ 
                                 
                                   a 
                                   3 
                                   L 
                                 
                               
                             
                             ) 
                           
                         
                       
                     
                   
                   
                     
                       
                         
                           = 
                             
                           ⁢ 
                           
                             r 
                             ⁢ 
                             
                               { 
                               
                                 
                                   2 
                                   ⁢ 
                                   
                                     
                                       r 
                                       . 
                                     
                                     ⁡ 
                                     
                                       ( 
                                       
                                         
                                           
                                             
                                               ω 
                                               _ 
                                             
                                             2 
                                           
                                           ⁢ 
                                           
                                             ω 
                                             2 
                                           
                                         
                                         + 
                                         
                                           
                                             
                                               ω 
                                               _ 
                                             
                                             3 
                                           
                                           ⁢ 
                                           
                                             ω 
                                             3 
                                           
                                         
                                       
                                       ) 
                                     
                                   
                                 
                                 + 
                                 
                                   r 
                                   ⁡ 
                                   
                                     ( 
                                     
                                       
                                         
                                           
                                             ω 
                                             _ 
                                           
                                           2 
                                         
                                         ⁢ 
                                         
                                           
                                             ω 
                                             . 
                                           
                                           2 
                                         
                                       
                                       + 
                                       
                                         
                                           
                                             ω 
                                             _ 
                                           
                                           3 
                                         
                                         ⁢ 
                                         
                                           
                                             ω 
                                             . 
                                           
                                           3 
                                         
                                       
                                     
                                     ) 
                                   
                                 
                               
                               } 
                             
                           
                         
                         , 
                         and 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   72 
                 
               
             
             
               
                 
                   
                     
                       e 
                       . 
                     
                     3 
                     h 
                   
                   = 
                   
                     
                       
                         - 
                         
                           r 
                           h 
                         
                       
                       ⁢ 
                       
                         ( 
                         
                           
                             
                               
                                 ω 
                                 _ 
                               
                               2 
                             
                             ⁢ 
                             
                               a 
                               2 
                               L 
                             
                           
                           + 
                           
                             
                               
                                 ω 
                                 _ 
                               
                               3 
                             
                             ⁢ 
                             
                               a 
                               3 
                               L 
                             
                           
                         
                         ) 
                       
                       ⁢ 
                       
                         e 
                         2 
                         h 
                       
                     
                     = 
                     
                       
                         - 
                         
                           r 
                           h 
                         
                       
                       ⁢ 
                       
                         a 
                         3 
                         h 
                       
                       ⁢ 
                       
                         e 
                         2 
                         h 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   73 
                 
               
             
             
               
                 
                   = 
                   
                     
                       - 
                       
                         
                           r 
                           2 
                         
                         h 
                       
                     
                     ⁢ 
                     
                       { 
                       
                         
                           
                             
                               ω 
                               _ 
                             
                             1 
                           
                           ⁡ 
                           
                             ( 
                             
                               
                                 ω 
                                 2 
                                 2 
                               
                               + 
                               
                                 ω 
                                 3 
                                 2 
                               
                             
                             ) 
                           
                         
                         + 
                         
                           ( 
                           
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 2 
                               
                               ⁢ 
                               
                                 
                                   ω 
                                   . 
                                 
                                 3 
                               
                             
                             - 
                             
                               
                                 
                                   ω 
                                   _ 
                                 
                                 3 
                               
                               ⁢ 
                               
                                 
                                   ω 
                                   . 
                                 
                                 2 
                               
                             
                           
                           ) 
                         
                       
                       } 
                     
                     ⁢ 
                     
                       
                         e 
                         2 
                         h 
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   74 
                 
               
             
           
         
       
     
     Substituting Eqs. 72 and 74 into Eq. 68 yields:
 
 {dot over (h)}=−r   2 {  ω   1 (ω 2   2 +ω 3   2 )+(  ω   2 {dot over (ω)} 3 −  ω   3 {dot over (ω)} 2 )} e   2   h  
 
+ r{ 2 {dot over (r)}(  ω     2 ω 2 +  ω   3 ω 3 )+ r (  ω   2 {dot over (ω)} 2 +  ω   3 {dot over (ω)} 3 )} e   3   h .  Eq. 75
 
     By comparing Eqs. 66 and 72, one obtains: 
     
       
         
           
             
               
                 
                   
                     a 
                     2 
                     h 
                   
                   = 
                   
                     
                       
                         
                           
                             ω 
                             _ 
                           
                           3 
                         
                         ⁢ 
                         
                           a 
                           2 
                           L 
                         
                       
                       - 
                       
                         
                           
                             ω 
                             _ 
                           
                           2 
                         
                         ⁢ 
                         
                           a 
                           3 
                           L 
                         
                       
                     
                     = 
                     
                       
                         
                           h 
                           . 
                         
                         r 
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   76 
                 
               
             
           
         
       
     
     By substituting Eqs. 65 and 76 into Eq. 61, the missile-to-target acceleration a becomes: 
     
       
         
           
             
               
                 
                   a 
                   = 
                   
                     
                       
                         ( 
                         
                           
                             r 
                             ¨ 
                           
                           - 
                           
                             
                               h 
                               2 
                             
                             
                               r 
                               3 
                             
                           
                         
                         } 
                       
                       ⁢ 
                       
                         e 
                         1 
                         h 
                       
                     
                     + 
                     
                       
                         
                           h 
                           . 
                         
                         r 
                       
                       ⁢ 
                       
                         e 
                         2 
                         h 
                       
                     
                     + 
                     
                       
                         a 
                         3 
                         h 
                       
                       ⁢ 
                       
                         
                           e 
                           3 
                           h 
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   77 
                 
               
             
           
         
       
     
     The missile command acceleration for the TPN is:
 
 a   M   =N{dot over (r)}e   1   L ×Ω,  Eq. 78
 
where N is the proportional navigation constant and:
 
                   Ω   =         r   ×     r   .         r   2       =       h     r   2       =         ω   2     ⁢     e   2   L       +       ω   3     ⁢       e   3   L     .                     Eq   .           ⁢   79               
Ω is the angular velocity of the LOS. With the help of Eqs. 51-53, 59, 60, and 79, Eq. 78 becomes:
 
     
       
         
           
             
               
                 
                   
                     a 
                     M 
                   
                   = 
                   
                     
                       
                         N 
                         ⁢ 
                         
                           r 
                           . 
                         
                         ⁢ 
                         
                           e 
                           1 
                           L 
                         
                         × 
                         h 
                       
                       
                         r 
                         2 
                       
                     
                     = 
                     
                       
                         
                           N 
                           ⁢ 
                           
                             rh 
                             . 
                           
                           ⁢ 
                           
                             e 
                             1 
                             h 
                           
                           × 
                           
                             e 
                             3 
                             h 
                           
                         
                         
                           r 
                           2 
                         
                       
                       = 
                       
                         
                           - 
                           
                             
                               N 
                               ⁢ 
                               
                                 r 
                                 . 
                               
                               ⁢ 
                               
                                 he 
                                 2 
                                 h 
                               
                             
                             
                               r 
                               2 
                             
                           
                         
                         = 
                         
                           
                             - 
                             N 
                           
                           ⁢ 
                           
                             r 
                             . 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             ω 
                             _ 
                           
                           ⁢ 
                           
                               
                           
                           ⁢ 
                           
                             e 
                             2 
                             h 
                           
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   80 
                 
               
             
             
               
                 
                   = 
                   
                     N 
                     ⁢ 
                     
                       
                         
                           r 
                           . 
                         
                         ⁡ 
                         
                           ( 
                           
                             
                               
                                 - 
                                 
                                   ω 
                                   3 
                                 
                               
                               ⁢ 
                               
                                 e 
                                 2 
                                 L 
                               
                             
                             + 
                             
                               
                                 ω 
                                 2 
                               
                               ⁢ 
                               
                                 e 
                                 3 
                                 L 
                               
                             
                           
                           ) 
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   81 
                 
               
             
           
         
       
     
     By assuming a non-accelerating target, the missile-to-target acceleration a is: 
     
       
         
           
             
               
                 
                   a 
                   = 
                   
                     
                       
                         { 
                         
                           
                             r 
                             ¨ 
                           
                           - 
                           
                             
                               h 
                               2 
                             
                             
                               r 
                               3 
                             
                           
                         
                         } 
                       
                       + 
                       
                         e 
                         1 
                         h 
                       
                       + 
                       
                         
                           
                             h 
                             . 
                           
                           r 
                         
                         ⁢ 
                         
                           e 
                           2 
                           h 
                         
                       
                       + 
                       
                         
                           a 
                           3 
                           h 
                         
                         ⁢ 
                         
                           e 
                           3 
                           h 
                         
                       
                     
                     = 
                     
                       
                         
                           N 
                           ⁢ 
                           
                             r 
                             . 
                           
                           ⁢ 
                           h 
                         
                         
                           r 
                           2 
                         
                       
                       ⁢ 
                       
                         
                           e 
                           2 
                           h 
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   82 
                 
               
             
           
         
       
     
     Eq. 82 leads to the following coupled nonlinear differential equations: 
                         r   ¨     -       h   2       r   3         =   0     ,           Eq   .           ⁢   83                   h   .     =       Nh   ⁢     r   .       r       ,   and           Eq   .           ⁢   84                 a   3   h =0.  Eq. 85 
     Assuming the solution for h is of the form:
 
 h=c   1   r   K ,  Eq. 86
 
where c 1  is an unknown to be determined. Differentiating Eq. 86 yields:
 
                     h   .     =         c   1     ⁢     Kr     K   -   1       ⁢     r   .       =         Kh   ⁢     r   .       r     .               Eq   .           ⁢   87               
By comparing Eqs. 84 and 87, it is apparent that K=N. Therefore:
   h=c   1   r   N .  Eq. 88 
     Rewriting Eq. 83 using Eq. 88 yields:
 
 {dot over (r)}   2   =c   1   2   r   2N−3 =0.  Eq. 89
 
Assuming the solution for {dot over (r)} is of the form:
 
 {dot over (r)}   2   =c   2   +c   3   r   M ,  Eq. 90
 
where c 2 , c 3 , and M are the unknowns to be determined. Differentiating Eq. 90 yields:
 
2 {dot over (r)}{umlaut over (r)}=c   3   Mr   M−1   {dot over (r)}.   Eq. 91
 
Substituting Eq. 89 into Eq. 91 yields:
 
2 c   1   2   r   2N−3   =c   3   Mr   M−1   {dot over (r)}.   Eq. 92
 
From Eq. 92, the unknowns are determined to be:
 
 M= 2 N− 2, and  Eq. 93
 
     
       
         
           
             
               
                 
                   
                     c 
                     3 
                   
                   = 
                   
                     
                       
                         2 
                         ⁢ 
                         
                           c 
                           1 
                           2 
                         
                       
                       M 
                     
                     = 
                     
                       
                         
                           c 
                           1 
                           2 
                         
                         
                           N 
                           - 
                           1 
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   94 
                 
               
             
           
         
       
     
     Rewriting Eq. 90 in view of Eqs. 93 and 94 shows: 
     
       
         
           
             
               
                 
                   
                     
                       r 
                       . 
                     
                     2 
                   
                   = 
                   
                     
                       c 
                       2 
                     
                     + 
                     
                       
                         
                           c 
                           1 
                           2 
                         
                         
                           N 
                           - 
                           1 
                         
                       
                       ⁢ 
                       
                         
                           r 
                           
                             
                               2 
                               ⁢ 
                               N 
                             
                             - 
                             2 
                           
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   95 
                 
               
             
           
         
       
     
     By defining r 0 , {dot over (r)} 0 , h 0 , and  ω   0  to be the initial values of r, {dot over (r)}, h, and ω, respectively, Eq. 88 can be rewritten as: 
     
       
         
           
             
               
                 
                   
                     c 
                     1 
                   
                   = 
                   
                     
                       
                         h 
                         0 
                       
                       
                         r 
                         0 
                         N 
                       
                     
                     . 
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   96 
                 
               
             
           
         
       
     
     By applying Eq. 96 and the above initial values to Eq. 95 and solving for c 2  shows: 
     
       
         
           
             
               
                 
                   
                     c 
                     2 
                   
                   = 
                   
                     
                       
                         
                           r 
                           . 
                         
                         0 
                         2 
                       
                       - 
                       
                         
                           
                             
                               h 
                               0 
                               2 
                             
                             / 
                             
                               r 
                               0 
                               
                                 2 
                                 ⁢ 
                                 N 
                               
                             
                           
                           
                             N 
                             - 
                             1 
                           
                         
                         ⁢ 
                         
                           r 
                           0 
                           
                             
                               2 
                               ⁢ 
                               N 
                             
                             - 
                             2 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           r 
                           . 
                         
                         0 
                         2 
                       
                       - 
                       
                         
                           
                             h 
                             0 
                             2 
                           
                           / 
                           
                             r 
                             0 
                             2 
                           
                         
                         
                           N 
                           - 
                           1 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   97 
                 
               
             
           
         
       
     
     Substituting Eq. 96 into Eqs. 88 and 95, the solutions for the angular momentum h and the range rate {dot over (r)} are thus: 
     
       
         
           
             
               
                 
                   
                     h 
                     = 
                     
                       
                         
                           h 
                           0 
                         
                         ⁡ 
                         
                           ( 
                           
                             r 
                             
                               r 
                               0 
                             
                           
                           ) 
                         
                       
                       N 
                     
                   
                   , 
                   and 
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   98 
                 
               
             
             
               
                 
                   
                     r 
                     . 
                   
                   = 
                   
                     - 
                     
                       
                         
                           
                             
                               r 
                               . 
                             
                             0 
                             2 
                           
                           - 
                           
                             
                               
                                 h 
                                 0 
                                 2 
                               
                               / 
                               
                                 r 
                                 0 
                                 2 
                               
                             
                             
                               N 
                               - 
                               1 
                             
                           
                           + 
                           
                             
                               
                                 
                                   h 
                                   0 
                                   2 
                                 
                                 / 
                                 
                                   r 
                                   0 
                                   
                                     2 
                                     ⁢ 
                                     N 
                                   
                                 
                               
                               
                                 N 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               r 
                               
                                 
                                   2 
                                   ⁢ 
                                   N 
                                 
                                 - 
                                 2 
                               
                             
                           
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   99 
                 
               
             
           
         
       
     
     By substituting Eq. 98 into Eq. 79, the magnitude of the LOS angular velocity Ω is: 
                   Ω   =       h     r   2       =         h   0       r   0   2       ⁢         (     r     r   0       )       N   -   2       .                 Eq   .           ⁢   100               
To maintain finite acceleration, N must thus be greater than 2.
 
     For Eq. 99 to yield a real solution for the range rate {dot over (r)}, the following condition must be satisfied for a successful interception: 
                         r   .     0   2     -         h   0   2     /     r   0   2         N   -   1         &gt;   0.           Eq   .           ⁢   101               
Using Eq. 52, Eq. 101 becomes:
 
     
       
         
           
             
               
                 
                   
                     
                        
                       
                         
                           r 
                           . 
                         
                         0 
                       
                        
                     
                     
                       
                         r 
                         0 
                       
                       ⁢ 
                       
                         
                           ω 
                           _ 
                         
                         0 
                       
                     
                   
                   &gt; 
                   
                     
                       
                         1 
                         
                           N 
                           - 
                           1 
                         
                       
                     
                     . 
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   102 
                 
               
             
           
         
       
     
     Returning to Eq. 47 and using Eq. 52, the magnitude of the missile-to-target velocity v is: 
     
       
         
           
             
               
                 
                   v 
                   = 
                   
                     
                       
                         
                           
                             r 
                             . 
                           
                           2 
                         
                         + 
                         
                           
                             r 
                             2 
                           
                           ⁡ 
                           
                             ( 
                             
                               
                                 ω 
                                 2 
                                 2 
                               
                               + 
                               
                                 ω 
                                 3 
                                 2 
                               
                             
                             ) 
                           
                         
                       
                     
                     = 
                     
                       
                         
                           
                             
                               r 
                               . 
                             
                             2 
                           
                           + 
                           
                             
                               h 
                               2 
                             
                             
                               r 
                               2 
                             
                           
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   103 
                 
               
             
           
         
       
     
     Similarly, the magnitudes of the angular momentum h and the range rate {dot over (r)} from Eq. 50 and  FIG. 1  are:
 
 h=∥r×{dot over (r)}∥=rv  sin α, and  Eq. 104
 
 {dot over (r)}=v  cos α.  Eq. 105
 
     The following dimensionless parameters are defined as the normalized range  r , the normalized angular momentum  h , and the normalized time  t : 
                       r   _     =     r     r   0         ,           Eq   .           ⁢   106                   h   _     =     h       r   0     ⁢     v   0           ,   and           Eq   .           ⁢   107                   t   _     =     t       r   0     /     v   0           ,           Eq   .           ⁢   108               
where v 0  and t 0  are initial values of v and t, respectively. Using Eqs. 106-108, Eqs. 98 and 99 simplify as:
     h =  h     0     r     N ,  Eq. 109 
     
       
         
           
             
               
                 
                   
                     
                       ⅆ 
                       
                         r 
                         _ 
                       
                     
                     
                       ⅆ 
                       
                         t 
                         _ 
                       
                     
                   
                   = 
                   
                     - 
                     
                       
                         
                           
                             
                               
                                 r 
                                 . 
                               
                               0 
                               2 
                             
                             
                               v 
                               0 
                               2 
                             
                           
                           + 
                           
                             
                               
                                 
                                   h 
                                   _ 
                                 
                                 0 
                                 2 
                               
                               
                                 N 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               ( 
                               
                                 
                                   
                                     r 
                                     _ 
                                   
                                   
                                     
                                       2 
                                       ⁢ 
                                       N 
                                     
                                     - 
                                     2 
                                   
                                 
                                 - 
                                 1 
                               
                               ) 
                             
                           
                         
                       
                       . 
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   110 
                 
               
             
           
         
       
     
     Using Eq. 110, the normalized time  t  for the normalized range  r  is: 
                     t   _     =     -       ∫   1     r   _       ⁢         d   ⁢     r   _                 r   .     0   2       v   0   2       +           h   _     0   2       N   -   1       ⁢     (         r   _         2   ⁢   N     -   2       -   1     )             .                 Eq   .           ⁢   111               
From Eqs. 104, 105, and 107, it is clear that:
 
                           r   .     0       v   0       =     cos   ⁢           ⁢     α   0         ,   and           Eq   .           ⁢   112                     h   _     0     =     sin   ⁢           ⁢     α   0         ,           Eq   .           ⁢   113               
where α 0  is the initial value of α. Eq. 111 therefore becomes:
 
     
       
         
           
             
               
                 
                   
                     t 
                     _ 
                   
                   = 
                   
                     
                       - 
                       sec 
                     
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       α 
                       0 
                     
                     ⁢ 
                     
                       
                         ∫ 
                         1 
                         
                           r 
                           _ 
                         
                       
                       ⁢ 
                       
                         
                           
                             d 
                             ⁢ 
                             
                               r 
                               _ 
                             
                           
                           
                             
                               1 
                               + 
                               
                                 
                                   
                                     
                                       tan 
                                       2 
                                     
                                     ⁢ 
                                     
                                       α 
                                       0 
                                     
                                   
                                   
                                     N 
                                     - 
                                     1 
                                   
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       
                                         r 
                                         _ 
                                       
                                       
                                         
                                           2 
                                           ⁢ 
                                           N 
                                         
                                         - 
                                         2 
                                       
                                     
                                     - 
                                     1 
                                   
                                   ) 
                                 
                               
                             
                           
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   114 
                 
               
             
           
         
       
     
     The normalized time-to-go  τ  is: 
                     τ   _     =       -   sec     ⁢           ⁢     α   0     ⁢       ∫   0   1     ⁢         d   ⁢     r   _           1   +           tan   2     ⁢     α   0         N   -   1       ⁢     (         r   _         2   ⁢   N     -   2       -   1     )             .                 Eq   .           ⁢   115               
If α 0 =0, then:
   τ =1, and  Eq. 116 τ= r   0   /v   0 .  Eq. 117 
     A real solution to Eq. 115 imposes the following requirement: 
                     α   0     &lt;       tan     -   1       ⁢         (       N   -   1       1   -       r   _         2   ⁢   N     -   2           )       .               Eq   .           ⁢   118               
As the normalized range  r →0, then Eq. 118 simplifies to:
 α 0 &lt;tan −1 √ {square root over ( N− 1)}.  Eq. 119 
     The normalized missile acceleration command ā M  is defined as: 
                       a   _     M     =         a   M         v   0   2     /     r   0         =       -       N   ⁢     r   .     ⁢   h         r   2     ⁢       v   0   2     /     r   0             =         -       N   ⁢     h   _           r   _     2         ⁢       ⅆ     r   _         ⅆ     t   _           =     N   ⁢       h   _     0     ⁢       r   _       N   -   2       ⁢       ⅆ     r   _         ⅆ     t   _                         Eq   .           ⁢   120               =     N   ⁢       h   _     0     ⁢       r   _       N   -   2       ⁢             r   .     0   2       v   0   2       +           h   _     0   2       N   -   1       ⁢     (         r   _         2   ⁢   N     -   2       -   1     )                     Eq   .           ⁢   121                 =         sin   ⁢           ⁢   2   ⁢           ⁢     α   0     ⁢   N   ⁢       r   _       N   -   2         2     ⁢       1   +           tan   2     ⁢     α   0         N   -   1       ⁢     (         r   _         2   ⁢   N     -   2       -   1     )               ,           Eq   .           ⁢   122               
when Eqs. 106-110 and 113 are used.
 
     The above results will now be used to compute an estimated time-to-go that accounts for the missile acceleration due to TPN guidance. Turning to Eqs. 115 and 117, the time-to-go τ is: 
                   τ   =           r   0     ⁢   sec   ⁢           ⁢     α   0         v   0       ⁢       ∫   0   1     ⁢         d   ⁢     r   _           1   +           tan   2     ⁢     α   0         N   -   1       ⁢     (         r   _         2   ⁢   N     -   2       -   1     )             .                 Eq   .           ⁢   123               
Note that for a given TPN constant N, the estimated time-to-go is dependent on the initial relative range and speed and the angle between the initial relative position and velocity vectors α. As the time-to-go is a function of both the TPN constant N and the angle α, Eq. 123 becomes:
 
                     τ   =         r   0     ⁢     f   ⁡     (     N   ,     α   0       )           v   0         ,           Eq   .           ⁢   124               
where:
 
     
       
         
           
             
               
                 
                   
                     f 
                     ⁡ 
                     
                       ( 
                       
                         N 
                         , 
                         
                           α 
                           0 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     sec 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       α 
                       0 
                     
                     ⁢ 
                     
                       
                         ∫ 
                         0 
                         1 
                       
                       ⁢ 
                       
                         
                           
                             d 
                             ⁢ 
                             
                               r 
                               _ 
                             
                           
                           
                             
                               1 
                               + 
                               
                                 
                                   
                                     
                                       tan 
                                       2 
                                     
                                     ⁢ 
                                     
                                       α 
                                       0 
                                     
                                   
                                   
                                     N 
                                     - 
                                     1 
                                   
                                 
                                 ⁢ 
                                 
                                   ( 
                                   
                                     
                                       
                                         r 
                                         _ 
                                       
                                       
                                         
                                           2 
                                           ⁢ 
                                           N 
                                         
                                         - 
                                         2 
                                       
                                     
                                     - 
                                     1 
                                   
                                   ) 
                                 
                               
                             
                           
                         
                         . 
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   125 
                 
               
             
           
         
       
     
     The function f(N,α 0 ) in Eq. 125 is the TPN guidance scaling factor for the time-to-go calculation that accounts for the missile acceleration due to TPN acceleration commands. Plots of f(N,α 0 ) vs. α 0  for N=3, 4, and 5 are shown in  FIG. 3 . 
     The following equation is a good approximation of Eq. 124 for N=3, 4, and 5. 
                     τ   =               r   0     ⁢     {     1   +         p   1     ⁡     (   N   )       ⁢     α   0       +         p   2     ⁡     (   N   )       ⁢     α   0   2       +                           p   3     ⁡     (   N   )       ⁢     α   0   3       +         p   4     ⁡     (   N   )       ⁢     α   0   4       +         p   5     ⁡     (   N   )       ⁢     α   0   5         }             v   0         ,           Eq   .           ⁢   126               
where p 1 (N), p 2 (N), p 3 (N), p 4 (N), and p 5 (N) are polynomials of the form:
   p   1 ( N )=2.5285−1.05197 N+ 0.1115 N   2 ,  Eq. 127A   p   2 ( N )=−31.6485+13.4178 N− 1.4236 N   2 ,  Eq. 127B   p   3 ( N )=134.5987−55.7204 N+ 5.8922 N   2 ,  Eq. 127C   p   4 ( N )=−220.3862+91.0563 N− 9.6156 N   2 , and  Eq. 127D   p   5 ( N )=127.9458−52.3959 N+ 5.5147 N   2 .  Eq. 127E 
     Eq. 125 can be rewritten as: 
     
       
         
           
             
               
                 
                   
                     f 
                     ⁡ 
                     
                       ( 
                       
                         N 
                         , 
                         
                           α 
                           0 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     sec 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       α 
                       0 
                     
                     ⁢ 
                     
                       
                         { 
                         
                           1 
                           - 
                           
                             
                               
                                 tan 
                                 2 
                               
                               ⁢ 
                               
                                 α 
                                 0 
                               
                             
                             
                               N 
                               - 
                               1 
                             
                           
                         
                         } 
                       
                       
                         - 
                         
                           1 
                           2 
                         
                       
                     
                     ⁢ 
                     
                       
                         ∫ 
                         0 
                         1 
                       
                       ⁢ 
                       
                         
                           
                             { 
                             
                               1 
                               + 
                               
                                 
                                   
                                     tan 
                                     2 
                                   
                                   ⁢ 
                                   
                                     α 
                                     0 
                                   
                                   ⁢ 
                                   
                                     
                                       r 
                                       _ 
                                     
                                     
                                       
                                         2 
                                         ⁢ 
                                         N 
                                       
                                       - 
                                       2 
                                     
                                   
                                 
                                 
                                   
                                     ( 
                                     
                                       N 
                                       - 
                                       1 
                                     
                                     ) 
                                   
                                   - 
                                   
                                     
                                       tan 
                                       2 
                                     
                                     ⁢ 
                                     
                                       α 
                                       0 
                                     
                                   
                                 
                               
                             
                             } 
                           
                           
                             - 
                             
                               1 
                               2 
                             
                           
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
                           
                             ⅆ 
                             
                               r 
                               _ 
                             
                           
                           . 
                         
                       
                     
                   
                 
               
               
                 
                   Eq 
                   . 
                   
                       
                   
                   ⁢ 
                   128 
                 
               
             
           
         
       
     
     When the initial angle α 0  is small, i.e.: 
                           tan   2     ⁢     α   0           (     N   -   1     )     -       tan   2     ⁢     α   0           &lt;   1     ,           Eq   .           ⁢   129               
Eq. 129 may be approximated by:
 
                       tan   2     ⁢     α   0       &lt;         N   -   1     2     .             Eq   .           ⁢   130               
This leads to the further approximation of Eq. 128 as:
 
     
       
         
           
             
               
                 
                   
                     f 
                     ⁡ 
                     
                       ( 
                       
                         N 
                         , 
                         
                           α 
                           0 
                         
                       
                       ) 
                     
                   
                   = 
                   
                     sec 
                     ⁢ 
                     
                         
                     
                     ⁢ 
                     
                       α 
                       0 
                     
                     ⁢ 
                     
                       
                         { 
                         
                           1 
                           - 
                           
                             
                               
                                 tan 
                                 2 
                               
                               ⁢ 
                               
                                 α 
                                 0 
                               
                             
                             
                               N 
                               - 
                               1 
                             
                           
                         
                         } 
                       
                       
                         - 
                         
                           1 
                           2 
                         
                       
                     
                     ⁢ 
                     
                       
                         ∫ 
                         0 
                         1 
                       
                       ⁢ 
                       
                         
                           { 
                           
                             1 
                             - 
                             
                               
                                 
                                   tan 
                                   2 
                                 
                                 ⁢ 
                                 
                                   α 
                                   0 
                                 
                                 ⁢ 
                                 
                                   
                                     r 
                                     _ 
                                   
                                   
                                     
                                       2 
                                       ⁢ 
                                       N 
                                     
                                     - 
                                     2 
                                   
                                 
                               
                               
                                 2 
                                 ⁡ 
                                 
                                   [ 
                                   
                                     
                                       ( 
                                       
                                         N 
                                         - 
                                         1 
                                       
                                       ) 
                                     
                                     - 
                                     
                                       
                                         tan 
                                         2 
                                       
                                       ⁢ 
                                       
                                         α 
                                         0 
                                       
                                     
                                   
                                   ] 
                                 
                               
                             
                           
                           } 
                         
                         ⁢ 
                         
                             
                         
                         ⁢ 
                         
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                   132 
                 
               
             
           
         
       
     
     The time-to-go τ under these small initial angle α 0  conditions is approximately: 
                   τ   =           r   0     ⁢   sec   ⁢           ⁢     α   0     ⁢     {     1   -         tan   2     ⁢     α   0         2   ⁢       (       2   ⁢   N     -   1     )     ⁡     [       (     N   -   1     )     -       tan   2     ⁢     α   0         ]             }           v   0     ⁢       {     1   -         tan   2     ⁢     α   0         N   -   1         }           .             Eq   .           ⁢   133               
Numerical Examples
 
     The results of several numerical examples for time-to-go calculations will now be discussed. In the first example, r=(5000, 5000, 5000), v=(−300, −250, −200), and a=(−40, −50, −60). The results are shown in  FIG. 4 . It is clear that Eq. 33 yields the exact solution while Eq. 7 returns a large error initially, though the time-to-go error is reduced as the simulation time draws closer to intercept. If a missile, which carries a warhead that must detonate when the missile is close to the target, used Eq. 7 to arm itself, the warhead would uselessly explode far beyond the target as Eq. 7&#39;s time-to-go is almost twice the actual time-to-go. 
     The second numerical example is a TPN simulation, with a proportional navigation gain N=3. The initial missile and target conditions are: 
     
       
         
               
               
               
             
               
               
               
               
             
           
               
                   
                   
               
               
                   
                 Missile 
                 Target 
               
               
                   
                   
               
             
             
               
                   
               
             
          
           
               
                   
                 Initial Position 
                 (0, 0, 0) 
                 (1000, 1000, 500) 
               
               
                   
                 Initial Velocity 
                 (100, 0, 0) 
                 (−10, −5, −5) 
               
               
                   
                 Initial Acceleration 
                 (0, 0, 0) 
                 (0, 0, 0) 
               
               
                   
                   
               
             
          
         
       
     
     The results for several time-to-go approximations are plotted in  FIG. 5 . It is clear that Eq. 123 provides substantially the exact time-to-go. Eq. 126 is based on curve fitting of Eq. 123, and the result is almost identical to Eq. 123. Eq. 133 is based on an approximation (Eq. 130) of the integral in order to obtain the closed-form solution. The result using Eq. 133 is good even when the initial angle α 0  between the relative velocity and the LOS used in this example is 44.7°. The acceleration used in Eq. 33 is based on half of the initial missile acceleration due to TPN guidance as the acceleration at intercept is assumed to be zero. In this numerical example, Eqs. 7 and 9 will produce the same results because the acceleration is perpendicular to the LOS, thus causing the mean acceleration along the LOS to be zero. Eq. 4 grossly underestimates the time-to-go. 
     In the third numerical simulation, the trajectories of three missiles and a target are shown in  FIG. 6 . For this simulation, the three missiles use proportional navigation (PNG), augmented PNG (APNG), and Eq. 34 in conjunction with Eqs. 30 or 33, respectively. The combined use of Eqs. 34 and 30 or 33 will be termed zero-effort-miss with acceleration compensation guidance (ZEMACG). The ZEMACG missile clearly provides the most direct interception trajectory, with the trajectory being nearly linear for most of the flight. The advantage of ZEMACG is that it accounts for the actual target acceleration properly and steers the missile toward the proper interception path as early as possible. 
       FIG. 7  illustrates the magnitude of the acceleration correction for each of the three missiles illustrated in  FIG. 6 . The PNG missile initially has no acceleration correction, but climbs rapidly and continues to have its trajectory corrected until the moment of interception. The APNG missile has some initial acceleration correction that increases during the course of the flight, but does not require as large an acceleration correction as the PNG missile. Lastly, the ZEMACG missile shows the greatest initial acceleration correction, but the magnitude rapidly decreases with virtually no acceleration correction required shortly before interception. Because of the higher acceleration required near the end of a PNG missile flight, it might not have enough acceleration to intercept the target. This problem may be exacerbated because the acceleration of the PNG missile can become saturated. The net result is a greater miss distance. This problem is greatest at high altitudes where the air is thin and missile maneuverability is low. Under these circumstances, it is desirable to make the acceleration corrections early, at low altitude, while the missile has high maneuverability. A ZEMACG missile, with its greater acceleration correction early in flight, thus has the advantage. 
       FIG. 8  illustrates the cumulative use of guidance energy due to acceleration correction as a function of flight time. As shown in  FIG. 8 , the PNG missile uses approximately three times as much guidance energy as does the ZEMACG missile, while the APNG missile uses more than twice as much. An additional advantage of the ZEMACG missile is that it requires less energy and thus less weight. The result is that a lighter missile is feasible. Alternatively, if the same weight is retained, a faster and/or more lethal missile is possible. 
       FIG. 9  shows the miss distance for a ZEMACG missile as a function of acceleration error. This simulation shows the ZEMACG missile will intercept the target even when the acceleration error is as large as ±15 m/sec 2 . The ZEMACG missile, even with target acceleration errors, still outperforms the PNG missile. 
       FIG. 10  illustrates the total use of guidance energy due to acceleration correction as a function of acceleration error. The energy used by the ZEMACG missile is a function of acceleration error with greater error leading to greater energy demands. An acceleration error of ±20 m/sec 2  is required before the ZEMACG missile requires as much energy as the PNG missile. 
     Implementation 
     Depending upon the time-to-go estimation implemented, various input values are required. In the simplest case, Eq. 33 requires inputs of the missile-to-target vector r, the missile-to-target velocity v, and the missile-to-target acceleration a. Even the most computationally complex time-to-go τ estimation scheme based on Eq. 123 requires the same inputs of r, v, and a. 
     These three inputs can come from a variety of sources. In a “fire and forget” missile system  100 , as shown in  FIG. 11 , the three inputs may be determined based upon an on-board radar  104 . A position unit  112  that determines the missile-to-target vector r processes a radar return signal  108 . A velocity unit  116  that determines the missile-to-target velocity v also processes the radar return signal  108 . Lastly, the radar return signal  108  is processed by an acceleration unit  120  that determines the missile-to-target acceleration a. A time-to-go unit  124  then determines the time-to-go τ based upon the three inputs r, v, and a. For guidance purposes, a processor  128  calculates an acceleration command A based upon Eq. 34 using the four inputs r, v, a, and τ. It should be noted that while the position unit  112 , the velocity unit  116 , the acceleration unit  120 , the time-to-go unit  124 , and the processor  128  are illustrated as separate elements, each could be implemented in software using a single processor. The time-to-go τ and the acceleration command A are iteratively computed during the course of the intercept trajectory, preferably on a periodic basis. The acceleration command A from the processor  128  is then fed to a control unit  132  that controls the trajectory of the missile system  100 . While this example uses an on-board radar  104 , use of an on-board optical system is also envisioned. 
     An alternative way to implement a time-to-go estimation scheme is to receive information from an external source as shown in  FIG. 12 . The missile system  200  in this case receives updated r, v, and a values from the external source, preferably on a periodic basis, and calculates revised time-to-go τ and acceleration command A values. The external source may be an aircraft  204  that launched the missile system  200 . The external source may alternatively be a ground-based tracking system  208 . The missile system  200  may alternatively be ground launched rather than air launched. 
     Yet another alternative way to implement a time-to-go estimation scheme is to store at least a portion of the information in a memory. This method applies when the velocity and/or acceleration profiles for both the missile system and the target are known a priori. The initial values of r, v, and a would still need to be provided to the missile system. 
     The control unit  132  in missile system  100  may include one or more control elements. These possible control elements include, but are not limited to, axial thrusters, radial thrusters, and control surfaces such as fins or canards. 
     While the above description disclosed application of the time-to-go method to a missile system traveling in air, it is equally applicable to other intercepting vehicles. In particular, the disclosed time-to-go method can also be applied to torpedoes traveling in water. 
     Accident Avoidance 
     The embodiments described above relate to the intentional interception of a target by a vehicle. In many situations, just the reverse is desired. As an example, an accident avoidance system may be implemented to guide a vehicle away from another vehicle or obstacle. By including velocity and actual or real time acceleration effects in an acceleration command, an automobile can more accurately avoid moving vehicles/obstacles, such as an abrupt lane change by another automobile. This is in contrast to most current automobile systems that typically warn only of fixed vehicles/obstacles, especially when reversing into a parking spot. After estimating the time-to-go from either Eq. 30 or Eq. 33, Eq. 10 can then be used to determine the closest distance between the two vehicles if the vehicles continue at their current velocities and accelerations. An accident avoidance system according to the present invention would thus provide for earlier detection of potential accidents. The sooner a potential accident is detected, the more time a driver or system has to react and the less acceleration will be needed to avoid the accident. Such an accident avoidance system could generate an acceleration command A′ that is the complete opposite of the acceleration command A generated by the system in which an interception is intended. As such an acceleration command A′ might be more abrupt than needed to avoid an accident, the accident avoidance system would preferably generate an acceleration command A″ only of sufficient magnitude to avoid the accident. The magnitude of this acceleration command A″ could also be determined by a minimum margin required to avoid an accident by, for example, a predetermined number of feet. For purposes of an accident avoidance system, an offset vector ψ is added to the original acceleration command equation, resulting in: 
                     A   ″     =       r     τ   2       +     v   τ     +       1   2     ⁢   a     +     ψ   .               Eq   .           ⁢   134               
The offset vector ψ can be a fixed vector that yields the margin required to avoid an accident. Alternatively, the offset vector ψ may be a variable, such that the margin required to avoid an accident is a function of the velocities or accelerations of the vehicle and/or obstacle. In the simplest case of an automobile accident avoidance system, the acceleration command A″ may be a braking command as many cars are equipped with automatic braking systems (ABS). The acceleration command A″ may alternatively be implemented by using a guidance unit that causes a change in direction. Such a guidance unit could include applying the brakes in such a fashion so as to change the direction of the automobile or overriding the steering wheel.
 
     Such accident avoidance systems may also be readily applied to other modes of transportation. For example, passenger airplanes, due to their high value in human life, would benefit from an accident avoidance system based upon the current invention. An airplane accident avoidance system could automatically cause an airplane to take evasive action, such as a turn, to avoid colliding with another airplane or other obstacle. Because the present invention includes velocity and acceleration effects in calculating an acceleration command, if the obstacle similarly takes evasive action, the magnitude of the action can be diminished. For example, if two airplanes have accident avoidance systems based upon the present invention, each airplane would sense changes in velocity and acceleration in the other airplane. This would permit each airplane to reduce the amount of banking required to avoid a collision. 
     While the above embodiments are based upon interactions between vehicles, the accident avoidance system could be separate from the vehicles. As an example, if an airport control tower included an accident avoidance system based upon the present invention, the system could warn air traffic controllers, who could relay warnings to the appropriate pilots. The airport control tower system would use the airplanes&#39; velocities and accelerations and calculate the closest distance between the airplanes if they continue their present flight paths. If the predicted closest distance is less than desirable, the air traffic controllers can alert each pilot and recommend a steering direction based on Eq. 134. A busy harbor that must coordinate shipping traffic could employ a similar accident avoidance system. 
     Vehicle Guidance 
     As yet another embodiment of the present invention, such a system could be used for vehicle guidance. In particular, a vehicle guidance system would be beneficial in areas of high vehicle density. The vehicle guidance system would permit vehicles to be more closely spaced allowing greater traffic flow as each vehicle would be more accurately and safely guided. Returning to the example of airplanes, airplane guidance systems would permit more frequent take-offs and landings as the interaction between airplanes would be more tightly controlled. Such airplane guidance systems would also permit closer formations of airplanes in flight. Similar to an accident avoidance system, the airplane guidance system could generate an acceleration command to keep one airplane within a predetermined range of another airplane, perhaps when flying in formation. 
     While many of the above embodiments have an active system that generates an acceleration command, this need not be the case. The system, especially if it is of the accident avoidance or vehicle guidance types, may be passive and merely provide an operator with a warning or a suggested action. In a simple automobile accident avoidance system, the system may provide only a visible or audible warning of another automobile or obstacle. In an airplane, a more sophisticated guidance system may provide the suggestions of banking right and increasing altitude. 
     Although the present invention has been described by way of examples with reference to the accompanying drawings, it is to be noted that various changes and modifications will be apparent to those skilled in the art. Therefore, such changes and modifications should be construed as being within the scope of the invention.