7(0.1)(4(0.1)n- yo(n)yo(n + 1)= yo(n) +(3 + 3(0.1)n - x。(n)2 + (4(0.1)n - y(n)A little algebraic clean up gives us the following equations:0.7(3+0.3n-xo(n)(8)Xo(n +I)= xo(n)+/(3 +0.3n- x(n)2 +(0.4n- y(n)20.7(0.4n-y。(n)(9)yo(n + 1)= y(n)+/(3 + 0.3n -x (n)* +(0.4n - yo(n)2with the initial condition that xo(0) = -3 and yo(O) = 0.Now we can iterate Equations (8) and (9) to determine the path of thechaser.We'll do oneby hand here,but continuingthis work is verytedious andtimeconsuming.We'll needtogetmore computational helptoperformmoreiteration.WebeginbysubstitutingOforninEquation(8) to get0.7(3 + 0.3(0)- x(0))xo(0 + 1)= X(0) +/(3 + 0.3(0)- xo (0)* + (0.4(0)- y (0)2Thenwe simplify,substitutethe initial conditions,andperformmoresimplifying to obtain:0.7(6)4.20.7(3 - (-3)xo(I)= -3 + 2.3V626/(3 -(-3)2Wedo thesameforEquation(9),wheremanypartsoftheequationevaluate to O and disappear in the calculations to obtain:0.7(0 - 0)=0+0=0 (1) = yo (0) +V(3 - (-3)2Therefore,thechasermovestothepoint (-2.3,0)duringthefirsttimestep ofthe chase.We continue thisprocedure for n =1,2,3,..,10toproduce the iterates given in Table 1. We determined these valuesusing a computertoperform all thetedious,butnecessarycomputations.114
114 y n y n n y n n x n n y n 0 0 0 0 2 0 2 1 7 0 1 4 0 1 3 3 0 1 4 0 1 ( ) ( ) ( . )( ( . ) ( )) ( ( . ) ( )) ( ( . ) ( )) + = + - + - + - A little algebraic clean up gives us the following equations: x n x n n x n n x n n y n 0 0 0 0 2 0 2 1 0 7 3 0 3 3 0 3 0 ( ) ( ) . ( . ( )) ( . ( )) ( .4 ( )) + = + + - + - + - (8) y n y n n y n n x n n y n 0 0 0 0 2 0 2 1 0 7 0 3 0 3 0 ( ) ( ) . ( .4 ( )) ( . ( )) ( .4 ( )) + = + - + - + - (9) with the initial condition that x0(0) = -3 and y0(0) = 0. Now we can iterate Equations (8) and (9) to determine the path of the chaser. We’ll do one by hand here, but continuing this work is very tedious and time consuming. We’ll need to get more computational help to perform more iteration. We begin by substituting 0 for n in Equation (8) to get x x x x y 0 0 0 0 2 0 2 0 1 0 0 7 3 0 3 0 0 3 0 3 0 0 0 0 0 ( ) ( ) . ( . ( ) ( )) ( . ( ) ( )) ( .4( ) ( )) + = + + - + - + - Then we simplify, substitute the initial conditions, and perform more simplifying to obtain: x 0 2 2 1 3 0 7 3 3 3 3 3 0 7 6 6 3 4 2 6 ( ) 2 3 . ( ( )) ( ( )) . ( ) . = - + . - - - - = - + = - + = - We do the same for Equation (9), where many parts of the equation evaluate to 0 and disappear in the calculations to obtain: y y 0 0 2 1 0 0 7 0 0 3 3 ( ) ( ) 0 0 0 . ( ) ( ( )) = + - - - = + = Therefore, the chaser moves to the point (-2.3,0) during the first timestep of the chase. We continue this procedure for n = 1,2,3,.,10 to produce the iterates given in Table 1. We determined these values using a computer to perform all the tedious, but necessary computations
nXo(n)Yo(n00-310-2.32-1.60.053-0.910.1540.30-0.2350.450.5061.100.7471.741.0382.361.3592.961.72103.542.11Table1.IteratesforEquations (8)and (9),providingthepathforthechaser.Whendo we stop our iteration?There is noneedtocontinue afterthe chaserhas caught thetarget.We need to refineour modelto includea stopping criteriaforthe iteration that reflects“catching"thetarget.This does notmean that thelocationofthechaserandthetargethavetobeexactlythesameattheendofatimeinterval.Thiswouldbeextremelydifficultorimpossibletoachieve.Sincewe don't have a continuous functionforlocation,we don't have an easymechanism to check locations during the time interval. Therefore,we willassume that"catching"the target means just being"close enough"at the end ofa timestep.First, we need to determine whatis"close enough."Ifthe chaserisanexplosivemunition with a large"kill radius,"then wemight use that radiustodetermine"close enough.”Ifwe need an impact of the chaser and the target,wemay say"close enough"is a very small radius. We usually call this“closeenough"distance or the radius of kill,the tolerance of the stopping criteria anddenoteitby.Wehavenumerouschoicesfordeterminingthistolerancevalue.Itcouldbeafixedvalue,liketheradiusofkill.Itcouldbeafunctionofthespeedsandtimeintervalt.Weknowthatinourdiscretemodelthechasermovesadistances△tovereachtimestep.Evenmoresophisticatedmodelscombiningthesetwocriteriaandothersarepossible.Forourexample,wewillusethedistance=0.5(s△t).Thismeanthatouriterationwillstopwhenthechaserandthetarget arewithin half the distance traveled by the chaser in a timestep. Let'sreturntoourproblemtocompletethecalculationswestartedinExample1.115
115 n x0(n) y0(n 0 -3 0 1 -2.3 0 2 -1.6 0.05 3 -0.91 0.15 4 -0.23 0.30 5 0.45 0.50 6 1.10 0.74 7 1.74 1.03 8 2.36 1.35 9 2.96 1.72 10 3.54 2.11 Table 1. Iterates for Equations (8) and (9), providing the path for the chaser. When do we stop our iteration? There is no need to continue after the chaser has caught the target. We need to refine our model to include a stopping criteria for the iteration that reflects “catching” the target. This does not mean that the location of the chaser and the target have to be exactly the same at the end of a time interval. This would be extremely difficult or impossible to achieve. Since we don’t have a continuous function for location, we don’t have an easy mechanism to check locations during the time interval. Therefore, we will assume that “catching” the target means just being “close enough” at the end of a timestep. First, we need to determine what is “close enough.” If the chaser is an explosive munition with a large “kill radius,” then we might use that radius to determine “close enough.” If we need an impact of the chaser and the target, we may say “close enough” is a very small radius. We usually call this “close enough” distance or the radius of kill, the tolerance of the stopping criteria and denote it by e . We have numerous choices for determining this tolerance value. It could be a fixed value, like the radius of kill. It could be a function of the speed s and time interval D t. We know that in our discrete model the chaser moves a distance s D t over each timestep. Even more sophisticated models combining these two criteria and others are possible. For our example, we will use the distance e = 0.5(sDt). This mean that our iteration will stop when the chaser and the target are within half the distance traveled by the chaser in a timestep. Let’s return to our problem to complete the calculations we started in Example 1