{VERSION 5 0 "SUN SPARC SOLARIS" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 96 " # Bernoulli equation\neq := (1/ 2)*rho[p]*(v-u)^2+Yp=(1/2)*rho[t]*u^2+Rt:\npen := solve(eq, u);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 " # We take the second solution\n # because of u < v and simplify it\nu_v[1] := simpli fy(pen[2]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "simplify(ser ies(u_v[1],rho[p]=rho[t]),symbolic):\nu_v[2] := subs(\{rho[t]=rho,rho[ p]=rho\},op(1,%));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 164 " \+ # Definition of initial value problem\ndeq := diff(l(v), v)=l(v)*rho[p ]*(v-u(v))/Yp:\nincond := l(v0) = L:\nlen[1] := simplify (rhs (dsolve( \{deq, incond\}, l(v))));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "subs(u=unapply(u_v[2],v),\{rho[p]=rho,rho[t]=rho\},value(len[1])): \nlen[2] := simplify (expand (%));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "u_v[3] := simplify (subs(Rt=Yp, u_v[1]), symbolic):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "subs(u=unapply(u_v[3],v),len[1]): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "l := simplify(%, symbolic): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "pe := simplify(int(l*u_v[3], v=0. .v0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "simplify(limit(pe ,rho[t]=rho[p]),symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "vm := sqrt(2*(m-1)*Yp/rho):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 " simplify (subs(Rt=m*Yp,len[2]*u_v[2])):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "pe := (rho/Yp)*int(%, v=vm..v0):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "pe1:=factor(simplify(expand(limit(pe,m=1))));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "pe3 := simplify(expand(subs( m=3, pe)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 295 "restart:\n \+ # Bernoulli equation for same densities rho=rho[t]=rho[p]\neqe := (1/ 2)*rho*(v(t)-u)^2+Yp=(1/2)*rho*u^2+Rt:\nu := solve(eqe,u):\ndeqe1 := r ho*l(t)*diff(v(t), t)=-Yp:\ndeqe2 := diff(l(t), t)=-(v(t)-u):\nincond \+ := l(0)=L, v(0)=v0:\nsol := dsolve(\{deqe1, deqe2, incond\}, \{l(t), v (t)\}, series):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 " # Polynomial a pproximation of v(t) and l(t)" }{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "v := subs(sol,v(t));" }{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 177 "l := subs(sol,l(t)): \n # Polynomial approxim ation of u(t)\nuappr := convert(series(v/2+(Yp-Rt)/(rho*v),t=0,6),poly nom):\n # First few coefficients of uappr\nseries(uappr,t=0,2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 269 "rho := 7810: # Target and p enetrator density\nL := 0.25: # Length of rod\n # Target resistance Rt = 900*10^6\n # Strength of projectile Yp = 900*10^6\nu := subs(Rt =900*10^6,Yp=900*10^6,uappr):\nplot3d(u,t=0..0.0004,v0=1000..2500,\n \+ axes=boxed,orientation=[-30,60]);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "restart ; assume(mp>0); \nde := -k(mu)*DP(P)*diff(DP(P), P)/(a + c(mu)*DP(P)^2 ) =1/(mp - mu*P);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "Dsu:= [diff(DP(P),P)=diff(P(t),t,t)/diff(P(t),t),DP(P)=diff(P(t),t),P=P(t)]; \nmap(u->simplify(u*denom(lhs(de))*denom(rhs(de))),de):\ndeq:=subs(Dsu ,%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "ic := DP(0) = vp:\n pen := \{dsolve(\{de,ic\},DP(P))\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "sol:=map(u->`if`(op(1,rhs(u))=-1,NULL,rhs(u)),pen)[]: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "su_1:=c(mu)/mu/k(mu)=ep silon(mu)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "bs_1:=epsil on(mu)=solve(su_1,epsilon(mu)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "t_1 := simplify(subs(su_1,mp=Mp,sol)): " }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 61 "su_2:=(Mp-mu*P)^(epsilon(mu))=A^epsilon(mu)*Mp ^(epsilon(mu));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "bs_2:=A= expand(solve(su_2,A)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "t _2:=expand((subs(su_2,numer(t_1)^2/a/c(mu)))):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 63 "dP:=subs(bs_2,sqrt(convert(t_2,horner,A^epsilo n(mu))*a/c(mu)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Pf:=si mplify(solve(dP=0,P));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 626 " # Projectile and target data and the final penetration depth\n rho [t] := 7810: Mp := 1.491: Yt := 9.70*10^8:\n rho[p] := rho[t ]: dp := 0.0242: Yp := 15.81*10^8: \n Ap := Pi*dp^2/4: v0 := 1457.9:\n pf := 0.245:\n\n # 1st step - Determination of \+ mup and graph penetration depth \n # vs erosion rate mu, the defini tion of constants ap, a, b \n # and initial penetration velocity\na p := 1/(2*rho[p]*Ap): a := 3*Yt*Ap:\nb := rho[t]*Ap/2: vp := v0/2+(Yt-Yp)/(rho[t]*v0):\n # The definition of functions k, c, ep s and Pf of mu\nk := unapply(1+ap*mu, mu): c := unapply(b-ap*m u^2,mu):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "epsilon:=unappl y(rhs(bs_1),mu):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "\n # - Graph of influence of mu on the penetration depth\nplot(Pf,mu=0..6) ;" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 122 "\n # Determination of limit \nPfzero := \+ limit(Pf,mu=0):\n # Solution of the equation Pf(mu)=pf\nmup := fsol ve(Pf=pf,mu):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 " # 2nd \+ step - Graph of the penetration velocity\n # along the penetration \+ path" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot(subs(mu=mup,dP ),P=0..pf); " }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 166 " # 3rd step - Graph of \+ the time dependence of the penetration tf := 0.0006:\n # The evalua tion of the time dependence \n # of the penetration and its velocit y" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "deq:=subs(mp=Mp,mu=mup ,deq):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "ini := P(0) = 0, D(P)(0) = vp:\nF := dsolve(\{deq, ini\}, P(t), numeric):\nplots[odepl ot](F, [t, P(t)], 0..0.0006);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 " # 4th s tep - The evaluation of the influence \n # of target strength and i mpact velocity" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "vp:=vi + \+ (y - Yp)/(rho[t]*vi):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "p lot3d(subs(mu=mup,Pf),vi=1400..2200,y=900*10^6..2000*10^6,orientation= [-125,60], axes=boxed,color=white);" }}}}{MARK "0 0 0" 8 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }