{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 291 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 292 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 297 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 299 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 301 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 302 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 303 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 304 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 306 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 307 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 308 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 309 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 310 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 311 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 312 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 313 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 314 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 315 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 316 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 317 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 318 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 319 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 320 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 321 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 322 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 323 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 324 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 310 17 "Rocket propulsion" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 242 "The prob lem of mechanical systems in one dimension with varying mass represent s an interesting application of Newton's law. It comes up in the rocke t problem (variable mass and velocity), the conveyor belt problem, and others. (cf. K. Symon: " }{TEXT 317 9 "Mechanics" }{TEXT -1 33 ", 3rd ed. p. 172ff, L.S. Lerner: " }{TEXT 316 7 "Physics" }{TEXT -1 52 ", p . 226 ff, or almost any first-year physics text)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "We consider a rocket wit hout external forces (no gravity). The rocket's momentum for motion in one dimension is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "P_R:=M(t)*v(t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Here " } {TEXT 257 1 "v" }{TEXT -1 1 "(" }{TEXT 258 1 "t" }{TEXT -1 141 ") is t he velocity of the rocket with respect to a stationary frame. The rock et is propelled by the exhaust of gases with a constant velocity " } {TEXT 256 1 "u" }{TEXT -1 110 " relative to the rocket. With respect t o the stationary (laboratory) frame the exhaust gas has a velocity of \+ " }{TEXT 259 1 "v" }{TEXT -1 1 "+" }{TEXT 260 1 "u" }{TEXT -1 292 ". \+ The total rate of change of momentum has to vanish, as there are no ex ternal forces. Thus, the total rate of change of the rocket's momentum is balanced by the rate of change of the exhaust gas momentum. The la tter is given by the change in mass times the exhaust velocity in the \+ lab frame." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "NEq:=diff(P_R ,t)-diff(M(t),t)*(v(t)+u)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "NEq:=simplify(NEq);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "It a ppears, as if there is an asymmetry in how the velocity " }{TEXT 276 1 "v" }{TEXT -1 1 "(" }{TEXT 277 1 "t" }{TEXT -1 372 ") is treated: in the rate of change of the rocket momentum derivatives are considered \+ both for the mass and for the velocity. They are both important as the y are multiplied by the other respective variable (as dictated by the \+ product rule). In the rate of change of the exhaust fuel momentum it a ppears as if the change in part of the velocity in the lab frame (name ly in " }{TEXT 279 1 "v" }{TEXT -1 1 "(" }{TEXT 278 1 "t" }{TEXT -1 164 ")) is ignored. This is indeed the case, but it is justified since the velocity is multiplied at each instant in time by a small amount \+ of mass: between time layers " }{TEXT 281 1 "t" }{TEXT -1 5 " and " } {TEXT 282 1 "t" }{TEXT -1 2 "+d" }{TEXT 283 1 "t" }{TEXT -1 55 " the o nly mass that counts on the exhaust gas side is d" }{TEXT 280 1 "M" } {TEXT -1 53 ". One is justified to ignore the change in velocity d" } {TEXT 318 1 "v" }{TEXT -1 17 " as the product d" }{TEXT 284 1 "m" } {TEXT -1 1 "d" }{TEXT 285 1 "v" }{TEXT -1 125 " is of higher order and does not come into play. It is in fact cancelled by a similar term ar ising in the rocket momentum at " }{TEXT 291 1 "t" }{TEXT -1 2 "+d" } {TEXT 290 1 "t" }{TEXT -1 45 " which expressed with differentials read s as " }}{PARA 0 "" 0 "" {TEXT -1 1 "(" }{TEXT 288 1 "M" }{TEXT -1 2 " +d" }{TEXT 289 1 "M" }{TEXT -1 3 ")*(" }{TEXT 286 1 "v" }{TEXT -1 2 "+ d" }{TEXT 287 1 "v" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 101 "We can solve the differential equation \+ either for the mass or for the velocity as a function of time:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solM:=dsolve(NEq,M(t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solv:=dsolve(NEq,v(t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 " In the solution for the mass " }{TEXT 269 1 "M" }{TEXT -1 1 "(" } {TEXT 270 1 "t" }{TEXT -1 57 ") the integration constant _C1 is clearl y a mass, namely " }}{PARA 0 "" 0 "" {TEXT 271 1 "M" }{TEXT -1 1 "(" } {TEXT 273 1 "t" }{TEXT -1 4 "=0)=" }{TEXT 272 1 "M" }{TEXT -1 4 "(0). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 218 "In t he second solution for v(t) some further thinking is required. It make s no sense to take the logarithm of a dimensionful quantity (e.g. a ma ss). Thus _C1 should really be reexpressed as a constant that represen ts -" }{TEXT 266 1 "u" }{TEXT -1 4 " ln(" }{TEXT 265 1 "M" }{TEXT -1 1 "(" }{TEXT 268 1 "t" }{TEXT -1 53 "=0)), in which case the solution \+ can be expressed as " }}{PARA 0 "" 0 "" {TEXT 264 1 "u" }{TEXT -1 4 " \+ ln(" }{TEXT 263 1 "M" }{TEXT -1 1 "(" }{TEXT 267 1 "t" }{TEXT -1 2 ")/ " }{TEXT 262 1 "M" }{TEXT -1 80 "(0)). Now the logarithm is taken of a dimensionless quantity, which makes sense." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{TEXT 261 1 "u" } {TEXT -1 102 " is the exhaust velocity (not just the speed), i.e., the direction is opposite to the rocket velocity " }{TEXT 274 1 "v" } {TEXT -1 1 "(" }{TEXT 275 1 "t" }{TEXT -1 2 ")." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "With specified initial co nditions the solution for the velocity is obtained as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "solv:=dsolve(\{NEq,v(0)=v0\},v(t)); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "The fact that the solution ca n be found either for " }{TEXT 292 1 "M" }{TEXT -1 1 "(" }{TEXT 293 1 "t" }{TEXT -1 9 ") or for " }{TEXT 295 1 "v" }{TEXT -1 1 "(" }{TEXT 294 1 "t" }{TEXT -1 51 ") should have made us suspicious. The solution for " }{TEXT 299 1 "v" }{TEXT -1 1 "(" }{TEXT 298 1 "t" }{TEXT -1 35 ") depends on the time evolution of " }{TEXT 297 1 "M" }{TEXT -1 1 "( " }{TEXT 296 1 "t" }{TEXT -1 317 ") and vice versa. This makes sense, \+ as we have specified the exhaust velocity (or speed) of the gas, but n ot the rate at which the fuel is being burned. The rocket velocity dep ends clearly on how much mass is being sent off in the opposite direct ion at a given exhaust speed. Let us assume a linear relationship with " }{TEXT 311 1 "m" }{TEXT -1 74 " being the mass of the rocket withou t fuel, and a valid time range of 0 < " }{TEXT 312 1 "t" }{TEXT -1 42 " < 10 (all quantities in MKSA (SI) units):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "m:=500;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "M:=t->m+1000-100*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " v_R:=unapply(rhs(solv),u,v0,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "plot(v_R(-100,0,t),t=0..10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "For " }{TEXT 319 2 "t " }{TEXT -1 73 "> t_f = 10 the solution is not meaningful since the fuel contribution to " }{TEXT 320 1 "M" } {TEXT -1 1 "(" }{TEXT 321 1 "t" }{TEXT -1 56 ") becomes negative accor ding to the mapping given above." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 111 "The acceleration of the rocket changes a s a function of time. It is calculated by differentiating the velocity :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "v_R(-100,0,t);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "a_R:=diff(%,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "plot(a_R,t=0..10);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "The strong rise in the rocket's accelera tion near the end of the fuel burn cycle is associated with the fact t hat according to the solution the increase in velocity d" }{TEXT 305 1 "v" }{TEXT -1 77 " is given by the product of the exhaust speed and \+ the log of the mass ratio (" }{TEXT 300 1 "M" }{TEXT -1 1 "(" }{TEXT 304 1 "t" }{TEXT -1 2 "+d" }{TEXT 303 1 "t" }{TEXT -1 2 ")/" }{TEXT 301 1 "M" }{TEXT -1 1 "(" }{TEXT 302 1 "t" }{TEXT -1 140 ")). The mass ratio is close to unity while a lot of heavy fuel is on board, but d eviates strongly from that value as the fuel is exhausted." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 427 "Another interest ing observation can be made from the graph of the acceleration at earl y times The rocket starts with a positive non-zero acceleration. If we add the gravitational acceleration that acts in the opposite directio n we may not have enough thrust for lift-off. We discuss thrust furthe r below: it is the magnitude of the equivalent force that results as a consequence of hurling out particles at a given flow rate (d" }{TEXT 313 1 "M" }{TEXT -1 2 "/d" }{TEXT 314 1 "t" }{TEXT -1 27 ") with some \+ exhaust speed |" }{TEXT 315 1 "u" }{TEXT -1 604 "|. The product of the se two quantities has the dimensions of a force and it is responsible \+ for the rocket's acceleration. For a constant rate of fuel burn-up and for constant exhaust speed this force is constant. Thus, the increase in the rockets acceleration during its fuel-burning phase is a result of the decreasing rocket mass. Multi-stage rockets have the advantage that the mass associated with a big motor (or entire booster rockets \+ on the Space Shuttle) can be jettisoned as the contained fuel is burne d up. For the subsequent stage(s) a more convenient fuel/gross payload ratio can be achieved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 523 "The design of rockets calls for a thrust that is se veral times larger than the initial weight in order to gain speed quic kly. When a rocket is ignited, however, it takes some time for the bui ld-up of full thrust: this is repsonsible for a short delay between fi ring of engines and lift-off. For the example given above the product \+ of fuel burning rate and exhaust velocity, i.e., the thrust is not eno ugh to overcome gravity (the acceleration does not exceed 9.8 m/s^2). \+ In free space, however, this issue is not important." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 386 "A similar phenomenon c an be experienced during take-off of jet airplanes, i.e., a time delay between the revving up of engines and actual forward acceleration (he re one needs to overcome friction to set the plane in motion). In a je t engine thrust is not achieved by just hurling burned fuel out of a n ozzle: air is taken in and compressed by burning kerosene using air fr om the intake." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 211 "If we do not like to rely on Maple's dsolve engine, we c an perform the required steps to find the solution ourselves. We resta rt the session and redefine Newton's equation as it is safer than to s imply unassign " }{TEXT 322 1 "M" }{TEXT -1 1 "(" }{TEXT 323 1 "t" } {TEXT -1 64 ") (we found this to be unreliable in early releases of Ma ple V):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "NEq:=simplify(diff(M(t)*v(t) ,t)-diff(M(t),t)*(v(t)+u)=0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "accel:=solve(NEq,diff(v(t),t));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sol:=v(s)-v(0)=int(accel,t=0..s);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 258 "To combine the solution to the Tsiolkovski formu la (named after a rocket pioneer who found it in 1897) we need some tr icks: it is necessary to assume the arguments of the ln function to be positive for the combine command to work; the assumption of positive \+ " }{TEXT 324 1 "M" }{TEXT -1 156 " doesn't work properly as it is a su bscripted variable, so we substitute two symbols for the initial and f inal mass, factor the expression and then combine:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "assume(M_f>0,M_i>0); interface(showassumed= 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "Delta:=combine(facto r(subs(M(s)=M_f,M(0)=M_i,rhs(sol))),ln);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "The interpretation of this expression for the change in \+ the rocket velocity Delta for given exhaust velocity " }{TEXT 306 1 "u " }{TEXT -1 15 " is as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 254 "The mass ratio that appears as the argum ent of the logarithm is less then unity, and thus the logarithm yields a negative number. This is consistent with the fact that the rocket a cquires a change in velocity of opposite sign to the exhaust velocity. If " }{TEXT 307 1 "u" }{TEXT -1 275 " were to stand for the exhaust s peed, the mass ratio would need to be inverted. Note that the formula \+ is valid for any finite time interval, as it represents the solution t o the differential equation. In particular, it is not important how th e fuel was burned, i.e., whether " }{TEXT 308 1 "M" }{TEXT -1 1 "(" } {TEXT 309 1 "t" }{TEXT -1 68 ") was linear in time as assumed for the \+ graph of the solution above." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }