HW4_solution

HW4_solution - Wily Brehlem 2:54 Starting at the top of the...

Info iconThis preview shows pages 1–19. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Wily Brehlem 2:54 Starting at the top of the vehicle, and using Table 7-5, the masses we need are #5: mfz = m* +11;82 == 291 kg iia»é£ern,§ moz = mf2 + mp2 = 706 kg mfl = m 2 + ms1 = 819 kg mol = mfl + mp1 2 1986 kg since we wish to consider two burn strategies, we will wor burnout speeds separately for each stage. The rocket law gives moi = .9 c ve1 ln [ 1 2447 m/se 1 1“f1 moz v2 2 vez 1n [ m 1 = 2449.3 m/sec f2 -,}h“ fact that this is an optimized rocket is shown by the fact that _i.cn stage shoulders an equal portion of the burnout speed. N7?f In the burn ~ burn — coast strategy, the total burnout velocity = 4897.2 m/sec. Working out the conservation of energy per —%— via = 1.199 x 10’ m2/5e02 = g smax 7 V ives a value of H = 1223 km, indicating that we have exceeded the t ‘ max &-%gpunds of the constant gravity assumption: this number is too large a ;-;§raction of one earth radius! Using inverse square law gravity, energy is given by —l— v2 — ~E— : -50.50 ka/sec2 2 ho ro f:starting from the surface of the earth, r0 = 6378 km. Since at apogee jév a o, the energy must equal -u/rmax, giVing 3 Value °f rmax ' 7893 skm, or Hmax n 1515 km. This value is larger than that using the .Q constant 9 assumption, since gravity drops off with altitude in the is real earth. V For the ,burn — coast - burn — coast method successive applications of energy conservation give, for the first burn —%— vi — —E— a -59.50 kmz/sec2 - ~ ~5— ro r1 » r1 = 6699.1km and, for the second burn 1 2 5 v2 .t — —E- = -55.500 km2/8602 - - —3— r1 r2 » r2 * 7055 km 4 “1ka new: .. This translates to a maximum altitude of Hmax - 677 km. This is much less than the burn - burn ~ coast method. since the VOhiCl‘ trade‘ _ _,.. _ . ... .L, Lu---u. ".1nc1tv results in k out the § W145 M WEE/Mam? E? WWW»¢«me/m .4. a—(W-«vN‘ M.“ WWMMW Awmwywlaww a‘céém E 2 igfldl w 33%« WNWWMWMWWMWWWMV ww*»uwmw., “NW A“ _ ,. (,WWWMM_ .J In met-Ad J/aca/ I'M/5647 0/4 facgd 4"- m {3 Fafi f 0%: [E [at J ' ‘ 1‘ 3:? , 4 V r ’I fit flag/m" V‘fif/ 3%; «fififhaf/igfw m/ We KC MAWfi Vééa‘r . M A “"3 f . if “$2 4%; z 6 3' m-S‘: (a? ff eff? /' may) #4 Th damfién 4%; 1} “cl? KL, aaéu} m3 W " ltéé AG ,(Xffwcbn 1}” gamwéaf‘ I'm/oféffi'k/ fad K1 no/féa 0/, d4: (‘1 [Jyémc 19am C/ajen. Fsém Q/fifi‘az/ «Va-51!: (sf/en «L (6&1 if fléaf go éldfidjrfi/ Z A: ,fl, Me,” 0/ lac/Q} xkrz‘éggg 43" 11/4 Anew/K J96 . NJ Mmflmwamaééé 5: £259? Jao/ a» Mafia; 5Q 404;? M 71¢ MIR/76165;: are I: 632%» % Velvet} dfifl fwcflflu f/m r‘é / daring,“ 6g mag}? ’54; Qw¢flm¢nb ém/ fiann’e/dco. £5 1%, Ema} 62%} 7%“ 0'7 4%; aw. flea/J faé’L yaks/g 2% 53/ @éééwg xi 1%; épfg/ L “(may/gaff m W5 [WQ' Mfi§£m (6" ftéza’s‘l/Z to %. [46ch gig/z) Co/ayonoé 4, an a mar/[mg Mk ILA/J49. III: 54 {xi/5a. . (at) I» 15/1 fang (la/3" rhgxiféc/ flame? 3&5; {@fififi Chm};- algae! I}; {(4 fléfim We Icfcz o v: If f f ,5?“ *3 r F3; ' .: . adv “fl/ 6’2,“ ifigi MM w .. flmwé M: hug” (aw; L95; 7h M07 J AL mar fob/“>1 5 A [max m, gin" w: 1/ :“ZJ’XF + F . Tlua u. aK/cm. KL 7,45%, “M65” P“ ML {4be In [it flCUnL: m5; ., M -= i; J {. m3" +J'5‘f'fi}; (We) a 5/ (a (Mg/,9: Wu (7:53 aézzmoyé an/L émd/«wj we. 13%; / 12¢ a “(9, A ém/z-w;). 1" #‘L [‘m/ gm‘ of [J‘wa 90¢ nun-é b éé x26 «(om/’A flan —' ('rz erfi‘éo/ accoérafiém . J u r : if f 2 ‘ Weft, i? «It ML fol/56,, 0/) A have; 1:6. Nb. Roe“ 41W” ,5 Ifik den/5r dfl KL M .. 9") F: R; + [i'+Z:Xj+ ixfil’xf‘V] k—‘N HM, aéflaonfiqfibh ('4 1:7 ,fl/m U l 733 d ( 1;; x4 xoa/ 5”". '~ “0" = w x 5): 7‘3 .3 3 4 ) 0410/ w 3 3:0 ( 4 (g If A (Im‘f 2-6;“? yea/"cf 0; A» 52/4 can/er [her/9AM Franz. ) . Hg}? 09 ,;, ,sa 7mm of ma, 0/ x. 4,06,} final “1% {flu/1‘: !}: K 7ad/ré’n/ 0/ Iva/'50 IA I4 40/ fl/[Jeaa #ML . HM: (4/ ’ “Maia/g #5; 504$)» (QM/j TL inf/22v] Canafi‘fibnj 4r lg: ale/4... Je/«a/Ién; we : a u .— r z .. ‘ f ' (0) Q "‘ ¢ 1;. c06’0(dlafca . (‘10): wX f 5d"? m(a) ’ m (0) 6L: A" INQ" .. 0 ¢: ‘ I of éunJ/(J. 2' t. ma a .. 3 c . It" a“); a .14. NJ“ “Afar/IL af Van Mlle 1“; Mali fer/c. WI flaf’ Va" ('4 AC Wfijnmafi RLQJOI’MLIL 2/ fir floéém k/m/ [g l)? I I A," fl/ 1 f m o/aQr Q [we] a [w Ilia; 0'! "Amt/2”! dflété af JW’aQ’ I; fog/,5», I Jay/er rad“ 0// MW}! la 45/: {M446 w: 7nz/«(c'mr a/g Inc/Sn] k oft! ICJ' & cf 1% k fame Can/424’; ylfim — “(9 {£144 nameca/ Vaéltj KL M¢*,}q“m «(6&4 a} Java» vou/ 0’ AM, 3 2%.? 4.. [flee/ad VA” 57%,. fl Ms.» .4. % éa/ [Jyaa 4am} 1% WW I} gzne/af/ a/zéhe/ 4; AQ z- (av/aghast , MK 5%,; fighi/ién/ / ~ ESE, § 12 //.-A // 9/: mfg/+4.. dccyé/é’x W w! .. Mam/«K # 5 Jam» my Tit [LJthJmhaaéleme/{Lnéaf ale/[g ragga] K: @601 /AM¢_ (1i / 711 Jan ov/ "/méf wig/f "an u/oca/ Iowa/Mm yak 312/ I3 Am. We email/col 1462M Max»; of A 4/ “£5. Mt (/eéa‘ éré fiJJ ’1' fine“ f4 4 A /aa/ ‘44:}. 5;“, (/2, [cu/L‘ Jud l» A liffeaafiéa 159 ,1'5 W Mya/jofia/ Ma, {AL (her/1M 1/6/04);— g, {4 I .02 A5,}; 1% w /& ar/h/amrma/ fa xf/L flax», Vac/5r ax/ea Cow‘céxg fié mafia» A, #6 M (ea/Er flaw / W44 1/: 0&5)?“— 5’" loL ’54 ‘ 4/ J45» /a/éae ,5 Kc age of A da/ Maw/Ma» a4 A: may. “fay/Rafi; F 350:“? (/1 43mg} 0 [to J . c; §§§M «M r . WA w 10 :— “aa ’ a%f——%—— a, gig o 711% filo I'e/a/ 64:! W I‘M/{é it half” I “f bkfrgh/‘f [and 1&5,” ‘ L ’5» f7ke/ m. w// m A ":Mfiéx/I’ Common/7: 505/ a/ fl, Wé/f/ am, Can olfaf/ 00/39: K: fan, 04“, 6 41% a! eta/fl)»; of 1% Krw/al/dc/xéfi 0/ M ma- xx'zer/fé/ aide/Va/xw- Mfg, a,er %% flat fa» 1- {eyuw K; can/rng acaé/f fggfl ' ' , Grid» a m a 4 g M mafia» 4/ r4 5% '4' I64 éfi/J/flme a/’ 44/4»: . W Emil w pea/J Me /a fat” 02% fl’af 2% (fie/73,, a" fla fir flat/5‘ I}: u Wei/M/ pad/Ila» o/A’e ficggf”; /4 A,/ [me fix $1? fa/é'ww“ VeK'oé. ‘1 garage; amweg a” gégéfiéim gage; 10000 90 r v I 9000 ‘ 3 2.3 7000 r 3% v C 89 - g 6000 ~ 7 > m ‘5 E E 5000_ r r 8 E 88.5 4000 - r r g 3000 - I 88 r' 0 20 40 60 ‘ time (s) altitude as a function of time 80 “ Orbit of the rocket in the inertial frame: (x,y) pm}. A _5520 . t t , . t . . , ,' . . _ _ V V , , . V . V . _ V V V . .I V _ _ . t . , . t . E 60 ‘ : a) _5540 L. t , . . . . t . , . . . V V . . , , . . . . t , V . . . . a 0 ,. i. ....................... ,. g 40 . —556 V “5 _5580.7; NHL V . . _ . . , , . V V t V V V ,V‘ V V V , . . _ t, 20 - 950 1000 1050 1100 1150 x (km) time (s) Orbith the rocket in the inertial frame: (x,z) taro}. Orbit of the rocket in the inertial frame: (y,z) pro}. 1 I I l ” ‘1 -, . V , V , . . _ _ _ V . , r . , . r , , , , _ V V , , ,‘_ 3070 -' E r r g 3070 v A ‘ L ‘ I A . V . , , _ ., N ‘ N 3050 ~ 1 t 3050 _ , _ V 1000 1020 1040 1060 1080 —5580 —5560 —5540 —5520 —5500 x (km) y (km) Egg??? gaff gfaageg as: total mass of the rocket as a function of tés 1e flight path angle (inertial) as a function of time e get, game; ea; gee? WW WWWBAWM,WWWWWW,WW¢WW,~W WW total mass of the rocket as a function of ttme flight path angle (local) as a function of time 10000 v r x a 90 r v I 8000 fg‘ 89-5 z E A Q) m _ '5 g 3’ 89 v 8 6000 m m E E 8. E 88.5 4000' g 1 : l 88 2000 ‘ * ‘ 0 20 40 60 time (s) altitude as a function of time Z " Orbit of the rocket in the local frame: (x,y) pro]. 3 .. , ' . , ' V . E 2 L. , ‘ V. é, : O i . . , . t . . . . , t 4 time (s) Orbit of the rocket in the local frame: (x,z) proj. Orbit of the rocket in the local frame: (y,z) proj. r“ f r f '“ l j ' ‘ ‘ ‘ 80 , , , , _ , , _, , ,, ,. ., .‘ . , , _ _ , ., : V I ' I , , _ , , . , ,: . _ , , , 20; I , I , , _ , , , 20 O O i n L J 400 -50 0 50 100 400 —50 50 100 x (km) %% Small script to solve HW4, Pb2, question 2 % _._ % Some conversion factors zeros(Ni,l}; % altitude and flight path angler d2r m pi/180.0; % degrees to radians r2d = 180.0/pi; % radians to degree m2d = l.0/60.0; % minutes to degrees % Some constants (units of km, kg, 3) Re = 6372; W = pi/(l2*60*60>; h = 0.003; lat x ( 28.0 + 28.0*m2d )*d2r; lon = (—80.0 + 32.0*m2d )*d2r; %% Initial conditions 96 ________________ .___ m0 = 10000.0; % Inertial space: X = [position, velocity, mass]; P030 = (Re+h)*[ cos(lon)*cos(lat); sin(lon)*cos(lat); VelO = W*[ -PosO(2); PosO(l); 0.0 1; X0 = [Pos0; VelO; m0]; % Local frame: x = [position, velocity, mass]; posO = [ 0.0; 0.0; 0.0]; velO = I 0.0; 0.0; 0.03; x0 = [pos0; velO; m0]; % Time span m_fuel = 8000.0; dmdt z -llS.0; Tspan = [0.0; -m-fuel/dmdt]; %% Trajectory integration % [T,X] = ode45(@PbZQZMFinertial, Tspan, X0); [t,x] = ode45(@Pb2Q2flFlocal, Tspan, x0); %% Altitude and flight path angle % % in the inertial frame [Ni,Mi] = size(X); alt-i = zeros(Ni,l); fpa-i = (inertial) for i = lzfii, % inertial altitude and flight path angle R = norm(i(i,l:3}); U =[X(l,4)‘r"§*}i(i,2), X<i,5)—W*X(i,l>; X(i,6)]; V = ncrm(U); alt-i(i§ : R—Re; if (V==0.0) fpa-i(i) = 90 3, else fpa0i(l) = 9‘0w acos( dot(X(i,l:3),U)/(R*V) )*r2d; sin(lat) end end % in the local reference [Nl,Ml} = size(x); ‘ alt_l ~ zeros(Nl,l); fpawl for i = lle, % altitude and flight path angle in the local reference frame v = norm(x(i,4:6)); altwl(i) = x(i,3); if (v:=0.0) fpa_l(i) N 90.0; 90 ~ W "0 SD H p [I end % Results at burn~out alt_i(Ni) fpa_i(Ni) alt_l(Nl) fpa_l(Nl) %% Trajectory plots % inertial figure(l) clf; subplot(321) plot( T, X(:,7), "b'>; grid on; title('total mass of the xlabel(’time (S)')i ylabel('mass (kg)'>; axis tight; subplot(322> plot( T, fpawi, "b’); grid on; title(’flight path angle xlabel('time (s>'}; frame it e A we ere; zeros(Nl,l>; % altitude and flight path angle acos( x(i,6)/v }*r2d; rocket as a function of time’); (inertial) ylabel(’flight path angle (deg)’>; axis tight; subplot(323) plot( T, alt_i, '“b’); grid on; as a function of time'); title(’altitude as a function of time'}; xlabel 'time (s)'); ylabel(’altitude (km)'); axis tight; subplot(324) plOt( X(:,1), X(:,2}, ’—b'); grid on; title('0:bit of the rocket in the inertial frame: (x,y) proj.'); (local) sag” 5555 i amass? safaris. We; title(’Orbit of the rocket in the local frame: (x,z) proj.’); xlabel(’x (km)’); ylabel(’z (km)'); axis equal; subplot<326 plot( x(:, grid on; title(’©rbit of the rocket in the local frame: (y,z) proj.’); xlabel(’v {km}'); Yiabelefkml); g’éfifigéfig (figfiwfié $3; wgéfigggfia ) 2)! x<213)r ,wb’); sea/sf 33: i giasif is if % Function defining the equations of motion for Pb2, Quest. 2 (inertial) function dth = Pb2Q2_Finertial(t,X) % dth must be a column vector, so we initialize it with zeros dth = zeros(7,l); %% Constants in (g, km, s O 6 __ % Central body: Earth G = 6.673e-20; Me = 5.9742e24; mu = G*Me; W = pi/(12*60*60); % Earth rotation rate % Engine Isp = 300.0; g = 0.00981; dmdt z ~115.0; % This was a missing parameter in the assignment... % Conveninent variables mu = G*Me; Th = wIsp*g*dmdt; r3 = ( X(1)“2 + X(2)A2 + X(3)“2 )Al.5; U = [X(4)+W*X(2); X(5)~W*X(l); X(6)]; if (norm(U) > le~3) V = U/norm(U); else V = X(l:3)/norm(X(l:3)); end %% Equations of motion 0 a position dth(l) : X(4); dth(2) : X(5); dthCS) = Xfé); % velocity dth(4) z —mu*X(l)/r3 + Th*V(l)/X(7); dth(5) z -mu*X(2)/r3 + Th*V(2)/X(7); dth(6) = —mu*X(3)/r3 + Th*V(3)/X(7); %mass dth(7) z dmdt: if 7.5... 5: see§i$%a§ ref,asssféaalwaséafiaasiaygo g“ éggie ggéfiaeeéefgiwéf g .£%£§2§%e % % Function defining the equations of motion for Pb2, Quest. 2 (local frame) function ont = Pb2Q2~Flocal(t,X) % dth must be a column vector, so we initialize it with zeros dth = zeros(7,l); %% Constants (units: kg, km, s) % % some conversion factors d2r = pi/180.0; m2d = l.0/60.0; % Central body: Earth Re = 6372; G = 6.673e—20; Me = 5.9742e24; mu = G*Me; W = pi/(12*60*60); % Earth rotation rate % Laucnh site h = 0.003; R1 = Re+h; lat = ( 28.0 + 28.0*m2d>*d2r; lon = (‘80.0 + 32.0*m2d)*d2r; erc = W*[ 0.0; cos(lat); sin(lat)]; local frame %R = Rl*[ cos(lon)*cos(lat); sin(lon)*cos(lat); sin(lat) ]; o\° coordinates of the angular velocity in the! % Engine Isp = 300.0; g = 0.00981; dmdt = —llS.0; % This was a missing parameter in the assignment... % Conveninent variables mu = G*Me; T = *Isp*g*dmdt; r3 = ( X(l)“2 + X(2)“2 + (X(3)+Rl)“2 )“l.5; % Thrust direction U = [X(4); Xi5); X(6)]; if (norm(U) > le—3) V = U/norm(U); else V = [0.0; 0.0; 1.0]; end % noninertial forces Cor = 2.0*cross( erc, X(%:6) ); Cen = cross( erc, cross< erc, [X(l); KiZ); X(3) + Rl] ) ); %# Equations of motion % velocity gxggil?1:nv<4\- dth<4> = “mU*X<1>/r3 + T*V<l>/X<?; ~COr(1) + Cen(1); i; ” dth(5) = -mu*X(2) r3 + T*V(2)/X(7E -Cor(2) + Cen(2); dth(2) = 0(5); 6 — — * Y(3‘+Rl‘/”3 + T*V(3)/X(73 —COr(3) — Cen(3)' = ) — mu (x j ,1 i. / v. v I 96103.33 dth(7) = dmdt; «£25 35% MM 14; .. liamwws #3 3.4%». (Ca/g) «3; ,3] TAR maHal Mfr"! 0" A g/émy/yu- @16%/. TL may? ma¢7¢caflm Ik/fl W560 0/ an Mar A7; /a 06}, Kc Plane/fir dumb camaralm/yén in him JM/fo/ fié //mém1 711mg». Ear/1L (mag gal} rank/ebb! {IL}; ‘44:. c} Pg“! «4 q J/E/i é A fl 1060'»; air/50%;»). Conmea/fg— A) KL hauae .4}?! Auto“ car/2?. thvglf/k/{mk/oaé/Vo‘tb‘m' J‘s/o Mfg % ,h/qué‘ thl/t'c‘fir [/50 «a I} % Ind/chain ,‘fmfi d file/eq/ 54"“ flan/Q sing/34V i] Wow Auto-01:1, 149m fi MW? flak, & r6564» éflk éarfl £34 I" 4/406! Jé 4135501 ( r“: (Inf/5d Vzéa/t/Z 13/“, m. 1'! q {fray/f aflu/a/a/ 7i} 0‘ Jain/Ar /; fig We flee/El 1/" f1 /. Ar any/on‘ma/Saz WM’ can we 5Q sow-04y kW 7mm. my a: m. 0/ xx: 404,,“ .2, H50 = ’iéi—me) - Jffigfimmw) gyms“; fl nam'ca/ “4‘4 gm ,+ m ) 2 a O 5 u / fly” ‘/v/?/ Mall/’44:»; 0/, \,,f §§§§§ egg gmeeg 5; meet meg; flight path angle (inertial) as a function fuel mass l f l I l I l l 88 >- a Q) E % 87.95% _, E: .C E g 87.85: 87.8? _ V l, , , . ., ._ ' l l i . l l l _. 5000 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 fuel mass (kg) altitude as a function of fuel mass 200*” 180.. V . _ _ _ ., 160*” _L 4; O l 120*" altitude (km) 100 r-” 60 i l l l l l I l l 5000 5500 6000 6500 7000 7500 8000 8500 9000 9500 10000 fuel (kg) altitude (km) flight path angle (deg) 00 Q a 5W? W flight path angle (inertial) as a function isp I i 03 CO CD 0”) CO 4:. CO N) 78 76] 74?? 72 200 250 300 350 400 450 500 isp (s) altitude as a function of Isp ,lt , ,.1, , ,. ,. VI 180 r 160 140 ~ .4 [\7 O 100 mg ,, $995995 5%; $8858. {85%; altitude (km) flight path angie (deg) 90 89.8 89.6 CD 89 4:. co 99 m 00 (D co .00 00 co .00 m 88.4: 88.2 98.5 98.49 98.48 © 9*.) A \J 9846 98.45 98.44 98.43 “AWN... w... w. flight path angle (inertial) as a function of latitude egg 55g- gegkfigée ~45? gag/0. % Small script to solve Hag; Phi. % fie: g W % Some conversion er : pi/180.0; % r to radians r2d : 180.0/pi; % radiane to degree mZd = l.0/60.0; % minutes to degree m (D31) Q 0 tore e s Q (D % Some constants (units of km, kg, 5) Re = 6072; W 2 pi/(l2*60*60); %>>>> Parametric study: variation of Isp <<<< % M Npts = 101; Isp = zeros(Npts,l); alt_i = zeros(Npts,l); fpa_i : zeros(Npts,l); % altitude and flight path anglez (inertial) for i = lszts, % Isp Isp(i) = 200 + (i—1)*3; Launch site do h : 0.003; lat = ( 28.0 + 28.0*m2d )*d2r; lon = (—80.0 + 32.0*m2d )*d2r; Initial conditions do 6 ________________ -_ m0 = 10000.0; % Inertial 3pace: X = [position, velocity, mass, Isp]; P050 = (Re+h)*[ cos(lon)*cos(lat); sin(lon)*cos(lat); sin(lat) ]; VelO = W*[ —P030(2); PosO(l); 0.0 1; X0 = [PosO; VelO; m0; Isp(i)]; % Local frame: x = [position, velocity, mass]; p080 = [ 0.0; 0.0; 0.0}; velO = [ 0.0; 0.0; 0.0}; x0 = [pon; vel0; m0]; % Time span m_fuel = 8000.0; dmdt = ~115.0; Tspan = [0.0; ~mflfuel/dmdt]; Trajectory integration o\° 0\° [T,X] = ode45 @PbEQBwFinertial, Tspan, X0); Altitude and flign: oath angle do opp l fiéé €55; i, gawng $5? g 6%: pm {KW U =[X(Ni,4}+W*X(Ni,2}; X(Ni,5)#W*X(Ni,l}; XiNi,6}]; V = norm(U); alt~i(i) = R-Re; if (Vm:0.0) fpa_i(i) = 90.0; else fpa_i(i) : 90- acos( dot(X(Ni,l:3),U)/(R*V) }*r2d; end end % Trajectory plots % ineztial figure<l§ clf; subplot<21l} plot< Isp, fpa_i, ’~b’); grid on; title(’flight path angle (inertial) as a function Isp’); xlabel('Isp (s}’); ylabel('flight path angle (deg)’); axis tight; subplot(212) plot( Isp, altmi, 'wb’); grid on; title(’altitude as a function of Isp’); xlabel(’Isp (S)’); ylabel(’altitude (km)’); axis tight; §unction defining the equations of motion for PbZ, Quest eection dth : Pb2©3_Finertial(t,X) X = [pos, vel, mass, Isp] (WW W9 % dth must be a column vector, so we initialize it with m dth = zeros(8,l); %% Constants in fig, km, s % Central body: Earth G = 6.673e—20; Me : 5.9742e24; mu = G*Me; W : pi/(l2*60*60); % Earth rotation rate % Engine Isp = X(8); g : 0.00981; dmdt = “115.0; % This was a missing parameter in the assiqt \ % Conveninent variables mu 3 G*Me; Th = ~Isp*g*dmdt; r3 = ( X(l)“2 + X(2)“2 + X(3)“2 )“l.5; U x [X(4)+W*X(2); X(5)“W*X(l); X(6)]; if (norm(U) > lefl3) V 2 U/horm(U); else V : X(l:3)/norm(X(l:3)); end %% Equations of motion % position dth(l) : X(4) dth(2) : X(5); dth<3> : X(6) % velocity dth(4) = “mu*X(l)/r3 + Th*V(l)/X(7); dth(5) 3 “mu*X(2)/r3 " Th*V(2)/X(7); dth(6) : “mu*X(3)/r3 " Th*V(3)/X(7); MM [45 - l/mm/é #5? 564%?» « 3%6ém g prggglgm 1:5; We are asking what payload a certain configuration can placze in a low earth orbit. Using the rocket Equation, the (required) 3" burx1out velocity is found from V = 2 * 31,000 ft/sec = Z 4 x 504600 + 35000 + 703000 + m* s 32‘2 X 290 1“ 4 x 81900 + 35000 + 703000 + m* 35000 + 703000 + m* 35000 + m* + 32.2 x 455 in [ It is not necessary to be careful in the argument of the logarithms, and use pounds. The arguments are dimensionless, and m* will come out in kg. Performing the operations above not requiring m*, we have 3076400 + m* 31,000 9338 1n 1065600 + m* 738000 + m* + 14651 In [ } ft/sec 35000 + m* (ii) Now, the payload mass is certainly much less than 3 x 10‘ kg. It is also probably negligible compared to 1 x 106 kg, and then should probably be neglected compared to 700,000 kg as well, for the sake of consistency. It is certainly not small compared to the dry tank mass, 35000 kg, since this is the nominal payload of the usual shuttle! Ignoring m* in these places, we have 31,000 ft/sec s 9900 ft/sec 738000 + 14651 In 35000 + m* Solving for m* yields the value of m* x 139,800 kg = 307,000 lb gar almost 5 normal shuttle payloads. This vehicle has a sflightly yzarger payload than the Apollo Saturn 5 moon rocket. (The exact solution is m*-= 167092 kg.) » 9 p» A, r ...
View Full Document

This homework help was uploaded on 04/07/2008 for the course MAE 146 taught by Professor Villac during the Winter '08 term at UC Irvine.

Page1 / 19

HW4_solution - Wily Brehlem 2:54 Starting at the top of the...

This preview shows document pages 1 - 19. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online