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HW1_solution - MAE I46 Homwar’i#I Sea/’5""a”...

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Unformatted text preview: MAE I46. Homwar’i #I Sea/’5" * "a” ’ K: J 5“ ”’fi‘l ”’7’!“ “"a/ 19““! an /.3‘ and/.8; fife/3'4}, i} but, [at/L fit.» at dimméqu/{J ’44 f/‘Mf/cd/naé'an [efygg a». MAM 4am (am: “.544 Maul J MA; /' m/ K. Ata/ g’almg a,” 4' /¢'?hlL 0!, {4 {any [#412, M% 1.711% (9%! [J‘M V ax—a xx.“ é ,;~ film § [{L 127:4th or" for/311»: (@ ice and afgé/é», W 4/ R " Rel.) ' RLCW' 23(‘91‘7"; (<9) (I L @ ‘1 36,0 2r (”6/07 t‘ I} McaJtt/Zo/ a}. {cu/r. Maw 4r fld/uwfilfi/ w. maul 1% Aw“! fl/am’“ 6' fl" ’hV/fé/ ’e’mg . #4 Ma: In“! I; ‘2 a” ”(a R" M 1% éco/ (cw/(AA . fl} W2 M 2 a m W -, - Wt. LIA/L 72 :: R: fa flay/— W02)” R- :. 72f 7 7- f A 3 (63) 723 (61/. Rt (.40 72; ‘ M e, r (jam/Er #42647)? QTY?) 3 72 (- 1F) / Or “L. Com/U #w‘é/ J.“ “C 1&7/ '0’, /? 4 Orbit of the spacecraft in the local frame: BD View spacecraft 1 4 x10 1 0000 x 104 Orbit of the spacecraft in the local frame: (x,y) projection l 1 1 l l l l l l _1 ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, E 5/ _1.5 ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,, >s ”2| ‘ l ,,,,,,,,,, I ,,,,,,,,,,, |,,. l H“. ,,,,,,,,,, . ‘ —4 —3 ~2 ~1 0 1 2 3 4 x (km) Orbit of the spacecraft in the local frame: (or, 5) plot 75 l‘ ' l g #1 , .. 70 ~ ‘ -~ 9? B 65 fl -‘ “O 60 L VVVVV _ ,_ 55 " l l L ‘ l " ~120 —100 —40 ~20 1/16/09 8:33 PM /home/Benjamin/Teaching/MAE146/Winter09/HW1/Maflabbe1”solutionm 1 012 % Solution to Homework #1, Pb 3: Re : 6378; % radius of the Earth in kilometers. d2r : pi/180.0; % conversion factor from degrees to radians. r2d = 180.0/pi; % conversion factor from radians to degrees. % Local frame of reference t = 0:24; W2 2 i 38089.3021667812 36020.8347389221 35974.2921684086 38949.8526485?85 35944.1418707471 k’ 35955.5720203286 35984.1911064967 360814502861860 36099.1695663230 36188.0321055068 l! 362960949182032 36417.8193996254 365439876887360 36662.6204300364 36760.7290989193 I! 368265107801930 36851.4849849880 36832.0940227161 36770.4163214803 36673.8359666123 I! 36553.7323506826 364234652892646 36296.0949182032 36182.3494789195 36089.3021657012; —69.5006395529 —67.0440787000 —64.2616854033 —61.6727568207 —60.0267881864 I! —59.8975055280 «613398306859 —639298309018 —67.0595164578 —70.1950962574 I! —72.9805683303 —75.2289320697 —76.8745781396 «779286463179 —78.4483032538 !( —78.5171549310 —78.2313691423 —77.6875422555 —769704930049 —76.1408023671 L’ —75.2228005253 —74.1942644494 «729805683303 —71.4605287642 —69.5006395529; 703852835887 726785203598 74.3732260049 753577745079 755974209411 I! 751215951294 739914703283 722804990979 70.797047279 675103634207 647273834646 I! 619111451386 592529454177 569388493944 55.1345028308 539720391672 L’ 535397328742 538749764968 549610616103 567279400303 590566962233 61.7870862164 i! 647273834646 676660239677 703852835887 1; Z1 = W2(1,2).*(cos(W2(2,:)*d2r)f*cos(W2(3,:)*d2r)); Z2 2 W2(1,2).*(sin(W2(2,:)*d2r).‘cos(W2(3,:)*d2r)); Z3 = W2(1,:).*sin(W2(3,:)*d2r); Z = [Z1; Z2; Z3]; figure(1) subplot<311) plot3(Z1,Z2,ZS,’—b’); %Labeling grid on; title(’Orbit of the spacecraft in the local frame: 3D view’); xlabel(’x (km)’); ylabeK’y (mi); zlabel(’z (km)’); axis square; axis equal; axis([0 10000 —22000 —8000 30000 350001); legend(’spacecraft 1’, ’spacecraft 2’); subplot(312) plot(Z1,Z2,’—b’); 38Labeling grid on; titiei‘Orbit of the spacecraft in the local frame: (x;y) projection’); xiabel(’x (kmfl; yiabeli’y (kmrl; axis square; axis equal; subploti313) plot(W2(2,:),W2(3,:),’—b’); %Labeling title(’Orbit of the spacecraft in the local frame: (\alpha, \delta) plot’); xlabel(’\alpha (deg)’); ylabel(’\delta (deg)’); grid on; axis square; axis equal; 0/2) Transformation to inertial frame 1/16/09 8:33 PM /home/Beniamin/Teaching/MAE146/Winter09/HW1/Matlabbe1_solution.m longO : —116.89*d2r; lat0235.35*d2r; altO = 0925; R0 : Re+alt0; for} = 1:25, % Position of the observing station in inertial frame long : longO + 2.0*pi*(t(i)/24); S : R0'*[oos{iatOYcoonng); cosUatorsinUong); sin(lat0)]; % Rotation transfrom the local frame into the inertial one. Rot_long ={ cos(—long) sin(—long) 0.0; —sin(—long) cosHong) 0.0 ; 0.0 0.0 1.0 ]; Rot_lat =[ cos(lat0) 0.0 -—sin(lat0) ; 00 1.0 0.0 ; sin(lat0) 0.0 cos(lat0) ]; Rotuflip:[001;100;010]; Rot : Rot_long * Rot_lat * Rot_flip; % Transformations of the coordinates of 8/0 U2(:,j) : Rot*(Z(:,j)) + S; d2p : norm(U2(1:2,j)); d2f = norm(U2(:,j)); V2(:,j) : [ d2f; r2d*sign(U2(2,j))*aoos(U2(1,j)/d2p); r2d*asin(U2(3,j)/d2f)] end; figure(2) clf; subplot(211) plot3(U2(t,:),U2(2,:),U2(3,:),’—-b’); % Plot the Earth (just for fun!) hold on; [Ex,Ey,Ez]=sphere(100,100); Earth : imread(‘ear0xuu2.jpg’); warp(Re*Ex,Re*Ey,-—Re*Ez,Earth); set(gca,’XDir’,’normal’,‘YDIR’,’normal’,’ZDir’,’normal’); % warp tends to flip the axis... hold off; % Labeling grid on; title(’Orbit of the spacecrafts in the inertial frame: 3D view’); xlabel(’x (km)’); ylabelry (kml’); zlabel(’z (km)’); legend(’spacecraft’); axis equal; subplot(212) V2-mod1 :[~180 V2(2,1:23)]; V2nmod2 : f 0 V2(3,1:23)l; plot(V2#mod1,V2‘mod2,’~b’); 9/5 Labeling grid on; title(’Orbit of the spacecraft in the inertial frame: (\alpha, \delta) plot’); xlabel(’\alpha (deg)’); ylabel(’\delta (deg)’); axis equal; 20f2 Orbit of the spacecrafts in the inertia! frame: SD View spacecraft 1 spacecraft 2 Orbit of the spacecraft in the inertial frame: (a, 6) plot 40... ______ , , , ,,,,,, ,,,,,,,,,,,, _ _ ' I Mdeg) if i! : v20 ”' 740 BW _ , , , i,” , , , Mi ,,,,,, , L i _ ,, , 4150 400 '50 0 50 100 150 1/16/09 8:47 PM /home/Benjamin/Teaching/MAE146/Winter09/HW1IMaHab/Pm.2Mdata1xt 1 of 1 WWW Homework #1, Problem 1, #2 results. First row: Range (km) Second row: Right Asc. (deg) Third row: Declination (deg) Midnight through 7:OOam 42152 42152 42152 42152 42152 42152 42152 42152 ~110.76 ~95.763 ~80.654 —65.45 —50.2 —34.973 —19.834 ~4.8268 19.547 21.62 23.214 24.216 24.554 24.203 23.192 21.595 8:00am through 2:00pm 42175 42153 42153 42153 42153 42153 42153 42153 9.8204 24.796 39.493 54.196 68.969 83.854 98.87 114 20.103 17.126 14.558 11.989 9.5885 7.5186 5.9215 4.9109 3:00pm through 11:00pm 42153 42153 42153 42153 42153 42153 42153 42153 129.21 144.44 159.63 174.73 —170.27 455.36 —140.51 —125.66 4.5611 4.899 5.9008 7.4947 9.5677 11.976 14.558 17.139 [L7 egg/3e, 72;; fifzééém. 53w» $7Mjw ységaf} 54% 2g; :22? fig flgiégf" 5 , ,, §~ ‘5 Q; g3 f ‘ f h . sf: ~33. i ( ? gyfi {5%}; £52; 5% 2 fig? gwméfiajé’?” gig: 5% {25¢ faggfffifif 959665} a}; J m i: 5 {fag xwygééékfiffiw:4:€ 9135:” . E L £43"; fwfigflx M, é’iwa 5“? “.3 r? ““3 Qfi : 5”: ’1; {Egg 5%; 3 fig“ j: £3? @52/‘(755‘1 fifiéfaa/ {If $5, {3/ a» &/®2W jé‘w/éék ”a figfégn 4:: _. W} W} ff 3a‘éégéaig. 2%qu %§;%:Wéi§ flijfé’fiaflfié ,1; gang”, 5:52,, egg; a?" g j} W (1/: fiffi/fiffimg) E ‘ f 4 262: / €55? Eém gang {jg y’f £$fi§£¢ fiat% é? w /{? figfi‘g “@le w ifiafig £3 ' / 5%? ~§ 5 5/ W 532' a “fig? ._ g}; a 5 /; {Egg/g *' gygfié (9/: W M6, Hammadé #( Jacki”- fiafém’j ~ Egghlem i:&; The acceleration A is the inertial derivative of {fit velocity vector V i -3... A—dtV However, since the vehicle is flying in a trimmed conditior velocity vector has a simple n in the unit vectors of the body frame b: V'— representatio Now, the first derivative relation (1.37) applied. to this 1 becomes ' b _d _d ehi A-.———dtV—---—dv+wxv The derivative of V in the body frame is just de/dt = V b1 unit vectors of the b frame are treated as constants when a 3 derivative is calculated. Also, by the right hand rule, the velocity of the b frame with respect to the inertial frame i is since 7 increases ’upwards’, ' b3 angular velocity. The cross product is then 3b1x V = i Vbq % V b2. So, the inertial acceleration of the vehicle can be A s V bl + 1 V b2 We will find this result quite useful in the chapters on re rocket performance. Enables i224 As in the previous problem, we could obtain the acceleration by calculating another derivative of the velocity vector, which was obtained for this system in pr: However, in this caSe we will use the acceleration expansia and calculate the acceleration from the beginning. Reca problem 4 that the angular velocity vector is esi _ ' ' ' w ~ 6 s1 + y(w$+l) sin 5 s2 + (wb+A) cos 5 s3 and that the position vector r is r = ( R + H ) 52 The inertial acceleration is given by the acceleration as i 2 s 2 s d2r= d2r+——a-Ed 538in dt dt esi sd esi esi +2 xdtr + x(w xr) .i. r = H s , 2 r = fi 82 The derivative of the angular velocity vector, the acceleration vector, is s d u ., w = 6 s + . “a? 1 ( A sin 5 + (“5+“) 3 cos a ) 32 + u II. 0 O ( A cos 6 (w$+A) 8 sin a ) 33 bf:£i we are now in a position to perform the cross products needec the acceleration formula. Without going into detail for each, intermediate results are 391 x r = ( ~(R+H)(wb+i) ccsa ) 81 + (R+H)$ s3 351 x (351 x r) (R+H)(we+A) 8 sin 5 s1 2 2 + ( -(R+H)(w$+i)2 cos 6 — (R+H) 5 ) s 2 + ( (R+H)(w$+i)2 cos 5 sin 6 ) 33 d o o o r 2 H (wg+h) cos a 31 + 2 H S 33 s . . d 351 x r ‘e ~(R+H)( 3 cos‘a - (w$+A) 5 Sin 5 ) 31 + (R+H) 8 33 Adding together these intermediate terms, the final in: acceleration vector is idz dt 2r = sl( -(R+H) X cosa + 2(R+H)([email protected]+i)3 sins — 2fi(w$+i)cos6 ) 2 ) + s3( (R+H) S + (R+H)(w$+i)2coss sin6 + 2&5 ) + sz( fi — (R+H)(w$+i)2 00525 - (n+3) é H45 [45- [x/F/a Moo/'e/Jw/ /Q)aafiéxz / ffoéémj OJWK/ an offer. A star, at midnight on a given day, has a dec 'nation and right ascension of ~149.7 deg and 155.2 deg in an Earth centered frame of reference defined by positive x and graxis crossing the equator at 0 deg and 90 deg longitude respectively. 1, From which observatory is it possible to observe this star: Goldstone, loouted at a longitude and latitude of «1169 and 35.3 deg respectively, Canberra, located at a longitude and latitude of 149.1 deg and ~35.3 deg respectively or Madrid, located at a longitude and latitude of -33? dog and 404 deg , respectively? (justify your answer). ‘2. What are the locai right ascension and decimation of the star from the observatory {seieoted in the previous question} View point? 3. Assuming the Earth rotation to be uniform about the North pole (latitude +90deg), with 21 ‘24h period, what would be the local right ascension and declination of the start at 4:00am (midnight + 4 hours) on the some day? ls it possible to observe the star at that time from the chosen observatory? J x4 J/Er 1'; Jana/L 5m an o/wwfa/a 2/ J, j,cad/o/A~f¢ a. {(1. Kid game d flfi74~ 71‘ij Covr Una/z. I)“ al/énha/ év It éf/flVéét/f 0% & ~50?! Vet/o; 0/ fl“ J/M‘ MM ' //:< ,f/«ca/ lat/r} dived/27 at}! m. com/a7. 1,, A («4/ cake/M Adm A #5, 1a, / (bowling/l /Cafo/ coJJ X A C‘” ‘9 Call 0 M . er 1". 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