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HW2_solution - fig(5 Harman/orig 452‘ Z Sadr/7}”...

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Unformatted text preview: fig (5- Harman/orig 452‘ Z Sadr/7}”. gojzém / Wit—=34 veLL ocity vector V i _ d A - ——————dtV However, since the vehicle is flying in a trimmed conditior veLocity vector has a simple representation in the unit vectors of the body frame b: V = Now, the first derivative relation (1.37) applied to this 1 becomes ' b A=~9~v=iv+ifibixv The derivative of V in the body frame is just de/dt = {7 131 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 1 is since 7 increases ’upwards’, and this is the direction of a 3 b3 angular velocity. The cross product is then ablx V '5 Vb1 f} V 132. So, the inertial acceleration of the vehicle can be A=Vb+1Vb We will find this result quite useful in the chapters on re rocket performance. The acceleration A is the inertial derivative 0? HM: Dig Mg“ HOmw/dfg #2 (/QDAt/mq , gogém 2 ‘ BrQbLem l-3. A rotation matrix can be thought of as assembled out of the unit vectors of one frame expressed components in the other frame. So, the matrix which perf rotation from the 3 frame to the 1 frame is RSi - s E s E s l g 2 g 3 The columns of a rotation matrix then behave as the basis veci reference frame: the sum of the squares of any row or column eg, I 81 l = l ), while if any column is multiplied, e1 element, with another column, and then added, the result is 2 8.05. = O, isj ). In particular, then, the rate of change of l J . be found by calculating the inertial derivatives of the vectors. This is most easily done by referring the inertial derivatives back to the 5 frame itself. Since the s fr; vectors are constant in their own frame, Sdsj/dt = 0. Their rate of change is then i s . d _ d +51 951 “HE Sj '- “‘a‘E Sj + (A) X Sj (I) If we interpret the cross product of a vector with a matrix H XSj Inatrix whose columns are the cross products with the columns original matrix, the three derivatives of the 8 unit vectors written as :4 50 l x 50 ; O y es r, z, and ¢, which together form a cylindrical coordinate system. The s frame in Figure 1-21 has an angular velocity 351 = ( w - é ) 82 since the 5 frame rotates not only with the colony ( w ), but also with the ball ( é ). The angular velocity of the 8 frame can change if with time, so there is an angular acceleration vector SdaSI “HE“ = ‘5’82 Once again, the problem is to write the inertial acceleration of the baseball in the 5 unit vectors. The acceleration can be written 1d2 sdz sdasi 2r: 2r+‘af‘xr dt dt esi Sd esi asi +2w >< dtr + w x(w xr) The position vector r from the origin on the rotational axis of the colony is '1 1! rs +ZS where, obviously, r t [r]. The 5 frame velocity and acceleration vectors are then We are now in a position to begin calculating cross products. The first cross product of the centripetal acceleration term is 931 51 $2. 83 ‘ w x r = 0 w — ¢ 0 = ~ r ( w — ¢ ) 53 r z The angular acceleration and Coriolis terms are virtually identical to the above: s s s . l 2 3 1351x(2381xr)= O m-¢ O o o -r(w—¢) - r (w — ¢ ) s1 The baseball is in free fall in a force free environment, so P 0. Newton’s second law then becomes We {45- How/M gr 2.. (some. - The angular velocity can be found by considerin using the right hand rule to find the and then adding together mm 1:3,; g each angular coordinate in turn, direction of a positive rate in that angle, the resulting individual rates. Since the angular velocity i351 is a vector, its individual component parts add like vectors. As the rates due to the rotation of the earth, “’9' shown in the sketch, and the longitude rate it add, and are aligned with the earth's polar unit vector . The axis, while the latitude rate 3 is along the 81 angular velocity vector can then be written 381 = S + ( w$ + i ) x Resolving the unit vector I: along in a hybrid set of basis vectors. ed in the 3 frame as 52 and 33, we have the angular velocity express asi- - - - m ~561+(w6+k)sin682+(m$+7t)coSv533 Now, the position vector of the spacecraft can be written as r== (1263+H)s2 where RG3 is the radius of the earth. The inertial velocity of the vehicle is given by the velocity rule (1.37) as i s (1 ‘EE 1‘ = —SE 1' + 381 X r The a frame derivative of r is simply calculated: sdr/dt = ii 3 , while rha necessarv cross product is 2 81 S2 83 esi - - ¢ x r = 6 (ma-rinsinfi (w$+A)cos5 O Re+ H O =—sl(R®+H)([email protected]+A)cos6 + 83(R$+H)<§ The total inertial velocity is then i d ~a—t-r —sl(R$+H)(m+A)cosa+sH II WWW:W”W ‘ t I]. , I} We {45’ - ;E;leem l-g, 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< :flowever, in this caSe we will use the acceleration expansio and calculate the acceleration from the beginning. Reca problem 4 that the angular velocity vector is esi _ v - . . w ~ 6 51 + (w9+A) Sin 6 $2 + ([email protected]+h) cos 6 33 and that the position vector r is r = ( R + H ) 52 The inertial acceleration is given by the acceleration as i 2 s 2 s d2r= d2r+—B—Ed 381x]? dt dt esi 5(1 Si 5' +20) xdtr +6 x(_(31x r) The 3 frame derivatives of r are given by s d . d .. ——— r = H s , r = H 5 dt 2 dtz 2 The derivative of the angular velocity vector, the acceleration vector, is .3? w = 6 el + ( A sin 6 + (w$+i) 6 cos 6 ) 32 + u — C O ( A cos 6 (w$+A) 6 sin 6 ) s3 PLQér/ /45’, {fiamruveng ~Zficg cgbéafgén v7 Egngénn gf [/Cnr"é(;/K We are now in a position to perform the cross products needec the acceleration formula. Without going into detail for each, inteermediate results are 351 x r = ( -(R+H)(we+i) cosa ) s1 + (R+H)$ 33 asi asi . . w x (w x r) = (R+H)(we+A) 6 sin 5 31 + ( —(R+H)(we+i)2 00526 — (R+H) 82 ) 32 + ( (R+H)(w$+i)2 cos 8 sin 6 ) s3 s 2 351 x gt r = - 2 H (wQ+A) cos 6 51 + 2 H 3 53 8d -}Si " s ‘ ' ~3E w x r ‘= -(R+H)( A cos 6 - (w$+A) 6 sin 8 ) s1 + (R+H) S 33 Adding together these intermediate terms, the final in acceleration vector is idZ dt 21: = sl( -(R+H) X cosa + 2(R+H)(we+i)é sins - 2fi([email protected]+i)cos5 ) + 82( fi - (R+H)(w6+i)2 c0326 - (R+H) 52 ) + s3( (n+3) S + (R+H)(w$+i)zcosa sins + 2&5 ) @5553“? [email protected] g5 gig; 3&5 Egg ”3 g wgid ; f fig: 55;; i it w gaff? a??? fig; 552,??? mémgéézg mééfxéég » a; , g»; f w? a»; a 5 5;? $5 5‘ §§§gg%f%£ gfmgf; fl Egg/jffiéfi % £3 3%%$% amjgf , Mfgffifiwéégf {$93 g; @M& gig? @593: 5;? » jgfaafég fifiag g {b 9 5 a {6%} fig? 2 1%; mi} 2* 3% . .A gaéé’é fifg N fiégggé :5? @ if éééw fig? 7%.? ffiifia 5 g§§§ fl fliggggfgg j m wgég % .- gs; fifgég a?“ éfgéazgéf g a " www % Egg; if 515%; gfigéfi ? $3; "’ 5‘5 Egg wag? W 3 mmgfigsggmg §§5§§v 4% $55 M 5000 2 (km) -5000 5000 v (km) Trajectory in inertial space ’5000 -5000 X(kn0 Trajectory in rotating frame (s-frame) 01 O O O I 01 O O O 2 (km) (53 direction) 5000 3; (km) (52 direction) ~5000 —5000 x (km) (51 direction) 3 {§ , / f i y ?K; ,r I; E 5 552%: 54:55 W 522555555 2553 555%?)55 5535555 53 £343 m Wmfiwwflwywvcm. Altitude as a function of time 800 I I I I I E x .E 600 :1: 400 I l 1 l L 0 2000 4000 6000 8000 10000 12000 time (3) 7i. as a function of time 200 (I) a: 9 g o .E << ~200 I I I I I 0 2000 4000 6000 8000 10000 12000 time (s) 5 as a function of time 50 I I I I I (D 0) 9 O) a) .. ‘C .E 00 _50 I I I I I 2000 4000 6000 8000 10000 12000 time (s) Mf¢ l of 2 XO(4) = dH*cos(delta)*cos(lambda) ~(Re+H)*( ddelta*sin(delta)*cos(lambda) w)*cos(delta)*sin(lambda)); XO(5) = dH*cos(delta)*sin(lambda) —(Re+H)*( ddelta*sin(delta)*sin(lambda) w)*cos(delta)*cos(lambda)); XO(6) = dH*sin(delta) + (Re+H)*ddelta*cos(delta); %% Numerical integration Tspan = [0.0:10.0:3.0*60.0*60.0]; %options=odeset(’RelTol’,1e—3’,’AbsTol’,l.O); [T,Y] = Ode45(@RZBP,Tspan,XO); %% Coordinate transformation to obtain H, delta, lambda [m,n] = size(T); for j = l: m Dt = norm(Y(j,lt2)); Ht(j) = norm(Y(j,l:3)) ~ Re; % H lambdat(j) = r2d*sign(Y(j,2))*acos(Y(j,l)/Dt) ~ w*T(j); % l deltat(j) = r2d*asin(Y(j,3)/(Re+Ht(j))); % end; %% Plot figure(l) 1/25/08 7:27 PM /home/Benjamin/Teaching/MAE146/Wi.../trajectory~plot_modified.m %% Small script to plot a trajectory around the Earth %% Angle conversion constants and other physical constants r2d = l80.0/pi; % radian to degrees d2r : pi/l80.0; % degrees to radians Re = 6371,- % km M = 5.9736e24; % kg G = 6.673e—20; % km“3/kg.s“2 mu = G*M w = 2*pi/(24.0*60.0*60.0); % rad/s %% Initial conditions % In H, delta, lambda system: H = 600; delta = 28.0*d2r; lambda = 10.0*d2r; dH = 0.0; ddelta = 0.0; dlambda = 0.06658*d2r; % Conversion to cartesian coordinates % Position (Earth centered, Cartesian, km) X0(l) = (Re+H)*cos(delta)*cos(lambda); X0(2) = (Re+H)*cos(delta)*sin(lambda); X0(3) = (Re+H)*sin(delta); % Velocity (inertial, km/s) (dlambda +K (dlambda +2 1/25/08 7:27 PM /home/Benjamin/Teaching/MAE146/Wi.../trajectory_plot_modified.m subplot(211) plot3(Y(:,l),Y(:,2),Y(:,3),’r~’); 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 tend to flip the! axis... axis equal; xlabel(’X (km)’); ylabel(’Y (km)’); zlabelC’Z (km)’); title(’Trajectory in inertial space’); hold off; subplot(212) Z = zeros(size(Ht)); plot3(Ht,Z,Z,’r—’); hold on; [Ex,Ey,Ez}=sphere(lOO,lOO); Earth = imread(’ear0xuu2.jpg’); warp(Re*Ex,Re*Ey,“Re*Ez,Earth); set(gca,’XDir’,’normal’,’YDIR’,’normal’,’ZDir’,’normal’); % warp tend to flip thez axis... axis equal; xlabel(’x (km) (sl direction)’); ylabel(’y (km) (s2 direction)’); zlabel(’z (km) (s3 direction)’); title(’Trajectory in rotating frame (s—frame)’); hold off; figure(2) subplot(311) plot (T,Ht, ’—b’ ); xlabel(’time (s)’); ylabel(’H in km’); title(’Altitude as a function of time’); subplot(312) -ot<T,lambdat,’-b’); label(’time (s)’); label(’\lambda in degrees’); title(’\lambda as a function of time’); ‘~<><'U subplot(3l3) plot(T,deltat,"b’); xlabel(’time (s)’); ylabel(’\delta in degrees’); title(’\delta as a function of time’); ...
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