Dynamics cheat sheet

More documents

Recommendations

Info

$my Latex and Tex4ht cheat sheet$

Now that µ 1 , µ 2 are found, the mass normalized eigen vectors are found. They are called Φ 1 , Φ 2 ⎧ ⎧ ⎫ ⎨ ⎨ 1 ⎬ Φ 1 = ϕ 1 √ µ1 = ⎩ ϕ 11 ⎧ ⎫ ⎨ 0.925 ⎬ = ⎩ −0.219 ⎭ ϕ 21 ⎫ ⎬ ⎭ √ µ1 = ⎩ −0.237 ⎭ √ 1.169 Similarly Φ 2 = ϕ 2 √ µ2 = ⎧ ⎨ ⎩ ⎧ ⎫ ⎨0.380⎬ = ⎩ 0.534 ⎭ ϕ 12 ϕ 22 ⎫ ⎬ ⎭ √ µ2 = ⎧ ⎨ 1 ⎫ ⎬ ⎩ 1.404 ⎭ √ 6.914 Therefore, the modal transformation matrix is [Φ] = [Φ 1 Φ 2 ] ⎡ ⎤ 0.925 = ⎣ 0.380 ⎦ −0.219 0.534 This result can be verified using Matlab’s eig function as follows EDU>> K=[3 -2;-2 2]; M=[1 0;0 3]; EDU>> [phi,lam]=eig(K,M) phi = -0.3803 -0.9249 -0.5340 0.2196 EDU>> diag(sqrt(lam)) 0.4380 1.8641 Matlab result agrees with the result obtained above. The sign difference is not important. Now step 5 is applied. Matlab generates mass normalized eigenvectors by default. Now that [Φ] is found, the transformation from the normal coordinates {x} to modal coordinates, called {η} , is obtained {x} = [Φ] {η} ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎨x 1 (t) ⎬ ⎩ x 2 (t) ⎭ = 0.925 0.380 ⎨η ⎣ ⎦ 1 (t) ⎬ −0.219 0.534 ⎩ η 2 (t) ⎭ 12
The transformation from modal coordinates back to normal coordinates is However, [Φ] −1 = [Φ] T [M] therefore {η} = [Φ] −1 {x} ⎧ ⎫ ⎡ ⎤−1 ⎧ ⎫ ⎨η 1 (t) ⎬ ⎩ η 2 (t) ⎭ = 0.925 0.380 ⎨ x ⎣ ⎦ 1 (t) ⎬ −0.219 0.534 ⎩ x 2 (t) ⎭ {η} = [Φ] T [M] {x} ⎧ ⎫ ⎡ ⎤T ⎡ ⎤ ⎧ ⎫ ⎨η 1 (t) ⎬ ⎩ η 2 (t) ⎭ = 0.925 0.380 ⎣ ⎦ ⎣ 1 0 ⎨x ⎦ 1 (t) ⎬ −0.219 0.534 0 3 ⎩ x 2 (t) ⎭ ⎡ ⎤ ⎧ ⎫ 0.925 −0.657 ⎨x = ⎣ ⎦ 1 (t) ⎬ 0.38 1.6 ⎩ x 2 (t) ⎭ The next step is to apply this transformation to the original equations of motion in order to decouple them. Applying step 6 results in ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎣ 1 0 ⎨η 1 ⎦ ′′ ⎬ 0 1 ⎩ η 2 ′′ ⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨ 0 ω2 2 ⎩ η ⎭ = 0 ⎬ [Φ]T ⎩ 2 sin (5t) ⎭ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤T ⎧ ⎫ ⎣ 1 0 ⎨η 1 ⎦ ′′ ⎬ 0 1 ⎩ η 2 ′′ ⎭ + ⎣ 1.8642 0 ⎨η ⎦ 1 ⎬ 0 0.438 2 ⎩ η ⎭ = 0.925 0.380 ⎨ 0 ⎬ ⎣ ⎦ 2 −0.219 0.534 ⎩ sin (5t) ⎭ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎣ 1 0 ⎨η 1 ⎦ ′′ ⎬ 0 1 ⎩ η ′′ ⎭ + 3. 47 0 ⎨η ⎣ ⎦ 1 ⎬ ⎨ 0 0.192 ⎩ η ⎭ = −0.219 sin (5t) ⎬ ⎩ 2 0.534 sin (5t) ⎭ 2 The EOM are now decoupled and each EOM can be solved easily as follows η ′′ 1 (t) + 3.47η 1 (t) = −0.219 sin (5t) η ′′ 2 (t) + 0.192η 2 (t) = 0.534 sin (5t) To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎨η 1 (0) ⎬ ⎩ η 2 (0) ⎭ = 0.925 −0.657 ⎨x ⎣ ⎦ 1 (0) ⎬ 0.38 1.6 ⎩ x 2 (0) ⎭ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎨η 1 (0) ⎬ ⎩ η 2 (0) ⎭ = 0.925 −0.657 ⎨0⎬ ⎣ ⎦ 0.38 1.6 ⎩ 1 ⎭ ⎧ ⎫ ⎨−0.657⎬ = ⎩ 1.6 ⎭ 13
Page 1 and 2: Dynamics cheat sheet Nasser M. Abba
Page 3 and 4: 18.7.2 Bi-Elliptic transfer orbit .
Page 5 and 6: are the natural frequencies of the
Page 7 and 8: Similarly, µ 2 is found ⎧ ⎫T
Page 9 and 10: Or in full form and ⎧ ⎫ ⎡ ⎨
Page 11: Hence The first eigen vector is Sim
Page 15 and 16: Where F n = 2 T F 0 = 2 T ∫ T 0
Page 17 and 18: Let z = Re { Ze iωt} , z ′ = Re
Page 19 and 20: Now, u = Re { Ue iωt} , u ′ = Re
Page 21 and 22: Hence at t = 0 the phase of the res
Page 23 and 24: The following table gives the solut
Page 25 and 26: 5.3 no loading (free) mu ′′ + c
Page 27 and 28: 5.5 Impulse sin function Input is g
Page 29 and 30: Solution to single degree of freedo
Page 31 and 32: 7 Some common formulas These defini
Page 33 and 34: 8 Derivation of the solutions of th
Page 35 and 36: 8.2.2 under-damped ζ < 1 The gener
Page 37 and 38: When ϖ ≈ ω but less than ω, le
Page 39 and 40: Hence the full solution is u p = p
Page 41 and 42: 8.4 Response to impulsive loading 8
Page 43 and 44: where A = B = F 0 m 2ω √ ξ 2
Page 45 and 46: where ∆ is very small positive qu
Page 47 and 48: at t = t 1 ˙u (t 1 ) = −ξωe
Page 49 and 50: 8.5 references 1. Vibration analysi
Page 51 and 52: 11 Formulas sin 2 x 1cos2x 2 cos 2
Page 53 and 54: 12 Velocity and acceleration diagra
Page 55 and 56: 12.3 spring pendulum with block mov
Page 57 and 58: 13 Velocity and acceleration of rig
Page 59 and 60: 14.2 3D Not Using vehicle dynamics
Page 61 and 62: 14.2.2 Derivation for τ = Iω in 3
Page 63 and 64:
15 Wheel spinning precession x Mg
Page 65 and 66:
18 Astrodynamics 18.1 Ellipse main
Page 67 and 68:
magnitude of ⃗r period T mean sat
Page 69 and 70:
18.3 Flight path angle for ellipse
Page 71 and 72:
This diagram below from Orbital mec
Page 73 and 74:
73
Page 75 and 76:
18.7.3 semi-tangential elliptical t
Page 77 and 78:
18.8.1 Rocket equation with plane c
Page 79 and 80:
18.10 interplanetary transfer orbit
Page 81 and 82:
2. Find speed of second planet V 2
Page 83 and 84:
1 mu = 324859; 2 alt = 1475.776; 3
Page 85 and 86:
18.11.3 Two satellite, separate orb
Page 87 and 88:
1 TOF:=hohmann_rendezvous_2(theta=0
Page 89 and 90:
89
Page 91 and 92:
18.14 Orbit changing by low contiuo
Page 93 and 94:
C r below is the core chord of the
Page 95 and 96:
1 C L = ∂C L ∂α α = C Lα α
Page 97 and 98:
1 2 3 C Lwb = ∂C L wb ∂α wb α
Page 99 and 100:
2. stall from http://en.wikipedia.o
Page 101 and 102:
http://en.wikipedia.org/wiki/Flight
Page 103 and 104:
http://www.grc.nasa.gov/WWW/k-12/UE
Page 105 and 106:
http://en.wikipedia.org/wiki/Lift_c
Page 107 and 108:
http://adamone.rchomepage.com/cg_ca
Page 109 and 110:
From http://www.solar-city.net/2010
Page 111 and 112:
Page 184, flight dynamics principle
Page 113 and 114:
Image from http://www.americanflyer
Page 115 and 116:
From FAA pilot’s handbook I do no
Page 117 and 118:
zero-lift angle The angle of attack
Page 119:
Image from http://www.aerospaceweb.
show all

Dynamics cheat sheet

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?