Dynamics cheat sheet

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Since t 1 is very small, it can be assumed that u changes is negligible as well as the change in velocity, hence the above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but with initial velocity u ′ (0) = F 0 /m and zero initial position. Initial conditions are u (0) = 0 and u ′ (0) = 0 then the solution is applying initial conditions gives A = 0 and B = u impulse = e −ξωt (A cos ω d t + B sin ω d t) ( F0 ) m ω d , hence u impulse (t) = e −ξωt ( F0 mω d sin ω d t If the initial conditions were not zero, then ( the solution for these are added ) to the above. From earlier, it was found that the solution is u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω ω d sin ω d t , therefore, the full solution is due to IC only { ( }} ){ { ( }} ){ u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω sin ω d t + e −ξωt F0 sin ω d t ω d mω d ) due to impulse critically damped with impulse input ξ = c c r = 1 with initial conditions u (0) = 0 and u ′ (0) = 0 then the solution is u (t) = (A + Bt) e −( cr 2m ) t = (A + Bt) e −ωt where A, B are found from initial conditions A = u (0) = 0 and B = u ′ (0) + u (0) ω = F 0 m , hence the solution is u impulse (t) = F 0t m e−ωt If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is u (t) = (u 0 (1 + ωt) + u ′ (0) t) e −ωt , therefore, the full solution is u (t) = due to IC only { }} { ( u (0) (1 + ωt) + u ′ (0) t ) e −ωt + due to impulse { }} { F 0 t m e−ωt over-damped with impulse input ξ = c c r > 1 With initial conditions are u (0) = 0 and u ′ (0) = 0 the solution is u impulse (t) = Ae λ 1ωt + Be λ 2ωt where A, B are found from initial conditions and λ 1 = −ωξ + ω √ ξ 2 − 1 λ 2 = −ωξ − ω √ ξ 2 − 1 A = u′ (0) − u (0) λ 2 2ω √ ξ 2 − 1 B = −u′ (0) + u (0) λ 1 2ω √ ξ 2 − 1 Hence the solution is ( −ξ+ √ ( ξ u impulse (t) = Ae 2 −1 )ωt −ξ− √ ) ξ + Be 2 −1 ωt 42
where A = B = F 0 m 2ω √ ξ 2 − 1 − F 0 m 2ω √ ξ 2 − 1 Hence u impulse (t) = F 0 ( m 2ω √ ξ 2 − 1 e −ξ+ √ ) ξ 2 F 0 ( −1 ωt − m 2ω √ ξ 2 − 1 e −ξ− √ ) ξ 2 −1 ωt If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it was found that the solution is u (t) = Ae p 1t + Be p 2t , therefore, the full solution is u (t) = Ae λ 1ωt + Be λ 2ωt + F 0 F 0 m 2ω √ ξ 2 − 1 eλ 1ωt m − 2ω √ ξ 2 − 1 eλ 2ωt 8.4.2 sin impulse input Now assume the input is as follows A = u′ (0) − u (0) λ 2 2ω √ ξ 2 − 1 B = −u′ (0) + u (0) λ 1 2ω √ ξ 2 − 1 given by F (t) = F 0 sin (ϖt) where ϖ = 2π 2t 1 = π t 1 undamped system with sin impulse ⎧ ⎨ F 0 sin (ϖt) 0 ≤ t ≤ t 1 mü + ku = ⎩ 0 t > t 1 with u (0) = u 0 and ˙u (0) = v 0 . For 0 ≤ t ≤ t 1 the solution is ( ) ( ) v0 u (t) = u 0 cos ωt + ω − u r 1 π st 1 − r 2 sin ωt + u st 1 − r 2 sin t t 1 43
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 and 12: Hence The first eigen vector is Sim
Page 13 and 14: The transformation from modal coord
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: 8.4 Response to impulsive loading 8
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.
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Dynamics cheat sheet

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