Dynamics cheat sheet

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Therefore the first shape vector is ⎧ ⎫ ⎧ ⎨ϕ 11 ⎬ ⎪⎨ ϕ 1 = ⎩ ϕ ⎭ = 1 ⎪ 21 ⎩ − ( k 11 −ω1 2m ) ( 11 k12 −ω1 2m 12) ⎫ ⎪⎬ ⎪⎭ Similarly the second shape function is obtained. For ω 2 , let ϕ 12 = 1 and solve for ⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎣ k 11 k 12 ⎦ − ω2 2 ⎣ m 11 m 12 ⎨ 1 ⎬ ⎨ ⎦ k 21 k 22 m 21 m ⎩ 22 ϕ ⎭ = 0⎬ ⎩ 22 0 ⎭ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎣ k 11 − ω2 2m 11 k 12 − ω2 2m ⎨ 12 1 ⎬ ⎨ ⎦ k 21 − ω2 2m 21 k 22 − ω2 2m ⎩ 22 ϕ ⎭ = 0⎬ ⎩ 22 0 ⎭ Which gives one equation now to solve for ϕ 22 (the first row equation is only used) ( k11 − ω 2 2m 11 ) + ϕ22 ( k12 − ω 2 2m 12 ) = 0 Hence ϕ 22 = − ( k 11 − ω2 2m ) 11 ( k12 − ω2 2m 12) Therefore the second shape vector is ⎧ ⎫ ⎧ ⎨ϕ 12 ⎬ ⎪⎨ ϕ 2 = ⎩ ϕ ⎭ = 1 ⎪ 22 ⎩ − ( k 11 −ω2 2m ) ( 11 k12 −ω2 2m 12) Now that the two non-normalized shape vectors are found, the next step is to perform mass normalization 1.4 Step 4. Mass normalization of the shape vectors (or the eigenvectors) Let µ 1 = ϕ T 1 [M] ϕ 1 This results in a scalar value µ 1 , which is later used to normalize ϕ 1 . Similarly µ 2 = ϕ T 2 [M] ϕ 2 ⎫ ⎪⎬ ⎪⎭ For example, to find µ 1 ⎧ ⎫T ⎡ ⎤ ⎧ ⎫ ⎨ϕ 11 ⎬ µ 1 = ⎣ m 11 m 12 ⎨ϕ ⎦ 11 ⎬ ⎩ ϕ ⎭ 21 m 21 m ⎩ 22 ϕ ⎭ 21 { } ⎡ ⎤ ⎧ ⎫ = ϕ 11 ϕ ⎣ m 11 m 12 ⎨ϕ ⎦ 11 ⎬ 21 m ⎩ 22 ϕ ⎭ 21 m 21 } = {ϕ ⎧ ⎫ ⎨ϕ 11 ⎬ 11 m 11 + ϕ 21 m 21 ϕ 11 m 12 + ϕ 21 m 22 ⎩ ϕ ⎭ 21 = ϕ 11 (ϕ 11 m 11 + ϕ 21 m 21 ) + ϕ 21 (ϕ 11 m 12 + ϕ 21 m 22 ) 6
Similarly, µ 2 is found ⎧ ⎫T ⎡ ⎤ ⎧ ⎫ ⎨ϕ 12 ⎬ µ 2 = ⎣ m 11 m 12 ⎨ϕ ⎦ 12 ⎬ ⎩ ϕ ⎭ 22 m 21 m ⎩ 22 ϕ ⎭ 22 { } ⎡ ⎤ ⎧ ⎫ = ϕ 12 ϕ ⎣ m 11 m 12 ⎨ϕ ⎦ 12 ⎬ 22 m ⎩ 22 ϕ ⎭ 22 m 21 } = {ϕ ⎧ ⎫ ⎨ϕ 12 ⎬ 12 m 11 + ϕ 22 m 21 ϕ 12 m 12 + ϕ 22 m 22 ⎩ ϕ ⎭ 22 = ϕ 12 (ϕ 12 m 11 + ϕ 22 m 21 ) + ϕ 22 (ϕ 12 m 12 + ϕ 22 m 22 ) Now that µ 1 , µ 2 are obtained, the mass normalized shape vectors are found. They are called Φ 1 , Φ 2 ⎧ ⎫ ⎨ϕ 11 ⎬ ⎧ ⎫ Φ 1 = ϕ ⎩ ϕ ⎭ 1 21 ⎨ √µ1 ϕ 11 ⎬ √ = √ = µ1 µ1 ⎩ √µ1 ϕ 21 ⎭ Similarly Φ 2 = ϕ 2 √ µ2 = ⎧ ⎨ ⎩ ϕ 12 ϕ 22 ⎫ ⎬ ⎭ √ µ2 = ⎧ ⎨ √µ2 ϕ 12 ⎫ ⎬ ⎩ √µ2 ϕ 22 ⎭ 1.5 Step 5, obtain the modal transformation matrix Φ The modal transformation matrix is the 2 × 2 matrix made of of Φ 1 , Φ 2 in each of its columns [Φ] = [Φ 1 Φ 2 ] ⎡ = ⎣ ⎤ √µ1 ϕ 11 √µ2 ϕ 12 ⎦ √µ1 ϕ 21 √µ2 ϕ 22 Now the [Φ] is found, the transformation from the normal coordinates {x} to modal coordinates, which is called {η} is found {x} = [Φ] {η} ⎧ ⎫ ⎡ ⎨x 1 (t) ⎬ ϕ 11 ⎩ x 2 (t) ⎭ = ⎣ ⎤ √µ1 √µ2 ϕ 12 ⎦ √µ1 ϕ 21 √µ2 ϕ 22 ⎧ ⎫ ⎨η 1 (t) ⎬ ⎩ η 2 (t) ⎭ The transformation from modal coordinates back to normal coordinates is {η} = [Φ] −1 {x} ⎧ ⎫ ⎡ ⎨η 1 (t) ⎬ ϕ 11 ⎩ η 2 (t) ⎭ = ⎣ ⎤ √µ1 √µ2 ϕ 12 ⎦ √µ1 ϕ 21 √µ2 ϕ 22 −1 ⎧ ⎫ ⎨ x 1 (t) ⎬ ⎩ x 2 (t) ⎭ 7
Page 1 and 2: Dynamics cheat sheet Nasser M. Abba
Page 3 and 4: 18.7.2 Bi-Elliptic transfer orbit .
Page 5: are the natural frequencies of the
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 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
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14.2 3D Not Using vehicle dynamics
Page 61 and 62:
14.2.2 Derivation for τ = Iω in 3
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15 Wheel spinning precession x Mg
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18 Astrodynamics 18.1 Ellipse main
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magnitude of ⃗r period T mean sat
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18.3 Flight path angle for ellipse
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This diagram below from Orbital mec
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73
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18.7.3 semi-tangential elliptical t
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18.8.1 Rocket equation with plane c
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18.10 interplanetary transfer orbit
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2. Find speed of second planet V 2
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1 mu = 324859; 2 alt = 1475.776; 3
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18.11.3 Two satellite, separate orb
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1 TOF:=hohmann_rendezvous_2(theta=0
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89
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18.14 Orbit changing by low contiuo
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C r below is the core chord of the
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1 C L = ∂C L ∂α α = C Lα α
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1 2 3 C Lwb = ∂C L wb ∂α wb α
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2. stall from http://en.wikipedia.o
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http://en.wikipedia.org/wiki/Flight
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http://www.grc.nasa.gov/WWW/k-12/UE
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http://en.wikipedia.org/wiki/Lift_c
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http://adamone.rchomepage.com/cg_ca
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From http://www.solar-city.net/2010
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Page 184, flight dynamics principle
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Image from http://www.americanflyer
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From FAA pilot’s handbook I do no
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zero-lift angle The angle of attack
Page 119:
Image from http://www.aerospaceweb.
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