Dynamics cheat sheet

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$my Latex and Tex4ht cheat sheet$

However, [Φ] −1 = [Φ] T [M] therefore {η} = [Φ] T [M] {x} ⎧ ⎫ ⎡ ⎤ ⎨η 1 (t) ⎬ √µ1 ϕ 11 √µ2 ϕ 12 ⎩ η 2 (t) ⎭ = ⎣ ⎦ √µ1 ϕ 21 √µ2 ϕ 22 ⎤ ⎧ ⎫ ⎣ m 11 m 12 ⎨x ⎦ 1 (t) ⎬ m 21 m ⎩ 22 x 2 (t) ⎭ T ⎡ The next step is to apply this transformation to the original equations of motion in order to decouple them 1.6 Step 6, applying modal transformation to decouple the original equations of motion The EOM in normal coordinates is ⎡ ⎣ m 11 m 12 m 21 ⎤ ⎧ ⎨ ⎦ m ⎩ 22 x ′′ 1 x ′′ 2 ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎬ ⎭ + ⎣ k 11 k 12 ⎨x ⎦ 1 ⎬ ⎨ k ⎩ 22 x ⎭ = f 1 (t) ⎬ ⎩ 2 f 2 (t) ⎭ Applying the above modal transformation {x} = [Φ] {η} on the above results in ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎣ m 11 m 12 ⎨η 1 ⎦ ′′ ⎬ [Φ] m 21 m ⎩ 22 η ′′ ⎭ + ⎣ k 11 k 12 ⎨η ⎦ 1 ⎬ ⎨ [Φ] k 21 k ⎩ 22 η ⎭ = f 1 (t) ⎬ ⎩ 2 f 2 (t) ⎭ pre-multiplying by [Φ] T results in ⎡ ⎤ ⎧ [Φ] T ⎣ m 11 m 12 ⎨ ⎦ [Φ] m ⎩ 22 m 21 η 1 ′′ η 2 ′′ 2 k 21 ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎬ ⎭ + [Φ]T ⎣ k 11 k 12 ⎨η ⎦ 1 ⎬ ⎨ [Φ] k ⎩ 22 η ⎭ = f 1 (t) ⎬ [Φ]T ⎩ 2 f 2 (t) ⎭ ⎡ ⎤ ⎡ ⎤ The result of [Φ] T ⎣ m 11 m 12 ⎦ [Φ] will always be ⎣ 1 0 ⎦. This is because mass normalized shape vectors m 21 m 22 0 1 are used. If the shape functions were not mass normalized, then the diagonal values will not be 1 as shown. ⎡ ⎤ ⎡ ⎤ The result of [Φ] T ⎣ k 11 k 12 ⎦ [Φ] will be ⎣ ω2 1 0 ⎦. k 21 k 22 0 ω2 2 ⎧ ⎫ ⎧ ⎫ ⎨ Let the result of [Φ] T f 1 (t) ⎬ ⎨ ˜f ⎩ f 2 (t) ⎭ be 1 (t) ⎬ ,Therefore, in modal coordinates the original EOM becomes ⎩ ˜f 2 (t) ⎭ ⎡ ⎤ ⎧ ⎣ 1 0 ⎨ ⎦ 0 1 ⎩ η 1 ′′ η 2 ′′ k 21 ⎫ ⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎬ ⎭ + ⎣ ω2 1 0 ⎨η ⎦ 1 ⎬ ⎨ ˜f 0 ω2 2 ⎩ η ⎭ = 1 (t) ⎬ ⎩ 2 ˜f 2 (t) ⎭ The EOM are now decouples and each can be solved as follows η ′′ 1 (t) + ω 2 1η 1 (t) = ˜f 1 (t) η ′′ 2 (t) + ω 2 2η 2 (t) = ˜f 2 (t) To solve these EOM’s, the initial conditions in normal coordinates must be transformed to modal coordinates using the above transformation rules {η (0)} = [Φ] T [M] {x (0)} { η ′ (0) } = [Φ] T [M] { x ′ (0) } 8
Or in full form and ⎧ ⎫ ⎡ ⎨η 1 (0) ⎬ ⎩ η 2 (0) ⎭ = ⎣ ⎧ ⎫ ⎡ ⎨η 1 ′ (0) ⎬ ⎩ η 2 ′ (0) ⎭ = ⎣ ⎤T ⎡ √µ1 ϕ 11 √µ2 ϕ 12 ⎦ √µ1 ϕ 21 √µ2 ϕ 22 ⎤T ⎡ √µ1 ϕ 11 √µ2 ϕ 12 ⎦ √µ1 ϕ 21 √µ2 ϕ 22 ⎣ m 11 m 12 m 21 ⎣ m 11 m 12 m 21 ⎤ ⎧ ⎫ ⎨x ⎦ 1 (0) ⎬ m ⎩ 22 x 2 (0) ⎭ ⎤ ⎧ ⎫ ⎨x ⎦ ′ 1 (0) ⎬ m ⎩ 22 x ′ 2 (0) ⎭ Each of these EOM are solved using any of the standard methods. This will result is solutions η 1 (t) and η 2 (t) 1.7 Step 7. Converting modal solution to normal coordinates solution The solutions found above are in modal coordinates η 1 (t) , η 2 (t). The solution needed is x 1 (t) , x 2 (t). Therefore, the transformation {x} = [Φ] {η} is now applied to convert the solution to normal coordinates ⎧ ⎫ ⎡ ⎤ ⎧ ⎫ ⎨x 1 (t) ⎬ √µ1 ϕ 11 √µ2 ϕ 12 ⎨ ⎩ x 2 (t) ⎭ = η ⎣ ⎦ 1 (t) ⎬ √µ1 ϕ 21 √µ2 ϕ 22 ⎩ η 2 (t) ⎭ ⎧ ⎫ ⎨ √µ1 ϕ 11 η 1 (t) + √ ϕ 12 µ2 η 2 (t) ⎬ = ⎩ √µ1 ϕ 21 η 1 (t) + √ ϕ 22 µ2 η 2 (t) ⎭ Hence and x 1 (t) = ϕ 11 √ µ1 η 1 (t) + ϕ 12 √ µ2 η 2 (t) x 2 (t) = ϕ 21 √ µ1 η 1 (t) + ϕ 22 √ µ2 η 2 (t) Notice that the solution in normal coordinates is a linear combination of the modal solutions. The terms ϕ √ ij µ are just scaling factors that represent the contribution of each modal solution to the final solution. This completes modal analysis 1.8 Numerical solution using modal analysis This is a numerical example that implements the above steps using a numerical values for [K] and [M]. Let k 1 = 1, k 2 = 2, m 1 = 1, m 2 = 3 and let f 1 (t) = 0 and f 2 (t) = sin (5t). Let initial conditions be x 1 (0) = 0, x ′ 1 (0) = 1, x 2 (0) = 1.5, x ′ 2 (0) = 3, hence ⎧ ⎫ ⎫ ⎨x 1 (0) ⎬ 0⎬ ⎧ ⎨ ⎩ x 2 (0) ⎭ = ⎩ 1 ⎭ and ⎧ ⎫ ⎧ ⎫ ⎨x ′ 1 (0) ⎬ ⎨ ⎩ x ′ 2 (0) ⎭ = 1.5⎬ ⎩ 3 ⎭ In normal coordinates, the EOM are 9
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: Similarly, µ 2 is found ⎧ ⎫T
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
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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
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Image from http://www.aerospaceweb.
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