美国麻省理工大学：《Aerospace Dynamics（航空动力学）》英文版 lecture 4

Lecture #4 16.61 Aerospace Dynamics Extension to multiple intermediate frames(two) Copyright 2002 by Jonathan How

Lecture #4 16.61 Aerospace Dynamics • Extension to multiple intermediate frames (two) Copyright 2002 by Jonathan How. 1

Spring 2003 16.614-1 Introduction We started with one frame(B)rotating w and accelerating w with respect to another(I), and obtained the following expression for the absolute acceleration B B 7=7m+p+2×p+d×p+d×(×p of a point located at However, in many cases there are often several intermediate frames that have to be taken into account. onsider the situation in the figure Two frames that are moving, rotating, accelerating with respect to each other and the inertial reference frame Assume that 2w and w I are given with respect to the first intermediate frame 1 o The left superscript here simply denotes a label -there are two w's to consider in this problem Note that the position of point p with respect to the origin of frame 1 is given by 7=2+3 nd the position of point P with respect to the origin of the inertial frame is given by P=17+27+37=17+ We want PlT and Plr

Spring 2003 16.61 4–1 Introduction • We started with one frame (B) rotating
ω and accelerating ˙
ω with respect to another (I), and obtained the following expression for the absolute acceleration ¨
r I = ¨
r I cm + ¨ ρ
B + 2
ω × ˙ ρ
B + ˙
ω I × ρ
+
ω × (
ω × ρ
) of a point located at
r =
rcm + ρ
• However, in many cases there are often several intermediate frames that have to be taken into account. • Consider the situation in the figure: – Two frames that are moving, rotating, accelerating with respect to each other and the inertial reference frame. – Assume that 2
ω and 2 ˙
ω 1 are given with respect to the first intermediate frame 1. ✸ The left superscript here simply denotes a label – there are two
ω’s to consider in this problem. • Note that the position of point P with respect to the origin of frame 1 is given by P1
r = 2
r + 3
r and the position of point P with respect to the origin of the inertial frame is given by P I
r = 1
r + 2
r + 3
r ≡ 1
r + P1
r • We want P I ˙
r I and P I¨
r I

Spring 2003 16.614-2 The approach is to find the motion of P with respect to frame 1 P1 1 7+("r+ P1 7+3 27+(372+2(2×3)+21×37+2×(2×) While the notation is a bit laborious, there is nothing new here this is just the same case we have looked at before with one frame moving with respect to another Steps above were done ignoring the motion of frame 1 altogether Now consider what happens when we include the fact that frame 1 is moving with respect to the inertial frame I. Again P1 +-T+ +1y Now compute the desired velocity P1 PI P+(2×PF) I 2 7+2×7)+1×(27+7) 1 2 F+27+2+1×27+(1+2)×7

Spring 2003 16.61 4–2 • The approach is to find the motion of P with respect to frame 1. P1 ˙
r 1 = 2 ˙
r 1 + 3 ˙
r 1 ≡ P1
v = 2 ˙
r 1 + (3 ˙
r 2 + 2
ω × 3
r ) and P1¨
r 1 = 2¨
r 1 + 3¨
r 1 ≡ P1
a = 2¨
r 1 + (3¨
r 2 + 2(2
ω × 3 ˙
r 2 ) + 2 ˙
ω 1 × 3
r + 2
ω × ( 2
ω × 3
r )) • While the notation is a bit laborious, there is nothing new here – this is just the same case we have looked at before with one frame moving with respect to another. – Steps above were done ignoring the motion of frame 1 altogether. • Now consider what happens when we include the fact that frame 1 is moving with respect to the inertial frame I. Again: P I
r = 1
r + 2
r + 3
r ≡ 1
r + P1
r Now compute the desired velocity: P I ˙
r I = 1 ˙
r I + P1 ˙
r I ≡ P I
v = 1 ˙
r I + P1 ˙
r 1 + (1 ω
× P1
r ) = 1 ˙
r I +
2 ˙
r 1 + (3 ˙
r 2 + 2
ω × 3
r ) + 1
ω × ( 2
r + 3
r ) = 1 ˙
r I + 2 ˙
r 1 + 3 ˙
r 2 + 1
ω × 2
r + (1 ω
+ 2
ω) × 3
r

o=5E INERTIAL

Spring 2003 16.614-3 And acceleration Pl I P1:I PI I dt 1 t r+ +1×+1×F . Now substitute and condense Pr“I I F+2( P1)+ 1 1×P+× I r+ {2+(2+2×号2)+2×+2x +2(d×12+3 22×°T {27+37}+5×(×{27+57} +272+32+2(+2×3号2)+2(1d “T+ 2×(2×37)+1×(×[+37)+21×(2×37

Spring 2003 16.61 4–3 And acceleration: P I¨
r I = 1¨
r I + P1¨
r I ≡ P I
a = 1¨
r I + dI dt
P1 ˙
r I = 1¨
r I + dI dt
P1 ˙
r 1 + (1
ω × P1
r ) = 1¨
r I + P1¨
r 1 + 1
ω × P1 ˙
r 1  + 1 ˙
ω I × P1
r +1
ω ×
P1 ˙
r 1 + 1
ω × P1
r • Now substitute and condense. P I¨
r I = 1¨
r I + P1¨
r 1 + 2(1
ω × P1 ˙
r 1 ) + 1 ˙
ω I × P1
r + 1
ω × ( 1
ω × P1
r ) = 1¨
r I  + 2¨
r 1 + (3¨
r 2 + 2(2
ω × 3 ˙
r 2 ) + 2 ˙
ω 1 × 3
r + 2
ω × ( 2
ω × 3
r )) +2 1 ω
×  2 ˙
r 1 + 3 ˙
r 2 + 2
ω × 3
r  +1 ˙
ω I ×  2
r + 3
r  + 1 ω
× ( 1
ω ×  2
r + 3
r  ) = 1¨
r I + 2¨
r 1 + 3¨
r 2 + 2([1 ω
+ 2
ω] × 3 ˙
r 2 ) + 2(1 ω
× 2 ˙
r 1 ) +1 ˙
ω I × 2
r + [1 ˙
ω I + 2 ˙
ω 1 ] × 3
r +2
ω × ( 2
ω × 3
r ) + 1
ω × ( 1
ω × [ 2
r + 3
r ]) + 21
ω × ( 2
ω × 3
r )

Spring 2003 16.614-4 This final expression looks messy, but it is really just two nested versions of what we have seen before The nesting can continue to more levels and can be automated Note that this type of expression is very easy to get wrong if not done carefully and systematically The hardest term to capture here is the 2wx(2wx3r)which comes from the2(1d×P11 )term Plr is the relative velocity of P with respect to the origin of frame 1 as seen by an observer in the rotating frame Example: acceleration of the tip of the tail rotor on a helicopter The helicopter body is rotating The tail rotor is rotating as well, but the base is attached to the odV

Spring 2003 16.61 4–4 • This final expression looks messy, but it is really just two nested versions of what we have seen before. – The nesting can continue to more levels and can be automated. • Note that this type of expression is very easy to get wrong if not done carefully and systematically. – The hardest term to capture here is the 21
ω × (2
ω × 3
r ) which comes from the 2(1
ω × P1 ˙
r 1 ) term. – P1 ˙
r 1 is the relative velocity of P with respect to the origin of frame 1 as seen by an observer in the rotating frame 1. • Example: acceleration of the tip of the tail rotor on a helicopter: – The helicopter body is rotating. – The tail rotor is rotating as well, but the base is attached to the body