Lecture #4
16.61 Aerospace Dynamics
• Extension to multiple intermediate frames (two)
Copyright 2002 by Jonathan How.
1
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
I
I
r¨ = r¨cm + ρ¨
B
+ 2ω × ρ˙
B
I
+ ω˙ × 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 2ω and 2ω˙ 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 = 2r + 3r
and the position of point P with respect to the origin of the inertial
frame is given by
PI
• We want
PI ˙ I
r
and
r = 1r + 2r + 3r ≡ 1r + P 1r
PI¨ I
r
Spring 2003
16.61 4–2
• The approach is to find the motion of P with respect to frame 1.
P1 ˙ 1
r
1
1
= 2r˙ + 3r˙ ≡ P 1v
1
2
= 2r˙ + ( 3r˙ + 2ω × 3r )
and
P 1¨ 1
r
=
=
2¨ 1
1
r + 3r¨ ≡ P 1a
2
2
2¨ 1
r + ( 3¨r + 2( 2ω × 3r˙ ) + 2ω˙ 1 × 3r + 2ω × ( 2ω × 3r ))
• 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:
PI
r = 1r + 2r + 3r ≡ 1r + P 1r
Now compute the desired velocity:
PI ˙ I
r
I
I
= 1r˙ + P 1r˙ ≡ P I v
I
1
× P 1r )
= 1r˙ + P 1r˙ + ( 1ω
1˙ I
2˙ 1
3˙ 2
= r + r + ( r + ω × r ) + 1ω × ( 2r + 3r )
2
3
I
1
2
+ 2ω ) × 3r
= 1r˙ + 2r˙ + 3r˙ + 1ω × 2r + ( 1ω
Spring 2003
16.61 4–3
And acceleration:
PI¨ I
r
I
I
= 1r¨ + P 1r¨ ≡ P Ia
dI P 1 ˙ I 1¨ I
= r +
r
dt
dI P 1 ˙ 1
1¨ I
1
P1
= r +
r + ( ω × r )
dt
I
1¨ I
P 1¨ 1
1
P1 ˙ 1
r + ω × r + 1ω˙ × P 1r
= r +
+ ω ×
1
P1 ˙ 1
r + ω ×
1
P1
r
• Now substitute and condense.
PI¨ I
r
I
1
1
I
= 1¨r + P 1¨r + 2( 1ω × P 1r˙ ) + 1ω˙ × P 1r + 1ω × ( 1ω × P 1r )
I
= 1¨r +
2¨ 1
3¨ 2
2
3˙ 2
2˙ 1
3
2
2
3
r + ( r + 2( ω × r ) + ω × r + ω × ( ω × r ))
+2
1
ω
×
I
+ ω˙ ×
1
2˙ 1
3˙ 2
r + r + ω × r
2
3
r + r + ω
× ( ω ×
2
3
1
1
2
3
r + r )
I
1
2
2
1
= 1¨r + 2r¨ + 3¨r + 2([ 1ω
+ 2ω ] × 3r˙ ) + 2( 1ω
× 2r˙ )
I
I
1
+ 1ω˙ × 2r + [ 1ω˙ + 2ω˙ ] × 3r
+ 2ω × ( 2ω × 3r ) + 1ω × ( 1ω × [ 2r + 3r ]) + 2 1ω × ( 2ω × 3r )
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 2 1ω × ( 2ω × 3r ) which
1
comes from the 2( 1ω × P 1r˙ ) term.
–
P1 ˙ 1
r 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.