Lecture #4
16.61 Aerospace Dynamics
? Extension to multiple intermediate frames (two)
Copy right 2002 by Jon at h an H ow.
1
Spring 2003 16.61 4–1
Introduction
? We started with one frame (B) rotating vectorω and accelerating
˙
vectorω with
respect to another (I), and obtained the following expression for the
absolute acceleration
¨
vectorr
I
=
¨
vectorr
I
cm
+
¨
vectorρ
B
+2vectorω ×
˙
vectorρ
B
+
˙
vectorω
I
×vectorρ + vectorω × (vectorω ×vectorρ)
of a point located at
vectorr = vectorr
cm
+ vectorρ
? 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
vectorω and
2
˙
vectorω
1
are given with respect to the first
intermediate frame 1.
? The left superscript here simply denotes a label – there are
two vectorω’s to consider in this problem.
? Note that the position of point P with respect to the origin of frame
1isgivenby
P1
vectorr =
2
vectorr +
3
vectorr
and the position of point P with respect to the origin of the inertial
frame is given by
PI
vectorr =
1
vectorr +
2
vectorr +
3
vectorr ≡
1
vectorr +
P1
vectorr
? We want
PI
˙
vectorr
I
and
PI
¨
vectorr
I
Spring 2003 16.61 4–2
? The approach is to find the motion of P with respect to frame 1.
P1
˙
vectorr
1
=
2
˙
vectorr
1
+
3
˙
vectorr
1
≡
P1
vectorv
=
2
˙
vectorr
1
+(
3
˙
vectorr
2
+
2
vectorω ×
3
vectorr )
and
P1
¨
vectorr
1
=
2
¨
vectorr
1
+
3
¨
vectorr
1
≡
P1
vectora
=
2
¨
vectorr
1
+(
3
¨
vectorr
2
+2(
2
vectorω ×
3
˙
vectorr
2
)+
2
˙
vectorω
1
×
3
vectorr +
2
vectorω ×(
2
vectorω ×
3
vectorr ))
? 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
vectorr =
1
vectorr +
2
vectorr +
3
vectorr ≡
1
vectorr +
P1
vectorr
Now compute the desired velocity:
PI
˙
vectorr
I
=
1
˙
vectorr
I
+
P1
˙
vectorr
I
≡
PI
vectorv
=
1
˙
vectorr
I
+
P1
˙
vectorr
1
+(
1
vectorω ×
P1
vectorr )
=
1
˙
vectorr
I
+
bracketleftBigg
2
˙
vectorr
1
+(
3
˙
vectorr
2
+
2
vectorω ×
3
vectorr )
bracketrightBigg
+
1
vectorω ×(
2
vectorr +
3
vectorr )
=
1
˙
vectorr
I
+
2
˙
vectorr
1
+
3
˙
vectorr
2
+
1
vectorω ×
2
vectorr +(
1
vectorω +
2
vectorω)×
3
vectorr
Spring 2003 16.61 4–3
And acceleration:
PI
¨
vectorr
I
=
1
¨
vectorr
I
+
P1
¨
vectorr
I
≡
PI
vectora
=
1
¨
vectorr
I
+
d
I
dt
bracketleftBigg
P1
˙
vectorr
I
bracketrightBigg
=
1
¨
vectorr
I
+
d
I
dt
bracketleftBigg
P1
˙
vectorr
1
+(
1
vectorω ×
P1
vectorr )
bracketrightBigg
=
1
¨
vectorr
I
+
parenleftBigg
P1
¨
vectorr
1
+
1
vectorω ×
P1
˙
vectorr
1
parenrightBigg
+
1
˙
vectorω
I
×
P1
vectorr
+
1
vectorω ×
bracketleftBigg
P1
˙
vectorr
1
+
1
vectorω ×
P1
vectorr
bracketrightBigg
? Now substitute and condense.
PI
¨
vectorr
I
=
1
¨
vectorr
I
+
P1
¨
vectorr
1
+2(
1
vectorω ×
P1
˙
vectorr
1
)+
1
˙
vectorω
I
×
P1
vectorr +
1
vectorω × (
1
vectorω ×
P1
vectorr )
=
1
¨
vectorr
I
+
braceleftBigg
2
¨
vectorr
1
+(
3
¨
vectorr
2
+2(
2
vectorω ×
3
˙
vectorr
2
)+
2
˙
vectorω
1
×
3
vectorr +
2
vectorω ×(
2
vectorω ×
3
vectorr ))
bracerightBigg
+2
parenleftBigg
1
vectorω ×
braceleftBigg
2
˙
vectorr
1
+
3
˙
vectorr
2
+
2
vectorω ×
3
vectorr
bracerightBiggparenrightBigg
+
1
˙
vectorω
I
×
braceleftbigg
2
vectorr +
3
vectorr
bracerightbigg
+
1
vectorω ×(
1
vectorω ×
braceleftbigg
2
vectorr +
3
vectorr
bracerightbigg
)
=
1
¨
vectorr
I
+
2
¨
vectorr
1
+
3
¨
vectorr
2
+ 2([
1
vectorω +
2
vectorω]×
3
˙
vectorr
2
)+2(
1
vectorω ×
2
˙
vectorr
1
)
+
1
˙
vectorω
I
×
2
vectorr +[
1
˙
vectorω
I
+
2
˙
vectorω
1
] ×
3
vectorr
+
2
vectorω ×(
2
vectorω ×
3
vectorr )+
1
vectorω × (
1
vectorω × [
2
vectorr +
3
vectorr ]) + 2
1
vectorω ×(
2
vectorω ×
3
vectorr )
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
1
vectorω×(
2
vectorω×
3
vectorr )which
comes from the 2(
1
vectorω ×
P1
˙
vectorr
1
)term.
–
P1
˙
vectorr
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.