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.