16.07 Dynamics Fall 2004 Lecture D11-Relative Motion using Translating Axes In the previous lectures we have described particle motion as it would be seen by an observer standing still at a fixed origin. This type of motion is called absolute motion. In many situations of practical interest ve find ourselves forced to describe the motion of bodies while we are simultaneously moving with respect to a more basic reference. There are many examples were such situations occur. The absolute motion of a passenger inside an aircraft is best described if we first consider the motion of the passenger relative to the aircraft, and then the motion of the aircraft relative to the ground. If we try to track the motion of aircraft in the airspace using satellites, it makes sense to first consider the motion of the aircraft relative to the satellite and then combine this motion with the motion of the satellite relative to the earth's surface. In this lecture we will introduce the ideas of relative motion analysis Types of For the purpose of studying relative motion, we will consider three different types of observers(or reference frames)depending on their motion with respect to a fixed frame observers who do not accelerate or rotate, i. e. those who at most have constant velocity. servers who accelerate but do not rotat servers who accelerate and In this lecture we will consider the relative motion involving observers of the first two types, and defer for he next lecture the study of relative motion involving rotating frames Relative motion using translating axes Consider two particles, A and B, which may have different curvilinear motions. We consider a fixed reference frame ryz with origin O and with unit vectors i, j and k, as before, and call the motion relative to this frame absolute. In addition, we consider another translating reference frame attached to particle B, r'y'z with unit vectors i', j and k. Translating means that the angles between the axes yz and r'y'z' do not change during the motion
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D11 - Relative Motion using Translating Axes In the previous lectures we have described particle motion as it would be seen by an observer standing still at a fixed origin. This type of motion is called absolute motion. In many situations of practical interest, we find ourselves forced to describe the motion of bodies while we are simultaneously moving with respect to a more basic reference. There are many examples were such situations occur. The absolute motion of a passenger inside an aircraft is best described if we first consider the motion of the passenger relative to the aircraft, and then the motion of the aircraft relative to the ground. If we try to track the motion of aircraft in the airspace using satellites, it makes sense to first consider the motion of the aircraft relative to the satellite and then combine this motion with the motion of the satellite relative to the earth’s surface. In this lecture we will introduce the ideas of relative motion analysis. Types of observers For the purpose of studying relative motion, we will consider three different types of observers (or reference frames) depending on their motion with respect to a fixed frame: • observers who do not accelerate or rotate, i.e. those who at most have constant velocity. • observers who accelerate but do not rotate • observers who accelerate and rotate In this lecture we will consider the relative motion involving observers of the first two types, and defer for the next lecture the study of relative motion involving rotating frames. Relative motion using translating axes Consider two particles, A and B, which may have different curvilinear motions. We consider a fixed reference frame xyz with origin O and with unit vectors i, j and k, as before, and call the motion relative to this frame absolute. In addition, we consider another translating reference frame attached to particle B, x ′y ′ z ′ , with unit vectors i ′ , j ′ and k ′ . Translating means that the angles between the axes xyz and x ′y ′ z ′ do not change during the motion. 1
A/B In the figure, we have chosen, for convenience, the axes rye to be parallel to the axes r'y'z', but it should be clear that one could have non-parallel translating axes The position vector rA/B defines the position of A with respect to point B in the reference frame x'y'z' The subscript notation"A/B"means"A relative to B". The positions of A and B relative to the absolute frame are given by the vectors TA and rB, respectively. Thus, we have If we derive this expression with respect to time, we obtain TA=TB+TA/B which relates the absolute velocities uA and ub to the relative velocity of A as observed by B. Differentiating again, we obtain an analogous expression for the accelerations TA=rB+r LA/B· If we reverse the roles of A and B and attach the reference frame z'y'z' to a, then we can observe b from A B/A The same arguments as before will give us rB=TA+rB/A
In the figure, we have chosen, for convenience, the axes xyz to be parallel to the axes x ′y ′ z ′ , but it should be clear that one could have non-parallel translating axes. The position vector rA/B defines the position of A with respect to point B in the reference frame x’y’z’. The subscript notation“A/B” means “A relative to B”. The positions of A and B relative to the absolute frame are given by the vectors rA and rB, respectively. Thus, we have rA = rB + rA/B . If we derive this expression with respect to time, we obtain r˙ A = r˙ B + r˙ A/B or vA = vB + vA/B , which relates the absolute velocities vA and vB to the relative velocity of A as observed by B. Differentiating again, we obtain an analogous expression for the accelerations, r¨A = r¨B + r¨A/B or aA = aB + aA/B . If we reverse the roles of A and B and attach the reference frame x ′y ′ z ′ to A, then we can observe B from A. The same arguments as before will give us, rB = rA + rB/A, vB = vA + vB/A, aB = aA + aB/A . 2
Comparing these expressions with those above, we see that T aB/A=aA/B One important observation is that, whenever the moving system, say A, has a constant velocity relative to the fixed system, O, then the acceleration seen by the two observers is the same, i. e if aA=0, then aB=aB/A. We shall see that this broadens the application of Newton's second law to systems which have a constant absolute velocity Example Glider in cross wind Consider a glider flying at the edge of a cloud. The glider is flying horizontally, and the cloud, which is below the glider, is stationary with respect to the ground. At the altitude of the glider fight, there is a wind locity, vu, of magnitude 52 knots, as shown in the sketch 210° 240°270 30° 3600 North Edge of Cloud The glider, on the other hand, is flying at a speed UG/w relative to the local wind. In order to determine the speed of the glider with respect to the ground, we consider a reference frame moving with the wind spee and write G=Uw t ug/u or graphically G By looking at the scaled diagram below, we can verify the following situations. If the glider flies with a heading of 330 and a speed (relative to the wind)of 90 knots, it will stay at the edge of the cloud with a
Comparing these expressions with those above, we see that rB/A = −rA/B, vB/A = −vA/B, aB/A = −aA/B , as expected. One important observation is that, whenever the moving system, say A, has a constant velocity relative to the fixed system, O, then the acceleration seen by the two observers is the same, i.e., if aA = 0, then aB = aB/A. We shall see that this broadens the application of Newton’s second law to systems which have a constant absolute velocity. Example Glider in cross wind Consider a glider flying at the edge of a cloud. The glider is flying horizontally, and the cloud, which is below the glider, is stationary with respect to the ground. At the altitude of the glider flight, there is a wind velocity, vw, of magnitude 52 knots, as shown in the sketch. North Edge of Cloud Cloud The glider, on the other hand, is flying at a speed vG/w relative to the local wind. In order to determine the speed of the glider with respect to the ground, we consider a reference frame moving with the wind speed and write vG = vw + wG/w, or graphically, By looking at the scaled diagram below, we can verify the following situations. If the glider flies with a heading of 330o and a speed (relative to the wind) of 90 knots, it will stay at the edge of the cloud with a 3
ground speed of 104=(26+ 78)knots north. If, on the other hand, the glider fies with a heading of 210 at 90 knots, the glider will stay at the edge of the cloud with a ground speed of 52 knots south. If the glider Hies into the wind at 240, with a speed equal to 52 knots, it will remain stationary with respect to the ground. Finally, the lowest speed at which the glider can fly to stay at the edge of the cloud is 45 knots and his will be possible for a heading of 270. The ground speed in this case will be 24 knots 210 240270° 330 45 90 Example Aircraft towing glider(Merriam) Airplane A is flying horizontally with a constant speed of 200km/h, and is towing a glider B. The glider is gaining altitude. The tow cable has a length r= 60m, and 0 is increasing at a constant rate of 5 degrees per econd. We want to determine the magnitude of the velocity, UG, and the acceleration, aG, of the glider for G The velocity of the glider will be UG=UA+UG/A. The velocity of the glider relative to A is best expressed using a polar coordinate system. Thus, we write UG/A=rer+reee Since the length of the cable is constant, vr=F=0, and 0= 5(2T/360)=0.0874 rad /sec, which gives vg=r0=5.236 m/s. We can now transform the velocity vector to cartesian coordinates to obtain v/A=5.236=5.236sin15°i+5.236cs15°j=1.3552i+5.0575jm/s v=((200/3.6)+1.3552)i+5.0575jm/s,vg=57.135m/s=205.7km/h For the acceleration aG =aG/A, since aA=0. We also know that 0=0, and, obviously, r=0, thus, -457enm/2,ac/s=4.57m/2
ground speed of 104 = (26 + 78) knots north. If, on the other hand, the glider flies with a heading of 210o at 90 knots, the glider will stay at the edge of the cloud with a ground speed of 52 knots south. If the glider flies into the wind at 240o , with a speed equal to 52 knots, it will remain stationary with respect to the ground. Finally, the lowest speed at which the glider can fly to stay at the edge of the cloud is 45 knots and this will be possible for a heading of 270o . The ground speed in this case will be 24 knots. Example Aircraft towing glider (Merriam) Airplane A is flying horizontally with a constant speed of 200km/h, and is towing a glider B. The glider is gaining altitude. The tow cable has a length r = 60m, and θ is increasing at a constant rate of 5 degrees per second. We want to determine the magnitude of the velocity, vG, and the acceleration, aG, of the glider for the instant when θ = 15o . The velocity of the glider will be vG = vA + vG/A. The velocity of the glider relative to A is best expressed using a polar coordinate system. Thus, we write, vG/A = ˙rer + r ˙θeθ . Since the length of the cable is constant, vr = ˙r = 0, and ˙θ = 5(2π/360) = 0.0874 rad/sec, which gives vθ = r ˙θ = 5.236 m/s. We can now transform the velocity vector to cartesian coordinates to obtain, vG/A = 5.236eθ = 5.236 sin 15o i + 5.236 cos 15o j = 1.3552i + 5.0575j m/s , or, vG = ((200/3.6) + 1.3552)i + 5.0575j m/s, vG = 57.135 m/s = 205.7 km/h . For the acceleration aG = aG/A, since aA = 0. We also know that ¨θ = 0, and, obviously, ¨r = 0, thus, aG/A = −r ˙θ 2 er = −4.57er m/s2 , aG/S = 4.57 m/s2 . 4
Example Aircraft collision avoidance Consider two aircrafts, A and B, flying horizontally at the same level with constant velocities u a and uB Aircraft B is capable of determining by measurement the relative position and velocity of A relative to B 1. e, TA/B and U A/B. We want to know whether it is possible to determine from this measurement whether the two aircraft are on a collision course TA Take the origin O at the hypothetical point of collision where the two trajectories intersect. If collision is to occur. the time taken for both aircraft to reach O will be the same. Therefore Now consider the velocity triangle formed by UA=UB+UA/B B A/B Clearly, UA is parallel to rA, and uB is parallel to rB. In addition, if collision is to occur, UA/B will be parallel to TA/B, since the position and velocity triangles will, in this case, be similar. We see, therefore that the condition for the two planes to collide is that the relative velocity between them is parallel to the relative position vector. If aircraft B uses a polar coordinate system(common for radar measurements) then the relative position of A with respect to B will be of the form TA/B=rer. Therefore, the velocity so has to be of that form, which implies that ve= re=0, or 6= constant. That is, the angle under which B sees A (sometimes called the bearing angle) should not change
Example Aircraft collision avoidance Consider two aircrafts, A and B, flying horizontally at the same level with constant velocities vA and vB. Aircraft B is capable of determining by measurement the relative position and velocity of A relative to B, i.e., rA/B and vA/B. We want to know whether it is possible to determine from this measurement whether the two aircraft are on a collision course. Take the origin O at the hypothetical point of collision where the two trajectories intersect. If collision is to occur, the time taken for both aircraft to reach O will be the same. Therefore, rA vA = rB vB . Now consider the velocity triangle formed by vA = vB + vA/B. Clearly, vA is parallel to rA, and vB is parallel to rB. In addition, if collision is to occur, vA/B will be parallel to rA/B, since the position and velocity triangles will, in this case, be similar. We see, therefore, that the condition for the two planes to collide is that the relative velocity between them is parallel to the relative position vector. If aircraft B uses a polar coordinate system (common for radar measurements), then the relative position of A with respect to B will be of the form rA/B = rer. Therefore, the velocity also has to be of that form, which implies that vθ = r ˙θ = 0, or θ = constant. That is, the angle under which B sees A (sometimes called the bearing angle) should not change. 5