0:06

Okay, so

Â at this stage we've kind of wrapped up a lot of the particle kinematics.

Â In particular, how to use rotating frames to get inertial velocities and

Â relative velocities I want to run through an example here to kind of illustrate

Â methodologies that we were discussing in class.

Â So in this particular dynamical system you've got here an origin,

Â inertially fixed, can have a piston expanding.

Â So this point A will moved along the n1 axis.

Â And then I have dual linked robotic system attached to that.

Â 0:36

Both these lengths of the same length.

Â Big R and the angles relative to n1 here is theta, and then the relative

Â displacement from the original length to the second length, that's called phi.

Â All these seems to be varying generally in time and

Â in this problem you've seen some frames attached already.

Â The inertial frame is given through n1 and n2.

Â And at the end point, that last link has a frame associated with it called SR and SV.

Â So let's see,

Â first thing we want to do is actually find the neuronal velocity of this point s.

Â 1:11

So, what we're going to do is we're going to start writing this r ends

Â up being from this point to get to s, I'm going to go from O to A, A to E, E to S.

Â That's kind of the easiest way to get distance times a direction.

Â Distance times a direction.

Â The first distance is just l times n one hat.

Â That gets me from O to A.

Â Then I want to go from A to E.

Â Now I want to go distance r and then you'll run into the first problem already

Â because there's no, this distance isn't defined.

Â So, what we'll want to do is we're actually going to give this

Â axis a direction.

Â I'm just going to call this Hat.

Â So now, I can say I'm going big R in the Hat direction, that gets me to E,

Â and then the final step is, again, the distance R But

Â now this direction already have a vector associated with it.

Â Good, that's already step one of the kinematics.

Â Right your position in the easiest way possible.

Â Now, what we have to do next is just we have an One vector by itself does not make

Â a complete coordinate frame.

Â So, we have to define all the frames.

Â And I'll be very explicit in this example.

Â The inertial frame, we're going to define to the triad n1, n2 and

Â then using the right hand rule, n3 comes out of the page.

Â Right, and then the next frame if we're going from here,

Â call that So without much imagination I'm just

Â going to call it the E frame The first axis is And

Â then the second one I need something orthogonal to And

Â I'm just going to make that one like this and

Â that would be called E goes with this angle theta so

Â we going to call it e theta And then the last one is out of the board.

Â I could call it e3, or in this problem,

Â because e3 is equal to n3, I'm just going to keep the letter n3.

Â 3:11

So good, this defines an E frame.

Â And then the last frame, we've got two of them already defined.

Â I'm just going to write out the rest.

Â So we have Sr as our first vector, S phi as the second unit vector,

Â and then n3 coming out the paper.

Â That completes the right handed coordinate system.

Â Good, so now we finished step two.

Â We completed all the frames for which we introduced some unit direction vectors.

Â The next thing before we differentiate is we have to actually

Â develop all the relative velocities.

Â So we have three frames, so

Â we need three different relative velocities that we might need to use.

Â 3:45

The easy one here starting out with is E relative to N.

Â So that's this frame moving relative to the N frame.

Â And the amplitude will simply be be theta dot, that's how fast you're rotating.

Â And this axis about which you're rotating.

Â In this case, it's just going to be n3, which sticks out of the paper.

Â 4:18

As the angular difference between S and E.

Â And the axis again, in 3, all the rotation axis are in and out of the paper.

Â And in case we did need the angular velocity of s relative to n,

Â that is simply going to be S relative to E.

Â Plus angular velocity of E relative to n.

Â Let me fix that, there we go.

Â All right, which will be nothing but

Â theta dot plus phi dot n3.

Â 4:59

Okay, so, now we have the angular velocities,

Â the last step is then always differentiate.

Â So, in this case, I do have r, and r in this case,

Â and just make it explicit is S relative to O.

Â So it's a written, the point s is taken relative to enough of frame so

Â I can find the inertial derivatives.

Â 5:38

And you can see it if r now has been written using a series of rotating frames,

Â so we can do it component by component, differentiation is the linear operator.

Â So I'm going to start out with the first one, while this already in the end frame.

Â So I just need the end frame derivative of L times and 1 hat.

Â That makes it easy.

Â This next term, though, is given in the E frame.

Â So instead of taking the inertial derivative, I would prefer to take the e

Â frame derivative of Rer.

Â 6:16

Then to get the inertial derivative you have to use the transport theorem and

Â I have to get the angular velocity of e relative to

Â n crossed with the vector itself again, which is Rer.

Â Now that's the second term, the third term, similarly this is in the s frame,

Â so I choose to differentiate this one as seen by the s frame.

Â 6:54

Good.

Â Now that we have this, let's start to break this down.

Â We have the derivative here N1 is a fixed vector as seen by N,

Â so the only thing that varies with time is L.

Â So I get L dot an 1 hat.

Â The derivative of Rer, R is a fixed length of these links.

Â is fixed as seen by the e frame.

Â So this thing just goes to 0.

Â Similarly our sr, the derivative as seen by the s frame will also go to 0.

Â A body fixed vector, as seen by that body, doesn't vary, it's body fixed.

Â So good, we've got that, now the next step is we have to put in these omegas.

Â So let's see, we had up here omega e n is theta dot.

Â 9:28

Theta dot phi dot S phi.

Â Okay, so that's the result we've just arrived.

Â And now, for part B, let's look at finding the inertial acceleration.

Â In this case, r double dot is just going to be another inertial derivative

Â of the velocity vector, we just treat this as another vector.

Â 9:50

And so we repeat as we did earlier.

Â We can do a little bit faster at this point.

Â So we going to see this one here as seen by the N frame and one doesn't vary.

Â All you would have is L double dot.

Â And one differentiating this one will do first and E frame derivative.

Â So, as seen by the E frame only theta varies.

Â That's a time-varying part.

Â R is a fixed length.

Â So I have this cross product,

Â e relative to n was theta dot e3.

Â So if I could take the cross product here,

Â you would have R theta dot squared E3 cross E theta,

Â that's the third vector across the second gives you minus the first.

Â So there will be a minus there.

Â 10:38

And for this term, very much the same thing.

Â I'm simply going to have another derivative as seen by the S frame.

Â R is fixed but these angular rates vary with time.

Â So, you have theta double dot plus phi, double dot s phi hat.

Â And now for the cross product we have omega

Â s relative to n which was theta dot plus phi dot so that repeats again.

Â So you end up with R theta dot phi dot squared.

Â And the third vector crossed the second, again,

Â is going to get you minus the first of that particular coordinate frame.

Â 11:16

That's it.

Â So you can see one method, I did it a little more verbose the second time.

Â Once you get good at this,

Â you can start to speed up a little bit, and take a lot of steps.

Â Do my one but that's it.

Â I've now written the inertia acceleration using a mixture of coordinate frames.

Â It turns out this is an actually very convenient way to write it for

Â many many a problem.

Â 11:36

Now the last part I want to look at is let's look at this problem.

Â Let's say the problem statement says what is

Â the motion of the point a as seen by an observer fixed to s.

Â So that task to this S frame the sink of a person sitting up here looking back and

Â wondering how to step motion of A look like E.

Â And here, this is actually very common thing we have to do when we have to

Â simulate the sensor signals.

Â What is the perceived rate that you'd have as a space craft in a rotating frame

Â approaches another space craft.

Â Also orbiting in its own plane and doing things.

Â So we're going to take a look at this.

Â So we're looking for the motion of A relative to S, as seen by the S frame.

Â What does that mean mathematically?

Â Well, the first step we do is we write the position vector of A relative to S, right?

Â We want We're interested in the motion of A relative to S, so

Â we have to write the position of A relative to S.

Â 12:50

Just start adding up vectors.

Â So, I'm going to write that out, that's the first one.

Â Minus Rsr hat and then minus Rer hat.

Â Great, now that we have the right relative position vector,

Â we have to find the velocity of this vector as seen by the S frame.

Â So that means this this time,

Â I am taking an S frame derivative of this relative position vector.

Â 13:48

So here, I need the s frame.

Â This first term is already an s frame component so

Â I'm just keeping it as an s frame derivative.

Â The second one is at an e frame so I choose to do an e frame derivative,

Â but then I have to add to the omega cross vector itself term again.

Â And the transport theorem I need an S frame derivative,

Â I'm choosing to do an e frame derivative.

Â I need Omega e relative to S.

Â Okay that wasn't too hard, now we start to simplify this first R is a constant it's

Â a constant length link SR hat is a constant as seen by the S frame so

Â that mean this term conveniently just drops out goes to 0 over here.

Â The r frame, the e frame derivative of where r is a constant,

Â is a constant as seen by the e frame, so this term too goes to zero.

Â Is we really begin the right to transport there.

Â because now we directly get down to this part where omega e rotate to S.

Â Here you notice that we have omega S rotate to E.

Â So if you needed omega e relative to s,

Â that is simply the negative of the original vector, right?

Â So I could take advantage of that.

Â So my angular velocity is going to be phi dot times n 3,

Â phi dot times n3,

Â with a minus sign there Sorry minus this, actually.

Â There's a minus here and then the angle velocity is minus phi dot times n3.

Â Can't draw.

Â And then the whole thing crossed with rer.

Â So those two terms actually cancel the minuses minuses,

Â that gives you a plus again.

Â The scalers, I can take up front R phi dot the third vector across the first gives me

Â plus the second, and

Â this is the A frame derivative,

Â 15:52

Okay, so that concludes this example and this will hopefully show you how you

Â can take a more complicated system with lots of different frames.

Â You can use the, you know, this is a lineal operator differentiation, so

Â you can always break up a whole series of things and

Â just do long pull the n stuff together.

Â Long pull the e stuff together.

Â Long pull the s stuff together.

Â And systematically go after it.

Â Also illustrates how we can take different angle velocities,

Â reverse directions as needed.

Â And particularly,

Â if you have to take a different kind of a derivative like an s frame.

Â Sometimes you have to reverse the ES versus SE kind of definition.

Â