0:06

Okay, so

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

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

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

methodologies that we were discussing in class.

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

inertially fixed, can have a piston expanding.

So this point A will moved along the n1 axis.

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

0:36

Both these lengths of the same length.

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

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

All these seems to be varying generally in time and

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

The inertial frame is given through n1 and n2.

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

So let's see,

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

1:11

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

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

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

Distance times a direction.

The first distance is just l times n one hat.

That gets me from O to A.

Then I want to go from A to E.

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

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

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

axis a direction.

I'm just going to call this Hat.

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

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

now this direction already have a vector associated with it.

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

Right your position in the easiest way possible.

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

a complete coordinate frame.

So, we have to define all the frames.

And I'll be very explicit in this example.

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

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

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

call that So without much imagination I'm just

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

then the second one I need something orthogonal to And

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

that would be called E goes with this angle theta so

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

I could call it e3, or in this problem,

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

3:11

So good, this defines an E frame.

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

I'm just going to write out the rest.

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

and then n3 coming out the paper.

That completes the right handed coordinate system.

Good, so now we finished step two.

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

The next thing before we differentiate is we have to actually

develop all the relative velocities.

So we have three frames, so

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

3:45

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

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

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

And this axis about which you're rotating.

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

4:18

As the angular difference between S and E.

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

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

that is simply going to be S relative to E.

Plus angular velocity of E relative to n.

Let me fix that, there we go.

All right, which will be nothing but

theta dot plus phi dot n3.

4:59

Okay, so, now we have the angular velocities,

the last step is then always differentiate.

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

and just make it explicit is S relative to O.

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

I can find the inertial derivatives.

5:38

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

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

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

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

That makes it easy.

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

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

frame derivative of Rer.

6:16

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

I have to get the angular velocity of e relative to

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

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

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

6:54

Good.

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

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

so the only thing that varies with time is L.

So I get L dot an 1 hat.

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

is fixed as seen by the e frame.

So this thing just goes to 0.

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

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

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

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

9:28

Theta dot phi dot S phi.

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

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

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

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

9:50

And so we repeat as we did earlier.

We can do a little bit faster at this point.

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

All you would have is L double dot.

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

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

That's a time-varying part.

R is a fixed length.

So I have this cross product,

e relative to n was theta dot e3.

So if I could take the cross product here,

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

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

So there will be a minus there.

10:38

And for this term, very much the same thing.

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

R is fixed but these angular rates vary with time.

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

And now for the cross product we have omega

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

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

And the third vector crossed the second, again,

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

11:16

That's it.

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

Once you get good at this,

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

Do my one but that's it.

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

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

many many a problem.

11:36

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

Let's say the problem statement says what is

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

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

wondering how to step motion of A look like E.

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

simulate the sensor signals.

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

approaches another space craft.

Also orbiting in its own plane and doing things.

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

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

What does that mean mathematically?

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

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

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

12:50

Just start adding up vectors.

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

Minus Rsr hat and then minus Rer hat.

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

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

So that means this this time,

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

13:48

So here, I need the s frame.

This first term is already an s frame component so

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

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

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

And the transport theorem I need an S frame derivative,

I'm choosing to do an e frame derivative.

I need Omega e relative to S.

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

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

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

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

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

Is we really begin the right to transport there.

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

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

So if you needed omega e relative to s,

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

So I could take advantage of that.

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

phi dot times n3,

with a minus sign there Sorry minus this, actually.

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

Can't draw.

And then the whole thing crossed with rer.

So those two terms actually cancel the minuses minuses,

that gives you a plus again.

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

plus the second, and

this is the A frame derivative,

15:52

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

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

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

you can always break up a whole series of things and

just do long pull the n stuff together.

Long pull the e stuff together.

Long pull the s stuff together.

And systematically go after it.

Also illustrates how we can take different angle velocities,

reverse directions as needed.

And particularly,

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

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