OK, we're going to continue our discussion of coordinate transformations. So if you remember in the last video
we talked about locating a position in space in terms of two different
coordinate systems. A Cartesian coordinate system, where the position is expressed in terms of components x, y, and z.
And a cylindrical coordinate system, where position is expressed in terms of components r, theta, and z.
So now I'm going to ask a kind of different question here. I want to ask how we can map
a velocity vector expressed in terms of Cartesian coordinates with components u, v, and w in the x, y, and z directions.
And map that between cylindrical coordinates
where the same velocity vector can be expressed in terms of
components v,r, v,theta, and v,z. in the r, theta, and z directions.
So as a starting point
remember that we can locate a point in space in two dimensions. So I'm looking at the xy plane here because
again the z direction is common between these two coordinate systems. I'm going to focus on the xy plane.
So I can locate this point in 2D in terms of its Cartesian coordinates x and y.
And its cylindrical coordinates r and theta.
OK, so now to begin talking about velocity.
I need to express at this point
the orientation of these coordinate systems. So at this point in space
I have locally, unit vectors
Cartesian unit vectors, i and j in the x and y directions.
Oriented parallel to the x and y directions globally.
But then I also have locally unit vectors in the r and theta directions. And so they are
oriented differently. So notice that the unit vector in the r direction is oriented parallel to r.
Since this is an orthogonal coordinate system, the unit vector in the theta direction has to be at a 90 degree angle to the r component.
So the unit vector in the theta direction is pointed as shown here.
OK, so that just is describing the point in space. So let's come over here and show this point.
So this is the same point in space that's located above.
And there's some velocity. The fluid has some velocity at that point. And that's expressed in terms of a velocity vector v.
OK, so now if I want to find, or identifiy, the components of the velocity vector in these different coordinate systems,
then I can do it as follows. So locally, remember this is the orientation of my Cartesian coordinate axes.
So the u and v components of this velocity vector
are given by these projections on these axes. And so that's shown here.
Pretty straightforward. I think we can all agree on that. But it gets a little more complex
when we start talking about r and theta components. And that's what I'm going to try to show here.
So remember that this point in space is located
by r and theta, which describe its location with respect to the origin.
So these r, theta, and z components and x, y, z components express the position of this point in space.
At this point there is a velocity vector.
And we want to express its components relative to these coordinate systems.
So you know that's straightforward for Cartesian coordinates.
But the problem comes in when we talk about cylindrical cordinates because notice that as r changes.
Because the unit vector in the r direction has to remain parallel. So as this point moves,
r moves. And so the orientation of these
axes is also going to change. So let's see how we would do it here. So we have the same velocity vector.
And so we want to map it in terms of the component in the r direction. So we can extend r forward.
And take the projection of v on r. So that's what is shown here.
And then the projection on the theta direction. These are normal, these are orthogonal directions.
So the projection of v on the theta axis.
The local theta direction axis is shown here.
So I put boxes around it so we can try to keep track of where they are.
So u and v are the projections on the Cartesian coordinate axes. And v,r and v,theta
represent the projections on the cylindrical coordinate axes.
OK so now I want to try to transform between these.
And you can probably guess that we can do it using geometry. But we've got a lot of triangles drawn here.
So it's kind of hard to see. So I'm going to try to break it down as best I can so that we can see
where some of these relationships come from. So let's talk about the u component.
And imagine that I want to express the x component of the velocity, u in Cartesian coordinates,
in terms of v,r and v,theta, the components in the cylindrical coordinate system.
So there's two terms I need to keep track of. I need to look at the projection of v,r on u,
and the projection of v,theta on u.
OK, so remember I can express this velocity vector in terms of two components in the r and theta directions.
So to transform those to the Cartesian system
I need to know what the projections of the r and theta components of the velocity vector are on u.
Both of them in general will have a projection because the coordiante systems are oriented differently in space.
So the projection of v,r on u, I'm going to call u1.
And the projection of v,theta on u, I'm going to call u2. So I can keep track of these two terms.
So let's try to take a look at where some of these terms would come from. So let me draw it here, just the terms that I need.
So the velocity vector v is drawn here.
And this is the r component v,r which is shown here in the box.
And the theta component v,theta is orthogonal to that.
And then I've also shown u, the component of the velocity in the x direction of the Cartesian system.
And so you can see that there are some right triangles involved here.
So let's look at these projections. So if I want to find the projection of v,r on u.
So this is v,r, the green line. And this is u, the red one. So the projection of v,r on u, on the u direction.
I'll just draw a line down here, orthogonal to u.
And this distance is the projection u1.
Similarly, now I need to find the projection of v,theta on u.
So notice that v,theta also has a projection on the u direction.
So again, if I draw my lines down perpendicular to u.
This little piece here is the projection of v,theta on u.
So if I want to express u in terms of v,r and v,theta, I need to express both of these components.
Both of these contributions to this projection.
OK so let me go to the next slide and redraw the geometry that I had so we can continue the discussion.
And remember that I want to express u in terms of v,r and v,theta. To find the projections of v,r and v,theta on to u.
So I've broken it down into these two components. So let me look at each one individually.
So let me look at u1 first. So let me just draw what this triangle would look like.
So u1, this is the projection of v,r on u.
And then v,r is here. And the angle between them is theta. This is just the projection on this area.
I just drew it down here so that you can see more clearly.
So v,r and u1, this is a right triangle. And the angle between them is theta.
So you can see from trigonometry
that cos(theta) is adjacent over hypotenuse. So that's u1 over v,r.
So I can express u1 as v,r * cos(theta).
OK, now for the other components. So now I want to find this piece.
So I can draw a triangle similarly here. This u2 is here. And then I end up with this triangle. So the bottom part is u2.
This is v,theta. And then this angle at the top of the triangle is theta. So how do I know that?
OK, so really this triangle you can imagine being right on the end here.
Instead of on this side. It may be easier to see.
So if I draw this triangle here. So r and theta are orthogonal. So there is a right angle between v,r and v,theta.
And this projection of v,r on u1 is also a right angle here.
So if you notice in the purple triangle, I have this angle is theta.
Then the opposite angle in this purple right triangle has to be 90 minus theta.
So now look at the outer triangle that combines this purple side and this brown side.
You can see that this angle is 90 minus theta.
And since this is a right angle, this has to be 90 minus this angle. So 90 minus (90 minus theta) is theta.
So this angle at the top has to be theta. And that's supposed to help show that.
So then if you believe that,
then you can notice from this triangle construction that relative to this theta, sin(theta) is opposite over hypotenuse or
u2 over v,theta. So u2 is equal to v,theta * sin(theta).
Now I can put those together.
So u, the x component of the velocity vector, has these two components: u1 and u2. So u1 is v,r * cos(theta), and
u2 is minus v,theta * sin(theta). Why is it minus? Well notice that the projection of v,theta on to u is in the opposite direction of u.
So therefore we have a minus sign here.
OK, so I think you can kind of see how this process goes. As long as you can keep track of the triangles you're working with,
and you're careful, you can apply it similarly for all the components. So you can express u as [Reading Equation],
v, the y component of the velocity is [Reading Equation],
and w is equal to v,z.
OK, and similarly we can also apply that just to express the relationship between the unit vectors because those are
we are already expressing the velocity in terms of components.
The velocity vector in terms of components in these directions. So it's straightforward to show
that we are basically getting a relationship between these directions also.
So the unit vector i can be expressed as [Reading Equation],
the unit vector j can be expressed as [Reading Equation],
and the unit vector k is the same as the unit vector ez in the cylindrical system.
OK, so we just showed how we could express
the u component, the x component of the velocity in the Cartesian system, in terms of
its components in the r and theta directions. Now let's ask the inverse question.
How can we express one of the velocity components in the cylindrical system, say v,r
in terms of its components in the Cartesian system u and v?
So we can follow the same construction. So v,r, we need to account for the the projection of u on v,r.
And the projection of v on v,r. The x and y components of the velocity vector
in the Cartesian system each have projections on v,r.
So again starting with this construction, we can take the projection of u on v,r. So that's given by this right triangle.
So this is the projection of u on v,r shown in purple.
So that the v, or y component, of the velocity is shown here in red. And so then its projection onto v,r is shown here in brown.
And so then its contribution, the projection of v on v,r, is represented by this distance vr2. So we have the v component of velocity
and we see its projection on v,r.
OK, so now again let's break out these triangles and see what the relationship is between these components.
So if we look first at the first term.
I can draw a triangle where one side is u, the x component of velocity in the Cartesian system.
And what is that relative to vr1?
So we find this right triangle, and again the interior angle is theta.
So cos(theta) is vr1 over u, adjacent over hypotenuse.
So vr1 is u * cos(theta).
Similarly, I can look at this triangle
where one side is the y component of velocity, v, in Cartesian coordinates. And this bottom part
is its projection on v,r, which I'm calling vr2. And again I can use the
same arguments that I used before. I can show that this angle at the top is theta.
And so therefore, sin(theta) is opposite over the hypotenuse. So vr2 over v.
So vr2 is equal to v * sin(theta).
So then I can put these together just like before. And I get that
[Reading Equations]
And I can also express the unit vectors, because again
I'm already mapping the velocity vector along these directions.
So if I just look only at the directional components, I get that
[Reading Equations]
So now with this construction, I'm able to map the velocity vectors expressed
in components from either coordinate system, back and forth between the coordinate systems.