I had seen how to define parameters of mechanism, the fix parameters and the variable parameters.

Variables are the joint angles of the joint displacement, which we called offsets in the

previous class.

And the fix parameters are the ling lenses, ling twist and either the offset of the joint

or the angular able joint.

Now we also had started looking at a six revaluate, robot manipulator and stared fixing reference

frames to that so as to define to find out what this ling parameters are so to take you

back to what we were doing?

This is a typical articulated robot configuration with a vertical revolute joint first with

two horizontal joints one in the shoulder, one in the elbow.

And there is a wrist here which as three revolute pairs, which intersect at a point.

And the adjust in revolute joints there are orthogonal, this is required for full orient

ability of the wrist.

Fine, so that is the kind of wrist that is there.

We had started setting up the robot, drawing the robot in a specific configuration stressed

out.

So that the various joint angles, turn out to be either 90 or 0 degree or 180degrees,

and in this particular case then the joint axis are aligned along three perpendicular

directions.

So that a description of the various, when we draw the reference frames here, it will

be fairly easy to figure out what the joint parameters are?

So the first joint is the vertical revolute joint, then at the shoulder you have a horizontal

revolute joint.

Then at the elbow you have a horizontal revolute joint so we have come up to this point.

Then there is a rotation along the longitudinal direction that this revolute joint, and then

there is a revolute joint which is perpendicular to that, and that is this.

And then there is the final revolute joint, which is perpendicular to this and in the

position we have shown.

This two are a line, so these three form what is called the concurrent wrist.

Concurrence meaning the axis of all these three revolute joints intersects at a point.

Which is that point, there is somewhere here, okay now let us fix the various reference

frames are associated with the lings.

In order to do that lets us briefly go back what we saw in the last class.

We have joint I, I+1, and I+2 here, this is ling 'I' between joint I and joint I+1, the

reference frame attach to ling (I) is on the (J+1) joint axis.

And the link is represented by the common normal between the two joints associated with

the leg.

So this is common normal fine.

And the reference frame is in this fashion; extend that common normal the origin of the

reference frame on ling 'I' is here.

So this is the XI axis, ZI axis is along the joint it could be in this direction or the

opposite direction, it does not matter.

So you have XI and 'ZI' like that so in the second joint of the link on that axis you

have the origin of the reference frame of the link fine.

And the Z axis is along the joint and X axis is perpendicular to the joint fine.

So that was the convention that was used, and the way to measure the various lings parameter

I have been set up here, we will see it later.

Let us fix up the reference frames first, so we will consider ling '0' to be the base.

Ling 0 is the base; ling 1 is this ling, 2, 3, and so on.

So ling 0 as reference frame X0,Y0, ZO attach to that, right it has to be on the origin,

has to be on this joint J1, so this as origin in that, along the axis of that joint so (z0

) fine is this, this is (x0 ) either this are this could be x0, I took this as x0, so

this is y0, so this is the global reference frame attach to ling 0.

Which is the base, how many ling does this manipulator have, is a six revolute manipulator.

It is seven lings right, the base ling onwards okay ,the second reference frame as to be

attached to, ling 2, fine it is origin as to be on joint 2, so which is joint 2?

This is joint 2 right, so I draw this line which represents the axis of joint okay, And

we can take the Z1 direction in any fashion we want to, the way I have taken is this direction.

So this is (Z1) so what about (X1), this is a crucial question?

What about (X1), so now look at the ling.

The first ling is the common normal between the two joints on the link, which are the

two joint on the link.

Joint 1 and joint 2, so it as to be perpendicular to the axis of joint 1 and axis of joint 2,

right the ling as to be perpendicular it is a common normal.

The common normal unfortunately here turn to have 0 left, so you can see the ling.

Nevertheless we can give an X1 axis that as to be perpendicular to Z0 and Z1 correct,

so we can take it either along this or along this, fine, so x1,z1,y1 I am not drawing y1,

because it will cladirup the diagram.

So x1, z1 is enough, now what about ling two, ling two is between joint two and joint three.

Right, so we have to first see what joint three is, so joint three is this and joint

three axes can be drawn.

Let us say we take this direction, now ling three is a common normal between joint two

and joint three.

Or the reference frame z1, and z2, so this is z2, right, so I would like you to go back

to this particular diagram which we do in the last class.

I would not keep bring it back here, you can look at that when I do that explanation.

So z2 and z1 or here, and we need the common normal between them, now where is the common

normal.

Some line joining both of them perpendicular to both of them which is that line a ling

itself.

But what do you mean by the ling itself?

If I am look at from above, let us look at this from above, let it be in this fashion.

I see this revolute joint that pose which is rotates, so the access of the joint is

like that.

And I am really offsetting this, like this fine, so this is revolute joint through that.

So this will move up and down.

Okay, I looking at it from above and then come the next joint and the next link, I am

talking about this link now.

There is a joint axis there, fine, so looking at it from above, you see these two joint

axis like this now we are trying to look at, what is this ling.

What is it appearing like, it is a common normal between these two.

But the point is that these two axes are parallel.

I can put the common normal anywhere I want to, although some of you said put it at the

physical link, where is the physical link?

The middle of the physical link, something which passes through the centre of mass of

the physical link, there are so many possibilities.

For kinematics it just does not matter, where you put it, there may be some convenience

attach to where you put it.

For purely kinematic calculations does not matter.

Fine, so let me put it at the, as you suggested somewhere in the middle of the link fine,

so somewhere in the middle of the link, let me assume that in this diagram, is this particular.

So this is what I choose as a common normal, okay, I could have chosen this or any other.

It is important to realize this, what you said his convenient so put it at the middle

of the lings, so that you can define points of the lings very quickly in this reference

frame.

Attach to that ling, okay, so why did we draw this particular thing in order to set up the

x2, z2 , z2 is already there, to set the x2 axis right, What is the x2 axis?

It is this extended beyond this so the origin of the x2 axis reference frame is here.

So extend this and this will give you x2, fine, if you have questions shoot, fine, do

not hesitate.

You have to understand this, and remember that this is a convention.

It is nothing sacrificed about this you can set up your own convention.

But before you do that understand this very clearly.

Fine, now you have the x2, z2 reference frame of ling two, now for ling three we need x3,

z3 reference frame.

And which is z3, z3 is the axis through joint four, right so let me draw that.

Now what about x3 now let me indicate one thing, the way I have drawn it I will located

the axis of this joint.

Right along this that is not necessary that it be there it could have been offset in this

direction.

When we set up the parameters we will allow you to vary, we allow it to be anywhere so

that we will give a general notation for that, just remember that this particular line could

have been shifted that side or this side.

Okay, so now this is z3 and this is z2, now the ling three is the common normal between

these two, and right now the way we have drawn it and set it up there is no length there.

These two are coincident these two are intersecting at this point.

But the common normal definitely we can draw fine, so this is a point on z3 axis, and z2

axis, they are coincident and so x3 could be this.

It could have been downwards also, does not make a difference.

Okay, so you have z3, x3 that is reference frame attach to link three, then through this

you have z4, the perpendicular between z3 and z4, is that is x4, attach to link four,

this is the point of concurrence.

Where the last three axis, revolute axis intersect whatever be the angle you give to this joints.

The last three axes always intersect, that particular part of the mechanism, founds what

is called as spherical mechanism.

The property of the spherical mechanism is that it is consisting of the revolute joints

and all the revolute joints intersect at a common point.

Such mechanism are called spherical mechanism, okay, so you have z4, z3 the common normal

between them is x4, so let us take x4 in this direction fine, And now we need z5 and x5,

so this is J6, the sixth joints to which you have x5, so let z5 be this x5 is the common

normal between, z5 and z4, z4 is this z5 is there both are coincident both are intersecting

here.

And I take x5 to be along this, what I drawn is visible to you, fine, so we have set up

reference frames up to link five.

What about the last ling, ling six so on ling six what the problem is that, there

is only one revolute joint.

Right, so we have to take the point a reference point that was the origin of the frame should

be, so I take it along this so let us extend this, take this particular point, and what

I did was I aligned z6 along this, it is not necessary to do that There is one convenient

way of doing it and x6 perpendicular to that

So in this sort of skeletal figure whether links are not really drawn we have just put

the reference frames , we have to now imagine virtually where the links and physical sensor

fine, but now let us set up the parameters , what are the values of the parameters?

Right now in position you are seeing they are coincidence the two axis the J or Z 3

or Z5.

Are coincident right now, but they will not remain coincident if I rotate this joint,

what will happen is that this will rotate fine, so when that happens, this two will

not be parallel but they will concurrent , they will be intersecting this point .

Imagine that is a way where is rest has been built, so are you saying that they are not

coincident or concurrent looking at this figure, so what you see here is,

So this is one and other is like this, so it is drawn it like an angle they are suppose

to coincide okay, now let us set up the parameters

So the parameters are AI, aI an ling length and ling twist the joint of the set DI and

the joint angle ?I okay, let us now look at the ling length and ling twist, the lings

1 to 6.

Now here the ling length and ling twist was defined as follows, AI and aI are obtained

are what a use to take ZI- 1 along X side, so that it eventually coincides with ZI, AI,

and aI are the translation and rotation you have to give to the ZI-1 axis, to make it

coincide with the ZI axis, by translation and rotation along x I axis okay,

So A1 and a1 would be obtained by taking Z 0 to Z1 right,

Along X1 axis in this direction so what s A1?

A1 is 0 and the way we have given the sense various vectors here axis here, what should

be a1?

90 degrees because, x1 I have to rotate so Y is that AI is 0, A1 is 0 it is because this

two intersect already right,

So I can rotate this particular line about his point without translating it by 90 degrees

along this axis to make it align with Z1 okay, so this is 90 degrees.

Here is a common normal between ZO and Z1, that is the length of the common normal between,

but you noted to give the sense to that we imagine that, we are bringing Z 0 in to coincide

with Z1.

By a translation and rotation right since they are intersecting no translation is required

to do that, so A1 becomes 0, but you need a rotation to align that so that is 90 degree

okay, so that is very clear . So let us proceed like that Z1 to Z2, A2 is

it 0?

It is not 0 right, it is some value will call it simply A2 and we can give any value for

A2 , that will change the distance between these two so let us just also indicate here

what A2 is,

That is the length we can see on the robot, and the distance between the shoulder and

the elbow and what about a2 else a road it is because they are already aligned, now A3

you bring Z2 into coincidence with Z3, they are already in the setting.

Right so A3 is 0

And then I rotate Z2 about in order to align it with Z3, X3 in this direction because of

that this is 90 degrees.

So this is the way we proceed ,so to get to X4 you take Z3 and bring it to alignment with

Z4 ,they are already intersecting A4 is 0, Z3 to align with Z4 you have to consider rotation

about X4 that is a common normal again is 90 degree.

Because Z 3 is in this direction,

Z4 in this direction, and directed X4 in this direction, so you have to rotate 90 degrees

not - 90 degrees, so it just happens so because I took the X4 in that sense, I could have

taken X4 in the opposite sense,

Now A5 and a5, take Z4 and make it align with Z5, then they are coincident they are intersecting

this is again 0 and the way I took X5, this is 90 degrees again,

so all very convenient.

I could have taken 1) any of the Z in the opposite direction.

2) X is also I could have taken in different direction , so you might get 90 degrees or

- 90 degrees depending on what you took, but important thing that none of these could be

Can other be - 9+0 degrees, cannot be 0 degrees or 180 degrees, that is particular

Articulated robot configuration that is a different articulated robot configuration.

This particular robot is a revaluate joint perpendicular to the next, parallel to the

next and the way we have taken then, perpendicular to the next perpendicular to the next and

again perpendicular to the next.

It is this parallel the fact that this is parallel that you see as 0o is here rest of

them are 90o so minus(-) that okay, now, A6 a6 so take is at 5 make it coincide with is

at 6 they are already coinciding nothing is required to do that, so zero(0) and zero (0).

Now, let us look at the (I) and theta (I) the (I) and theta (I) are the offsets at the

joints so, what is required is the One and theta One is obtain by talking zero(0) axis

and making it coincide with X one axis by movement along is at zero(0) axis right, so,

X zero to X one along is at zero you may have to move it as well as rotate it translate

it as well as rotate it.

In this particular case you have only move it translate it no rotation required because

these are parallel.

So, D1 is this and this is joint one so, D1 is this so, we will give it a simple rather

than a number any number if a substitute will be fine, except in some these cases if you

use zero(0) then you have troubles in this case only okay, physically what do we see

here as D1 is this length this particular length from here to here right, from here

to here is D1 fine, you can see that physically the way it has been drawn so D1 is there what

about theta one(1) theta one(1) is a variable here it appears to be zero(0) but actually

speaking I can rotate this ling with respect to this so it will be aligned some other way

if I had rotated it.

I just shown this in a particular configuration position that is all , in this position it

is zero degrees( 0o ) but it is actually a variable so we need to call it theta one (1).

To indicate that this is the variable we know that in this particular case all are revolute

joints so this is the way.

So, the value for that is zero degree in that position so, now D2 theta (2) or joint (2)

so, we take X1 to X2 right, along what along is at one(1) right, so the way I have put

it that is here that will have some negative value right, this is the distance so, that

is D2 we call it D2 so that we can give any value to that.

But here it is this particular this is what is needed physically and theta (2) D2 is a

variable, what is the value in this position?

Zero right, when will it change from zero? when we rotate this joint if I rotate this

joint X1 will remain there and X2 will rotate okay, X1 will remain there, X1 will move when

I rotate J1 okay, X2 also would move if I would have fixed everything else and move

J1 it will rotate in that direction.

If I fix this and rotate this X1 will remain the with this rotating okay, now D3 theta

(3) so D3 is the distance between X2 and X3 , X2 is this and X3 is this fine, X3 is this

done?

So distance between them so we have done is that put it back on this line.

So, in magnitude D2 and D3 will turn out to be the same it is not necessary to be so but

simply call it D3.

They are fixed values but we can give any values for them that w will be changing the

dimension of the robot okay, and then I will have to rotate it above is at (2) in order

to make this align with is at X3, now what is the direction of the rotation?

90o but this is a variable it is just in that position it turns out to be 90o remember that

it is a variable okay, so, that is theta (3) all this will be variables everything in that

column is variable in this particular position in turns out to be 90o that is all.

So, similarly we proceed so this is again magnitude wise this is D3 okay, you can see

that physical dimensions here D1 A2 then D3, D2 and D3 okay, there are just two more ling

dimensions from here to here and from here to here.

X3 Z3 is the reference frame adjust to ling(3) is at (3) the question was why was X3 placed

here rather than on joint (4) it is on joint (4) actually what is joint (4) ? is this axis

which is actually on that axis.

So now D4 theta (4) is any way is theta (4) D4 what is it?

X3 to X4 right, X3 to X4 okay, so here to move it by this margin along is at (3) you

have to move it by this mansion that is D4 and so D4 is this distance.

If you make A2 or D4 zero (0) than you have trouble with the manipulator it will not be

able to do what region in space and 6o of freedom and from things like that.

Okay, now D5 theta (5) is theta (5) is theta (6) D5 you have to X4 to X5 along is at 4

so they are already coinciding so this is zero (0) right, the fact that this zero is

important because if this is not zero (0) you do not have a concurrent wrist and without

a concurrent wrist you will see that out inverse kinematics cannot be done easily.

That we will see, so this is zero(0) all of you recognize that so this the final thing

is D6 this has some value and what is see is D6 is this particular length.

So, to put it on the manipulator this lengths on the manipulator, let me quickly show this

particular distance from here to here if I set up the this particular here is the distance

from this point to this point that is D1, the distance between these two axis is A2

right, now if you look at this depending on where you really set up this actually if you

look at this axis and extended backwards then, you have a D2 taking it in this direction

and D3 taking in back in this direction.

Okay, so, what is fixed is really look at this axis which is this particular joint four

axis, joint four axis with respect to this particular vertical axis that is the crucial

thing, D2 and D3 can actually take whatever appropriate values but finally.

When you come back this particular offset should be there.

Okay, so this D2 this is D3 and then this particular distance which again is something

you see on the mechanism from here to the concurrent wrist is this particular ling D4

and from the concurrent wrist to where you put the reference point on the end effectors,

what is the reference point on the end effectors, it is a special point you define for determining

the work space of eh mechanism typically that is what it is.

You use the reference point to define the work space of the manipulator so, we have

put the origin of the reference frame let see at the reference point.

So, the distance from the concurrency point of eh concurrent is to the reference point

is this distance D6 so those are the lengths that you see on the manipulator.

So fact that these are 90 zero 90,90,90 is because of this perpendicular parallel perpendicular.

Okay, so this is what these are the parameters joint angles and fixed parameters of the revolute

joined manipulator.

Now, let us do forward kinematics, forward kinematics is now appears to be trivial to

do turns out to be trivial to do,

Let us define the problem, now, we can define the problem of forward kinematics is very

simple terms well, the problem is the following given ,

the mechanism and is fixed parameters what are the fixed parameters?

ai alpha (I) and Di in our case in this particular mechanism these are the fixed parameters and

the joint variables for all the lings and joints so, in our case there is (6) what is

that required to be found out?

The position and orientation of every ling right, what gives the how did we define position

and orientation of a ling ? using the rotation metrics of the ling and the offset of eh origin

of the reference frame on the ling with respect to the global frame.

There are twelve parameters which define that we had put all of them into what we call the

homogenous transformation metrics if you remember.

So, what we need to find are the elements of this transformation metrics for the Kth

ling with respect to the zero ling that is the global frame for all the lings

this is what is to be determined okay, and based on the transformations that we done

this is the trivial problem TK with respect to zero (0) is nothing but, T1 with respect

to zero (0) multiplying T2 with respect to one and so on and finally TK with respect

to (k-1).

K is similar to that,

So then for our manipulator, what is the relationship

between reference and frames I(i-1).so look at the

I reference frame form for the element of the matrix you have to set up jump equation

for that, if we do that and what is left for us is to determine this.

The position of the reference frame in the previous link reference frames okay, so that

turns out to be the following.

If you fix up the reference frame if you look at the reference frames

so this (Xi-1 ) (Yi-1) and (Zi-1) this ZI (xi)and YI okay, and the way we have been

defining this parameters this is joint (I) the axis is (i-1) is along joint (I) and (DI)

is this distance theta I is this angle look, along the(I =1) look at the angle between

this (XI) and(XI is 1)that is theta I okay, and (AI) is this distance, the distance between

(ZI) and (ZI= 1) the common normal between them that is this and alpha (a) (I) is the

angle between the (ZI) and (ZI= 1).

So, we can make the determination of what we need to get is Ti with respect to i-1.

We make things easy we set up in the intermediate reference frame.

So let me call it z term along this the origin here.

And Xn along that and Ym the cross product of the z xm.

If I do that then I can write this as Tm is respected to i-1 multiplying Ti will respect

to M. I set up in the intermediate reference plane Xm Ym Zm.

Xm along this Ym along this.

With that we can Pevely see and derived.

That distance out to be equal to the following.

This is the 4/4 terms almost transmission maintains.

The last column is aIcos?I This particular position, this point in this reference frame.

That is AI cos?I and sin?I Then along Zx is that is di, and1.

This something you can easily derived.

Fine, just multiplied set of these two matrix in multiply.

Or even without that is not all the differential of the software.

Remember, that this three the vectors form by these three numbers, is the Xi direction

in this global reference frame.

And this is the Yi direction, and this is the Zi direction.

This is the position of the origin global reference frame.

So with that doing the forward kinematics of the mechanism we saw is very simple.

Right, all the transformation matrixes and multiplying them in

the chain.

Oaky, yeah, AI cos?I AI sin?I and di Well without asking me you could have just taken

the position of this in the reference frame.

Okay, so, I am not going to the substitute can remember some of the a iis 0.

Those particular matrixes can the return for any I to I minus 1.

Many of the as are 90.

So the responding cause a it will be 0.

Sin a you will be 1, also simplification will be take place.

And you can refer those matrixes there is.

So, just before we leave one question.

Suppose, we normally do this forward kinematics in notice to answer not

problems question following type when the mechanism moves would some part of a link

hi an ups and down.

Fine, so you at in some position one of those links is like this and, the reference frame

attached to this is lesser seem like this.

The global reference frame is here and there is some obstrucle out here.

And it is going very closed to that.

We need the now figure out.

whether any part of the physical link would hit this obstrucles.

How do you check that?

So you imagine that this is cuboids.

This particular link is in cuboids.

Yeah, is this not very easy actually, so I will for example, what you said, what the

suggestion there was look at the end of the links.

Let say the corner points.

Suppose it is a corner point also lets us.

Okay, take corner points and see whether it is inside the cube of cuboids.

If any of those corner points are inside this already hits.

Right, it is something we can check.

Now none of these points corners of inside this obstrucles then we conclude it is not

hitting we cannot.

Okay, another thing we can check is whether any of these corners of this obstrucles is

inside this cuboids.

Because this cuboids is a symbol thing to define in space.

Those check and better.

Now let me don't look at that this is not enough.

You remember that we have to also that check whether any of edge is in contact with this,

or any of the face is in contact with this.

It needs to be check.

The certain ways to be can be done for felled in out of space.

Fine, now let us look at the first thing that we are going to check whether any of the corners,

let's take this particularly corner is inside the cube.

We know the position of the obstructers.

So, now you to know the global position of this particular point,

Do we have that derived now, We only have the local position of that in this reference

frame.

But we also know the position and orientation of this reference frame in the global frame.

Right, so this particular point is P, then the global position of the P can be obtained

xp, yp, zp,1 that is the augmented vector.

In the global frame equals the transformation from I to 0.

Fine, sorry we normally put it inside.

In local position of this points xp,yp,zp,1 local.

Fine, This particular position the local frame is

this augmented matrix 4/1 matrixes.

The local position.

3 multiplied by the how much in the transmission matrix which takes points 1 to 0.

That will give us the global position.

So, once you get the global location we will get.

Whether the points inside okay, so, this is why we did this is, this is one of the reason

why we did forward kinematics.

Okay, we need to able to check inside the slide.

And that also is easy to yeah, so that the crucial thing is to get this.

And we just saw how to do this.

Okay, so tomorrow we will discuss the next class will discuss how to do the inverse kinematics.