OK, last time we have seen an example of getting the age parameters, assigning the age friend and
getting the age parameters.
And in this video, we will do one more example, which will also contain spherical Greece, which is
interesting for us.
OK, here, as you can see, I have summarized the the and the age parameter.
What are the meaning of the age parameters?
We have seen these in the last video.
So I have written all of them here again, you know, for reference.
And so key to the eight and I fossil be did for four parameters.
Two of them rotation.
So Theta and alpha and two of them translational parameters, which is the and a.
So let's let me show you the whole robot to you, and you can see we will get the parameters for this
robotic manipulator.
So we will we will look forward kinematics for this robot manipulator.
If you watch this robot manipulator, it consists of two parts.
Okay, let me show you these two parts.
The first one is this one.
OK?
This is called cylindrical robot.
This is cylindrical robot manipulator.
OK.
So this part is only responsible for positioning the positioning in Cartesian space.
OK, I presume space space.
This is our end effector.
OK, this is in the victor and the effector and this part.
OK.
This Part A is only responsible for positioning of the end effector.
It doesn't affect the rotation, OK of the end effector.
It's only used for positioning.
OK, so but as you know, it's very it's not interesting for us.
In order to be able to control only the position of the end effector, we want also to be able to control
its rotation.
Yeah, in three dimensions how we will do that.
Indeed, if you've got your arm, you will see that the rotation part for your hand is due to the wrist.
Your wrist.
Yes, indeed, your wrist is a spherical wrist, and it consists of three rotational joints.
Yeah, indeed we have seen before.
During mentioning joints, we have said the spherical joint and this is nothing but a spherical wrist.
OK, it's very cool breeze.
And what we have said is that it consists of spherical joint spherical spherical joint, which was three
degrees of freedom here because it provides three rotation in three Dimension X Y Z.
So this part so this B part, which is spherical wrist, is only responsible for rotation, in part
not positioning, but OK.
So what we can do in any manipulator, you can attach spherical wrist in order to give dexterity to
your robot's manipulators and effective so it can rotate OK in three degree of freedom over three axis.
What we will do in order to calculate the age parameter, the H parameters on forward kinematics for
this remote manipulator, we will first calculate we will separate these two parts A and B, OK.
In order to make our job easier.
What we will talk with VS will calculate the parameters for a part.
Then we will calculate the age parameters.
And then we will calculate the age parameters for the B part.
It is OK for the B part after we will join these two together in order to obtain or obtain final homogeneous
transformation matrix or so.
Final forward kinematics for the whole robotic manipulator.
Okay, so let's continue first.
As you can see, we have taken the first part of the robot manipulator.
This is the part a what we have said.
Yeah, this is the part.
So let's do.
OK.
As you can see, we have from here, we took out the spherical wrist and connected the victim directly
to the robot manipulator.
You can sing it like that in your wrist stuck, and it doesn't move.
So it's something like this.
Yeah, OK, so let's continue.
Let's try to get the parameters.
First of all, yeah, I forgot.
First, we will just assign frames you important in the convention.
Let's assume it was easy to assign.
First, we assign a set axis as yes, this is for zero zero.
How we would assign.
Does it axis it axis has to be in the direction of motion.
Yet as in our case, this joint is rotating in this way here in the rotation angle is teetering on.
In this way we are doing the year.
We are choosing our rotation axis in this way, OK?
So in this case, the rotate, this is a prismatic joint, so its motion is in this way.
So we will choose this, as is it one?
Yes.
And then the motion is like that, OK, we will rotate the two, OK?
Then what we will do, we will talk, then we will choose that axis for this frame.
OK, for this joint.
OK, this is it to, in this case, all the extreme prismatic.
So it will be something that these three and as we have said before, just could be the last one for
the in the vector.
This will be the three.
Yes.
OK, now let's assign x axis a smoking.
Let's assign x axis in this way for this automatic manipulator, OK?
And as we have said before, for the first run, we can assign in any way, we can also do this in this
way.
We can also do it in this way, but we have chosen it in this way.
So out of the page and there you will see why, because it will make it our publication of parameters
easier.
You will see what.
And now let's choose this one for the first frame.
We can choose it in this way yet.
What was the law for choosing the x axis?
So this x one axis has to be perpendicular to the z zero axis and also has to be has intersected, OK?
If we choose, as you can see x one in this way.
So if it was x one in this race or endroit, it will surely be perpendicular to the Z zero and cut it
so we can choose it.
In this way, we can choose it.
Also it in this way.
Yeah, we can choose also it in inside the page.
All of these cases, it will cut the z zero axis.
But as we have said before, we always tried to do what we always tried to try to be as much similar.
Um oh, sorry, yeah, we always tried to be.
Um.
Similar to the last previous frame.
Yeah.
So that's one of you will choose it in this way and will minimize the number of rotations.
OK, so we will not choose it.
Well, to the right or the left or inside the page.
OK.
In order to minimize the changes, we will choose it as exceeded.
We can do the scene.
OK, and one more thing for choosing the center, as we have said as at zero and that one is parallel
to each other.
We can choose the center for frame one or the Z one axis and in order to make it easier.
And there is no meaning to put in offsets to put the center here.
We will choose the center as the center of the joint, OK, the center of the frame as the center of
joint for this one.
And what will happen?
What will be the center?
As we have said, we have to first make it make for this axis longer.
Is it one longer?
And also this is it too.
We have to continue and the intersection is the center.
As you can see, the intersection is here, so.
So so this will be the Frames center.
OK, so we will choose again our X to it in this way because in order to minimize the number of minimize
the changes, OK?
And also X2 will be perpendicular and also cut or intersects with that one.
OK.
And we will copy the last one year.
We will copy, as we have said, the previous frame for the end effector.
And this will be.
This will be our X three.
OK.
As you can see, we have finished assigning axis.
I will not do y axis because, as you know, it will make only complex the drawing and also even not
use it.
But a year after assigning X and Z, it's very easy to assign y axis.
You can just need to follow right hand rule.
OK.
So let's continue now to calculate T dot title parameter in what was the T done T towards.
Let's remember it was rotation around zip ie minus won't be careful to match x i minus one and X I.
So for the first joint, OK, if I equals to one.
And so this for the second drawing, but for the frame, for the frame one.
So how much we have to rotate both the left and put the rotation angles I have forgotten this year,
we have put it in this way this is these three.
Oh, OK, we have done.
Yeah, we have put it the rotation and there's no need to do that.
OK.
So so how much we have to rotate frame is zero with respect to z zero.
Yes.
In order to match x one and x zero at the first time, you can see they are parallel to each other.
So you can say that we don't need to anything because they are parallel and x one and it's zero parallel
to each other.
We don't need to do any rotation, but it's not correct.
Yeah, as we have said before, you have to and you have to consider the angle.
And if you watch from talk to the robot, if this is the first joint, OK, and this is OK, this will
be our X0.
Let's throw it at our X0.
In this way, and this is our digits zero.
Yes, this is our digits zero.
And what will be the first frame first frames?
It will be the same again because as the first joined, the rotation is about the Zetec says it will
not change.
It will be that one, but X1 will be change and it will change in this direction as the rotation is
like that here.
As the rotation is in this way, the X1 will become like that.
As you can see, the difference between them is nothing but theta one.
So we have to make T to one of rotation in order to match x zero and x one.
So let me write Theta one here.
So what about Frame two or how much we have to rotate frame one with respect to Z volume in order to
match X two and X1 as joint two is prismatic?
It will not provide any rotation.
So X2 and X1 will be always parallel to each other, so we don't have to do any rotation yet.
This will be zero two.
So they are on the same plane and as they will not, they never leave the same plane.
You will never leave this plane because the joint is prismatic.
It's not.
It doesn't provide rotation.
They will be always parallel to each other.
OK, now let's come for the last one in the last one, also it's prismatic joints, so how much we have
to rotate frame to with respect to the two in order to match X to X three?
We don't have to do anything because it is the prismatic joint, so they will be always parallel to
each other.
We have finished with data.
OK, let's continue with the what was Dee Dee was translations through that I'm on this one to merge
origins of origins of sequential or subsequent frames.
OK.
Um, let's do that.
Okay.
How much we have to move.
Origin of frames zero.
OK.
In the direction of z zero.
Yes.
Or the direction of Z zero is upward in the direction of Z zero.
In order to match it with all this origin, how much we have to do.
As you can see, the offset is once, so we need to do it.
And one.
Yes, it was easy.
So this split me.
Oops.
One.
OK.
Let me clear this one, this one.
So for this, uh, second frame, how much we have to move the frame?
OK, this frame so frame one in the direction of Z one, which is also upward in order to match it with
the origin or frame two.
As you can see, this is nothing but L to the offset is able.
OK.
People write it here L two and I want to remind I want to say one thing not once into you.
As you can see, L2 is not only the length of the robot manipulator, for example.
Here L1 is constant because this is a rotary joint, but in L2, you have to consider also the two.
Yeah, because they're the prismatic joint moves upward and downward.
And there is a link connected to it.
The links of the link and these together, which is the two, is coincided at L2.
Yes, OK, I am writing you by considering both of them.
OK.
So while in the first joint, I mean, in the rotary joint L1 is constant in a prismatic joint, L2
is not constant because of the two changes yet.
So no, let's go for the last one, for the last one, as we have said, how much we have to move the
the frame to in the direction of the two, which is in this direction in order to match two with origin
to the origin.
Three.
And as you can see, this is nothing but L3 again in L3 considers the three prismatic joint movement.
Yeah.
Okay, perfect.
Now, as we have done this one also, we have finished with D.
Also, let's go to parameter eight now.
How we will do parameter h parameter h was translation through X arm to match the origins.
Yeah.
So let's go for the frame one how much we have to move the original frame.
Zero.
Yes.
In the direction of X one, the direction of X one is in this way, yes.
In order to get that, in order to match the origins of frame zero and frame by frame zero and frame
one, as you can see, we don't have to do anything in this direction yet.
We don't have to move them in this direction.
We have to move them in this direction, but not in this direction because there is no movement in this
direction.
There is no offset in this direction.
They are, um, they don't need to be displaced in direction of x one.
So we it is zero.
Yes.
And let's two for water.
For, um, yeah.
Uh, for the second joint or the second frame.
Look at how much we have to move frame one in the direction of x one.
Excuse me, X2, which is again in this direction in order to match the original frame to frame one
with original frame two.
As you can see again, we don't have to do anything because there is no offset in this direction between
these two.
Origins of these two frames.
And this is also zero.
And for the last one, as you can see how much we have to move frame in the second frame origin in the
ex-city direction, which is again in this phase of of the page OK x direction in order to match the
origin of frame to with original frame three.
And as you can see again, there is no offset between them.
OK.
There is.
Offset in the $2 addiction, but there is no offset in X to, excuse me, x direction, OK?
So it will be also zero and the last parameter for IFR rotation around excitement is it ie minus one?
And is it all right?
So let's do this one.
Also, when you have to be careful a bit with the rotation parameters because you need to take the joint
rotations, yes, you have to visualize in 3D.
So OK, that's in a fast way to this one.
Also, so how much we have to rotate frame zero frame zero years frame zero over which axis over x one
axis yes.
X one axis um, in order to match z one zero zero.
As you can see, there is no need to match because it one two zero zero because they already match even
the robot rotates to two degrees in this way.
Yeah, this is, excuse me, two to one degrees.
In this way, they will always get one and the two will always match with each other, OK?
The X, the x one and x zero will be different.
Excuse me, this is not is that this is a zero four x one and X0.
As we have said before, there will be um, there will be offset between them.
OK, there will be a difference between them, but there will not be anything different between that
one.
And it's, you know, they will always be parallel to each other.
So and it will be the first and there will be nothing but zero.
So what what about the second frame in terms of second frame, how much we have to rotate frame one
in the direction or so through also now through the X two OK x two axis in order to match that one?
And is it two OK, OK, as you as you can see, we are rotating in x two direction.
Yes.
And so in the external direction, the rotation angle is positive.
In this way.
If you put your finger in extra direction the same before extra direction and curl your fingers, the
rotation positive rotation is this way.
So however, in order to match it one with the two, we have to rotate frame one in this direction.
Yes, along x x tour in order to match is that one and the two?
This is 90 degrees because the difference between that one and the two is 90 degrees.
But however, it is opposite to the positive direction, so it will be minus PI over two.
OK.
Okay,
so let's see what will happen for the last joint.
OK, last frame.
So as you can see, there is nothing to do here.
Because why?
Because the frame we don't have to rotate frame two in x three axis in order to merge the two ends its
three because they all already match with each other.
And as the Z two, excuse me, the joint two of the three assert joint is nothing but prismatic joint.
It will not provide the rotation, so the Z two and the the three will always be parallel to each other,
so we don't have to do any rotation.
As you can see, we have obtained each parameter table for this cylindrical part, OK, for part, and
from here, what we can do from this is using these parameters, I can find this edge one zero.
Yes, I can find H2 one from here and I can find it three two from here.
So how can I find the h, uh, uh, zero and three?
So the relation between this frame, the base frame and the end of the frame, how can I do that?
I can do that by multiplying each one zero by age two one and this by age three two.
Okay, this will give us age zero three.
That's perfect.
Let's go with the second part, which was cylindrical arm, a wrist, OK?
And as we have seen, the cylindrical wrist is nothing but the combination of steering a joint rotational
joints.
OK, now let's try to find the
the parameters for this also.
Okay, first, let's assign frames.
OK.
We will try to assign the frame for it.
OK, let's do how we can.
First, we will choose.
Again, does it access this and as we have said, that exercise will be in which direction that access
is will be in the direction of motion, OK?
And also for in this case, then it will be in this way.
This will be that one and four.
In this case, this will be what is it?
Excuse me.
Let me start from before we start from not one, but instead it will be three.
Why?
Because anyway, this will.
This reset will be connected to the end effector saw these two friends will coincide with each other.
As you can see, this third frame instead of this third frame, there will be this frame.
So we are right.
And is it three two here?
Is it four to here, Z five two here and the copy of the last one?
It will be that six here.
Perfect.
Now there's an interesting scene here.
As you know in, though, it is interesting seeing is that choosing the origin for the frames if the
we can choose the first frames origin as we want.
OK.
We have said this one arbitrarily.
We you can choose.
OK, let's keep it.
Then this joint destroying this frame like that.
OK, for now, we will not touch it.
But what will happen for this joint, this frame?
OK.
In order to find its center, we have to make this in this way and we have to make this in this way
and its intersection will be the center, OK?
As you can see, this is the same to the point.
So.
Okay, let me make this again back, go back, think OK in this way.
So the center will be for the fourth free will be here.
Let's check for the fifth frame how we will find for the fifth frame.
We will again let me do it in this way so that the data will be easy.
We will make it longer like that.
We will also make it longer life.
That intersection point is again here.
So we have to bring this frame from here.
We have to take frames from here, OK, and bring it to back to here.
Yes, because its center will be here.
No, I think it's much clearer to you how we find the same.
OK, so what about the Z6 in order to find the center of Z six?
Let's take it also.
Let's make it longer.
And as you can see, it doesn't intersect with Z five.
It's parallel to that.
Five So we can choose as it seeks alongside the frame of the original frame frame six alongside the
is it six OK like distortion in any point and in order to make our job easier, as we have said before,
we will choose it to stay here.
OK.
However, in order to make the calculation of the H parameters much more easier.
What we will do, we will take this frame.
OK, the first frame, the short frame, I mean, OK.
And we will, as we have said, we can assign it center arbitrarily.
It will take it and we will put it also here.
Yes.
This will be also in this way.
This will be the Z three.
OK.
As you can see, this is now how we assign the D.H parameters.
OK, now let's calculate, excuse me, how we assign the frames.
Now that's calculated the Oh, I have forgotten the x axis.
Now that's the x axis.
OK.
So as we have said, and let's assign, does it exist for the, um, third frame because it is the first
one and we can assign it arbitrarily.
As we have said earlier, this is x three.
Let me go that this one, I will put it here.
OK, this is our thought.
Look at yeah, for itself.
Or indeed, yes, it's in the therefore.
Yes.
OK.
So this is extreme.
OK, what will happen with X four?
Then X four has to be in such a direction that it will intersect X3.
OK.
So we can g we can use this direction.
We can put it also in the direction of X three in order to minimize the changes.
So a calculation of the H parameters will be easier for us.
And as you can see, X four will intersect z three axis and OK, which we which is required for the
H parameter assigning frame assignment.
OK.
And we can do the same.
Indeed, we can choose the same direction for X x five also y because x five have to, it has to intersect.
Z four OK.
And if we choose it, it as a as part of the page in this way.
Yeah, the same way as you can see, if you make it longer, it will cut again.
Is it for Intersect at Fort, which is required for the frame assignment and we can copy the last frame
for them, as we have said.
And so if we have a, we can copy X five to here in order to get X six.
OK, now we have frames assigned, we can easily calculate the H parameters.
OK, let's calculated the H parameters.
Let's start with to.
So Peter, was this one how much we have to rotate frame?
Three.
Yes frames.
Three.
How much we have to rotate frame three.
In our view, in so and through Z, the three, how much is through that?
Three.
In order to make X three and x four to match each other.
So we want to match X three and four with each other.
Yes.
And we have to do a rotation in which along which direction is it?
Three.
Yes.
So let's see.
Um, the first time x three and X forward seems that they are match with each other.
Yes, they seems that they are matching with each other, however.
You have to take into account which rotation the angle of the again, you have to take the angles into
consideration.
So four x three let me at right angles.
Also, I forgot to put them x three.
The positive rotation will be in this way for this one.
This will be five this one.
This will be three to six.
Yes.
This will be Theta six.
Perfect.
No.
So how much we have to rotate is so are if there is a full rotation.
Yes.
Oh my gosh, what would I have done?
Put it for the x axis.
I'm sorry.
I forgot.
Okay, they they should be x axis.
Yeah, that was stupid mistake.
So for us, it's three.
It will be in this way.
So the positive rotation is in this way three to four for Z.
For the positive direction will be in this way T to five and for Z five, the positive direction will
be again in this way.
If this is Theta six.
Yeah, perfect.
No.
If the if the first joint will rotate, if you watch from for direction, you will see that, um, it
will be in this way and this will be.
So we try to match X three and explored the there will be a rotation between X three and X4 because
X three will come like that.
Excuse me, x four will become a will become like that and X3 will be in this way.
Yes.
And in order to match these together, in order to merge this together, we need, as you can see,
two to four angle.
OK.
We need data for angle.
However, we have.
So our first key to this will be data for so we need in data for rotation in order to match x with X
four.
That's perfect.
Now let's go with, um, table two.
OK.
And in order to find it.
So the title parameter for the second frame.
So how much we have to rotate the frame one with respect to an excuse me, I say frame one.
But as we have said from frame three, how much we have to rotate frame four.
Yes, with respect to Z four, in order to match X4 and S5, as you can see again, you can see that
exponent X-File are the same, but you have to take into account the rotation.
OK, let's take into rotation account.
And in this case, if joined this joint, OK, which is the second joint rotates, OK, if it rotates
theta five degrees, what will happen if it rotates to five degrees, then this will happen.
OK.
Um, the X for X four will be in this way.
But X five will become in this way.
OK, x five.
There is angle of three to five between them, so we have to rotate in positive direction.
Among the how long is it?
Four in order and three to five degrees in order to match x for with x five.
OK.
So we have three to five here and in the same way and four in the same reasoning.
You can you can understood that this will be a six in order to match with in order to match what in
order to match x five with x six year excuse me.
X five with a six book and rotation will be among which axis.
It will be a moment alongside a Z five access through Z five axis.
Yes, because if frame if joined in, if this sort of joint rotates by an angle of T to six, then as
you know, the X x five will be in this way.
X five will be in this way, OK?
But again, X will become like that pick six.
So there will be three to six angle between them that we have to consider.
OK, if we consider this angle, also, it will be status.
Six.
OK, I want you.
Um, if you cannot, if you couldn't catch it, please stop and go step by step by pondering and trying
to imagining.
Yes, everything.
It's very easy.
You have to just watch it by step, by step and.
I want you also drove it in your paper and tried to make you try to follow me in each step.
OK, so we have finished with Theda, then we can go with the parameters and indeed this translation
problem.
This will be very easy for us.
They will be all of them will be zero.
I know from where surely I know it because all the frames are out on top of each other.
See, it makes it easier to do the calculation.
So what was?
But anyway, let's see this.
What was d d was that how much we have to move if the centre of frame three OK in the direction of Z
three, which is like that, OK?
This is like that in order to match origin of three and fours frame.
OK.
As you can see, their origin matches with each other and there is no prismatic joint that will separate.
These make distance between these two sensors.
So it will be zero.
OK.
So the same way how much we have to move the origin or frame forward in order to move out of the frame
for in order to match it with the original frame.
Five Among which direction z for the direction?
Yes.
As you can see, there's four directions upward and we don't have to do anything.
OK?
Because there is no offset in this direction between these two centers, so it will be zero and the
four the last one, how much we have to move.
Origin of frame five.
OK, which is here.
Origin of frame five in the direction of Z five.
Yes.
How much we have to move the order in the direction of five in order to obtain in order to make what?
In order to match that five vs six.
Excuse me, origin of five.
Origin of six news how much we have to do that.
As you can see, this is nothing but the sum of L5 and L6.
Okay, these two things, because the offset in this dilution in Z590 chain is like that.
So we have to ride these two five plus IL six.
OK.
We have to move like that.
OK, now let's.
Oh my gosh, don't do that.
Yeah.
By clicking OK.
Not necessarily.
Yeah.
No.
Let's do this one.
We have finished with the parameters.
OK, now let's continue with parameter e OK for parameter A. What we will do for parameter A. It's also
translation parameter how much we have to translate the center of frame three.
Central Frame three.
OK, here center of frames three.
In the direction of which direction x four by addiction.
OK, in the direction of x four, which is in this way in order to match the origins.
OK, and the original frames leave its original frame for.
And as you can see, there is no need to do anything because they are on top of each other in the way.
OK.
This is zero.
And again, what we will do for the frame and the other frame, OK, how much we have to move the original
frame forward in the direction of X five, OK x five, which is also in this direction in order to match
the original frame for with original frame.
Five.
As you can see, there is no offset in the direction of X five.
So we don't need to do and it will be, but it's certainly OK.
It will be zero.
The last one for the last one is, as you can see, how much we have to move in the direction of which
x six direction, which is like that in order to match the original frame five, which is here with
the original frame six, which is here.
As you can see, there is no offset in the direction of X six.
There was an offset in the direction of Z six, which was Influentials six, but in the direction of
X six, there is no offset, so it will be zero.
OK.
As you can see, much of the translation parameter is R zero.
OK.
Because they are, um uh, on top of each other.
OK.
And the last parameter, which is rotation.
OK.
I find how much we have to rotate the frame.
Uh, three.
OK.
Uh, alongside so through X in the direction of X forward in order to match what Z is three and is thought
OK.
As you can see, there is an offset already between the three and Z four.
OK, and this is how much this is 90 degrees.
Yes, Pi over two.
But we want to know if it is positive or negative.
So as you know, we have to rotate, we have to rotate which frame sort of frame in order to match the
Z four.
So we have the angle of rotation is PI over two and it should be in this direction.
But in a alongside which axis, scroll through which axis we are providing the rotation X for and alongside
X for direction.
If you put your index for direction and curl your fingers, the positive rotation direction is like
that and we are turning in this direction here, as you can see.
So it will be positive.
If it was like that, the rotation, it was like that, then we would take it as negative.
OK.
So the first one is positive pi over two.
That's perfect.
Now let's check for the other one.
What will happen?
Let's say for the other one.
Cool.
OK.
So how much we have to move the frame, how much we have to rotate frame for with respect to yes x five
in order to match.
Z four and a Z five, as you can see it again, there is an offset of PI over two between the four and
the five, and we have to rotate Z four in this direction.
Pi over two.
In order to match.
Is it fast?
But let's again see what is the sign of the rotation via rotating alongside x five axis, which is,
as you can see, the positive direction is like that, and in this case, as you can see, our rotation
is opposite to this positive direction.
So it will be minus PI over two.
Yes, it will be at minus PI over two.
Perfect.
We have finished for this one also and the last one, how much we have to rotate frame five in order
to match in order more alongside execs.
OK, along six axis in order to match.
Is it five with.
And as you can see, we don't need to do anything in execution because of these five and six already
match with each other.
OK.
It will be so zero.
We don't have to do it in the rotation.
OK.
As you can see, we have finished our the table for the cylindrical vest also.
But what we can?
Let me just change these numbers because these are not correct.
This should start from I'm sorry, this should start from Portlaoise.
This should start from Ford this far.
This should be six.
So this first one will give us what this will give us h four.
Three.
This will give us age five four and this will give us Page Six.
Five.
OK, now how can we find the relation between the center of this frame, which is here and this one?
OK?
We have two front Page Six three, which will be given by H four to three times.
Page five four times Page Six four.
OK, now we have in our hands h x age six three.
OK, and we have eight zero three.
Let me make it in this together.
As you can see, we have eight zero three in this part and we have eight six three in this part.
And for this robotic manipulator, in order as we have seen to find in order to find the relation between
these base frame and the fact that we have to find what Page six zero, because the frame attached to
this one is six frame splits.
Let's draw the frame.
The frame was a six frame.
Yes, this was it six.
And this was x six.
And this was.
Is it zero and this was X0.
Yes, that's perfect.
Now, in order to find the relation between these two frames, we have to find age six zero.
And how we will find age six zero we have in our hands.
Age three zero.
And we from this is from a.
And we have in our hands eight six three.
Yes, which is won't be.
If you multiply these together, I will get the relation between age is six zero and this is our forward
kinematics for this cylindrical robot arm vs spherical joint, which is very important.
Yes.
Keep in mind that they always try to remove the spherical restock to complete her, and he can't wait
for the age parameter.
Um, for this list and for the robot and then combine them in this way.
OK.
And also, let me say to you one thing this point for the spherical wrist is called wrist center or
wrist point.
OK, this is the wrist centre or wrist point.
This is called for this vertical wrist.
That's perfect.
I think that's enough for it.
As an example to the parameters, we have seen many things how to calculate it.