In the last lesson, we have seen how to assign two frames to the robot manipulator in the convention.
Now we can, after assigning the frames in the correct way, we can calculate the H parameter table
easily.
So what are the parameters that we have to calculate?
There are four parameters to calculate for each of the joints.
Yeah, this is due to the AI yet to AI, the AI for AI and A.I., so too does shows how much trotted
along zit AI minus one axis in order to align x AI minus one with inside the AI shows how much the trust
leads alongside AI minus one axis to align or AI minus one with or AI.
What's all?
All means origin or frame AI minus one with respect or a frame central authority or frame arm alpha
is how much to rotate along x y axis.
Be careful about axis along x axis to align the timeline this one does.
The tie and the air shows ha shows how much to translate our exposure to align or ie minus one and or
eye.
As you can see, we have two rotational parameters, which is the high and III for I am two translational
parameters, which is to the excuse me and a I.
OK.
And the order of calculating these parameters is we will coordinate first one, then second, then this
one.
And I have written here first, but these should be three, then this should be four of them.
So then hopefully we will use these parameters in order to find the relation between the frames consecutive
frames, OK, and how we will describe the relation between consecutive frames if this is using homogeneous
transformation matrix.
OK, how we will get homogeneous transformation matrix.
We have four parameters to translation um two rotation.
So we first calculate TITA, which is rotation around Z-axis.
Yes, then we have translation, which is the eye along the z axis and translation again along X and
the rotation along the X I degrees.
So we can write this for these mattresses as a homogeneous transformation matrix.
In this way, as you can see, the first one is what the first one is just rotation matrix so that this
it's transitional part is zero.
Excuse me, there should be one here, OK?
Yeah, it's a transitional part is zero, but it's a rotational part.
Is rotation among a lot of Z-axis?
Yeah, because z rotation matrix with angle of Tita I.
The second one is translation, so its rotation has to be identity because there is no rotation.
But there is translation along the z axis with the eye, as it can be seen here.
Then we have again translation along X with a I am sorry there should be zero here.
OK, as you can see, we have translation along x eight amount and the rotation.
There is no rotation, so identity perfect.
Then we have rotation along x OK around X with a degree of four eyes.
So we have rotation.
This is x rotation matrix, as you can see with degree of over ie and there is no translation.
OK.
And at the end, we get our homogeneous transformation metrics, which indicates the relation or relation.
Let me draw the relation between and between frames I and I minus one OK, I and I'm minus one.
Um, so as you can see here, h I, I mean soon after we calculate this for each frame.
OK, then for each joint, excuse me, then we can multiply them and get the relation between any desire
to frame OK.
We will see more clearly.
So let's go to the calculation of the homogeneous transformation.
Excuse me, the parameters before doing it.
I want you to be careful about one thing.
OK.
Try to imagine everything in 3D.
OK.
Finish with 2D.
There is no to the.
We are working in three dimension.
OK?
You have to use your 3D.
The imagination and here is what this is here, I want to just last time I showed you how we can choose
the rotation.
OK, rotation direction, but how we determine the rotation direction.
But it is more clear picture again, as what we have said, even rotation is, for example, in this
direction, OK, we curl our fingers.
As you can recall, our fingers in the direction of rotation and our sun by automatically shows us the
positive rotation.
No, that's OK.
Because the rotation that if it shows us the rotation direction, OK, axis of rotation is OK.
So we curl in the direction of rotation and this show's axis of rotation, our sun.
That's perfect.
So let's go to this calculation of parameters.
OK, first, we will start with Tita Eye.
OK, let's go with the first joint.
OK, excuse me, first frame rate was Tita Ed.
I'm here.
You can read again.
Tita is how much trotting along is it?
I mean, this one.
Be careful along the time minus one we rotate.
We are rotating in order to align x minus one with X.
OK, perfect.
Let's see how much we have to return frame zero along zero zero.
OK, because we have seen the time minus one, so we are co-creating for the first one, as you can
see here.
One we are calculating for first one.
So it should be along z zero zero minus one set minus one minus one, which is at zero.
OK.
And B how much we have to return it turn, excuse me, or rotate the frame zero in order to match x
zero width x one x zero x one.
OK.
First, let me not wanting to say that, as you can see, I have removed the y axis because they will
not.
You will not need them in order to calculate the parameters so y to make complex problem complex by
adding them also.
No need to add them.
Just it is enough to write only the the X and said accesses and then by L I show, as you can see,
this purple color l are nothing, but
they show just the link lengths.
OK.
So let's continue.
Let's continue, OK?
You can see that.
You can see that, OK, there are in one plane, OK?
So we do not need to rotate frames zero a manga, excuse me, along the Z zero in order to match x x1.
They already match.
But it's not correct because we don't take into account rotation.
There is rotation of, as you can see, T to one.
So in order to be to know it much better in order to visualize it much better, let's look at the robot
from upside.
OK.
This is our first joint, OK?
And then let's throw its axis is, as you can see, this is nothing but z zero.
This is nothing but x zero.
And if we rotate it, tita yeah.
Excuse me, not like that.
But if you rotated like Tatooine, then the second joint will be in this way from the top side, OK?
And it's it will be rotated like that.
So the X1 will be like that and the Z1 will be like that.
As you can see now, there is an angle between
now there is an angle between the zero, excuse me, zero and X1.
OK, sorry for the delay.
There is a delay.
I think I I'm because projecting the iPod screen to the computer, there is a delay.
So as you can see, there is an angle between between the zero and X1.
If you Project X zero here again, you will see easily to one angle.
OK.
Zero.
You have to take into account the rotation and you have to look at from upside of it or if it is needed
from the down side, or even if this is needed to watch from their sides.
OK, but you have to take into account a rotation and also translation if needed.
OK.
So we have seen that we, you know, we have to rotate frame zero one alongside Z zero in order to get
on with the.
Spectre's did zero in order to match teeth, one angle in order to match zero is x one.
So it's right, then here T to one.
OK, but let's go for the second point.
OK, let's go for the second one and see what will happen again.
For the second joint, this is the same.
How much we have to return frame one.
In order to match X, I'm on a longer through z one axis in order to match x one with what?
In order to match x one with X two OK.
In order to again, we have to take into account rotation because at first you can see that OK, there
is no need to rotate because they are on the same plane an x one, an x two parallel to each other.
So there's no need in the rotation, but it is not correct.
You have to take into account these rotation angle of 2.2 also.
So if we draw the robot again from top, for example, this is the joint OK from topside.
This is x one.
This this is z one.
And we rotate it as he thought to OK.
We have rotated into two, so our transitional joint become like that from the top.
OK, from the top side, that's perfect.
Now its frame will be like that again.
OK, if we project this is X to this is excuse me, I didn't do that in this.
That X is the correct way.
OK.
The z axis should be in this way.
OK.
This is it, too.
As you can see, if you project one again here, there is little tool angle between them.
We have to take it into account, so we have to rotate the frame.
One.
OK.
Along Z one, excuse me, sort of in in the direction of Z1 in order to obtain you on the other to make
X1 coincides with X2.
That's perfect now.
So let's write it.
Let's first erase all of these and then let's write it here.
OK?
We have to rotate by 2.2.
That's perfect.
No.
Let's go with the frame.
Three.
Now, what should we do?
What we know in order to find it three.
It shows its meaning is that how much we have to rotate frame two in order and more in the direction
of Z tool in order make X2 and X3 coincide with each other.
OK.
As you can see now, in this case, there is no translator, excuse me, rotation, but translation.
So what does it mean as there is no rotation?
The X2 and X3 will always be OK.
We always coincide with each other because this joint will only up will move up and down.
So X2 will not move with respect to X3 in terms of the two in the two direction, OK?
So we don't have to rotate frame two in the direction of Z two in order to make X2 and X3 coincide with
each other, they anyway coincide with each other.
OK, so this will be zero and you can now see that y always we try to make the rotations change as small
as possible.
We have seen in the last video that I have said that tried to make axis is coincide with each other
as much as possible.
OK, tried.
If it is possible, choose the consecutive axis is the same with respect to each other.
So the H two calculating the H parameters will be easier for you.
OK, so now let's call good for the last one.
For the last one, for the fourth.
OK.
What it will be again is the have rotation angle of three to four.
We have to take it into account.
So this means that it often means that how much we have to rotate Keita's excuse me, frame three in
the direction of Z three in order to make it three and X4 coincide with each other.
At first, they can be seen that they are on the same plane, so there is no need to rotation.
Excuse the monitor rotation, but take into account little for rotation angle.
Be careful the robot will return when it tries to in the column to give them to eat.
So X3 and X4 will not coincide with each other, so we need a titov for.
Station and all these rotations will be a plus.
Why now?
Let me just explain over to four, and you can apply the same for others angles because if you try to
rotate it in every frame tool, we have to rotate it in this way in order to match X3 and X4.
And in this direction of rotation, if you call your fingers in the direction of rotation, as we have
shown in this picture, your thumb will show this direction, which is positive.
That's the direction, OK?
So the rotation angle will be positive.
If otherwise, if we would require to rotate it in this way, OK, in order to match X and export,
then our son blue the point in this way.
So rotation axis would be in this way.
The rotation direction would be in this way.
So the rotation axis direction would be in this way, which is opposite to the Z three, so it would
be negative.
OK, be careful about that.
OK, so let's continue.
Let's continue.
We have finished 48th out.
Let's continue with the let's see what what this deep.
These shows us how much to translate.
OK, this is not rotation along the Z Eye, minus one to align origins of two consecutive frames.
OK, let's go.
Let's go with the one.
OK with this one.
This means that how much we have to move this order, Jim.
OK.
Over the z zero axis, over the z zero axis, OK, or the z zero axis.
In order to make this origin OK to match this origin, do we have to do something?
Surely, no, surely not.
Yeah, we don't have to do anything because they in the way coincide with each other.
OK.
And these joints don't move like that up and down, so we don't need any uh, we don't need to translate
this orygen in the Z zero direction.
So which is up or minus zero direction, which is known in order to match it with Z one because they
will on anyway, they anyway coincide with each other.
There is no need for translation again.
So the idea would be zero for the first one.
So let's check for the second one, OK, for the second one, is it the same?
How much we have to translate origin this origin alongside that one?
OK?
In the direction of Z five in order to match this origin with this origin, as you can see again, there
is no need to do that kind of thing because this joint is rotational.
It will not change.
I won't say it will not change in a transitional way.
I mean, up or down.
So the origins will always coincide and we don't need any kind of translation.
So this will be also zero.
OK, perfect.
So let's check for the assert frame, OK?
How much of this have to translate or origin of this joint in the direction of Z two?
So which is done in order to match this origin?
OK?
Origin of frame two vs origin of frames three.
Note zero or note.
Surely it is not zero.
OK.
Because in the direction of the two, which is.
OK, we have to slide this year.
We have to make it to come to here and how much this does.
We have to make it to come.
This is L.
Three.
Yes.
We have to make this origin to come here L three amount in order.
Um, these origins to coincide with each other.
Perfect.
So that's right.
It also so this will be L three.
OK.
This will be our.
No.
Let's go with the last one.
With the last one.
What we have to do this is our origin.
This is our origin.
So how much we have to translate the origin of frames three in order to match?
Yeah, just frame.
Three.
In order to match it with the original frame forward.
Well, in the direction of what Z three do, we have to do something?
Yes, surely we have to do.
We have to translate because Z three duration is Don and we have to slide it down in L four amount in
order to match the origin of frames three of its origin of frame for yes.
So this is L four.
That's perfect.
OK.
As you can see, it is very easy, you have to have just the imagination of 3-D.
Let's go with a walk in the is a as you can see here, because the sorts that this is sort of parameter
we have to calculate is that how much we have to translate again, translation along IXI, I'd be careful
along.
Not that I might miss one in order to match again the consecutive frames origins.
Let's check with the first joint book how much we have to translate alongside x one.
OK.
Alongside what X1?
Because it's alongside, it should be in direction of ex-spy, which in this case is one so x1 in order
to obtain in order to make the origin of frame zero.
Yeah, coincide with the original frame one how much we have to do.
As you can see, the X1 direction shows this way, so we have to translate it in the direction of X1
and one amount.
Yeah, in order to match these two centers.
So this has to be L1.
OK, that's perfect.
This has to be L1.
That was easy.
Now let's go with the circle joy into how much we have to translate this center in the direction of
X2, which is to the right X2 in order to match origins OK with each other.
So original frame one with origin of frame two how much we have to do.
As you can see, this is L2 because the offset in the direction of X2, so in the right direction is
nothing but L2.
So this has to be L2.
Let's write it again.
That's perfect.
So then I can make it in this way.
Oh my gosh.
Why you're doing it in this way?
Yeah, it's very, very precise, I think.
Yeah.
Anyway, let's continue.
OK.
So we have done this one.
OK, we will go with the third joint.
OK?
The third one saw how much we have to move the origin of frame two in the direction of X three.
Yes, in the direction of X three.
In order to match X to another to merge these origin.
Three original frame three with, excuse me, ordinal three and two with alternate frames.
Three.
And as you can see, if we don't have to do any translation because X3, the addiction is in this way,
we don't have to translate this in this direction in order to match it, because in this direction,
over this duration, they all rate the match with each other.
OK.
That's very obvious.
So let's let's try it.
It's zero.
We don't have to do any translation.
There is nine offset.
So the last one, how much we have to translate the original frame three in order to match it with the
frame of the original frame forward in the direction of X four, as you can see it again in the direction
of X squared, which is like that, we don't have to do any translation.
There is no offset in that direction, so it will be zero.
Let's write it zero.
So as you can see, the last one became zero.
So now we the last is over.
OK.
The last parameter is alpha, which is how much to rotate along x y axis.
OK?
Be careful along x axis in order to match z i minus one to.
I.
So let's do that.
So how much we have to rotate frame zero in the direction of X one?
OK, in the direction of X1 in order to match.
Is it zero bit?
Is it one?
Let's see.
Should we do something like this one?
No, because in this is it zero and this is it.
One way matches with each other, even though there is little one rotation.
This will not affect Z zero or that one because the rotation is about z axis as OK.
And if you watched from the top side, is the for the first after the rotation, this is a zero zero
zero.
This is zero after the rotation of T to one.
OK.
This will be from the top side.
This will be x one.
OK, this x1.
But again, z axis will be like that, OK, because it will not rotate.
OK.
So there is no need.
To rotate a longer election in X One Direction in order to match Z zero with that one, so let's write
it, it will be zero.
Yes.
So the first one is zero for the second joint.
Let's check for the second joint how much we have to rotate in, of how much we have to rotate frame
one in order in the direction of X two.
In order to match Z one.
Read it to look at that one with Z.
Should we have, should or should we rotate?
Indeed, yes.
As you can see, they are opposite to each other.
And again, the rotation as the rotation is with respect is with respect to the z axis, even interpolated
to rotation and the directions of Z one and Z two, even after the rotation will be opposite to each
other.
The ratios of X2 and X3 will change on the X2 will be in this direction and excuse me and extend X1,
X1 will be in this direction, OK, after rotation of the rotation of T2.
However, Z one and Z two are how much we are, how much we have to rotate frame one.
We have to rotate frame one in this direction, OK, if we take in in extra direction, OK, if we take
an external direction, we have to rotate it.
How much degrees?
If you put your put your thumb in the direction of X2, OK and rotated, curl your fingers.
Once the positive direction of rotation surely seen this way, should we?
And in order to rotate the Z1 in order to rotate frame in order to match that one with the two, we
have to rotate the frame f two with, excuse me, a frame f one with respect to X2 width.
How much?
Um, minus excuse me, plus peak angles.
Yet because they are one hundred eighty degrees from each other and the rotation is a static rotation,
so always will be.
They will be apart from each other by 180 degrees and the rotation is in the positive direction.
OK, we have to rotate in positive directions, so let's right it.
The rotation will be a plus pi.
Let's write it OK, the rotation will be plus pi.
OK, then let's do this one again.
Let's do this one how much we have to rotate, how much we have to rotate.
Yes.
Frame two with respect to Axis X three.
In order to Match Z two with Z three, as you can see Z two and the three already matched each other.
And as this is prismatic joint, there is no rotation that we have to take into account, so we don't
need any rotation.
They will always match with each other.
So we need zero rotation and the last one for the last one.
Again, how much we have to rotate frames three along x in the direction of export in order to match
Z three, we visit for Z three and Z four will always match with each other.
Why?
Because the rotation for the for this joint, OK for the first joint will be always in this direction.
Yes.
So this ring will not change its direction from its launch from top side.
This is extra.
Yes, this is Z three.
Yes.
And the end effector.
Even if this joint rotates, the India factor axis will be the same as this one here.
It will not change.
It will be Z for its x axis will only change.
Yes, this will be X4 will only change by how much to the for.
OK.
Think about it.
Be careful about this.
Okay, so we don't need any rotations.
It's V and Z Fold will anyway coincide with each other, so we need zero degrees of rotation.
That's perfect.
We have finished with the D.H table again, so you can understand it better by.
If you missed something along side, please draw it and by stopping and playing VIDEO again, OK?
Step by step, try to follow me so you can understand and imagine everything it is indeed.
Very easy.
Not that if.
A gold, if you can imagine in 3-D.
Okay, this is over the edge table.
OK.
And we will use each of these entries, OK in order to calculate a homogeneous transformation mattresses.
OK.
And then to get final, how can I say a final relation between the base frame and in the fact that a
frame?
Let's see how we can do that.
We have seen the formula for the M between H and H by minus one.
Yes.
You just think this is c t de I.
Maybe I have to write it.
I forgot to say it.
C Tita I means question Tita to yeah, if it is also I, this means question for I.
OK.
And s tita I means the same thing, but with seamless theta pi.
OK?
And if it is alpha I, this is since I I.
OK, perfect.
So you just need to take Tita.
I I for I am a I and D and just plug in this in this formula.
Yes, and you will get H I and I mine this one, for example.
We will not do it.
We will not plug it because you can do it easily in MATLAB.
OK.
But I will show you the the procedure for the first one.
You take it and you've what you've given I for the first one by putting these parameter, these each
of these parameters in the h i i minus one, if I equals to one.
Yeah, for the first one, but we will get we will get h one zero.
OK, this will give by just plugging in these yeah, these OK parameters you will get in this formula.
OK, in this final formula, you will get H.
Well, zero.
OK.
As you can see how easily you get it.
This procedure can be automated easily.
OK.
Yeah.
If you take the second one, so I equals the two, what you will get, you will get h two one four.
I equals to three.
You will get h three, two and four.
Pi equals four.
You will get h for three.
As you can see, you have the relation between each frame.
OK.
Between each consecutive frame.
OK.
For example, no, you want to know what will be the position of and affect the position and orientation
of and the vector with respect to the vote, with respect to that base frame so that you want to frame
the you run to find four zero, what each four zero will give for is our and the vector frame of is
zero is our base frame.
This is, as you can see, four is in the vector frame.
And this is our base frame.
You want to know what is the what is this distance between them?
Yeah, this is distance.
What is this?
For example, let's see.
This is the s.
We are interested for this one.
And what is the OK orientation difference between these two?
OK, so and what we want to know are four is zero.
Yes.
What is the orientation difference between?
What is the orientation of the vector frame with respect to the base frame, we want to know of these
two things how we can calculate it.
Surely it's very easily, uh, we will do it by just multiplying what homogeneous transformation methods
in each other h one zero with each h two one with h as well and with age four three.
And this will give you age for zero.
OK.
The end effector frame with respect to their base frame.
OK.
And then it will have in this.
It will have this kind of shape.
This will be R four to zero.
Yes.
Zero and D s and one.
OK.
This is three by one vector.
OK.
This is three by one vector, which is x y z saw.
So it will show this OK in how much the distance between this origin and this origin occurs or and the
vector and the base frame in x y z axis is yes, that's perfect.
And this will show this is four by four matrix.
Excuse me this for my foot.
Well, three by three matrix, which shows us the orientation.
Of falls in the sector of rain with respect to their base frame.
That's perfect.
So but what we have said, we have also said that we can find that any fraying the relation between
any frames so we can find what is, for example, the joint tools.
Yeah, for example, these joint tools orientation.
And but this does with respect to a base frame, you can find it, you can find it easily.
So what we want to find, we want to fight age to zero, how we can find by multiplying just age one
zero with age two one.
As you can see, you can find the relation between any two frames, any two frames, and that is the
best the perfect site of forward kinematics.
And you can complete it using the age correlation D.H method very easily.
I hope that you have understand we will do one more example in order to solidify the concept to make
it more robust for you.
So let's solve.
One more example also.