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GIX MSTI TECHIN 517 Inverse and Differential Kinematics Lecture by Haonan Peng

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This content introduces inverse and differential kinematics for robot arms, building upon forward kinematics. It emphasizes the importance of understanding a robot's workspace before tackling inverse kinematics and explains how to analyze robot workspace, solve inverse kinematics problems using algebraic, geometric, and numerical methods, and delves into differential kinematics, including Jacobian matrices and singularities.

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All right, welcome to the class today.

Today we are going to introduce inverse

and differential kinematics. And this

would be assuming that you are already

familiar with forward kinematics. If you

need a refreshing, please uh review the

materials we have on the forward

kinematics. So uh long story short,

inverse kinematics is trying to answer

the question that if we have a desired

end factor location and orientation, how

can we find the the joint rotations or

joint translations that can achieve the

target or desired any factor poles? And

differential kinematics is trying to

answer the question on finding the

relationship between the the joint

velocities and and any factor angular

But before we move on to uh inverse and

differential kinematics, let's take a

quick look on what is the workspace of

the robot arm first. This is because

inverse kinematics solving inverse

kinematics can sometimes be complicated

and sometimes it can be computationally

expensive. So before we move on to solve

inverse kinematics, we want to make sure

that first the desired manufacturer

location is within the robot workspace.

So we have a solution for inverse

kinematics and then it will worth our

time and effort to move on to solve it.

So we should first look at the robot

workspace first. We should check if the

target location we want to uh we want

any factor to go is actually within the

robot workspace.

So robot workspace is actually is

defined as a um it can be a a space in

3D uh it can be a 3D space it can also

be a um a space in a in a planner uh

view. So it can be 2D, it can be 3D um

depending on how we define the uh the

working environment of our robot. And

there are uh typically two types of the

um of the workspace. The first one is uh

is dextterious workspace which uh means

that uh inside inside that workspace we

we can have our robot uh any factor to

reach the a target location with all

possible orientations

which means that if I reach a certain

location I can actually reach this

location in various angle and and in all orientations

orientations

And another one is of course the

reachable workspace which means that we

can we can still reach

um any points in this workspace but we

might not be able to reach a certain

Let's look look at a quick example here

that let's assume we have a very uh

simple planner robot that we have two

joints and two links. So we have join

one here which is set theta 1 and we

have join two here set theta 2 and we

have two links one is 10 cm and one is 8 cm

and if we assume that uh we have no

joint limits which means that all the

joints they can rotate for uh 360°

and then if we draw out the workspace

it's actually um pretty straightforward

that We will we will have a workspace

that the uh the

the furthest that the the robot can go

is to reach as far as 18 cm. It will be

10 uh cm plus 8 cm which means that we

have our robot arm to be fully uh stretching

stretching

and the closest point that we can reach

is 2 cm. We cannot reach anywhere

within two centimeter because even if we

fully folded back the second um even if

we fully folded back joint two this will

be the e uh this will be the uh the

closest point that we can reach and we

can we can no longer reach uh anywhere

closer within two centimeter.

And we can also see that for this um

workspace all the points inside this uh

yellow workspace they are reachable um

they are reachable workspace because uh

if we want to go to a certain point for

example the point here

it either it will be either we can uh go

through this uh we can have orientation

of this or orientation of this one. It

is not we are not able to uh reach this

certain point with all possible orientations.

And if we further take a small

modification to our robot, now the two

links they are no longer eight and 10.

Instead they are all 9 cm.

And then what we can find is that the

robot workspace now

becomes a full circle that we can reach

all the point all the points inside this

circle and the diameter would still the

the radius would still be uh 18 cm. But

now we can we can reach the center of

the circle at this point.

And we should also pay attention to that

for all the other points for example for

the points

except for the center of the circle

these points they are still reachable

workspace but for this

center of the

circle that specific point is is

dexterious workspace because we can

actually fully folding the robot arm

like this

to reach the center of the circle. And

we can actually rotate

this configuration all the round so that

we can actually reach the center of the

uh center of the circle within with all

And furthermore, if we make another

modification that we uh we now have the

the two links um change back to 10 cm

and 8 cm, but now we've add a new joint

joint three right at the end factor.

That would immediately change the entire

workspace into dexterious workspace

because we can use join one and join two

to reach any point in the robot

workspace and then we can use join three

to change the orientation of the end

factor with all possible orientations.

So adding a join three uh here

immediately change all the robot

workspace into dexterious robot workspace.

And now let's consider joint limits

because for most of the um most of the

robot arms in in real world we might not

be able to have joints that can rotate

for 360°. We might have some joint

limits. For example, let's consider this um

2D planer robot. We have similarly we

have two join. We have two joints and

two links.

But this time

instead of being able to rotate for 360°,

360°,

we can we have joint one that can only

rotate from zero to

180 degree. And the SATA SATA 2 the

joint two can only rotate from 90

negative 90° to 180°

180°

which means that we can have uh joint one

one

to rotate from zero all the way to

180 and we have

join two that can rotate from negative

90. So we will we will often consider

this as zero uh rotation for joint two.

So negative 90 would be here and then it

can rotate all the way to 180° which

means that it can actually rotate to be

fully folded back in this way.

So this is roughly the u the mechanism

of the robot arm. And then let's take a

look on how we can find the robot

workspace of this robot arm.

So this is um the this is how we can uh

solve for it. Uh

there are a bunch of colors uh we should

pay attention to. So if it is labeled as

as blue, it means that um at this moment

this this link or this link is uh is

resulted by we have a we have the joint

uh to be at its minimum um rotation and

similarly if it's showing red it is

maximum rotation and we will we will use

orange color to describe if the robot

arm or if if this joint is actually

fully stretching the next link to be

with the previous link. For example,

um we can see here that um at this point

this blue bar means that we have joined

one to be at its minimum which is zero degree.

degree.

And this blue bar means that we have

drawing two to be at minimum rotation.

And similarly um we have here you can

see at this location joint two is still

in its minimum while joint one

showing red which means that it's as it

it is at its maximum rotation.

And the orange color here it is showing

that um at this moment we have drawing

two to rotate to be rotating as such as

such that the robot arm is fully

stretching. It's is it is reaching the

longest distance it can reach and the

green color shows fully folded which

means that we have the we have drawing

two to fully folded folding the next

link back to the previous link. So

that's how we label it here because

these are these would be very important

situations that we need to consider once

we are uh analyzing

the u the robot arm workspace.

So first we should look at um if we have

joined one to be at minimum and to be at

maximum. Which means that we have joined

one have that that has been minimum or maximum

maximum

and then um

at this moment we will consider we have

draw two rotating. So we can see that if

we have if we if we have the uh joint

one to be at the minimum

position and then we can rotate joint

two. So join two can rotate from

positive

180 degrees. So this is what we can do

the uh for the the range of rotation

that um that joint two can do. So this

will give us the arc that I marked as

red here. And then similarly once we

have the

once we have the uh joint one to be at

its maximum then we can similarly also

rotate joint two from its minimum 90 all

the way to

180. So these two arcs they are resulted

by we have joint one to be at minimum uh

or maximum

and then we rotate join um then we

rotate joint two. Now we can further

consider joint one can also rotate.

Joint one can rotate from here

from zero all the way to

180. And that can result in we are

actually using this

we are actually using this arc to

kind of a scan and paint this space. So

eventually that will result in the two

And eventually we also need to consider

that if we have

joint one at minimum which is here and

join two also

also

at minimum which is here. And then

if we fix joint two to be at its minimum

which is 90 negative 90° and then we rotate

rotate

joint one. So which means that we are

actually rotating this line and it will

draw the arc here.

This red arc will be uh generated by if

we rotate this

this line and with the with a with this

um with this radiance. So you can then

find out that this

is actually the the closest

uh distance that we can reach under this

configuration. And we can actually we

can reach these this area but we cannot

So furthermore, if we uh if we combine

all the previous um arcs that we draw

from the u from from from rotating the

joints then we can the these are all the

arcs that we draw and we can clearly

find out that um

this arc is actually within the uh the

workspace so we don't need it and also

this arc is also So within the workspace

that we don't need it. So furthermore if

we u remove all the unnecessary arcs

this closed shape would eventually be

the workspace of the robot arm that we

uh described before.

Um this is not a this is a even if we

have a clear pipeline on how to draw the

robot workspace. This is definitely not

a trivial task and it actually requires

um a number of practice to be familiar

and to to master uh how do we find the

robot workspace.

But even if we practice a lot, the the

ability that human can uh can just

manually draw the robot workspace is

still very limited. We are not we can we

can we might be able to do it well if

the robot arm is uh is is work is

working in a planer space. But assuming

that we have a huge number of drawings

that and the robot arm can work in 3D

space, this can actually be very very

complicated because we are then not

drawing a a shape on 2D plane. Instead,

we will draw a volume in 3D space which

can be very complicated.

So this conventional way may not be a

very very good uh approach if we are

really working with for example an

industrial robot arm with six joints and

that can it has a pretty large and

complicated shape of of workspace in 3D space.

space.

Um so this would be um this would be the

uh the pipeline and the procedure that

we can uh we can draw the robot

workspace manually.

we need to carefully follow these steps

to u to draw the robot robot uh robot

workspace accurately and correctly. However,

However,

in real practice u as we mentioned for many

many

industrial robot nowadays industrial

robot arms nowadays

it might be very difficult to apply

these methods. Instead a more feasible

way is to use Monte Carlo visualization

or Monte Carlo simulation of the robot workspace.

workspace.

Um by using this method what we

eventually can get will be the point

clause that can represent the workspace

of the robot arm. So this is uh for the

WAM robot arm um that researchers have u

have performed the Monte Carlo uh

visualization of the robot workspace.

And this is what I did for the Raven 2

surgical robot on the robot workspace.

And uh furthermore, I've I've used some

special colors to kind of represent the

the edges of the robot workspace to just

to provide a better visualization.

Um I I did this where uh when we have no

um AI for coding nowadays I believe it's

not difficult that we can ask uh

language model to generate code that can

give us a very good Monte Carlo

visualization of the robot workspace. Um

so this is overall what we can get from

Monte Carlo visualization but how we can

do it and why what is mont monteardo visualization.

visualization.

So overall the idea is not difficult. It

is it is from forward kinematics.

From last week we've learned forward

kinematics and once we've we get the

forward kinematics of the robot arm

it will mean that we get this function.

Basically forward kinematics give us

this function f that we can simply just

input all the um we can input all the

joint positions and then we will get any factor

factor

if uh position and orientation.

So what then what we can do to generate

that point cloud is fairly

straightforward that

we first need to get all the joint

limits. Similar as before that for

example we have the joint we have joint

one that can rotate from zero to 180

degrees and then we have joint two that

can rotate from one angle to another

angle and so on and so forth. And then

we can randomly sample

the joint positions within the joint

limits. For example, uh I can have

hundreds of examples, hundreds of

samples that uh I have joined one equal

to one degree, two degrees, three

degree, so on and so forth. It can be

random. It can also be with a fixed

interval. um it uh as long as we have a

a large enough numbers we we should be

be able to get a a good point cloud. So

we just need to randomly sample a bunch

of point points that we have uh

different uh joint positions and then

we we simply have all of these uh

to get the target um to get the factor

uh position and orientation.

And then

we just need to plot all of these um all

of the any factor positions in the 3D

space and that will give us a point

cloud. For example, if I have 1 million

um random uh random joint positions and

we plug plug in all of these uh 1

million joint positions into forward

kinematics and then we get 1 million 1

million antifactor poles and then we

just need to draw the antifactor poles

in the 3D space and eventually what we

get will be a point cloud like this. Uh

as far as I remember this is more than

one meter. There's probably um a couple

of meters um of points that give us a

very uh solid uh representation of what

the Raven 2 surgical robot workspace

would look like.

And once we have this point cloud, it is

also not difficult to use it to check if

one certain point is within

the robot space or not. For example, if

I have one point here, then I can easily

check this point uh if it's within the

workspace and then we if we have another

point that is outside the work

workspace, we can also easily check it.

Uh one of the simple method that that

can be uh we just need to find some

closer points the closest points of the

um of the workspace point point cloud

the the the target point that we want to check.

check.

For example, here if the point is

outside the workspace, we can see that

all of these uh vectors they will have

similar uh orientations. Well, if

the point is inside the workspace, we

might be able to find out that these

vectors they kind of point in in all of

Now we've uh we've introduced how do we

find the robot workspace and then we can

move on to work on inverse kinematics.

We've learned forward kinematics before

in the previous week and eventually what

forward kinematics give us is the

function or the relationship that we we can

can

input all the joint positions and get

the any factor poles.

So inverse kinematics literally is

inversing the function so that the input

will this time be

the factor pose and the output will be

the joint positions that can achieve

But for forward kinematics what we often

have is is a standard pipeline that um

as as we mentioned in the previous

lecture we just uh we need to perform

link frame assignment and then

uh and then we perform uh then we find

the DH parameters and then we find the

transformations uh for each each row in

the DH parameter table and eventually we

combine the the transformations from

drawing uh from from frame zero to one

one to two so on so forth and get to the

any factor. So you can see uh even if

it's even if there's a a bunch of work

that we need to do it's kind of a a

pretty fixed pipeline that to to do it.

But for inverse kinematics there's um

there is unfortunately unfortunately

unfortunately no um fixed pipeline or

very standard pipeline that we can work

on it. So inverse kinematics is

generally considered a bit harder than

forward kinematics and there are some

questions we need to consider. First of

all do we have a solution? We might have

a point we might want to control the

robot to reach a certain point but that

point is actually outside the workspace.

So there's act so there is actually no solution

solution

and if the solution exist how can we

find the solution and eventually if we

can find the solution do we have

multiple solutions and how we can choose

Let's first take a look at uh the

solutions of inverse kinematics. We

might have finite solutions, we might

have infinite solutions or we might have

no solutions.

Uh again we we can consider a simple uh

two joint planer robot that we only have

two joint uh joint joint one here and

join two here. And

And

we have two joints which means that this

robot arm has uh two degree of freedom.

And for a planer robot if we only

consider the location of the manifactor

which is X and Y here it will be um you

can see that we have two degree of

freedom robot arm and we are also

working in a two degree two degree of uh

freedom workspace. So joint one and join

two versus X and Y two degree of freedom

on both side. This often will result in

finite solutions.

But we can see that even if we have

finite solutions, we don't have one

single solution. Instead, we will have

two solutions as shown in the figure here.

here. Um

in this if we have the if the degree of

freedom of the robot uh workspace and

the degree of freedom for the robot

joints they are uh equal to each other

we will we will uh say uh this work uh

this task is uh completely specified and

we will often get finite solutions out

of this case. Um and in this uh specific

scenario we will often call this is this

configuration as elbow up and this

configuration as elbow down.

So very likely uh even if we have

completely specified the task we we

might we might not be able to get only

one single solution. Instead we will

often get f uh multiple solutions and

often we need to pick one of the best

solution that we want to use. And long

story short uh once we are choosing

among multiple solutions

a general recommendation is we will

choose the solution that result in the

minimum movement of the robot arm in for

example if I'm already in this

configuration I might prefer just to

move the robot a little bit instead of

move the robot all the way to elbow down situation.

But furthermore if we add one more

joints to the robot arm. So that this is two

join but now here we have three joints

but we are we are also just considering

the position X and Y. So we have three

joints versus two uh dimension workspace

which is just x and y because we are uh

here let's assume we are not considering

the orientation we are not considering

the orientation of the factor instead we

just consider the position

position

and then if we have more degree of

freedom in joints and less degree of

freedom in the task in the robot workspace

workspace

then we will what we will offer can get

is infinite solutions. We can see that

from this figure we can see we all we

can reach this uh target location with

with infinite uh possibilities on the

robot arm configuration.

So um this is often called kinematic

redundancy and actually many robot arm

nowadays they actually designed for the

kinematic redundancy because this can

actually provide us better dexterity in

the robot arm control. Um

in general for for a 3D space if we want

if we want full control on both position

and orientation which means that we have

full control on both XYZ on position and

RPY angle on orientation that would

would result in six degree of freedom.

So if we if our robot arm has uh six

joint six joints which which often means

that we will have um uh completely

define a specified uh task which we will

get finite solutions but many many uh

robot arm today they might have seven

joints or even more. So they will have

some kinematic redundancy and it

actually improve the robot arm uh with

better dexterity but that also the

trade-off is that the control might be

uh more complicated. So it really

depends on uh do we want simple control

or do we want better dexterity of the

And for sometimes the inverse kinematics

may not have a solution which means that

the target location uh that we want to

reach uh cannot be achieved by the robot

arm. There are two general uh situations

that can result in no solution. The

first one is we don't have enough degree

of freedom or is too far or too close to

reach the uh to reach the uh that point.

For example, this point

is clearly outside the the robot

workspace is too far to reach. And this point

point

obviously is too close to reach. Even

even if we fully

even if we fully fold back the second

joint, the probably the closest point

that we can reach is here.

So we can reach no clo no no closer uh

points. So this point is probably is

very likely uh too close to reach. And

for example, if we want to reach this

point but not with

this direction or this direction,

instead we want to reach with this

direction. then we might also get no

solution because the you can see that

the only two configuration elbow up and

elbow down configuration we can reach

is through the uh is either through this

orientation or this orientation. So this

is um this is typically um a classic

problem on we don't have enough degree

of freedom to reach this uh x uh three

orientation but we can then if we add

one more joint

the any factor if we add one more joint

here we will be able to

reach this uh location with our desired

orientation x3 three.

So this is these are the typical

examples on we don't have uh enough uh

degree of freedom or the points is uh

the points are too far away or too

close. It's outside the they are outside

the robot workspace but it can also be

outside the joint limits which is this

orange point. So

in for this problem um

it is similar uh in practice it is

similar issue that the robot just

similar uh just cannot reach this point.

Um but that would be uh still be uh

technically different because uh if the

if this for this yellow um for this uh

orange point it is basically that uh we

cannot reach because we cannot rotate

joint one

beyond beyond this yellow uh be beyond

this blue uh limitation. So if somehow

we can break this limitation this point

is actually reachable with the robot arm links.

links.

So the difference here is is that um we

can find the first situation

during the solution of inverse

kinematics during the um the solution

method that we will uh introduce later

on. So if we have the first uh situation

what we will get is we might be able to

find out that um we want to use ark sign

but uh uh we want to we want to compute

ark sin x but x

is larger than one or we want to take

square root of x but x is smaller than

zero then they will result in imaginary

part in the solution. And that generally

means that we have no solution.

But if the case is the second case then

the then we will not be able to find uh

similar clues like we are taking square

root of a negative number. Instead we

might be able we might be able to get a

get a solution angle out of the solving

procedure. But eventually we still need

to check if these joint positions or

joint angles they are actually uh within

the joint limit. So which now you'll see

the difference is that if we have the

first uh situation that we have no

solution then

we will we will be able to find it out

during solving procedure. But for the

second uh case we must double check the

joint limits after we get the solution

because we will get solution. It's just

that solution is outside the joint limit.

So overall the methods of solving

inverse kinematics uh generally we have

three major types of uh solution. The

first type is closed form solution.

Closed form solution means that we will

get a symbolic solution. Um that um for

example what what we will get is that we

will get say the one the joint one that

is that is a function of uh x y z and

orientations of the factor. So once we

get uh once we tell the robot the XYZ

location we want the manifactor to go

this function will give us um theta uh

angle of the join rotation. So we can

then control the joints to reach the target.

target.

So this is closed form um solution.

Um another major

type of solution is the numerical

methods. Instead of getting a symbolic

solution um numerical methods will will

often iteratively solve for equations

instead of using symbolic ways to u to

solve the uh to solve the equations. So

it will uh generally numerical method

they are uh they are often easier to be

implemented because once we have the

uh once we have the equation uh

described we can then u write write some

program to solve to numerically solve

the equations. Well for closed form

solution we often need to manually solve

uh the solve the equations by ourself.

Um today we will focus more on uh how to

obtain closed form solution because

uh first of all uh numerical method they

are they often follow

a straightforward or fixed pipeline that

we just need to um code for the uh

numerical solver for the equations. Well

closed form solution it actually

requires some uh tricks or techniques to

do it. Uh so

now the question is

if we have numerical method they are

very easy to use and they can also give

us solution. Why do we still want closed

form solution

that we have solved to we have to do all

the math and we have to solve all the

equations by our using our hands and our brain.

brain.

The difference in the computational

efficiency is that because numerical

solver numerical method they are all

they are they are often iterative. So we

might need the solver to run thousands

of iterations to find one valid

solution. So that can be pretty slow if

we want to control the robot arm in real

time with a very low-level controller.

Well closed form solution is more more

like once for all. If we have a symbolic

solution and then in the uh in the

real-time control we just need to plug

in all the numbers and then get the uh

get the joint angle that that can go to

the target. So closed form uh solution

will generally be way faster in

computing than numerical method. So

that's a benefit on why do we still want

the closed form solution and actually in

uh once you are uh once you use uh Ross

move it package they are they provide IK

function and IK fast function. So IK uh

inverse IK indicate for uh inverse

kinematics. So IK function generally use

numerical method and you can find out

that if you have a complicated

environment to to do the robot arm pass

planning the uh the default IK function

in Ross move it can be pretty slow. Well

IK fast use closed form solution and

that can be way faster than default IK.

But the the question now is if we want

to use IK fast we need to provide

symbolic solution to the uh to Ross move

it package and then it will be way

faster. So that's the benefits and

motivation why we still want to learn

and use closed form solution because it

can be way faster in real time control.

And the third method is hybrid method

which means that um we kind of take

advantage of both closed form solution

and the numerical solution.

So the closed form solution we will use

closed form solution for some joints

that we can easily solve and then they

will be way faster in real time control.

But some other joints might be very

difficult to solve it symbolically and

then we can for these joints we can use

numerical method. It's kind of like a

combination of a closed form and

Today we are uh we are going to mainly

look at numerical solution. Um and first

since we are working on inverse

kinematics we might also need to work on

the inverse of matrix and specifically

here for robot manipulation we are

looking for homogeneous transformation.

Uh again if you are uh if you need a

refreshing on homogeneous transformation

please go to our materials in 516 on

coordinate system and homogeneous transformation.

transformation. Um

generally once when we talk when we talk

about matrix uh the uh inverting matrix

matrices they are not always invertible.

However, uh specifically for homogeneous

transformations and the rotation matrix,

they are always invertible because

because

physically we can understand this as a

transformation or a rotation. They can

always be on down. If I can move one

cube, if I can move the cube from the

base to this location, which definitely

means that I can also move it back.

So uh

homogeneous transformation matrix and

rotation matrix they are always invertible

invertible

and here gives the the formula on how to

do the u inverse of the of a homogeneous

transformation. Um

for rotation matrix it's actually pretty

uh simple to uh invert rotation matrix

that that is the the transpose of the

rotation matrix is the inverse of the

rotation matrix and please make sure

that um please keep in mind that it's

not it not it does not apply to all the

matrix um specifically it mainly applies

for rotation matrix if I have a random matrix

matrix

an arbitrary uh matrix. It it does not

mean that I I can use a transpose as inverse

inverse

and this is the inverse of the um of the

translation vector and we will not go

through the um uh the mathematical

details on how to derive this. Um but

this formula here is uh is pretty uh

straightforward and easy to use.

And we will also quickly review uh that

we have s and cosine uh and we will also

have arc sign and arc cosine. So

basically arc sign is that if we give a

sign value of an angle it will return us

that angle. So it's inverse of s and

similarly arc cosine is inverse of cosine.

And then um we have the uh the the

formula of the sum of angles. If we have

cosine theta 1 plus theta 2.

Um and again in our note

here in robot manipulation we will use

c12 to represent cosine uh theta 1 plus

theta 2. So you can see this is a very

uh clear and a simple notation because

in the future we might see a very long

equation so that if we write all of the

uh signs and cosine in this way the

equation will be super super long. So

it's of our interest to represent the uh

the signs and cosine in the in a very uh

concise way that we are using here.

And the equation is uh the formula is

basically given here that if we have the

if we have cosine theta 1 plus theta 2

it basically equal to c1 c2 cosine theta

1 cosine theta 2 minus s1 s_2 sin theta

1 sin theta 2

and uh these are the uh the equation for

the sum of angles and we also have the

law of cosiness.

Um we will not do all the math details

today but this law of cosine cosiness

can sometimes also be very very useful

And besides the inverse of s and cosine

we sometimes or actually more frequently

we will use the inverse of tangent.

But in terms of using arc tangent there

is problem that arc tangent is ambiguous

of quadrant which means that if one if I

have uh if I'm using arc tangent I'm

actually not

not able to tell whether that angle is in

the first quadrant or the third

quadrant. And similarly we we might also

not be able to differentiate

the second quadrant and the fourth

they actually have the same um tangent

value. So if we use arc tangent we

actually don't know where is this angle.

So instead of using arc tangent

in robot u manipulation inverse kinematics

kinematics

it is highly recommended that we use arc

tangent 2.

Um so arc tangent 2 is basically trying

to preserve the you can see that for arc

tangent we only have one input argument

instead for arc tangent two we have two

argument and such that the uh the

positive and negative sign of the uh of

the two inputs they are preserved. So we

can actually find out that if we are in

first quadrant then both y and x they

are positive while if we are in the

third quadrant they are both negative.

So that's a benefit of using arc tangent 2.

Um if you want to look uh more about uh

arc tangent versus arc tangent 2 uh you

can take a pause here and then uh search

In terms of solving inverse kinematics,

we will first take a look at algebra

bracket solution. Um it might looks a

little bit complicated but overall uh

what we are doing in algebraic solution

is basically solving equations. So

as we've learned from the forward kinematics

for a robot arm we can um the forward

kinematics will give us the homogeneous

transformations and in which let's

assume if we have all the uh drawings to

be rotational then uh the the variables

will be um set theta one and sorry this

should be theta 2 will be set theta 1

theta 2 theta so on and so forth. Um,

including all of our joint positions

and that would eventually uh if we

compute all of that we will what we will

get is um is a combined homogeneous

transformation in which we can find our

rotation matrix which is which can also

be converted to rpy angle or zyx oiler angle.

angle.

Um if you are if you need a refreshing

of what is RPY angle and ZYX oiler angle

you can also refer to um uh 516 material

and also uh we we get the translation

vector on XY

and Z 3D position. So

in forward kinematics we we often in

most of the cases we consider they are known

known and

and

the combined t 0 to n

this is on no

But

now we are working on inverse kinematics which

which

we will assume

that this

is known. the the combined t from zero

to n this is known while

and here the diff the differences will be

in the previous um in the previous

lecture once we are working on forward

kinematics we simply just need to plug

in those theta values to get

the to get t to get t from zero to n.

But now

since we have t 0 to n to be to be known

while all of these theta n theta theta

rotations joint position they are

unknown. We then need to solve for these

status and this task is nontrivial

and uh most of the difficulty of inverse

Um so generally what we can do is now

we've uh we've write all of these uh

values out. So basically on the uh

on the left side of the

of the equation all of these values are known

known

and the forward kinematics values they

of the equation.

So these R112 R33 and X uh XD YD ZD they

are known uh and oh so the XD YDZD here

D um indicated for desired. So often

what we want to do in inverse kinematics

is that we have a desired and factor

location and we want to solve for the

joint positions that can achieve the

certain location.

so generally uh what we will get is that

here um we will keep these theta values

as u symbolic values and then after

multiplying all of this we will get

another big matrix here

which in in in it we will have a lot of

like signs and cosiness. and so on and

so forth. And

eventually we will get a a bunch of

equations that for example here u what

we get is R11 equal to C1 S2

and so on and so forth we will get get

R12 equal to uh this entry and R13 equal

to this entry and so on and so forth. So

we kind of get a bunch of equations and

we then want to solve these equations.

This this can sometimes be very very

difficult. Um but there are some tricks

that we can use and we will introduce

later on. But overall u a general

procedure is that uh for example if we

can get

theta one or theta n solved let's say

somehow we combined some equations and

we solved theta 1 and theta n first and

then we can move these two

terms because they are now solved they

can be considered as known. So we can

move these two terms to the left hand

side of the equation.

We can move them to the left hand side.

So that now the right hand side will be

uh will be simplified because uh

previously it has n unknown variables

from the one to set the n. But now it

only has n plus n minus two unknown

variables starting from setta 2 to n uh

to set the n minus one. Sorry for the

mistake here. So this will should be

theta n uh plus uh minus one.

So again and again we will we will do

this again and again. So once we get one

uh angle solved we then consider that is

known and then we simplify the equations

and then we do it so again and again

until we solve all the angles.

So this is a general uh workflow of uh

algebraic solutions on inverse

kinematics. Uh let's look at one example.

example.

So this is actually a pretty easy

example for most of the industrial robot

arms uh which has more than uh six uh

joints. The in the inverse kinematics

algebraic solutions can be very very

complicated. But here uh we are looking

at this simplified robot arm which only

has three uh orientations

and we are only considering the uh the

orientation uh sorry. So this uh

simplified robot arm has only three

joints and we are only considering the

orientation of the manifactor and we are

ignoring all the we are ignoring all the

So what we get here is again

all of these values they are known and

all of these values

they are on known

known

all of the uh specifically it's actually

all of the theta values they are unknown

but uh generally on the left hand side

uh of the equation they are all known

from R11 to R33 they are known and

on the right hand side um

the cosine uh angle a cosine angle b and

c they are unknown because we do not

know the joint rotation uh a b and c.

So for this this example instead of

using joint one two three uh we are

Now let's take a look at how we can

solve it.

So um on the right hand side is actually

the forward kinematics equation. Um we

have our we have uh skipped uh link link

frame assignment and also the forward

kinematics part. So we we directly give

the forward kinematics sol u result

here. So we can focus on the uh solution

of inverse kinematics here.

So on the again on the left hand side of

the of the equation we have uh these

R112 R33 values to be known and on the

right hand side we have ABC to be

unknown and we want to solve for them and

and

a general procedure is to look at the

these equations and try to find out um

which equations

or which what combinations of equations

can can can isolate some joint angles

out so that it's it will be easier for

us to solve it.

So by looking at this this uh this huge

matrix here

um intuitively for as for now I do not

want to touch these

these

entries because they are uh they look

pretty much more uh complicated and we

might need a lot of computation to to

really trying to isolate one uh one angle

angle

of these complicated equations. Instead,

So what we uh what we can

work on first is that we can try to find

out how we can combine because um for

these two terms

they they not only contain one uh

unknown variable but instead they

contain two unknown variables A and B.

They both have unknown variable A and B.

So what we want to do here is we want to

is that we want to isolate the um

we want to isolate the uh the unknown

variables so that like we have only one

variable in one equation which can be

we can first look at these two terms.

So that we have r11 equal to

ca cb cosine a and cosine b and we have

r21 equal to s a c b which we will write

the equation down here. And

And now we have two equations. We have two

now we have two equations. We have two unknown variables. But we have two for

unknown variables. But we have two for each single equation we still have two

each single equation we still have two unknown variables we cannot solve. Um

unknown variables we cannot solve. Um but here we have one trick to use that

but here we have one trick to use that is

from middle school. We probably all learned that sin square a + cosine

learned that sin square a + cosine square a equal to 1.

square a equal to 1. This is very nice because you can see

This is very nice because you can see that we have even if a might be unknown

that we have even if a might be unknown but as long as we can somehow get sin

but as long as we can somehow get sin square a plus cosine square a we can we

square a plus cosine square a we can we cancel out a to be a constant number one

cancel out a to be a constant number one which is very easy to use. And luckily

which is very easy to use. And luckily we can see that for these two equations

we can see that for these two equations we get here

for these two equations we get here.

here. If we just simply take the square of the

If we just simply take the square of the two equations and then add them up, then

two equations and then add them up, then we will have R11 square plus R21 squared

we will have R11 square plus R21 squared equal to CB² time CA 2 + S A square.

equal to CB² time CA 2 + S A square. This is sin A square plus sin this is

This is sin A square plus sin this is cosine A square plus sin A squar which

cosine A square plus sin A squar which will be one.

will be one. So eventually simplify this equation we

So eventually simplify this equation we will get CV²

will get CV² equal to R11 square + R 21 squared

equal to R11 square + R 21 squared and as we mentioned R11 and R21 they are

and as we mentioned R11 and R21 they are both known variable they are both

both known variable they are both constant act constants actually so here

constant act constants actually so here we get cosine B square and we just need

we get cosine B square and we just need to take the square root to get cosine B

to take the square root to get cosine B that is equal to plus and minus square

that is equal to plus and minus square root of r11 square + r21 squared.

root of r11 square + r21 squared. And we should notice that here once we

And we should notice that here once we are taking square root we are getting

are taking square root we are getting two solutions one positive and one

two solutions one positive and one negative.

So now we have cosine B. In terms of finding the angle B, we want to use arc

finding the angle B, we want to use arc tangent 2. I will mention later uh we

tangent 2. I will mention later uh we don't want to use cosine uh arc cosine

don't want to use cosine uh arc cosine uh here. Instead, we want to use arc

uh here. Instead, we want to use arc tangent 2. We have cosine B. So we still

tangent 2. We have cosine B. So we still need sin B to use arc tangent to use arc

need sin B to use arc tangent to use arc tangent 2.

tangent 2. And luckily we can find out that we have

And luckily we can find out that we have sine b which is equal to r

sine b which is equal to r uh 31. Uh I'm sorry I missed a

uh 31. Uh I'm sorry I missed a a negative sign here. So we actually

a negative sign here. So we actually have negative sign b equal to r 31.

Uh I will move the negative side here. All right. So then

now we have both cosine B and sin B and then we can use arc tangent 2 to solve

then we can use arc tangent 2 to solve for the angle B which is here. So B

for the angle B which is here. So B actually the angle B actually equal to

actually the angle B actually equal to arc tangent 2

arc tangent 2 and what's inside is s and cosine. So

and what's inside is s and cosine. So sine comes first and cosine comes next.

sine comes first and cosine comes next. So this is basically sin B and this is

So this is basically sin B and this is cosine B

cosine B and we get sin B here

and we get sin B here and we get cosine B here. So we just

and we get cosine B here. So we just need they are both known values. So we

need they are both known values. So we just need to plug in the values. All the

just need to plug in the values. All the R values they are known. So we now can

R values they are known. So we now can use arc tangent 2 to solve for the angle

use arc tangent 2 to solve for the angle B. You can see we actually work through

B. You can see we actually work through it to eventually just to get one joint

it to eventually just to get one joint angle out which is B for now. And you

angle out which is B for now. And you can see that at this moment we already

can see that at this moment we already have

cosine B. And technically we can use arc cosine on it to get the angle B out. But

cosine on it to get the angle B out. But in general practice I do not recommend

in general practice I do not recommend um using arc sine or arc cosiness um if

um using arc sine or arc cosiness um if if we can find both s and cosine. This

if we can find both s and cosine. This is because if we directly use arc sine

is because if we directly use arc sine or arc cosine, we might get fake

or arc cosine, we might get fake multiple solutions because

because uh if we only get arc sine or arc cosine

if we only get ark sign let's say if I only get ark sign I might get two

only get ark sign I might get two solutions one is here and one is here

solutions one is here and one is here and if I only get arc arc cosine

and if I only get arc arc cosine similarly it will be one here one here

similarly it will be one here one here and one here but but actually if we know

and one here but but actually if we know both cosine value and sine value we

both cosine value and sine value we actually know that it is one angle

actually know that it is one angle instead of two. So directly using arc

instead of two. So directly using arc sign or arc cosine it can result in uh

sign or arc cosine it can result in uh fake multiple solutions.

fake multiple solutions. So um it is not uh it it will not

So um it is not uh it it will not eventually give us wrong answer but it

eventually give us wrong answer but it will uh require extra effort on checking

will uh require extra effort on checking the answers. For example, if sometimes

the answers. For example, if sometimes if the equations are they are super

if the equations are they are super super complicated and we simply just

super complicated and we simply just cannot find both s and cosine value of

cannot find both s and cosine value of one angle one joint angle then we have

one angle one joint angle then we have to use ark sign or arc cosine. Then uh

to use ark sign or arc cosine. Then uh my general recommendation is that once

my general recommendation is that once you use uh arc sign or arc cosine at any

you use uh arc sign or arc cosine at any where in the procedure of solving

where in the procedure of solving inverse kinematics then after

inverse kinematics then after after we solved uh get the inverse

after we solved uh get the inverse kinematic solution we should plug in all

kinematic solution we should plug in all the inverse kinematic solutions back to

the inverse kinematic solutions back to forward kinematics forward kinematics

forward kinematics forward kinematics kinematics to check if these solutions

kinematics to check if these solutions they are valid or not. So you can see

they are valid or not. So you can see that once we if we are if we use arc

that once we if we are if we use arc sign or arc cosine we will then need one

sign or arc cosine we will then need one more step on checking the inverse

more step on checking the inverse kinematic solutions to verify if they

kinematic solutions to verify if they are valid solutions or not.

Sorry for missing that uh negative sign. So now we have B angle B solved. And

So now we have B angle B solved. And then

then we still need to work on angle A and

we still need to work on angle A and angle C.

But similarly we can still use these two equations because now we have angle B

equations because now we have angle B solved. So these two equations now only

solved. So these two equations now only have unknown variables as

CA and SA. So only A is unknown in these two equations. Now so this is uh because

two equations. Now so this is uh because we have already solved for B.

we have already solved for B. So similarly we will use the two

So similarly we will use the two equations R11 equal to CACB and R21

equations R11 equal to CACB and R21 equal to S A CB. And now

equal to S A CB. And now because CB is known we can rewrite the

because CB is known we can rewrite the equations to get CA and SA out. So um

equations to get CA and SA out. So um cosine a and sin a and both and all of

cosine a and sin a and both and all of the r11 cb r21 cb they are all known

the r11 cb r21 cb they are all known values because b is solved and similarly

values because b is solved and similarly we have cosine a and sin a we can then

we have cosine a and sin a we can then use uh arc tangent 2 to get angle a so

use uh arc tangent 2 to get angle a so that we have sin a come first and cosine

that we have sin a come first and cosine a come

a come second. We just need to plug in SA and

second. We just need to plug in SA and the CA and use arc tangent two to get

the CA and use arc tangent two to get angle A.

angle A. And of course here because we need to

And of course here because we need to divide uh R21

divide uh R21 uh sorry this will be um

uh sorry this will be um sorry for the mistake this will be uh

sorry for the mistake this will be uh R21 CB uh this will be R11 and CB sorry

R21 CB uh this will be R11 and CB sorry for the mistake.

for the mistake. So

because we need to divide R11 and R21 by CB, we have to we must assume that CV is

CB, we have to we must assume that CV is not equal to zero. And there is actually

not equal to zero. And there is actually a special condition if we do have CV

a special condition if we do have CV equal to zero

equal to zero and that is called singularity. We will

and that is called singularity. We will mention that later on.

mention that later on. So we should now also notice that a

So we should now also notice that a equal to tangent arc tangent 2 of uh of

equal to tangent arc tangent 2 of uh of this equation here. And we should notice

this equation here. And we should notice that a also has two solutions according

that a also has two solutions according to b

to b because b has two solution solutions.

because b has two solution solutions. one for the choices of whether uh we

one for the choices of whether uh we take the positive value of this square

take the positive value of this square root or negative value of this of this

root or negative value of this of this square root.

So B has two solution and A also has two solutions.

And now we have both A and B solved. And we want to solve for angle C now. So we

we want to solve for angle C now. So we can similarly find two equations that we

can similarly find two equations that we can use now. It would be these two

can use now. It would be these two equations because CB are solved. So only

equations because CB are solved. So only left unknown in these two equations they

left unknown in these two equations they are they are SC and CC sin C and cosine

are they are SC and CC sin C and cosine C.

C. So we have these two equations and then

So we have these two equations and then we just need to uh plug in these two

we just need to uh plug in these two equations into the arc tangent equation.

equations into the arc tangent equation. So we have um so this is basically um

So we have um so this is basically um we can use similar solution that uh here

we can use similar solution that uh here we have SC

we have SC equal to R32 over CB and we have um here

equal to R32 over CB and we have um here we have C C equal to uh R33 over C B

and we have We just need to plug in s first and cosine next and then arc

first and cosine next and then arc tangent 2 will give us the solution of

tangent 2 will give us the solution of angle C. And similarly we also have two

angle C. And similarly we also have two solutions here.

solutions here. uh because of there's two solutions for

uh because of there's two solutions for C B on choosing positive or negative on

C B on choosing positive or negative on uh on the square root same square root

uh on the square root same square root here.

So now we've solved all the joint positions A, B and C. And we can see

positions A, B and C. And we can see that we have two solutions depending on

that we have two solutions depending on whether we choose CB to be larger than

whether we choose CB to be larger than zero or CB to be smaller than zero.

zero or CB to be smaller than zero. Again CB equal to positive and negative

Again CB equal to positive and negative R11 + R 21 square we get from the

R11 + R 21 square we get from the previous steps.

previous steps. So because we need to we we took the uh

So because we need to we we took the uh the square root at that moment. So it

the square root at that moment. So it result in two multiple solutions

result in two multiple solutions positive and negative. And then

positive and negative. And then depending on whether we choose CV to be

depending on whether we choose CV to be larger than zero or smaller than zero

larger than zero or smaller than zero and then we will have two uh uh we have

and then we will have two uh uh we have two multiple solutions. Um and we can

two multiple solutions. Um and we can generally draw a tree of multiple

generally draw a tree of multiple solutions like this.

um we will uh we will not have something like B1 will lead to uh kind of A1 and

like B1 will lead to uh kind of A1 and A2 here. So it will have no more uh

A2 here. So it will have no more uh leaves because

leaves because all of the multiple solutions of both of

all of the multiple solutions of both of all angle A, B and C they are all

all angle A, B and C they are all dependent on whether we choose CB to be

dependent on whether we choose CB to be larger than zero or smaller than zero.

larger than zero or smaller than zero. So this is overall an example of

So this is overall an example of algebraic solution. Uh we used a simple

algebraic solution. Uh we used a simple example here. You can see that we we

example here. You can see that we we actually did not perform a lot of super

actually did not perform a lot of super complicated math to get the solution of

complicated math to get the solution of all the joint positions A, B and C. But

all the joint positions A, B and C. But in real uh in real practice for for

in real uh in real practice for for those uh six or more joints industrial

those uh six or more joints industrial robot arms um the in uh the inverse

robot arms um the in uh the inverse kinematics algebraic solution can

kinematics algebraic solution can actually be very very complicated. But

actually be very very complicated. But just for um for uh for simple demon uh

just for um for uh for simple demon uh demonstration here we are not using a

demonstration here we are not using a very complicated example.

very complicated example. So in the real practice if we find out

So in the real practice if we find out that the algebraic solution of those uh

that the algebraic solution of those uh six joint robot arms are super super

six joint robot arms are super super difficult then we can on the other hand

difficult then we can on the other hand refer to numerical methods which are

refer to numerical methods which are easier to use but again they might

easier to use but again they might requires uh they might require larger uh

requires uh they might require larger uh computational cost.

So uh we've introduced algebraic solution and now we will uh introduce

solution and now we will uh introduce geometric solution. Uh in real practice,

geometric solution. Uh in real practice, geometric solution uh in my opinion it

geometric solution uh in my opinion it it will be less um

it will be less um it will be less possible to be

it will be less possible to be implemented because geometric solution

implemented because geometric solution uh often works better uh for planner

uh often works better uh for planner robot or for some robot uh robot

robot or for some robot uh robot subspace

subspace which means that somehow we can we can

which means that somehow we can we can further find a a subspace of the robot

further find a a subspace of the robot arm and in that subspace is is planner

arm and in that subspace is is planner robot for example if I have one robot

robot for example if I have one robot arm that can rotate around but if we if

arm that can rotate around but if we if we do not consider the first joint is

we do not consider the first joint is basically just to do the peaching in one

basically just to do the peaching in one uh in as a planner uh robot so that that

uh in as a planner uh robot so that that is called robot subspace in that case uh

is called robot subspace in that case uh geometric solution might be uh might

geometric solution might be uh might might be a more handy tool to use but

might be a more handy tool to use but overall uh algebraic solution I believe

overall uh algebraic solution I believe uh is more uh generally applicable

So now let's take a look on how do we use uh geometric solution. Basically

use uh geometric solution. Basically that's um so it it is uh um we are then

that's um so it it is uh um we are then utilizing uh the geometric relationships

utilizing uh the geometric relationships that we can observe on the robot arm um

that we can observe on the robot arm um the robot arm configuration and then try

the robot arm configuration and then try to figure out what the joint positions

to figure out what the joint positions can be. So it will be easier to look

can be. So it will be easier to look this through uh an example

this through uh an example a lot of equations here but uh we will

a lot of equations here but uh we will we will go through them one by one.

we will go through them one by one. So if we take a look at this robot arm,

So if we take a look at this robot arm, we have three joints. The first joint is

we have three joints. The first joint is a rotational joint say the one that can

a rotational joint say the one that can rotate in this way.

rotate in this way. And then we have a second joint that is

And then we have a second joint that is a translational joint which is this kind

a translational joint which is this kind of like green rod that it can translate

of like green rod that it can translate through this direction.

through this direction. And the third joint 3 is rotational

And the third joint 3 is rotational again which is here.

again which is here. So

if we look at this robot arm um we can see that it's a planer robot arm and now

see that it's a planer robot arm and now we can try to use geometric solution to

we can try to use geometric solution to solve it. Um

solve it. Um according to this uh if we look look at

according to this uh if we look look at this robot arm one easy relation we can

this robot arm one easy relation we can find out is that the the factor

find out is that the the factor orientation ga

orientation ga is equal to theta 1 plus theta 3 because

is equal to theta 1 plus theta 3 because there's only two rotation we can get

there's only two rotation we can get which is theta 1 and theta 3. So this

which is theta 1 and theta 3. So this gma angle will equal to theta 1 plus

gma angle will equal to theta 1 plus theta 3

and but this one equation is not enough to solve three unknown uh unknown

to solve three unknown uh unknown variables. We at least need uh three

variables. We at least need uh three equations to solve for three unknown

equations to solve for three unknown variables in general. So we need to find

variables in general. So we need to find more relationship.

more relationship. And now what we can see is we what we

And now what we can see is we what we can find is that now um let's look at

can find is that now um let's look at this point

this point for this point

for this point what we can get is that if we look at

what we can get is that if we look at this distance

it may not be clear. So uh if we look at this distance it is actually

this distance it is actually cosine

theta 1 times d2 because this is d2 and this is theta 1. So this distance

this is theta 1. So this distance distance distance is actually cosine

distance distance is actually cosine theta 1 times d2 which is here.

theta 1 times d2 which is here. But we can also try to find this

But we can also try to find this distance in from the other direction

distance in from the other direction that is here.

that is here. So overall

We have this distance to be X because that's a that's the location of the any

that's a that's the location of the any factor.

factor. And then we can also look for this

And then we can also look for this distance.

This distance is also not the not difficult to get which should be L3

difficult to get which should be L3 which is

this distance. This will be L3. So this distance will be equal to L3 times

distance will be equal to L3 times cosine

So you can see now we can use these two distance to get the same distance here

distance to get the same distance here which is x - l3

which is x - l3 cossine

cossine gamma

gamma which is here.

which is here. Now you can see this give us one more

Now you can see this give us one more equation on uh d2 * cosine theta 1 equal

equation on uh d2 * cosine theta 1 equal to x - l3 cosine gamma

to x - l3 cosine gamma and similarly now we've this time we are

and similarly now we've this time we are using the x

using the x uh the relationship in x axis but we can

uh the relationship in x axis but we can do it similarly for yaxis I will not do

do it similarly for yaxis I will not do the details for now due to the time

the details for now due to the time limit but you can work work for that for

limit but you can work work for that for by yourself and you can then find out

by yourself and you can then find out this second equation on D2 sin theta 1

this second equation on D2 sin theta 1 equal to Y - I3 sin gamma and now

equal to Y - I3 sin gamma and now now we have overall three equations

now we have overall three equations one

one one two and I will label this as three

one two and I will label this as three so one two three three equations

and let's take a look on how we and solve it. So now we have three equations

solve it. So now we have three equations and we have three unknown variables

and we have three unknown variables theta theta 1 d2 and theta 3 to solve.

theta theta 1 d2 and theta 3 to solve. So what we want to solve is actually

So what we want to solve is actually theta 1 theta 3 d2 and theta 1. So all

theta 1 theta 3 d2 and theta 1. So all of these uh underlined variables they

of these uh underlined variables they are unknown. Well, we actually know

are unknown. Well, we actually know Garma X and Y because and and also L3

Garma X and Y because and and also L3 because L3 is one link of the robot. We

because L3 is one link of the robot. We know the we know the length of of this

know the we know the length of of this link

link and also uh we know X and Y because that

and also uh we know X and Y because that is the target location we want to go and

is the target location we want to go and we also know the garma that is the

we also know the garma that is the target orientation we want to achieve.

So now three equations for three unknown variables. What we can do is we should

variables. What we can do is we should take a look on the three equations and

take a look on the three equations and then try to figure out can we isolate um

then try to figure out can we isolate um one uh joint position out so we can

one uh joint position out so we can solve for it first. You can see that

solve for it first. You can see that from now on it's sol is solving equation

from now on it's sol is solving equation again similar to algebraic solution. The

again similar to algebraic solution. The difference is is that for algebraic

difference is is that for algebraic solution we actually use forward

solution we actually use forward kinematics to get all of the equations

kinematics to get all of the equations and for geometric solution we are using

and for geometric solution we are using these geometric relationships to get the

these geometric relationships to get the equations. But once we get the equation

equations. But once we get the equation the following steps will be the same on

the following steps will be the same on solving the equations.

solving the equations. Now if we look at these equations

intuitively I find out that this equation may not be useful at the very

equation may not be useful at the very beginning because that's garma equal to

beginning because that's garma equal to theta 1 plus theta 3 two equ two unknown

theta 1 plus theta 3 two equ two unknown variables in one

variables in one equation. So it it might not be able to

equation. So it it might not be able to help. Instead

help. Instead let's look at these two equations. For

let's look at these two equations. For these two equations, we can see that um

these two equations, we can see that um the only unknown variables will be d2

the only unknown variables will be d2 and theta

and theta one. There's no 3 fortunately.

one. There's no 3 fortunately. So it's very nice that we can then take

So it's very nice that we can then take a look here and

a look here and we can use the same trick again that we

we can use the same trick again that we know

know cosine

theta 1 squared plus sin theta 1 square = 1. If we can somehow get into this

= 1. If we can somehow get into this format, we will be able to cancel out

format, we will be able to cancel out theta 1. And you can then see that it's

theta 1. And you can then see that it's actually not it is not difficult to get

actually not it is not difficult to get that. we just need to take take the

that. we just need to take take the square of both uh equation one and

square of both uh equation one and equation two and just and then add them

equation two and just and then add them up. So on the left hand side it will be

up. So on the left hand side it will be d2 squar times cossine theta 1 square

d2 squar times cossine theta 1 square plus sin 1 theta 1 squar and this term

plus sin 1 theta 1 squar and this term is actually one

is actually one and on the right hand side it's just a

and on the right hand side it's just a bunch of mass that we need to take

bunch of mass that we need to take square of these two terms and then add

square of these two terms and then add them up.

them up. Then we do all the mass and then we

Then we do all the mass and then we simplify it. This will be the equation

simplify it. This will be the equation we get. And we can then take a look at

we get. And we can then take a look at this equation. There's only one unknown

this equation. There's only one unknown variable that is D2 squared.

variable that is D2 squared. We know X, we know Y, we know L3, we

We know X, we know Y, we know L3, we know garma because

know garma because they are uh X, Y and garma. They are the

they are uh X, Y and garma. They are the any factor location and L3 is a robot

any factor location and L3 is a robot arm link length. They are all known

arm link length. They are all known constant.

constant. So in terms to solve in terms of solving

So in terms to solve in terms of solving d2 we just need to take the square root

d2 we just need to take the square root of d2

of d2 and again because we are taking the

and again because we are taking the square root so we will have two

square root so we will have two solutions one positive and one negative

solutions one positive and one negative and that give us the solution of d2 and

and that give us the solution of d2 and and then let's come back to see these

and then let's come back to see these two equations again we can see that if

two equations again we can see that if now we consider D2 is known then these

now we consider D2 is known then these two equations

two equations only has one unknown variable that is

only has one unknown variable that is theta 1.

theta 1. So and also it it is very nice that we

So and also it it is very nice that we get cosine theta 1 in equation one and

get cosine theta 1 in equation one and we get sin theta 1 in equation two. So

we get sin theta 1 in equation two. So we now get get both s and cosine of

we now get get both s and cosine of theta 1 and now we can use arc tangent.

theta 1 and now we can use arc tangent. We can use arc tangent 2 so that um

so that we can have this equation here theta 1 equal to arc tangent 2 and s

theta 1 equal to arc tangent 2 and s comes first and cosine comes next

comes first and cosine comes next and simply the d2 will cancel out here

and simply the d2 will cancel out here and eventually we just need uh we just

and eventually we just need uh we just need to plug in the two uh equations

need to plug in the two uh equations here that s comes first and the cosine

here that s comes first and the cosine comes next. But here they they are

comes next. But here they they are actually

should also be divided by d2. But we will um we will simplify that

will um we will simplify that and just plug in the two equations

and just plug in the two equations equation one and two here.

equation one and two here. But we should not forget that d2 has two

But we should not forget that d2 has two solutions one positive and one negative.

solutions one positive and one negative. So that can result in theta 1 also has

So that can result in theta 1 also has two uh two solutions depending on

two uh two solutions depending on whether whether d2 is positive or d2 is

whether whether d2 is positive or d2 is negative. And if d2 is negative we will

negative. And if d2 is negative we will get um a different solution on theta 1

get um a different solution on theta 1 which is actually uh if theta if theta 1

which is actually uh if theta if theta 1 one

one is pointing in this direction and then

is pointing in this direction and then theta one two

theta one two will be pointing in this direction by

will be pointing in this direction by adding pi.

So that give us the um the solution of theta 1. So we have two solutions on D2

theta 1. So we have two solutions on D2 and two solutions on theta 1.

and two solutions on theta 1. So now we can finally use garma equal

So now we can finally use garma equal equal to theta 1 plus theta 3. And

equal to theta 1 plus theta 3. And because we have already solved theta 1

because we have already solved theta 1 so only theta 3 here is unknown

so only theta 3 here is unknown variable. So it's very simple that um

variable. So it's very simple that um theta 3 equal to gamma equal to uh gamma

theta 3 equal to gamma equal to uh gamma minus theta 1. But because we have two

minus theta 1. But because we have two solutions on theta 1, so we also have

solutions on theta 1, so we also have two solutions on theta 3. And now we

two solutions on theta 3. And now we have all the solutions of d2,

have all the solutions of d2, theta 1's

theta 1's and

and theta 3. They are all solved now. And

theta 3. They are all solved now. And they all have two solutions depending on

they all have two solutions depending on whether d2 we choose d2 to be positive

whether d2 we choose d2 to be positive or to be negative.

And here we can we can see that for the two solutions we always assume either D2

two solutions we always assume either D2 to be larger than zero or D2 to be

to be larger than zero or D2 to be smaller than zero.

smaller than zero. We haven't got discussed one condition

We haven't got discussed one condition that is D2

that is D2 equal to zero. What happens if D2 equals

equal to zero. What happens if D2 equals equals to zero?

equals to zero? So if D2 equals to zero, we can further

So if D2 equals to zero, we can further draw the robot arm now because we no

draw the robot arm now because we no longer have uh

longer have uh and non Z a nonzero D2 which means that

and non Z a nonzero D2 which means that D2 is basically zero distance uh right

D2 is basically zero distance uh right at this

at this right at this center.

right at this center. So the robot arm will look pretty short

So the robot arm will look pretty short like what we are what we draw here.

like what we are what we draw here. So if D2 equal to zero what we can find

So if D2 equal to zero what we can find out is actually very interesting that

out is actually very interesting that now theta 1 and theta 3 they are they

now theta 1 and theta 3 they are they are actually all like joined one and

are actually all like joined one and join three they actually they are all at

join three they actually they are all at the center here

the center here all at the center here. So we can then

all at the center here. So we can then see that if we rotate theta 1 and if we

see that if we rotate theta 1 and if we rotate theta 3 they result in the same

rotate theta 3 they result in the same movement of the Annif factor the factor

movement of the Annif factor the factor will will always just move

will will always just move on this arc. We are kind of like using

on this arc. We are kind of like using this robot robot arm like a compass.

this robot robot arm like a compass. So in this case we actually lose one

So in this case we actually lose one degree of freedom of this robot arm

degree of freedom of this robot arm because we have joint one and joint

because we have joint one and joint three. They have the same movement. So

three. They have the same movement. So they basically can be considered as one

they basically can be considered as one joint.

joint. So we kind of lose one degree of freedom

So we kind of lose one degree of freedom and this this configuration is called

and this this configuration is called singularity and we will take a look on

singularity and we will take a look on singularity later on when we are

singularity later on when we are introducing differential kinematics

and we will quickly look for how we can do how we can solve inverse kinematics

do how we can solve inverse kinematics using numerical solution.

using numerical solution. with numerical solution. Uh what we will

with numerical solution. Uh what we will what we have is um instead of solving

what we have is um instead of solving all the uh all the mass in in the

all the uh all the mass in in the symbolic way as we did before this time

symbolic way as we did before this time we will just use uh equation solvers uh

we will just use uh equation solvers uh numerical equation solvers like Newton

numerical equation solvers like Newton Rabson or Rangakuda there's a bunch of

Rabson or Rangakuda there's a bunch of equation solvers available um in Python

equation solvers available um in Python or in MATLAB that we can easily find uh

or in MATLAB that we can easily find uh u proper tools that we want to use.

u proper tools that we want to use. So overall uh what a general procedure

So overall uh what a general procedure is we first need still need the forward

is we first need still need the forward kinematics.

kinematics. That is why forward kinematics is so

That is why forward kinematics is so important. We still need the link frame

important. We still need the link frame assignment. We still need DH parameters

assignment. We still need DH parameters so that we can have the forward

so that we can have the forward kinematics ready because u forward

kinematics ready because u forward kinematics actually represent how the

kinematics actually represent how the robot can move. So even if we are using

robot can move. So even if we are using the uh numerical solver the solver will

the uh numerical solver the solver will still need to iterate on uh how to move

still need to iterate on uh how to move the robot and find out the inverse

the robot and find out the inverse kinematic solution

kinematic solution and then we need to define an objective

and then we need to define an objective objective function. This is often

objective function. This is often required for most for many equation

required for most for many equation solvers and um we will not go go through

solvers and um we will not go go through details on what can be a good objective

details on what can be a good objective function. So if we want the numerical

function. So if we want the numerical solver uh to be stable to be efficient

solver uh to be stable to be efficient there are actually some tricks to get

there are actually some tricks to get the objective function but uh our focus

the objective function but uh our focus today is on the uh symbolic solution. So

today is on the uh symbolic solution. So we will not go through detail here

we will not go through detail here and then we need to define u inverse

and then we need to define u inverse kinematic kinematics function. So that

kinematic kinematics function. So that would basically we need uh we need a

would basically we need uh we need a need an inverse kinematics function and

need an inverse kinematics function and equations so that our um our numerical

equations so that our um our numerical solver will will try to solve it and

solver will will try to solve it and eventually we need to find an optimizer

eventually we need to find an optimizer which is uh which is uh uh typically the

which is uh which is uh uh typically the equation solvers um as I mentioned we

equation solvers um as I mentioned we can use rangakuda we can use newton

can use rangakuda we can use newton repson and there's a bunch of other ways

repson and there's a bunch of other ways um uh for example also like SLSQP then

um uh for example also like SLSQP then different optimizers they might have

different optimizers they might have different performances. Some will um

different performances. Some will um some will uh have more stable

some will uh have more stable performance but some will have like a

performance but some will have like a faster computation there. There could be

faster computation there. There could be a lot of trade-offs in between different

a lot of trade-offs in between different uh numeric numerical solvers. So it's

uh numeric numerical solvers. So it's not there's no uh in my opinion there's

not there's no uh in my opinion there's no uh gold standard uh on the of the

no uh gold standard uh on the of the optimizer. It's basically depending on

optimizer. It's basically depending on um the trade-offs we want to consider on

um the trade-offs we want to consider on whether whether we want faster computing

whether whether we want faster computing or whether we want uh want to deal with

or whether we want uh want to deal with some uh cases that we might have no

some uh cases that we might have no solution but we still want to get to the

solution but we still want to get to the closest point. For example, if I want to

closest point. For example, if I want to reach a point, reach a point, but

reach a point, reach a point, but actually my robot arm is not long

actually my robot arm is not long enough, but I still want to reach the

enough, but I still want to reach the closest point uh as possible. So that

closest point uh as possible. So that some optimizer will give us performance

some optimizer will give us performance like that. So which sometimes can be

like that. So which sometimes can be very handy.

very handy. So this is overall uh the numerical

So this is overall uh the numerical solution. uh nowadays with the uh with

solution. uh nowadays with the uh with AI language models uh it should not be

AI language models uh it should not be very difficult to um to ask uh AI to

very difficult to um to ask uh AI to generate a numerical solution uh for

generate a numerical solution uh for kinematic uh for inverse kinematics for

kinematic uh for inverse kinematics for a robot arm but we again still need to

a robot arm but we again still need to provide the DH parameters as that that

provide the DH parameters as that that is required for the forward kinematics.

So a general workflow uh with inverse kinematics on the robot is we will start

kinematics on the robot is we will start with

with the joint position. We can read the

the joint position. We can read the current joint position often from the

current joint position often from the encoders on the motor. So these encoder

encoders on the motor. So these encoder sensors will give us feedback on the

sensors will give us feedback on the current joint position and then we use a

current joint position and then we use a current joint position and forward

current joint position and forward kinematics to get the current and factor

kinematics to get the current and factor position or pose uh here because this

position or pose uh here because this robot can only control uh position. So I

robot can only control uh position. So I will I will use position here.

will I will use position here. uh in in six degree of freedom um robot

uh in in six degree of freedom um robot uh industry robot arm we can control

uh industry robot arm we can control both position and orientation which

both position and orientation which combined uh that is called pose.

combined uh that is called pose. So with the current manifactor position

So with the current manifactor position we can then take cartition control

we can then take cartition control command for example if the user want the

command for example if the user want the robot arm the robot grasper to move up

robot arm the robot grasper to move up or move down for a certain distance and

or move down for a certain distance and then we will add for example add uh

then we will add for example add uh we will have z uh which is going up and

we will have z uh which is going up and down we have z at for example 10 cm.

down we have z at for example 10 cm. So that would be something uh that the

So that would be something uh that the user can command the robot to move uh up

user can command the robot to move uh up and down, left and right or rotate uh as

and down, left and right or rotate uh as um incremental control. This is very uh

um incremental control. This is very uh common in uh tally operation that you

common in uh tally operation that you will be using in your uh project.

will be using in your uh project. So um we have once we have the

So um we have once we have the cartitioning uh control command then we

cartitioning uh control command then we add the control command to the current

add the control command to the current we we add the control command to the

we we add the control command to the current factor uh

current factor uh position and then what we get is the

position and then what we get is the desired factor position and once we have

desired factor position and once we have the desired factor position we will use

the desired factor position we will use inverse kinematics to get the desired

inverse kinematics to get the desired joint position.

So what we learned is how to get this step down today. And eventually once we

step down today. And eventually once we get the desired joint position, we just

get the desired joint position, we just need to run the motor control and then

need to run the motor control and then the robot arm will move to will rotate

the robot arm will move to will rotate to that certain configuration that user

to that certain configuration that user want it to go.

want it to go. So this is general control workflow of

So this is general control workflow of inverse kinematics.

Now let's move on to differential kinematics.

kinematics. For differential kinematics um as we uh

For differential kinematics um as we uh worked on before on from forward for

worked on before on from forward for forward kinematics it actually give us

forward kinematics it actually give us the relationship from the joint position

the relationship from the joint position to the any factor pose.

to the any factor pose. And for inverse kinematics reversely it

And for inverse kinematics reversely it give us the relation from any factor

give us the relation from any factor pose to the joint position. But

pose to the joint position. But sometimes besides the position control

sometimes besides the position control we might also want the velocity control.

we might also want the velocity control. So differential kinematics basically

So differential kinematics basically give us the relationship

give us the relationship u not between the any factor position

u not between the any factor position and joint position. Instead it will be

and joint position. Instead it will be any factor velocity and uh joint

any factor velocity and uh joint velocity. It can be um it can be

velocity. It can be um it can be represented in continuous way as uh the

represented in continuous way as uh the joint velocity and effective velocity.

joint velocity and effective velocity. It can also be uh joint increment and

It can also be uh joint increment and and factor increment as a discrete

and factor increment as a discrete system. Um

system. Um a quick re a quick reminder uh review

a quick re a quick reminder uh review that if we have theta dot that means

that if we have theta dot that means that we are taking the derivative of

that we are taking the derivative of theta with respect to time which means

theta with respect to time which means that if I have theta dot if theta is

that if I have theta dot if theta is joint position then theta dot is joint

joint position then theta dot is joint velocity

velocity and furthermore if we have theta double

and furthermore if we have theta double dot that mean that generally means that

dot that mean that generally means that we uh the acceleration of the of the

we uh the acceleration of the of the joint but we will not work on

joint but we will not work on acceleration. We will just work on uh

acceleration. We will just work on uh velocity in differential kinematics

besides the relationship from the joint velocity to the any factor velocity. We

velocity to the any factor velocity. We might also want to do some something

might also want to do some something like inverse differential kinematics

like inverse differential kinematics that is from the

that is from the um from the uh the factor velocity to

um from the uh the factor velocity to the joint velocity

the joint velocity and this matrix J theta is called

and this matrix J theta is called Jacobian matrix because we are going to

Jacobian matrix because we are going to take the the uh the Jacobian matrix of

take the the uh the Jacobian matrix of the forward kinematics matrix.

the forward kinematics matrix. So this is often labeled as J. We will

So this is often labeled as J. We will mention how to uh how to get Jacoba

mention how to uh how to get Jacoba matrix later.

this equation what we get is actually here. So this differential kinematics

here. So this differential kinematics specifically Jacobian matrix will be

specifically Jacobian matrix will be able to give us the relationship that we

able to give us the relationship that we if we provide all the joint velocities

if we provide all the joint velocities we will get the any factor velocities

we will get the any factor velocities and they xyz dot they are linear

velocity and this omega xyz they are angular

and this omega xyz they are angular angular velocity. I think you are fairly

angular velocity. I think you are fairly familiar with linear and angular

familiar with linear and angular velocity once you are controlling the

velocity once you are controlling the turtlebot in the previous quarter.

So generally if we have u six uh if we have the full six degree of freedom of

have the full six degree of freedom of the any factor which is xyz and rpy or

the any factor which is xyz and rpy or this xyz uh x doty dotz dot and omega

this xyz uh x doty dotz dot and omega xyz then the jacobian matrix will have

xyz then the jacobian matrix will have the size of six times n where n will be

the size of six times n where n will be our number of joints. But sometimes if

our number of joints. But sometimes if we are working on a planer robot which

we are working on a planer robot which means that we are only uh we only uh

means that we are only uh we only uh want XY position and one garma angle as

want XY position and one garma angle as we used as example before then the

we used as example before then the Jacobia matrix can be 3 * N where n is

Jacobia matrix can be 3 * N where n is similarly the uh the number of joints.

similarly the uh the number of joints. So uh then the the dimension of the

So uh then the the dimension of the Jacob Jacobia matrix is is also

Jacob Jacobia matrix is is also determined by uh are we working in 3D

determined by uh are we working in 3D space or are we working in 2D space um

space or are we working in 2D space um and some other type of spaces as well.

and some other type of spaces as well. Um so once we have the differential

Um so once we have the differential kinematics and the chakovia matrix we

kinematics and the chakovia matrix we can then answer interesting questions on

can then answer interesting questions on for example on question one if um if we

for example on question one if um if we rotate all the all the joints with uh

rotate all the all the joints with uh with 0.1 rad per second what will be the

with 0.1 rad per second what will be the velocity of the any factor in that case

velocity of the any factor in that case what we can do is we can use a jacobian

what we can do is we can use a jacobian matrix Um and we just need to use a jug

matrix Um and we just need to use a jug matrix to multiplying all these uh joint

matrix to multiplying all these uh joint uh the a vector of joint velocities with

uh the a vector of joint velocities with all 0.1s and then we just need to do the

all 0.1s and then we just need to do the computation and eventually we will find

computation and eventually we will find the any factor velocity and then

the any factor velocity and then reversely if we have the inverse of the

reversely if we have the inverse of the Jacobian matrix

Jacobian matrix what we can do is uh let's assume the

what we can do is uh let's assume the user want the uh the factor to move uh

user want the uh the factor to move uh to move in abs and y direction which

to move in abs and y direction which kind of like move uh diagonally uh with

kind of like move uh diagonally uh with a with a 10 mm uh per sec speed per

a with a 10 mm uh per sec speed per second

second and then what we can do

and then what we can do and then uh we want to know what how how

and then uh we want to know what how how fast we want to rotate the joints to

fast we want to rotate the joints to achieve that end factor um velocity. So

achieve that end factor um velocity. So what we can do is uh we just need to

what we can do is uh we just need to take the inverse of the Jacobian matrix

take the inverse of the Jacobian matrix and then

and then we will use that to multiply uh so this

we will use that to multiply uh so this is this will be x dot and y dot and this

is this will be x dot and y dot and this is z dot and omega x omega y omega z. So

is z dot and omega x omega y omega z. So because we only want the uh the factor

because we only want the uh the factor to move in x and y direction with 10 mm

to move in x and y direction with 10 mm second speed. So we can we just need 10

second speed. So we can we just need 10 and 10 here and all the other entries

and 10 here and all the other entries will be zero and we do the math and we

will be zero and we do the math and we will get the joint velocity that we need

will get the joint velocity that we need to control so that the robot and factor

to control so that the robot and factor will move with the target uh or desired

will move with the target uh or desired velocity that we want.

So now let's take a look on how do we get that Jacobian matrix. We we already

get that Jacobian matrix. We we already find uh we already show that Jacob

find uh we already show that Jacob matrix can be very very useful. But then

matrix can be very very useful. But then how can we find it?

So one uh one straightforward way is to use

one uh one straightforward way is to use uh is to uh get the chakoba matrix by uh

uh is to uh get the chakoba matrix by uh taking the differentiation of the

taking the differentiation of the forward kinematics uh equations with

forward kinematics uh equations with respect to time. um we will we will only

respect to time. um we will we will only introduce this method today. There might

introduce this method today. There might there are there's also other uh methods

there are there's also other uh methods such as velocity propagation but that

such as velocity propagation but that one would would be more complicated. Um

one would would be more complicated. Um if you are interested um you can uh you

if you are interested um you can uh you can message me and I will share more

can message me and I will share more materials with you but today we will be

materials with you but today we will be focusing on uh obtaining Jacobian matrix

focusing on uh obtaining Jacobian matrix by differentiation.

by differentiation. So let's take a look at this uh this

So let's take a look at this uh this example here that we have um

example here that we have um three joints join one

three joints join one joint two and join three.

So all the three joints they are rotational

rotational and we will um we will skip the forward

and we will um we will skip the forward kinematics part and part and directly

kinematics part and part and directly give the result of forward kinematics

give the result of forward kinematics that we already solved for

that we already solved for x y and alpha. So alpha here stand alpha

x y and alpha. So alpha here stand alpha here please notice that alpha here is

here please notice that alpha here is not the alpha in the dh parameters

not the alpha in the dh parameters instead alpha is is the angle of

instead alpha is is the angle of this

this the

the uh the orientation of the factor.

uh the orientation of the factor. So we we have x and y and alpha

So we we have x and y and alpha represented by um the the joint

represented by um the the joint positions

positions theta 1 theta 2 and theta 3. So

theta 1 theta 2 and theta 3. So these are the forward kinematics

these are the forward kinematics functions. With these forward kinematics

functions. With these forward kinematics functions we will then take derivatives

functions we will then take derivatives to get the jacobian matrix.

if we take the if we take the derivative with respect

if we take the derivative with respect to time for this first equation,

to time for this first equation, what we will get is

what we will get is x will becomes x dot because it's taking

x will becomes x dot because it's taking the uh the derivative with respect to

the uh the derivative with respect to time. So x position will be x velocity.

time. So x position will be x velocity. now

now and

this term will be here and this term will be here and this term will be here.

will be here and this term will be here. Um we will again we will not go over all

Um we will again we will not go over all the math. If you uh if you need a review

the math. If you uh if you need a review on how to take the uh take the

on how to take the uh take the derivatives then uh you can you can

derivatives then uh you can you can search for it or you can ask a AI on how

search for it or you can ask a AI on how to take the derivatives with respect to

to take the derivatives with respect to time. Um

time. Um and then we we can then rewrite this

and then we we can then rewrite this function to isolate uh say the say the

function to isolate uh say the say the one dot say the two dot and the three

one dot say the two dot and the three dot. So you can see here they are not

dot. So you can see here they are not isolated but in terms of format of

isolated but in terms of format of chobia matrix we need to

chobia matrix we need to factor out all the theta 1 theta 2 and

factor out all the theta 1 theta 2 and theta 3. So these are the va uh these

theta 3. So these are the va uh these are the uh the joint velocities that we

are the uh the joint velocities that we will be um we will be uh given as input

will be um we will be uh given as input and these terms

and these terms they will eventually be the entry

the entry entries in the jacobian matrix.

entries in the jacobian matrix. So again if you uh here we the the

So again if you uh here we the the reason uh we get these terms is because

reason uh we get these terms is because we we need to take the chain rule of the

we we need to take the chain rule of the derivative. Again if uh we will not do

derivative. Again if uh we will not do the math today here. So if you uh want

the math today here. So if you uh want to take um look at more details of how

to take um look at more details of how to take these uh derivatives then you

to take these uh derivatives then you can search for it or you can ask AI.

So this first equation is what we get from the previous uh slide and that was

from the previous uh slide and that was on the first equation here.

on the first equation here. Now we still need to take the derivative

Now we still need to take the derivative for the second and the third equation.

for the second and the third equation. So similarly

So similarly we can take the derivative for the

we can take the derivative for the second and the third. The third one is

second and the third. The third one is fairly simple enough. We if we have the

fairly simple enough. We if we have the alpha equal to theta 1 plus theta 2 plus

alpha equal to theta 1 plus theta 2 plus theta 3 then the derivative is basically

theta 3 then the derivative is basically just alpha dot equal to theta 1 dot plus

just alpha dot equal to theta 1 dot plus theta 2 dot plus theta 3 dot but taking

theta 2 dot plus theta 3 dot but taking derivative of ss and cosiness they are

derivative of ss and cosiness they are still they are still a bit more trickier

still they are still a bit more trickier and but similarly we can use a

and but similarly we can use a uh use uh use a sim similar uh procedure

uh use uh use a sim similar uh procedure that we did in the previous slides and

that we did in the previous slides and eventually what we get will

eventually what we get will this equation

this equation and again we will factor out all the 1,

and again we will factor out all the 1, 2 and theta 3 dots.

2 and theta 3 dots. And now with these three equations we

And now with these three equations we can then

can then factor out all theta one two three dots

factor out all theta one two three dots and

and and what we we will get is

and what we we will get is so xy

so xy and alpha

and alpha they are

they are xy uh sorry x dot y dot and alpha dot

xy uh sorry x dot y dot and alpha dot they are actually q dot and

they are actually q dot and theta

theta Theta 1 dot, theta 2 dot and theta 3 dot

Theta 1 dot, theta 2 dot and theta 3 dot they are actually theta dot here

they are actually theta dot here and what is left

and what is left will be the jacobian matrix. So if we

will be the jacobian matrix. So if we factoring out all of that eventually

factoring out all of that eventually what we get is this Jacobian matrix and

what we get is this Jacobian matrix and it's actually not that difficult because

it's actually not that difficult because um

because you basically can find out that uh what we want.

uh what we want. So this equation here will then be um

so what we will get is actually uh theta 1 dot theta 2 dot and theta 3 dot

1 dot theta 2 dot and theta 3 dot this this is here and this is here and

this this is here and this is here and the Q dot will again be um be x dot y

the Q dot will again be um be x dot y dot and alpha dot. So

dot and alpha dot. So in how do we find these terms is

in how do we find these terms is actually you can see the terms are

actually you can see the terms are actually here. So this term

actually here. So this term comes here and this term

comes here and this term comes here

and this term comes here because the first row

because the first row is about x dot and then y dot and then

is about x dot and then y dot and then alpha dot. So we can have we can we can

alpha dot. So we can have we can we can see that the first row is all about the

see that the first row is all about the equations of X.

equations of X. And similarly

for the second row it will be here

here and here because that will be the second row of the

second row of the because the second row is about y dot.

because the second row is about y dot. So we have the second equation. These

So we have the second equation. These terms will comes as a second row of the

terms will comes as a second row of the of the Jacobian matrix. And eventually

of the Jacobian matrix. And eventually because R of a dot is just theta 1 dot

because R of a dot is just theta 1 dot plus theta 2 dot plus theta 3 dot. So on

plus theta 2 dot plus theta 3 dot. So on this on the third row it will be one one

this on the third row it will be one one and one.

So this jacobian matrix here will be able to provide us the relationship

able to provide us the relationship between the joint velocities and any

between the joint velocities and any factor velocities.

So now let's take a look at another example. This is actually the example

example. This is actually the example that we used for the geometric solution

that we used for the geometric solution for the inverse kinematics. But now we

for the inverse kinematics. But now we will use it use it again for

will use it use it again for differential kinematics.

differential kinematics. And

And these are the forward uh these are the

these are the forward uh these are the forward kinematics equations. Again we

forward kinematics equations. Again we will skip on we will skip how do we find

will skip on we will skip how do we find all of this. We assume that we already

all of this. We assume that we already have the forward kinematic result. If

have the forward kinematic result. If you are interested in uh doing the

you are interested in uh doing the forward kinematics for it, you can you

forward kinematics for it, you can you can then try to do it by yourself on

can then try to do it by yourself on finding what's the relationship of x y

finding what's the relationship of x y and gamma with the with theta 1 d2 and

and gamma with the with theta 1 d2 and theta 3. These are the force these are

theta 3. These are the force these are forward kinematics.

But now what we need to do is to just take the derivative respect to time.

take the derivative respect to time. So if we take the first

If we take the first equation and take the derivative with with respect to time

the derivative with with respect to time then what we will get is that this this

then what we will get is that this this term will result in these two terms

term will result in these two terms we will skip the math. um this the one

we will skip the math. um this the one uh one term here result in two terms

uh one term here result in two terms here for of after the derivative of time

here for of after the derivative of time is because we have two unknown variables

is because we have two unknown variables d2 and theta 1 here.

d2 and theta 1 here. So if we take the derivative of it we

So if we take the derivative of it we will uh we will get two terms and this

will uh we will get two terms and this term if we take derivative it comes

term if we take derivative it comes here.

here. Again, we will not do the math today,

Again, we will not do the math today, but if you are interested in all of

but if you are interested in all of these math, you can take a pause and

these math, you can take a pause and work on it by yourself.

And what we need to do is is simply just to take the

take the equations and take the

with the uh derivative with respect to time.

time. So

So actually finding Jacobian matrix in my

actually finding Jacobian matrix in my opinion is easier than finding inverse

opinion is easier than finding inverse kinematic solution because we what we

kinematic solution because we what we need to do is to just get the forward

need to do is to just get the forward kinematics which is a fixed pipeline and

kinematics which is a fixed pipeline and workflow and then after we get all these

workflow and then after we get all these forward kinematics kinematics equations.

forward kinematics kinematics equations. We just take the uh derivative with

We just take the uh derivative with respect to time and then we get the we

respect to time and then we get the we get these equations and we also remember

get these equations and we also remember that we also we always want to factor

that we also we always want to factor out the joint

out the joint positions so that eventually we can then

positions so that eventually we can then factor out to get the uh the jacobian

factor out to get the uh the jacobian matrix.

matrix. So now if we do it and factor out the

So now if we do it and factor out the jacobian matrix what we will get is

jacobian matrix what we will get is here. So um recall that uh what we have

here. So um recall that uh what we have is what we have here is um

is what we have here is um theta 1 d2 and

theta 1 d2 and theta 3 and on this side is x y and

theta 3 and on this side is x y and gamma.

So you can see that eventually we can easily find the

easily find the um easily find the terms that we have.

um easily find the terms that we have. Um

Um first of for theta 1 dot um we need uh

first of for theta 1 dot um we need uh we have negative d2 s1

we have negative d2 s1 this is here and we have l3 sin theta 1

this is here and we have l3 sin theta 1 plus theta 3. So s13 here

plus theta 3. So s13 here is actually

is actually sin

sin theta 1

theta 1 plus theta 3.

plus theta 3. So you can see this is how we get the um

So you can see this is how we get the um this is how we get the uh the entries in

this is how we get the uh the entries in the jacobian matrix and the second the

the jacobian matrix and the second the second term is about d2. So for the

second term is about d2. So for the first equation we only have one term of

first equation we only have one term of d2 that is here. So what we need? So d2

d2 that is here. So what we need? So d2 is factor uh sorry these are the theta 1

is factor uh sorry these are the theta 1 dot d2 dot and theta 3 dot. So

dot d2 dot and theta 3 dot. So eventually what we get here is we just

eventually what we get here is we just need this cosine theta 1 here to be c1.

uh we will not do it one by one for this example but uh we just need to keep

example but uh we just need to keep following this these these steps and

following this these these steps and then we will be able to eventually get

then we will be able to eventually get the the entire Jacobian matrix from

the the entire Jacobian matrix from these three equations.

Now let's quick uh let's take a look on the inverse of Jacobian matrix.

the inverse of Jacobian matrix. The Jacobian matrix

The Jacobian matrix what what a Jacobian matrix can do is to

what what a Jacobian matrix can do is to map the joint velocity to the n factor

map the joint velocity to the n factor velocity. But sometimes what we need is

velocity. But sometimes what we need is to map the factor velocity back to joint

to map the factor velocity back to joint velocity. In that case we will need to

velocity. In that case we will need to take the inverse of Jacobian matrix.

But as we mentioned the inverse of the of a matrix does not

the inverse of the of a matrix does not always exist. So sometimes we if we find

always exist. So sometimes we if we find out that the determinant of the Jacobian

out that the determinant of the Jacobian matrix is equal to zero which means that

matrix is equal to zero which means that the Jacobian matrix has no inverse and

the Jacobian matrix has no inverse and there's no solution

there's no solution for the equation that we want to map the

for the equation that we want to map the antifactor velocity back to joint

antifactor velocity back to joint velocity.

velocity. In this configuration, it is called a

In this configuration, it is called a singularity where the Jacobian matrix

singularity where the Jacobian matrix has no inverse.

has no inverse. There's a there's a uh a number of

There's a there's a uh a number of phenomenon that we will we can we can

phenomenon that we will we can we can find in uh in singularity and we will

find in uh in singularity and we will mention that later on. And here uh the

mention that later on. And here uh the determinant if you are not familiar with

determinant if you are not familiar with this math again search for it or ask AI

this math again search for it or ask AI for it.

So now let's take a deeper look at the singularity because it's actually very

singularity because it's actually very important in robot arm control. We long

important in robot arm control. We long story short we generally want to avoid

story short we generally want to avoid operating our robot arm around

operating our robot arm around singularity configuration.

Uh one one way to find uh whether we are at a singularity uh configuration of the

at a singularity uh configuration of the robot

robot is to check the Jacobia matrix the rank

is to check the Jacobia matrix the rank of the Jacobia matrix. Um

of the Jacobia matrix. Um so overall singularity of a robot arm uh

so overall singularity of a robot arm uh is that is that the robot arm sometimes

is that is that the robot arm sometimes will be at a a certain configuration

will be at a a certain configuration like a a certain pose. either look like

like a a certain pose. either look like this or look like this or look like

this or look like this or look like this. Sometimes it might be in a special

this. Sometimes it might be in a special uh situation that the the inverse

uh situation that the the inverse kinematics u control of the robot arm

kinematics u control of the robot arm might not work very well. And there

might not work very well. And there might be al there can also be other

might be al there can also be other phenomenons that that singularity is

phenomenons that that singularity is very special case that the control

very special case that the control at singularity or near the singularity

at singularity or near the singularity area would be um would be trickier or

area would be um would be trickier or would have some special um

would have some special um considerations that we we have to make.

considerations that we we have to make. But first of all how do we check if we

But first of all how do we check if we are at singularity uh configuration? One

are at singularity uh configuration? One straightforward way is to check the rank

straightforward way is to check the rank of the Jacobia matrix

of the Jacobia matrix and uh we will quickly introduce what is

and uh we will quickly introduce what is rank of a matrix.

rank of a matrix. So rank of the matrix is a number of

So rank of the matrix is a number of linearly independent columns of or rows

linearly independent columns of or rows of a matrix. So if we have a if we have

of a matrix. So if we have a if we have uh an uh an m byn matrix, we will call

uh an uh an m byn matrix, we will call the matrix that is full rank if the rank

the matrix that is full rank if the rank is equal to uh the equal to m or n

is equal to uh the equal to m or n whichever one is smaller.

whichever one is smaller. Which means that we for a full red

Which means that we for a full red matrix we need uh we need all the rows

matrix we need uh we need all the rows or all the columns that that they are

or all the columns that that they are all linearly independent.

all linearly independent. For example,

For example, if we have some matrix like uh uh like 1

if we have some matrix like uh uh like 1 1 0 2 0 and 0 0 1. This is not a full

1 0 2 0 and 0 0 1. This is not a full rank matrix because you can see that the

rank matrix because you can see that the first two rows they are actually linear

first two rows they are actually linear uh linear dependent because if we if we

uh linear dependent because if we if we label this as R1 this is R2 then we have

label this as R1 this is R2 then we have a linear relationship that R1 equal to 2

a linear relationship that R1 equal to 2 R2. So this is an example that this is

R2. So this is an example that this is not

not full

full rank

and eventually we should notice that the matrix matrix is only invertible when it

matrix matrix is only invertible when it has full rank. So we must check if the

has full rank. So we must check if the Jacobian matrix is full rank or not

Jacobian matrix is full rank or not before we can actually take inverse of

before we can actually take inverse of it. Um and if the Jacobia matrix is not

it. Um and if the Jacobia matrix is not full rank, our robot is at singularity

because it is defined that if we have no inverse of Jacobia matrix then we are at

inverse of Jacobia matrix then we are at singularity. So which means that not

singularity. So which means that not full rank result in no inverse of

full rank result in no inverse of Jacobia matrix result in singularity.

So if we go back to uh Jacobian matrix um so we no longer have Jacobian matrix

um so we no longer have Jacobian matrix inverse because it is at at singularity.

inverse because it is at at singularity. If we are um if we are in the

If we are um if we are in the singularity configuration we no longer

singularity configuration we no longer have the inverse of Jacobian matrix.

have the inverse of Jacobian matrix. Instead

Instead what we will get is we can find out that

what we will get is we can find out that for some

for some joint rotations or join velocities at

joint rotations or join velocities at singularity

at singularity we might find a situation that we have some nonzero

we have uh some nonzero joint velocities but

joint velocities but the any factor velocity is zero which

the any factor velocity is zero which means that at singularity we might move

means that at singularity we might move we might rotate some joints but the any

we might rotate some joints but the any factor is not moving at all which

factor is not moving at all which somehow seems like the robot arm is

somehow seems like the robot arm is locked there even if we are rotating

locked there even if we are rotating some joints but the robot the but the

some joints but the robot the but the robot any factor is not moving

uh we will skip all the math here. Um if you are interested you can take a deeper

you are interested you can take a deeper look. But overall what I want to mention

look. But overall what I want to mention is that um

is that um if we are approaching

if we are approaching the singularity

If we are approaching the singularity, what we can find is that

we can sometimes get a finite and effective velocity. If we for example if

effective velocity. If we for example if we are very close or if we are on the

we are very close or if we are on the singularity configuration of the robot

singularity configuration of the robot arm and sometimes we just need we just

arm and sometimes we just need we just want a very small velocity or very small

want a very small velocity or very small movement of the any factor and if we are

movement of the any factor and if we are using inverse kinematics that can result

using inverse kinematics that can result in a super super large or even infinite

in a super super large or even infinite joint velocity. So you can see here that

joint velocity. So you can see here that is why I mentioned that we want to avoid

is why I mentioned that we want to avoid operating the robot using inverse

operating the robot using inverse kinematics around singularity because if

kinematics around singularity because if we are around or at singularity

we are around or at singularity configuration

configuration even if we just want a very small and

even if we just want a very small and effective velocity we can sometimes get

effective velocity we can sometimes get a very very fast join velocity which is

a very very fast join velocity which is unsafe.

So let's look at one example here. Um we will use this robot arm with three

Um we will use this robot arm with three joints. Join one, join three, join two

joints. Join one, join three, join two and join three. Um

and join three. Um and we will skip all the forward

and we will skip all the forward kinematics and the the procedures that

kinematics and the the procedures that we find the Jacobian matrix. Instead we

we find the Jacobian matrix. Instead we directly give the Jacobian matrix here

directly give the Jacobian matrix here and then we consider the poles that we

and then we consider the poles that we have set theta 1 equal to 45° theta 2

have set theta 1 equal to 45° theta 2 equal to zero theta 3 equal to 180°.

equal to zero theta 3 equal to 180°. This is actually not uh it will not

This is actually not uh it will not actually looking like this. I will not

actually looking like this. I will not draw it for now. Instead I will show it

draw it for now. Instead I will show it later. But if you're interested, you can

later. But if you're interested, you can try to draw this configuration out for

try to draw this configuration out for the robot to to take an observation on

the robot to to take an observation on what the robot look like under this

what the robot look like under this configuration.

configuration. But if we are under this configuration,

But if we are under this configuration, now we can plug in the numbers for the

now we can plug in the numbers for the uh for theta 1, theta 2 and theta 3. So

uh for theta 1, theta 2 and theta 3. So these n these numbers now they are no

these n these numbers now they are no longer unknown variable. Instead we

longer unknown variable. Instead we already know the one two three. So we

already know the one two three. So we can actually plug in the numbers here

can actually plug in the numbers here and then we can simplify the Jacobian

and then we can simplify the Jacobian matrix into this one.

matrix into this one. Um so we still keep

Um so we still keep S1 and C1 as theta ones but uh actually

S1 and C1 as theta ones but uh actually set theta 1 here is known for 45

set theta 1 here is known for 45 degrees. We do this because um we will

degrees. We do this because um we will use that later on. But overall these um

use that later on. But overall these um out of these numbers they are now known

out of these numbers they are now known known values and constant values.

So all the values in these in this matrix are known and constant.

And now let's take a look um is this a is this a singularity configuration or

is this a singularity configuration or not.

And also if you want to uh further consider uh you can try it by yourself

consider uh you can try it by yourself that uh this is

that uh this is this exam this configur configuration

this exam this configur configuration set theta 1 equal to 45° theta 2 equal

set theta 1 equal to 45° theta 2 equal to zero and theta 3 equal to 180 is what

to zero and theta 3 equal to 180 is what we are going to show next but you can

we are going to show next but you can also try yourself on can you find

also try yourself on can you find another singularity pose of this robot

another singularity pose of this robot arm

arm but let's take a look on the current

but let's take a look on the current configuration. If we draw the

configuration. If we draw the configuration out, we can actually find

configuration out, we can actually find out that the robot will look like this

out that the robot will look like this because you can see that you can see

because you can see that you can see that SATA one is 45,

that SATA one is 45, SA 2 is zero and SATA 3 is 180.

SA 2 is zero and SATA 3 is 180. And then

And then if we look at the Jacobian matrix we we

if we look at the Jacobian matrix we we should pay attention to that sin 45

should pay attention to that sin 45 equal to equal to cosine 45 = to 0.707

that actually means that we have s1 equal to c1.

equal to c1. So now if we look at the first two rows

So now if we look at the first two rows of the Jacobian matrix, we can label the

of the Jacobian matrix, we can label the first row as R1, the second row as R2.

And what we can find out is that R1 and R2, they are not linearly independent.

R2, they are not linearly independent. Instead, we actually have R1 equal to

Instead, we actually have R1 equal to negative R2. They have a linear

negative R2. They have a linear relationship which means that this

relationship which means that this Jacobian matrix is not full rank and we

Jacobian matrix is not full rank and we cannot take inverse of it and our robot

cannot take inverse of it and our robot is in singularity for this

is in singularity for this configuration.

on the robot arm, no matter how I rotate joint one, joint two, and no matter if I

joint one, joint two, and no matter if I rotate join one, join two or joint

rotate join one, join two or joint three, they always they all just result

three, they always they all just result in the manufacturer to move in these two

in the manufacturer to move in these two green

green directions. We cannot move in the in

directions. We cannot move in the in this red direction

this red direction at this moment.

at this moment. So that you can see that then we lose

So that you can see that then we lose one degree of freedom of the factor

one degree of freedom of the factor movement since we can only move to the

movement since we can only move to the green direction instead of the red

green direction instead of the red direction. And well if our robot is in

direction. And well if our robot is in other configurations let's for example

other configurations let's for example do it here

do it here then at this configuration we it's very

then at this configuration we it's very likely that we can move both

likely that we can move both horizontally and vertically. But here at

horizontally and vertically. But here at this singularity configuration,

this singularity configuration, we can only move along the green

we can only move along the green direction instead of the red.

And furthermore, if we re if we remove

if we re if we remove this

this definition,

definition, instead we just keep setta 1 as a ar as

instead we just keep setta 1 as a ar as an arbitrary value and we only keep

an arbitrary value and we only keep theta 2 equal to zero and set theta 3

theta 2 equal to zero and set theta 3 equal to 180. And we will still be able

equal to 180. And we will still be able to find out that um

to find out that um for the row one and row two they still

for the row one and row two they still have a relationship that row one over

have a relationship that row one over row two equal to tangent theta 1 and

row two equal to tangent theta 1 and tangent theta 1 will be a like

tangent theta 1 will be a like regardless what number we choose for

regardless what number we choose for what rotation we choose for uh joint one

what rotation we choose for uh joint one this will be a constant number which

this will be a constant number which means that even if we even if say the

means that even if we even if say the is at other angle other than 45° we

is at other angle other than 45° we still have row one and row two they they

still have row one and row two they they are they have a linear relationship. So

are they have a linear relationship. So actually the singularity here is that as

actually the singularity here is that as long as we have set theta 2 and theta 3

long as we have set theta 2 and theta 3 equal to 180 degree then

equal to 180 degree then regardless of where joint one is this

regardless of where joint one is this robot arm is at singularity. It's

robot arm is at singularity. It's actually very easy to understand like we

actually very easy to understand like we can overall rotate this configuration

can overall rotate this configuration uh while fixing uh theta 3 and theta 2

uh while fixing uh theta 3 and theta 2 only rotates theta one like this big rod

only rotates theta one like this big rod uh rotating like here or here. You can

uh rotating like here or here. You can actually see that we are facing a very

actually see that we are facing a very very similar uh situation and the robot

very similar uh situation and the robot arm is in singularity.

And another uh easy example here is also that if we have all the

that if we have all the if we have the entire robot arm to be

if we have the entire robot arm to be fully stretching

fully stretching then under this configuration no matter

then under this configuration no matter if we rotate joint one or rotate joint

if we rotate joint one or rotate joint two or rotate joint three they all

two or rotate joint three they all result in the green direction of the

result in the green direction of the factor movement and the antifactor

factor movement and the antifactor cannot move upward or

cannot move upward or cannot move up or lower. So this

cannot move up or lower. So this direction

direction of movement will be lost if we are under

of movement will be lost if we are under this configuration. It might be less

this configuration. It might be less intuitive that of course you will agree

intuitive that of course you will agree with me that we cannot move up

with me that we cannot move up but you might argue that we might be

but you might argue that we might be able to move here. This is definitely

able to move here. This is definitely within the robot workspace. But

what we will see here is that at this specific configuration right as right as

specific configuration right as right as this moment all the three joint

this moment all the three joint rotations will only result in horizontal

rotations will only result in horizontal movement.

movement. If somehow we controlled

If somehow we controlled the robot arm to a lower the robot any

the robot arm to a lower the robot any factor to a lower location, it's very

factor to a lower location, it's very it's almost certain that our trajectory

it's almost certain that our trajectory will be a curve like this. We cannot

will be a curve like this. We cannot move vertically down because you can

move vertically down because you can imagine the rotation of the joints in

imagine the rotation of the joints in your mind and you can you can see that

your mind and you can you can see that vertically moving down at this configur

vertically moving down at this configur configuration is impossible. So we are

configuration is impossible. So we are losing one degree of freedom and our

losing one degree of freedom and our robot arm is in singularity

and we will then quickly take a look on velocity mapping and force mapping.

velocity mapping and force mapping. So for velocity mapping um we here we

So for velocity mapping um we here we are using a very very simple example

are using a very very simple example here

here that uh we just have one joint

that uh we just have one joint and one link. So very very simple

and one link. So very very simple example

example and

and then we can find out that the any factor velocity V will be equal to R the

factor velocity V will be equal to R the length of the link

length of the link times theta dot which is the velocity of

times theta dot which is the velocity of the joint

and and then we can project the factor velocity

the factor velocity V

V into X direction and Y direction. We can

into X direction and Y direction. We can project V into Vx and V Y.

project V into Vx and V Y. And because this angle because this

And because this angle because this angle is theta

angle is theta so that this angle will be theta and

so that this angle will be theta and this angle will be 90° minus theta.

Sorry. Uh this angle will be 90° minus theta.

theta. And actually then we will have

this is theta. So vx will be uh because it's pointing to the left so it's in the

it's pointing to the left so it's in the negative direction of x. So vx will be

negative direction of x. So vx will be negative r

negative r uh times uh sin uh sin theta time theta

uh times uh sin uh sin theta time theta dot and similarly y here uh v y here

dot and similarly y here uh v y here will be r um will be r theta dot times

will be r um will be r theta dot times cosine theta dot because it's uh it's

cosine theta dot because it's uh it's actually this vector

and we simply can just uh write the write the these two equations into a

write the these two equations into a matrix form

matrix form as this.

as this. So you can actually see that this is

So you can actually see that this is actually the um

actually the um the Q

dot. Uh here I I use a bad notation but this

Uh here I I use a bad notation but this this should be labeled as Q dot uh just

this should be labeled as Q dot uh just just to be uh less ambiguous. Um

just to be uh less ambiguous. Um and then on the right hand side it's

and then on the right hand side it's theta dot times this

vector or it can still be considered as a as a as a 2x1 matrix. So actually by

a as a as a 2x1 matrix. So actually by velocity mapping we can also find the

velocity mapping we can also find the Jacobian matrix because this is Q dot

Jacobian matrix because this is Q dot this is theta dot. So this is actually

this is theta dot. So this is actually the Jacobian matrix.

So we can use this Jacobian matrix to map between the joint velocity and any

map between the joint velocity and any factor velocity.

And similarly we can also do the force mapping.

mapping. For the force mapping uh it will be a

For the force mapping uh it will be a bit more uh complicated and due to the

bit more uh complicated and due to the limited of time we will not go through

limited of time we will not go through the force mapping uh here uh for the

the force mapping uh here uh for the details at this moment. But what I want

details at this moment. But what I want to what I want to show that uh if we do

to what I want to show that uh if we do all the force mapping if we do all the

all the force mapping if we do all the math in this page

math in this page eventually what we can get for

eventually what we can get for singularity is one phenomenon that uh

singularity is one phenomenon that uh can help us to to find the singularity

can help us to to find the singularity that is

that is at singularity

it I'm not saying all forces there will be a certain force or a couple of

be a certain force or a couple of certain forces

certain forces on the any factor that will be passively

on the any factor that will be passively fully resisted by the

fully resisted by the robot arm structure

mentioned that is passively and fully resisted. It's very important that uh is

resisted. It's very important that uh is passively and fully resisted. This might

passively and fully resisted. This might look look like a very complicated

look look like a very complicated statement but

statement but in fact it is pretty intuitive to feel.

in fact it is pretty intuitive to feel. That is

That is if you fully stretch your arm and you

if you fully stretch your arm and you try to push on the wall. What you will

try to push on the wall. What you will feel is that your elbow is not is not

feel is that your elbow is not is not actively resisting it because the force

actively resisting it because the force d the force directly transmitted through

d the force directly transmitted through the bones. My elbow is not going to need

the bones. My elbow is not going to need any force to resist the external force.

any force to resist the external force. Instead if I have my elbow to be not not

Instead if I have my elbow to be not not uh to be not fully stretch stretched and

uh to be not fully stretch stretched and if I have external force and my elbow

if I have external force and my elbow actually need to fight against it. So

actually need to fight against it. So you can see that here it's actually

you can see that here it's actually intuitive one intuitive one intuitive

intuitive one intuitive one intuitive way to find singularity that at

way to find singularity that at singularity there will be a certain

singularity there will be a certain force on the factor that we push on the

force on the factor that we push on the wall. So the wall is giving us a force

wall. So the wall is giving us a force that is passing our arm and this force

that is passing our arm and this force is fully passively resisted.

is fully passively resisted. So this actually indicate that for many

So this actually indicate that for many robot arms as long as the robot arm is

robot arms as long as the robot arm is fully stretched

then the force that is pointing vertically

vertically pointing directly through the robot arm

pointing directly through the robot arm will be passively resisted. So for most

will be passively resisted. So for most of the robot arms

of the robot arms if we fully stretch the robot arm it is

if we fully stretch the robot arm it is very likely to be a singularity as you

very likely to be a singularity as you if you work with any robot you fully

if you work with any robot you fully stretch it you try to push on it if

stretch it you try to push on it if there's no joint

there's no joint if there's is no joint that you you find

if there's is no joint that you you find out that is actively resisting the force

out that is actively resisting the force then this configuration will be a

then this configuration will be a singularity.

singularity. So this is also one of the reason that

So this is also one of the reason that for many robot arms you see their uh

for many robot arms you see their uh their default pose or their default

their default pose or their default configuration or their homing

configuration or their homing configuration is not fully stretching

configuration is not fully stretching out. Instead it's kind of like folding

out. Instead it's kind of like folding in a certain configuration instead of

in a certain configuration instead of fully stretching out because that is

fully stretching out because that is singularity and we generally we do not

singularity and we generally we do not want to uh operate the robot around

want to uh operate the robot around singularity configuration.

All right, that is all of the lecture today. Uh there we covered a lot of

today. Uh there we covered a lot of details today. Um and this is a very

details today. Um and this is a very fast pace. Um

fast pace. Um it will be uh you are you are only

it will be uh you are you are only required to um to conceptionally

required to um to conceptionally understand the inverse kinematics and

understand the inverse kinematics and the differential kinematics and the

the differential kinematics and the singularity. But if you are interested

singularity. But if you are interested in uh all the mathematical details,

in uh all the mathematical details, please message me and we can have

please message me and we can have further discussions and we I can give

further discussions and we I can give you and share I can share with you more

you and share I can share with you more uh materials. Thank you.

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YouTube Transcript:GIX MSTI TECHIN 517 Inverse and Differential Kinematics Lecture by Haonan Peng