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Robot Dynamics Overview
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0:05
good morning this is Boaz Eidelberg and today's discussion will be robot dynamics we'll take a look at physical environment of robot we look at the kinematic diagrams will define its coordinate system will define its variables will define the problem and we look at the general solution to a dynamic problem here's an example of an articulated robot which you see here that the robot has a revolute joint at the base it has another revolute for the
0:44
first link and another revolute joint for the second link it has a gripper with few degrees of freedom here and a to Center Point which holds the part
1:05
if you look at the description of this robot in kinematic diagram we see similar description with revolution for the base revolution for one arm revolution for another arm to Centrepointe and also we have some reference scale to determine the size a general size we call this robot triple R which is standing four revolute revolute and revolute the kinematic diagram is showing an inertial frame x0 y0 z0 fixed to the base it shows an axis x1 y1 and
1:57
z1 which is rotating with respect to x0 in the direction of z0 a certain angle and this is the yaw and then between x1 to x2 we have a pitch angle and then this represents x2 and then in coordinate system - we have not on x2 was a little bit offset we have the coordinate origin of system 3 where the excesses along the link and Z and y shown over here and coordinate system and basically this is the representation of the robot coordinate system over here
2:55
we see a different view to get oriented pretty much a front view of the same configuration to simplify the definition of various variables we remove the robot and show just the coordinate system and here's an example of it may get complicated to describe the various parameters and variables and we like to define the kinematic variables position velocity acceleration over here we see the position theta 1 theta 2 and theta 3 of the angles that we saw
3:34
earlier as well as their derivatives velocity and acceleration for each one of them we would like to note the positive direction of these variables the theta is positive in the direction shown because it's rotating extra world towards y0 in a direction of z0 which is exactly the right hand rule of right hand screw however if we look at x2 x2 is rotating in a positive direction of X of Y 2 as shown but it is actually opposite to the right hand
4:23
screw it will not advance right hand screw and this rotation as shown here with this will advance in a direction that's why we have a minus sign here and in this case where we showing a positive rotation in this direction for going from X 2 to X 3 it is going again in a positive direction of Y 2 and this is the right hand screw we move in this direction and the screw will advance in the direction of Y 2 other parameters that we have over here is various
5:03
position vectors such as R 1 R 2 or 3 you see to get to the still set point we have this vector 1 2 plus R 3 they are oriented in space each one in its own coordinate system can be easily defined and we also have in a similar way the definition of position of the center of gravity's finally we need the inertial parameters of the system such as mass of link three and mass of link 2 and the inertia of link 2 and the inertia of link 3 so these are
5:45
the parameters and let's take a look at the problem definition in general what we're looking for is the robot is doing some work and he has some forces and loads on a gripper and we'd like to know what are the required motor torques in each one of these joints and the reaction on a bearing in order to do some sizing so finally required joint motor torques and resulting joint forces and torques reactions for a given load and robot parameters the robot
6:20
parameters as we saw a size inertia and motion and the loads the gripper force and a gripper torque as a result of the work that were what is doing now in order to find a solution we have to start looking at the free body diagram of a typical link we see a typical link here this link has a center of gravity and has two joint this joint is where link 0 neighboring link is acting on the in quantities we call this one link 1 this joint over here are they connecting
6:58
to the two center point or to enable link which is linked to so the the solution involved in first setting up the the dynamic equations and we have two set of equations for each link one is the linear equilibrium of forces for this link the other one is the rotational equilibrium of moments now let's take a look at the summation of forces now the forces that act on this link is f01 the fourth link still acting on one we also have similarly f21 the fourth link to is acting on the link one these
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are these two guys then we have acting at the CG an inertial force and we have acting on a CG a gravitational force so we have these two four forces acting on link 1 and the equilibrium equation sets them to 0 now the inertial force is equal to minus 1 times the absolute acceleration of the CG of link 1 this is so-called Dalambert principle we take Newton's law F equals MA we plug it into the summation of forces equal to 0 and we treat it as just a static force with
8:43
a minus sign here the gravitational forces and 1 times the gravity and this basically is one equation the second equation is the Euler rotational equation if similar we take the summation of torques about the CG equal to zero it's a little bit more complicated but let's take a look at the details first of all we have the torque that link 0 is acting on 1 this is the torque that links that link 0 is acting on 1 at joint 1 similarly we have the link that linked
9:19
the torque that link 2 is acting on 1 this is the link 2 is acting on one it could be the gripper but for now let's just assume that let's say another link then we have the torque that this force is acting on the CG force 0 1 has a radius vector from the CG to the point of application of the force now this vector over here you see is equal to the minus of R 1 plus R 1 CG so the reason why it's - is because the sum of these two vectors will be a
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vector from here to here but we interested in a vector from the c.g. to the point of application of force that's why we have this minus sign here and we have to cross it with this force over here similarly we have a vector from the CG to the point of application of this force and this is minus r1 CG so that's why we have a minus sign here very important to remember that and we have to cross it with this force and finally we have the torque of the inertial torque
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that has to all sum up to zero now the inertial torque has two components the inertial torque basically is the rate of change in time of the angular momentum the angular momentum equals to the moment of inertia times the angular velocity so if we take the derivative of this expression the first term will be the moment of inertia times the acceleration and this is this term is over here moment of inertia I of link 1 times the absolute angular acceleration
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of this link is equal to the inertial torque as we usually know it another term which is not that familiar with is the second term where we take Omega the rate of change of velocity I mean the velocity the rate of change position velocity of this link in space and we have two absolute the last thing we have to cross it with the momentum because the momentum may change direction and therefore it has to be taken as part of the derivative as we saw in earlier
12:01
presentations and we call this particular term the gyroscopic torque so this the total initial torque and we have to remember to put a minus sign when we plug it in here so to get a solution we would have to take the equation of summation of forces and the first thing that we want to do is to solve for the unknown force which is over here we know these forces because they're coming in from previous solutions of previous length we're starting with the one that is engaged with the gripper where we
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know the forces and torques and then we come down to find the reaction torque and reaction force and once we go through the chain this will be known and what we'll be looking at this particular case is for F 0 1 for this particular force here so what we do we solve for it from this equation and this means that we have to take everything else to the other side with a minus sign so we take this is minus F 2 and this is minus F initial but we remember that it has a
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minus sign here and this is minus the the gravitational force so then we take care of this we move this you see that this term becomes plus so we have to be careful with this we go step by step and eliminate these redundant signs to end up with the final size for all these stairs now similarly if we want to solve for the torque of the of this neighboring link on on the link of interest then what we have to do we have to take this equation or the equation
14:11
and solve for the talk of interest so this is the 1 and then everything else moving to the other side with a minus sign you see we take in t2 one with the minus sign we're taking this whole expression with the minus sign and we have to remember that it was before minus sign so it's minus minus we have to take care of that again we have minus minus and then the inertia force we have minus and again two minuses so when we add them all up and
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we take care of these duplicate minus s we get minus t21 and this two masses will become plus and these two masses will come plus and then over here we'll have pluses so this will be the only minus sign after we converted to the other side of the equation now this was for one link the generall solution for robot with n links will be n times two sets of equations one linear and one rotary for each link so what we have to do is to just keep on going through the
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chain from the gripper to the first link find the reactions then know in this reaction go to the second link find the reaction and then go to the base and find the reaction but when we do the solution for a link we have to keep in mind that all the vectors which are engaged in that particular equation must be resolved to components in that coordinate system see this link one has its coordinate system and everything all the vectors you no matter where the
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coming from this guy may be coming in from a different coordinate system we have to use the transformation matrix to resolve it to this coordinates otherwise we will end up with an error so here's an example just took one one example the the Newton's equilibrium equation with the solution F 2 1 was probably represented in coordinates system to this force over here where it will serve but now we have to use it in coordinate system 1 therefore we have to hit it with the transformation matrix
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from coordinate system 2 to 1 a vector which was represented in coordinate system 2 and this will bring it as a happy camper in this equation and it could actually be added and subtracted to all the other components which will be represented in coordinate system 1 now we have here absolute acceleration of the CG and this also must be in coordinate system 1 but when we look at its expression then we can see that the absolute acceleration of this point is
17:19
equal to the absolute acceleration of a reference point and let's call this joint 0 1 so this is the absolute acceleration of the CG in coordinate system 1 equals to the absolute acceleration of joint 1 ok when it was actually presented in coordinate system one that's where we calculated it and then we have to hit it with a transformation method from 0 to 1 in order to be a happy camper over here in this equation otherwise this coordinates
18:03
is not going to add up to the other one and this is an example how we do that now when we talk about this particular transformation of Omega Omega 1 is the absolute angular velocity of link 1 but it is equal to the sum of the two relative velocity 1 velocity of 0 1 respect to inertia and 1 velocity of system 1 with respect to system 0 which is Omega 1 remember theta 1 was the angle of rotation of coordinate system 1 respect to zero so Omega 1 is a relative angular
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velocity so we have to take this Omega 0 which was probably conveniently represented in coordinate system 0 and hit it with the transformation matrix from coordinate system 0 to 1 and similarly we will have to do a so this get a little bit complicated and the only way to get through is number 1 to be careful with carying over all these signs a number 2 at the end of a solution we like to check it graphically to make sure that everything is correct
19:38
and many times with the graphical solution you find mistakes so in conclusion the general solution of robot dynamics problem includes drawing kinematic diagram which has all the degrees of freedom defining coordinate system for the joints and links define variables and parameters write two Newton and Euler vector equations and is the number of degrees of freedom add to each equation all the required transformations define how the
20:09
transformation matrices will look like and then solve these equations one at a time using computerized tools such as mathcad that we saw before in previous presentation or MATLAB or C+ C++ Excel etc and finally check the results graphically with a CAD system just to make sure that there's no errors in a presentation of solution so this pretty much concludes this presentation and in upcoming presentations we take a look at some example problems thank you
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