Motion Madness

I was having one of our new Sales Engineers work on a timing study for movement of a table and he said he would get a much faster move with a triangular motion profile vs. a traditional 1/3, 1/3, 1/3 trapazoidal motion.

The triangular move has a constant acceleration for the first half of the move time and a constant deceleration for the second half of the move time. In the traditional 1/3, 1/3, 1/3 trapazoid move, there is a constant acceleration for 1/3 of the move time, then movement at a constant velocity for 1/3 of the move time and finally, a constant deceleration for 1/3 of the move time.

These motion profiles are shown in the graphs below:



For the system we were working on, the actuator has a maximum acceleration of 0.3g where g=9.81 m/s^2. The move in question was a 24" move.

1. Using the maximum acceleration and distance given, how much faster is the triangle move vs. the 1/3 trapazoid?
2. What is the general formula for the time of a triangular move given an acceleration and distance?
3. What is the general formula for the time of a 1/3 trapazoidal move given an acceleration and distance?

Assume the deceleration is the same as the acceleration for all cases.


Bonus Questions:

1. What is the ratio of Time required for a 1/3 Trapazoid move vs. the time required for an equivalent Triangular move (i.e. same acceleration and distance)?
2. What is the ratio of maximum velocity achieved in the Triangular move vs. maximum velocity achieved in the 1/3 trapazoidal move?
3. For the Trapazoidal move,
   • What percentage of the total distance is traveled during the acceleration part of the motion?
   • What percentage of the total distance is traveled during the constant velocity part of the motion?
   • Explain how you can determine the above two percentages without using the equations: x=(1/2)*a*(t^2) or x=v*t. (where x is distance, a is acceleration, v is velocity and t is time).

This should be fun.

I can think of a lot more questions.
I am assuming Norm knows the answers and this thread is a test so I shouldn't ruin the fun.

I can see there are no one that knows the answers or can calculate these answers quickly.

I guess with all of Ron's math questions I wanted to get in on the fun. Yes, I believe I know the answers... and there are many more questions that can be asked but it appears I've already overloaded the input buffer

If you want fun...

ndzied1 said:

I guess with all of Ron's math questions I wanted to get in on the fun.

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You should have tried the rollercoaster loop diameter problems.

OK, time for a shot:


1) It takes 0.05523 seconds longer for the trapaziodal move
2) t = 2*SQRT(d/a)
3) t = 3*SQRT(d/(2a))

BONUS

1) (3/2)*SQRT(0.5) = 1.06066
2) SQRT(2) = 1.414213
3a) 1/4 of the total distance
3b) 1/2 of the total distance
3c) Use geometry. The area under a velocity versus time curve is the distance traveled. A standard 1/3-1/3-1/3 trapaziodal move splits the move time equally between the move phases. The end velocity of the acceleration is the constant velocity. Given that, the area under the constant constant velocity portion of the curve is the area in a rectangle. The area under the accel (and decel) phase is the area in a right triangle, which is one half that of a rectangle with the same base and height. This gives us a distance traveled relationship of 2:1 constant speed to accel. Given the 2:1 distance ratio and that there are two accel phases to the move we can deduce that the constant speed distance must be 1/2 the total distance.

ndzied1 said:

a traditional 1/3, 1/3, 1/3 trapazoidal motion.

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It must be very region specific tradition. Never heard of.

I'm also intrigued by the 1/3, 1/3, 1/3 specification. Most of the motion applications I've dealt with want to complete the move as rapidly as possible while ending the move at the target position within the specified tolerance band.

The trapezoidal motion profile happens when you have to place a limit on how fast the axis is allowed to move. You accelerate up to the maximum velocity and maintain that velocity until you have to start decelerating so that you reach the target position at zero velocity.

If the move is a short enough distance, you get halfway to the target position before you reach the maximum velocity so you have to start decelerating. That's the triangular velocity profile.

I've never seen an application where the profile was specified to be 1/3 of the time accelerating, 1/3 at constant velocity, and 1/3 decelerating.

The 1/3-1/3-1/3 trapaziodal profile is one of the better 'simple' profiles to use because it tends to limit velocity while maximizing power utilization for a given motion.

The initial example limited the available peak acceleration. While this does happen on occasion it is more often the case that either velocity or RMS torque (accleration) are the limiting factor. In these cases the 1/3-1/3-1/3 profile helps on both counts.

As an example, take the triangular profile from above, figure out the total move time, divide it by three and come up withe the acceleration rate needed to complete the trapaziodal move in the same time as the triangular move. You will find that the peak torque in this case is only 12.5% higer than the torque in a triangular profile of the same time and distance.

Furthermore, if you do the RMS torque (acceleration) calculation you will find that the trapaziodal move requires roughly 8% less RMS torque to accomplish the same move in the same time.

And, since the max velocity is lower, you can increase the axis gearing, which will increase your available peak acceleration rate.

As a side consideration, most of the high performance servo motors use rare earth magnets for high torque density. These magnets will develop eddy currents as they rotate through the stator magentic field, generating heat. This is why most servo motor torque/speed curves drop off so bad as speed increases. By keeping your speeds lower, the 1/3-1/3-1/3 profile further aids in RMS torque capacity, allowing you to increase the accel rate further yet.

Keith

All right you guys -- where the XXXX did I put that college physics book?

Wish I worked with these calcs to stay on top of em. I admit I can calc acceleration and get velocity AND I can calculate distance with constant velocity BUT I get all screwed up when there is changing velocity (acceleration).

The other one I just cannot get my mind to accept is constant torque. ON many motor driven loads (excepting pumps and some blowrs) torque is constant. HP rises with increasing speed but not torque - I just cannot get my head wrapped around that one and get my mind to accept it automatically.

Dan Bentler

leitmotif said:

The other one I just cannot get my mind to accept is constant torque. ON many motor driven loads (excepting pumps and some blowrs) torque is constant. HP rises with increasing speed but not torque - I just cannot get my head wrapped around that one and get my mind to accept it automatically.

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I hate to make your day worse, but positive displacement blowers (Roots rotary lobe type) and positive displacement pumps (gear or vane types) operate as constant torque devices.

Let's take a gear pump as an example. Look at it this way - the discharge pressure is constant on the gear teeth at the outside diameter of the pump.

That creates a constant force resisting gear rotation (Force = Pressure x Area)

That creates a constant torque on the pump shaft (Torque = Force x Radius)

That creates a constant torque load on the motor.

Now, since hp = Torque_lbft x rpm / 5252 the faster you rotate the pump the more horsepower you need. That makes sense, because you are moving more fluid.

Tom
I meant centrifugal pumps. Thanks for info on roots and gear pumps.

I think the problem is I cannot "see" or put my hands on torque.
I have to keep going back to HP = torque x RPM div'd by 5252
to make my dang mind BEHAVE and think right.

most frustrating.

Dan Bentler

jacekd said:

It must be very region specific tradition. Never heard of.

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The only place where I have heard of 1/3 1/3 1/3 profiles is from hydraulic guys. Although Norm was specific about the thirds being time I have normally heard of the thirds referring to distance. I wouldn't call 1/3 1/3 1/3 profiles traditional. I don't think most people can calculate the time it takes to make the full move knowing just the acceleration. Most of our customers just increase the velocity and acceleration rates until the control gets sloppy and back off a bit. Most servo motion controllers express accelerations in counts/sec^2 or engineering units / sec^2.

The trapezoidal and triangular motion profiles are not achievable. One can be close if the power to mass ratio is large and you like abusing machinery. I normally like to use a sinusoidal motion or a 5th order polynomial to keep the motion as smooth as possible. We have commands that allow one to just specify the destination and the time and the controller calculates a smooth motion profile to get there in the allotted time.


Well Norm, did Keith get the right answer? Mixing unit types is not good. I usually use G = 386.4 in/sec^2.

Originally posted by Peter Nachtwey:

Although Norm was specific about the thirds being time I have normally heard of the thirds referring to distance.

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Interesting. Peter's reference is the first time I've heard of the thirds referring to distance.

Peter's point about the triangular and trapaziodal profiles not being achievable is a good point. But I always thought it was more of a compliance issue than a mass issue.

Originally posted by Peter Nachtwey:

Mixing unit types is not good.

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Only if you're NASA.

Keith

kamenges said:

Peter's point about the triangular and trapaziodal profiles not being achievable is a good point. But I always thought it was more of a compliance issue than a mass issue.
Keith

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It takes force to acclerate a mass. The greater the mass the greater the force. Metal deforms or gives roughly proportional to the force applied. What is you definition of compliance. Give units. I use natrual frequency as a measure. The lower the natural frquency the more compliant.

Natural frequency = sqrt(k/m) where k is the spring constant with units of force / distance. m is the mass. The result is 1/sec which is a frequency in radians per second.

Originally posted by Peter Nachtwey:

What is you definition of compliance. Give units.

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When I refer to compliance, rightly or wrongly, I am talking about the reciprocal of stiffness or spring constant. So the units would be distance per force.

What threw me a little in your post was you seemed to say that high power to mass ratio will allow you to run infinite jerk profiles more effectively. What good is a high power to mass ratio if then power is transmitted through a rubber band (highly compliant)? You still have a system with a low natural frequency.

Keith