Feeder Accel,Decel calculations

johnfarrugi

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Join Date
Mar 2008
Location
London
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I'm working on a press feeder and I'm trying to figure out the calculations to determine what accel and decel rates I need.
I know the press runs at 20 strokes per minute. I know the Feed length is 12 inches and the feed angle is 60 degrees.

60sec/20 strokes= 3 seconds per stroke.
3sec/360 degrees= 0.008333 sec per degree * 60 degrees= 0.5 seconds

So now I know I have 0.5 seconds to feed 12 inches

I have a display which allows the operator to select the accel and decel percentages of the entire move. Meaning, if the operator selects 20% accel and 10% decel then the first 20% of the feed move is used to accelerate untill full speed. Travel at full speed for %70 and then decel for the last %10 of the feed.

I'm not sure how to calculate the acceleration/Deceleration and how to determine what the max speed will be.

Can anyone help me with the calculations?
 
johnfarrugi said:
So now I know I have 0.5 seconds to feed 12 inches

I have a display which allows the operator to select the accel and decel percentages of the entire move. Meaning, if the operator selects 20% accel and 10% decel then the first 20% of the feed move
By time or distance?

is used to accelerate untill full speed. Travel at full speed for %70 and then decel for the last %10 of the feed.

I'm not sure how to calculate the acceleration/Deceleration and how to determine what the max speed will be.

Can anyone help me with the calculations?
Hmmm, two unknowns. This makes a good Friday night quiz.
 
Peter Nachtwey said:
By time or distance?

Either one, its the same. The distance is a fixed 12 inches and the entire move most be completed in 0.5 seconds.

If they set the accel for 20% then 20% of 12inches is 2.4 and 20% of 0.5 seconds is 0.1 seconds. That means it must accelerate from 0 inchs to 2.4 inchs in 0.1 seconds.

Also,

If they set the decel for 10% then 10% of 12inches is 1.2 and 10% of 0.5 seconds is 0.05 seconds. That means it must decelerate from max speed to 0 in 0.05 seconds.

So now

We know the accel is 20% and the decel is 10% that leaves the constant velocity move at 70%. So, if 2.4 inches is used on the accel and 1.2 inches on the decel, that leaves 8.4 inches for the constant velocity move that must be completed in 0.5-0.1-0.05 = 0.35 seconds


I just dont understand how I determine the accel and decel rate and what the constant velocity will need to be for the feeder to move 12 inches in 0.5 seconds given the accel and decel percentages.

Hope that clears things up!
 
johnfarrugi said:
Peter Nachtwey said:
By time or distance?

Either one, its the same.
How can you be sure if you can't do the calculations?
Does any body else think that specifying the percentages of time or distance will yield the same results?
However, since you used distance we should probably specify the percentages by distance. Oops. There are three unknowns! That is 50% more fun!

I solve it. I doubt anybody will solve except maybe Norm and that is because he has Mathcad. I am going to try use Maxima.

John, I computed at typical Velocity, AccelRate and DecelRate will be on the order of 40 in/sec, 200 in/sec^2. I assumed the AccelFraction was 1/3 and DecelFraction was 1/3 of the total distance. That is pretty fast.
 
Last edited:
I did this with Maxima but I actually did most of the optimizing by hand.

/* Inputs
(%i13) TotalDist: 12;
TotalTime: 0.5;
AccelFrac: 1/3;
DecelFrac: 1/3;
/* These are the formulas for velocity, AccelRate and DecelRate */
Velocity: TotalDist*(1+AccelFrac+DecelFrac)/TotalTime;
AccelRate: Velocity^2/(2*TotalDist*AccelFrac);
DecelRate: Velocity^2/(2*TotalDist*DecelFrac);
(%o13) 12
(%o14) 0.5
(%o15) 1/3
(%o16) 1/3
(%o17) 40.0
(%o18) 200.0
(%o19) 200.0

It actually wasn't that hard. It just required solving three equations and three unknowns.
 
Peter,

Looking at your calculation, The 1/3 accelfrac and decelfrac look like the percentage I am trying to adjust. 1/3 being 33% of the total travel. Am I correct? If I change the accel percentage to 20% and the decel to 10% then the equation should look like this:

velocity = (12 x(1+0.2+0.1))/0.5 = 31.2 inch/sec
Accel Rate =31.2^2/(2 x 12 x 0.2) = 202.8 inch/sec^2
Decel Rate =31.2^2/(2 x 12 x 0.1) = 405.6 inch/sec^2

Does that seem right?
 
You are correct.

Notice that the deceleration rate is very high. That is over 1 g.
I would not give the operator the ability to change the ramp fractons or percentages. I think it is up to you to find those values for acceleration and deceleratin that will not push the limit. I like my suggested fractions but it may result in a speed faster than what the servo can go. It is much easier to control the lower derivatices ( speed vs acceleration or position vs speed ) so I would favor more speed and less acceleration and deceleration. There is no need to apply excess force ( friction, wear and tear ) on the machine if the motion will still occur in 0.5 seconds.
 
This problem must have blown every one away

No one tried and no one even asked how to derive the equations.

John, it does make a difference if the percentages are expressed in time or distance. Norm had a thread a long time ago that is related to this. Basically it suggest using 1/3 of the time for ramping up, 1/3 of the time at constant velocity and 1/3 of the time for decelerations. In this case 1/4 of the distance is used to ramp up. 1/2 of the time is spent at constant velocity and 1/4 of the time is spent ramping down. I found that Norm's thread had merit so I worked out the details so I could have general rules that for quick calculations.

The basic formulas I worked out these.
Velocity=1.5*distance/time
Acceleration=4.5*distance/time^2

This assumes the acceleration and deceleration rates are equal.
Using your distance and time results in a velocity of 32 in/sec and a acceleration and deceleration of 216 inches per second^2. I have an old SWEO drive manual the compares the efficiency of a different motion profiles for fast point to point moves assuming the motion profile was capable of infinite jerk. The parabolic profile was the most efficient. Next was the 1/3,1/3,1/3 profile and last was the 1/2,1/2 or triangle profile. The question I have is about the use of linear ramps in these calculations instead of s-curves. Linear ramps assume the motor can change acceleration rates instantly. This can't happen in reality. A good motion controller will compute s-curves for your point to point move given the time and distance.
Norm showed how to compute 7th order motion profiles for point to point moves a few months ago.
 
CharlesM said:
I was planning on looking at your flying shear program that you posted a while back. Seems to me this is almost the same type of application.
johnfarrugi says his application is timed based and not geared so johnfarrugi's application is very simple once the formulas are computed. We can do these calculations just before executing a simple Move Absolute command. If we don't didn't need different accelerations and decelerations then we could do this with a Time Move Absolute command where all that is required is the destination and the time to get there.

If johnfarrugi can get back the press rate from and encoder then the feeder can be synchronized to the press rate. We can do that using a Clutch-By-Distance command or Advanced Gear Move command. If you ever need to do something like that just let us know. It should be simple.
 
So in my application you would suggest an MAM move? Would I use an MAG move to make sure the following rolls moved with the master roll?
 
johnfarrugi said:
So in my application you would suggest an MAM move?
Yes, if you want it timed base.

Would I use an MAG move to make sure the following rolls moved with the master roll?
This is a gearing move isn't it? MAG=motion axis gear. You didn't mention gearing before. Are you thinking that gearing may be a better way to go? I take it you have a Control Logix with a M02AE or M02AS. I suggest that you gear a cam profile that moves 12 inches to the machine. Gerry may know a better way.
 
There are three servos on this feeder. The one closest to the prss is the master and the one behide that is a slave and then the steel roll is controlled by the cradle servo. There is no encoded feedback from the press to the feeder. The Strokes per minute is a manual input from the operator. I know its a bad way to do it but that upgrade is for a future date. I planed on using the calculations you provided to figure out the velocity, accel and decel rates and use an MAM to control the first roll. The other two roll I planned on using an MAG command referencing the first roll. I am using a CLX PLC. The servo is driving the outside of the steel roll so I dont need to figure out the roll diameter. Most products will feed 12 inches, the press will run at 30 SPM, and the feed angle is 40°. Is the Mag command the corrct one to use so the slave drives follow the master drive?
 
Don't gear. Issue three MAM commands to all three drives.

johnfarrugi said:
Is the Mag command the corrct one to use so the slave drives follow the master drive?
I think so since they are just following roll one. I don't have the documentation in front of me to say for sure. Perhaps some one else can help and say for sure.

If all axes are making exactly the same move why not issue the same MAM to all three axes at the same time? This way you know that all three axes will have the same target position, velocity and acceleration targets at every millisecond. Never gear if you can rely on the target generator of the motion controller or PLC in this case. This system is two dynamic for gearing to work well.

Think about this. Your acceleration and deceleration rates are very high. This means velocity and acceleration feedforwards are required. The motion controller may be able to differentiate the master position to get velocity but I doubt the control logix can calculate an accurate acceleration. The jitter due to quantizing non-linearities will make it hard for the slaves to use high gains.
 
Just a little OT

johnfarrugi said:
There is no encoded feedback from the press to the feeder.

So are you using a PLS system to trigger the feed advance. If so how are you doing speed compensation? Are you just inching the press around until the pilot pins or top tooling clears the material? Are you using pilot pins? If so I would look at how you are releasing the material. Free wheeling the drive or opening the pinch rollers? Just a few OT questions that I would watch out for.
 

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