Distance to stop formula

[need answers]

We are winding a real lightweight sheet paper that tears easily. I to am trying to use a footage to stop formula on a control logix processor. We typically run 3000fpm, with a acceleration rate and deceleration rate of 128 ft per sec squared. Our jerk rate is 1 fps cubed. They go into an s curve block that creates the reference to the drive.
We tried the formula that was posted in a older thread X = [V^2 / (2 * a)] + [(V * a) / (2 * J)] , it does not seem to work. From the old thread, we end up in the condition of a^2 > vj as described further into that thread. If I increase J, our paper tears as the drive reacts too quickly.

Does anyone have a good formula that can be used in ladder logic to accomplish this.
Thanks

 

[Peter Nachtwey]

I am assuming you are using 3rd order s ramps and the a^2>v*j condition exists and you are ramping down to 0 velocity.
The total ramp distance = sqrt(v*j)*v/j. So the ramp distance = 353.55 ft. The total time is 2*sqrt(v*j)/j = 14.142 sec. The peak deceleration is is -sqrt(v*j) = -7.07 ft/sec^2.
I have all of these equations worked out. They only require solving for 8 unknowns using 8 equations.

 

[kamenges]

First of all, are you sure about that 128 fps² value for accel rate? That is a serious accel rate for most paper winders. If that is right, without jerk to keep you reasonable the accel rate would very likely become unattainable given the motor torque you likely have available. In effect you are limiting your maximum accel rate using jerk, which I wouldn't recommend. As Peter shows below you are only getting to 7 ft/sec² anyway so why set an accel rate any higher than that? You can't reach it.

On the web lines we make here we seldom run decel rates higher than 3ft/sec². We just don't see the need. Our roll runs are long enough that the accel/decel time isn't a big factor. If you are running relatively short rolls and are trying to get more rolls through per minute this may not apply to you. But you would still never need anything like 128 ft/sec².

If you operate with a reasonably attainable decel rate you will be able to use a higher jerk rate. The basic idea is set a decel rate your system is capable of achieving based on torque capacity and use the jerk rate to allow the system to keep up with the torque changes needed to reach the decel rate. Lower jerk rates result in slower changed in required torque/force and slower changes in control loop outputs.

Keith

 

[Peter Nachtwey]

Keith brought out a point that I forgot to mention. 128 ft/sec^2 is 4g. This is unreasonable in most motion applications. Keith mentioned 3ft/sec^2 as a reasonable deceleration rate. It is less than 1/10th g.

I like to equate accelerations and decelerations to g because then it is easy to estimate the acceleration and deceleration forces.

 

[shooter]

can the motors keep up with the speedchange?
and the controls need very fast update to be in sync.

 

[Peter Nachtwey]

can the motors keep up with the speedchange?
and the controls need very fast update to be in sync.

It should be easy.

 

[kamenges]

As Peter said, the command side of this is not the issue...with the right tools, as Peter would also tell you. However, I don't think need_answers is being limited by his command generation side. I would suspect he is running out of torque somewhere, very likely at the winding roll.

The OP didn't give us any information on the power transmission side of this thing so it is hard to tell what the machine is capable of. He also didn't define the winding technology (center, surface or center/surface). Both of those play into the capacity of the system. However, we can say, at least anecdotally, that 7ft/sec² is where the system probably runs out of torque.

Keith

 

[kamenges]

I hope this guy comes back. I'd be interested in finding out what he has going on.

Keith

 

[Peter Nachtwey]

I would like to see a trend or plot of the distance, velocity and control output as a function of time.