Encoder count changes due to wheel diameter

Hello,
I was hoping someone could help me out with a bit of physics.

Currently I have a cart that has 5 inch (diamter) wheels on it. On the drive shaft of the cart, there is an encoder whose resolution is 360 pulses per rev. The Encoder is connected to a Rockwell Compact Logix fixed unit. This fixed unit (1769-L23E-QBFC1) has an input, output, analog in/out and high speed counter on it. Encoder is connected to the High speed counter.

My issue: When the cart does not have a load on it, the wheel diameter is 5". When it is loaded with a heavy weight, the wheel diameter can change by as much as 1/16th of an inch. I dont think the diameter really changes as much as the wheel becomes out of round. I command the cart to travel to a set point based on encoder counts. So lets say the set point is 1000 pulses at a speed of 30 F/M. Based on my little chart below, there are two roll diameter. One is the original 5 inch and the other is 4.9375 which should be 1/16 change when loaded. So when the cart is commanded to the destination, depending on if it is loaded or unloaded, the cart will not travel to the same point because of the diameter change.

PIRoll DimeterDistance per Rev (Circumferance)Conveyor Speed (fpm)Conveyor Speed (Inch/sec)Encoder Pulses/RevPulse/Sec (Freq)Pulse RatePPI RateDistance/Pulse (InchesDistance Traveled for One Wheel Revolution3.1415926544.937515.51161373306360139.2510.0071813030.0035906510.0430878215.511613733.141592654515.70796327306360137.5100.0072722050.0036361030.0436332315.70796327

My Question: I want to try and correct this error based on encoder pulses captured at a certain distance. So If the cart travels 24 inches with out a load, I should see a count of 550 encoder pulses. If the diameter changes by a 1/16th then I should see 557 pulses for a travel distance of 24inches. In a perfect world I would then compute (557/550)* 1000 (setpoint) = 1012 for the new set point due to the change in wheel diameter. Lets say I have a timer in the PLC where after 4 seconds I capture the encoder counts to execute the calculation shown above. If my speed is a constant set point of 30F/M or 6 Inches/Second, should I be able to capture the increase in pulses. There was some discussion at work, that I would need to have a laser PE that looks at the side rail when the cart is traveling. The side rail has two holes spaced 24 inches apart. So when the PE sees the first hole I start capturing encoder counts, when the PE sees the second hole in the rail, I capture the count again. Then I would subtract the two and I should see around 550 pulses. Cant I do the same as I described without the Photo eye? I understand the complications of using timers and so forth in a PLC, my question is if the PLC were perfect, can I get the increase in counts with a timer.

Thanks in advance for any help you can provide

My chart did not come through, It was an excel sheet that showed the Roll Dimeter, Distance per Rev (Circumferance), Conveyor Speed (fpm), Conveyor Speed (Inch/sec), Encoder Pulses/Rev, Pulse/Sec (Freq), Pulse Rate, PPI Rate, Distance/Pulse, and Distance Traveled for One Wheel Revolution. I have to figure out a way to put it back in.

Sorry about this

Have you actually seen this difference in counts between the loaded and unloaded state? Or are you predicting it?

Regardless of the diameter or shape of the wheel (unless it folds over on itself) the perimeter is the same. One full revolution of the drive shaft should move it the same perimeter distance. You may be overcomplicating this.

I have not personally seen the difference. The company I am working for originally had the encoder on the shaft and they changed it to a measuring wheel mounted on the side of the cart. They claim that the wheel diameter change was affecting the positioning of the cart. Thats why they went to the measuring wheel. Recently I troubleshot the measuring wheel becasue they could not get that to work either. The mount for that measuring wheel is not correct. I told them lets try changing that out, but the director wants it back on the shaft with a error correction. All of this was before I was hired. So, now that I am done complaining.....I am preparing myself for the anticipated error they claim. There were some programming issues that I corrected that may have been the whole problem.

I have wondered about this also. However to travel X distance tire must rotate Y RPM. Increased load on tire does change that radius from shat to ground and for sure is proportional to rolling resistance ie tire flexes more. I theorize it does not change the circumferance. If it did then the tire must slip on the rim.

Under loaded and unloaded conditions
Mark with chalk on tire. Referance that mark to a "zero distance" mark on floor.
Using tape measure measure distance traveled on floor.
Count revolutions
Compare that to encoder count if you can.

Now if the wheel is slipping on the shaft all bets are off. This may be a possiblity but you should be able to determine that with the floor and tape maeasure.
Dan Bentler

I attached the encoder table that I was unable to before

Leitmotif, I thought about performing the same test. However I have to travel out of town and wont be able to do this for another week or so.

The tire is not slipping on the rim. It is a very hard plastic that is somehow chemically bonded to a steel hub. You cant even put a dent in the wheel with your finger. It takes about 1000 lbs of weight to make the error where it is slightly out of round....so I am told.

Thanks

I see the table but wonder you have several columns labeled conveyer.

In original post you said cart.

On a very heavy load the conveyer belt can slip on the driver. Same thing for the cart especially so on wet slippery floor or maybe even rolling over a rock or other junk on floor.

I think I would do the measurements on floor and make sure all is correct to the wheel. If your problem is then the tire slipping on the floor that is another issue - your only solution is to increase friction factor between tire and floor.

OR sense cart position ??? That is what you are trying to control.

The other thing that comes to mind - maybe your error is real small ie 0.000005. After many trips this error will add up and the totalized error will throw all out of whack. You may want to rezero every time cart is at one end or the other ie rehome it.

Dan Bentler

The change in diameter doesn't matter.

It's just like the wheel on your car. The circumference of the tire is fixed byt the "free" diameter. Ever time you make a revolution the cart must move one circumference. The change in the diameter is just taken up by more contact area between the wheel and the ground. On an undeflected wheel it is theoretically point contact.

If the tire is driving the cart then Dan is right - you are experiencing slip. Try putting the encoder on a non-driving wheel.

I would start off by coming up with the most precise possible way to verify the difference. Travel distance should be as large as possible, and manually measured position should be done to exacting marks. Use raw counts from the encoder, if possible, to eliminate any rounding that may take place.

What Tom points out about the circumference is certainly true in theory...in reality, though, there may be other variables that distort the true circumference. That hard plastic may indeed compress and not stretch at the edges of the contact patch like a pneumatic tire does.

I would repeat a test at least ten times unloaded and then ten times loaded. It would not surprise me to see that you are slipping when unloaded and not slipping as much when loaded. Repeating the test ten times may reveal a repeatability problem in one case and/or the other.

Prove there is a problem and measure its magnitude before you get too heavy into working out a solution. If the thing isn't repeatable in the first place, then don't waste time fixing a mechanical problem with code.

I vote for steel wheels

No encoder expert here by any measure but isn't a 360 pulse/rev encoder rather coarse? I usually see people suggesting encoders that have much higher pulse counts.

Shawn

Shawn
The encoder is on the motor. ASSUMING a gear ratio of 10:1 then encoder does 3600 counts per wheel rotation. I would expect a higer gear reduction - picked 10 to make math and concept easy to grasp

Dan Bentler

Thanks everyone for the responses.

Dan, the sheet I attached says conveyor because I created that sheet when I used to do airport conveyor systems. Every belt is tracking on an Airport system so due to cost, all encoders are around 3 pulses/rev with a 6 inch roll. The calculations should still hold on this application I believe.

Also I dont think the drive is slipping beacause from the gearbox, there is a shaft coming out of both sides with a connection point (round disc with holes from the factory) our company machines a matching disc and attaches is to a shaft and bolts it directly to the gear box. The shaft is straight to the tire on both sides. The tire has a key way that slips on to the shaft and there is a snap ring that holds it in place. So based on that I dont physically see how or where it would slip, but Im no mechanical engineer. So I cant 100% confirm this. I can confirm that the tires are not slipping when running. These carts ride on rails in a pharmaceutical enviorment. Besides dust build up, there isnt any debris on the rails. Maybe a pill or two occasionally. I have watched the carts extensively to ensure the wheels are not slipping.

Also the comment about the steel wheels, this company had a prototype system installed in their warehouse that has steel wheels, the design has changed dramatically since then, but the operation is the same. Those carts dont have this issue.

OOPS got encoder on wrong shaft - it is on wheel shaft. 360 counts per rev at 10 in/rev gives about 36 counts per inch.

What is total travel distance? What is the tolerance for starting position and stopping positions?

What is the error after several operations? I do not think you will be able to visually detect 1% wheel slip on those rails.

Think I would go look at the other system and find any differances and determine if any changes need to be made.