PLC Coveyor Question

[kevilay]

Hello, Im building a small conveyor to move dowels. Their between 1 and 2 inches long. The length of the conveyor will be around 1 foot. Each end will have a steel cylinder for the belt to go around on. I was wondering what kind of DC motor I Would need. I noticed most are rated in in/lbs but im not really sure what to relate that to. The maximum amount of dowels would weight somewhere around one pound. I was also wondering I have a micrologix 1000 plc. Can I run a 24v motor right off an output or does it draw to much ampage? Im an electronics student, In class we learned about the programming only, So im trying to learn how to take it a step further, any help is much appreciated.

Kevin

 

[bkottaras]

Definetely NOT.
Use a small contactor to run the motor.
Use the output to turn on the contactor whenever the motor needs to run.
Logic should be pretty straigh forward including the conditions to turn motor on and off.
Not sure about the in/lbs, sounds like torque to me.

 

[kevilay]

Definetely NOT.
Use a small contactor to run the motor.
Use the output to turn on the contactor whenever the motor needs to run.
Logic should be pretty straigh forward including the conditions to turn motor on and off.
Not sure about the in/lbs, sounds like torque to me.

Thanks, yeah its definitely a torque, but I have no idea what it relates to and how much i need. I was thinking of maybe taking a motor out of a cordless drill

Kevin

 

[bkottaras]

Buy a cheap DC Motor driver board.
Dart, KB, Minaric, they all make them.
You can set Min-Max speed, CL and such on the board.
Makes life easier.
Then use the Inhibit on the board to turn the motor On and Off.
The Inhibit will be controlled by your PLC output.

 

[kevilay]

this is for my final project Id probably be better off building a board like that myself. I understand the concept and how they work, but I don't know what kinda motor I need.

Kevin

 

[milldrone]

I noticed most are rated in in/lbs but im not really sure what to relate that to.

The in/lbs rating refers to the amount of torque or twisting force the motor has. In the case of in/lbs imagine a one inch long lever attached to the shaft, the lbs refers to how many lbs the one inch lever can lift.

The maximum amount of dowels would weight somewhere around one pound.

Not enough info. We do not know the diameter of the pulley, the friction of the bearings and the friction of the belt against any framework. The old fashioned way is to attach a lever to the shaft and use a measuring device like a fishing scale to measure the force required to move the belt with a load on it. To make things easy the lever should be 1 inch or 1 foot from the centerline of the shaft to the scale attachment point.

Here is a link to a member of PLCs.net that has passed on but his legacy remains. There is some motor data and torque information here also.
http://www.patchn.com/index.php?option=com_content&task=view&id=43&Itemid=74

I was also wondering I have a micrologix 1000 plc. Can I run a 24v motor right off an output or does it draw to much ampage? Im an electronics student,

Probably not. Too much power required.

In class we learned about the programming only,

Shame on the instructor!

 

[leitmotif]

Thanks, yeah its definitely a torque, but I have no idea what it relates to and how much i need. I was thinking of maybe taking a motor out of a cordless drill
Kevin

Kevin
You said you do not comprehend torque and have no idea what it means to you when building equipment.
I repeat what Mildrone said - shame on your instructor.
OK so here we go.
Force X distance = work ie HP Kw whatever
Torque is a force on a cylindrical object to make it turn.
Tie a piece of rope to a 2' diameter wheel (radius = 1 foot) so it will not slip. Mount wheel above floor and tie a 100 pound weight to it. Put a torque wrench on wheel shaft and apply force to wrench. A 100 lb force you put on wrench at one foot will raise the load. The weight applies if you want to call it that "counter torque" better yet think of it as torque imposed by the load.
IF you keep twisting the wrench and applying 100 lb force to wrench you will keep raising the weight.

Work is proportional to time so the faster you go the more work you accomplish ie feet raised per unit time
NOT how much sweat you are producing.
HP = 550 pounds raised one foot in one second.
HP (work) = torque X RPM / 5252

This example is a constant torque load
it will always impose 100 ft lb torque. Your conveyer is a constant torque load provided the number and size of your spindles is constant.

HOMEWORK
get an inch pounds torque wrench and measure the torque of the drive pulley shaft.
You can also get a piece of heavy string or wire and wrap around shaft and then hang weights.
YY pounds weight X shaft diameter / 2 X 12 gives you foot pounds.

RPM of drive sheave = desired speed of conveyer (fpm) / sheave circumferance.
Now plug this value into HP = torque X RPM / 5252
and you will get motor HP

We will do the gearbox next lesson.

You may argue but I am only going to do the programming. I will save you the Peter Nachtway et al lecture
- programming cannot overcome poor machine design.
If you cannot distinguish poor vs good machine design you are going to have problems.

Dan Bentler

 

[kevilay]

thanks so much guys, sorry I feel kinda stupid asking these questions. Our teacher comes in and gives us an assignment then leaves... it kinda sucks... guess thats community college for you.

Kevin

 

[leitmotif]

thanks so much guys, sorry I feel kinda stupid asking these questions. Our teacher comes in and gives us an assignment then leaves... it kinda sucks... guess thats community college for you. Kevin

Kevin
I went to both Community College and University of Washington. You can find that type of instructor at both.
In the past I have complained about them and it did sometimes work - proportional to how squeky a wheel I was.

What Comm Coll just for curiosity.
It is your tuition money
the instructor is your employee
are you getting satisfactory results for the wages you pay.

You should have seen the look on the PhD types when I told them they worked for me - the good ones shut up for a few seconds and then agreed. Older students were more respected and valued by the good instructors because we were not afraid to cry foul.

Dan Bentler

 

[Bitmore]

Since no one else pointed out the facts of your design, I will.
Previous post stated using a contactor, at these low hp requirements you could also use a relay. To protect the relay from motor EMF you might want to add a capacitor accross the motor leads to prevent relay open arching.
Now about the torque requirements, the amount of the weight is slight and the moment torque forces required to start the conveyor are more important. Next you stated the conveyor is 1 foot long, this is not that important either, I would consider the sustained torque is inconsequential.
Your main concern is the moment of movement as that will require the most demand of torque.
Inch pounds are just another way to describe the moment of Torque, with this arrangement it appears your moment of torque can be determined with a test. If you have the conveyor built already, load it up with dowels, wrap a string or thread around the drive shaft. Then attach a scale to the string or thread and pull. View the highest setting on the scale, bet you it's when you try to move it intially, then the required force will decrease. This maximum force required to start motion is called the "moment of torque". I would suggest you use a fish scale or postal scale that measures ounces.
All motors specify the "moment of torque", in A.C. motors it is called FLA or full load amperage this is the maximum torque available for the A.C. motor at any time, in DC or servo motors it may be refered to as maximum torque and it is rated over time. In other words a 5 inch pound motor may exibit a moment of torque of 20 inch pounds for 5 seconds.
As far as the conversion needed from foot/lbs to inch/lbs it is widely available on the internet.

bitmore