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Torque calculation needed for unique application

by provideo
Tags: application, calculation, torque, unique
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provideo
#1
May7-14, 05:02 AM
P: 5
For several years I have restored antique movie theater popcorn machines from the 1940's. I design and manufacture obsolete parts for these machines for restorers across the nation. I am currently prototyping a reproduction "popcorn stirrer flexible drive cable" and am in the process of building a stress test unit to measure and test several parameters concerning the flexible cable.
Please look at my attached drawing. I found a unique way of measuring the stress applied to the "test cable" by monitoring the force and movement of the main drive motor's "shaft end play" by using a "strain gauge". When the worm gear is turning under "no load" the motor's shaft is fully retracted and applies no force to the strain gauge. But when the worm gear is under a load it forces the motor's shaft to extend slightly and applies force to the strain gauge.

I retrofitted a fully restored popcorn machine with my test device for the purpose of establishing a known set of operating parameters. Here were my results:

1. With no popcorn or oil in the kettle I calibrated the strain gauge to measure 0 oz. of force.
2. I poured in the normal oil and popcorn mix and the readout started measuring force.
3. As the popcorn started popping the force increased and eventually displayed 15.2 oz. of force with a fully loaded and popped kettle of popcorn.
4. The test results established a maximum force of 15.2 oz. (under normal operating conditions) applied to my flexible stirrer drive cable.

Here is where I need help. Can the 15.2 oz. force measurement be used to calculate a "torque" measurement? What would be the amount of torque on my cable at full load and would I express it as PSI or another standard unit of force? Do I need more data to make the conversion (gear sizes, motor RPM, shaft size, etc.) ?

Thanks,
Mike Pruitt
Attached Thumbnails
load drawing.jpg   FINISH5.jpg  
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Simon Bridge
#2
May7-14, 08:24 AM
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Off the first picture - the torque about the axle A is "driven" by the worm gear.
What you need is to be very clear about what exactly the strain-gauge is measuring.
From your diagram it seems unlikely that the way you have set up the gauge will be very useful for measuring the torque about A.

Aside: your use of quotes around some words or phrases is a little puzzling - it appears ironic or skeptical: was that the intention?
http://www.dailywritingtips.com/3-er...-scare-quotes/
provideo
#3
May7-14, 10:38 AM
P: 5
Sorry about the quotes sir. I was just placing emphasis on the word groupings.
My main goal in measuring the test machine's drive under its maximum load is to hopefully construct a test jig for exercising newly constructed cables before shipping them out to restorers. I want to place that maximum load on the new cable for a long period of time as a stress test.
I observed that under no load the motor shaft and worm gear were always at their fully retracted position and as the load increased the motor shaft and worm gear were forced outward due the 3/16" motor shaft endplay. In my drawing this would be under no load the shaft is extreme right and under a load the shaft moved to the left in a proportional manner to the amount of load. I saw this as a measureable feedback point for my strain gauge. So now that I know the amout of force applied to the motor shaft and worm gear under a full load (15.2 oz) I can simulate the load in an extended time test without wasting loads of popcorn and oil. I plan to load the test kettle with hard, uncooked brown beans until the desired load is achieved. I will keep the stirrer motor energized for an hour and subject the newly manufactured cable to a load force of 15.2 ozs.
So my main question is can I convert my results into a more standardized torque number? The torque on my shaft is equal to 15 ozs. of force at full load at the motor shaft and worm gear.
Thanks,
Mike

Simon Bridge
#4
May7-14, 11:03 AM
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Torque calculation needed for unique application

I observed that under no load the motor shaft and worm gear were always at their fully retracted position and as the load increased the motor shaft and worm gear were forced outward due the 3/16" motor shaft endplay. In my drawing this would be under no load the shaft is extreme right and under a load the shaft moved to the left in a proportional manner to the amount of load. I saw this as a measureable feedback point for my strain gauge.
So the strain gauge is measuring how hard the shaft presses against it's contact point?

So my main question is can I convert my results into a more standardized torque number? The torque on my shaft is equal to 15 ozs. of force at full load at the motor shaft and worm gear.
The reply is the same as before.

If I have understood you correctly - the retraction/extension of the shaft will be proportional to the load ... if the shaft pushes to the left harder with increased load, then the motor must be pulling to the right (to retract the shaft under reduced load) and the strain gauge is, therefore, measuring the difference between these two forces.

Note
- if the stirrer turns at constant speed, the net torque must be zero.

It sounds like you are trying to find out something by sub-optimal methods.
What exactly are you trying to find out here?
provideo
#5
May7-14, 12:23 PM
P: 5
Am I not reverse engineering the force created by the popcorn in the stirrer kettle? In my mind I see the motor shaft movement as an indicator of the increased force required by the stirrer moving through the mass of the popcorn in the kettle. (See attached drawing to this reply) I am just trying to put a measurement label on the amount of force created by a full kettle of popcorn during the popping process. Again, my goal is to establish a control for testing newly manufactured cables applying the same amount of maximum stress on each cable.
Thanks,
Mike
Attached Thumbnails
final calc.jpg  
Simon Bridge
#6
May7-14, 12:29 PM
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That's what I figured. Please note: I am familiar with popcorn machines.

Reply is still the same.
You seem to be having trouble understanding it.
I'll spell it out - the number you want cannot be obtained from the information you have.
provideo
#7
May7-14, 12:33 PM
P: 5
Actually I am just trying to measure force. Did I confuse torque with force? Is torque only the moment of energy needed to start the cable rotation and force the consistent amount of energy to keep it rotating under a load? If so I am sorry for the confusion.
provideo
#8
May7-14, 12:39 PM
P: 5
Thanks for taking the time to communicate with me on my attempt to understand cable stress and testing.
Simon Bridge
#9
May7-14, 08:27 PM
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Quote Quote by provideo View Post
Actually I am just trying to measure force.
Force of what? Where?
Can you be real specific?

Did I confuse torque with force? Is torque only the moment of energy needed to start the cable rotation and force the consistent amount of energy to keep it rotating under a load?
Seems reasonable:
"moment of energy" has no meaning that I am aware of and force is not an energy anyway.

"Force" is defined in Newton's second Law to be the rate of change of momentum. It is thought of as a push or a pull along a line. In terms of energy - force is the gradient of the potential energy function.

"Torque" is the description of a force acting on something constrained to move in a circle - it is defined as the product of the applied force and the moment arm - the moment arm being the perpendicular distance between the line-of-action of the force and the pivot point. It is thought of as a "twisting force" or the "force of turning". It is the same as the rate of change of angular momentum.

The result of a force is an acceleration, the result of a torque is an angular acceleration. No force or torque is needed for uniform motion. The reason you experience the need to push constantly to maintain a constant velocity is that your applied force is opposed by some form of drag or friction. The friction is usually proportional to the velocity - so you experience that you have to push harder to maintain faster speeds. Whatever you are pushing accelerates until the friction is equal and opposite the applied force.

Bearing all this in mind:
The shaft holding the worm-gear is a bit loose in it's housing - so you can pull it back and forth right?
If you pull it out with your hand and let go, does it stay there or pop back in?

When the motor turns on, the worm gear engages the toothed wheel - pushing it in one direction, causing it to accelerate. As a result, the wheel delivers an equal and opposite reaction force back to the worm-gear, pushing it laterally to one side until something brings it up short.

Bringing a force transducer to the end of the work-gear and clamping it so that the gear does not move laterally when the motor is switched on, then the force transducer is providing a balancing force ... which appears on the readout.

What is important here is that the shaft not move laterally once the force-transducer is in place, otherwise there may be internal components (spring, padding, magnets, whatever) which can act to reduce the force needed to balance the motion. I cannot tell from your description if this is the case or not.

The force you are interested in should, then, be close to the force on the readout ... the torque at A will be (approximately) the value of this force multiplied by the radius of the wheel.

I don't know if the approximation is good enough because I don't understand what you are trying to achieve or what you hope to do with the number.

Anyway: the stirrer accelerates until friction is the same as the torque - after which time it rotates at a constant angular velocity.

The arrangement basically means that a higher load (more popcorn, heavier oil etc) results in a slower rotation speed. The losses may not behave in a nice way - so you will need to plot a curve of rotation-speed or torque (if you prefer) vs load for varying load to see what is going on. If you are very lucky, the curve will be close to a straight line in the machine's operating range.

That should give you enough to go on with.


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