# Calculating torque required to drive crank and slider

1. Sep 27, 2013

### Koth

Hi all

The title pretty much says it all.....

The slider is required to push and pull a load of 70N. Radius between centre of crank axle and connecting rod is 12mm.

I have been unable to find a formula to calculate torque required to enable selection of an appropriate motor to drive the crank shaft.

Cheers
Koth

2. Sep 27, 2013

### Simon Bridge

Welcome to PF;
Basically you are missing a whole lot of information.
i.e. you need the friction in the system and the max angular speed you want to for crank. You also need the dimensions and masses for the entire coupled system.

It would help to know how fast you want it to be able to get to the speed.

With no friction, any torque will accelerate the rank - depending on the masses and moments of inertia.

3. Sep 28, 2013

### Koth

I don't want to overcomplicate this, i don't think it's necessary for this particular application. I would like to assume no friction as I will use roller bearings on the connecting rods and maybe on the slider to avoid the need to harden everything.

I need between 20mm and 40mm of stroke in the slider. Not particularly concerned about acceleration and deceleration as I should be able to control this with a step motor.

Last edited: Sep 28, 2013
4. Sep 28, 2013

### Koth

Also, should be able to control reciprocating speed by increasing or decreasing motor rpm.

5. Sep 28, 2013

### Simon Bridge

Well - with no friction, zero torque is required to maintain a particular speed of rotation.

6. Sep 28, 2013

### Koth

Disagree. There is a 60N load at the end of the slider. Surely, assuming no friction, there is a basic formula to calculate how much torque is required to overcome the force exerted on the slider? The radius between the centre of the crank axle and the centre of the connecting rod end, and the connecting rod length are known values.

7. Sep 28, 2013

### Simon Bridge

Depends on the geometry ... i.e. if the 60N points along the direction of the slider, then it will exert a torque on the crank which varies with the position in the rotation. You can work out when that will be a maximum - which will tell you the torque required to exactly balance the load.

If you said which direction the load points, I didn't spot it.
You did say it had to "push and pull a load of 60N" which suggests to me that the load is 90deg to the action of the slider. The load does accelerate - so you are right that this requires some torque.

My main point holds though - not enough information supplied.
You are clearly unwilling to supply the needed information so... good luck.

8. Sep 28, 2013

### Koth

I agree that torque depends on the position of the crank, I am interested in the maximum torque. I think this occurs when the angle between the centre of the crank axle and centre of the connecting rod end (imaginary line between these two points) is perpendicular to the connecting rod. I am not interested in the torque curve assosiated with varying crank positions.

Yes, the load is perpendicular to the slider.

I am trying to come up with a concept to enable creation of a lab test jig that will engage and separate a plug and socket 10000 times. I have not designed the crank system yet, for this reason I am not able to supply the friction and mass details that you require. However, I need a short stroke, so would like to base any simple calculations on a crank radius of around 12mm (24mm of resiprocating movement at the slider).

I am trying to establish a ball-park torque figure so that I can get a good idea of what type of drive I will need for this system, and therefore establish some budgetary cost details for the system before I proceed to the next stage of this project (cost dependant).

I am not unwilling to provide details, previous posts have been brief due to the restrictions of having to post via a mobile phone.

I will do a diagram of what I am trying to achieve if that helps?

9. Sep 28, 2013

This sounds like a great job for a pneumatic actuator. Is 60N the insertion force of the plug or some equipment that you have to move? Great thing about pneumatic actuators is that you can get the length you want and just change the air pressure to change the speed and force applied you can also add control orifices to control the maximum actuation speed while still being able to apply large forces at low speeds. You can probably find something on ebay for under 100. 25mm actuator - http://www.ebay.com/itm/SMC-Pneumatic-Guided-Cylinder-MGPM16-20/140895001950?rt=nc 10. Sep 29, 2013 ### Koth Hi, thanks for the advice. Yes 60N is the insertion/extraction force on the plug. I had initially considered doing exactly what you have suggested. However, I have written the idea off for two reasons as follows: 1. I want to be able to carry out the test on 5 samples at once (not previously mentioned), this would mean that I would need to buy 5 small cylinders or one large cylinder and an elaborate method of aligning the 5 samples. I was hoping to get around this by using 5 connecting rods on a crank shaft, each of which will be staggered so that each of the samples engages in turn. 2. I had assumed that there would be a requirement for some sort of motion control system such as a PLC device or programmable relay. Would this not require the use of proximity sensors and the like to check that the actuator is fully engaged/retracted and fed back into the control device? My crankshaft idea seems more simple if I could just find the torque formula that I need. Can a pneumatic piston be controlled using some sort of plunger actuated valves to verify piston position and reverse piston actuation direction, rather than using PLC control? What device is required to control linear speed of the actuator? Your advice is much appreciated on the subject. 11. Sep 30, 2013 ### ScienceGeyser Wow! The crank shaft idea sounds very cool, fun to build and fun to watch run operate. For a simple single pin crank shaft with a pin eccentricity of 12mm your max torque calc is just force*distance (60*0.012=.72Nm). This is roughly the torque available with a typical NEMA 23 stepper motor. Assuming that the insertion and extraction force are the same, you will have the maximum force twice in your rotation at 180 degrees from each other. If you add four more pins at different angles (72 degrees for even spacing) then you won't need 5 times the torque since the torque for each pin is roughly the sine of the angle. Depending on how long the push rods are, you will have some overlap of forces which will add together to increase the peak torque required. You must also not neglect the extraction force of the plugs which will be acting upon the crank pins at 180 degrees from the maximum insertion torque for each pin, which will also add to the required torque. Since the insertion/extraction force of the plug is not likely to be same during the entire stroke in both directions, doing the real calculation will be complicated and heavily dependent on the mechanical characteristics of the plug. For a rough approximation though you can apply the sine of the angle of each pin to calculate force at some arbitrary angle. Something like this: Max T = (sin(90)+sine(18)+sine(18)+sine(54)+sine(54))Tm Where Tm is the max torque on a single pin @ 90 degrees and one pin is at 90 degrees and the pin angle increment is 72 degrees. Max T = 3.24Tm (only true for 5 pins) Since Tm= 0.7Nm Max T = 2.27Nm This of course ignores all sorts of details but will get you in the ball park for figuring out the minimum torque your motor should be capable of. As far as pneumatic controls go, you can definitely control them without PLC's Some simple switches along with a solenoid valve manifold would put in position to control the cylinders with no controller at all. Air pressure determines the maximum force applied and air volume determines the rate of change in the cylinder. As far as feedback for positioning goes, you will not need any if your mechanical system is designed to constrain the limits of movement just like in your crank shaft design. If you need 20mm of movement, get a cylinder that moves only that much or design in a mechanical stop for the cylinder stroke. I hope that's not too confusing. Sometimes I don't spill my thoughts in a completely coherent manner. 12. Sep 30, 2013 ### Koth Thank you very much for this. I am going to cost-up both options. I have a minor quandary, I'd like to have a short dwell after engagement of the plug (1 second). If using the camshaft method, this means that I will need to incorporate a spring at the end of the slider in an attempt to maintain pressure on the plug even though the connecting rod has started to retract the slider. Alternatively, in the pneumatic system, I am thinking of using a limit switch to trigger a delay timer relay that in turn triggers a solenoid valve to retract the cylinder(s). Thoughts? 13. Sep 30, 2013 ### ScienceGeyser Do you need to keep pressure on the plug when it has reached full insertion? The crank shaft will have some inherent dwell, backlash and such will cause the plug to sit still for a moment before it changes direction. If you need more dwell time, you can put a small amount of play in any joint on the push rod. If you have play in the push rod though, you will need to increase the pin eccentricity to maintain stroke distance. The LM555 can make a nice and cheap delay timer. I love these little chips and use them all the time in my projects. Chip can be found for under1 and the rest of the components should be less than \$5 with the protoboard probably being the most expensive part.

14. Oct 1, 2013

### Koth

Yes, I need a dwell in the fully engaged position, prior to disengagement. Now decided to go down the pneumatic route for this reason.

Changed my mind about the solenoid valves! Will now attempt to use an air timer in conjunction with a two way air pilot valve. Hoping to be able to get the function I want without the need for electricity.

15. Oct 1, 2013

### ScienceGeyser

Great! Mead makes some stuff you might be interested in. http://www.mead-usa.com/ Good luck with your project. Would you mind posting pictures when you are done? I'm sure it will be interesting.

16. Oct 1, 2013

### Koth

I'll be needing pictures for an engineering report so uploading them here thereafter shouldn't be a problem. Thanks for your input!