Calculating Radial Piston Input Torque: An Interactive Spreadsheet Tool

[deckart]

My math ability is very limited. How would I determine the input torque required to rotate this radial cylinder arrangement? I started an editable Google Sheet here:

https://docs.google.com/spreadsheets/d/1w9VebhPT0S78buN_33hN0ekUggX26lyB4BVugedR1zM/edit?usp=sharing

I'm hoping to have a calc sheet to where I can change the parameters and know the final torque.

Thank you!

 

[sophiecentaur]

Why not start by calculating the force on one piston due to pressure as the crank moves round (gas laws)? Then calculate the torque about the crank axis. Then add the effects of the other cylinders in the appropriate phases.

 

[deckart]

When a single cylinder is at right angles times distance to the center, I get that. At any other angle, which they all are most of the time, I just don't know the calcs for that. It really is beyond my education.

 

[Tom.G]

When a single cylinder is at right angles times distance to the center, I get that.

Then you have got a good part of it. The torque is proportional to the sine of the angle you are referring to. When the piston is at top or bottom dead center the angle is Zero and the the torque is Zero. As the crank pin rotates away from Zero degrees, the torque increases proportional to sin(θ). Watch out though because θ is defined as the angle between the connecting rod and the crank throw. Crank torque = (force on connecting rod) x sin(θ).

It gets even trickier when you realize that the (force on connecting rod) due to the piston is also proportional to the sine of the angle between the piston travel and the connecting rod.

So you need to know the piston force, and two angles; piston pin to con-rod and con-rod to crank throw. Multiply the two sines then multiply by the piston force to get force on the crank throw. Multiply the force on the crank throw by the distance of the throw from the rotational axis and you have torque.

Doing it in a spreadsheet is recommended.

Note: If the wrist-pin to con-rod angle gets too steep, the side force on the piston may exceed the capabilities of the lubricating oil, causing piston seizure to the cylinder wall.

Cheers,
Tom

 

[sophiecentaur]

Google can be handy. I did a search for " piston crank calculation" (Images) and this link emerged. Any use? Image searches can be good because they can eliminate adverts.

 

[Dr.D]

As is so often the case, this particular problem is 90% kinematics and 10% other. You need to look on the net for a kinematic analysis of a slider-crank system. Your system is just 5 slider-cranks. You might also find it helpful to look at the principle of virtual work for this problem; it makes it quite easy once you have the kinematic tools to deal with the changing geometry.

 

[Dr.D]

@deckart You need to be aware that with all of this mass moving around (the five hydraulic cylinders), there will be an inertia reaction torque proportional to the square of the speed. This is sometimes called the "centripetal term," not because anything is moving in a circle but simply because it is proportional to speed squared. If the only motions of interest are slow, this is not a problem; if the system is to move at all fast, you will need to take it into account. If you need help with this, send me a message and we can talk about it.

 

[deckart]

@deckart You need to be aware that with all of this mass moving around (the five hydraulic cylinders), there will be an inertia reaction torque proportional to the square of the speed. This is sometimes called the "centripetal term," not because anything is moving in a circle but simply because it is proportional to speed squared. If the only motions of interest are slow, this is not a problem; if the system is to move at all fast, you will need to take it into account. If you need help with this, send me a message and we can talk about it.

Thanks, Dr.D. The example system will rotate at a low velocity by design, inertial forces should not be a factor. I will definitely be messaging!

 

[deckart]

Then you have got a good part of it. The torque is proportional to the sine of the angle you are referring to. When the piston is at top or bottom dead center the angle is Zero and the the torque is Zero. As the crank pin rotates away from Zero degrees, the torque increases proportional to sin(θ). Watch out though because θ is defined as the angle between the connecting rod and the crank throw. Crank torque = (force on connecting rod) x sin(θ).

It gets even trickier when you realize that the (force on connecting rod) due to the piston is also proportional to the sine of the angle between the piston travel and the connecting rod.

So you need to know the piston force, and two angles; piston pin to con-rod and con-rod to crank throw. Multiply the two sines then multiply by the piston force to get force on the crank throw. Multiply the force on the crank throw by the distance of the throw from the rotational axis and you have torque.

Doing it in a spreadsheet is recommended.

Note: If the wrist-pin to con-rod angle gets too steep, the side force on the piston may exceed the capabilities of the lubricating oil, causing piston seizure to the cylinder wall.

Cheers,
Tom

I had to look it up, I was not sure what a wrist pin is. I don't believe there is one (*edit-there is no wrist pin) at the piston in this arrangement because the barrel of the the cylinder pivots at the rear. The barrel is not in a fixed position, it moves normal to the crank for the most part though there will be an offset to make room to mount the rod ends together. The offset forces will be at the rod end because we cannot easily join all the cylinders axially at the eccentric crank.

One of the cylinders rod ends will be fixed to the center yoke (term?) at the crank that the other cylinder rod ends attach so that the whole arrangement won't simply twist ineffectively. This is typical of a radial piston aircraft motor, which is where I got the idea. Though a radial piston motor has fixed barrel positions with wrist pins, this arrangement does not. I depict the pivot at that back of each cylinder in the schematic though it doesn't completely describe how it all attaches at the crank.

I will model it later when I have more access to the software at work. Conceptually, the math will be easier simply looking at one cylinder at time, but the offset at the yoke will likely increase the complexity of determining a final torque.

 

[Tom.G]

Ahh, I obviously didn't pay enough attention to the details of your drawing. I had never run across your configuration of pivoting cylinders. Thanks for the update, I learned something new!

Tom

 

[Dr.D]

One of the cylinders rod ends will be fixed to the center yoke (term?) at the crank that the other cylinder rod ends attach so that the whole arrangement won't simply twist ineffectively. This is typical of a radial piston aircraft motor, which is where I got the idea. Though a radial piston motor has fixed barrel positions with wrist pins, this arrangement does not. I depict the pivot at that back of each cylinder in the schematic though it doesn't completely describe how it all attaches at the crank.

In the typical radial aircraft engine, there is one Master Rod that connects directly to the crank. The other rods connect to the crank end of the master rod and and are called Slave Rods. A nine cylinder radial will have 1 master rod and 8 slave rods. This is because making the crank pin long enough to accept all 9 connecting rods makes for a very long pin, subject to bending. The Master Rod/Slave Rod arrangement makes for even more complicated kinematics.

 

[deckart]

I've come to the conclusion that determining the torque about this point (the small red xy arrows) is simply a summation of the forces of these torque arms. Does anyone concur? I believe I was making it more difficult than it is.

 

[Dr.D]

Well, yes, but then again no. What will you do when the geometry shifts slightly, as it certainly must in operation>

 

[deckart]

Well, yes, but then again no. What will you do when the geometry shifts slightly, as it certainly must in operation>[/QUO
.

I expect there will be an oscillation but I'm interested in the maximum torque. I will have to plot various positions to determine what that is

 

[deckart]

Well, yes, but then again no. What will you do when the geometry shifts slightly, as it certainly must in operation>

The shifting of each cylinder is accounted for in the model. With the shaft driving the cylinders at 2000 psi the combined forces average to approximately 4900 lb-in of torque at the shaft. +/- 8% throughout the entire rotation. This was a great exercise!

With multiple radial cylinder sections on a common crankshaft, this could make a nice compact pump unit to be used in wind turbine head.

W

 

[Dr.D]

Congratulations! Apparently you did not need any help after all!