Motor driven hydraulic cylinder design

[Wale]

I am trying to design a mechanism which has a hydraulic cylinder driven by a step motor. The motor shaft is connected to the hydraulic cylinder piston via two rods of lengths r and L as shown in the below figure. I want the cylinder's piston to be driven at constant velocity v, so I am trying to solve for the rotational speed profile w (omega) that I need to drive the step motor with in order to achieve the constant linear velocity v.

Any help?

 

[anorlunda]

The arrangement you sketched will not produce constant velocity V. It will move back and forth in a sinusoidal motion, like the wheels of a steam engine.

But what do you mean constant velocity? If v moves to the right at one mile per hour, then 24 hours later it will be 24 miles away. Surely, that's not what you want.

 

[Wale]

The arrangement you sketched will not produce constant velocity V. It will move back and forth in a sinusoidal motion, like the wheels of a steam engine.

View attachment 219112

But what do you mean constant velocity? If v moves to the right at one mile per hour, then 24 hours later it will be 24 miles away. Surely, that's not what you want.

Yes, you are correct. It will move back and forth, which is what I want because it will be driving a a hydraulic cylinder. What I want is for the hydraulic cylinder strokes to be actuated at constant velocity (aka constant flow rate for the fluid that the cylinder is pushing).

 

[Asymptotic]

It will move back and forth,

Isn't velocity zero at each turnaround? For example, at top dead center and bottom dead center for a piston moved by a crack mechanism in an internal combustion engine.

 

[Wale]

Isn't velocity zero at each turnaround? For example, at top dead center and bottom dead center for a piston moved by a crack mechanism in an internal combustion engine.

Yes. That is correct.

 

[Asymptotic]

Isn't velocity zero at each turnaround? For example, at top dead center and bottom dead center for a piston moved by a crack mechanism in an internal combustion engine.

Yes. That is correct.

Then it can't be this

constant velocity (aka constant flow rate for the fluid that the cylinder is pushing).

 

[Wale]

Then it can't be this

Constant (or near constant) flow rate for each stroke. In other words for a given stroke, I would like for the velocity of the piston to be constant or near constant.

 

[Asymptotic]

From Wikipedia Piston Motion Equation article. Velocity is sinusoidal.

 

[Wale]

From Wikipedia Piston Motion Equation article. Velocity is sinusoidal.

View attachment 219116

The velocity is sinusoidal if the driving shaft is rotating at constant RPM.

I am interested in solving for what the driving shaft rotational speed profile would need to be to drive piston at constant horizontal linear speed.

If we focus on just one stroke of piston from left to right (not continuous rotation of shaft). We should be able to derive a profile for the shaft angular speed that would result in a constant linear speed of piston for one stroke.

 

[Asymptotic]

Velocity formula is in the piston motion article.

 

[Wale]

Velocity formula is in the piston motion article.

Again, velocity formula in the piston motion article is for constant rotational speed. Not what I am looking for.

 

[Mech_Engineer]

Have you considered using a ballscrew for this instead? I'm not sure modulating the motor's speed will allow for a fine level of control, and directly coupling to a slider mechanism doesn't allow for torque multiplication.

 

[Asymptotic]

Again, velocity formula in the piston motion article is for constant rotational speed. Not what I am looking for.

What if you work backwards from slider position, and figure crank rotational speed by using the angular form?

I may be thinking about this incorrectly, but to maintain a nearly constant slider speed the crank will need to accelerate nearly instantaneously to a speed several times higher than the mean as it comes off of BDC, decelerate and maintain a more or less constant slower speed through most of the stroke, accelerate again while approaching TDC, then decelerate nearly instantaneously to zero when reaching TDC. If such a system were to be built it would tear itself apart at an alarming frequency.

@Mech_Engineer 's ballscrew idea has a lot going for it, or how about a rack and pinion arrangement?

 

[anorlunda]

@Mech_Engineer 's ballscrew idea has a lot going for it, or how about a rack and pinion arrangement?

Can you clarify your question? Are you seeking a theoretical calculation or are you looking for a practical way to get constant velocity?

 

[Wale]

What if you work backwards from slider position, and figure crank rotational speed by using the angular form?

I may be thinking about this incorrectly, but to maintain a nearly constant slider speed the crank will need to accelerate nearly instantaneously to a speed several times higher than the mean as it comes off of BDC, decelerate and maintain a more or less constant slower speed through most of the stroke, accelerate again while approaching TDC, then decelerate nearly instantaneously to zero when reaching TDC. If such a system were to be built it would tear itself apart at an alarming frequency.

@Mech_Engineer 's ballscrew idea has a lot going for it, or how about a rack and pinion arrangement?

The challenge with ball screw or rack and pinion is that you can't get to high enough speed/frequency.

 

[Wale]

Can you clarify your question? Are you seeking a theoretical calculation or are you looking for a practical way to get constant velocity?

Both. I am looking for a theoretical formulation which might then allow me to calculate a numerical solution to program my step motor.

 

[Wale]

To clarify, I only care about one given stroke. So for example a solution where the motor would drive one stroke clockwise at constant velocity, stop, then drive the return stroke counterclockwise, would also work.

 

[Asymptotic]

The challenge with ball screw or rack and pinion is that you can't get to high enough speed/frequency.

What linear speed, move distance, and move time are you shooting for?

 

[anorlunda]

Do you understand that all things with mass take time to go from zero velocity to a specified velocity V. Acceleration can not be infinite.

So if you include starting stopping, then the theoretical answer is no, constant velocity is impossible.

You should think more carefully about your question.

 

[Wale]

What linear speed, move distance, and move time are you shooting for?

Speed range of 0.5-4 inches/second. Stroke length range of 0.5-4 inches.

 

[Asymptotic]

So about 1 inch/second?
Ball screw and rack & pinion are both viable options. Don't recall exactly how much weight was involved (probably around 100 pounds), but I've worked on servo-driven ball screw systems that could readily make a 9 inch move in under 1/2 second.

 

[Dr.D]

To clarify, I only care about one given stroke. So for example a solution where the motor would drive one stroke clockwise at constant velocity, stop, then drive the return stroke counterclockwise, would also work.

What you want is kinematically impossible. You have already acknowledged that at both BDC and TDC the piston velocity must be zero (this is true). If it starts and stops at zero, and must also satisfy the condition of being constant velocity, then the only constant that will satisfy this condition is zero. Thus you require the crank to "turn" at zero angular velocity, which is no motion at all. Sorry 'bout that ...

 

[Mech_Engineer]

It seems like the easiest way to approach this is to:

1. Use the piston motion equation's cosine law to solve for θ
2. Assume θ and x are the only two variables, defining them as θ(t) and x(t)
3. Define equation for x(t)
4. Take partial derivative of resulting equation (d/dt)

A quick run through a symbolic math tool gets the following for θ and x:

Use piston motion cosine rule equation to solve for piston position vs crank angle:

Alternatively, rearrange the equation to find crank angle as a function of piston position:

You will need to assume an equation to describe x(t) with a constant velocity (maybe something like x(t)=(I-r)+v*t where I and r are the crank and connecting rod lengths and v is the piston speed) you should be able to take the derivative w.r.t. time and have a solution.

 

[Mech_Engineer]

Speed range of 0.5-4 inches/second. Stroke length range of 0.5-4 inches.

These requirements should be achievable with a ball screw by the way. It's easy to find ballscrew-driven stages which achieve faster speeds than this.

 

[Dr.D]

Consider two facts:

1) I think we have established that the velocity is zero at TDC and BDC.

2) The design requirement is that the velocity be CONSTANT.

If the velocity is zero anywhere (let alone two places), and if it is constant, then it is necessarily zero everywhere. What is difficult about that? There is no need to consider the kinematics of the slider-crank; it simply cannot be done. (By the way, it cannot be done with a ball screw or any other mechanism either.)

 

[Baluncore]

@ Wale.
By driving a hydraulic cylinder you are making a pump that will have to return at the end of the stroke.
The rod volume will make a difference between the extend and retract flow rates.

There is an alternative solution.
You could use a stepper motor to drive a small fixed displacement pump.
Use a gear pump for high pressures, or a peristaltic pump for low pressures.

 

[Wale]

It seems like the easiest way to approach this is to:

1. Use the piston motion equation's cosine law to solve for θ
2. Assume θ and x are the only two variables, defining them as θ(t) and x(t)
3. Define equation for x(t)
4. Take partial derivative of resulting equation (d/dt)

A quick run through a symbolic math tool gets the following for θ and x:

View attachment 219170

Use piston motion cosine rule equation to solve for piston position vs crank angle:
View attachment 219173Alternatively, rearrange the equation to find crank angle as a function of piston position:
View attachment 219172

You will need to assume an equation to describe x(t) with a constant velocity (maybe something like x(t)=(I-r)+v*t where I and r are the crank and connecting rod lengths and v is the piston speed) you should be able to take the derivative w.r.t. time and have a solution.

This is very helpful. Thank you very much!