Scotch Yoke Actuator Torque Output 5

[Glix]

Hi, I'm at the end of my rope on this problem. As I deal with different actuators, I thought it would be important to theoretically/mathematically predict the torque curves for them.

The principle of the scotch-yoke actuator is that two pistons move linearly and cause the shaft to rotate. Each piston has an arm. The end of that arm slides down the gap as the circle rotates until it hits the mid-point, and then slides back up (from a radial perspective).

For simplicity and consistency, we'll focus on one side. The linear force from the piston would be applied in a vector with a 0 degree angle. The initial arm position is at 45 degrees. It rotates the shaft through to 135 degrees (giving the 90 degrees needed for quarter turn valve).


I initially got the same result as the rest of the world, high start and end torque due to the changing length of the effective radius. But then I realized that the torque is a function of the radius and the force perpendicular or tangent to the circle. When I took this into account, I end up with a flat torque profile.

My general approach is as follows:

T (torque) = R (effective radius) x F (tangential force)

Starting and end positions - 45(135) degrees

R = r (the entire length of the circle)
F = F sin 45
T = F*r*sin45

Mid position - 90 degrees

R = r sin 45
F = F (force is perpendicular to the radius)
T = F*r*sin45

When trying other angles, I get the same result.

When changing the starting angle (for example to 60 degrees instead of 45), you get a higher breakaway torque, but the end torque is lower than the mid point. I am not getting a parabolic result. I feel like I'm missing an important factor, but I just can't figure out what it is. Please help.

 

[handleman]

Yes, it is Statics 101.... you're pulling a C minus right now.

Draw some FBDs. Solve some equations. Don't skip anything.

-handleman, CSWP (The new, easy test)

 

[dvd]

Work = force x displacement = torque x angular displacement

 

[FeX32]

more like statics 100 ....
Now how about that linear piston accelerates at a constant 20m/s^2, how much faster (or slower) is the rotary rod accelerating when it is at theta=45deg. compared to when it is at 90deg.? .... now we are starting dynamics 100 (even though what I stated is a kinematics problem...which then is usually mated with a dyn force analysis...hence dynamics 100.)

Cheers


Fe

 

[Glix]

To Handleman: Your post is the epitome of unhelpful. Thank you for your condescending attitude.

To Dvd and Fex32: I want to thank both of you from the bottom of my heart. This problem has been eating at me for about a week. I used the work equation and finally achieved the result that everyone else has been spitting out matter-of-factly. Thank you.

 

[handleman]

Did you draw any FBDs? I didn't say you are pulling a C minus because you are mentally deficient or something. It's because you recognized that it's a statics problem but you are not using the very first thing you are taught in statics class: a free body diagram. Draw a free body diagram of the pin. Write the equilibrium equations. Draw a free body diagram of the rotor. Write the equilibrium equations. Solve them as you were taught in Statics class. You are trying to skip these steps and just solve it intuitively. Any statics professor would give you low marks for your current methods.

-handleman, CSWP (The new, easy test)

 

[handleman]

Because you neglected to create a proper FBD, you missed the torque component due to the Y direction reactions.

Please see the attached FBD.

Modifying the equations to solve for torque relationship as a function of plunger displacement is left as a trigonometry exercise for the pupil.

-handleman, CSWP (The new, easy test)

 

[FeX32]

Np.
But no one answered my question


Fe

 

[zekeman]

"I mean, think of a brick on a hill at a 45 degree angle. The weight of the brick is your applied force. The parallel component (Fx) drives the brick down the ramp. The perpendicular component (Fy) gives you the normal force the ramp is acting on the ball... which can in turn give you friction if you have the factor. Those components do not exceed the bricks weight".

Glix,
This is not your problem. The following is:

Take the frictionless brick of weight, W and let it be constrained to slide vertically onto the inclined frictionless plane at angle @. Now find the reaction force of the inclined plane against the brick. You should find, contrary to your opinion that that force is greater than the weight of the brick. Statics 101.
When you solve this you solved the problem.
I did and got W/sin@ as I showed you earlier with your problem.



"To Dvd and Fex32: I want to thank both of you from the bottom of my heart. This problem has been eating at me for about a week. I used the work equation and finally achieved the result that everyone else has been spitting out matter-of-factly. "

A lot of us have been trying to help you on this but nothing is penetrating. To say that we have been "spitting out matter-of-factly" is not realistic.

Also, while Dvd has given you an average answer, it does not give you the instantaneous result you need and doesn't show the variation of torque with theta.

You need:

F*dx= T*d@ as the energy equation.

 

[zekeman]

Correction
I wrote:

.....Take the frictionless brick of weight, W and let it be constrained to slide vertically onto the inclined frictionless plane at angle @. ...

It should read:

.....inclined plane at angle @ to the vertical....


;:

 

[Glix]

You're right. I apologize. I do thank everyone for trying to help, it's just the concept of displacement and angular movement was the push I needed to see how I was viewing the problem wrong.

I'm done thinking about this for a long time, my mind needs to rest.

 

[zekeman]

For what it's worth I am appending the energy solution

 

[BrianE22]

glix -

You can't ignore the spring force. The force available to turn the shaft equals F(from air pressure) - F(spring). That would cause the force from the actuator to be greatest when at one stop (air pressure force high, spring force low).

Looks like there are some things done to get the higher torques at the ends:

http://www.qtrco.com/sites/default/files/papers/scotchyoke-paper.pdf

 

[chicopee]

See my attachment for a general equation for CCW rotation only. What you will need are the displacement, velocity and acceleration diagrams to use in this equation. Also, moment of inertias,spring constant,and exteral load on output torque will need to be determine in this general equation.