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Scotch Yoke Actuator Torque Output
5

Scotch Yoke Actuator Torque Output

Scotch Yoke Actuator Torque Output

(OP)
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.
 

RE: Scotch Yoke Actuator Torque Output


The "arm" ( perpendicular distance from the force vector line to the center of rotation) is constant, so, the torque is the force times that arm.
If the force remains constant , so does the torque.

 

RE: Scotch Yoke Actuator Torque Output

(OP)
The "arm" or what I called "effective radius" is not constant. It changes throughout the rotation of the circle.

The linear force applied is constant, but the angle of application changes throughout the rotation of the circle. This should mean that tangential force changes as well.

I'm uploading an animation of the scotch-yoke. I know this is spring-return, just ignore that. This is just for visualization purposes.

RE: Scotch Yoke Actuator Torque Output

"The "arm" or what I called "effective radius" is not constant. It changes throughout the rotation of the circle."

Not in your animation.  Zekeman called it correctly, as the slide in your example is attached to the crank, thus the perpendicular distance (effective moment arm) from piston to shaft remains constant.

If the slide was attached to the piston instead (see example below), you would have a variable moment arm and torque would reach a peak at/near mid-stroke, for a typical 1/4-turn actuator.

http://en.wikipedia.org/wiki/Scotch_yoke
 

RE: Scotch Yoke Actuator Torque Output

(OP)
Ok, yes, I understand that the perpendicular distance from the piston arm to the center of the crank is constant. And also, linear force is constant.

The point of my original question is why am I getting a constant torque in my calculations when it is accepted world-wide that scotch-yoke has higher torque at the start and end of the stroke. What is the factor I am missing?

RE: Scotch Yoke Actuator Torque Output

"Ok, yes, I understand that the perpendicular distance from the piston arm to the center of the crank is constant. And also, linear force is constant.

The point of my original question is why am I getting a constant torque in my calculations when it is accepted world-wide that scotch-yoke has higher torque at the start and end of the stroke. What is the factor I am missing?"  


 
 

RE: Scotch Yoke Actuator Torque Output

What you are missing is that for the linear actuator the torque would be constant, your case, and you proved it and I and Trueblood confirmed it.

The variable torque you mention is  for many applications , like the gheneva mechanism, the input motion is rotary, so the torque would vary over the stroke.

RE: Scotch Yoke Actuator Torque Output

(OP)
This cannot be correct. Look at any company that sells scotch-yoke actuators... any literature on them. They all say the same thing.

Scotch Yoke actuators have high breakaway and closing torque, with low running torque. That means the torque is NOT constant.

Just type it into google.

RE: Scotch Yoke Actuator Torque Output

Read the second part of my post, and the linked example of a scotch yoke mechanism, but note the difference in where the slide is located.  The mechanism in the link I posted would have a variable torque, and (I think) is a more common mechanism for actuators than the one in the animation you posted.

RE: Scotch Yoke Actuator Torque Output

OK, I googled it and it shows  a linear actuator mechanism in which the slot is on the linear part of the actuator and the pin engaging the slot causes the "arm" to vary from zero to r during the cycle just as Trueblood points out.

So, your mechanism is a constant torque mechanism whereas the more common  ones are variable.

RE: Scotch Yoke Actuator Torque Output

hi Glix

If your force is constant and the radius arm is constant then your torque is constant, however as the actuator operates its compressing a spring so the torque output will vary because as the spring force increases it decreases the amount of availble force of the actuator.

desertfox

RE: Scotch Yoke Actuator Torque Output

(OP)
To Zekeman and Trueblood: The mechanisms you are describing have a variable torque that increase at the midpoint (at 90 degrees, when the arm is r). Correct? Constant force and the arm gets larger, maxing at the midpoint. The torque curves described in literature have high start and end torques (at 45 and 135 degrees) and the lowest torque is close to the midpoint. This is the opposite of what you are describing.

Here are some sites for you to check out.

http://www.instrumentation.co.za/article.aspx?pklarticleid=4659

http://www.kcicms.com/pdf/factfiles/actuation/Vw0811_actuation_oxler.pdf?resourceId=66  (This link even shows the exact same mechanism I posted)

http://www.k-controls.co.uk/Training%20material/1.01%20K%20Controls%20e-training%20-%20Operating%20characteristics%20and%20sizing%20of%20pneumatic%20actuators.pdf     (This link shows the mechanism with angled slides, but the motion is still linear)


To Desertfox: The animation I posted was just for visualization. I'm only talking about double-acting actuators with no springs. The only animation I had with no springs was an aerial view and didn't show the mechanism as well. Please ignore the springs.

RE: Scotch Yoke Actuator Torque Output

(OP)
By the way, thank you for your help, this has been frustrating me for about a week now.

RE: Scotch Yoke Actuator Torque Output

My bad, I got it all wrong.Elementary statics.

In fact, the normal force of the piston on the slot is

F/ sin@,and the "arm" is h/sin@, where h is the distance from the line of action to the centerline and @ is the  arm angle to the horizontal

So, the torque is

Fh/sin^2(@)

If @ starts at 45 degrees,and ends at 135 degrees, the torque is


Fh/.5= 2Fh


since sin@=.707, sin^2@=.5


In between , the sin^2@ >.5 and becomes 1 at 90 degrees.

the torque falls. At the midposition, @=90 degrees


T=Fh

 

RE: Scotch Yoke Actuator Torque Output

(OP)
Finally, some math. Ok, maybe I have a gross misunderstanding of statics (my background is actually in Chemical Engineering).

I understand the h/sin@. That's what I've been using.

Why are you dividing F/sin@. Shouldn't it be F*sin@? If @ = 0 degrees (linear force directed straight to the center of the crank), your equation would mean the crank is experiencing infinite torque, when in fact isn't it experiencing zero torque? Sorry, this is what's confusing me. How can a component of an applied force be larger than the applied force?

RE: Scotch Yoke Actuator Torque Output


"Why are you dividing F/sin@. Shouldn't it be F*sin@? If @ = 0 degrees (linear force directed straight to the center of the crank), your equation would mean the crank is experiencing infinite torque, when in fact isn't it experiencing zero torque? Sorry, this is what's confusing me. How can a component of an applied force be larger than the applied force?

I am trying to append a force diagram that shows the slot reaction, Fn and its 2 components, F and Fy which act on the cylinder; the F sustains the horizontal force and the Fy  introduces a vertical componenr  component to the cylinder.

RE: Scotch Yoke Actuator Torque Output

Note that unlike the usual conception of a scotch yoke where the drive slot is perpendicular to the piston axis, in this design, the slot is part of the rotating member, and assumes the angle that the shaft does.  If the shaft is positioned at a plus or minus 45 degree angle, the slot will assume the same angle, and the piston pin will be pushing against that ramp angle.  The pin will exert an equal force against or away from the shaft axis, making the applied force on the ramp surface 1.414 times the actual axial force.  Because the resultant force vector is normal to the slot face, it all contributes to the shaft moment.  (Neglecting friction of course.) The pin engagement radius, and consequently the moment arm is greatest at the end of travel.  (This wouldn't be true if the total rotation exceeded 90 degrees.)

If you examine the phantom view of the scotch yoke actuator, you will see bearing surfaces intended to handle the in/out forces created by the drive pin's reaction with the angled engagement face of the slot.  The non-linear torque is created by the wedging action of the rotation of the slot.

RE: Scotch Yoke Actuator Torque Output

(OP)
Yes, I agree with everything except the "1.414 times the actual axial force" that you stated with no proof.

Isn't this Statics 101?

When you have any object (in this case a pin) acting on a surface (the ramp/slot/slide wall), you have 2 components that can be drawn. One component is parallel to the surface (Fx) and one component is perpendicular to the surface (Fy). These components are never larger than the initial Force.

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.

RE: Scotch Yoke Actuator Torque Output

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)

RE: Scotch Yoke Actuator Torque Output

Work = force x displacement = torque x angular displacement

RE: Scotch Yoke Actuator Torque Output

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  

peace
Fe

RE: Scotch Yoke Actuator Torque Output

(OP)
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.

RE: Scotch Yoke Actuator Torque Output

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)

RE: Scotch Yoke Actuator Torque Output

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)

RE: Scotch Yoke Actuator Torque Output

Np.
But no one answered my question thumbsup2

peace
Fe

RE: Scotch Yoke Actuator Torque Output

"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.
  

RE: Scotch Yoke Actuator Torque Output

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....


;:



 

RE: Scotch Yoke Actuator Torque Output

(OP)
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.

RE: Scotch Yoke Actuator Torque Output

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

RE: Scotch Yoke Actuator Torque Output

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.
 

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