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.

 

[zekeman]

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.

 

[Glix]

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.

 

[btrueblood]

"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

 

[Glix]

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?

 

[Glix]

http://www.eng-tips.com/viewthread.cfm?qid=143012&page=130#post

The second to last post of this thread is exactly the question I am asking about. Who is right? Unfortunately, no one explains it and the link is dead.

 

[zekeman]

"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?"

 

[zekeman]

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.

 

[Glix]

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.

 

[btrueblood]

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.

 

[zekeman]

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.

 

[desertfox]

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

 

[Glix]

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/Trainin...ristics and sizing of pneumatic actuators.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.

 

[Glix]

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

 

[zekeman]

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

 

[Glix]

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?

 

[zekeman]

"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.

 

[Glix]

This might help clear things up for me.

Please look at this attachment and tell me which point has the highest torque and why. Both points are experiencing the same Force vector.

 

[potteryshard]

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.

 

[Glix]

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.