Help on Scotch Yoke

[SERVOCAM]

I have a problem involving a Scotch Yoke and calculating force and was hoping someone could layout the formulas and calcuations for this problem?

I have a metal bar that weights 150 lbs and 2.5 inches square x 34 inches long. The bar pivots in the center. I need to calculate the force required to rotate a bar

When the bar is at a +45 degree angle and I need to go to -45 degrees, The linear actuator will need to move 34 inches (if I calcuated right). So, I need to make the 34 inch move in 450msec with a 50msec dwell. I could calcualte the Force needed by my actuator if I understood the relation ship of the bar to it.

I can calculate the inertia of the bar from the center pivot (20.44 lb-in-sec^2) and the aceleration torqe to rotate it from the center (650 lb-in based on a 200msec accel time). But now how does that relate to the force on the actuator with the Scotch Yoke?

The Force should vary depending on the angle of the bar to the linear actuator, right?

Someone told me that you can take the Accel torque and divide it by the length of the lever arm (lenght of bar from center pivot to the actuator). If I did this, at 45 degrees, the length is 24 inches. So 650/24 = 27 lbs. At 0 degrees, the length is 17 inches, so 650/17 = 38 lbs. If this is correct, it gives me the rough amount. I could calcuate say at every inch then do RMS of that, but there should be a formula for this.

Any help would be much appreciated.
-cam

 

[SERVOCAM]

Ok, so I have thought about this a bit more and going through my memory banks on magnitude of a moment. Also looking at simple terms of Force = Torqe / Radius. So when perpendicualr, the 650 lb-in / 17 inches = 38 lbf would be correct.

In this case when the angle of the bar changes from 0 degrees, the moment does not get shorter, but longer, so I would agree that at 45 degrees the torque would only be 27 lbs.

I guess if anyone has a formula that can calculte the RMS force for this, please share?

 

[desertfox]

Hi SERVOCAM

This bar of yours is it just oscillating back and forth between +45 degrees to -45 degrees ie 90 degrees?
Where is the linear actuator attached ie radius and angular position relative to the bar at some given position?
With a bit more information we might be able to help a bit more. Is the dwell at the end of stroke which is what I am assuming?

regards desertfox

 

[SERVOCAM]

Hi, thank you for the response. Yes, the bar is just rotating back and forth (+90, then -90)I should of stated the location of the actuator to the bar. The Actuator base is fixed and carriage is Perpindicular to the bar (when the bar is at 0 degrees). When at the bar is at 0 degrees, the radius is 17 inches (distance from Yoke to Bar Pivot). When the bar is at +/-45 degrees, the radius is 24 inches (distace from yoke to bar pivot).

I'll try to roughly illistrate:
A is Actuator, traverses up/down on the page (will be linear servo motor)
[A] is the carriage where scotch yoke / bushing assembly is
- is the slider-rod for the yoke
= is the bar, rotates CW/CCW on the page +/-45 deg (90 total)
O is the pivot location.

A
A
A
A
--[A]=======O=======
A
A
A
A


Move Profile is:
Move 34 inches (90 deg of bar) in 450 msec. Accel = 200msec, Traverse = 50 msec, Decel = 200 msec.
Dwell = 50 msec

I know how to calculate RMS force when the Accel, Traverse/Slew, and Decel forces don't change, so on most of these applications like this and cranks & cams, I generally over size or take a calcualtion every inch or 10 degrees then calculate RMS from that. Good news is that in 8 years, I have never misized any servo/stepper system. This same mechanical system lends it's self to S-Cure Move profiles, that the torque is changing during accel/decel, and I just usually double my linear profile to compensate.

Do you have the magic formula / polynomial equation for calculating RMS force? I am applying the correct logic to the system on force, right?


Thanx,
-cam

 

[SERVOCAM]

Ok, so, here is my new delima, as there is a diference in oppion from two friedns of mine.

Ken says that the Scotch-Yoke / Lever Arm gives you advantage because the lever-arm increases in length and the force stays constant/perpendicular, so you gain a mechanical advantage, Force curve is Sinusoidal. (Oh, and I forgot, if the force/torque is sinusiodal, RMS is .707 of the peak)

Brian, says, which was my initial take on it, is that yes the lever arm increases, but the vector force is no longer perpendicular, so the Force decreases, and lever arm increases, the result mechanical advantage is 1:1.

Who is corect and why?



-cam
"Your just jealous that the voices don't talk to you"

 

[SERVOCAM]

Found my answer.

http://www.qtrco.com/images/Binder Lit PDF Files 041205/7-K2-SY Vs RP Torques 2.pdf