Pneumatics Math Help

[Chris86]

So this is a torque balance problem about your hinge point:

The scalar form of this is:
Tcyl = Fcyl*distance to hinge point.

But what we really want to work with is the vector form:
Tcyl = Fcyl x distance (where x represents the vector cross product)

Letting x represent horizontal distances (along the stationary base) and y representing vertical distances and the origin be at the hinge point where your bar is rotating about.

I'm going to assume you know a the value A1 which is the angle above the horizontal from the hinge point of the piece to the cylinder's attachment point on the link and the distance between these two points. (In your extended images its around 30 deg, and retracted is ~120 deg)

We can find the x, y locations of the attachment point as:
xa = R*cos(A1), ya = R*sin(A1)

Let A2 be the angle above the horizontal from the rear hinge point (xh, yh) of the pneumatic cylinder to its attachment point to the link and L1 be the distance between this point Now, we also have
xa = xh + L1*cos(A2), ya = yh + L1*sin(A2) (two equations, two unknowns - L1 and A2)

So now we know the direction of the force which is what we needed. The torque it generates is a combination of the torques from its horizontal and vertical components: (define clockwise as negative - since I worked this out in the scalar instead of vector form, you need to handle the signs yourself)
Horizontal Component Fcyl,h = Fcyl*cos(A2)
Tcyl,h = Fcyl*cos(A2)*ya = Fcyl*cos(A2)*R*sin(A1)
(As shown, this component should always be positive)

Vertical Component Fcyl,v = Fcyl*sin(A2)
Tcyl,v = Fcyl*sin(A2)*xa = Fcyl*sin(A2)*R*cos(A1)
(As shown, this component should be negative for A1>90 deg)

Tcyl = Fcyl*R*(cos(A2)*sin(A1) + sin(A2)*cos(A1));

For solving how big the cylinder should be, you'll need to do a similar problem with the weight of your actuated member/any external forces applied and solve for the torque that those forces generate.

[s_forbes]

For a linkage like this, you need to work out how much torque the cylinder applies to the member that moves. The diagram would look like this:





The blue line shows the axis of the pneumatic cylinder. The red line in the sketch shows the distance between the axis of the cylinder and the point the mechanism rotates around. The red and blue lines are perpendicular no matter what angle the mechanism is at (that's important! and possibly the confusing part).

The torque applied to the moving member is equal to the force from the cylinder times the distance indicated by the red line. Torque = Force x Distance. It looks like you are using Solidworks to model your mechanism, I'd recommend making a simple sketch like the one above that has the correct dimensions for all of the parts, then you can measure the distance of the red line. With that you can find the torque on the moving member at any orientation you put it in.

Chris86 posted the mathematics behind this, but diagrams are a bit easier to understand if you are getting used to the concepts. Note that when the red line is the longest, you have the maximum torque applied to the mechanism! And when the red line becomes very small, you have very little torque applied to the mechanism.

[Sam Skoglund]

Thanks to everyone for the replies! I will be sure to post again here if I get it figured out, so that others can work through this calculation too.

[GeeTwo]

Not meaning to contradict any of the notes above, but in designing such a joint, I find it convenient to first look at the amount of work to be done. In the case of the 90 degree rotating arm, this would be π/4 (90 degrees in radians) * arm length * [average] force required at end of arm. The result will be in ft-lb or N-m, or other similar units. Then, divide this amount of work by the working pressure on the cylinder (perhaps 50 lb / in²) which is the smallest possible displacement (that is, area times stroke) to do this amount of work. Don't forget to convert feet to inches!

Multiply this minimum displacement by a nice safety factor (1.5 if you don't mind the stroke being rather stately, or perhaps 5 to 10 if you want fast action). This will allow you to come up with a fairly small set of reasonable "stock" cylinders. If you use Bimba cylinders, the "power factor" (first two digits of original line cylinders part number) is the area of the piston in tenths of a square inch, and the remainder is the stroke length, so to calculate the displacement of a Bimba cyldinder, just multiply power factor * stroke / 10. (Example: an SR 1715 cylinder's displacement is 17 * 15 / 10 = 25.5 in³.)

THEN, for each candidate cylinder, figure out the appropriate mount points. Shorter, thicker cylinders will mount closer to the pivot but will require larger forces, possibly meaning thicker plates and bolts and certainly better precision. Longer, thinner cylinders will mount further from the pivot (requiring more maneuvering room), but the forces will be smaller and probably more reliable, especially if your manufacturing is subject to error.

[Ari423]

I love the complicated physics in this problem as much as the next guy, but it might be easier if you can mount something like this rotary actuator next to the pivot point. This will also have the added benefit of providing a constant force throughout the rotation as opposed to a dynamic force with the linear actuator. Most pneumatics suppliers sell some version of this.

Disclaimer: I have not actually used one of these (because I haven't had the need) so I don't have any real world experience to say whether or not it's easier.

[ollien]

Originally Posted by Ari423

I love the complicated physics in this problem as much as the next guy, but it might be easier if you can mount something like this rotary actuator next to the pivot point. This will also have the added benefit of providing a constant force throughout the rotation as opposed to a dynamic force with the linear actuator. Most pneumatics suppliers sell some version of this.

Disclaimer: I have not actually used one of these (because I haven't had the need) so I don't have any real world experience to say whether or not it's easier.

Pretty sweet. A bit pricey, but nothing too crazy. Would love to see a place I could purchase these other than Bimba, because of their crazy turnaround times.

McMaster sells them, but at a uh.. rather hefty price.