Pneumatics Math Help

I am trying to design a joint that will actuate from 90 to 180 degrees with a pneumatic cylinder. Here is a link to what I have designed so far.

I am now trying to figure out the forces involved in the system in order to choose the right bore size for the application. This is simply an offseason design concept, so there are no force requirements for this specific design, however I would like to know how to calculate the forces for future use of the design. I understand how to calculate the force exerted by the cylinder from the bore size, that is not what I am confused about. I am wondering how to calculate the force exerted perpendicular to the arm throughout the different angles of motion, at different distances from the hinge. This might be very simple but I am getting confused because of the linkage connecting the cylinder to the actuated arm.

Let’s see if I understand the question:

Given that you are applying a given force F to the arm at a distance X (the location of the link point) from the hinge, you want to find the force that is being applied in the direction of motion at any given point?

I keep thinking this could be solved just by using torque, but then I think about it and realize that I’m not correct.

Are you familiar with the concept of a Free-Body Diagram? If you are, try drawing one. If not, then you may want to learn–essentially, it’s a diagram that shows all the forces acting on an object as arrows.

Yes that is exactly what I am wondering! Thanks for clarifying. Yes I am familiar with free body diagrams but I have only been taking physics for 1 trimester in school so I can’t seem to figure this one out yet.

I think I can help. I’m a little rusty, so I’m talking to Messrs. Beer and Johnston to brush up.

[Edit] BTW, for those that don’t get the reference: That would be the statics textbook written by said authors.

I can’t say that I have a solution ready to hand to you, and even if I did I’m not sure that I would. So you’ll need to do a little geometry.

You actually need TWO free body diagrams. One for the arm, with a force for the cylinder coming in, and one for the cylinder itself to determine the force that the cylinder is putting in. The geometry gets messy on that one, though.

Arm diagram, sum moments around the hinge. In order to make this thing “stop”, you’ll need to add a moment going in the opposite direction–net result is a relationship between cylinder force, angle of arm, and distance from hinge point that the force is applied.

The fun part is when you have to figure out the applied force from the cylinder in the direction of motion, given that the cylinder is changing angles the whole time. If I’m not mistaken, that would involve knowing both distances from the arm’s hinge point to the cylinder attach points (one on the arm, which is variable, and one on the support structure, which is not), and then figuring out the angle.

I think you can boil it down to one equation, too, with one variable (the angle of the arm) and several “we’ll engineer these when we design the robot” numbers.

That is where you’ll find the answer, but it won’t be waiting for you to copy-and-paste it. I have never met anyone who learned to solve statics problems without spending many evenings with Beer and Johnston. I know Eric is aware of this, and has put that time in already.

Back in the day at Georgia Tech, the level of commitment (read: hours of homework) required by that class was a wake-up call for engineering students – either you do the work needed to learn statics, or you find another major. I will be truly astonished if someone tells me it is not like that today.

Back to the OP: Question – can the clevis on your cylinder be mounted further from the pivoting member?

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 = Rcos(A1), ya = Rsin(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 + L1cos(A2), ya = yh + L1sin(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 = Fcylcos(A2)
Tcyl,h = Fcyl
cos(A2)ya = Fcylcos(A2)Rsin(A1)
(As shown, this component should always be positive)

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

Tcyl = FcylR(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.

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.

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.

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.

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.

Having graduated 3.5 years ago, yep it’s still the case :). Though at Clemson Mech Eng had to take a combined Statics&Dynamics class. Twice the credit hours of the statics class that other major took. So many ex-engineering students after that class, heh.

If you plan on orienting this vertically, you’ll want some sort of counterbalance otherwise your cylinder will need to be unnecessarily hefty to work against gravity. A coil spring or elastic material (surgical tubing) would approximately work. Or you could make a true counterbalance by extending the end of the actuated arm and adding a weight such that the torques cancel out :slight_smile:

If you’re looking at repeatedly lifting a load and leaving it there, using a counterbalance (whether gravity, spring, or otherwise) will buy you a factor of two in the amount of work you need to do in each half stroke. That is, you can do work both on the upstroke (lifting the load with spring/counterbalance help) and the down stroke (adding energy to the spring/counterbalance). When using pneumatics for a lift, deciding whether the complexity of the counterbalance is worth having a smaller cylinder is usually a real engineering question (meaning that the answer is “it depends”). In my experience, counterbalancing a pneumatic actuator is less frequently an improvement than in the case of an electric actuator. This is especially true when you do not need to pressurize both sides of a cylinder, but get a “free” light spring or gravity return.