Calculating torque and angular acceleration 1

[cwsink]

Hello,

I've got a project that utilizes a diverter mechanism that will operate very quickly (one stroke in 0.094s). I need to calculate the cylinder size that will activate the diverter assembly fast enough. The diverter is basically a glorified flapper that pivots on one end and the other end moves up and down approximately 1". I've completed my calculations and wanted to attach them, but I'm not sure how to do that???

But, my "real" problem is that the force seems rather low. My methodology is as follows:

1) Calc cyl velocity from max time and cyl stroke (.094s/1")
2) Calc angular Vel. from cyl velocity (omega=V/r)
3) Calc angular acceleration from ang. vel. and total time (.094s) (alpha=omega/t)
4) Calc torque from ang. accel and Mass Mom. Inertia (T=I*alpha)
5) Calc cylinder force needed from torque and radius of moment arm.

Am I missing something?

I appreciate any help on this,




Craig Sink
Mechanical Engineer
Force Design, Inc.

 

[rb1957]

step 3) looks a little dodgy to me.
from steps 1) and 2) you're assuming a constant velocity
which implies an infinite acceleration at both ends of the travel.

maybe you need to assume a real acceleration value and check that you get the required displacement in the time allowed.

does this flipper flip one way (from a starting point to a finsihing point) or both ways (start-end-back to start) in one activation ?

is anything being deflected by the flipper ?

 

[israelkk]

What about the time to fill the cylinder with the fluid?
You need to write the complete differential equations for the system (control valve dynamics, fluid flow dynamics, instant pressure inside the cylinder, system inertia's, friction etc.) and solve it.

You assume constant cylinder force but this is not the real thing. While the cylinder moving a fluid flow in and the pressure in the cylinder varies. The valve that command the fluid to the cylinder has a time constant too.

Depends on the flow rates and the fluid pressure the valve time constant may even be larger than the cylinder time constant.

By the way it should be 1) Calc cyl velocity from max time and cyl stroke (1"/.094s).

 

[cwsink]

rb1957,

Yes, step 3 is a problem. I realized that I need to calculate the acceleration from the final velocity, not the average. E.g. my avg V would be 1/2 of the final V if accel is constant.

EX:
s = 1" (travel)
t = 0.094s
V_avg = 1"/0.094s
V_initial = 0in/s
V_final = (1/.094)*2
Accel = V_final/t

I will also take your suggestion to assume a real acceleration and work backwards as a check. In regards to the flipper, my calculations are for start to finish, no return as an activation.

Thank you,



Craig Sink
Mechanical Engineer
Force Design, Inc.

 

[dvd]

Another aspect that you may need to address (depending on your diverter design) is what happens to the kinetic energy of the object being diverted. If the object being diverted would tend to open the diverter, then you will have to size the cylinder to handle all of that energy. The kinetic energy of the object will need to be absorbed in some distance by the stop, so that Wcyl=FcylxD can be equated to the kinetic energy of the object.

 

[chicopee]

You should be able to graph your motion on a vertical scale and time on the horizontal scale just as you would do with cams. From that plot you can get your velocity and acceleration graphs with derivatives.

 

[handleman]

It sounds like you're probably spending too much time looking at the body mechanics of the flapper and not enough considering the fluid mechanics of the system. I believe israelkk has discerned the real limiting factor of your design's actuation time. You didn't mention whether you're using pneumatics or hydraulics. I'd assume pneumatics. Since you're dealing with 1/10s actuation time you are really going to have to consider the fluid flow, valve actuation time, etc. I'd wager that the cylinder that will give you your calculated force at steady state will not actually produce that output force within 0.1s. The reasons are many. Here's an actuation timeline for this motion, assuming this is a standard, dual acting cylinder operated by a standard, 5-port air-piloted pneumatic solenoid valve. Each of these steps takes a finite amout of time that you will need to consider.

1. The solenoid receives a voltage that energizes and magnetizes a coil. This coil takes a certain amount of time to magnetize due to the coil's inductance.
2. The magnetized coil pulls open the pilot valve.
3. Air rushes through the pilot, building pressure on the valve spool.
4. Once the pressure is great enough it pushes the spool to the opposite end of its travel, opening both the path from the supply port of the valve to the inlet port of the cylinder and the path from the exhaust port of the cylinder to the exhaust port of the valve.
5. Pressure in the rod side of the cylinder begins to vent out the exhaust port of the valve while pressure builds in the piston side. Fluid mechanics dictate the speed with which the air can exhaust from the rod side and fill the piston side.
6. When pressure in the piston side rises above the falling rod side pressure the cylinder begins to move. During this time the output force rises as the differential between piston and rod sides. Even if the rod side started out with zero (gage) pressure, the piston side still has to force all the atmospheric air out as it travels. With the speed you're looking at, this will likely be a factor.

There are a few ways you can attempt to reduce the impact of these factors:

1. Use a spring return cylinder if possible, and drill out its rod side port to a rather large size.
2. Choose a cylinder that has large ports to start with.
3. Choose a large valve with low restriction.
4. Minimize all piping runs, especially between valve and cylinder.
5. Make sure supply lines are large up to the valve.
6. Use as high of pressure as you can.

I'm guessing this is a low-volume, one-off type design rather than a consumer product design. Rather than setting up all the equations as suggested by israelkk, in your shoes I would probably do as much real-world testing as possible. If you're using off-the-shelf components they are all rather inexpensive compared to design time labor hours.

 

[cwsink]

Thanks for all the input.

In regards to the additional factors such as valve delay and fluid flow, etc. Those factors don't necessarily have to happen within the specified time. The need to actuate the diverter will be determined ahead of time by the controls so the actuation could begin[\i] early-if that makes sense! My calculations are to determine the proper cylinder size such that the cylinder can operate in the given time constraints. As Handleman mentioned, there will be real world testing and adjustment done rather than modeling the entire system and calculating from there (we don't have the privilege of doing that, unfortunately).

That said, I appreciate all of the input.

Craig Sink
Mechanical Engineer
Force Design, Inc.

 

[israelkk]

As once one smart man said: "there is never enough time and money to do a good design but there is always all the time and money to fix a bad designed system".

 

[handleman]

So... how fast would a cylinder with no load operate? Infinitely fast? Steps 1-4 of my previous post are usually fairly negligible, and, as you say, could be taken care of by sending the actuation signal "ahead". However, steps 5 and 6 occur during the actuation, and will have a significant impact on how much of the cylinder's steady-state force will be available to move your load. Now that you know what that load is, if you have a good relationship with a cylinder vendor I'd go ahead and get them involved. They're used to figuring out stuff like this.