# Newbie help!



## Razaray

Hey guys. Found this site last year and it was a great help as a newbie to the Halloween overhaul seen. But now I have a slightly off topic project. Looking to make a softball pitching machine out of the portable air compressor with a simple catapult arm. The idea is to fully charge the tank at home, drag it out somewhere without AC plugs and have it fire off a couple hundred pitches in 10 second intervals. I got some free components off of a buddy and looking to make it work any suggestions on other components needed or design would be greatly appreciated.

I have: 
20mm bore x 50mm travel cylinder dbl acting
25mm bore x 50mm travel cylinder dbl acting
3 way solenoid valve
Timer to energize solenoid for .5 sec then de-energize for 10 sec. 
Flow control valve
ON/OFF Switch

Was basically going to mount cylinder about 2 or 3 inches up arm from pivot point and hoping to launch the ball 30-40ft. 

The rest of the functions like arm return and ball loading are going to be mechanical or electrical. And to save air I only want to use air to launch the arm not retract or anything else. One worry I have is if the cylinder is slamming full extention if that will damage the cylinder over time. And if it will how might I arange to avoid it from slamming open or closed. 

Any help is greatly appretiated


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## gadget-evilusions

How large of a tank do you plan on filling at home? That will determine the number of actuations you can get before having to refill.


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## Razaray

8 gallon


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## Razaray

I don't want to restrict flow cause I think I'll need the cylinder to extend as fast as possible. But I also don't think its good to be slamming the cylinder to full extension. The valve timer I have only is adjustable down to half a second which I'm thinking will be too long to avoid the cylinder from slamming full extention. Do you think the cylinder can handle this bottoming out? And if not, any suggestions on how to avoid bottoming out using this half second timing.


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## Razaray

Another issue I will have is that it has to Throw at a consistent distance. Which means consistent pressure. The tank size is 8 gallon and max psi is 125. I'm thinking I'll have to try to get it to work at like 60psi so that by the time tank pressure gets down to 60 I'll have been able to get off a couple hundred shots hopefully. I've been trying to calculate it but I don't know the math enough to be sure. So it's gonna be trial and error for me. But if anyone could calculate how long my .75" bore by 2" travel cylinder will last using the above numbers I would love to know.


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## mikkojay

What are the specific make and model of your cylinders?
You might research them to see if they have internal air dampening in them to avoid the slamming you describe. Many of the used cylinders I buy on ebay have this. You can feel it if you try to extend the cylinder to its full range by hand very quickly. Many cylinders have this, so yours may- if they do, it will be one less thing to worry about with your plans. Sounds like a fun project- good luck!
-Mike


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## Razaray

No, it's double action. I suppose I could plug one end to act as a damper. Could even control the damping by plugging it with cylinder in various positions. I may have solved that problem myself lol.


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## bfjou812

I think some of the other variables that need looking into are what is the size and weight of the ball? What is the desired distance? What is the desired speed? Is this going to be a true catapult? If it is going to be a true catapult I would look into a solenoid that would release a cable of some sort that would be attached to a catapult. There are lots of solenoids that will run on 12 volts dc so the portability issue isn't one. I think that your pneumatic idea is a good one , but I don't think you'll get the desired results from such a small tank. Especially if you're wanting to shoot several hundred balls per session. I think the air tank will be depleted faster than you would think. Just my 2¢ worth..........


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## spinman1949

Probably be a better idea to use spring for the catapult and use the pneumatics to cock the spring. That way you don't have to worry about damaging the actuator. I tend to agree with BF on this. I honestly think you are going to find that pneumatics to actually move a ball at say even 50 MPH is not an easy task. What you need is stored kinetic energy. IE a spring. You could even use a short throw actuator that fires more than once with a ratcheting cocking mechanism. Think car jack.


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## gadget-evilusions

Out of that small of a tank, you'll also only get a few shots at full power.


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## Razaray

50 MPH seems fast, I was thinking more like 30-40 depending on loft. It's just for slowpitch. Ya it is a smaller tank. From what I figure most mechanisms are going to use similar amounts of air. Whether it's a catapult or a retract against an opposing force, in my thoughts it is basically a force needed vs travel problem. The closer the force point is to the pivot point of any mechanism means less travel but more pressure needed which translates into X amount of air. The further the force is to the pivot point of any mechanism means more travel needed but less pressure which I would think translates into about the same amount of air. Otherwise you'd get free energy from somewhere. I think the only way I'll be able to save air is to test and regulate to the minimum PSI and arm travel actually needed to get the distance I'll be happy with. But right now I'm assuming 60psi and full cylinder fill to 60psi.

These are my calculations. Please let me know if they make any sense cause the actuations seem high.

8 gallon tank = 1,848 Cubic Inches

.75"bore x 2" travel cylinder = .885 Cubic Inches

Therefore 1,848/.885 = 2,091

Tank PSI = 125

Cylinder PSI = 62

Therefore 125 / 62 = 2

Therefore 2,091 / 2 = 1, 045 - 1,045 actuations at 60 PSI

I'm not sure if pressure is calculated this linear though. And even if this is drastically wrong, and I say divide this by 4 I still get my 260 or so actuations.

If I were to use stored kinetic energy I would probably just use a motor and gears with 3/4 of the gears shaved off and that would give me my timing and my retract and release all in one like this. (See link) 



 I just had this compressor laying around and some free components so I thought I'd go the pneumatic route lol.

Here is something I threw together lol. (See link).


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## spinman1949

Razaray,

Key will be the required force to accelerate the ball to the desired velocity. 

3/4 cylinder = .44 square inches of surface.

.44 X 62 = 27 lbs of force. 

Ratio of expected movement. 

Based on your drawing I am estimating that the cylinder you are showing is much closer to a 4 inch throw then 2 inches, and it appears that your ratio is about 6 to 1. 

Thus you need to double your cubic inch of air per stroke figure and your resulting force should be around 4 lbs. 

So as long as the 62 lbs force will overcome the initial weight of the ball and mechanism to achieve your desired terminal velocity you should be good to go.

The other issue will be how long the cylinder will maintain the static 62 lbs.


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## Razaray

No no, that video is just for fun to show the concept. Nothing is to scale or standard. The only constants I have are that the cylinder is .75" x 2" throw and the tank maxes out at 125psi. The rest can be adjustable ie: distance from pivote point to release point, release trajectory(should be 35-40 degrees), ball speed at release, cylinder connection point to the arm, pressure regulation and amount of cylinder force and travel needed. All of these factors effects one another but my main concern is conserving air. I know this can all be tested mathematically but I haven't received enough insight on it to have concrete numbers. I'm just gonna have to put a prototype together and start testing. Shoot, I'll start at 40psi if that's all it takes. What do you mean exactly in saying how long the cylinder will hold the static 62psi. I'm assuming if this cylinder is adequate at the most it will use the entire 2" stroke filled to a required psi(whatever that ends up being) the trigger I have keeps the valve open for .5 seconds so I'm pretty sure the cylinder will be fully pressurized in that time as I'm not limiting flow to achieve fastest extention. 

Here's a question: Does the volume of air used in a 1 cubic inch cylinder at 100psi equate to a 2 cubic inch cylinder at 50psi? In other words, does twice the pressure equate to twice the volume?


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## Razaray

I think I answered my question. See Boyle's Law


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## spinman1949

Razaray,

Each time you fire the cylinder the volume in the tank will drop and the available pressure will drop. Your regulator will maintain your static setting as long as the pressure in the tank exceeds the desired setting. So somewhere around half tank your 62 lbs will start to drop. 

As far as your design. I personally think a 2" throw is too short for what you are attempting to achieve. Best thing is to just try it. Bottom line is velocity needed. 

Let us know how your testing works out please.


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## spinman1949

Razaray,

m*a F = .185 kg * 6 m/s^2 F= 1.11 N

Softball weighs on average around .185 Kg 6 meters a second is around 20 Mph. A Newton is kinda hard to figure what it exactly means, but my uneducated guess is that it represents 1000 grams of force. Which would be 2lbs 7 oz. 

Not sure if this helps, but I would set the leverage ratio to end up with around 3 lbs of force. So the arm ratio should be about 10 to 1. This should give you an arm throw distance of around 20 inches.


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## Razaray

What do you mean when you say"arm throw distance"? I should end up with 120N of force at cylinder end at 62 psi. But bottom line is that I'm sure there is a formula out there for calculating all the factors. Problem is finding it and understanding it. It's easier to use software to test it out. The program I used is very basic but it has enough parameters to get me close I think within 5% of values needed. I should be able to allow for that 5% in small adjustments throughout the system, easiest being pressure regulation. I'll update soon with a software test using realistic numbers to my design. Then if all seems well on the screen I'll build a prototype.


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## spinman1949

Razaray,

What is the velocity you are attempting to achieve? That will help. Also it will be important to establish the actual time it will take for the cylinder to make it's full 2" inch travel.

So allow me to give you some figures to look at.

60 MPH = 1 Mile per minute. Thus 88 Ft per second. 5280 / 60 
Now lets say you want to toss the ball at 10 MPH. 88 / 6 = 14.7 FPS
So the arm that is going to throw the ball must at the end of it's release point be moving 14.7 FPS. 
Now here it the tough part. How long will it take for your cylinder to travel the 2 inches? 
Lets say it will be very quick. .1 second. If that is the case then the cylinder piston will be travelling at 20 IPS. But you need 14.7 FPS. So your arm ratio will need to be around. 7 to 1. I really think .1 sec for the duration of the cylinder travel is pretty optimistic. My guess is it will likely be closer to .2 seconds at best. Thus your arm ratio will need to double to 14 to 1. Keep in mind that some of the time needed to accelerate the ball will be eaten by some static compression within the cylinder to overcome the inertia. You are dealing with air here not fluid. If you were using hydraulics there would be no delay.

Bottom line. It all comes down to how fast the cylinder will travel the 2 inches and if the air pressure you are providing is sufficient to overcome the inertia of the ball and mechanism quickly. If your cylinder throw was longer, you would gain travel that would allow for more time to allow for acceleration to the required pace.

Can you provide a link to the software you are using to model this? There are many projects that could benefit from such software.

It may be useful if you check out some sites that talk about pneumatics. Like this one.

http://www.womackmachine.com/engine...h-influence-the-speed-of-an-air-cylinder.aspx

How fast a cylinder fills has many factors to consider. Inlet pressure of course, but valve orifice size is equally critical. Not to mention exhaust size and also exhaust side pressure. The piston as it moves compresses the air on the other side of the piston which provides resistance. Knowing what the airflow through your valve will help. Use that to figure based on the PSI you are basing your calculations on and then reduce that be about 20% to cover exhaust loss. That should give you a flow rate that will provide you with the expected time of actuation.

Tom


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## Razaray

http://www.algodoo.com

It's free too! Only down side is it doesn't have many pre made components such actuators.


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## Razaray

Ya that does help me. But as I should be able to easily calculate the exit velocity and angle needed for the pitch, there are still factors like you mentioned that can't be estimated cause they are significant


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## Razaray

I will get concrete numbers soon. 

Thanks again


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## spinman1949

Razaray,

That is cool software. Looks like it includes force calculations as well. Going to go through the tutorial. 

Tom


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## spinman1949

Razaray,

How are you doing with that software? Have you figured out how to set the force and mass figures?

Tom


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## Razaray

Yes. The only thing I'm not sure on what values to use are the friction values. I've just been setting them relatively low for cylinder the cylinder actuation. I only put together a very simple cylinder. If you yourself made one I'd be interested in seeing it


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## Razaray

To overcome the static compression loss and to achieve close to maximum cylinder extention speed I'm gonna try to add a type of release that holds the arm for a split second to allow for pressure buildup and hopefully more torque and faster extention. I'll only do this if my initial tests fall short.


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## spinman1949

Razaray,

Go here. You can install this software and design the pneumatic portion of your design.

http://getintopc.com/softwares/electronics/festo-fluidsim-pneumatic-and-hydraulic-free-download/

Tom


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## spinman1949

Razaray,

It will not help to hold the arm. Everything has to do with PSI and flow rate through the system. The key is going to be flow rate for you. A force of 120 N should be adequate to throw the ball at a 10 to 1 ratio. So far velocity of the actuator with the flow I am using is about .62 m/s. This is based on 30 gpm flow rate. So with a 10 to 1 ratio on your arm, your ball velocity will be around 12 MPH. If you can provide the specs on the valve Ican give you more accurate numbers.


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## spinman1949

I took some flow control out of my test system. m/s is up to .79. So that brings up your ball velocity to 17 MPH based on 10 to 1 arm ratio. So arm needs to be about 22 inches long. So 120 N or 62 PSI with your .75 cylinder should get what you want. The only factor I am guessing on is the flow rate. I am guessing at 30 GPM. The velocity figure is hard to catch as it only displays during the actual movement of the piston. I will increase the length of the cylinder to give more time to see the max velocity.


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## spinman1949

Do you have an idea as to the total mass you will be moving? Ball. Ball holder. Arm. Friction is nominal to consider. You are not pushing a block of steel on a surface. The key is total mass and linear momentum. I used 5 lbs as a baseline.


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## spinman1949

Just so we are clear. The arm ratio will increase the mass being seen by the actuator. So the ball that weighs .4 lbs now weighs 4 lbs when you consider the 10 to 1 ratio I suspect you will need to achieve the desired velocity for the ball. Add to that the weight of the arm and ball holder. The ball holder weight will increase the mass 10 to 1. The arm mass increase will be variable. Average that at 5 to 1. So obviously you need to use lightweight materials if possible.


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## Razaray

Want to say thanks for all the input. Having 3 young kids doesn't give me much time to work on this but I have a full speced design at least. I'll post a link to that. Hopefully I will get it put together for trials before the end of the season. I'll keep you posted.


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