Formula question and gun design

[thedeathofall]

So I have another question for the math geeks out there.

If I were to make a hand pump. How could I tell how much force it would take to pump it at a certain PSI.

Say the cylinder is 0.74 inches ID and is 6.5 inches long. How much force would it take to bring it and the chamber to 200PSI?

[jeepkahn]

that's actually multiple questions.... to figure out how much effort for a certain bore pump, just multiply 200psi by the square inches of the bore... but you are also asking how many strokes to reach 200psi, because the effort required will remain the same no matter how big your chamber is...

[thedeathofall]

ahah.... I see. So how do you figure that out then?

It will take about 85 lbs of force to push 200 PSI of air. But how many strokes? do I just divide the chamber volume by the pump volume?

[daberno123]

Ok I think I know how to do this but someone correct me if I'm wrong.

You didn't mention a specific chamber volume so lets say you have a 2 cubic inch chamber to be filled to 200 psi. Pump volume is approximately .43 cubic inches.

Ideal Gas Law
P1*V1=P2*V2

Rearranged its P1*V1/V2=P2

Plugging in your variables thats 200*2/.43=P2

Using that I get 930 pumps. You definitely need to lengthen your pump stroke.

Again, I'm not sure if this is right or not. I'm sure if it isn't someone more knowledgeable will correct me.

[Technician1002]

A quick note on the force needed to push the pump. The force on the piston is figured from the pressure pushing on the area of the piston.

Say the cylinder is 0.74 inches ID
chamber to 200PSI

These are the numbers you need to balance the force of the pump stroke to the pressure of the air pushing back (balanced) In reality a slightly higher pressure is needed to MOVE the plunger down with the friction. Anyway the balance force is;

Pi X radius squared. Radious is 1/2 the diameter.

D of 0.74 provides R = 0.37
Radius squared is radius times radius. 0.37 X 0.37 = 0.1369
Pi times Radius squared is 3.14159265 X 0.1369 = 0.430084034 square inches.

Force is pressure times area. Pressure is 200. Area is 0.430084034

200 times 0.430084034 = 86.016806757 Lbs.
This is the balanced pressure. Higher pressure is required to move it.

[far_cry]

i prefer the math like this picture that i did for some one

[thedeathofall]

Using that I get 930 pumps. You definitely need to lengthen your pump stroke.

I think something is wrong with your math (no offence). That is way to many strokes to be accurate.

My chamber volume will be around 11.5 cubic inches. The reason why it is so short, is that it will be a built in hand pump for a portable gun. I cant reach out 20 inches.

@ Tech,

So over 86 lbs of force will be required to move it, but how many strokes? how do can that equation be set up?

This will be pumped with one arm. Will this be to hard to use as a built in pump? Should I go with a smaller dimeter bore?


Thanks for the input guys!

[suburban spudgunner]

Ok I think I know how to do this but someone correct me if I'm wrong.

You didn't mention a specific chamber volume so lets say you have a 2 cubic inch chamber to be filled to 200 psi. Pump volume is approximately .43 cubic inches.

Ideal Gas Law
P1*V1=P2*V2

Rearranged its P1*V1/V2=P2

Plugging in your variables thats 200*2/.43=P2

Using that I get 930 pumps. You definitely need to lengthen your pump stroke.

Again, I'm not sure if this is right or not. I'm sure if it isn't someone more knowledgeable will correct me.

Did you take into account the amount of pressure generated by one pump? You seem to have solved for a quantity of pressure and not pumps (I'm not sure about this, of course). Sorry if I'm mistaken.

[far_cry]

if you go to small diameter pump you will get a higher pressures but,
you will need more strokes to get to the wanted pressure

so to make it easy for you .what is the amount of pressure you run after
400 - 500 - 700 psi etc...........
and if you want just 200 psi 2.5 cm diameter will be just fine according to your wight
this will allow us to tell you what the best diameter to get the wanted pressure in less effort


and one more thing
the force you will apply on the pump not only your wight
i had build hand pump ,according to my wight i can get 500 psi but i get 650 psi

[Technician1002]

I think something is wrong with your math (no offense). That is way to many strokes to be accurate.

My chamber volume will be around 11.5 cubic inches. The reason why it is so short, is that it will be a built in hand pump for a portable gun. I cant reach out 20 inches.

@ Tech,

So over 86 lbs of force will be required to move it, but how many strokes? how do can that equation be set up?

This will be pumped with one arm. Will this be to hard to use as a built in pump? Should I go with a smaller diameter bore?


Thanks for the input guys!

Veterans, bear with me. I'm trying to keep it simple and in the ballpark. I'm not adding compression heating and expansion to the mix. That is an advanced subject you may wish to learn later.

You need to know a few things at first. 1 is the volume of your pump. You gave a pump cylinder length, but didn't give a stroke length. For this example I'm making the assumption the stroke is the 6.5 inch length given and the cylinder is longer to contain the length of the piston.

Volume of a cylinder is the base area times length.
0.430084034 square inches times 6.5 = 2.795546221 cubic inches.

When compressing gas, it does not care what the pressure on the gauge is. The ABSOLUTE pressure in relation to a vacuum is needed to find the amount of air moved. So we start with 1 atmosphere in the pump which is pushed out on the stroke. Lets call this volume of air 2.795546221 cubic inches at standard temperature and pressure (room temp at atmospheric pressure) Each stroke will deliver this much air in a "Perfect" pump, but alas, there is dead space in the pump at the end of the stroke to the check valve. We will cover that later.

Now on to your chamber. It has a volume and at start it has some air in it at atmospheric pressure. Don't forget it is there even though your gauge reads zero.

To continue the volume of the chamber needs to be known. In the first post this is not given. For sake of argument assume it is 10 times the volume of the pump. Remember the first amount already in the chamber. Add that pump stroke to it and the chamber which had 10 pump volumes of air now has 11. In 10 strokes it would be at 20 and so forth. So what is the pressure now? Simple take the atmospheric pressure and double it. Using a rounded number of 15 PSI for atmospheric (absolute) that's 30 PSI (absolute). Since the gauge is measuring the relation to the pressure of atmosphere, the gauge will see a difference of 15 PSI.

Working backwards, you want 200 PSI. That's 200 PSI over the 15 PSI atmospheric or a final pressure of 215 PSI absolute.
Pressure is directly related to the volume of the chamber and amount of air pumped in. 10 strokes got us to 15 PSI guage. To make it easier, this pressure is 1 bar. Which is the pressure of the atmosphere doubled or 15 psi. Every 10 strokes gives us 1 bar. We need a total of 13-1/3 bar (200 PSI) so 133 strokes is needed with a perfect pump.

Perfect pump?? yes. Looking at the roughly 13 bar, this is a problem. When we take a stroke and push the plunger all the way down, not all the air goes out. Some remains. How much.. Is it a problem?

Let's assume the space between the piston and check valve and around the piston to the seals is 1/20th the volume of the pump (very believable).

As we pump near the end of pumping up the chamber and near 13 bar, that space has 13X it's volume in that space that doesn't get out of the pump. On the return stroke it returns back to it's original volume as it goes back to 1 atm. That 1/20th of the volume is now 13 times bigger for 13/20ths of the pump volume.

It means at higher pressure you deliver less than 1/2 the volume each stroke so your perfect pump 133 strokes become closer to 300 strokes real world.

A long skinny pump can have a much smaller space because it is skinny and the longer stroke provides a higher ratio of cylinder volume to dead space. This is why shock pumps are long and skinny.

I hope this helps. It is a little long and didn't cover the effects of heating, the force needed to open the check valve, the volume of the hose or pipe to the chamber, moisture condensation, ets. This is a start. :blah5: Why is the math so much work?

As for pumping it one handed, hit the gym and do about 200 reps doing 1 arm curls pulling towards your chest.

The original design is for Popeye or someone working on his biceps to get arms like Popeye. I used to lift 80 Lb bales of hay. I have a point of reference.

Smaller bore and longer stroke is a really good idea.

Feel free to try the math when you have your pump design and chamber volume figured.

[far_cry]

I think something is wrong with your math (no offense). That is way to many strokes to be accurate.

My chamber volume will be around 11.5 cubic inches. The reason why it is so short, is that it will be a built in hand pump for a portable gun. I cant reach out 20 inches.

@ Tech,

So over 86 lbs of force will be required to move it, but how many strokes? how do can that equation be set up?

This will be pumped with one arm. Will this be to hard to use as a built in pump? Should I go with a smaller diameter bore?


Thanks for the input guys!

Veterans, bear with me. I'm trying to keep it simple and in the ballpark. I'm not adding compression heating and expansion to the mix. That is an advanced subject you may wish to learn later.

You need to know a few things at first. 1 is the volume of your pump. You gave a pump cylinder length, but didn't give a stroke length. For this example I'm making the assumption the stroke is the 6.5 inch length given and the cylinder is longer to contain the length of the piston.

Volume of a cylinder is the base area times length.
0.430084034 square inches times 6.5 = 2.795546221 cubic inches.

When compressing gas, it does not care what the pressure on the gauge is. The ABSOLUTE pressure in relation to a vacuum is needed to find the amount of air moved. So we start with 1 atmosphere in the pump which is pushed out on the stroke. Lets call this volume of air 2.795546221 cubic inches at standard temperature and pressure (room temp at atmospheric pressure) Each stroke will deliver this much air in a "Perfect" pump, but alas, there is dead space in the pump at the end of the stroke to the check valve. We will cover that later.

Now on to your chamber. It has a volume and at start it has some air in it at atmospheric pressure. Don't forget it is there even though your gauge reads zero.

To continue the volume of the chamber needs to be known. In the first post this is not given. For sake of argument assume it is 10 times the volume of the pump. Remember the first amount already in the chamber. Add that pump stroke to it and the chamber which had 10 pump volumes of air now has 11. In 10 strokes it would be at 20 and so forth. So what is the pressure now? Simple take the atmospheric pressure and double it. Using a rounded number of 15 PSI for atmospheric (absolute) that's 30 PSI (absolute). Since the gauge is measuring the relation to the pressure of atmosphere, the gauge will see a difference of 15 PSI.

Working backwards, you want 200 PSI. That's 200 PSI over the 15 PSI atmospheric or a final pressure of 215 PSI absolute.
Pressure is directly related to the volume of the chamber and amount of air pumped in. 10 strokes got us to 15 PSI guage. To make it easier, this pressure is 1 bar. Which is the pressure of the atmosphere doubled or 15 psi. Every 10 strokes gives us 1 bar. We need a total of 13-1/3 bar (200 PSI) so 133 strokes is needed with a perfect pump.

Perfect pump?? yes. Looking at the roughly 13 bar, this is a problem. When we take a stroke and push the plunger all the way down, not all the air goes out. Some remains. How much.. Is it a problem?

Let's assume the space between the piston and check valve and around the piston to the seals is 1/20th the volume of the pump (very believable).

As we pump near the end of pumping up the chamber and near 13 bar, that space has 13X it's volume in that space that doesn't get out of the pump. On the return stroke it returns back to it's original volume as it goes back to 1 atm. That 1/20th of the volume is now 13 times bigger for 13/20ths of the pump volume.

It means at higher pressure you deliver less than 1/2 the volume each stroke so your perfect pump 133 strokes become closer to 300 strokes real world.

A long skinny pump can have a much smaller space because it is skinny and the longer stroke provides a higher ratio of cylinder volume to dead space. This is why shock pumps are long and skinny.

I hope this helps. It is a little long and didn't cover the effects of heating, the force needed to open the check valve, the volume of the hose or pipe to the chamber, moisture condensation, ets. This is a start. :blah5: Why is the math so much work?

As for pumping it one handed, hit the gym and do about 200 reps doing 1 arm curls pulling towards your chest.

The original design is for Popeye or someone working on his biceps to get arms like Popeye. I used to lift 80 Lb bales of hay. I have a point of reference.

Smaller bore and longer stroke is a really good idea.

Feel free to try the math when you have your pump design and chamber volume figured.

i think this guy who want help he will not need it any more

your posts allwas long and a lot to read (but true stuff)

[thedeathofall]

Hmm I see I may have a problem. 300 pumps it about 275 too many for me

Well after looking at my design, i realised I had another problems; it was WAY too heavy. I was planning on using steel, so I could reach higher pressures, but I may not need to.

On GGDT (if I programed the QEV right) I will be getting way more velocity and range than i can safely want. So i decided to go with a lower pressure, which enables me to use something like PVC for the larger parts. (I will still probably use steel for the fittings, to help prevent it from cracking if it is dropped.)

I guess by now I should probably let you guys in on my design.

Its a hand held grenade launcher. It will fire 5-6 oz water balloons and paint grenades. I wanted it to have a built in pump so I wont have to carry Co2 or HPA. I also wanted it to be easy for a littler kid (or idiotic family member) to use.

Here is my paint design



The barrel will be SDR 21, either 2 or 2.5 inches (not sure yet). The chamber will be 3/4 fittings necked up to 1 inch for the long part. This is to give it more chamber volume.

The valve will be a QEV, and the exhaust I am still working on...

I want to lob (yes lob, NOT LAUNCH) grenades about 200-450 feet. Absolutely Not Any More Than That!

A 5 oz water balloon traveling at 100 FPS can leave a serious bruise and I don't want to hurt anyone.



Anyway, I may need some input from you guys about the design....

Oh and BTW, thanks a bunch Tech! You can explain algebra way better than my teacher
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[Technician1002]

All my cannons up to this year were hand pumped. No compressor. Hand pumps work in the field. A 50 cu in tank is not too bad to pump to 100 PSI. The 700 cu in tank is a workout to get above 60.

My wife liked the hand pump. Kept me in shape and I tended to limit the shots to something sensible.

It's frustrating when reaching higher pressure. The work is getting much harder as you get tired and the gauge moves less and less on each stroke.

You take a shot when it feels right, not when it's at insane pressure.

At a men's retreat, I was asked how high they could take the cannon. I smiled and said how ever far you wish. He aimed for 100 PSI but quit at 45.. Somehow I expected that. My high volume hand pump has a bore of 1.65 inches.

Bye the way, Nice drawing. Well done. It would look fantastic in polished copper.

A note on lobbing grenades.. I've used a tennis ball barrel 36 inches long on one of my 1 inch valve air cannons and it has a max reach of about 325 feet at 100 PSI if that helps. The 3/4 inch QEV should keep water balloons well within your maximum distance limit.

[daberno123]

I think something is wrong with your math (no offence).

None taken. I've never even taken a basic chemistry class and all that stuff was what just seemed logical to me. Even I thought that seemed like way too many pumps but after typing all that out I wasn't just going to not post it.

Reading through Tech's explanation I see that I was wrong. Listen to him not me.

Thanks Tech I learned something new too.

[thedeathofall]

Okay well thanks for all the help. Especially you Tech.

I should increase bore size, but by how much? I could extend the pump out to 8 inches (max) which would be about a 7.5 inch stroke. I would really like to try to keep the gun as one piece, which is why the built in pump. Trying to carry a stirrup pump on top of everything else may be a problem. (then again, It may be the only solution)

Bye the way, Nice drawing. Well done.

Thanks! As wierd as it sounds, I find making gun designs on microsoft paint, is very relaxing. I can use the system really well, its not complicated, and while it can take a while, it usually turns out okay.

@Daberno

Me neither, which is why I ask

I was kind of hoping Tech would answer because he seems to know how explain things. Maybe he should change his name to Teach