Pumpkin cannon

I need help with a gas expansion problem.

A pumpkin is put on a sabot ( approx. 10 kg total )

the sabot fits in a 250 mm diameter 3 metre long barrel.

The base of the sabot must be held a distance X from the end of the barrel, and the space under the sabot is then pressurised to 7 bar ( gage ) at 20 C

The sabot is released.... the pumpkin hopefully exits in a hurry.

I need to know how to calculate the exit velocity of the pumpkin to find the best value of X. To small an X and there is not enough air to propel the load, to large and there is not enough barrel left to accelerate the load.

P1 x V1 = P2 x V2 is not going to work, as energy is not conserved within the gas.

Is the state of the gas " charge " indeterminate for the problem as stated ? if so what additional information is required ?

Possibly significant factors that might complicate things are the gas mass, seal friction, and ambient temperature. If I have a simple model I can then see how important they are likely to be.

Once I have a prediction of the exit velocity I can work out how far I can fire the pumpkin... over 200 m would be nice :-)

Reply to
Jonathan Barnes
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"Jonathan Barnes" wrote in news:dn3pu4$i3h$ snipped-for-privacy@nwrdmz03.dmz.ncs.ea.ibs-infra.bt.com:

Oh good, dangerous experiments.

I think you can assume that the expansion will be adiabatic, so p*v^gamma is constant, gamma is 1.4.

However, the air in the cannon is unlikely to start at ambient temperature, so you'll have to figure that out, or, preferably, measure it.

Gas mass won't matter.

Or at least, I'd be very impressed if it does.

I have a ballistics program on my website, it will do a reasonable job of predicting the range for you, once you know the muzzle velocity. The Cd of a sphere is well known.

formatting link
then look on the downloads page I think.

Cheers

Greg

Reply to
Greg Locock

Yes but FUN... I am belive that air guns delivering more than 6 ft / lb need a licence in England...BUT, if it's a cannon rather than a gun it dous not need a licence...

Leave it to cool and top it up for best performance..

If you assume that the X value is 750mm, the mass of gas ( air ) will be about 0.37 kg for the charge the portions of which nearest the pumpkin will be traveling almost at the same speed,and about 0.15 kg of free air ahead of the pumpkin will need to be kicked out . I would say ignoring air mass could lead to an error of 5% which is irelevent first time round the loop, but could be compensated for in later models by adding to the payload mass.

Reply to
Jonathan Barnes

one thing you have to bear in mind as you work thru this is that you can't accelerate the pumpkin too much or you'll be firing pumpkin pie filling , not a pumpkin, out of the muzzle......

Reply to
Michael

L = 300mm

Pile up enough simplifying assumptions so that you can assert that the exit kinetic energy of the projectile plus the sabot will be equal to the integral of the pdV work of the air charge during expansion.

The pdV work will be proportional to the mass of the original charge, which is proportional to initial charge volume, which is proportional to (L - X).

The final volume is proportional to L.

Assume adiabatic expansion of the charge.

Integrate pdv from L - X to L and get expression for muzzle kinetic energy as a function of X. Bore diamemter and masses ought to drop out.

Find the maximum(a).

Post videos.

Fred Klingener

Reply to
Fred Klingener

Initial acceleration should be about 35 g... Maybe to much ? but the end of the trip won't do it much good either :-)

Reply to
Jonathan Barnes

Doh!

L = 3000mm

Reply to
Fred Klingener

That's why you fill the sabot with pumpkin pie filling.

Hth, Fred Klingener

Reply to
Fred Klingener

"Jonathan Barnes" wrote i news:dn3pu4$i3h$ snipped-for-privacy@nwrdmz03.dmz.ncs.ea.ibs-infra.bt.com:

Oh good, dangerous experiments

I think you can assume that the expansion will be adiabatic, so

p*v^gamma is constant, gamma is 1.4

However, the air in the cannon is unlikely to start at ambient

temperature, so you'll have to figure that out, or, preferably measure it

Gas mass won't matter

Or at least, I'd be very impressed if it does

I have a ballistics program on my website, it will do a reasonable jo

of predicting the range for you, once you know the muzzle velocity

The Cd of a sphere is well known

formatting link
then look on the downloads page I think

Cheer

Gre

Reply to
Greg Locock

I might have slipped a decimal but I get 345 g's

cheers Bob

Reply to
BobK207

I don't think so. 7 atmos pressure on a 0.25m diameter sabot is an initial pressure of 7 X 14.7 psi on a sabot area of pi X 10 X 10 /4 = 78.5 sq in is a force of 8000 lb(f) or 36000 newtons on 22 lb mass or 10 kg so the initial acceleration is about f/m = 36000/10 = 3600 m/s/s or 367 g

I don't hold out much hope for the pumpkin! To let 7 bar down to 1.2 bar gage at the muzzle (as an arbitrary selection), it looks like you allocate about 66 cm to the power reservoir.

Your numbers may vary......

Brian Whatcott.

Reply to
Brian Whatcott

.

O.K., metric units....

P1 = 800,000 Pa, ( absolute ) assume X = 600 mm, V1 = 0.0295 m^3 V2 = 0.1473 m^3

P2 = P1 x V1^1.4 / V2^1.4 = 84000. ( below atmospheric need to increase X :-) )

Work done ( energy to pumpkin :-) ) = P1 x V1 - P2 x V2

Energy to pumpkin = 11kJ

Mass of pumpkin 10 kg,

muzzel velocity = 47 m/s.

That's one fast pumpkin ( or pie filling equivelent )

Range ( 45 degree, no air resistance ) about 220 m.

As a first run through, the basic dimentions for a 200 m range cannon are looking good, but the conversion of the pumpkin to pie filling needs thinking about.

Thanks for your help so far.....

Reply to
Jonathan Barnes

Ouch.. a pesky decimal point must have jumped over a numeral while I was not watching :-)

Thanks for the correction.

Reply to
Jonathan Barnes

About hi-g launches....how about embedding the pumpkin completely in special pumpkin paste filling the sabot, loaded to an SG slightly denser than pumpkin (with sugar? salt? sand?). Then at launch, the pumpkin will ride up slightly on its bed of paste. The sabot will dump the paste all over at launch, and with luck a mostly complete pumpkin might sail away??

Brian Whatcott

Reply to
Brian Whatcott

hey man over there you want to know only the exit velocity? can i get the exact picture what do want to ask maybe i can try to help you and your problem.so if you have enough time then try to have a picture and send to me peter

Reply to
peter

I suggest BEANO

Reply to
Bob Wilson

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