Working out the power of this air gun?
11 May 2005, 11:24 AM
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Working out the power of this air gun?
Just a few plans of mine.. 
So, imaging having several litres of compressed air (145 bar ).. blatted out, *all at once*, behind a 100(ish) grain 9MM lead slug, accelerated w/use of a nice long barrel .. 
Any idea what power it could possibly acheave (ft/lbs) ?
This is purely theoretical btw .
So, imaging having several litres of compressed air (145 bar
).. blatted out, *all at once*, behind a 100(ish) grain 9MM lead slug, accelerated w/use of a nice long barrel .. Any idea what power it could possibly acheave (ft/lbs) ?
This is purely theoretical btw .
11 May 2005, 11:48 AM
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Originally Posted by OllyK
I suspect your bike valve will give up the ghost long before you get to 145 bar
We'll theoretically use a tractor innertube valve.
And we'll also replace the water pipe for high pressure gas pipe.
11 May 2005, 12:08 PM
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Originally Posted by Leslie
And your 18 volt solenoid valve may well burn out with the 27 volts from your stated 3 x 9 volt batteries in series.
Les
Les
stop being critical, actually i have no idea what the power of the valve would be as i dont have one, i said its theoretical.
and obviously it would only need 2 9V batteries, my bad.
11 May 2005, 12:12 PM
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The actual power of the air gun is measured by the energy carried by the projectile when leaving the barrel of the gun.
With the correct ammunition, it's more than possible to do some serious damage with a "legal" air rifle.
Wouldn't it be simpler to modify a "normal" air gun(s) and have them on an electric. If you go for one of the new gas powered auto-loaders you should be able to get 5 shots out of it before you have to go to your shed to re-load
With the correct ammunition, it's more than possible to do some serious damage with a "legal" air rifle.
Wouldn't it be simpler to modify a "normal" air gun(s) and have them on an electric. If you go for one of the new gas powered auto-loaders you should be able to get 5 shots out of it before you have to go to your shed to re-load
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11 May 2005, 12:53 PM
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Rough and ready theoretical calculation...
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
11 May 2005, 01:39 PM
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Originally Posted by ajm
Rough and ready theoretical calculation...
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
I knew
it was a beast Now I have to find where Transco are doing some repair work on their gas mains and nab some pipe
11 May 2005, 01:44 PM
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Originally Posted by 03-CTR
no offence chaos but i think its just as well your design is so full of holes that it will be difficult to get working
11 May 2005, 04:52 PM
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Originally Posted by David_Wallis
also Id reconsider how your wiring that switch!!
11 May 2005, 05:14 PM
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Originally Posted by ajm
Rough and ready theoretical calculation...
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
145bar = 14,500Kpa (14,500,000 N/m^2)
100 Grain = 0.0065 kg
Say Barrel Length = 1m
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Assuming negligable pressure drop in air pressure as projectile moves up the bore (i.e. large air reservoir) then force acting on projectile = 14,500,000 * 0.000254 = 3689.8N
(also assuming no friction/air resistance)
Force = mass * acceleration
so acceleration = 3689.8 / 0.0065 = 567662 m/s/s
distance = ut + 1/2 at^2
so 1 = 1/2 * 567662 * t^2
so t (time to reach muzzel) = 0.00188 secs
Muzzle velocity = acceleration * time = 1066 m/s
Muzzle energy = 1/2mv^2 = 0.5 * 0.0065 * (1066)^2 = 4432 Joules = 3269 ftlbs!!!
A .308 winchester is about 3000 ftlbs at the muzzle!!!
In practice friction/air resistance up the bore will come into play, plus your air supply will drop in pressure as the projectile moves up the bore. Even so, you will still have a very dangerous weapon on your hands, and one that would require a section one firearms certificate!
Id also suggest that the little bollocks puts his mothers pressure cooker on his head when he tries it out!
Simon
11 May 2005, 06:56 PM
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Originally Posted by GC8
Id also suggest that the little bollocks puts his mothers pressure cooker on his head when he tries it out!
Simon
Simon
A REFINERY
11 May 2005, 08:00 PM
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Originally Posted by ajm
Love the .22 pellet dents in the side of it! 
if I ever get the bits to build this thing Ill post a piccie of said pan..WITH A CLEAN HOLE RIGHT THROUGH IT!!
11 May 2005, 09:21 PM
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Originally Posted by ajm
Cross Sectional Area of bore = Pi * 0.009^2 = 0.000254 m^2
Which puts things out by a factor of 4, and then comes down to a factor of 2 from the squareroot in the speed calculation.
Still a lot though
John.
11 May 2005, 09:40 PM
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Originally Posted by john_s
You've done pi * diameter^2, should be pi * radius^2
Which puts things out by a factor of 4, and then comes down to a factor of 2 from the squareroot in the speed calculation.
Still a lot though
John.
Which puts things out by a factor of 4, and then comes down to a factor of 2 from the squareroot in the speed calculation.
Still a lot though
John.
12 May 2005, 12:56 PM
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If you put up a diagram and state the same thing in your text, then you don't have the right to have a "hissy fit" when someone points out a glaring mistake. Especially when it was endangering your solenoid by using too high a voltage.
Time you grew up and learned how to accept a bit of constructive criticism which was meant to be of help to you!
Nitro tastes a bit like Tequila-without the worm.
Les
Time you grew up and learned how to accept a bit of constructive criticism which was meant to be of help to you!
Nitro tastes a bit like Tequila-without the worm.
Les
Last edited by Leslie; 12 May 2005 at 12:58 PM.
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