Calculating Speed?

If you have an object with a mass of 2000 grams, and it has an average force of 635.029 kg pushing it for a distance of 6.096 meters before it becomes a free-flying projectile.  How fast is the object traveling after the 6 meters and how do you calculate it.

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frollard4 years ago
To start I don't want to do your homework for you, but please take the time to actually read this and perhaps learn a tidbit from it.

force = mass times acceleration
You have a problem here...a force should be in newtons, kilograms is mass, and doesn't inherently have a force.  a kilogram of mass exerts a force due to gravity of 9.81newtons, because of the acceleration due to gravity; not because kilogram is a force.

So you could mean a force of 635.029 newtons...
or you could mean the force 635.029 kilograms exerts due to gravity, which would be 635.029kg*9.81m/s2 = 6229.63449 kg-m/s2, the units for newtons.
I'll assume you mean 635.029 newtons for the remainder.

Vf2= Vi2 +2ad
Final velocity squared = initial velocity squared + 2 (acceleration) * (distance)

unknown = acceleration, but since we have a force and a mass, we can say
F = ma
a = F/m
a = 635.029kgm/s2 /2kg
a = 317.5145 m/s2

Assuming initial speed of zero, and rearranging the formula, we have
Vf = SquareRoot(0 + 2 * (a d))
Vf = SquareRoot(2 * (317.5145m/s2 * 6.096m)
Vf = SquareRoot(3871.136784m2/s2)
Vf = 62.218 m/s
jj.inc (author)  frollard4 years ago
Just so you aren't concerned, this is not for homework, it is an air cannon I am theorizing. I have been trying to find these equations for a very long time and I have not known what to look for so thanks for your help. I do have one problem however. I came up with 635kg from p.s.i. Through some more simple mathematics I have found that at the very end of the barrel the remaining pressure will equal roughly 72 psi which I then converted to SI kg across the whole surface of the projectile. So I guess what I need to find is the newton force of expanding air at 72 psi.
What is the inside diameter of the barrel? In cm.
he replied to me...aparently 5 inches diameter
at 72psi
1413 pounds force...
or 6288 newtons.
ten times the force I had in my initial equation...

mother of god.
At sizes and pressures like that the friction and air resistance start to become a REALLY big deal
Without losses that's 282.6 m/s or 632 mph. The speed of sound is ≈350 m/s. I very much doubt he manages to fire an air cannon at Mach 0.8.
makes perfect sense to have mach 0.8 with those forces and NO resistance, but even the backpressure of the static air in the barrel would be a huge pressure to overcome (it couldn't get out of the way fast enough)

Also assumes the valve will open from zero to full bore instantly, while in reality most valve solutions (except burst disk) only open to about 25-45% by the time the projectile has left the barrel. This results in a MUCH lower average psi, especially when considering the huge volume of a 5" diameter barrel.
jj.inc (author)  bwrussell4 years ago
I did it all out and calculated huge losses due to friction air resistance ect. The speed will be 300-400mph
frollard jj.inc4 years ago
See, pounds is a force, not a mass. Pounds are oddly enough already a force like newtons.

for example

60 pounds in newtons works


60 kilograms in newtons doesn't.

So, if you have 110psi at start, and 72 psi at end of barrel, you can LOOSELY say that you have an average of 91psi

If your barrel is say...2 inches diameter, 1 inch radius, the rear surface area is pi r squared,

which is pi 1 squared

which is 3.14 square inches.

91psi at 3.14 square inches is 285ish pounds of force 'average', more at the start, less at the end.

1 pound force = 4.44822162 newtons

285 pounds force = 1 267.74316 newtons

http://www.spudguns.org/calculators.html  will do ALL the math for you :D

It came up with 272 pounds of force with the same info because I said the barrel took up some room in the chamber, about 10%, and the barrel pressure drop was 100 down to 73, average 86psi...the rest of the calculations are easy based on my first post
jj.inc (author)  frollard4 years ago
Thank, I guess the proper measurements are a 5" barrel 20 feet long, the average psi will be 72, and the rest is all the same
What material is the cannon being made of? What material and dimensions does the projectile have? Your barrel has a large value which is going to force you into a very large pressure vessel. Lets look at Boyle's Law:
P1·V1= P2·V2
If we take 1 to be the moment before firing and 2 to be the instant before the projectile breaks the seal on the end of the barrel we can calculate the necessary volume for the pressure vessel.
P1= 110 psi, P2= 72 psi,
V2= (pi/4)*(5in)2 +V1= 19in2*240in + V1= 4712 in3+V1
110psi·V1=72psi·(4712in3+V1) → 110psi·V1=339264 in·lbf + 72psi·V1
38psi·V1= 339264 in·lbf → V1=8928 in3
That is the size of pressure chamber you need, pressurized to 110 psi, in order to get 72 psi at the end of the barrel.

If you were to fire with these properties then drag and probably heat, along with other factors, would have a huge effect on the final velocity. In order to achieve anything close to the theoretical velocity ( 280 m/s) you are going to have to do some serious engineering.
jj.inc (author)  bwrussell4 years ago
My original calculations were if all the air was relocated to just the barrel I forgot to include the chamber in the original calc so the average became 70
Using the same kinematic equation as frollard and assuming an average constant pressure of 72 psi (496.422 kPa) and a perfect seal on the projectile, while neglecting air resistance and friction, the final velocity as a function of the inside diameter of the barrel in cm is:
VfID 6.244·d3/2 m/s
(This gives a 4.631 cm diameter when you input frollard's velocity. If that is close then you know the assumption that the 635 really was in newton's. If it's not close then it means you need to run the equation above with the correct diameter so that your units are correct.)

The general function in terms of pressure (P, in Pa, N/m2), inside diameter (d, in cm), and mass (m, in kg) is:

Vf.0125·sqrt((P·d3)/m) m/s
This is not that straight forward. You do not have a constant force on the projectile because as volume increases (the object is pushed down the barrel) pressure decreases. You could take an average pressure and then get an approximate average force. You're also mixing units with PSI and everything else. Make sure you made the correct conversions.
jj.inc (author)  frollard4 years ago
All though the final velocity is very practical and likely to be correct minus friction and air resistance.
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