- #1

- 1

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Additional Info:

Mass of Projectile: ~51g

Thanks in advance!

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- Thread starter RMorrow
- Start date

- #1

- 1

- 0

Additional Info:

Mass of Projectile: ~51g

Thanks in advance!

- #2

- 778

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.51g = .00051 kg

F=MA so F=(.00051 kg)(120 m/s^2) --> F= (.0612 N)

Now Force and Pressure are two different things, so you won't be able to convert directly between the two, but Pressure=[itex]\frac{Force}{Area}[/itex] so if you know the cannon's specs you can figure out the pressure required.

Note 1: This is the minimum pressure and it is assuming that the projectile will be accelerated over a period of 1 second, considering it will most likely be far less time than that, you will have to adjust the pressure accordingly.

Note 2: If I did my math right, the pressure you would get out of these equations will be in Pascals, not PSI, but any computer converter can translate between the two.

- #3

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Using the two equations for position and velocity as well a known parameters

x = .5a*(t)^2

v = a*t

v = 120m/s

m = .051kg

x = (distance to be accelerated over)m

We can determine the acceleration required

(t = v/a) -> (x = .5a*(t)^2) = (x = .5a*(v/a)^2)

x = .5*(v^2)/a

a = .5*(v^2)/x

Since F = m*a, then a = F/m

Putting that in we get, F/m = .5*(v^2)/x

So, F = .5*m*(v^2)/x

Force is in Newtons so then convert to pounds; After that you'll want to divide that by the area (in, in^2) that the pressure is being applied over to determine the needed PSI.

That's the best I can do. You'll will need to tweak the final pressure since I'm assuming that this an ideal situation in my head.

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