# Finding speed of projectile propelled by pressurised air

• Pharrahnox
In summary, a 4 gram projectile was launched from an air tank at a pressure of 120 psi through a barrel with a length of 80 centimetres and a diameter of 8 millimetres. Using Pascal's Law and the equation F=ma, it was determined that the projectile would experience an acceleration of 10342.5 m/s. To find the muzzle velocity, kinematic equations were used to find the time of acceleration and then solve for the final velocity. The resulting estimate for the muzzle velocity was 126.5 m/s.
Pharrahnox
I have a projectile of 4 grams, an air tank of 120 psi, a barrel length of 80 centimetres and a barrel diametre of 8 millimetres. I would like to find the ideal speed (muzzle velocity) that the projectile would be launched at, not taking into account the pressure drop, friction atmospheric pressure and stuff like that.

What I did was start by finding out how much force the air from the tank would exert on the projectile. So I used Pascal's Law: F (Newtons) = Pressure (pascals) * Area (metres^2).
120 psi is roughly 827371 pascals
The area (m^2) - 0.004^2*pi = approx. 5*10^-5 (1/20000 or 0.00005)m^2
So F = 827371*0.00005
F = 41.37N

Then I put that into F = ma to find the acceleration:
a = F/m
a = 41.37N / 0.004kg
a = 10342.5 m/s

So I have the acceleration, but how can I find how lond it accelerates for? It travels (under pressure) down the barrel for 0.8 metres, is there a calculation I can use to find the speed?

Thanks for any help.

Pharrahnox said:
I have a projectile of 4 grams, an air tank of 120 psi, a barrel length of 80 centimetres and a barrel diametre of 8 millimetres. I would like to find the ideal speed (muzzle velocity) that the projectile would be launched at, not taking into account the pressure drop, friction atmospheric pressure and stuff like that.
Fire it through a photogate.

What I did was start by finding out how much force the air from the tank would exert on the projectile. So I used Pascal's Law: F (Newtons) = Pressure (pascals) * Area (metres^2).
120 psi is roughly 827371 pascals
The area (m^2) - 0.004^2*pi = approx. 5*10^-5 (1/20000 or 0.00005)m^2
So F = 827371*0.00005
F = 41.37N

Then I put that into F = ma to find the acceleration:
a = F/m
a = 41.37N / 0.004kg
a = 10342.5 m/s

So I have the acceleration, but how can I find how lond it accelerates for? It travels (under pressure) down the barrel for 0.8 metres, is there a calculation I can use to find the speed?

Thanks for any help.
If the volume of the barrel is small compared with the volume of your reservoir (so the pressure does not drop much with the increased volume) then you can just use kinematic equations and the length of the barrel as an approximate method.

If the projectile is light - you could figure that the projectile quickly reaches the same speed as the escaping air and work that out. If you have a pressure gauge on the reservoir you can measure the pressure before and after firing once and work out the change in energy that suggests... guessing that most of that energy goes into the projectile. Be aware that anything you do will be approximate though.

You will be much better working empirically - since you've built the thing!

The air released in each shot is small compared to the tank's volume (24 litres). I was unfamiliar with the kinematic equations, so I had a look at them. None of them solve for time. Can one of them be rearranged to solve for time or speed with the known variables?

Pharrahnox said:
The air released in each shot is small compared to the tank's volume (24 litres). I was unfamiliar with the kinematic equations, so I had a look at them. None of them solve for time. Can one of them be rearranged to solve for time or speed with the known variables?
That's right.

Acceleration is in metres per second squared. So if you re-arrange...

Seconds=sqrt(metres/acceleration)

@JustinRyan that would give me a value 89.44, which seems like a reasonable estimate.

Thankyou both for your help. I will refer to that method in the future, I had never even thought of that relationship before.

##s=ut+\frac{1}{2}at^2 \Rightarrow t=\sqrt{2s/a}##
(s=barrel length, t=time in the barrel, a=acceleration, u=initial velocity)

But you don't need the time of the acceleration to get the muzzle velocity.
Since the initial velocity is 0, you have the two equations:

(1) ##s=\frac{1}{2}vt##
(2) ##v=at##

you know s, and a ... you want to know v.
solve for t in (2) and substitute into (1)... then make v the subject.

Sorry if this sounds like a stupid question, but what do v and s represent? If it is velocity and speed, then in equation (2), I still only know 1 of 3 variables.

Pharrahnox said:
Sorry if this sounds like a stupid question, but what do v and s represent? If it is velocity and speed, then in equation (2), I still only know 1 of 3 variables.
me said:
(s=barrel length, t=time in the barrel, a=acceleration, u=initial velocity) ...you know s, and a ... you want to know v.
... what is it you wanted to know? muzzle velocity right? That's v.

You told me earlier that you had looked up kinematic equations.
The standard set are called "suvat" equations - the standard notation is:

s: displacement
u: initial velocity
v: final velocity
a: acceleration
t: time

BTW: you know the relationship between area and diameter for a circle is ##A=\frac{1}{4}\pi d^2## ?

Last edited:
Simon Bridge said:

Apologies.
Call it what it is...wrong. By a factor of two.

Had to look through the derivation of the suvat to where the 2 came from.

JustinRyan said:
Simon Bridge said:

Apologies.
Call it what it is...wrong. By a factor of two.

Had to look through the derivation of the suvat to where the 2 came from.
It was accurate dimensional analysis ;) Just remember that the displacement is the area under the v-t graph - and that, for u=0, that graph is a triangle: the 2 comes from the formula for the area of a triangle.

I just used pi*r^2 instead, which gives the same result.

Oh I see what you mean now. My brother helped me, I haven't done this sort of stuff yet in school...

So it would then be: s = 1/2*at^2
Which when rearranged to make t the subject gives: t = sqrt(2s/a), which I know see in the second part of the equation.

I then multiplied by the acceleration -approx. 10000- and that gave me 126.5 m/s.

Is this correct?

That would be it.

In general it is better to avoid lots of steps when you put the numbers in by doing the algebra earlier - like this:

Equations:
(1) v = at
(2) s = (1/2)at^2 (since you insist - but see eq1 post #7)

from (1) you get: t = v/a
put that into (2) to get: s = (1/2)a(v/a)^2 = (v^2)/2a
rearrange to give: v = sqrt(2as) <--<<< which is what you wanted.

The scientific bit would be to check this using an experiment.

That may be somewhat close, but that is a significant fraction (a bit under 40%) of the speed of sound in air, so the airflow down the tube is a factor that must be considered. My guess is that (assuming that your derivation and calculations are correct) your value should be used as an upper limit, when in reality, the exit velocity will probably be a bit slower.

Yeah, I was just looking for an ideal velocity, thankyou all for your help.

## 1. How do you measure the speed of a projectile propelled by pressurized air?

The speed of a projectile propelled by pressurized air can be measured using a high-speed camera and a distance measuring tool. The high-speed camera captures the motion of the projectile and the distance measuring tool calculates the displacement of the projectile over time. By dividing the displacement by the time, the speed of the projectile can be determined.

## 2. What factors can affect the speed of a projectile propelled by pressurized air?

The speed of a projectile propelled by pressurized air can be affected by several factors such as the air pressure, the size and shape of the projectile, the angle of launch, and the surface it travels on. Other factors that can affect the speed include the temperature and humidity of the air, as well as any external forces acting on the projectile.

## 3. Can the speed of a projectile propelled by pressurized air be calculated using formulas?

Yes, the speed of a projectile propelled by pressurized air can be calculated using the formula v = √(2P/ρ), where v is the velocity of the projectile, P is the pressure of the air, and ρ is the density of the air. This formula is known as the Bernoulli's equation and is derived from the laws of fluid dynamics.

## 4. What is the relationship between the speed of a projectile and the pressure of the air?

The speed of a projectile is directly proportional to the pressure of the air. This means that as the pressure of the air increases, the speed of the projectile also increases. However, it is important to note that this relationship is not linear and other factors such as the size and shape of the projectile can also influence the speed.

## 5. How can the speed of a projectile propelled by pressurized air be used in real-world applications?

The speed of a projectile propelled by pressurized air has several practical applications. It is commonly used in sports such as paintball and airsoft, where the speed of the projectile determines its accuracy and impact. It is also used in industries such as aerospace and defense, where the speed of projectiles can have critical implications on the performance of weapons and vehicles.

• Engineering and Comp Sci Homework Help
Replies
1
Views
387
• Engineering and Comp Sci Homework Help
Replies
14
Views
1K
• Mechanics
Replies
1
Views
1K
• Mechanics
Replies
5
Views
3K
• Mechanics
Replies
4
Views
2K
• Mechanics
Replies
11
Views
2K
• Aerospace Engineering
Replies
2
Views
2K
• Mechanics
Replies
1
Views
1K
• Mechanics
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
31
Views
4K