Re: Air canon, sanity check on my calculations

(thomas.greenleaf@xxxxxxxxx) writes:
I made a few quick calculations consering an air canon with the
folowing specks.

"cannon"; "specs" ( = specifications)

Barrel length = 2m long with inside diameter = 1cm
Air pressure = 8kg/cm^2
^^^^^^^
units??

Ammunition is a steel bal with diameter = 1cm and mass=4.08g and a
frontal area (air resistance) of 0.000079m^2

I assume no friction between ball and barrel and I assume no drop in
air pressure as the ball travels down the barrel. I allso assume an
instantanious valve which puts ful air pressure on the ball right from
the start. I also assume that the barrel is not filed with air in
front of the ball which resists its movement. That may be the problem?

I suspect so--air in the barrel would reduce your muzzle velocity by
6 percent at the very least.


I did a simulation which told me that if this contraption was to fire
the bal at an inclination of 35 degrees, it would travel over 1200
meters. The mussel velocity would be arround 233m/s = 838 km/t
That sounds fun, but also a little too optimistic.

I think the reason it sounds high is that it's hard to produce a supply of
gas that would generate that sort of pressure instantaneously and keep
it constant while the ball is in the barrel--and still be capable of firing
a large number of shots and remaining portable. (You'd probably need a
teflon-lined barrel too.) In other words, an air cannon like yours might
work as a demonstration but wouldn't be useful otherwise--and a practical,
portable air cannon would have a lower performance.

I have made quite a few assumptions along the way. Some maybe more
serious than I suspect, so I would like to hear where I have perhaps
miscalculated badly...

Given your assumptions the results look reasonable.

------

What i did was the following.

The ball has a front (and back) area of 0.79cm which gives it push
from the air of 0.79cm*8kg/cm^2=6,32kg=62N

(Using kg as a unit of force isn't standard or recommended...)

This means an acceleration of 62N/4.08g = 15196m/s^2
That means that it takes the ball sqrt(2*2m / 15196m/s^2)=0.01622s to
move through the barrel at a constant acceleration.
That again gives the ball a velocity of 0.01622s * 15196m/s^2 = 247m/s

From that point on I solved the ODE with matlab which again used RK4.
Using the above values the range is 1284m after a flight taking
arround 18s.

The function y=f(t,x) used in the RK4 step looks like the folowing.
Note that the statevector is [X Y VX VY]^T

----

%state vector is [x y vx vy]'
%therefor the derivative is [vx vy ax ay]'
function derivative = ballThroughAir(t,stateVector)

derivative = zeros(4,1);
cd = 0.1; %drag
m = 4.08 / 1000; %mass in kg
p = 1.225; %airDensity in kg/m^3
A = 0.000079; %area in m^3
g = -9.82; %gravitational aceleration in m/s^2

xPos = 1;
yPos = 2;
xVel = 3;
yVel = 4;

velocity = norm(stateVector(xVel:yVel),2);
frictionForce = cd * 0.5 * p * A * velocity^2;

motionVector = stateVector(xVel:yVel);
motionVector = motionVector ./ norm(motionVector,2); %normalized
motion

frictionForce = motionVector.*frictionForce; % friction distrubuted on
x and y now

derivative(xPos) = stateVector(xVel);
derivative(yPos) = stateVector(yVel);
derivative(xVel) = -frictionForce(1)/m;
derivative(yVel) = g - frictionForce(2)/m;

------

A bonus question is what would hapen if I would make the ball spin
quicky arround the horisontal axis so that it would generate lift due
to the rotation and velocity. I had a vagua idea about placing strong
magnets along the top of the barrel to make the steel ball tend to
roll along the top and give it rotation. Silly idea?

Well, something similar apparently works for golf balls...

--John Park
.

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