Welcome To P8ntballer.com
The Home Of European Paintball
Sign Up & Join In

how a paintball marker works (math project)

phoenix pete

paintballing karteka
Jan 12, 2008
99
0
0
you can simplify s=ut+0.5a(t^2)
as assuming no are resistance a=0
therefore s=ut

t= time for ball to drop, can be found with same equation

s=ut+0.5a(t^2)
however this time there is no initial speed (u) so
s=0.5a(t^2)
re-arrange
(2s/a)^0.5=t

but you proberbly knew that right?
 

Skeet

Platinum Member
Sorry, I think everyone is missing the point, including our "desperate to relate any school work to paintball", initial poster.

I do not see how you can do a Maths project, based on "how" a DM7 or any other marker, works. It's a Physics thing. Yes, Maths is a part of Physics, but the reverse is not true. If I handed a physics project in to a Maths lesson, I would expect to get a stern telling off.

You can use maths to calculate trajectory, acceleration etc, but that isn't HOW the marker works.

The marker works, due to a number of mechanical and electro-mechanical events.
 

Skeet

Platinum Member
OMG:O
This thread is ver confusing too me:plol:cool:
Did we need to know that?

Your only contribution to the thread, (which is as you say confusing, but only because people are trying to be clever, without actually addressing the fact that you can't do what has been suggested) is to state that you do not follow it.

No request for an explanation, no offer of advice. Nope, all you have done, is increase your post count.

Harsh? Yes, probably:D
 

GrimAssasin

New Member
May 3, 2008
1
0
0
Sorry to up an old thread, but I need help with this as well. I have a project due in under a week, and need someone to explain to me some of the equations that have been listed. Basically, tell me what each variable represents, since i will have to explain it.

Specifically:
you can simplify s=ut+0.5a(t^2)
as assuming no are resistance a=0
therefore s=ut

t= time for ball to drop, can be found with same equation

s=ut+0.5a(t^2)
however this time there is no initial speed (u) so
s=0.5a(t^2)
re-arrange
(2s/a)^0.5=t
That IS the simple version.
u=100m/s
a=0
m=.002kg (approx)
acc. due to gravity=g= 9.8m/s/s
horizontal distance traveled=s=ut + 1/2a[t^2] => s=100t
just pick a height and apply the same formula, putting the chosen height as s, and using g for the acceleration which gives you the time (t) for it to hit the ground, then stick that into the horizontal formula: that's how far a paintball will travel when fired parallel to the ground. The forces get a little more complex (not much) when you angle it because then you have to split it into vectors and you have an initial speed (u) for the vertical movement with -g as the acceleration. You'd have to do a bit of calculus if you wanted to take air resistance into account because the acceleration would change with velocity (a=-kv)
and
another reasonably easy one to work out is how far will the ball travel before it hits the ground.
assume a hight (measure it) you are holding the gun and assume that you are holding it horizontal....you know the speed the ball is moving horizontally and you can use affect of gravity to work out how long the ball will take to drop to the groud.......then use this time to work out how far the ball will travel in this time. if you want to get a little more involved you can do the same thing but assume the barrel is angled up at 5*.
Thanks.
 

Devrij

Sex-terrorist
Dec 3, 2007
1,341
2
63
37
Bristol
That last quote you have there explains the first two: assuming for simplicity's sake that there's no resistance to motion, the speed the ball leaves the marker horizontally will not change. acceleration due to gravity = 9.8 m/s so if it's not angled upwards it will have an initial vertical velocity of 0, and the height is then the distance travelled vertically, and you justplug those values into the equation below for both vertical (to get the time, t which is the same for both equations) and horizontal (to get the distance travelled horizontally) values. Hope that helped

s (distance travelled) = u(initial velocity)t(time) + 1/2 a(acceleration)(t squared)