I know nothing about guns but I've always wondered like how the bullet starts dropping when it comes out the pipe? So how come they don't make upside down pipes and guns so the bullet actually goes up. You could shoot very far away and aeroplanes and space
(You)
bump
Gravity
Honestly OP, favorite thread of the week. Someone give this man an over complicated answer please.
>I know nothing
succinct
That is why you shake the gun while firing. You pull the trigger right before you reach the top. It balances out the drop when firing long distances. You can also reverse magnitize them in a tesla coil, but the effect only lasts about 30 min.
Go rest your head under that apple tree Newton and the awnser will come to you
But instead of an apple it will be a cinder block
There's a reason why they banned guns in Australia anon, that's because when people fired guns the bullets would fly upwards and hit aircraft.
Is this a joke? In Australia aircrafts fly underground, above people's heads.
Ok let's analyze the forces acting on a projectile
we'll assume t=0 is the instant where the powder detonates which has a inferior calific power h and a mass m. The explosion gives us Q = mh now we apply the first principle of thermodynamics so Q' + Lx' + dV = dU/dt + m'(hin + ek + ez) - n'(ho + eko + ezo) now given the fact the bullet is a solid object with thermal coefficient k and the fact several terms are null we substitute U = mkdT and we're able to get the kinetic energy of the bullet with some algebra. Now we consider the bullet in it's path and write down all the forces acting on it: projecting on the x axis (in a bullet cartesian system) Rx = -kx^2 - mgcos theta Ry = -mgsintheta + Lift where lift is given by 1/2 rho Cl x'^2 A with Cl being the coefficient of life and A the area of the bullet. Finally we substitute the resultants in the Newton's first law (or second idk) and we get Rx = mx'' and Ry = my'' you can have fun solving the differential equation for x substituting 0.5 m x'^2 = ek V where V is the volume of the bullet as ek is a intensive quantity and as you may find out, the lift term is an order of magnitude smaller than the gravity of the bullet ending up with a resultant headed to the negative y which you can port into the real word by writing y = y' cos theta and thus the main component of the acceleration is headed downward. This is justified by the analytical solution of the equation of motion
You could've just said the coriolis effect or spin drift. But thanks for putting this here, I'm gonna use it to frick with some flerfers I know.
You didn't even considered drag, how is OP supposed to understand anything if you don't inject drag in Newton's second law x-axis projection. Also, a simpler way to do it would have to use Lagrangian or Hamiltonian mechanics. Fricking pleb.
-kx'^2 is quadratic drag, k is a coefficient you can use to simplify the notation, generally 0.5 rho Cd S.
Given drag is always opposed to motion if you project on the bullet ex ey drag is only present on the first.
Also I didn't use the Lagrangian because OP probably doesn't have a physics background so I wanted to keep it simple.
I typed all of that in 1 minute and didn't want to bother with anything more complex than a material point being subject to forces
Congrats OP. You figured out Jerry’s secret technique.
If you spin sideways while shooting, you can get the bullet to spin around corners
If you spin the barrel at high speed while firing, you will negate the need for a rifled barrel