The subject line says it all, but, nonetheless...
I need a little help regarding constant acceleration along a half-meter length of barrel. How would I calculate the kinds of accelerations a projectile would be subjected to if its final velocity is in the neighborhood of 1900 to 2000 meters a second if it were shot down said barrel? Also, if time is a significant factor in calculating these accelerations, how so?
Much thanks.
A few physics calculations...
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Re: A few physics calculations...
(fiddles a bit with definitions of velocity and acceleration)
OK, now I remember the equation; I have to rederive this one from scratch every time because I can never remember it:
v2 = 2ax,
where x is distance traveled and v is final velocity, assuming constant acceleration. This can be derived directly from the constant-acceleration formulae for final velocity and position:
x = (1/2)at2 and v = at
So... ~4*106 m/s2, or roughly four hundred thousand gravities. This is far in excess of what you can get out of chemical propellants, which is why tank guns are way more than half a meter long: they increase the barrel length to increase the duration of acceleration.
Unless this is an infantry rifle, there's no reason to use such extreme accelerations. Is it one?
OK, now I remember the equation; I have to rederive this one from scratch every time because I can never remember it:
v2 = 2ax,
where x is distance traveled and v is final velocity, assuming constant acceleration. This can be derived directly from the constant-acceleration formulae for final velocity and position:
x = (1/2)at2 and v = at
So... ~4*106 m/s2, or roughly four hundred thousand gravities. This is far in excess of what you can get out of chemical propellants, which is why tank guns are way more than half a meter long: they increase the barrel length to increase the duration of acceleration.
Unless this is an infantry rifle, there's no reason to use such extreme accelerations. Is it one?
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Re: A few physics calculations...
x=x(t)
x'(t)_0 = 0
x'(t)_t` = 2000m/s
integral_0_t`_(x'(t))=0.5m
suppose x''(t) = constant, e.g. linear acceleration
then x'=integral(x'')= const*t => t`=(2000m/s)/const => const
integral(x')= const/2*t^2 => t`=(2*0.5m/const)^0.5
integral(x')=(x'_avg)t => t` = 0.5m/(1000m/s)
Solve:
t` = 0.5m/1000m/s = (5*10^-4)s
x'' = const = (2000m/s)/t` = (4*10^6)m/s^2
Now you know everything about this system.
This requires the initial assumption x''=const. This is a good approximation for some things, like a rocket, an object sliding or rolling with friction, an object in a constant gravitational field, a charge in a constant electric field, or a magnet in a constant magnetic field (notably, raliguns and coilguns, since you seem to be talking about a weapon). For a nonlinear accelerator (for instance, a bullet in a gun, anything else fired by an explosive charge, or an object subjected to a (velocity-dependent) drag) it only describes the average acceleration over the length of the accelerating path, and the actual acceleration of the object may not be derived without further information about the system.
x'(t)_0 = 0
x'(t)_t` = 2000m/s
integral_0_t`_(x'(t))=0.5m
suppose x''(t) = constant, e.g. linear acceleration
then x'=integral(x'')= const*t => t`=(2000m/s)/const => const
integral(x')= const/2*t^2 => t`=(2*0.5m/const)^0.5
integral(x')=(x'_avg)t => t` = 0.5m/(1000m/s)
Solve:
t` = 0.5m/1000m/s = (5*10^-4)s
x'' = const = (2000m/s)/t` = (4*10^6)m/s^2
Now you know everything about this system.
This requires the initial assumption x''=const. This is a good approximation for some things, like a rocket, an object sliding or rolling with friction, an object in a constant gravitational field, a charge in a constant electric field, or a magnet in a constant magnetic field (notably, raliguns and coilguns, since you seem to be talking about a weapon). For a nonlinear accelerator (for instance, a bullet in a gun, anything else fired by an explosive charge, or an object subjected to a (velocity-dependent) drag) it only describes the average acceleration over the length of the accelerating path, and the actual acceleration of the object may not be derived without further information about the system.
Re: A few physics calculations...
I have physics right now and these r the 3 equations my physics teacher has given us:
Xf = Xo + Vo t + 1/2 a t^2
Vf = Vo + a t
Vf^2 = Vo^2 + 2 a x
("Xf" is final distance, "Xo" is the initial distance, "Vo" is initial velocity, "t" for both of them is time and "a" is acceleration, and "x" is the distance covered)
Hope it helps.
Xf = Xo + Vo t + 1/2 a t^2
Vf = Vo + a t
Vf^2 = Vo^2 + 2 a x
("Xf" is final distance, "Xo" is the initial distance, "Vo" is initial velocity, "t" for both of them is time and "a" is acceleration, and "x" is the distance covered)
Hope it helps.
Re: A few physics calculations...
Yeah, all of it helps a lot. Thanks, guys. It's really appreciated.
By the way, (yes, the coilgun guess was right) any projectile would have to have some sort of either iron or steel jacketing around the main projectile, like modern high-powered rifle rounds, for them to be capable of being used in an electric rifle, right? Unless tungsten is affected by magnets...
Oh, and does anyone know a formula or a method (other than experimental, of course) of determining the amount of heat that an object moving through the atmosphere would generate through sheer friction? And correct me if I'm wrong, but to make a bullet of any useful size out of tungsten, you'd need at least 20 grams of the stuff, right? Also, does anyone know the heat of fusion and the specific heat capacity of DU? For that matter, does anyone else know of any high density, high strength metals that could be useful as bullet material in an infantry coilgun?
Again, much thanks.
By the way, (yes, the coilgun guess was right) any projectile would have to have some sort of either iron or steel jacketing around the main projectile, like modern high-powered rifle rounds, for them to be capable of being used in an electric rifle, right? Unless tungsten is affected by magnets...
Oh, and does anyone know a formula or a method (other than experimental, of course) of determining the amount of heat that an object moving through the atmosphere would generate through sheer friction? And correct me if I'm wrong, but to make a bullet of any useful size out of tungsten, you'd need at least 20 grams of the stuff, right? Also, does anyone know the heat of fusion and the specific heat capacity of DU? For that matter, does anyone else know of any high density, high strength metals that could be useful as bullet material in an infantry coilgun?
Again, much thanks.
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Re: A few physics calculations...
Yes, you can do fluid dynamics in a supersonic regime analytically to an extent, but it is much more complicated than the subsonic case, and given your level of physics knowledge and likely math skill, I doubt you would be able to make much use of the equations. I would instead look for a good online calculator if you want numbers.Caiaphas wrote:Oh, and does anyone know a formula or a method (other than experimental, of course) of determining the amount of heat that an object moving through the atmosphere would generate through sheer friction?
Re: A few physics calculations...
Any suggestions on where to look or what to google?I would instead look for a good online calculator if you want numbers.
Re: A few physics calculations...
A railgun will work with any conductive material. A coilgun requires a ferromagnet.Caiaphas wrote:Yeah, all of it helps a lot. Thanks, guys. It's really appreciated.
By the way, (yes, the coilgun guess was right) any projectile would have to have some sort of either iron or steel jacketing around the main projectile, like modern high-powered rifle rounds, for them to be capable of being used in an electric rifle, right? Unless tungsten is affected by magnets...
2000m/s isn't that fast. You can ignore heat from drag. You will have heat from the firing mechanism, especially if you use a railgun, which has a habit of turning much of the projectile (and potentially the rails of the weapon itself!) into plasma on account of the current being pumped through it. Uranium melts at around 1400K and has a specific heat of around 28J/(mol*K) at room temp. Specific heat increases with temperature, but the rate of increase falls off towards zero above a few hundred kelvin; you can probably get away with taking the specific heat to be around 30J/(mol*K) across the whole regime for back-of-envelope calculations.Oh, and does anyone know a formula or a method (other than experimental, of course) of determining the amount of heat that an object moving through the atmosphere would generate through sheer friction? And correct me if I'm wrong, but to make a bullet of any useful size out of tungsten, you'd need at least 20 grams of the stuff, right? Also, does anyone know the heat of fusion and the specific heat capacity of DU? For that matter, does anyone else know of any high density, high strength metals that could be useful as bullet material in an infantry coilgun?
Again, much thanks.
Re: A few physics calculations...
Which is why I prefer coilguns over railguns--sure, they're a helluva lot more inefficient, but at least you can get more than one shot off per gun, and your projectile will actually hit something. Still, do you think the unfortunate "Oops my gun just vaporized" situation could be avoided by using some sort of superconductive material for the rails? That is, in a railgun.Feil wrote:You will have heat from the firing mechanism, especially if you use a railgun, which has a habit of turning much of the projectile (and potentially the rails of the weapon itself!) into plasma on account of the current being pumped through it.
Thanks for the data. It'll help a lot.Feil wrote:Uranium melts at around 1400K and has a specific heat of around 28J/(mol*K) at room temp. Specific heat increases with temperature, but the rate of increase falls off towards zero above a few hundred kelvin; you can probably get away with taking the specific heat to be around 30J/(mol*K) across the whole regime for back-of-envelope calculations.