Science Question: Ballistics and Acceleration

[darthbob88]

First, apologies if this has already been posted. I could not find anything on this subject.

I'm working on a personal project, involving ballistic trajectories, and I have run into a problem. Using the 2nd law of motion, a=F/m, I have worked out an acceleration of umpteen meters/second-squared from a known force and mass. The problem is that this assumes acceleration over a full second, and the acceleration will only last a small fraction of a second. Actually, all I need this for is the muzzle velocity, so if the muzzle velocity can be directly derived, that would be quite nice.
The problem is this, then: given a known acceleration and a known muzzle length, how much time in barrel can I expect, and/or what muzzle velocity can I expect? Either one should be sufficient.

[Hawkwings]

hmm... Assuming my firearms knowledge is correct, the bullet accelerates pretty much the entire way through the gun.

So you have constant acceleration, distance, and you want to find time.

d=V(initial)*t + .5at^2
v(initial) is zero, so

d=.5at^2

t^2=d/(.5a)
t=sqr(d/(.5a))

[darthbob88]

hmm... Assuming my firearms knowledge is correct, the bullet accelerates pretty much the entire way through the gun.

So you have constant acceleration, distance, and you want to find time.

d=V(initial)*t + .5at^2
v(initial) is zero, so

d=.5at^2

t^2=d/(.5a)
t=sqr(d/(.5a))

To confirm that I understand:
d=(Vinit * t) + ((a*t^2)/2), or for Vinit of 0,
d=(a*t^2)/2, and by algebra,
t^2=d/(a/2), or t^2=2d/a, so
t=sqrt(2d/a), where
d=distance traveled while accelerating,
Vinit= initial velocity,
a=acceleration, and
t=time
Correct?

[Darth Wong]

This is why you need calculus in order to do physics. Otherwise you just try to memorize formulas without understanding where they came from.

[Kuroneko]

Assume that the propellant gas is roughly ideal and undergoes adiabatic expansion with index γ. Then, since the volume will be proportional position along barrel x, the acceleration on the bullet will be a = F/m = [(Area)(Pressure) - F_{friction}]/m ≅ A/x^γ - Bv² for some constants {A,B}. Now, if one assumes that the friction force is negligible over the length of the barrel (B/v²≅0), then integration (v = dx/dt, a = dv/dt) gives v² = 2Ax^{1-γ}/(1-γ), yielding the velocity in terms of barrel distance. Realistically, friction should start to dominate in rifles, but for relatively low muzzle velocity hand-held guns, this should be a reasonable approximation.

[darthbob88]

This is why you need calculus in order to do physics. Otherwise you just try to memorize formulas without understanding where they came from.

Which is why I am taking calculus, part I now and part II next semester. Until then, I'll settle for memorizing formulae. Thank you for the assistance, May the Force Be With You.

[darthbob88]

Assume that the propellant gas is roughly ideal and undergoes adiabatic expansion with index γ. Then, since the volume will be proportional position along barrel x, the acceleration on the bullet will be a = F/m = [(Area)(Pressure) - F_{friction}]/m ≅ A/x^γ - Bv² for some constants {A,B}. Now, if one assumes that the friction force is negligible over the length of the barrel (B/v²≅0), then integration (v = dx/dt, a = dv/dt) gives v² = 2Ax^{1-γ}/(1-γ), yielding the velocity in terms of barrel distance. Realistically, friction should start to dominate in rifles, but for relatively low muzzle velocity hand-held guns, this should be a reasonable approximation.

Ah, yes, right. Could I have that again in English? I can understand most of it, but those constants {A,B} are new to me and I have not seen integration yet in my calc class. Also, this will be a very low-velocity weapon: I calculated an approximate muzzle velocity of 54 meters per second for an approximate range of 300 meters.

[Kuroneko]

I'm assuming you are at least familiar with differentiation. Velocity is the time-derivative of position: v = dx/dt; acceleration is the time-derivative of velocity, i.e., the second derivative of position: a = dv/dt = d²x/dt². For most purposes, you can think of integration as the inverse of differentiation: velocity is the integral of acceleration with respect to time, position is the integral wrt time, etc. "What's the integral of f?" is essentially equivalent to "what should I differentiate to get f?" (there are actually infinitely many answers, since the derivative of a constant is zero, so that if F is an integral of f, so is f+c for any constant c, but the "constants of integration" all turn out to be zero for this particular problem).

As for physics, 'adiabatic' means 'no heat transfer'--the gas simply expands, which is a reasonable approximation (although there is some conduction, as evidenced by barrels getting hot), and 'ideal' refers to assuming that that particles are non-interacting and of negligible size, which is actually questionable, because the gas starts out in a rather dense state (but other equations of state, e.g., van der Waals, would make this problem overly complicated). For the moment, just assume that the pressure is proportional to 1/V^γ, where γ is the adiabatic index of the propellant gas and is a constant determined by the composition of the gas only. The area of the barrel should be constant, so the Force = Area * Pressure should be proportional to 1/V^γ as well, but volume is itself proportional to the bullet's position x, so acceleration ∝ force ∝ 1/V^γ ∝ 1/x^γ. I simply called the constant of proportionality "A", so that a = A/x^γ if friction is neglected (that's the second term).

Now, have you covered the chain rule [d/dt][f(g(t))] = [df/dg][dg/dt]? That means, in particular, [d/dt][v(t)²] = 2v dv/dt = 2va. Consequently, a = A/x^γ is equivalent to 2v dv/dt = [2A/x^g] dx/dt. Integrating the left-hand side gives v² and the right-hand side gives 2Ax^{1-γ}/(1-γ) (check both with the chain rule and power rule). If you know the muzzle velocity of the projectile at a certain barrel length, you can find the constant A, and thus extrapolate to the muzzle velocity for different barrel lengths using this formula. It is not reasonable to assume that the acceleration will be constant throughout the barrel (the propellant gas, after all, will lose pressure by expanding), which is why this is a bit more complicated.

[darthbob88]

I'm assuming you are at least familiar with differentiation. Velocity is the time-derivative of position: v = dx/dt; acceleration is the time-derivative of velocity, i.e., the second derivative of position: a = dv/dt = d²x/dt². For most purposes, you can think of integration as the inverse of differentiation: velocity is the integral of acceleration with respect to time, position is the integral wrt time, etc. "What's the integral of f?" is essentially equivalent to "what should I differentiate to get f?" (there are actually infinitely many answers, since the derivative of a constant is zero, so that if F is an integral of f, so is f+c for any constant c, but the "constants of integration" all turn out to be zero for this particular problem).

Derivatives and differentiation I'm familiar with, and I had a hunch that differentiation and integration were "reciprocal" like that, but I wasn't quite sure and didn't want to risk the wrong answer in a ballistics problem.

As for physics, 'adiabatic' means 'no heat transfer'--the gas simply expands, which is a reasonable approximation (although there is some conduction, as evidenced by barrels getting hot), and 'ideal' refers to assuming that that particles are non-interacting and of negligible size, which is actually questionable, because the gas starts out in a rather dense state (but other equations of state, e.g., van der Waals, would make this problem overly complicated). For the moment, just assume that the pressure is proportional to 1/V^γ, where γ is the adiabatic index of the propellant gas and is a constant determined by the composition of the gas only. The area of the barrel should be constant, so the Force = Area * Pressure should be proportional to 1/V^γ as well, but volume is itself proportional to the bullet's position x, so acceleration ∝ force ∝ 1/V^γ ∝ 1/x^γ. I simply called the constant of proportionality "A", so that a = A/x^γ if friction is neglected (that's the second term).

I was familiar with adiabatic expansion, but only in passing, and your figures seem reasonable to me. However, you're making the assumption that this is due to a chemical explosion, like in a rifle, where the moles of gas is a constant, but volume and therefore pressure are variable. What about a constant force, like an airgun fed off an air compressor? How does that impact the calculations?

Now, have you covered the chain rule [d/dt][f(g(t))] = [df/dg][dg/dt]? That means, in particular, [d/dt][v(t)²] = 2v dv/dt = 2va. This means that a = A/x^γ is equivalent to 2v dv/dt = [2A/x^g] dx/dt. Integrating the left-hand side gives v² and the right-hand side gives 2Ax^{1-γ}/(1-γ) (check both with the chain rule and power rule). If you know the muzzle velocity of the projectile at a certain barrel length, you can find the constant A, and thus extrapolate to the muzzle velocity for different barrel lengths using this formula. It is not reasonable to assume that the acceleration will be constant throughout the barrel (the propellant gas, after all, will lose pressure by expanding), which is why this is a bit more complicated.

I have seen the chain rule before, and know how to apply it. The formula 2Ax^(1-γ)/(1-γ) seems like it would come out to 2Ax^1, so I must have misunderstood. Do you mean (2A*x^(1-γ))/(1-γ)? Also, again, would it change the figures if the force and acceleration was in fact constant?

[darthbob88]

Scratch most of that; your assumptions match with reality quite well enough. I'm still unsure what you mean by 2Ax^{1-γ}/(1-γ) Do you mean (2A*x^{1-γ})/(1-γ), or something completely different.

[Kuroneko]

However, you're making the assumption that this is due to a chemical explosion, like in a rifle, where the moles of gas is a constant, but volume and therefore pressure are variable. What about a constant force, like an airgun fed off an air compressor? How does that impact the calculations?

Yes, I've assumed that the mechanism is that of a typical firearm. If that's not the case, then the model doesn't apply (this should have been made clearer in OP). Hawkwings already gave the answer assuming a constant force, although he mistakenly attributed this to be typical in a firearm. In that case, you have a constant acceleration a, so you simply have to integrate twice wrt time to get the position a → v = at + v_0 → x = at²/2 + v_0 t + x_0 (v_0 = x_0 = 0 for your problem).

Do you mean (2A*x^(1-γ))/(1-γ)? Also, again, would it change the figures if the force and acceleration was in fact constant?

Yes and yes.

[darthbob88]

However, you're making the assumption that this is due to a chemical explosion, like in a rifle, where the moles of gas is a constant, but volume and therefore pressure are variable. What about a constant force, like an airgun fed off an air compressor? How does that impact the calculations?

Yes, I've assumed that the mechanism is that of a typical firearm. If that's not the case, then the model doesn't apply (this should have been made clearer in OP). Hawkwings already gave the answer assuming a constant force, although he mistakenly attributed this to be typical in a firearm. In that case, you have a constant acceleration a, so you simply have to integrate twice wrt time to get the position a → v = at + v_0 → x = at²/2 + v_0 t + x_0 (v_0 = x_0 = 0 for your problem).

Do you mean (2A*x^(1-γ))/(1-γ)? Also, again, would it change the figures if the force and acceleration was in fact constant?

Yes and yes.

Definitely should have been clearer. Mea culpa, my bad. However, I was apparently incorrect in correcting you; the plan is to use a pressurized reservoir rather than a constant force, so your original figures are correct, and I was in error. Thanks for the assistance, MTFBWY.

[Kuroneko]

Note that the model assumes that the entire gas is in the barrel. If it is actually in a pressurized tank of volume V, and the barrel cross-sectional area is A and length d, then you can adjust for it by having the position coordinate x go from x = V/A to x = V/A+d, thus treating the tank as if it was part of the barrel with length V/A. If you're fitting this model to data points (just one is enough in this case), it will probably be much easier to fit to muzzle energy instead of velocity: ΔK = Δ[mv²/2] = F(V/A+d) - F(V/A), where F(x) = k[x^{1-γ}]/(1-γ) for some constant of proportionality k. Interestingly, if your gas has adiabatic index γ = 2 (air has γ = 1.4), then the projectile behaves qualitatively as if it was in gravitational freefall.

[darthbob88]

Note that the model assumes that the entire gas is in the barrel. If it is actually in a pressurized tank of volume V, and the barrel cross-sectional area is A and length d, then you can adjust for it by having the position coordinate x go from x = V/A to x = V/A+d, thus treating the tank as if it was part of the barrel with length V/A. If you're fitting this model to data points (just one is enough in this case), it will probably be much easier to fit to muzzle energy instead of velocity: ΔK = Δ[mv²/2] = F(V/A+d) - F(V/A), where F(x) = k[x^{1-γ}]/(1-γ) for some constant of proportionality k. Interestingly, if your gas has adiabatic index γ = 2 (air has γ = 1.4), then the projectile behaves qualitatively as if it was in gravitational freefall.

So we treat this as a pipe of cross-section A, attached to a tank of volume V, so the actual volume/pressure is derived from (A*x)+V, yes? And, actually, I just want to work out some factors to obtain a given muzzle velocity, which leads to a given range. To do so, though, I need to know how to convert from non-constant acceleration to actual velocity, which your formulae show how to do. Thank you for your assistance.

[Kuroneko]

So we treat this as a pipe of cross-section A, attached to a tank of volume V, so the actual volume/pressure is derived from (A*x)+V, yes?

(I shouldn't have reused 'A' for pipe/barrel cross-section area; previously, it was just a proportionality constant.) Close, but not quite--x is a position coordinate that no longer represents how far the projectile traveled. Instead, x is rescaled as if the tank was part of the barrel, so that the projectile is considered to start at x=V/A. If the projectile travels a distance ε, then its position is x = V/A+ε, so that the volume of the gas behind the projectile is Ax = V+Aε.

And, actually, I just want to work out some factors to obtain a given muzzle velocity, which leads to a given range. To do so, though, I need to know how to convert from non-constant acceleration to actual velocity, which your formulae show how to do.

If you have a known adiabatic index γ, tank volume V, pressure P, the barrel cross-sectional area A, and projectile mass m, then you initially have ma = PA = k/(V/A)^γ, so that k is found immediately. Then the solution is F(x) = k x^{1-γ}/(1-γ), with kinetic energy change over length d as Δ[mv²/2] = F(V/A+d) - F(V/A).

[darthbob88]

So we treat this as a pipe of cross-section A, attached to a tank of volume V, so the actual volume/pressure is derived from (A*x)+V, yes?

(I shouldn't have reused 'A' for pipe/barrel cross-section area; previously, it was just a proportionality constant.) Close, but not quite--x is a position coordinate that no longer represents how far the projectile traveled. Instead, x is rescaled as if the tank was part of the barrel, so that the projectile is considered to start at x=V/A. If the projectile travels a distance ε, then its position is x = V/A+ε, so that the volume of the gas behind the projectile is Ax = V+Aε.

I heard about a teacher who never reused a variable during a single lecture-by the end he was writing things like ":) = * ". The tank being V/A seems quite obvious now that I see it. Thank you.

And, actually, I just want to work out some factors to obtain a given muzzle velocity, which leads to a given range. To do so, though, I need to know how to convert from non-constant acceleration to actual velocity, which your formulae show how to do.

If you have a known adiabatic index γ, tank volume V, pressure P, the barrel cross-sectional area A, and projectile mass m, then you initially have ma = PA = k/(V/A)^γ, so that k is found immediately. Then the solution is F(x) = k x^{1-γ}/(1-γ), with kinetic energy change over length d as Δ[mv²/2] = F(V/A+d) - F(V/A).

To confirm I understand: k = PA *((V/A)^γ), with velocity at point x being F(x) = k * (x^(1-γ))/(1-γ)?

[Kuroneko]

To confirm I understand: k = PA *((V/A)^γ), with velocity at point x being F(x) = k * (x^(1-γ))/(1-γ)?

Yes, k = PA[V/A]^γ. As for F(x), it's not velocity. It's actually the potential due to the force of the gas--take the difference of the value of F between two different points to find the change in kinetic energy: ΔK = Δ[mv²/2] = F(x_1) - F(x_0). To get the velocity, use the standard velocity-kinetic energy relationship.

Edit: whoops. Corrected.

[darthbob88]

To confirm I understand: k = PA *((V/A)^γ), with velocity at point x being F(x) = k * (x^(1-γ))/(1-γ)?

Yes, k = PA[V/A]^γ. As for F(x), it's not velocity. It's actually the potential due to the force of the gas--take the difference of the value of F between two different points to find the change in kinetic energy: ΔK = Δ[mv²/2] = F(x_1) - F(x_0). To get the velocity, use the standard velocity-kinetic energy relationship.

Edit: whoops. Corrected.

So, the KE at a point x is the limit of F(x) at x, with the velocity being v = sqrt(2/m*that)?
KE = .5*mv^2, so
v=sqrt(2*KE/m)?

[Kuroneko]

No, the kinetic energy at x is the difference F(x) - F(V/A); F(x) by itself is just a potential. For a projectile accelerated across a barrel of length d, the kinetic energy is F(V/A+d) - F(V/A). Your formula for getting the velocity is correct, however.

[darthbob88]

No, the kinetic energy at x is the difference F(x) - F(V/A); F(x) by itself is just a potential. For a projectile accelerated across a barrel of length d, the kinetic energy is F(V/A+d) - F(V/A). Your formula for getting the velocity is correct, however.

Of course it would; what is wanted is delta-KE from start to finish, not KE at a point nor delta-KE from part-way. Thank you for your patience, sir, and I hope I finally got it right.