calbiterol

I'd be more worried about the weight of the engine than the prop speed, personally - if you can't get off the ground, then you're SOL, but at least something that chops props LOOKS cool while dive-bombing shrapnel...

Naturalist, this is true. Good point. Nice forum resurrection, by the way.

Exactly. The heat is doing work on the projectile (actually not a bullet but no worries) - hence the term. I never said it was positive...

So does the temperature change of the gas matter then?

... Is that valid? It seems like it might be a little... counterintuitive... if it is. If it's valid, though, I'll throw in a term for drag and a term for friction and call it a day.

It seems I might have overanalyzed the problem. Here, how does this sound? [math]R_{air}[/math] is the specific gas constant of air and approximately equal to 287.05. [math]C_p[/math] is specific heat, in terms of mass, and [math]T_i[/math] and [math]P_i[/math] are initial temperature and pressure, respectively. V is volume, v is velocity. Adjustment of [math]B_k[/math]: Manipulation of Ideal Gas Law: [math]PV=nRT[/math] [math]\frac{PV}{T}=nR[/math] Density equations: [math]\rho_i=\frac{m}{V}[/math] [math]\rho_{i_{air}}=\frac{P}{R_{air}T}[/math] [math]B_k=\frac{m_{air} C_p}{nR}+1[/math] [math]B_k=\frac{m_{air} C_p T_i}{P_i V_i}+1[/math] Substitution [math]B_k=\frac{\rho C_p T}{P}+1[/math] Substitution [math]B_k=\frac{C_p}{R_{air}}+1[/math] Definition of density, simplification [math]B_k~\frac{1.0035}{287.05}+1[/math] Derivation: [math]P_i V_i = m_{air}C_p(T-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math] [math]P_i V_i = m_{air}C_p(\frac{PV}{nR}-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math] Substitution [math]P_i V_i + m_{air}C_p T_i = PV (\frac{m_{air} C_p}{nR}+1)+\frac{m_{projectile}v^2}{2}[/math] Distribute, rearrange [math]P_i V_i + m_{air}C_p T_i = P (A x) (B_k)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i + m_{air}C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution (from above) [math]P_i V_i + \rho_i V_i C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Definition of density [math]P_i V_i + (\frac{P_i}{R_{air} T_i}) V_i C_p T_i = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i + \frac{P_i V_i C_p}{R_{air}} = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Substitution [math]P_i V_i (\frac{C_p}{R_{air}} + 1) = P (A x) (\frac{C_p}{R_{air}}+1)+\frac{m_{projectile}v^2}{2}[/math] Rearrange [math]P_i V_i = P (A x) + \frac{m_{projectile}v^2}{2(\frac{C_p}{R_{air}} + 1)}[/math] Division I suppose the question is, is that valid?

My point is, energy density doesn't do you any good if the energy capacity is low. It'd be cheaper (in the short run) to just power the thing off of AA or AAA or C or D batteries (in the US, that is - not sure what the foreign equivalent is) and use caps for bursts of power. In actual helicopters, two full-sized rotors will counter-rotate (opposite directions) to preserve angular momentum AND to provide even-er lift (there's a fairly complicated effect in here that I neither remember the name of nor can explain. It involves vortices, if I remember correctly).

I would be pretty surprised if anything RC is powered by a capacitor. Capacitors are meant to either smooth over the flow of electricity or to provide spikes of energy. Capacitors might be used under high-torque-requiring situations in order to give some extra juice to the motors, but powering the entire thing off of capacitors is fairly unlikely - unless you want a 2-5 second flight. You'll need batteries (or a fuel cell haha) somewhere. Even AA or AAA or 9V might work - it all depends on the circuitry.

That's fine, I was just making sure that you weren't speculating on what the cyclone is.

Yes, you can you ideal gas laws, you just have to treat temperature as a variable. If I had to guess, I'd say the temperature drop is around 60 degrees F.

No, heat transfer out of the system isn't an issue. I'm ignoring that. However, I do need to know the rate the projectile is flying down the barrel, for a couple reasons. First, I need to know what the velocity of the projectile is at the moment it leaves the barrel. Second, I need to know the rate at which the gas is expanding, because I am NOT making the assumption that all of the gas has been depressurized by the time the projectile leaves the barrel. Third, instantaneous velocity is needed to include the requisite drag and friction terms, which I will incorporate sometime soon. But here's the biggie, which is also why I quoted John. Temperature change is big. I'm talking pressures ranging from 300 to 800 psi, down a 6-inch to 3-foot barrel. I've done similar experiments without any maths, and dropping only 120 psi down a 2.5-foot barrel produces enough of a temperature drop to create ice fog at the end of the barrel. Oh, and for what it's worth, I'm perfectly comfortable with calculus, as long as I take time to remember it. I certainly appreciate the help and the input - and I definitely think there is a certain kind of beauty to simple solutions - but I also went about this problem the way I did for a reason. I should have clarified that earlier; sorry. Cheers, Calbit

Okay, fixed the problem, and I'm stuck here: [math]B\frac{dP}{dx}(Ax+V_i)+BAP=-m_{proj}a[/math]. I suppose I should integrate with respect to x, but I'm frankly too tired to figure out the integral of the first term.

Scratch that, I took an integral instead of a derivative. I'll fix that later, as of right now, that equation is false.

CPL.Luke, that ignores energy lost or gained due to temperature change. There's something else I'm forgetting along those lines... Also, somewhere along the line of deriving the equation I have for acceleration in terms of x I had equations that were very friendly to adding in terms to accommodate friction and drag. So long story short, I want to stick with my equation, I just need to turn acceleration in terms of x to velocity in terms of x - which is more complicated than just integrating with respect to x, isn't it?

I think I might have figured it out. When I know for sure, I'll throw up the end equation, the derivation, and then start to test it. Cheers. Edit: I suppose I might as well update you on my progress. I have an equation that defines acceleration in terms of x. Before stating that, though, I'll describe the system more fully. Re-Edit: Okay, here goes. First, for readability, I'll define the constant [math]B_k[/math] as [math]\frac{m_{air} C_p}{nR}+1[/math], or reconfigured in terms of density and simplified, [math]\frac{C_pT_i^2R_{air}}{P_i^2}+1[/math], where [math]R_{air}[/math] is the specific gas constant of air and approximately equal to 287.05. [math]C_p[/math] is specific heat, in terms of mass, and [math]T_i[/math] and [math]P_i[/math] are initial temperature and pressure, respectively. Now, derivation ignored because it's long and I want to wait until I finish the equation, I've gotten to this equation: [math]a = -\frac{B_k A x + V_i}{A (B_k + 1)}[/math], where a = acceleration, A = area of barrel and x = displacement of projectile (see above). This equation started as: [math]P_iV_i = m_{air}C_p(T-T_i) +\frac{m_{projectile}v^2}{2}+PV[/math]. (Not the best available symbols, but v is velocity and V is volume.)

Edit: And If I don't know entropy?

Is there any law for rates of gas expansion? I got to an equation that would be solved by a gas expansion law. It's come a LONG way from my last post.

Could one use energy transfer to relate the two? In other words, could one take the potential energy of the gas under pressure, [math]U_{compression} = PV [/math], and the kinetic energy transferred to the projectile, [math]E_k = \frac{mv^2}{2} [/math], and set the two equations equal? Only, from that point, I'm not sure how to find an equation for change in pressure with respect to x. Gah.

But gas density is dependent on temperature and pressure. If I had an equation for what the density was at each point, I could figure it out. I suppose a better question would be, when volume varies directly with distance x, how do pressure and temperature vary with distance x?

I need a way to relate pressure and volume without temperature. I've tried deriving an equation myself but I can't get one that (seems to) work. Long story short, I need to determine the initial volume of a gas (air) needed to bring a projectile to a specific velocity, without knowing or measuring temperature change. Diameter/radius and length of barrel are both constant in each case - but a general format is needed that keeps barrel length and diameters as variables. In other words, for the purposes of differentiation/integration, diameter and length are constants, but for final equation form, must be variables. I'm perfectly willing to derive the final equations myself, I just need a place to start. Thanks much!

Cyclone is a trademark. http://www.ultratwistedpaintball.biz/mm5/animations/tippmanncycloneanimation.gif The driveshaft you see in that animation is connected to a sprocket that forces paintballs into the chamber.

GutZ, it really depends on what the pneumatics are.

Sam, are you looking to do something like a cyclone feed on a Tippmann? Long story short, they use a waste gas-driven sprocket to force feed paintballs into the chamber. With very few exceptions, it doesn't skip or chop paintballs, and it can be "tweaked" to feed up to and in excess of 25 bps. I'm a paintball guy and an engineering guy; modding and tweaking is a bit of a hobby. I'm currently working on building a (rather unique) gun from the ground up. In my spare time, that is (hahahahahaha).

First of all, read DH's post. I won't touch on the automotive example because he did that already. Second, no, I do not know what MTBF is. I'm not a Nasa employee. I don't have a degree in aerospace engineering (yet), or a pilot's license (yet), or am even in college. I'm a high school senior (grade 12, since "KPH" tends to mean "high school" is not used as a term). I was in a bad mood when I responded, and for that initial frustration, I apologize. I did not think you were flaming. I just despise being thought less of because I am (physically) younger than others - which is why I like online forums so much. Again, I may have been angry in the last post, but it was not totally because of you, and so I apologize. Now, back to the meat. All I was saying is that files can be corrupted fairly easily. You had previously said that files are very difficult to corrupt; I was refuting that specific argument. I was not applying this to the Mars mission concept, and I should have stated that at the time. Actually, Microsoft will, if you want me to send you a link. I had a known software issue, for which Microsoft offered limited help. I did not say that NTFS itself was unstable, I said that my build of it was. This was certainly not (directly) because of Microsoft. In rare cases, a certain series of circumstances can lead to corruption of the file system. That is what happened. For 4 months, every tim I tried to boot Windows I got the blue screen of death. When I finally managed to force Windows to boot, the file system was still unstable (albeit useable), and my system would frequently crash and was permanently running at the half-processor-speed associated with power saving on laptops. I was not hit with a virus or other form of malware. I finally got around to reinstalling Windows and repartitioning my hard drive over the Christmas hols, and now my computer is again working fine. Well, yes I did, because that's exactly how it felt to me. Whether it was from your specific choice of words, for a misconception or misunderstanding, or just by dumb luck, is irrelevant, because neither of our actions can be undone. Again, I am not arguing that it cannot be done. I am arguing that there are times when these things have failed. There is, after all, a reasion that the people who designed and programmed these craft are "sitting on the edge of their seats" until the mission was more or less successful. To be totally honest, I understood very little between this and the last quote. I'm guessing that MTTDL means "minimum time to data loss," but the other acronyms are beyond me; therefore, so is the math, etc. I have no argument there; again, I haven't been arguing that it cannot be done. I am very much an advocate for extraterrestrial exploration and I find no reason to believe that a straight Mars shot would fail, given enough time and research for preparation. I think that a moon base would be a valuable asset in that preparation, as well as a more cost-effective manufacture and launching point than the surface of Earth. _____ Though I have enjoyed our debate, Ndi, I must point out that we have both gone far beyond the scope of the original argument. The question was originally "Why the moon?", and we now sit arguing over the viability of hardware systems for extraterrestrial exploration. As hypocritical as this may sound, let's make an effort to bring this debate back to the topic. Cheers, Calbit

That's what I think, even though I've been arguing about the small stuff here, too...