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calbiterol

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Posts posted by calbiterol

  1. It seems I might have overanalyzed the problem. Here, how does this sound?

    quickgecopyub2.gif

     

    [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?

  2. 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).

  3. 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.

  4. I don't think the temperature change will be big.
    mc delta t is only necessary if you are alowing heat to enter or leave your system, and considering the rate at which the projectile is going to fly down the barrel I don't think that much heat is going to be allowed out of the system, and even if it were, the quantity would be immeasurable.

     

    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

  5. 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.

  6. 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?

  7. 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.

     

    quickgecopyub2.gif

     

    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.)

  8. 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.

  9. 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?

  10. 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!

  11. 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).

  12. 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.

     

    Then you also realize this has nothing to do whatsoever with a Mars mission, they don't use the same architecture... This is relevant how? "Can be destroyed" and "will not self destruct for an estimated amount of time" are different concepts... Multi-mirrored systems are much more difficult than non-mirrored, but certainly not impossible. Viruses take out systems - even multi-mirrored ones - on a regular basis.

    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.

     

     

    I'm sorry, I really try to stay flame free, but nobody will *ever* convince me you are an experienced professional when you claim that a self-healing, stable file system that runs servers with huge uptime and load is just a barrel of fun with you. Millions test that FS every day and it's been deemed stable. When a million people walk downstream and you are the only one going the other way you are either a visionary or you should take a look at your strategy again.

    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.

     

     

    Quit twisting my words. I never said a mirror is unkillable, I'm saying that given sufficient attention it will not fail on it's own to a point where data is unrecoverable KNOWING it's not a PC, it's not a commercial product and they don't use failing disk. I'm saying that a NASA-designed storage CAN make it safely and reliably to Mars and back. Not that it never made it before and much farther away on deep space crafts.[/qote]

    If I twisted your words I apologize, but it was due to a misunderstanding, not willful intent.

    I'm not saying that NASA-designed storage cannot make it to Mars, or the Kuiper Belt, or what have you. I'm saying that it has the potential to fail, and that Nasa-designed hardware, software, and/or firmware have all failed in the past. For examples: Apollo 13 was a hardware issue, as was the fire on the Gemini craft that killed all inside (while still on Earth, I might add); The Mars Polar Lander was deemed a software issue (lack of consistency with units). I will admit that I cannot think of a firmware failure off the top of my head; hence "and/or."

     

    You somehow made it look like I'm accusing you of being incompetent because your NTFS gets mangled data when you work from different unreliable and/or virused computers.

    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.

     

    The company where I work has huge databases, gigabytes in size, being accessed continuously for over 10 years now, with logs going back even more years. No catastrophic failure ever occurred, nor are any of those mangled. It CAN be done... No virus, either. There are such things as routing, DMZ, firewalls and bridges.

    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.

     

     

    What's the MTTDL now?

    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.

     

    So it's settled. Mirrored flash disks as well as processing units are reliable enough for the trip.

    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

  13. Desktop computers crash. Dedicated systems are much more resilient and given enough time can become predictable. When was the last time you saw a video camera crash? Or a DVD player? They are computers.

    To be honest, my DVD-player crashed a few weeks ago, and it's not some piece of crap from Wal-Mart. Every component has the possibility of crashing - whether or not it is a software issue. Overheating, miscalculating, manufacturing defects. Even Mars rovers. I quote myself: "The numerous failed orbiters sent to Mars could have averted destruction/failure had they human pilots or commanders on-site - a conversion error isn't a problem when you're flying by feel." The Mars Polar Lander was certainly a failure. The fact that computers crash was an example, not my entire case.

     

     

    Files don't "get corrupted" if you know what the heck you are doing. Do you have any idea how resilient and robust storage can be

    Frankly, this kind of offended me. I work with computers and file storage every day, and I am very well aware of its limitations and issues. I very much know what I'm doing, and at least one of my files is corrupted per year. I also know that this is due to the fact that I have a number of different working copies spread across multiple media and computers, some of which go through a hefty everyday beating. I certainly know what I'm doing - I've been programming computers since I was 11, much less using them. Also, I know exactly how to force a file to corrupt itself. It's extremely simple and takes about 30 seconds of work, even for multi-mirrored systems.

     

     

    Ever tried to think up a way to destroy a multi-mirrored NTFS/ext/whatever?
    Yes. Multi-mirrored systems are much more difficult than non-mirrored, but certainly not impossible. Viruses take out systems - even multi-mirrored ones - on a regular basis. As far as non-mirrored goes, I'm currently running Windows on an extremely unstable NTFS file system, and I'm waiting for it to catastrophically fail (again). I have more experience with dead file systems than I would ever want to have. If there is one thing I've learned in my life, it is that EVERYTHING has the potential to fail.

     

     

    Please, guys, don't turn this into a flaming contest. I'd rather not see a useful thread (that I have learned from) become nothing more than a bashfest.

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