Re: A small rocket-science brain teaser
From: Androcles (dummy_at_dummy.net)
Date: 10/18/04
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Date: Mon, 18 Oct 2004 04:00:30 GMT
"sal" <pragmatist@nospam.org> wrote in message
news:pan.2004.10.18.02.02.56.342106@nospam.org...
> On Sun, 17 Oct 2004 15:13:44 +0000, Androcles wrote:
>
>>
>> "sal" <pragmatist@nospam.org> wrote in message
>> news:pan.2004.10.17.03.04.10.117080@nospam.org...
>>> On Sun, 17 Oct 2004 01:52:56 +0000, Androcles wrote:
>>>
>>>
>>>> "sal" <pragmatist@nospam.org> wrote in message
>>>> news:pan.2004.10.16.21.22.05.333457@nospam.org...
>>>>> On Sat, 16 Oct 2004 19:13:04 +0000, Androcles wrote:
>>>>>
>>>>>
>>>>>> "sal" <pragmatist@nospam.org> wrote in message
>>>>>> news:pan.2004.10.16.17.14.44.663219@nospam.org...
>>>>>>> This occurred to me during a recent argument over rockets and I
>>>>>>> thought I'd post it, in case anyone else is also amused by it.
>>>>>>>
>>>>>>> Let's suppose I've just invented a new gun. Instead of firing
>>>>>>> ordinary bullets, it shoots little rockets -- sort of like a "pocket
>>>>>>> bazooka".
>>>>>>>
>>>>>>> The little rocket engines have an exhaust velocity of 900 feet per
>>>>>>> second. (I'll throw in some photons later on just to keep this "on
>>>>>>> topic" for a relativity group.)
>>>>>>>
>>>>>>> One question came up immediately, and after I realized how to answer
>>>>>>> that one, a few other questions cropped up.
>>>>>>>
>>>>>>> 1) I want the rockets to hit the target with as much momentum as
>>>>>>> they can. How fast should they go?
>>>>>>
>>>>>> Well, the only possible answer is "as fast as possible".
>>>>>
>>>>> Really? ;-)
>>>>>
>>>>> Please, show us your proof -- I don't seem to see it in your post.
>>>>
>>>> See next line.
>>>
>>> That's not a proof, it's an assertion.
>>>
>>> Prove, mathematically, that the momentum of the rocket, at the moment of
>>> impact, is maximized by letting it go "as fast as possible".
>>>
>>> Hint: The exhaust velocity was specified.
>>
>> Hint: The thrust was not specified.
>
> Rather than just saying "There's no proof here" I'll spell out what I mean
> by a "proof".
>
> I've asked for the velocity at which to shut off the engine in order to
> maximize the momentum carried by the projectile.
Yeah, I already covered that when I asked if the "rocket" was a pellet of
propellant or a jacketed pellet. You really should look further into all
I've
written before jumping to conclusions. If it just a propellant, the terminal
mass is zero.
>
> You are asserting that depends on the burn rate (=> thrust) and the fuel
> load.
>
> If you want to prove that, it should be easy: you can do it with a single
> pair of examples. Pick a reasonable value for the burn rate, and pick a
> reasonable value for the total mass (fuel + projectile). Figure out the
> velocity at which the engine must be shut off (or must run out of fuel!)
> in order to maximize the momentum carried by the projectile (along with
> any leftover fuel).
>
> Then, pick _different_ values for the burn rate and/or total mass, and do
> it again, and determine that the "ideal velocity" is different in the
> second case.
>
> If you can do that, you've proved your assertion.
>
> I happen to think this is an interesting problem. If you don't agree,
> nobody's forcing you to continue responding.
>
> Point of clarification: I posted this as a "brain teaser". I'll
> eventually post my own solution, but that won't be for a while yet --
> someone else might find it interesting enough to actually tackle the math,
> and I'm not going to preempt that. So don't hold your breath waiting for
> me to fill in the gaps for you here. If you really think there's not
> enough information, either prove it, or just ignore the problem.
>
>
>> We do not know if a single molecule is
>> ejected at 900 fps through a miniscule aperture at a rate of one per
>> year,
>> or the entire fuel load is burnt in a microsecond, or something in
>> between. If the former, then you have an ineffective rocket that will not
>> overcome the friction of the pocket bazooka tube and the Armed Forces
>> will
>> not be interested. If the latter, you have an ineffective bullet that
>> risks more damage
>> singeing the eyebrows of the soldier behind you than it does to the
>> target, and the Armed Forces will not be interested. Whatever I do, the
>> terminal momentum will still be mv. 900 fps is meaningless information.
>>
>>
>>>
>>>>>> It depends on the
>>>>>> burn time, the thrust and the mass. The smaller the mass the greater
>>>>>> the terminal velocity. The longer the burn the greater the terminal
>>>>>> velocity. I'd suggest they should burn until they hit the target. The
>>>>>> problem you run into is that the propellant is part of the mass and
>>>>>> that reduces the initial acceleration. However, increasing the mass
>>>>>> at the target also increases the momentum. If your little rockets are
>>>>>> ALL propellant, they end up with zero mass at a high velocity. SRB's
>>>>>> are an attempt to approximate this condition and as you know a seal
>>>>>> gave out resulting in the Challenger disaster. You can indefinitely
>>>>>> increase the length of the SRB until the mass of the propellant
>>>>>> exceeds the thrust needed to lift it against gravity, under which
>>>>>> conditions propellant would be wasted with a huge flame at ground
>>>>>> level, until the mass reduces to less that the constant force the
>>>>>> thrust exerts, at which moment the rocket will commence lift-off. So
>>>>>> we need to know the thrust. The same SBR in orbit would commence
>>>>>> acceleration instantly, of course, since it would not be held back by
>>>>>> gravity. Changing the diameter of a SRB changes the thrust. Changing
>>>>>> the length changes the mass and then burn time.
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>> 2) The Army hears about my new design, and they like it; they buy
>>>>>>> it.
>>>>>>> But they don't care about momentum -- "stopping power", in modern
>>>>>>> parlance. All they care about is _energy_. So, to make the army
>>>>>>> happy, how fast should the rockets go?
>>>>>>
>>>>>> I'd suggest they should burn until they hit the target and then flame
>>>>>> out,
>>>>>> having used all the propellant. Without knowing the mass of the
>>>>>> "bullet", the mass of the propellant and the thrust, it is impossible
>>>>>> to quote a speed.
>>>>>
>>>>> Really? Are you quite sure?
>>>> No. It depends on the criteria the Army is looking for.
>>>
>>> You're still implicitly asserting that you can't solve part (2) of the
>>> problem without more information. If that statement is true you should
>>> be able to prove it.
>>>
>>> I sure don't see a proof here!
>>
>> Given that the Army likes your design and has purchased your patent (a
>> reasonable assumption, it is doubtful that you would not have patented
>> the
>> design and it is the design that has been purchased) it follows that the
>> Army has more information than I do. I would not purchase your pocket
>> bazooka and miniature rocket without seeing the full design. As I said
>> above in answer to the question "Are you quite sure?", "No". Therefore I
>> can neither prove nor disprove your assertion that the Army
>> has purchased your design. Within the constraints of logical debate,
>> your
>> assertion is taken as a given. Therefore you must have given the Army
>> more
>> information than you have given me. I assert that I do not have
>> sufficient
>> information to solve part 1 or part 2, and that must serve as a proof
>> that
>> I cannot.
>>
>>
>>
>>>
>>>> Usually a rocket
>>>> has a warhead. I was only discussing the flight, not the explosive
>>>> energy of the warhead.
>>>>>
>>>>> How about a proof of that assertion?
>>>> The quantity of energy is finite and depends on the mass of the
>>>> chemical charge.
>>>> If the burn ends prematurely, i.e. before the target is reached,
>>>> kinetic energy will
>>>> be lost to atmospheric friction.
>>>
>>> Good point!
>>>
>>> I admit to neglecting air resistance in solving this.
>>
>> In that case I have solved part 1.
>
> No you haven't.
>
>> mv is maximized at the end of the burn,
>> whether it be 1 molecule per year ejected at 900 fps or Avogadro's number
>> of molecules ejected at 900 fps.
>
> Prove it. It's a maximization problem and you haven't even started on it,
> as far as I can see.
Point of clarification: You posted this as a "brain teaser".
So far I have raised two points that you had overlooked.
It should be obvious that mv is maximized when m is
maximized and v is maximized. Yet to increase v we must reduce m.
If there is mass that has a minimal value and cannot be used as fuel
(a payload).... Do I really need to say more?
As I said, I have solved part 1. I didn't say I would disclose the solution.
>
>> I am still uncertain whether your rocket
>> has
>> any mass left at the end of the burn. It might just be a pellet of
>> propellant
>> snugly fitting in the pocket bazooka tube. In that case the terminal
>> momentum
>> will be zero at the end of the burn,
>
> That's correct, and in fact the entire solution is embodied in that
> statement, but you need to use calculus to dig it out.
>
>
>> and your purchasers want your device
>> for the 4th of July without harming the public. You should place a
>> warning
>> in the pocket bazooka alerting the user as to which end of the pellet is
>> to be ignited, or it could "backfire", like your puzzle. If, however, the
>> propellant is jacketed with a closed end leading, then we can consider m
>> as the mass of the jacket and mv is is maximized at the end of the burn
>> when the mass has the greatest velocity. I'm still amazed that anyone
>> would purchase this device in preference to a cartridged bullet and
>> handgun, unless silencing is the main issue.
>>
>>
>>>
>>>> If the projectile reaches the target before the burn is over, then
>>>> again energy is wasted, too much propellant was used. Ergo the burn
>>>> should end when the missile reaches the target.
>>>>
>>>>
>>>>
>>>>>>> 3) The Special Forces want some, too. But they want them
>>>>>>> _silenced_.
>>>>>>> Can they do that, if they use the Army version of the gun?
>>>>>>
>>>>>> Don't fire them straight away. Throw them with a catapult, air rifle
>>>>>> or spring
>>>>>> cannon. You need a new gun. If you fire them at 900 fps from the
>>>>>> weapon with a timed fuse, then (neglecting any loss of velocity from
>>>>>> air resistance)
>>>>>> they'll still accelerate when the fuse time out. The exhaust will
>>>>>> fall straight
>>>>>> down to the ground, which, in the frame of the rocket, is moving
>>>>>> backwards
>>>>>> at 900 fps. As the rocket accelerates, the exhaust will fall
>>>>>> increasingly forward
>>>>>> as well as down.
>>>>>
>>>>> Perhaps you've overlooked something.
>>>>>
>>>>> Or, perhaps not. An estimate of sound levels from various sources,
>>>>> compare with a conventional gun, might help, but I don't see one here.
>>>>> ;-)
>>>> I'm making the assumption that the reason for the desired silence is
>>>> that the
>>>> enemy cannot detect, aurally, the source of the missile. Silence at
>>>> the target would be unnecessary, the target has been hit and has
>>>> knowledge of the strike. The whole purpose of stealth is to give no
>>>> warning of impending and imminent strike, but once the strike has been
>>>> made, something must have cause it. I also assume the Special Forces
>>>> are
>>>> reasonable and intelligent, or they would not be "special".
>>>> If the velocity of the missile exceeds the speed of sound, its approach
>>>> will not be heard by the target.
>>>
>>> Good point! And one I hadn't thought of.
>>>
>>> But, realistically, a bullet moves too fast for anyone to "duck", so it
>>> doesn't do the target any good to hear it coming. The reason for the
>>> silencer in this case is assumed to be so that non-targets won't hear
>>> the shot.
>> Realistically the Army wouldn't buy it except for bazooka training in a
>> classroom. I didn't think we were attempting realism here. Besides, this
>> is not a bullet, it is a rocket. Only the exhaust is travelling at 900
>> fps,
>> the rocket may only be travelling at 1 fps, sliding along the ground. I
>> can't determine that without knowing the mass and the thrust, which is
>> information you have not provided.
>>>
>>>
>>>> What have I overlooked, besides
>>>> explaining the laws of physics to people who should already know them
>>>> but apparently do not?
>>>> Do you still maintain that the derivative of (1/2)mv^2 is not mv?
>>>
>>> Care to tell me what you're taking the derivative with respect to?
>>>
>>> Velocity? Then d(1/2 mv^2)/dv = mv
>>
>> Remember when I quoted my school math teacher? "dy/dx : a small change in
>> y at y divided by a small change in x at x" What do you think dv means
>> here?
>
> Leibnitz would say it's an infinitesimal change in velocity.
>
>
>>> Or time? Then d(1/2 mv^2)/dt = mva = v * f
>>
>> f would be the thrust, which you have not provided. m would be the mass,
>> which you have not provided.
>
> Right on both counts.
>
>
>>> It makes a difference.
>>>
>>> dE/dv = p
>>>
>>> dE/dt = rate of energy transfer
>>
>> Uh huh.. my electricity meter has kilowatt-hours written on it which it
>> integrates and a function is applied to my bill. What do we call that?
>> Dang, I've forgotten. How powerful is your rocket, BTW?
>
> For a rocket, that number depends on the velocity, and isn't of much
> interest. Thrust is the primary measure used for rocket engines.
>
>
>> If the exhaust is 900 fps at one molecule a year, it's not very powerful
>> at all, although I'm sure the propellant will still provide a fixed
>> amount
>> of energy.
>> Let's see if we can estimate it. This may be what the Army wanted your
>> pocket bazooka for. We'll put a pellet of high burn rate propellant in,
>> then a bullet at the front and a bullet at the back. Both bullets leave
>> the tube at |mv|, and since mv + (-mv) = 0, momentum is conserved. The
>> energy of each bullet is 1/2mv^2, so the energy of the charge must be
>> fixed at mv^2.
>> Without changing m, and closing the rear end of the tube, the one bullet
>> would leave with velocity V, and 1/2 mV^2 = mv^2, right?
>
> No. Not right.
I have shown that the energy of the charge is mv^2, and that the velocity
at which the projectile of mass m emerges has to be SOME velocity > v
relative to the closed tube that I've called V.
> You've just posed an interesting problem in
> thermodynamics which doesn't have an easy answer. The relationship
> between V and v depends on the behavior of the gas in the tube, closed at
> one end, with two bullets inside rattling around (did you forget you left
> them both in place?). Force on the bullet drops as the gas expands as
> the bullet moves down the tube, and you haven't even doubled the effective
> length of the tube, unless you neglected to mention that -- you've still
> got two bullets starting in the middle of the tube. All you did is close
> one end.
I stated in an earlier post, prior to this puzzle of yours, that V was v+u
relative the weapon, to which you agreed. I did, however, overlook
the velocity of the gas following the projectile which will diminish
until the pressure inside the tube reaches equlibrium with the pressure
outside, and I've also overlooked any blow-by of gas reaching the end
of the tube before the projectile emerges. This will depend on any
rifling the tube may have, besides random flaws. You didn't state that
the tube was not rifled, and for all we know it could be of a n-sided
polygonal cross-section with an n-1 polygon projectile.
>
>
>> So 1/2 V^2 = v^2.
>> V = sqrt( 2v^2)
>> mV = m.sqrt(2v^2) is now maximized.
>>
>> It would be interesting to know what V was for your photon rocket, since
>> it cannot exceed c and you have mc^2 worth of energy to "burn". Oh.... I
>> forget, m is supposed to increase with velocity, right? Well, just
>> convert
>> the extra mass to energy and have a perpetual motion machine.
>>
>>
>>>> Or should I ask, the indefinite integral of mv is not (1/2)mv^2 ?
>>>
>>> Only if the variable of integration is "v". If it's time, then no, the
>>> indefinite integral of mv isn't 1/2 mv^2 ... as I'm sure you know.
>>>
>>>
>>>> Or that I have forgotten the chain rule?
>>>
>>> If you evaluate d(mv^2)/dt and get 2mv, then you have left out an "a",
>>> and indeed, you forgot to apply the chain rule. Whether you have
>>> forgotten it globally, or just forgot to apply it, I cannot say.
>>
>> I was quite happy with
>> d(1/2 mv^2)/a = mv,
>
> What on Earth is that supposed to mean?
I knew that would be foreign to you. Ces't la vie.
> You've got a differential on
> the left and the momentum on the right. What kind of definition are you
> using for "d" that makes you think you can equate those?
>
> d(anything) is infinitesimal. "a" isn't and neither is "mv". You
> can't equate those terms like that.
Don't see why not. You seem the think
½[tau(-vt,0,0,t)+tau(-vt,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v))
is identical to
½[tau(0,0,0,t)+tau(0,0,0,t+x'/(c-v)+x'/(c+v))] = tau(x',0,0,t+x'/(c-v)),
for which I have your assertion. I rather feel that makes v "infinitesimal",
but that's another story. My grandmother used to say "Sauce for the goose
is sauce for the gander.
Think of it as a brain teaser.
Androcles.
>
>> or d(1/2 mv^2) = mva = vf, just as I am with dv = a. "dx/dv : a small
>> change in x at x divided by a small change in v at v"
>>
>> Do you want to disclose the values of f and m, or do you still claim: "I
>> found this entertaining partly because it seems like there's not nearly
>> enough information here to answer the questions -- but there apparently
>> is. (Except maybe #4 -- I'm not sure about that one...)", or would you
>> prefer to snip and/or ignore it?
>
> Do as I said up top.
>
> There is enough information to solve the problem. If you disagree, prove
> it. So far all you've done is assert that there's not enough information
> present.
>
>>
>> I think it is your turn to prove "there apparently is", don't you? I
>> think
>> there obviously isn't.
>
> I know -- you already said that.
>
> Wait a week or so and I'll post my answer, but as I said, it's a brain
> teaser and I'm not going to give it away too soon.
>
>
>
> --
> I can be contacted through http://www.physicsinsights.org
>
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