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Energy Dissipation Of A Sphere Passing Through a Medium

9 years 6 months ago - 9 years 6 months ago #1 by con20or

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  • I was very impressed with the previous debates on firing angles etc, and have a similar question to pose to our resident experts.

    Yes it's a bit morbid, but purely academic!

    Assuming a 12lb'er hits a marching line of men - perfectly parallel, i.e. it will contact each soldier, it isnt dropping or rising, and hasn't hit the ground yet, how many men would it take out before slowing to a non-lethal velocity?

    I dont mind what size soldier is chosen, just mention it in the reply.

    You can ignore any air resistance, or deceleration as the shot passes trough the gap between marching troops.
    Last edit: 9 years 6 months ago by con20or.

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    9 years 6 months ago #2 by born2see

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  • Replied by born2see on topic Re: Momentum....
    Con20or,

    I think it would depend on the relative density of each individual as the shot passed through him, what he was wearing, equipment, etc...

    I don't think you could ignore deceleration as it passed through the individual.

    More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

    B

    "Those in whose judgment I rely, tell me that I fought the battle splendidly and that it was a masterpiece of art.” - George McClellan to his wife describing the battle of Antietam

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    9 years 6 months ago - 9 years 6 months ago #3 by con20or

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  • Replied by con20or on topic Re: Momentum....

    born2see wrote: Con20or,

    I think it would depend on the relative density of each individual as the shot passed through him, what he was wearing, equipment, etc...

    I don't think you could ignore deceleration as it passed through the individual.

    More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

    B


    Lol - looking forward to seeing the empirical data - MP does need spicing up some times but thats going a bit far :)

    I'll clarify this - deceleration between troops is as the shot passes from one soldier to the next... i.e the momentum is the same when it leaves one soldier as when it reaches the next...
    Last edit: 9 years 6 months ago by con20or.

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    9 years 6 months ago #4 by born2see

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  • Replied by born2see on topic Re: Momentum....
    Where are MTG and HTS when you need them?


    B

    "Those in whose judgment I rely, tell me that I fought the battle splendidly and that it was a masterpiece of art.” - George McClellan to his wife describing the battle of Antietam

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    9 years 6 months ago - 9 years 6 months ago #5 by Marching Thru Georgia

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  • born2see wrote:

    More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

    And indeed you shall. This is actually a straightforward calculation that yields an interesting result. The result is happily confirmed by the historic record.

    This thread should actually be titled "Energy Dissipation Of A Sphere Passing Through a Medium." Please change it Con20or. I am quite sure it will garner 10K hits within a week if you do. :laugh:

    We'll use the law of conservation of energy to solve this.
    The kinetic energy of the sphere is: E = 1/2mv**2
    m=mass of sphere, v=velocity of sphere

    The cannonball will lose energy as it passes through the medium, (people). That is a messy calculation, no pun intended. So we'll use an approximate solution that is valid for spheres which move slowly through a medium.
    The drag force on such a sphere is: f = 1/2Cpav**2
    C=drag coefficient, p=density of medium, a=cross sectional area of the sphere

    The energy loss due to drag is just: E = fd d=distance sphere travels in the medium
    So 1/2Cpadv**2 = 1/2mv**2
    Canceling out like terms, we end up with: d = m/Cpa

    That's rather remarkable when you think about it. The distance the cannonball travels is independent of the initial velocity. Extra credit will be awarded if you can explain why this is so.

    Time for numbers:
    m = 5.6kg (12.3 lbs ball)
    a = 0.01 M**2 (4.5 in. dia.)
    C = 0.1-0.47 depending on the smoothness of the sphere
    Human are not much more than bags of salty water.
    p = 1027 kg/m**3

    Plugging in the numbers we get
    d = 0.55/C or d = 1.16 - 5.5 M

    Human torsos are about 34cm wide and 26cm thick, (if you significantly exceed these measurements you are part of the obesity epidemic)

    So 4 - 21 humans would be killed if the ball strikes head-on and 3 - 16 in enfilade.

    The largest number of casualties from solid shot I have seen recorded is 14 at Shiloh and 12 at Gettysburg. Both figures fit nicely into the calculated ranges.

    I can make this march and I will make Georgia howl.
    Last edit: 9 years 6 months ago by Marching Thru Georgia.
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    9 years 6 months ago #6 by Jim

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  • More commonly, the cannonball was descending at some significant rate depending on the range and elevation of the gun relative to the infantry. There was an example cited from GB IIRC where a single cannonball killed four men, the first was hit in the head, the 2nd in the chest, the 3rd in the hips and the fourth lost both legs at the knees. The reaction of the 5th man in that line was not recorded. :sick:

    -Jim

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    9 years 6 months ago #7 by con20or

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  • very interesting! good work MTG.
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    9 years 6 months ago #8 by Group Captain Mandrake

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  • Marching Thru Georgia wrote: That's rather remarkable when you think about it. The distance the cannonball travels is independent of the initial velocity. Extra credit will be awarded if you can explain why this is so.


    That is quite remarkable. I am going to venture that it is because both KE and the retardive force vary with the square of velocity.

    Even so, I would prefer to get hit by a 12 lbr rolling along the ground 2 mph

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    9 years 6 months ago #9 by Marching Thru Georgia

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  • Group Captain Mandrake:

    I am going to venture that it is because both KE and the retardive force vary with the square of velocity.
    Even so, I would prefer to get hit by a 12 lbr rolling along the ground 2 mph

    Gutsy, your first post and you venture into a very deep pool. Yes, that's the correct answer. Both the kinetic and the dissipation of that energy are dependent on the same power of the velocity. So if the sphere is moving 1cm/sec or 500M/sec it will travel exactly the same distance when the kinetic energy will drop to 0. The reason that doesn't happen on the battlefield, is that we bags of salty water have a rather thin, but tough membrane surrounding us.

    Welcome to the forum.

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    9 years 6 months ago #10 by Group Captain Mandrake

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    9 years 6 months ago #11 by Barrow

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  • Hello,

    Sorry if what I'm about to write is TMI to some people...

    I agree with the approach presented, but I think I disagree with a couple
    of the assumptions:

    >So we'll use an approximate solution that is valid for spheres which
    >move slowly through a medium.
    >The drag force on such a sphere is: f = 1/2Cpav**2

    2 comments:

    1: I think this is the approximation for high velocity drag, not low velocity
    drag, i.e. Reynolds number > ~1000 (from Wikipedia en.wikipedia.org/wiki/Drag_(physics) .
    For low velocity the drag is proportional to the velocity,
    not the velocity squared (same wiki link).

    2: Regardless of comment 1, as the projectile slows down the drag force will
    decrease. The drag force is not a constant. It
    relates to the instantaneous velocity, not the initial velocity. Assuming constant drag
    force might be a good assumption at high velocities, but at low velocities
    this assumption makes a big difference.

    If the high velocity drag term is used, I think this is the resulting differential
    equation (dropping constants):

    d2x/d2t = (dx/dt)^2 , which would yield a natural logarithm solution x(t) = ln(c1 + t) +c2

    But anyway, I think the low velocity drag is better, which is proportional to velocity,
    not velocity squared, and gives an exponential solution (dropping constants):

    v(t) = 1 - exp(-t)


    Happy Thanksgiving...

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    9 years 6 months ago - 9 years 6 months ago #12 by Marching Thru Georgia

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  • Born2see and Con20or live for TMI. So now let's fill their plate. :laugh:

    No, I disagree. Stokes Law is valid when the Reynolds number, R, is less than 1. Actually R<=0.1 is more precise. That represents true laminar flow, which in our case is a sphere moving <1cm/sec.

    Now if we were discussing how air drag affects the trajectory of a bullet, then I would endorse using Stokes Law, since it greatly simplifies the calculation. Although Raleigh's Law would still be more appropriate, the solution is not nearly so elegant. And physics is all about elegance. :)

    Just to be clear, by low velocity, I meant subsonic speeds. Although Raleigh's Law can be used to determine the drag coefficient, C, Eulers equation is more appropriate in the supersonic realm.

    Now let's give Born2see and Con20or some desert to chew on. We all know that it takes more gas to go the same distance at 60MPH versus 50MPH. It is more than can be explained by the velocity squared. Why? Hint: Road friction is not the answer.

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    Last edit: 9 years 6 months ago by Marching Thru Georgia.

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    9 years 6 months ago #13 by Barrow

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  • >The drag force on such a sphere is: f = 1/2Cpav**2

    In the current situation the projectile is coming to a complete stop.
    The drag force is not a constant in this case, regardless of which
    approximation is used. I'm disagreeing with the step

    Fd=1/2mv^2, where a constant force is used to solve for d.

    Assuming a constant force works well in a situation where the
    projectile's deceleration is much less than its velocity, but
    not as it's about to stop moving.
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    9 years 6 months ago #14 by Marching Thru Georgia

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  • You are quite right. To be precise, we would need to take into account the transition from one regime to the other. If I was approving semester problems for a student to work on in a mechanics class, this would be a good one. But in this forum, with this audience, I think trying to keep things reasonably approximate and simple is best.

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    9 years 5 months ago #15 by Group Captain Mandrake

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  • Marching Thru Georgia wrote:
    Now let's give Born2see and Con20or some desert to chew on. We all know that it takes more gas to go the same distance at 60MPH versus 50MPH. It is more than can be explained by the velocity squared. Why? Hint: Road friction is not the answer.


    I am going to venture the extra fuel consumption beyond that can be ascribed to increased aerodynamic drag and road friction is explained by the higher engine revolutions required. Almost all modern cars will be geared such that 60 mph requires more revs that 50 mph. More revs equals more work expended compressing the gas in the cylinders and and reversing the direction of cyinder motion and fan speed and even perhaps some contribution from more internal friction. There may also be some contribution from driver error if he is trying to eat an egg sandwich at 60 mph.

    Also note: Fuel consumption is not solely a function of engine revs. Try going up hill in a BMW and watch the fuel cosnmption gage. :)

    So I bought this game. How on earth do you control anything larger than a brigade?

    Also, I have this question. Does the AI give intself any fire advantages at the default level?

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    9 years 5 months ago #16 by Marching Thru Georgia

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  • Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:

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    9 years 5 months ago #17 by con20or

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  • Marching Thru Georgia wrote: Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:


    Keep waiting - you don't get to examine the examiner!

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    9 years 5 months ago #18 by Group Captain Mandrake

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  • Marching Thru Georgia wrote: Engine and rolling friction are only minor components. There is one big contributor to the gas consumption. Still waiting to hear from the physics aficionados Born2see and Con20or. :laugh:


    Ah, I get it. This is like a Physics question not an automotive engineering question. From an energy balance standpoint most of the energy expended in the operation of an internal combustion engine in a moving vehicle is lost to heat (radiative and hot exhaust gasses).

    The heat losses are inadvertent "waste" resulting from the imperfect conversion of the chemical energy in the fuel. Not sure what the ratio of useful mechanical energy to heat loss is but I doubt it is much better than 50%.

    My point about the mechanical efficiency still applies. Most cars will have declining fuel efficiency between 50 and 60 because of drag and gear ratios.

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    9 years 5 months ago #19 by Hancock the Superb

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  • Air drag / wind resistance would be the answer you are looking for born2see.

    In addition, the solution to the original problem may need to account for rotational energy of the projectile, depending upon whether it is a smoothbore or rifled cannon.

    Hancock the Superb

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    9 years 5 months ago #20 by Marching Thru Georgia

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  • Hancock The Superb wrote:

    In addition, the solution to the original problem may need to account for rotational energy of the projectile, depending upon whether it is a smoothbore or rifled cannon.

    Please show your work to justify this statement.

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    9 years 5 months ago #21 by Hancock the Superb

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  • Simply that rotational kinetic energy recieved at the time of firing still needs to be dissappated before the projectile will truely be stopped (the previous solutions only account for translational kinetic energy). This rotational energy results in greater penetrating power, as evidence by the destruction of Fort Pulaski.

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    9 years 5 months ago #22 by Marching Thru Georgia

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  • Please show how rotational energy adds to the forward momentum of a projectile. This is a phenomena I am not familiar with. Rotation stabilizes a projectile in flight. In other words, it increases it's accuracy through the law of conservation of angular momentum. This is why it's easy to stay upright on a moving bicycle and hard to do so on one that is stationary. Draw yourself a vector diagram to show which way the angular and forward momentum point.

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    9 years 5 months ago - 9 years 5 months ago #23 by Hancock the Superb

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  • MTG: You are damn good at twisting my words around or my style of communication must be more confusing than jibberish.

    The rotational kinetic energy affects the friction, not the initial momentum. At this point, my limited knowledge of physics dissipates (pun intended), and I will proceed off of pure conjecture, based in logical reasoning.

    The projectile possesses a certain amount of kinetic energy. This energy must be matched by an equal reduction in energy from friction. So we cannot only take away translational kinetic energy, for the object will still be rotating. Friction being the only acting force, the rotational energy must dissipate through friction as well.

    Friction over a distance equals work and tangential force over a distance (as distance = THETA / radius) is also equal to work. Since the friction can be applied to any point on the object and have the same effect (Since the basic identity of friction is f=uFn), I will argue that the work friction does reduces that total energy of the object, not just the translational energy.

    This should make sense for anyone who tries to put a nail through tough container board. You may be able to push the nail in a little bit by just shoving it in, but you can push the nail through if you add swivel it back and forth. I ask: if you cannot push down on the nail any harder without twisting, why should it go through with twisting? I respond: you are adding energy, as torque and force can be mathmatically transformed into to work in this context (multiplying force by distance, etc). So the total energy has been increased, but not the downward force (for as MTG says, rotational energy does not affect velocity or forward acceleration). This energy is combated by the work of the friction force, but there is some energy to push the nail through the container board, unlike the first example, where the force you apply and the friction are equal (thus you are unable to push it in any more).

    Of course, what I just wrote could be entirely bogus, but then explain to me why when I try to pull a nail out of wood with the tail of a hammer, if I cannot get it out by just pulling, the nail will come out when I pull and twist it simultaneously.

    Hancock the Superb
    Last edit: 9 years 5 months ago by Hancock the Superb.

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    9 years 5 months ago #24 by Marching Thru Georgia

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  • Hancock The Superb wrote:

    At this point, my limited knowledge of physics dissipates (pun intended), and I will proceed off of pure conjecture, based in logical reasoning.

    And again, this is where you get yourself into trouble. As I said before, hand waving arguments are fine in fluff subjects like history or social studies, but in the physical sciences you have to do the calculation to justify your position. The rotational momentum is orthogonal to the translational momentum. I think part of your answer shows you know this. So the question arises, how can forces at right angles to each other, influence each other? The answer is they can't. This should have been one of the first things you learned in high school physics.

    So now let's examine the nail and the wood puzzle. Since we know the twisting does not add any force to the nail being moved through the wood. What is going on? In the case you describe, there are two kinds of friction, static and kinetic. This you may very well not be aware of. Static friction must be overcome to get an object to begin moving. Once it does, kinetic friction must be overcome to keep an object moving. The two have different magnitudes. So in the case of the nail, the twisting simply overcomes the static friction. The pushing on the nail at the same time overcomes the kinetic friction.

    Do you see how that is a different problem from the case of an already moving projectile?

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    9 years 5 months ago - 9 years 5 months ago #25 by Hancock the Superb

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  • You logic seems completely correct. Yes, I do seem to have forgotten to take static friction into account. Such are the problems of a high school senior who can only draw upon his current physics class to help him. Out of curiosity, may I further ask where the rotational energy goes?

    As far as my hand waving arguments, my sentences are merely physics equations in sentance form. If you wrote each sentence in equation form, you would notice that the units line up correctly as well as the majority of the vectors (I am rarely willing to say I am 100% certain). My mathematical work was done. Static friction is one thing I forgot about. And what is the point of physics if you cannot make some argument from it? We don't have electron orbital models to needlessly fill miserable students minds with information, we have them because now we can make an argument about bonding patterns and such.

    Finally, I award you full points ;) for keeping a non-condescending tone in your second paragraph. It might be interesting to see what would happen if you could hold the disparaging comments in the first paragraph as well as previous posts. I think most people would agree that I have been as polite as possible to everyone on the forum, and I would hope that the same would be returned. I suggest this helpful advice because you have helpfully corrected my mistake this time (last time I think you misunderstood my study).

    On the bright side, next year I will have little free time (as I will be running and studying nuclear engineering), so you will not have to suffer through to have interesting physics debates with me.

    Hancock the Superb
    Last edit: 9 years 5 months ago by Hancock the Superb.

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