Gunning for Newton's Third Law

By Phil Plait | October 28, 2008 12:30 pm

BABloggee Vernon Balbert tipped me on this little potentially fatal application of Newton’s Third Law…

Obama Pictures and McCain Pictures

I suppose if the gun were pointing into the water it would be marginally less dangerous, though he’d lose a bit of backward momentum transfer due to the cosine of the depression angle, but still… might be worth it.

CATEGORIZED UNDER: Humor, Science

Comments (63)

  1. Thomas Siefert

    At best might cancel out to outboard motor, we can only hope that his gun mount is a bad as the rest of the design so that he won’t breed.

  2. though he’d lose a bit of backward momentum transfer due to the cosine of the depression angle

    I’ll be taking physics next semester to fulfill the requirements for p-chem, so in the meantime-

    Huh?

  3. Schwa

    People tend to overestimate the momentum changes when bullets get fired; the mass of the bullets is certain to be a lot smaller than the mass of the boat, motor, gun, ammo belt, and dude, so no matter how fast the bullets come out, he’s unlikely to go backward very fast. It’s the same way that bullets can’t knock down a person who gets shot; even if it transfers all of its energy into the victim, the shooter didn’t get knocked down to begin with, so it can’t have enough momentum to knock a person down.

    in b4 .700 Nitro Express

  4. TheWalruss

    With gas prices being what they are…

  5. Jim Shaver

    Chemist:

    In a right triangle with sides a and b and hypotenuse c, a*a + b*b = c*c.

    It’s easy to work out from there. :)

  6. Brando

    Exactly. That’s a “Ma Deuce” .50BMG machine gun. The receiver and barrel weigh a combined 75lbs or so. The average load for it is ball ammo which tends to be a 650-700 grain bullet at around 2600 fps. Sure, it’s got a lot of recoil compared to a .223 or .308, especially without a muzzle brake to counter it…however there’s a lot of mass in that weapon to begin with.

  7. Trebuchet

    He’s not going to go forward very fast either, with that motor!

    The Mythbusters episode on people flying through the air when shot was, to me, one of their better efforts. I think about it every time I see that in a movie now.

  8. guestwork

    Kids these days…

    I remember, back in our day, this photo was circulated with the tagline “Join the Canadian Navy! Now with a boat!” …

  9. I know what a cosine is, it’s the rest I find kind of gibbershy. I should note that I have never taken a formal physics class in my life, so everything I know about it is from pop-sci and playing around with electronics.

    I know, thanks to both Phil and the folks over at Astronomy Cast, all (okay, not all) about black holes and space-time. I know all about voltages, charge, impedance etc. But start talking about basic mechanics and it’s easy to stump me.

  10. Thomas Siefert

    Once saw a clip on the news of a Taliban soldier (back when they were the good guys, fighting the evil Soviets) sitting on a hillside firing a similar gun down into a valley. When he noticed the TV camera was pointed in his direction he stood and picked the gun up off the ground a started firing it from his hip. He only got a few rounds off before the recoil set him off balance and he fell and rolled down the hill. It was the funniest thing I ever saw from a war, yet I doubt that the TV crew dared even crack a smile.

  11. tai

    With the whole cosine bit, he’s saying since the gun would be pointed slightly downwards, some of the momentum would be in an upwards direction, not all of it backwards, so he wouldn’t go back quite so quickly as with the gun pointed straight ahead.

  12. Hah- would it help any if he had the motor running at the same time? It looks like that’s what he’s about to do….

  13. rob

    @ the chemist:

    do you know about vectors?

    momentum is p=mv. since velocity, v is a vector and m mass is a scalar, the momentum is a vector. if you decline the gun a bit to shoot down towards the water, you can resolve the momentum vector into a component vertical (into the water) and a component horizontal (parallel with the water surface). if you make a vector diagram of the momentum and its components, and choose the same angle as Phil, you find that the horizontal momentum component will be proportional to the sine of the angle and the horizontal momentum component will be proportional to the cosine of the angle. it is the horizontal component of the bullet momentum that contributes to the recoil that can push the boat backward via application of newton’s third law.

    what Phil refers to, is that in order to minmize the danger of the recoil of the gun pushing the boat backward, you would need to aim the gun down towards the water, so less of the bullet momentum is horizontal. this horizontal component is proportional to the cosine of the angle.

    you get the least recoil in the horizontal direction if you aim straight down, since the cosine of 90 degrees is zero. however, only submerged foes would be in danger if you aimed straight down.

  14. Yoeman

    Brando beat me to it, and said it better than I could. With a muzzle brake he would be okay, unless someone actually returned fire that is. :)

  15. Craig

    Do I see a possible Mythbusters episode coming out of this?

  16. MH

    Even if he doesn’t move backwards, his groin is still going to be taking a pounding if the gun works like I imagine it does.

  17. Davidlpf

    Another tagline “Things not to do with a dingy.”

  18. @ the chemist:

    do you know about vectors?

    Now my friend, we are speaking my language.

  19. Reid

    We’re learning about this stuff in physics right now.

  20. IVAN3MAN

    Quick calculation. . .

    Mass of gun (M2HB) with tripod and T&E: 58 kg

    Mass of dude: ~90 kg

    Mass of motor & boat: ~90 kg

    M33 (.5BMG) bullet ballistics: Velocity — 882 m/s; Mass — 52 g; p = mv = 45.864 kg-m/s

    Rate of Fire (M2HB): 450-575 rounds/minute; approx 9 rounds/sec.

    Combined mass of gun, dude, and boat: ~238 kg

    p / m = v

    Therefore: 45.864 / 238 = 0.1927 and 0.1927 x 9 (rate of fire) = 1.734

    So the recoil velocity of the boat would be ~1.734 m/s.

  21. Mitch Miller

    “what Phil refers to, is that in order to minmize the danger of the recoil of the gun pushing the boat backward, you would need to aim the gun down towards the water”

    Why not just not shoot the gun then? Usually when trying to shoot someting the position and angle of the gun relative to the target is not arbitrary.

  22. Mu

    My guess would be 9.8 m/s^2, downwards, after the recoil breaks his little plastic tub.

  23. Craig

    “…however, only submerged foes would be in danger if you aimed straight down.

    According to the Mythbusters, you don’t have to be very far under water to be safe from gunfire. The shell impacting on the water tends to fracture into pieces. They even fired a .50 calibre sniper rifle into water and demonstrated that it likely would have not been lethal. The only caveat to this is I’m not sure if they fired vertically into the water. My guess is that the water would still shred the shell before it got too deep.

  24. Steve B

    Why is this guy wearing a helicopter pilot’s helmet (see https://www.militaryitems.com/shop/product.php?productid=19712)?

    Anyway, I’d think the danger would be in firing the gun sideways. I think there’d be real risk of rolling and capsizing the boat.

  25. Ha! What could possibly go wrong!

  26. Cheyenne

    That Mythbuster’s episode was crazy! I couldn’t believe to see how quickly the bullets shredded. Who would ever have guessed that you are in more danger (under a few feet of water) from a handgun than a .50 caliber?

  27. chaboyax

    on mythbusters they shot with gun outside the water. i think if you angled this gun down it would be in the water what effect would that have on the shell?

  28. chaboyax

    ooops had another look maybe not

  29. Vector, schmector, I just wanna see him do it! If it’s on YouTube, somebody link to it!
    Rich

  30. Radwaste

    Guys – even IVAN3MAN – you’re naked.

    Why? The major factor in firearms recoil is NOT the acceleration of the bullet. It’s variable and never more than a hundredth of the total recoil.

    Where does recoil really come from? Barrel blowdown.

    Combustion gases have a final STP volume which depends on the reactant mass and peak gas temperature. The combustion event is so rapid that even though barrel heating saps energy from the exothermic reaction and some of the products is water vapor, the majority of excess volume – that bove the volume of the cleared chamber and barrel – is still vented after the bullet has fled.

    If any of you load your own ammunition, you’ll see that this is one of the reasons felt recoil does not follow bullet weight or powder charge weight closely.

    It is also the reason muzzle brakes work. All a brake does is disrupt the uniform flow of gases from the barrel during blowdown, thus removing the vector of thrust as the firearm turns into a solid rocket motor.

  31. He’s in the water, not on ball-bearings. There’s still a fair amount of friction, and I almost guarantee that something is installed to arrest recoil, e.g. by transferring some of the recoil force equally to both sides of the weapon.

    He’ll move, but not all that fast, and as long as he can keep a bead on his target, could still be very effective. Just because it looks silly doesn’t mean it is silly.

  32. rob

    @IVAN3MAN

    conservation of momentum implies mv=MV, so

    V=mv/M

    where m is mass of bullet, v is velocity of bullet M is mass of crazy person, gun and boat etc. and V is the velocity one round would give the crazy person in the boat:

    using your numbers V=(52 g)*(882 m/s)/(238 kg) = .1927 m/s

    this is less than a foot per second, which doesn’t *seem* very spectacular.

    to find the force exerted by continuous fire of the gun by the crazy boat person you can use newton’s second law

    F=dP/dt

    so on average, the force is the change in momentum for one bullet divided by the time between shots.

    using m,v and rate of fire the force is:

    F=(mv)*(r.o.f)= (52 g)*(882 m/s)*(9 1/s) = 412 Newtons or nearly 100 lbs force.

    to find the resulting acceleration of the boat use newton’s second law again, this time F=ma,

    a=F/m so

    a=(412 N)/(238 kg) = 1.7 m/s^2

    since the acceleration due to gravity is about 9.8 m/s^2, this means firing the gun gives the boat 1/5 of a gee of acceleration.

    Firing the gun would be quite exciting!

    (unless my math is wrong…)

  33. Mu

    chaboyax – if you put the muzzle under water you still have enough air in front of the bullet to get a air/water transition. If you fill the whole barrel with water, you need to accelerate the whole slug of water with the bullet. This in turn would most surely lead to excessive gas pressure in the chamber, blowing up your gun. Gun powder does not instantaneously combust but is carefully timed to give you a controlled release of gas during the passage of the bullet through the whole barrel (which is why you get excessive muzzle blast firing a stubnose revolver, lots of powder still burning after it leaves the muzzle).

  34. Radwaste, I suggest that you read the article on Recoil in Wikipedia. Click on my name for the link.

  35. Radwaste

    IVAN3MAN, I have, and it’s incomplete. I’ve corresponded with the late Jeff Cooper on the subject, as well.

    Do the Charles-and-Boyle’s Law thing for properties of an ideal gas, and apply that. Blank charges recoil when touched off, but the key is peak combustion temperature so that all of the reactants become products before anything leaves the barrel. In normal operation, gas escaping is restricted by the bullet, which offers resistance not only by barrel friction but by inertia.

    For instance, a quick look at a .308 Winchester (7.62mm NATO) round shows that to propel a 180-grain bullet to just 2400 ft/sec in an 18″ barrel, neglecting friction/bullet deformation forces in the throat and barrel, the average barrel pressure will be above 20 thousand PSIG (the bullet’s average acceleration is on the order of 59 thousand G!). And these are just ballpark. Pressure is higher than that, but not more than by a factor of 3 or so – it’s been awhile since I looked this up. Pressure is typically limited by the chamber’s/bolt’s ability to support the cartridge case.

    Hint: if barrel blowdown didn’t occur, muzzle brakes wouldn’t work at all. That is how they work, above.

  36. Radwaste

    Addendum to the above: “average”, in barrel pressure, is not with respect to bullet position, but to time (pressure does not fall off abruptly until after the bullet has actually lost contact with the barrel – the bullet is accelerated the whole time, still confining the charge). The bullet leaves the barrel in the example in under two-thousandths of a second from first motion. “Lock time”, the duration between trigger pull and ignition, makes this a bunch longer and depends on the gun.

  37. Byron

    A similar calculation for battleships:

    http://www.navweaps.com/index_tech/tech-022.htm

    and, of course, the awesome video:

    http://www.youtube.com/watch?v=nsNlmiLJGIw

  38. Mass of gun (M2HB) with tripod and T&E: 58 kg

    Mass of dude: ~90 kg

    Mass of motor & boat: ~90 kg

    M33 (.5BMG) bullet ballistics: Velocity — 882 m/s; Mass — 52 g; p = mv = 45.864 kg-m/s

    Rate of Fire (M2HB): 450-575 rounds/minute; approx 9 rounds/sec.

    Combined mass of gun, dude, and boat: ~238 kg

    Creating a machinegun-propelled ship: priceless.

  39. Thomas Siefert

    Pieter Kok, you said it best :-)

  40. Bobcloclimar

    chaboyax Says:
    on mythbusters they shot with gun outside the water. i think if you angled this gun down it would be in the water what effect would that have on the shell?

    Depends on how deeply into the water the barrel is inserted, since there will be water in the barrel. If it’s just the tip, it’s essentially the same as with the barrel out of the water. Further up the barrel, and the bullet may not have enough time to accelerate before striking the water, and thus may not fragment. Of course, the escaping gases will then tend to push the water out of the barrel, so there could be some impressive recoil coming off that thing – perhaps even an overpressure if too much of the barrel is underwater?

  41. Radwaste, thanks for that. If I ever find myself teaching thermodynamics I will make that an exam question!

  42. I think he is going to have a bigger problem with vibration and materials failure and have the pintle rip out and land in his lap. Most of the recoil is back towards the stern of the boat, on the other hand, when the barrel slaps forwards against its stops and the action closes on the next round, the whole gun moves forwards with a fair amount of energy, and from the look of its design, he has nothing to absorb it, no biggie on a tank or humvee, or even on M3 tripod spaded into the dirt. As soon has he closes the triggers, things are going to get very interesting.

  43. Radwaste, you say that the Wikipedia article on Recoil that I referred you to is incomplete; then you must have neglected to follow the link on the physics of firearms in that article. I’ll save you the trouble, just click on my name again.

  44. robomonkster

    I applaud this person. This is how great discoveries are made.

    Everyone else can go back to their chalkboards and study science. This guy is going to BE science.

  45. The heck with Mythbusters going after it. I wanna see Jeff Foxworthy explain it!
    ;)

  46. MarkH

    As a former member of the U.S. Coast Guard this boat is only marginally smaller than the one that I was stationed on ;) .

    Actually this is some navy squids attempt at dissing the Coast Guard. You see they are still jealous that the Coast guard is the oldest continuous sea going service in the United States. :)

    Also even with the motor that boat can’t wiegh more than 15 maybe 20 kilos.
    I think the thing is made of plastic. :P

  47. IVAN3MAN, Radwaste—according to my modest collection of useless pieces of knowlege, I also do think the recoil isn’t just a function of bullet mass, acceleration, and stuff. Gas pressure has indeed a lot to do with it. But what I’d like to throw in here is that you can’t calculate the recoil as long as you don’t actually fire the gun and measure it—quite a lot depends on how the weapon is designed, how the gas is dissipated, rerouted in the opposite direction, and/or reused for operating the lock and cooperating with the feeding mechanism. At least, that’s how a gunsmith explained it to me some years ago. I can’t remember all the details, but that design’s got a lot to do with how hard it will kick, that has stuck.

    ^_^J.

  48. @ Radwaste

    Addendum: I think that what you were referring to is secondary recoil; there is a link (click on my name yet again for it) at the bottom of the Wikipedia article on Recoil to an external web-site which mentions that fact.

    I did not include that factor in my “quick calculation” above because I don’t have the propellant mass data for the M33 (.5 BMG) cartridge.

  49. IVAN3MAN

    @ gyokusai

    I have already acknowledged that fact in my post immediately above this one, but, like I said, I don’t have all the necessary data to do a fully comprehensive calculation to take into account propellant gas momentum, so I left it out.

  50. Randy A.

    I appreciate all the learned discussion of recoil, etc. But I just want to see what happens if he actually fires the gun!

  51. Jeffersonian

    @gyokusai Says:
    October 28th, 2008 at 7:44 pm

    ” you can’t calculate the recoil as long as you don’t actually fire the gun and measure it”

    plus there’s the unknown variable on the amount of friction the boat has in the water. Don’t some of these models assume the gun is free-floating when fired?

  52. @Jeffersonian: Good point!

    Cheers,
    ^_^J.

    p.s. all our miscellaneous mathematics aside, the picture’s caption is just lovely.

  53. Unfortunately, the timing of this post may be a bit “off”, given this recent actually-fatal consequence:

    Boy, 8, killed while firing Uzi at Mass. gun show

    The boy, Christopher Bizilj of Ashford, Conn., was with a certified instructor and “was shooting the weapon down range when the force of the weapon made it travel up and back toward his head, where he suffered the injury,” a police statement said.

  54. What, exactly, is an 8 year old doing with an Uzi submachine gun on full auto in the first place—certified instructor or not? I don’t think that this here topic becomes necessarily “unfortunately timed” in the light of this event; rather, the event raises all sorts of other questions on top of this topic.

    ^_^J.

  55. How the hell can an 8 year old be a certified instructor? I’d suggest there is something wrong with the certification process.

  56. Oops my bad. Misread that, with a certified instructor. Ok, rephrase… Why the hell was a certified instructor handing over Uzi to an 8 year old? Shouldn’t they be using a 22 or something?

  57. alfaniner

    I think the picture would have been a lot funnier if he had a cannon instead.

  58. chaboyax

    thanks for putting me right guys knew some of you clever people would know and make me look silly AGAIN

  59. Nigel Depledge

    Interesting – I was thinking that the principal danger was to other users of the waterways.

    Then again, to maintain the acceleration, he would need to keep up continuous fire of the MG, and after perhaps 1000 rounds the barrel would start to distort as it heats up (would thermal expansion of the metal cause the bore to narrow?).

    I know that the SA80 in the sustained fire role has a cooling jacket on the barrel, and LMGs such as the Bren from WWII would normally be carried with a spare barrel, which needed to be changed after about 500 – 600 rounds of prolonged fire.

  60. Ken B, I feel a Darwin Award coming up…

  61. Nigel Depledge:

    would thermal expansion of the metal cause the bore to narrow?

    Given the fact that heating a too-tight lid on a jar can loosen it by expanding the metal, I would have to say “no”.

  62. This is increasingly becoming esoteric.

    ^_^J.

  63. Nigel Depledge

    After a bit more thought, it occurs to me that Ken Clark has a point about the vibration and about the working parts of the gun moving forward as they engage the next round and hitting their end-stop.

    Firing a Bren in fully automatic mode, it vibrates a good deal, but doesn’t kick the way a high-velocity rifle kicks. If anything, it pulls forwards. I think this is a consequence of the design of the gun.

    When a Bren gun is fired, the gas propels the bullet along the barrel, up to the gas regulator, at which point a portion of the gas is applied to a piston that drives the working parts backwards. The working parts are being driven back against a spring that, when the forces balance, brings the working parts to a halt and then, when the (diminishing) gas pressure exerts less force than the spring, drives the working parts forwards to load the next round into the chamber. As the working parts slide forwards, there is no spring that brings them to a gradual halt. Instead, they slam against their end-stop (which exerts a large force for a short period of time, on account of f = ma) and the firing pin (assuming the trigger is held down) is the only part that is free to move forward any further.

    Since the weight of the working parts is considerably larger than the mass of the bullet and the propelleant gases combined, I think the biggest change of momentum involved is that engendered by the working parts hitting their end-stop at the end of their forward motion.

    I think that considering the problem from the point of view of ideal collisions and momentum conservation is not the right approach – instead, it should be considered from the point of view of forces and accelerations.

    From this perspective, the relative masses become extremely significant, as does the drag of the water.

    If we arbitrarily say the force applied by the gas pressure in the barrel is 10 N, and the mass of the bullet is 0.01 kg, and the bullet is accelerated for 0.1 s, then it gains a velocity of 100 m/s at an acceleration of 1000 m/s2.

    If the mass of the boat etc. is 100 kg, and we assume no friction, it is accelerated at 0.1 m/s2 for the same time to end up moving at 0.01 m/s in the opposite direction. Given friction from the water, I cannot imagine how you would ever notice this imparted velocity.

    Now, of course my figures are chosen for convenience and only approximate the reality. However, if I’m right to within an order of magnitude (are there any machine guns that generate a muzzle velocity of 1 km/s?), you would never notice the momentum imparted from firing one bullet above the vibration of the gun. From 100 rounds, and still assuming no friction, you might impart 1 m/s to the boat, but in reality we do have friction.

    The boat is accelerated backwards at 0.1 m/s2, applying 10 N to the water. I have no idea how much friction water can apply to a boat, but a frictional drag of 10 N does not sound outlandish.

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