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Homework Help: What is the force of a shotgun blast?

  1. Jun 5, 2008 #1
    I found this thread already:

    However, it really doesn't quite solve what I need.

    If I were to put a ship into space and attached a 12 gauge shotgun to the back (so that the recoil went entirely into moving the ship) and then fired it, how much force would it produce to move the ship in the opposite direction?

    The equations are simple. I know the mass of the bullet (1 oz). I know the velocity when it leaves the shotgun (1610 feet per second).
    m = 1 oz
    v = 1610 ft/s
    a = vt
    F = ma

    I just need to know t. The problem is that I have no idea how long it takes for the bullet (a slug) to go from 0 up to the 1610 ft/s it is travelling when it leaves the barrel.

    One of the responders on the linked page simply assumed 0.1s. Is that a real value? Where would I find out the real value?
  2. jcsd
  3. Jun 5, 2008 #2
    Oops. I obviously meant
    a = dv/dt = v/t
  4. Jun 6, 2008 #3


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    Since the force is only applied for a brief time it is more useful to think of conservation of momentum so mv of the bullet will be added to mv of the ship.
    Basically the ship will get faster by 1610ft/s * ratio of the mass bullet/mass ship.
  5. Jun 6, 2008 #4
    I think a conservation of momentum approach would be kind of inaccurate because it fails to account for the gas released from the shotgun barrel which would also have a momentum. Therefore, there are really 3 elemnts of the system the gun itself, the gas released when the charge is ignited and the bullet spray. Which is a mess, especially if you consider that the bullet spray will likely not be straight and have symmetric trajectories. That assumes the gun would even fire in outer space
  6. Jun 6, 2008 #5


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    I hadn't considered a shotgun, but in a normal gun ideally the gas expands to reach nornal pressure at the point the bullet leaves the barrel

    Is there any reason the gun wouldn't fire in outer space? Unless it had cooled to the point that grease had frozen and seized. Explosives don't need air to burn, they contain their own oxidiser.
  7. Jun 6, 2008 #6
    Sneezing is not twice as forceful as a shotgun blast.

    I'm using this comparison to compare sneezing with shotgun blasts. The end result I got was that if X sneezes are required to get up to a certain speed then X*2 shotgun blasts are required to push the ship to the same speed.

    Obviously, I must have messed up either my calculation of the shotgun momentum or the force of the sneeze.

    According to this page, sneezes generate 2.9G worth of force.
    That works out to be 28.439285 newtons of force.

    So, the way to get the required number of sneezes compared to the number of shots is:

    numSneezes = shipMass * maxVelocity / sneezeForce * sneezesPerSecond

    numShots = shipMass * maxVelocity / momentumPerShot

    But, that simply doesn't work out to reasonable values. Assume the ship mass is 1kg and the max velocity is 1 m/s and the sneezes per second is 1. That results in

    numSneezes = 1 / sneezeForce = 1 / 28.439285

    numShots = 1 / momentumPerShot = 1 / 13.9119048

    So, the end result is that 1/28 of a sneeze is required to get a 1 kg ship up to 1 m/s. Whereas 1/13 of a shotgun blast is required. That's silly.

    Can anyone see what the heck I'm doing wrong?
  8. Jun 6, 2008 #7


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    I think this whole thing is silly. sneezeForce? Sneezing and shotgun blasts are both complicated events. You are going to have make a reasonable simplified model of each one to compare them and state what the parameters of those models are. I have no idea why a sneeze is 2.4g, is it acceleration on your head? You can't just scramble all of these different numbers together and get anything reasonable. And you didn't.
  9. Jun 6, 2008 #8
    It is silly. :smile: But I don't see why it can't be done.

    The shotgun part should be easy. Shooting a slug (a single projectile) means a certain amount of mass is pushed away from the gun at a certain rate. I'm pretty sure those numbers are correct. (13.91 kg*m/s) In space there won't be any air in the barrel, so that shouldn't matter.

    I'm not so sure about the sneezing stuff. I've figured that 5.421 grams of mass are expelled in a 1 second sneeze. That seems reasonable. But then what? What information do I need to calculate the recoil momentum of the sneeze?

    FWIW, the reason I'm doing this is because I want to add sneeze propulsion as one of the comparisons on my site mymindblewup.com. I've already got things like cow stacking (how many cows are required to reach the moon) and relativistic travel to reach a party at a particular date in the future.

    The entire point of the site is to have fun with crazy comparisons. Sneeze propulsion should fit right in if I can just get the calculations correct. :biggrin:
  10. Jun 6, 2008 #9
    Actually, I guess recoil is a force, not a momentum. It would be the force against the head. If the back of the head was positioned against a space ship, the force would serve to propel the ship. For the purposes of the though experiment, I'm assuming the sneezer is using bottled air to fill up their longs (since there is no air in space) but they are then sneezing into the vacuum to create propulsion for the ship.
  11. Jun 6, 2008 #10


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    As long as you agree it's silly, then ok. The thing to compare between two events is total impulse. Mass expelled times the velocity it's expelled at. I.e. change in momentum as mgb_phys was saying. Divide impulse of a sneeze (mass sneeze snot+air*velocity sneeze) into impulse shotgun (mass bullet*velocity bullet). Estimate the first anyway you want. But that will give you sneezes/shotgun blast.
  12. Jun 6, 2008 #11
    BTW, if it's not possible to figure this out using information available online then I was thinking of measuring it myself. Here is how I was thinking of doing this:

    I'd take a tube (say, a paper towel roll) and stick a ping pong ball in the other end. I'd then turn on a video camera and then sneeze into the tube. I could then use the video to determine what the rate of movement of the ping pong ball was when it started moving. I'd weigh the ball on a postal scale and voila, I'd have the momentum produce by a sneeze!

    Will that work? If not, can anyone suggest a better way to do it?
  13. Jun 6, 2008 #12
    But, I don't think I know the velocity of a sneeze. I know the volume of air per second that is expelled. But I can't use that to get the velocity.
  14. Jun 7, 2008 #13


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    If you're an engineer, i suggest the first paper roll experiment. It's actually not a bad suggestion. I'm a theorist, I'd just guesstimate everything. Besides, I've noted a huge variety in the strength of sneezes. I don't recall a 'standard sneeze unit'. I would guess a 'standard shotgun blast' would exceed a 'standard sneeze blast' by a factor of around 10^6. The velocity is on the order of a thousand times faster and the mass, uh, well, quite a lot.
    Last edited: Jun 7, 2008
  15. Jun 7, 2008 #14
    Ok, I'll give the experiment a try. I'd be happy to just guesstimate everything, but my first attempts at that produced my original silly result. I don't know what else to try there, so just doing the experiment should at least give me a ballpark number.

    I'll have to go buy a ping pong ball first though, so I unfortunately can't do it right now. I'll post back here after I get one and give it a try.
  16. Jun 7, 2008 #15


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    Guesstimate first. Velocity of a sneeze is 10-20m/sec. MAX. An ounce of snot would be a lot. All you need is mass and velocity. Good luck.
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