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Homework Help: Realistic bullet drop calculation

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A little while ago I made a thread about the drop of a bullet due to gravity on a bullet with a constant velocity. Constant velocity is unrealistic, so I am going for a little more realism here.

    I am doing this assuming that I have detailed information about distance traveled as a function of time (represented by x(t) ).

    The distance an object with 0 initial vertical velocity fall is
    [tex]
    d_{drop}=\frac{1}{2}gt^{2}
    [/tex].

    And a rough estimate for x(t) could be [tex]\sqrt{t}[/tex], and velocity v(t) would be [tex]\frac{1}{2\sqrt{t}}[/tex].

    Now that first equation requires some argument for time. Right now I have:

    [tex]x(t)=\sqrt{t}[/tex]
    and
    [tex]v(t)=\frac{1}{2\sqrt{t}}[/tex].

    So would I use the value of x(t) or v(t) in either of the last two equations, and rearrange to solve for t, and solve from there?

    [tex]x(t)=\sqrt{t}{\rightarrow}x^{2}(t)=t[/tex]
    [tex]v(t)=\frac{1}{2\sqrt{t}}{\rightarrow}\frac{1}{4v^{2}(t)}=t[/tex]

    Thus making the equation

    [tex]
    d_{drop}=\frac{1}{2}g{\cdot}x^{4}(t)
    [/tex]
    and
    [tex]
    d_{drop}=\frac{1}{2}g{\cdot}\frac{1}{16v^{4}(t)}
    [/tex]

    Is any of this incorrect?
     
  2. jcsd
  3. Jul 12, 2010 #2

    kuruman

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    Only the first equation is correct. In fact it represents x(t),

    [tex]x(t)=\frac{1}{2}gt^{2}[/tex]

    Given that, it is plain wrong to assert that

    [tex]x(t)=\sqrt{t}[/tex]
     
  4. Jul 12, 2010 #3
    the first equation is y as a function of time. the distance it has fallen.

    the second equation is how far the bullet has traveled horizontally. and like i said, sqrt(t) is just an estimate, a rough idea of how the graph of x(t) might look. obivously if i had data points for distance as a function of time, I could make a better x(t) equation
     
  5. Jul 12, 2010 #4

    kuruman

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    OK, I see. This is a two-dimesional projectile problem. You need two equations that describe the vertical and horizontal position as measured from the firing point. Note that in the horizontal direction there is no acceleration. This means that bullet covers equal horizontal distances in equal times. An equation that expresses this is

    x(t) = v0t

    where v0 is the muzzle speed. For the vertical direction you have

    y(t) = (1/2)gt2

    Now that you have the equations, what question do you wish to answer or what are you looking for?
     
    Last edited: Jul 12, 2010
  6. Jul 13, 2010 #5
    The velocity along the bullets path is not constant. If it was constant, the graph of x vs. t would be linear, and the graph of v vs. t would be a flat horizontal line. This is not the case. I figured x vs. t would be represented roughly by square root of t. x starts off increasing rapidly as the bullet moves fast, and then x increases less and less and velocity decreases.

    So if you had detailed information about x vs. t, you could use a computer or a graphing program to determine an accurate equation for x as a function of t. But for now, lets just keep it as square root of t. You can solve this equation for t. And in the case of square root of t, t = x2. So in the equation (1/2)gt2, you can substitute and get (1/2)g(x2)2, giving you (1/2)gx4.
     
  7. Jul 13, 2010 #6

    kuruman

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    How do you know it is not the case? Do you have such a graph?
    I agree. What is the information that you do have? Have you made measurements of the bullet's position as a function of time or are you just guessing?
    Your equations are unusable; t=x2 is not dimensionally correct so you cannot plug in numbers and expect a reasonable answer.
     
  8. Jul 13, 2010 #7
    Well it's definitely not linear. Bullets slow down over time. If the graph of x vs t were linear, the graph of velocity would be a straight line, implying constant velocity, which is not the case.

    And if I had the means to do a measurement, I would. I've been doing internet searched for position vs time graphs and havent got anything so far

    edit: its almost what im looking for here, http://www.bized.co.uk/images/bulletgraph.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Jul 13, 2010 #8

    kuruman

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    I have been looking at ballistics charts. They give speed as a function of distance in 100 yard intervals from zero to 500 yards. You could estimate the time per interval by using the average velocity.

    [tex]100 yards=\frac{1}{2}(V_{initial}+V_{final})t[/tex]
     
    Last edited by a moderator: May 4, 2017
  10. Jul 14, 2010 #9
    okay, it took some searching, but i finally found a chart of velocity vs. time

    http://www.inpharmix.com/jps/_images/t_ht_vel.gif

    so i used my graphing calc, found that the equation is roughly 2.5t4-30t3+135t2-307.5t+350

    that v(t), so if you integrate it, you get x(t)=.5t5-7.5t4+45t3-153.75t2+350t+Constant

    then if you integrate that from 0 to 4 seconds, you get 412 feet. i jus thave to figure out how to put that into the equations above. unfortunately, theres no easy way to solve for t, and its not dimensionally correct......

    edit: actually, what you would do, it integrate from whatever time frames you want to find out the total horizontal distance in that time frame.
     
    Last edited by a moderator: Apr 25, 2017
  11. Jul 14, 2010 #10
    but then how to use horizontal distance to solve for time......

    how would you find time given variable velocity and horizontal distance?
     
    Last edited: Jul 14, 2010
  12. Jul 14, 2010 #11

    kuruman

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    I have a question and a suggestion for you.

    What does the graph that you found show? It looks like the vertical component of the velocity as when the bullet is shot straight up. Is that what you are interested in?

    If so, then instead of a polynomial, I think you might try approximating the v(t) vs t graph as a decaying exponential of the form

    v(t) = v0e-k*t

    You can read v0 from the graph. To get k, you make a semi-log plot, i.e. you plot ln(v(t)/v0) vs. t. Note that this should give you a straight line with slope -k. You can then integrate to get x(t) that involves another e-k*t term. That is easier to invert to get t(x).
     
  13. Jul 14, 2010 #12

    ehild

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    There is a realistic situation for a projectile which is easy to solve and takes drag into consideration. Assume that the magnitude of drag is proportional to the magnitude of velocity , the force is parallel and opposite to the velocity.

    [tex]\vec F_D = K \vec v [/tex]


    [tex] \frac{dv_x}{dt}=-kv_x[/tex]

    [tex] \frac{dv_y}{dt}=-g-kv_y[/tex]

    At t=0, the velocity components are vx0 and vy0

    The solution for the velocity components:

    [tex] v_x=v_{x0} \exp{(-kt)}[/tex]

    [tex] v_y=(g/k+v_{y0}) \exp{(-kt)}-g/k[/tex]

    g/k is the terminal velocity. The coordinates can be obtained by integrating the velocity components.

    ehild

    Edit:

    [tex]F_D = - k m \vec v[/tex]
     
    Last edited: Jul 15, 2010
  14. Jul 14, 2010 #13
    damn, i overlooked that, i think this is a graph of the bullet being shot straight up
     
  15. Jul 14, 2010 #14

    kuruman

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    I take it that's not what you were looking for.
     
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