Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. Subscriber joe shmo
    Strange Egg
    01 Dec '11 00:05 / 1 edit
    I'm trying to find position as a function of time for a fired bullet, but I'm missing something because I end up with a dimensionalized exponent in the final result...

    I start with

    -F_d = ma (1)

    F_d = drag force
    m= mass of bullet
    a= acceleration

    assume F_d is porportional to the square of the velocity in (2)

    F_d = C*rho*A*v^2/2 (2)

    C = coefficient of drag
    rho = density of medium
    A= crass sectional area of bullet
    v = velocity of bullet

    by eq(1&2) with initial condition that @ t=0, v= v_o

    C*rho*A/(2m)*INT(t,0)[dt] = -INT(v,v_o)[dv/v^2]

    which yeilds

    v= [C*rho*A/(2m)*t + 1/v_o]^(-1) eq(3)

    for the sake of simplicity let

    z=C*rho*A/(2m)

    then

    dx/dt = v

    INT(x,0)[dx] = INT(t,0)[ v dt]

    x= 1/z*ln( 1 + z*v_o*t)

    Edit: Nevermind I exponentiated it after the previous step and just became confused...Its funny how writing it out a second time made me realize that.
  2. Standard member smw6869
    Granny
    04 Dec '11 05:26
    Originally posted by joe shmo
    I'm trying to find position as a function of time for a fired bullet, but I'm missing something because I end up with a dimensionalized exponent in the final result...

    I start with

    -F_d = ma (1)

    F_d = drag force
    m= mass of bullet
    a= acceleration

    assume F_d is porportional to the square of the velocity in (2)

    F_d = C*rho*A*v^2/2 (2)

    C ...[text shortened]... ep and just became confused...Its funny how writing it out a second time made me realize that.
    Very simply joe, you forgot to take into consideration the bullet's ogive when you made your calculations. I made the same mistake when i was your age.


    GRANNY.
  3. Subscriber joe shmo
    Strange Egg
    04 Dec '11 15:27
    Originally posted by smw6869
    Very simply joe, you forgot to take into consideration the bullet's ogive when you made your calculations. I made the same mistake when i was your age.


    GRANNY.
    The shape of the bullet is accounted for in the coefficient of drag. The equation predicts the velocity of the bullet at a hundred yards within 0.5% of the manufacturers experimental results at that range...so i'm happy with it.
  4. 04 Dec '11 20:35
    Originally posted by joe shmo
    I'm trying to find position as a function of time for a fired bullet, but I'm missing something because I end up with a dimensionalized exponent in the final result...

    I start with

    -F_d = ma (1)

    F_d = drag force
    m= mass of bullet
    a= acceleration

    assume F_d is porportional to the square of the velocity in (2)

    F_d = C*rho*A*v^2/2 (2)

    C ...[text shortened]... ep and just became confused...Its funny how writing it out a second time made me realize that.
    I have nothing to contribute on this but have a question.

    Does the drag coefficient differ depending on whether the bullet velocity exceeds Mach 1? The charts I have looked at indicate that muzzle velocity is >Mach 1 generally. So, MV would decrease to below Mach 1 at some point, and then, do drag considerations change at this point?

    http://en.wikipedia.org/wiki/External_ballistics#The_transonic_problem
  5. Subscriber joe shmo
    Strange Egg
    04 Dec '11 22:20
    Originally posted by JS357
    I have nothing to contribute on this but have a question.

    Does the drag coefficient differ depending on whether the bullet velocity exceeds Mach 1? The charts I have looked at indicate that muzzle velocity is >Mach 1 generally. So, MV would decrease to below Mach 1 at some point, and then, do drag considerations change at this point?

    http://en.wikipedia.org/wiki/External_ballistics#The_transonic_problem
    Yeah, the coefficient of drag is a function of the Mach number. The coefficient is pretty constant for a projectile at M<.9, around Mach = 1 a sharp increase occurs, and with M>1 it decreases non-linearly, appearing to sort of leveling off after M>6.

    The equation I derrived doesnt take this curve into consideration(the Reynolds number was > 10^4, and M>2 througout the range distance). So I just used the maximum C around M =1.