- 01 Dec '11 00:05 / 1 editI'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. - 04 Dec '11 05:26

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.*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.

GRANNY. - 04 Dec '11 15:27

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.*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. - 04 Dec '11 20:35

I have nothing to contribute on this but have a question.*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.

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 - 04 Dec '11 22:20

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.*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

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.