```From: sfaber@intgp1.att.com
Subject: Re: [physics] ballistics of a bullet in flight
Organization: AT&T

From article <SWB.93Oct26191859@marvin.ae.utexas.edu>, by
swb@cfdlab.ae.utexas.edu (Steve W Bova):

# I ended up with a second order, nonlinear, ordinary differential
# equation for the downrange distance as a function of time.  I was able
# to solve this equation analytically. I then inverted that relationship
# to get flight time as a function of distance.  If anyone is interested
# in the details, I can provide them, but email is a bit awkward for
# this sort of thing.  I worked it out for any air density, and drag
# coefficient, but the following assumes a standard day, and a drag
# coefficient of 0.38, which is a reasonable value for a hemisphere.
# YMMV.
#
# Let
# D = bullet diameter
# M = bullet mass
# R = Range to target
# V = muzzle velocity
# T = flight time
#
# I found the relevant dimensionless group with respect to the
# importance of drag to be the ratio R/D.  If R/D is less than around
# 4,000 then the assumption that the bullet speed is constant is pretty
# good.

I like the idea of converting to a dimensionless variable.
The same thing was done to solve the internal ballistics equations
in the books I was reading.

# For my wife's M1 carbine this gives a "critical range" of around
# 100 ft.  After 100 ft, the effect of drag should be included.
#
# Now, introduce the dimensionless quantity
#
# k = 0.0926 (D^3)/M grains/inch^3.               (1)
#

I got .02926 for the constant, could this be a typo?

# Note that D should be  in inches, and M in grains for the above
# formula to be dimensionless.  The time of flight is then given by
#
# T = (exp(k*R/D) - 1.0)*D/(V*k)                  (2)
#

I see how you derived it and I got the same result.  Recently I was
playing around with a similar idea about doing an analytic solution,
only perhaps it is the next step up in your model.
Since the coeff. of drag varies quite a bit from 0 to 1500 ft/sec
then slopes down slowly from 1500 ft/sec on.  It can be fit to a line
pretty well (from my barrel twist posts).  I thought of substituting
Cd = a - b*v  and solving the diff. eq. analytically.
It can be integrated to the velocity (time) level but I haven't gone
further yet.  I don't think you would be able to explicitly solve for
the time in the end, but you could probably find it fast enough with
a couple iterations.  This might make a good pocket calulator algorithm
for finding bullet drop as accurate as the Ingalls tables in the region
of 1500 ft/sec to 3600 ft/sec.

Steve Faber

```

```From: sfaber@intgp1.att.com
Subject: Re: [physics] ballistics of a bullet in flight
Organization: AT&T

From article <SWB.93Oct29100531@marvin.ae.utexas.edu>, by
swb@cfdlab.ae.utexas.edu (Steve W Bova):

# #>>>> "SF" == sfaber  <sfaber@intgp1.att.com> writes:
#
# SF>   #From article <SWB.93Oct26191859@marvin.ae.utexas.edu>, by swb@cfdlab.ae.utexas.edu (Steve W Bova):
#
# SF>   #
# SF>   # Now, introduce the dimensionless quantity
# SF>   #
# SF>   # k = 0.0926 (D^3)/M grains/inch^3.               (1)
# SF>   #
#
# SF>   I got .02926 for the constant, could this be a typo?

Sorry, you were right, I found I left out a factor of "pi".

# SF>   I see how you derived it and I got the same result.  Recently I was
# SF>   playing around with a similar idea about doing an analytic solution,
# SF>   only perhaps it is the next step up in your model.
# SF>   Since the coeff. of drag varies quite a bit from 0 to 1500 ft/sec
# SF>   then slopes down slowly from 1500 ft/sec on.  It can be fit to a line
# SF>   pretty well (from my barrel twist posts).  I thought of substituting
# SF>   Cd = a - b*v  and solving the diff. eq. analytically.
# SF>   It can be integrated to the velocity (time) level but I haven't gone
# SF>   further yet.  I don't think you would be able to explicitly solve for
# SF>   the time in the end, but you could probably find it fast enough with
# SF>   a couple iterations.  This might make a good pocket calulator algorithm
# SF>   for finding bullet drop as accurate as the Ingalls tables in the region
# SF>   of 1500 ft/sec to 3600 ft/sec.
#
# Accounting for the variation in Cd during flight is an interesting
# idea.  Surely it could be done on a programmable pocket calculator, if not
# analytically.  But before I did that, I would look for a better
# value of Cd. The big weak spot in my (Cd=const) model is the value
# used for Cd.  This number can vary considerably, depending on the shape of
# the bullet.  For example, most round-nosed bullets have  a
# considerable cylindrical "shoulder" attached to the nose that will
# alter the Cd from that of a hemisphere.  And then there are
# boattails...I haven't had a chance to do any library work yet, but
# there has got to be some tables of experimentally determined Cd's for
# ballistic shapes in our engineering library.

My idea did not pan out too well.  The equation could be integrated OK
to velocity, but when you then solve for the velocity you get a pretty
complicated expression (with the help of a symbolic math program) that
involves a "W" function after which there is no hope of integrating again
to get distance vs. time.  One might as well fit the whole drag curve
to a function that is simple to integrate and you would be just as well
off.
You could get a good value of Cd for your formula by taking the
Cd at the average velocity from my fit to the Ingalls data, and then
using the usual ballistic coefficient defined for the bullet to find the
form factor "i".  Cd would = Cd standard * i

Cd (standard) = 1.146 - 1.047E-4 * v   v= 1500 to 3600 ft/sec

i=m/((2*r)^2*BC) , r in (in.), m in (lb)

Of course then you lose some "purity" from your formula.
If you can find some methods to predict some Cd's from pure theory
based on shape of the bullet I'd be interested, or what would be
really interesting is if you could get us going with some CFD
to compute the drag :).

Steve Faber

```