From: sfaber@ihlpb.att.com (Steven R Faber)
Newsgroups: rec.guns
Subject: Re: Rifling Twist Angles - stability
Date: 30 Sep 92 01:58:35 GMT
Organization: AT&T

Just noticed a couple dropped variables in that post so here are
the corrections.  The results are the same except there was
a mistake in the scaling of length by "L" part  - the
twist should be increased by L^3/2 instead of L^2.
-----------------------------------------------------
# Came across the book "Methods of Exterior Ballistics" written 
# just after WW1 that is a detailed mathematical treatment of 
# the subject, and has the answer we've been looking for.

Author: Forest Ray Moulton, Dover 1962 ed. reprinted from 1926 ed.

# 
# Instead of using the angular momentum over the transverse moment
# as a measure of stability, we should be looking at the rate of 
# change of the yaw angle and considering the moment of inertia around
# the transverse axis.
# 
# rate of change in yaw angle (y) = torque around trasverse axis/ It
  actually angular acceleration (y)

# 
# where "It" is transverse moment of inertia.
# 
Longitudinal is along its rotating axis and transverse is the "bad spin"
axis perpendicular to the direction of travel.

# Then the assumption is made that all the force acting on the projectile
# is due to the drag on the front cross sectional area and that is
# independent of yaw angle assuming small yaw angles.
# 
# Then scaling a given projectile by radius "r" including length.
# 
  y*It* sin(angle) =
     const (c) * dens of air (p) * area (r^2) *v^2* moment arm (r) sin(angle)


# 
# (It) is proportional to r^5  (1/2*m*r^2)
# 
# so:   y = c*p*v^2/r^2
# 
# 
# Then after an involved derivation he finds that a projectile is stable
# when 
# 
# 		y < 1/4*w^2*(Il/It)^2
# 
# where Il is the longitudinal moment of inertia and w is the spin rate.
# 
# so the stability ratio is:  substituting w=2*pi*v*twist(in^-1)
# 
# 		(v*tw)^2*(Il/It)^2*r^2/(p*v^2)  =
# 		
# 
# 		r^2*(Il/It)^2/p
# 		

 		(r*tw)^2*(Il/It)^2/p

# 		
# Note that velocity drops out!
# Of course this assumes all the velocity is obtained from the gun.
# 
# now substitute rifling angle (r.a) = 2*pi*r*twist   so the stab ratio =
# 
# 		(r.a.)^2*(Il/It)^2/p
# 		---------------------
# 		
# So stability is independent of velocity or the caliber! 
# Don't you love it when everything drops out:)
# Guess Bart's post really summed things up.
# 
# The above ratio implies that for greater stablity you want to
# concentrate the mass in the center of the bullet, so a longer projectile
# gives poorer stability, and denser air gives poorer stability.
# Stability is independent of the density of the projectile since it
# has the ratio of moments of inertia along the different axes.
# 
# The book mentioned that you can be over stabilized since the bullet
# has to follow an arcing trajectory, which could then lead to forces
# that could disintegrate the bullet if it is overstable.
# 
# 
# If we increase the length of a projectile by factor "L"
# and assume a ogival projectile  which has approximately the same
# moments of inertia as a cylinder:
# 
# the Il will be proportional to L and
# the It will be proportional to L^3
# 
Unfortunately we can't just substitute this since we have to go back
to the beginning and separate "r" from length "L"  when we scaled things
so the result is the rifling angle or twist must be increased by a
factor of L if all the force is on the front of the bullet or
L^3/2 if it all on the side of the bullet.

# so to keep the same stability we must increase the rifling angle
# by L^2.
(wrong)
# 
# 
# Steve