4.2
Yaw of Repose and Resulting Crossrange Deflection
Referring to Figure 4.11, as the bullet
flies, the principal aerodynamic force on the bullet acts directly
opposite to the velocity vector. The projection of this force along
the longitudinal axis is the drag force on the bullet. The drag
force acts through both the center of pressure and the center of
mass, and so it does not create a torque on the bullet. With a tiny
yaw angle (i.e., the yaw of repose) a very small component of the
aerodynamic force acts horizontally and sideward on the bullet.
This small sideforce, acting at the
Figure 4.11 Bullet Flight Characteristics
center of pressure, creates a torque on
the bullet equal to the force multiplied by the moment arm, that
is, the distance between the center of mass and the center of pressure.
The direction of this torque is downward for a right hand spinning
bullet, or upward for a lefthand spinning bullet.
To visualize this situation refer to Figure
4.21, which is Figure 4.11 viewed from directly above the bullet.
Note that there is no wind acting in Figure 4.21. The principal
trajectory parameters are displayed with the correct
relationships for a bullet with a right hand spin. The velocity
vector V is
in the trajectory plane and tangent to the trajectory path. The
aerodynamic force Faero
is directed opposite to the velocity
vector. The bullet has a spin angular momentum
H that
is directed along the longitudinal axis and forward for a righthand
spin. The yaw of repose is the small angle between the H
vector and the V
vector. This angle causes a small
component of the aerodynamic force, called Fside,
to act on the side of the bullet, and which can be thought of
as acting at the center of pressure of the bullet. This sideforce
creates a torque vector M on
the bullet. The torque vector M is
the vector cross product of the moment arm r,
which extends from the center of mass to the center of pressure,
and the sideforce Fside.
The direction of the torque vector M
is downward,
that is, perpendicular to the plane containing r
and Fside.
The torque vector does not point
exactly vertically downward, because of the inclination angle of
the trajectory, but it is exactly perpendicular to the plane of
r and
Fside.
Now consider the angular motion of the
bullet, which is governed by the equations of angular motion. A
key parameter in these equations is the angular momentum of the
bullet, which consists of two components. The first component is
the spin angular momentum H shown
in Figure 4.21, which is large in order to guarantee stabilization
of the bullet. Because the bullet rotates downward in the pitch
direction as it flies, a second component of
angular momentum is directed horizontally
in the direction opposite to Fside.
This component is so small that it
can be considered negligible compared to the spin angular momentum.
The magnitude of the spin angular momentum
of a bullet is nearly constant as the bullet flies. It changes very
slowly because the rotational frictional force and torque acting
on the bullet are small. Consequently, the change in the vector
angular momentum of the bullet as it flies is very nearly limited
to a change in direction of
the spin angular momentum vector H,
with no change in the magnitude of that quantity. Under this condition,
the equations of angular motion tell us that the angular momentum
vector H rotates
toward the torque vector M applied
to the bullet.
Consequently, the spin angular momentum
vector H,
which is always along the central axis of the bullet, rotates downward
toward the torque vector M caused
by the sideforce Fside.
So, in essence the sideforce causes the bullet to
rotate downward in the pitch direction to keep the axis of the bullet
almost exactly tangent to the trajectory curve as the bullet flies
along the trajectory arc. Of course, as the axis of the bullet and
the vector H rotate
downward, the vector M also
rotates at exactly the same rate, so that H
always remains perpendicular to M.
This entire situation is almost a steady state motion; everything
changes very slowly as the bullet flies. The yaw of repose angle,
the spin angular momentum magnitude, the torque magnitude, and the
sideforce are nearly, but not quite, constant as the bullet flies
from muzzle to target.
Because the sideforce Fside acts throughout
the flight of the bullet, a horizontal (crossrange) deflection of
the bullet will result. This deflection is generally small, but
it can be noticed, especially by longrange target shooters. This
is because the deflection increases as time of flight to the target
grows longer. Usually the observation comes about as follows. A
rifle is sighted in at point of aim, say, at 200 yards. Then the
range to the target is changed to 400 yards. The shooter makes an
elevation correction to the rifle sights for the longer range, and
a sighting shot (or group) is fired. The shooter notices that the
shot (or group) is deflected to the right (for a RH twist barrel)
by a few inches, but there is no crosswind to account for this deflection.
The shooter can apply a windage correction for the 400yard range,
and everything goes well at that range distance. Then, if the range
is changed to 600 yards, the shooter has the same experience. A
satisfactory sight elevation correction can be made, but shots will
be deflected a few inches to the right, necessitating a windage
correction even in the absence of a crosswind. The sideforce arising
from the yaw of repose is the cause of this unexpected crossrange
deflection of bullets. [Here we assume that the crosshairs in the
telescope (and the adjustment axes) are aligned precisely vertically
and horizontally, so that the sight adjustments are precisely vertical
and horizontal.] Crossrange deflections occur also for bullets with
lefthand spin, but the deflections are toward the left rather than
toward the right. For a bullet with lefthand spin, the spin angular
momentum vector is directed out of the tail of the bullet. To cause
the bullet to rotate downward in pitch, an upward vertical torque
is necessary so that the angular momentum vector will rotate upward.
This in turn requires a sideforce directed from right to left across
the trajectory plane, and this can result only from a yaw of repose
angle to the left of the trajectory plane (a small, noseleft angle
of the bullet as it flies). Consequently, the sideforce is directed
to the left, and the bullet deflects in the crossrange direction
to the left as it flies downrange.
