THE BASICS OF RISER ROTATION DURING THE POWER STROKE
An idea that seems to float around in archery is that the force the bow hand exerts on the riser when at full draw
still exists after the draw hand has released the string. So on release the bow hand generates a forwards force on the riser.
In this case to minimise any riser rotation the bow centre of gravity should coincide with the grip (pivot point) or at
least be level with it. This view is incorrect. The force on the bow hand at full draw results from the combination of the
string draw force and the gravity force (weight) of the bow. When the string is released the draw force component disappears
and what's left is the gravity force acting on the bow hand. An additional force on the bow hand is generated by
the recoil of the bow.
The following discussion describes the initial (instantaneous ) behaviour of the riser (ignoring gravity) after release of the bow string.
We have a stationary unconstrained riser on which a (recoil) force acts above the grip position.
The starting point will be Mass x Acceleration = Force which I hope most people will be happy with.
We have some object of overall mass M to which a force F1 is applied at some distance b from the object's Centre of Gravity at X. The obect is fitted with a spindle (at right angles to the applied force) at a distance a from the object's Centre of Gravity. When the force F1 is applied the object will try to rotate about some axis. The object at the spindle will in general try to move and so a force
F2 will need to be applied to the object by the spindle to prevent any movement. The force is assumed to act for a short enough time dt that the force F1 doesn't change in magnitude, position or direction. (e.g. with an arrow, as it rotates both the drag force and the effective position of where the drag force acts both change).
If you integrate Mass x Acceleration = Force with respect to time you get:-
Mass x Velocity = integral over time of the Force
Mass x Velocity is called the (linear) Momentum. The linear momentum that the object gets comes from the integral over time dt of the force F1 (call it J1) and the integral over time dt of the force F2 (call it J2). The linear momentum the object ends up with (mass x velocity) has to equal J1+ J2, the linear momentum you put in. The velocity were talking about is the velocity of the object's
centre of mass (COG). The object has to rotate around the spindle so the COG travels tangentially to an arc of a circle of radius a. The velocity of the COG is therefore aA where A is the angular velocity of the object around the spindle. So what we end up with as far as the linear 'conservation of momentum' goes is:
MaA = J1 + J2 (equation 1)
The rotational equivalent of Mass x Acceleration = Force is
Moment of Inertia x Angular Acceleration = applied Torque
If you integrate the above with respect to time (same way as the first case)
you get:
Moment of Inertia x Angular Velocity = integral over time of the Torque
The torque is the force x distance to the point of rotation i.e
the torque = (a+b) x F1
The integral over time dt of the Torque is therefore equal to
(a + b) J1
The moment of inertia of a mass around a rotation axis is the mass multiplied
by the square of the distance to the rotation axis. Suppose the overall moment
of inertia for rotation of the object around the spindle is Is.
We define a value Ks as:-
Ks2 = Is
/M
Ks is called the Radius of Gyration. If all the
object's mass were located at a distance Ks from
the spindle then the Moment of Inertia (the rotational properties) would
be the same as the object.
The Moment of Inertia x Angular Velocity = integral of time of the Torque
can, for the object, be therefore written as:-
MKs2A = (a + b)J1
(equation 2)
rearranging equation 2 you get:
A = (a + b)J1/MKs2 (equation 3)
Substituting the expression for A in equation 1 you get:-
Ma(a + b)J1/MKs2 = J1 + J2 (equation 4)
Rearranging equation 4 you get:-
a(a + b) = Ks2(1 + J2/J1) (equation 5)
Equation 5 is what you could call a 'weathervane' equation. It tells you
what force you get between the object and the spindle for a given applied
force/geometry. (In this case it describes the force exerted on the bow hand as a result of the bow recoil).
As you move the spindle about the value of J2 (i.e.F2) will vary. For some position of the spindle J2 (and
hence F2) will be zero i.e. the object rotates about the spindle with
no force between the spindle and the object. In this case removing the spindle
will make no difference to how the object rotates. The condition J2
= 0 in equation 5 therefore defines the axis of rotation for a free
object (no spindle). Putting J2 = 0 into equation 5 gives you:-
a(a + b) = Ks2 (equation 6)
As Ks for any realistic object cannot be zero it follows
from equation 6 that a can never be zero i.e. under a single applied
force for a free body the COG can never be the natural axis of rotation. You will often hear that "the object"
rotates at the centre of mass but this is always qualified by something like "and the centre of mass travels in a
curve". i.e. if you define a rotation axis to be at the centre of mass then to make the numbers add up there has to be
a second rotation axis about which the centre of mass rotates. The centre of mass can never be the true rotation axis.
You can tidy equation 6 up a bit (with a bit of sleight of hand) by applying
a theorem related to Moment of Inertia called the Parallel Axis Theorem.
This gives you that :-
Ks2 = Kg2 + a2 where Kg is the (constant) object radius of gyration calculated with respect to rotation around the COG.
Substituting for Ks in equation 6 gives a simple definition
of where the rotation axis of free object under a single applied force is
located i.e.
a = Kg2/b
In summary the recoil force (with no pivot/bow hand) acts to initially rotate the riser about some axis not at the
centre of centre of gravity. The bow grip is normally located between the centre of gravity and the point at which the
recoil force acts. The sense of rotation acts to drive the riser into the bow hand so the rotation axis of the bow shifts
(nominally instantaneously) to being the grip pressure point. The force exerted by the riser on the bow hand (F2 in the above discussion) is that required to
shift the bow rotation axis to the grip.
