4 Mathematical models

The first model developed is Hickman's model [3, pp.13-17] of a straight-end bow, which consists of a straight rod of wood, bent by placing the string. The position of the bow is specified in an Cartesian coordinate system (x,y), the line of symmetry of the bow coinciding with the x-axis and the origin O coinciding with the midpoint of the bow, analogous to the situation for the working-recurve bow in Figure 3. The limbs of the bow are assumed to have a uniform thickness and their width tapers to the tips. In that case the bending stiffness is given by EI(s)=EI(0)(L-s)/L where E is Young's modulus, I is the second moment of inertia and s is the length parameter along the limb with s=0 in the origin and s=L at the tip. The mass per unit of length is formula, where symbol is the density and C(s) is the area of the cross-section. The mass of one limb is then given by

formula 2

The relationship between the loading and the deformation is derived by use of the Euler-Bernoulli equation:

formula 3

where M is the bending moment at s. When the limb is deformed by a force F perpendicular to the limb at the tip, while it is clamped in the center of the bow, integration gives for small deformations, after substitution of I(s) and M=F(L-s):

formula 4


formula 5

In order to get a simplified model for the dynamics we assume that the shape of the moving limb equals at each moment the shape of the limb deformed by the force F giving the same deflection of the tip given by equation (4). In that case there is a simple relationship independent of time between the position of the tip x(L,t) and of all the other points along the limb, see equation (5). The kinetic energy is

formula 6

and the potential energy equals

formula 7

Hence we end up with a one degree of freedom system of which the state variable is the deflection x(L,t) of the tip. This is because of the relationship between x(s,t) and x(L,t) which is independent of the time t. The equation of motion can be derived using the Lagrangian formalism, with the Lagrangian function
for the conservative system:

Formula 8

formula 9

formula 10

This result shows that the equations of motion can be interpreted in terms of a lumped model. The mass of the limb is represented by a mass formula at the tip of the limb and the flexible limb is replaced by a massless rigid rod of length L, connected with an elastic hinge to the grip at s=0. The elastic hinge is a composition of a smooth hinge with a clock spring with elastic constant k=2EI(0)/L.

Now Hickman did not place the hinge at the clamped end (s=0), but at s=(1/4)L because he noticed that the trajectory of the tip is an arc of a circle with its centre at s=(1/4)L on the unbent limb. Then the elastic constant becomes k=(18/16)EI(0)/L. To derive the place of this point he used the fact that for small deflections the form of the bent limbs is nearly an arc of a circle. Let the radius of this circle be r, then we have the following parametric representation for the curve of the tip:

formula 11

The radius R of this curve is given by

formula 12

where for instance x' denotes dx/dr. For formula we have R=(3/4)L, and this shows that the trajectory of the tip and the trajectory of the tip in the hinge model are in good agreement when the position of the hinge is placed at s=(1/4)L.

Geometry of the bow in Hickman's model
Fig.7: Geometry of the bow in Hickman's model.

We are now in the position to derive the governing equations for the lumped parameter model. From Figure 7 we find for the x coordinate symbol of the midpoint of the string, where the arrow with mass formula is placed:

formula 13

The half length of the string is l, L0 is the half length of the grip, equal to (1/4)L in Hickman's model, and L1 is the length of the rod, equal to (3/4)L. Finally phi is the angle of rotation of the rod. When the unstrung bow is straight, we take the unloaded hinge for formula. The Lagrangian for the upper half of the bow is, using the coordinates and xi:

formula 14

where J is the moment of inertia of the limb with respect to pivot point S. Due to equation (13) we have one superfluous coordinate, which should be eliminated. Equation (13) gives

formula 15


formula 16

Substitution of this result in Lagrange's equation yields the equation of motion:

formula 17

The initial conditions are fixed by the solution of the static problem, where xi equals the draw length |OD| .

formula 18

where the subscript f indicates the fully drawn situation. The equation of equilibrium becomes:

formula 19

where F(|OD| ) is the weight of the bow. The initial velocity of rotation is zero: formula

In this model the arrow leaves the string for  phi=phi0, where the subscript b indicates the braced situation, when

formula 20

In [5] it was shown that in this model the bow converts all of the deformation energy of the elastic hinge into kinetic energy of the arrow. The energy integral of the system gives the solution of the equation of motion:

formula 21

(Note that for phi=phib , but that * is finite.) This completes the derivation of Hickman's model with a massless string. The mass of the string can easily be incorporated in the model by placing one third of its mass at the midpoint of the string and the rest at the ends. The extension of this model with an elastic string, obeying Hooke's law, introduces an extra state variable, so the model becomes two degrees of freedom, for instance the half-length of the string and the position of the arrow. The obtained equations look similar to those derived by Marlow [6] and are not reported here. They should be solved numerically, with a Runge-Kutta method for example.

Marlow [6] described a lumped mass model for the longbow. Unfortunately he let the limb rotate as a rigid body about the beginning of the limb at the place where it meets the grip. In that case for the bow described above the tip mass representing the mass of the limb, becomes mb/16 instead of mb/15, which makes the dynamic behaviour less realistic [5, p.111]. On the other hand Marlow introduced an elastic string to get a more realistic model as to the efficiency. He treated the string as a rod rigid with respect to shear and bending but elastic in the length direction as given by Hooke's law. This explains why in his model the arrow leaves the string while the string is already through its stretched position. This can be considered as an artefact of the model. A string is not capable to withstand shear forces, so it seems better to use a simple lumped mass model for the string.