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At first, we shall follow the theory of the static deformation of the bow limb as introduced by Kooi . Let us assume that the limbs of the compound bow are inextensible, they are symmetric as in [5, 8], and can be represented by an elastic line of zero thickness. Then it will be sufficient to consider the upper part of the bow only. Let $$L_tot$$ be the total length of the limb measured from the riser end (the bottom) of the limb to the axle point along the limb. As a part of the limb is inside the modified part of the riser (the limb pocket), let L be the length of the free (elastic) limb measured from the point where the limb touches the riser to the axle point along the limb. Let us assume that the part of the limb which is inside the limb pocket is straight and rigid. This rigid part of length $$L_tot-L$$ is also seen in Fig.1 as the line GO.

The upper part of the compound bow in drawn position with related variables and forces. The size of the wheel system is exaggerated. In balance, the supporting force $$-\overlineK$$ prevents the limb from moving. The force $$\overlineK$$ acting on the limb tip T is the sum of components $$\overlineF_s$$, $$\overlineF_cl$$ and $$\overlineF_cu$$. The line segment GO is the rigid part of the limb inside the limb pocket, whereas the curve OT is the free limb. Note that the angle $$\alpha$$ is here negative

Let $$O=(x(0),y(0))$$ be the point where the free limb meets the limb pocket. Let us call the point G in Fig. 1 as the bottom of the limb, whereas the point O may be called as the bottom of the free limb. The eccentric system is fixed on the other end of the limb. Let us call the axle point $$T= (x(L),$$ y(L)) of this eccentric system as the tip of the limb, as noted in Fig. 1. Let us also introduce the angle

where x and y are the Cartesian x- and y-coordinates of the infinitesimal segment of the limb measured from the bottom of the free limb towards the tip of the limb, and l the length coordinate along the limb measured from the bottom of the free limb with length L towards the tip of the limb. The boundary conditions at $$l=0$$ are

In , it was supposed that the limb bends as in Hickman model with a slight modification, which is quite different case as here. However, we can still use the compound bow model introduced in  in order to find the limb tip y-coordinate and the angle $$\tau$$, for these values can be calculated without the Hickman assumption. From Fig. 1 we find that the y-coordinate of the limb tip with respect to the bottom of the free limb is

For straight limbs with constant width and thickness, for small deformations the path made by the tip of the (free) limb of length L can be approximated by an arc of a circle whose radius is 5L/6 and whose center is located at a distance of 5L/6 from the tip of the undeflected limb. Hence we have an estimate to parameter A used in ,

Now we assume that $$(x_H,y_H)$$ is the hinge point of the undeflected limb from where we can bend the straight rigid limb in such a way that the coordinates of the bottom of the free limb will satisfy Eqs. (34). The path of the tip is then on the arc of the circle, which center is at the point $$(x_H, y_H)$$ and radius is some constant C as illustrated in Fig. 5.

The EB model (curve) and the modified H model of the deflected limb. In the modified H model, the hinge point $$(x_H, y_H)$$ divides the limb in two rigid parts of length C and B. The hinge point is on the tangential line (dashed line) of the end point $$(x_B, y_B)$$ of the curve. Note that here the tip of the limb is in the point (0, 0) and the point $$(x_B, y_B)$$ is the bottom of the free limb

Thus for a straight even-wide limb of length L, the path made by the limb tip for small deformations is an arc of the circle whose radius is $$C = 5L/6$$ and whose center is located at a distance of 5L/6 from the tip of the undeflected limb. In this derivation we have assumed that the limb can deflect from the tip to the bottom of the free limb. So if we want to use this approximation in the model of , the constant A can be estimated as 041b061a72