Bend a bow slightly

Let us describe the bow with the $y=y(x)$ function, where the $x$ axis runs along the string, and the $x=0$ point is at the center of the string. In this frame of reference $d=y(0)$ .

The inverse of the radius of the curvature of the bow is proportional to the torque created by the tension force, $1/\rho =\alpha \tau$ , where $\alpha$ depends on the rigidity and the cross section of the bow material. The torque is $\tau=Fy$ , where $F$ is the tension in the string. According to differential geometry, the inverse radius of curvature can be expressed as $\frac{1}{\rho}= \frac{y''}{(1+y'^2)^{3/2}}$ . Therefore we get

$\frac{y''}{(1+y'^2)^{3/2}}=\alpha Fy$

If the bow is only slightly bent $y'^2<<1$ and the equation simplifies to $y''=Fy$ . The solution is $y=d \cos \pi x/\ell$ , where $\ell$ is the distance between the end points of the bow (the length of the string). At the center of the bow

$\frac{\pi^2 d}{\ell^2}=\alpha F d$ or

$F=\frac{\pi^2 }{\alpha \ell^2}$

The force does not depend directly on the distance $d$ , but there is an indirect dependence, since $\ell$ is a function of $d$ . To obtain that, we express the original length of the bow as and integral for the arc length:

$s=\int_{-\ell/2}^{\ell/2} \sqrt{1+y'^2} dx=\int_{-\ell/2}^{\ell/2} \sqrt{1+a^2 \sin^2 x} dx\approx\frac {\ell}{\pi} \int_{-\pi/2}^{\pi/2} (1+\frac{1}{2}a^2 \sin^2 \eta) d\eta= \ell(1+a^2/4)$

where $a=\frac{\pi d}{\ell}$ , we introduced the integration variable $\eta = \frac{\pi x}{\ell}$ and we expanded the square root for $a<<1$ . Finally we get

$s \approx \ell + \ell \left(\frac{\pi d}{2\ell}\right)^2$ .

We can re-arrange this to

$\ell \approx s - s \left(\frac{\pi d}{2s}\right)^2$

when $d/s$ is small. We can see that the difference between $s$ and $\ell$ is quadratic in the small parameter $d/s$ . Accordingly, the force is, in a good approximation, independent of the distance $d$ for $d<<s$ .

Here is an example: Take a straight bow that is 1.5m long. We need to set the string to a length of $\ell\approx 1.48$ m if we want the bow bend so that $d=10$ cm. In order to get $d=20$ cm, we need to shorten the string by another 5cm. The force will change by 3.3%.

Note: for the calculation of the arc length we can also use $\int_{-\pi/2}^{\pi/2} \sqrt{1+a^2 \sin^2 \eta} d\eta =2 \frac {\ell}{\pi} E(-a^2)$ . The result is an elliptic integral of the second kind, that can be expanded for small values of $a^2$ , yielding $E(-a^2)\approx\frac {\pi}{2}(1+a^2/4-....)$ and the final result is the same.

Consider one half of the bow as a straight rod of length L clamped at one end. This end is the middle of the bow. Call this end C, and the free end of the rod A.

At A apply a force at right angles to CA to displace the end of the rod a distance d. For values of d small compared to L we can assume that the force required is proportional to d: F = kd. This force would cause a torque at C:

tau = kd.L.

However, in this situation the bending torque on the half-bow is not supplied by the imagined force at right angles to the bow but by the tension T in the string:

tau = T.d, d being the perpendicular distance from the string to C.

Therefore Td = kdL and T = kL, which is independent of d (for small d).

Let alpha be the angle between the end of the bow and the string. Approximate the bow as a blade spring (for which the amplitudo is linear to the perpendicular force, for small alpha). The amplitudo itself is also proportional to alpha, for small alpha. The perpendicular force caused by the string tension is proportional to tan(alpha) times the string tension (for small angles tan is proportional to alpha).

So

• the force needed to deform the bow and
• the effectiveness of the direction of the string tension to do that

are both linear to alpha (in first approximation) and their proportionalities cancel out for small alpha.