# 9.3.1 Schelleng’s bow force limits

In this section we derive the expressions for minimum bow force and maximum bow force used in Schelleng’s diagram. These build on earlier work by Raman, but for brevity I will call them “Schelleng’s” limits.

The first step is to calculate the slipping speed $v_s$ during an ideal Helmholtz motion. If the bow speed is $v_b$ and the bowing point is at position $\beta L$ on a string with vibrating length $L$, we can calculate $v_s$ easily from the condition that the integral of the string velocity over a complete period must be zero, in order that the string has no net sideways displacement. In an ideal Helmholtz motion with period $P$, the string is sticking for time $(1-\beta) P$ and slipping for time $\beta P$ in every cycle. So the condition is

$$v_b (1-\beta) P + v_s \beta P = 0 \tag{1}$$

so that

$$v_s=-v_b \left(\dfrac{1-\beta}{\beta} \right)=-v_b \left(\dfrac{1}{\beta} -1 \right) \tag{2}$$

and the velocity jump from sticking to slipping is $v_b-v_s = v_b/\beta$.

Schelleng’s formula for the maximum bow force now follows immediately from the graphical construction relating to frictional hysteresis, explained in section 9.2. In order for Helmholtz motion to be possible under the constraints of the hysteresis rule, the slip speed $v_s$ must be outside the range of the hysteresis. The limiting case is illustrated in Fig. 1, for a case with $v_b=0.1$ m/s and $v_s =1$ m/s. The sloping line passes through the point corresponding to $v_s$, and if it were any further to the left, $v_s$ would be inaccessible. We thus obtain the force criterion for the maximum bow force $f_{max}$ by equating two slopes:

$$\dfrac{f_{max} (\mu_s – \mu_d)}{v_v/\beta}=2 Z_0 \tag{3}$$

so that

$$f_{max}=\dfrac{2Z_0 v_b}{\beta (\mu_s – \mu_d)} \tag{4}$$

where $Z_0$ is the characteristic impedance of the string, $\mu_s$ is the maximum coefficient of friction during sticking (1.2 for the case plotted in Fig. 1) and $\mu_d$ is the coefficient of sliding friction at the slip speed $v_s$.

The condition for minimum bow force needs to take energy dissipation into account. Schelleng modelled the body as a mechanical resistance (or “dashpot”) with impedance $R$, assumed to be much greater than the impedance of the string, $Z_0 = \sqrt{Tm}$ where $T$ is the tension and $m$ the mass per unit length. This dashpot is driven into motion by the force at the bridge. For ideal Helmholtz motion, we know that this is a sawtooth wave. From the geometry, we can deduce the details: for a note with fundamental frequency $f_0$ it is a ramp with slope $Tv_b/\beta L$, interrupted by jumps of magnitude $-Tv_b/\beta L f_0=-2v_b Z_0/\beta$ using the standard formula

$$f_0=\dfrac{1}{2L}~\sqrt{\dfrac{T}{m}} . \tag{5}$$

The body dashpot will respond to this force with a velocity waveform of the same shape, divided by the factor $R$. We can integrate this to give the displacement at the bridge: one cycle can be written in the form

$$y=\dfrac{T v_b t^2}{2 \beta L R} + K, \mathrm{~~~~~}-\dfrac{1}{2f_0} < t < \dfrac{1}{2f_0} \tag{6}$$

where $K$ is a constant of integration. We have chosen an interval of time which makes the displacement waveform symmetrical, by choosing $t=0$ to occur in the middle of one ramp phase of the Helmholtz sawtooth. This ensures that successive cycles of the displacement waveform join up neatly, with a discontinuity of slope but no discontinuity of displacement.

This displacement at the bridge creates additional force at the bow. Schelleng used a simple approximation at this stage, by assuming that the bowed point is close to the bridge. The short section of string between bridge and bow can then be treated quasi-statically, because it is approximately straight for nearly all the time. It follows that the perturbing force at the bow is given by

$$f_{pert} \approx \dfrac{Ty}{\beta L} = \dfrac{T^2 v_b t^2}{2 \beta^2 L^2 R} + \dfrac{TK}{\beta L} . \tag{7}$$

The constant of integration is determined by noting that the perturbing force must be (approximately) zero during slipping because the total force is then uniquely determined by the friction curve. Slipping occurs at the moment when the displacement waveform has its slope discontinuity, corresponding to the flyback of the sawtooth, so we set $f_{pert}=0$ at $t=1/2f_0$. The result is

$$\dfrac{TK}{\beta L} \approx -\dfrac{T^2 v_b}{8f_0^2 \beta^2 L^2 R}= -\dfrac{Z_0^2 v_b}{2 \beta^2 R} . \tag{8}$$

For the Helmholtz motion to be possible, the force $f_{pert}$ must evoke an equal and opposite component of the friction force, which from equation (7) will be a parabolic arch with peak value $-TK/\beta L$ at time $t=0$. So finally, the condition for minimum bow force is that this value, added to the steady friction force $f_b \mu_d$, just reaches the limiting value $f_b \mu_s$. Thus

$$f_{min} \approx \dfrac{Z_0^2 v_b}{2 R \beta^2 (\mu_s-\mu_d)} . \tag{9}$$

This is Schelleng’s result.

Combining this with equation (4), we can deduce

$$\dfrac{f_{max}}{f_{min}}=4 \beta \dfrac{R}{Z_0} . \tag{10}$$

It follows that, within these approximations, there is a value $\beta=Z_0/4R$ below which the minimum bow force is bigger than the maximum bow force. In other words, there should be a limit to how near the bridge the bow can be placed, if Helmholtz motion is to be possible.