Trebuchet

The range of the projectile is $R=v^2 \sin 2\theta/g$ , where $v$ is the initial velocity and $\theta$ is the angle of the velocity vector to the horizontal. The initial velocity is determined by energy conservation, $\frac{1}{2} mv^2 =Mgh-mgh'$ , where $m$ is the mass of the projectile, $M$ is the mass of the counterweight, $h'$ is the elevation of the projectile before it is launched and $h$ is the drop of the counterweight. Therefore $R= 2\left(\frac{M}{m}h-h'\right)\sin 2\theta$ . When the lengths are multiplied by a factor 4, each mass is multiplied by $4^3=64$ , but the ratio of the masses does not change. The heights are multiplied by 4 and the range is multiplied by the same factor.

Let $\alpha$ be the factor around which all lengths are scaled such that $l '= \alpha l$ . Since all masses are proportional to their volume, the mass scales according to $m '= \alpha^3 m$ . Besides lengths and masses, the gravitational acceleration $g = \text{const}$ is the only important quantity.

Solution 1: Dimensional analysis

In simple terms, the trebuchet consists of two masses $m_1$ and $m_2$ and a lever arm with the lengths $l_1$ and $l_2$ . The only dimensionless quantities that can be formed from this are the ratios $\mu = m_1/m_2$ and $\lambda = l_1 / l_2$ , which are independent of $\alpha$ . Thus, the maximum range $s$ of the catapult can be represented by $s = l_1 \cdot f(\mu, \lambda)$ where $f(\mu, \lambda$ ) is an dimensionless function. In particular, $s \propto l_1 \propto \alpha$ So if you build a catapult, which is larger by a factor of 4, the result is a factor of 4 greater range.

Here, we assumed that there is no air resistance. If, for example, there is a frictional force $F_r = - c v^2$ with the coefficient of friction $c$ with the unit kg / m, then another dimensionless quantity $\gamma = c l_1/m_1 \propto \alpha^{-2}$ can be formed, which explicitly depends on the dimension $\alpha$ of the catapult. In this case, without a physical model, no statement about the range of the catapult would be possible.

Solution 2: Physical modelling

The energy source of the trebuchet is the potential energy of the counterweight with the mass $m_1$ . If $l_1$ is the length of the short lever arm, this results in a potential energy of $E_\text{pot} \propto m_1 g l_1 \propto \alpha^4$ When the projectile is thrown, the counterweight is practically in free fall, so that practically all the potential energy is converted into kinetic energy of the mass $m_1$ .(Provided, that the masses of the lever arm and the projectile are negligible small.) $\begin{aligned} E_\text{kin} &= \frac{1}{2} m_1 v_1^2 \approx E_\text{pot} \\ \Rightarrow \quad v_1 &\propto \alpha^{1/2} \end{aligned}$ The movement of the counterweight is transmitted via the lever arm to the stone of the mass $m_2$ . Thus, for its speed also applies $v_2 \approx v_1 \frac{l_2}{l_1} \propto \alpha^{1/2}$ with the length $l_2$ of the long lever arm. If the projectile is thrown with the initial velocity $v_2$ , then with an optimal throw angle of $\theta = 45^\circ$ we get an distance $\begin{aligned} s &= 2 \frac{v_2^2}{g} \\ \Rightarrow \quad s & \propto v_2^2 \propto \alpha \end{aligned}$ (Here we have assumed that gravitation is the only force acting on the projectile and in particular that there is no air friction.)