So we have also derived the solution to the constrained problem. optimal solution is also a catenary. Substitute, the given parameters in the above equation. $$R_{Max}=\left [\frac{P_t \sigma {A_e}^2}{4\pi \lambda^2 S_{min}}\right ]^{1/4}$$. This method gave better insight into the design problem graphically, but the equations were a little more convoluted. The economic limit will be 20-40% offset from the singularity. The potential energy of a given shape of the chain between the pulleys is: We have integrated only along the section between the pulleys, because also be a catenary. construct an unconstrained-problem such that gravitational potential energy. Of note, I assumed 400 W*hr/kg specific energy for the battery pack. In general, Radars use directional Antennas. The amount of power, which is reflected back towards the Radar depends on its cross section. $$K_e=\frac{c_b}{g} \frac{C_L}{C_D} \eta_i \eta_m \eta_p (1-f_{we}-\frac{gR}{c_b\frac{C_L}{C_D}\eta_i\eta_m\eta_p})^2$$. The space is interesting: If the chain is taut, We know the following formula for operating wavelength, $\lambda$ in terms of operating frequency, f. Substitute, $C=3\times 10^8m/sec$ and $f=10GHZ$ in above equation. We also care about the "cost" of increasing the range requirement of an aircraft (longer range aircraft need to be heavier). The one-variable method allows us to look at the impact of hybridization on the maximum takeoff weight (MTOW) of the airplane, which is important because MTOW is a key indicator of complexity and cost. $$P_r=\left (\frac{P_tG}{4\pi R^2}\right )\left (\frac{\sigma}{4\pi R^2}\right )A_e$$, $$\Rightarrow P_r=\frac{P_tG\sigma A_e}{\left (4\pi\right )^2 R^4}$$, $$\Rightarrow R^4=\frac{P_tG\sigma A_e}{\left (4\pi\right )^2 P_r}$$, $$\Rightarrow R=\left [\frac{P_tG\sigma A_e}{\left (4\pi\right )^2 P_r}\right ]^{1/4}\:\:\:\:\:Equation\:6$$. we can define at ground level, and ignore constants (the energy of the (no Lagrange Next, I considered a hybrid-electric aircraft where power is supplied exclusively by electricity but power is generated by either a turbine with generator or a battery pack. techniques for dual-energy aircraft. Intriguingly, there is a term that appears identically in the denominator of both expressions, which I will call $$K_e$$. We will get the following equation, by substituting $R=R_{Max}$ and $P_r=S_{min}$ in Equation 6. radar range equation represents the physical dependences of the transmit power, which is the wave propagation up to the receiving of the echo signals. $$A_e=\frac{G\lambda^2}{4\pi}\:\:\:\:\:Equation\:10$$, $$R_{Max}=\left [\frac{P_tG\sigma}{\left (4\pi\right )^2 S_{min}}(\frac{G\lambda^2}{4\pi})\right ]^{1/4}$$, $$\Rightarrow R_{Max}=\left [\frac{P_tG^2 \lambda^2 \sigma}{\left (4\pi\right )^2 S_{min}}\right ]^{1/4}\:\:\:\:\:Equation\:11$$. any constraints but resulting in lower energy. Applying Euler-Lagrange directly, we find: We could solve this, but the simpler approach is to use a theorem: Where the last step is applying the original form of the Euler-Lagrange