# How does the lift-to-drag ratio depend on absolute sizes?

A dependency is provided here $$(L/D)_{\max} = \frac{1}{2} \sqrt{\frac{\pi \varepsilon ~ \mathrm{AR}}{C_{D,0}}}$$

AR significantly affects the outcome in this dependency. AR varies significantly among different aircraft. AR doesn't depend on the overall absolute size.

Does $$\varepsilon$$ or $$C_{D,0}$$ depend on absolute sizes? Is there a size threshold beyond which $$\varepsilon$$ or $$C_{D,0}$$ becomes bad?

The Reynolds number is approximately proportional to the sizes and velocity. For small sizes, it's easier to achieve laminar flow without turbulence. At first glance, a smaller chord seems to be better than a larger one. Is there possibly a limitation on the maximum value of the velocity-chord product?

However, the opposite is claimed in this answer to the similar question Is there a maximum lift to drag ratio? . There suggesting that increasing a span is necessary rather than reducing a chord. Is there probably an optimal range for a chord size?

This question is about the interaction of surfaces with air. This question is not about wing strength. This question is not about wing deformation under its own weight.

Does $$\varepsilon$$ or $$C_{D,0}$$ depend on absolute sizes?

$$\epsilon$$ is in the range 0 to 1 and mainly depends on the spanwise lift distribution, so it does not directly depend on the global size of the airplane. Values >0.8 are common. $$C_{D,0}$$ does depend on the size of the airplane since it represents its global aerodynamic drag (except the drag due to lift which is captured by the other terms in that equation).

The Reynolds number is approximately proportional to the sizes and velocity

Reynolds number is not "approximately" proportional to speed and size rather it is directly proportional to them being defined as:

$$Re=\frac{\rho Vc}{\mu}$$

where $$V$$ is the speed and $$c$$ the length of the body.

For small sizes, it's easier to achieve laminar flow without turbulence. At first glance, a smaller chord seems to be better than a larger one.

Not really. Laminar flow doesn't always imply less drag. The ubiquitous plot of the drag of a sphere versus Reynolds shows that quite vividly (plot source):

At higher Reynolds number the boundary layer becomes turbulent and that actually helps the airflow stay attached to the body and globally reduces drag. This is in general true also for an airfoil with the added benefit of obtaining a delayed stall i.e. more lift at higher AoA.

However, the opposite is claimed in this answer... suggesting that increasing a span is necessary rather than reducing a chord.

That answer states (correctly) that increasing the span (i.e. the aspect ratio) is a good way to reduce the (induced) drag. It also states that if this increase is done maintaining the wing surface constant then it brings a reduction in the chord and therefore in the Reynolds number with the consequences just said before about the sphere.

• The Reynolds number is not directly but is approximately proportional to velocity. Because reynolds number depends on viscosity and density. Viscosity depends on temperature and pressure. Density, temperature, pressure of gas depend on velocity, because gas undergoes compression near the wing. Commented May 20 at 16:14
• @Imyaf: yes and know. When those plots are traced, they always consider the freestream condition, i.e. the fluid far from the aerodynamic body. Otherwise the Reynolds number (or the Mach number or any other number) would change according to which precise point on the stream you are considering and that would make things simply impossible to be compared. Commented May 20 at 18:12