# Does a high “wing loading” require a high approach speed or does it cause a high approach speed?

Raymer says:

Landing distance is largely determined by wing loading. Wing loading affects the approach speed, which must be a certain multiple of stall speed (1.3 for civil aircraft, 1.2 for military aircraft). Approach speed determines the touchdown speed, which in turn defines the kinetic energy which must be dissipated to bring the aircraft to a halt. The kinetic energy, and hence the stopping distance, varies as the square of the touchdown speed.

My question is : In which manner does wing loading affect approach speed?

I'll explain my question better:

1. The first option:
Higher wing loading implies less wing surface. Therefore, you have a higher value of stall speed (because, for the same lift needed and the same CL and the same density, with a small wing surface area, you have to increase speed to compensate for less surface), so you need a higher value of approach speed. In this case : a higher wing loading REQUIRES a higher approach speed.

2. The second option:
Higher wing loading causes less wing surface. Therefore, you have less exposed surface, so you have less drag, so it’s easier to maintain a higher speed. So if this last sentence is correct (I’m not sure), in this case it’s not only the required speed, but the CAUSED speed.

What do you think?

• SOURCE: DANIEL P: RAYMER “Aircraft Design: A Conceptual Approach”, Chapter 5 “THRUST-TO-WEIGHT RATIO AND WING LOADING”, Paragraph 5.3 WING LOADING, Subparagraph: “Landing distance”

The lift equation: $$L = C_L \cdot ½ \rho V^2 \cdot S$$, with S the wing area. Wing loading is defined as L/S, equalling: $$\frac{L}{S} = C_L \cdot ½ \rho V^2$$.
With a given $$C_{Lmax}$$, a higher wing loading implies a higher speed. The first option is the correct one.