What is the angle of incidence of a propeller in the terminology of the Wright Brothers and how did they obtain it?
In Feb. 1903 (see Screw Test 1903, page 8), using a 0.58 hp engine, the Wright brothers tested a propeller characterized by:
Diameter, D = 8.5 ft
Pitch = 15 deg
Surface, $S = 2\ ft^2$
RPM = 245
They measured what appears to be the static thrust and got:
Thrust = 18.75 lbf
Based on the assumption that the propeller can be modeled by a wing (Pitch = 15 deg, Surface = $2\ ft^2$, Lift = 18.75 lbf) having its entire area concentrated in a single point (called the Center of Pressure, C.P.) traveling at RPM = 245 on a circle with the diameter 0.824 x D, $V_{wing} = RPM \times \pi \times 0.824 \times D$, the Wright brothers evaluated its Drag and from Lift, Drag and Pitch using trigonometric relations they obtained the Normal (perpendicular to the chord) and Axial or Tangential (along the chord) components of Lift + Drag (the resultant force) and some angles related to them.
Now comes something I do not quite understand.
The two brothers calculated a reference force defined as:
Total Normal = $k \times S \times V_{wing}^2$ = 24.8 lbf,
where
$V_{wing} = RPM \times \pi \times 0.824 \times D = 61\ \mathrm{mph}$
$k =$ Smeaton’s coefficient $= 0.0033\ \mathrm{lbf} \cdot \mathrm{ft}^{-2} \cdot \mathrm{mph}^{-2}$
Further, the Wrights divided the Lift by the Total Normal
$\frac{\mathrm{Lift}}{\mathrm{Total\ Normal}} = 0.756$
and from this 0.756 they drew the conclusion that the angle of incidence is 7¼ deg. The question is why? What is this angle of incidence and how can it be calculated?
Update
The correspondence between the percentage 0.756 and the angle of incidence, 7¼, comes likely from a table like this (which is for a slightly different wing curvature):
Percentage of air pressure at various angles of incidence
However, as the propeller did not move, the incidence angle of its equivalent wing appears to be identical with the pitch of the same wing and they are not. Why?