# Regarding the 1902 Wright glider. Why is the airspeed, Va, calculated adding the ground speed, Vg, and wind speed, Vw?

Regarding the 1902 Wright glider. Why is the airspeed, Va, calculated adding the ground speed, Vg, and wind speed, Vw?

In August 1903, Octave Chanute published in L'Aerophile a table (see attached) showing the measured quantities:

(1) distance traveled along the ground (col. 2),

(2) flight time (col. 3),

(3) wind speed (col. 5),

(4) glide angle (col. 7),

corresponding to a few flights performed on Oct. 8, 1902. From these parameters and the total mass of the machine and pilot, O. Chanute calculated that the 1902 glider lifted about 62 kg/hp (see the last column).

It is not quite clear for me why Va = Vg + Vw (see col. 6). The airspeed should be constant, independent of the headwind speed, as long as the glide angle is the same because the thrust is the same. It is true Va = Vg + Vw but if Vg added to Vw does not give always a constant Va, for the same glide angle, then something is wrong with the measurements. The water speed of a boat traveling upstream depends only on its thrust and is independent of the river speed.

The lift in the table (~62 kg/hp) is quite big in comparison to the lift of the best gliders of the time, mentioned by O. Chanute a few lines below the table, that lifted only 45 kg/hp.

The precise value of the airspeed, Va, is essential in calculating the lift. If ones takes it only 1-2 m/s lower, due to miscalculations, then the evaluated lift grows artificially.

Update

There are a few sources of errors that could have altered substantially the results in the table.

(1) O. Chanute wrote, above the table (see the link), the Wrights did not drift, more than 1/4 in a arc of a circle from the straight line, to always have good headwinds. This drift increases the traveled distance.

(2) The glider was first lifted as a kite to a few meters above the ground with the help of two men. This height made the glide angle to be higher than that of the slope, which can be estimated from a picture (see the attached image where the glide angle appears to be 9.01 deg). If instead of 7 deg 20' the slope angle was 9.01 deg and the 1902 machine started from a few meters above the ground then instead of 58.1 kg / hp it would have lifted less than 114.4 kg / (114.4 kgf * sin(9 deg 20 minutes) * (54.7 m/12 s + 5.68 m/s) ) = 47.4 kg/hp.

(3) The third factor of incertitude is the wind speed that varies with altitude being stronger a few meters in the air than close to the ground. Wright glider, 1902. Undated picture.

Question: Do you know a modern replica of the 1902 glider that lifts ~62 kg/hp? If such a thing exists then the 1902 machine could have lifted 62 kg/hp.

• As an open cabin plane of very light weight, the drag from the pilot's body and clothing could vary substantially flight-to-flight based on minor positioning details, so I would not expect that airspeed would be constant for given glide angle and thrust. – Russell Borogove Oct 2 '15 at 21:16
• Russell Borogove, the drag, D, in kgf (col. 9) is calculated as D = m * sin(a) = Thrust, where a = the glide angle. As you see, D has nothing to do with changes in the position of the pilot or other things as long as the glide angle is constant. Whatever the pilot does reflects implicitly in the glide angle. As long as a = ct. the pilot induces the same drag. – Robert Werner Oct 2 '15 at 23:53