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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.

1903/08, O. Chanute, "Table parameters glided flights, Kitty Hawk, 8 Oct. 1902", L'Aerophile, pag. 180 Source: http://gallica.bnf.fr/ark:/12148/bpt6k65534693/f190.item.r=wright

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 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.

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  • $\begingroup$ 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. $\endgroup$ – Russell Borogove Oct 2 '15 at 21:16
  • $\begingroup$ 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. $\endgroup$ – Robert Werner Oct 2 '15 at 23:53
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The gliders had no instruments of any kind, let alone an airspeed indicator. Thus the only way to measure airspeed was to measure the windspeed and add that to the speed over the ground, itself calculated from distance flown and time aloft for each flight.

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  • $\begingroup$ Airsick, yes, you are right but the distance measured along the ground is always the shortest theoretically possible flight distance. In practice the glider flies more. For example, if instead of 57.4 m (first flight in the table) the glider flew in reality 80 m its lift would have dropped form 58.1 to 47.6 kg/hp. There are also serious question about the accuracy of the glide angle. How did the Wright brothers measure it with such a precision? A quite small error in measuring the glide angle leads to a significant change in thrust. $\endgroup$ – Robert Werner Oct 2 '15 at 23:51
  • $\begingroup$ @Robert Why would the glider fly more? And if it did, why wouldn't that be accounted for in the table? It's quite likely that details of the experimental method are covered elsewhere in the document you've linked to. Don't forget, this relates to the early days of aviation. Much of what you take for granted now was only poorly understood then. $\endgroup$ – user11516 Oct 3 '15 at 0:42
  • $\begingroup$ Airsick, I will update the question with more explanations. (1) O. Chanute wrote, above the table, 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 adds traveled distance. (2) The glider was first lifted as a kite to a few m with the help of two men. This height makes the glide angle to be higher than that of the slope, which can be measured. For example, a change from 7deg20' to 9deg20' leads to only 114.4 kg / (114.4 kgf * sin(9 deg 20 minutes) * (54.7 m/12 s + 5.68 m/s) ) = 45.8 kg / hp instead of 58.1 kg / hp. $\endgroup$ – Robert Werner Oct 3 '15 at 7:07

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