## Octave Chanute to Wilbur Wright

**Chicago, September 25, 1901**

I learn from various sources that some of the Engineers were much
interested by the address which you gave us. When will you give us the
manuscript? Do not be afraid of making it too technical.

Since you were here I have received Mr. Huffaker's notes taken during your
experiments. Shall I send them to you?

In looking over an old number of the *Revue de l'Aéronautique* I
find that after describing his machine as containing 14 square meters (151 sq.
ft.) with an arching of 1/12, it says:

Experiment showed that in calm air, the wings being practically horizontal,
it followed a trajectory inclined 9° below the horizon, with a velocity
of 9 meters per second [20.12 miles pr hour.] Previous experiments of Mr.
Lilienthal showed that for similar surfaces, with an angle of 9°, the
vertical component of the air pressure was 0.80 of the pressure on the same
surface, struck at right angles with the same speed. Admitting for the
pressure the formula P = 0.13V^{2}, where all is expressed in
kilograms, meters and seconds, we find in this actual case a vertical force of
118 kg., while the whole weight, machine and aeronaut, was really 100 kg.

Now 100 kg. is 220 lbs., and if the reported glide is correct we have with
Lilienthal's formula:

Lift 151 × 0.79 × (20.13)^{2} × 0.005 = 241 lbs.;
while by the Duchemin formula:

Lift 151 × 0.30 × (20.13)^{2} × 0.005 = 101 lbs.
Now this seems to corroborate Herring's statement that he descended in calm
air at a trajectory of 10° with a speed of 22 miles an hour, and the
puzzle remains why all the glides approximate an angle of 10°
irrespective of the wind velocity. Can you throw any light on this?

*Reply from Wilbur to Chanute,
September 26, 1901*