*Octave Chanute to Wilbur Wright*

*Chicago, July 12, 1903 *

I send you herewith a copy of my article for the French paper. You will readily
identify the 10 photos I sent of your glides.

I enclose also a revised schedule of some of the glides of Oct. 8th '02, in
which I have corrected the wind speeds as per Marvin's rating of the anemometer.

I have been trying to decompose the various elements of the resistance, i.e.,
drift, tangential, and head resis[tan]ce, to agree with the observed facts of
angles and speed, and find that they do not agree very well.

The first discrepancy is in the speed. You say that your apparatus needs a
relative speed of 18 to 20 miles an hour for support but all the observed speeds
are greater than this.

The next discrepancy is in the head resistance. You say that it amounts to
3 or 4 pounds at 20 miles per hour, but you give 8/3 as the effective area,
and using 0.0033 as coefficient of air resistance we have (8 x 202 x 0.0033)
/ 3 = 3.52 lbs.; while you said at Kitty Hawk that the head resistance was 30%
of the whole, which on 30 lbs. would amount to 9 lbs.; on this basis the effective
area would be 9 / 1.32 = 6.81 sq. ft. It is my impression that the latter figure
is more nearly correct than the first.

I think this can be determined by computing individual flights from the actual
observations and working backward from the propulsion. In doing this I will
use your own coeff[icient] of 0.0033.

Take glide No. 10, Oct. 8th. Speed 26.68 miles per hour. Pressure per square
foot 26.682 x 0.0033 = 2.35 lbs. Angle of descent 6 20' + 2 of chord = 8 20'.

Assume angle of incidence of wind 8 . As I do not know the coefficient of lift
of your surface I have to compute the normal, backward.

Normal 251.5 + x cos 8 or 0.99 = 254 lbs.

Drift 254 lbs. x sin 8 or 0.139 35.30 lbs.

Tangential 305 sq. ft. x 235 x 0.035 (Lilienthal) 25.08 "

Omitting head, leaves net resistance 10.22 "

But the propulsion is computed at 27.74 "

Hence left for head resistance 17.52 "

This gives equivalent area 17.52 2.35 = 7.4 square feet.

I would be glad to have the computations of glides 10, 12, 17, 18, 19, &
20, in which we guessed at the angle of the chord with the horizon, if you have
made any of them.

*Wilbur Wright to Octave Chanute,
July 14, 1903*