## Wilbur Wright to Octave Chanute

Dayton, December 3, 1900

Yours 2nd rec'd. The form of designation used in referring to us is immaterial but in so far as we have any preference it is for "Messrs. Wilbur and Orville Wright."

I think one half would be better than one third in stating the reduction in resistance of our machine and operator's position as compared with your machine and upright position. Our figures are only one third but I think that we have underestimated our resistance or you have slightly overestimated yours, and that the real reduction is only to one half instead of to one third the head resistance of your trials.

Our reasons for believing that the equivalent sq. ft. of the operator's body in the horizontal position is only 1/2 sq. ft. are as follows.

1. We tested the resistance of the machine with log chains upon it; also with a boy of equal weight and though the log chains offered almost no cross section we found the results almost the same. The resistance of the boy could not have exceeded a pound.

2. A bicycle rider has ridden 220 yards, without pace, in 12 seconds. This is 3,300 ft. per minute and 37 1/2 miles per hour. The resistance at 37 1/2 miles is 7 lbs. per sq. ft. so that each sq. foot requires 7 X 3,300 = 23,100 ft. pounds per minute. At 80 gear the rider makes 520 pedal strokes per mile, or 325 strokes per minute. The length of stroke is 1 1/6 ft. and the weight of rider 150 lbs. The extreme limit of power is 150 X 1 1/6 X 325 = 56,875 ft. pounds or nearly two horsepower. This is undoubtedly more than the man actually exerts. The friction of tires, bearings and chain requires a pull of 3 lbs. to overcome, or 3 X 3,300 = 9,900 ft. pounds per minute. Subtracting this from 56,875 leaves 46,875 [46,975] as the extreme power available for overcoming wind resistance. But the bicycle offers a cross section of 2 1/4 sq. ft. whose coefficient is probably 1/5, so its equivalent area is about 1/2 sq. ft. and the required ft. pounds 23,100 ÷ 2 = 11,550. But 46,875 - 11,550 = 35,325 is all that is now available for the resistance of the man; and his equivalent area is 35,325 ÷ 23,100 = 1 1/3 [1 1/2] sq. ft. But the cross section of a man, in racing position is a little over 3 sq. ft. so that his coefficient of resistance is less than one half. If no reduction be made in the area exposed it would require nearly four horsepower to drive him through the air at 37 1/2 miles per hour. If the coefficient for the bicycle racing position be less than 1/2 I see no reason for estimating the equivalent area of a man in the horizontal position at more than 1/2 sq. ft., and though I may be mistaken I doubt whether the resistance in the upright position is more than equivalent to 2 1/2 sq. ft.