Chicago, September 13, 1901

I received your letter of 10th yesterday and the slides are now being made. I will procure and enclose a few cards of invitation to your address, in case you wish to invite some friends.

As our Superintendent at the tie works was away I have had to run them myself. This took all my time. He got back yesterday and I now take up your very interesting letter of August 29th.

I may not have time today to discuss all your computations, but I will take up a few of them.

The most puzzling thing to me is the very low head resistance which you report. We measured the pull on our own unloaded machines in 1896, and got results confirming in a general way the estimate given by Mr. Herring. It did not seem to me possible that the embedding of the spars in the cloth, as you have done (we did not do this), should so greatly have reduced the head resistance, but perhaps the sharp curve at the front has also some influence.

You say in your Note 9: "The dynamic efficiency of the unloaded machine appears to be greatest at 10 meters per second. The reasonable inference would seem to be that at this point the drift & head resistance are equal, each being 7.5 lbs."

This seemed reasonable to me, but it was not proven. I have just made calculations to get at it otherwise.

The first thing which appears is that all the angles were over measured, and the lift is in excess, even by the Duchemin formula. I therefore assumed reduced angles which brought the lift nearer the weight, and from this new angle computed the drift. This brought it down below the "pull," which includes both drift and head resistance, and I figure the latter as follows:

The average is 3.3 square ft., which for 10 meters pr second, or 22.37 miles per hour, gives 3.3 × 2.5 = 8.25 lbs., a gratifying agreement. If we add 1 sq. ft. for the man (and it seems to me this is small enough), we will have at 24 miles per hour 4.3 × 2.88 = 12.38 lbs. head resistance, which agrees pretty well with the 10.4 lbs. you have figured.

I cannot agree equally well with your estimates of this. You figure separately the angle of descent required to overcome the head resistance, and to overcome the drift, and then you add them together. You thus obtain an aggregate angle of 11°, which will lead to the following result:

But even with the 40 lbs. the dynamic results come out askew. 24 miles an hour is 35.2 ft. per second, and the work done on the air is 40 × 35.2 = 1,408 ft. lbs. The speed over the ground is 19 ft. per second and the angle 10°, hence 19 × sin 10° (or 0.17368) = 3.30 ft. fall per second; and the work done by gravity comes out 240 × 3.30 = 792 ft. lbs., instead of 1,408 ft. lbs.

Moreover, if I understand you correctly, you seem to have taken the angle of descent as the angle of relative incidence. Now the machine falls 3.30 ft. per second, and the speed is 35.2 ft. per second, so that the relative angle of incidence would be: 3.30 ÷ 35.2 = 0.09403 or sine of 5° 23' instead of 10°.

To get back to this angle of 10°, which seems to fit the case with the Duchemin formula, we have to assume that the machine made a further angle of about 4 1/2° with the relative wind, and was thus horizontal itself. This agrees fairly well with the photographs.

I have tried to figure out this glide with the Lilienthal coefficients, but they do not fit at all.

I note what you say that the speed of greatest efficiency is between 31 & 32 miles per hour, yet the glides of August 9th were at relative speeds of 31 to 36 miles per hour, and were no flatter.

I will figure out my own glides, and discuss them with you when you come to my house. Please let me know by what train to expect you.

Another letter from Chanute to Wilbur, September 15, 1901