ATMOSPHERIC FRICTION. 
275 
friction on an elementary band of radius r and of width 
dSj into components at right angles and parallel to the axis, 
and integrating the latter, we have 
R = 2 k J" r f a ds. ~ = 2 iz f s rdx = nfA, 
in which f is the average unit friction and A is the area of 
the longitudinal section of the solid of revolution. 
Let us now apply this formula to compute the resistance 
of the Zeppelin balloon at a speed of, say, 10 feet a second. 
The balloon is a cylinder, with ogival ends of 1.5 calibers; 
the length is 390 feet; the diameter is 39 feet. Hence the 
longitudinal section may be taken as roughly equivalent to 
a rectangle 39 feet wide by 350 feet long, the area being 
13,650 square feet, approximately. Now, the average fric¬ 
tion on a plane surface 350 feet long, at 10 feet a second, is 
0.000366 of a pound per square foot. Hence, by the formula 
just established, R = n f A, the skin-friction on the entire 
convex surface is 0.000366 X 13,650 X 3.1416 = 15.7 pounds. 
The pure head resistance of prow and stern is about 61.6 
pounds, as determined by the writer’s unpublished experi¬ 
ments on spindles. Hence the total resistance of the balloon 
is 77.3 pounds, approximately, and thus the friction is about 
20 per cent, of the whole resistance. 
The value just computed of the ratio of friction to total 
resistance seems very small, but that is because the balloon 
is so blunt-ended. If, however, the cylindrical part be pro¬ 
vided with a two-caliber prow and nine-caliber stern, the re¬ 
sistance, figured as in the last paragraph, would be: Fric¬ 
tion, 16.5 pounds; pure head resistance, 15.6 pounds ; total 
resistance, 33.1 pounds. Thus the friction is about one-half 
of the entire resistance. 
Analyzing in a similar way the resistance of street cars 
and railway trains, it is seen that for a short, blunt car the 
skin-friction is of small consequence; for a long train it may 
equal, or exceed, the head resistance. When cars are run at 
