Woodward—Air-Ship Propeller Problems. 7 
That is to say, a complete mechanism consisting of propel- 
lers and a 60 horse-power motor, which when anchored can pro- 
duce a thrust of 650 tbs.—that being the thrust required when 
a certain air-ship is moving 15 miles per hour —can actually 
drive that air-ship only 13.3 miles per hour, unless the limit of 
60 horse-power is exceeded.* 
7. NUMERICAL RESULTS. 
The following table is of value in estimating the power required 
with propellers of various sizes for pulling or lifting different 
amounts when the jrame ts anchored in still air. The propellers 
are supposed to be ideally perfect in design and construction, 
and no allowance is made for cross currents and for friction. 
TABLE SHOWING HORSE-POWER WHEN P THE THRUST, PULL 
OR LIFT, AND THE RADIUS OF THE PROPELLER, OR THE 
TOTAL PROPELLER AREA ARE GIVEN. 
r=radius |A = total area 
Tifelnibs | ofeauvalent ofall propeller wer required 
+ 1 1 3.14 0.029 
4 1 3.14 0.23 
100 1 3.14 29.00 
1 5 78.53 0.006 
100 5 78.53 5.8 
400 5 78.53 46.4 
400 10 314.12 23.2 
650 8.1 206 .00 59.5 
900 10.4 339 .93 voc0 
* Throughout this paper I mean by one ‘“horse-power” 550 foot-—lbs. 
of real “‘work” per second, I make no use of a so-called “‘nominal horse- 
power.” 
+ Inarecent number of “Motor” (London), Mr. Rankin Kennedy says: 
“‘It would be a simple matter to prove by calculation that the power 
required of a propeller to sustain one pound weight in the air is 0.03 B.H.P. 
in any case, theoretically, 0.03 B.H.P. must be allowed for every pound 
weight to be lifted.””’ Mr. Kennedy then goes on to say, that it would take 
only 12 HP. to lift or sustain 400 lbs.!_ The statement is dangerously loose. 
It would be true only on condition that the effective area of the propeller be 
also increased 400 times! With the same propeller, it would take 240 horse 
power to lift his 400 pounds! See Formula [X]. 
