as Floats for Paddle-wheels, 355 



And the moment of the resistance on them 

 = i / o / ft)V 5 {-573869-(l+2A;)x-513256 + |(l + 2/:) 2 x -493765 



-Fx-10746-£ 4 x -05372 + £ 5 x -01035} 

 = ip y a)V{-184054--£x-532747 + Fx -493765 -k 3 x -10746 



-£ 4 x -05372 + Fx -01035}. 



1. When the vessel is going at a uniform speed, the amount 

 of useful work done by the resistance in a unit of time is Rv, 



Uv 



and the amount done by the engines is Mo>, therefore ^— is a 



measure of the efficiency of the paddle-wheel. 



(1) In the case investigated in my former paper k = 0- } therefore 



Uv _ -149092 l_. n . 

 Ma> ~ -184054* 2 



(2) Now, if we suppose that all the portions of the floats which 

 are within a distance of half the radius from the centre of the 

 wheel to be cut away, then k = ^. Substituting this value for k 

 we get 



Resistance = J p, co q r 4 x -04266, 



Moment = J p t wV x -05674. 

 Now 



^(1 + 2*). 



And when the vessel is going at full speed, the measure of the 

 efficiency is 



B f __l ^04266 5 4266 _ 



Mo, - 2 X -05674 X {l + % "> ~ 6 X 5674 ~ b ^' 



(3) If we suppose that the innermost portion of the float is 

 at a distance from the centre equal to two-thirds of the radius of 

 the wheel, then k = ^ } and we get 



Resistance =2 Pi ^^ x '01856, 

 Moment = A 0. (o*r b x -02465. 



Also 



V T 



-I QZf* 



therefore the efficiency = ^ . - fi ■ ='753. 



If the velocity of the vessel be 10 knots an hour or 1000 feet 



