PROPULSION OF VESSELS 243 



reduced speed will, as a general rule, be in excess of the cal- 

 culated consumption, by reason of the decrease in economy of 

 the engine induced by reducing the developed horsepower. 



EXAMPLE. A steamer consumes 20 T. of coal per day at 

 a normal speed of 10 knots; the distance to the nearest port 

 where coal can be had is 600 mi., and the estimated quantity 

 of coal in the bunkers is but 35 T. Find what speed should 

 be maintained in order to reach the coaling station with the 

 coal supply on hand. 



SOLUTION. The best way to proceed in a case of this kind 

 is to assume a lower speed, as 8 knots, and calculate the new coal 



consumption for that speed; thus, c = 3 = 10.24 T. per 



da., or .43 T. per hr. The time required to cover a distance 

 of 600 mi. at a speed of 8 knots is 600-^8 = 75 hr., and at a 

 coal consumption of .43 T. per hr. the total quantity of coal 

 required at that speed is 75 X. 43 = 32.25 T. Hence, if a speed 

 of 8 knots is maintained, the supply of coal on hand, or 35 T., 

 will suffice to reach the coaling station under ordinary weather 

 conditions. 



Relation Between Engine Speed and Ship's Speed. On 

 taking charge, the number of revolutions to produce a given 

 speed is often desired. In that case the pitch of the screw, or 

 the effective diameter of the paddle wheel (taking the effective 

 diameter for this purpose from center to center of buckets) 

 must be measured and a fair slip value assumed. Then, to 

 find the revolutions per minute, multiply the pitch of the 

 screw in feet, or the circumference in feet of the effective 

 diameter circle of a paddle wheel, by 60, and by the difference 

 between 1 and the assumed apparent slip expressed decimally. 

 Divide the speed of the ship in feet per hour by this product. 



EXAMPLE. The pitch of a screw propeller is 16 ft.; how 

 many revolutions per minute must it make to drive the ship 

 at the rate of 10 knots, the apparent slip being estimated 

 at 10%? 



SOLUTION. Applying the rule and taking the knot at 

 6,080ft. 10X6,080 



