244 ' REPORT — 1841. 



may therefore relieve itself from as much atmospheric resistance as is equal 

 to the gravitation of the plane by slackening its speed. If its speed be 

 slackened so as to render the resistance equal to that which it would have 

 upon a level, then the engine would have to work with a less evaporating 

 power than on a level, inasmuch as the motion would be slower. In practice, 

 therefore, it never can be needful to slacken the speed so much as to equalize 

 the resistance with that upon the level. Supposing the evaporating power 

 to remain the same, the speed need only be slackened, so that with the same 

 evaporation an increased resistance can be overcome at a speed less than 

 the level, but not so much less as would render the resistance equal to the 

 level. This, in fact, is what takes place in practice, as is apparent from the 

 results above given. 



By comparing the average evaporation effected in the above experiment 

 with the average speed, an approximate value of the mean pressure of steam 

 on the pistons may be obtained. 



Let L = the stroke of the piston in feet, 

 A = the area of the piston in square feet, 

 n = the number of Strokes of the piston per minute, 

 ••• 2 ?i A L = the number of cubic feet of space through which the piston 

 moves per minute. 



Let c L A = the clearage, or the space between the steam-valve and the 

 piston at each end of the stroke, 



•.• the volume of steam admitted to each cylinder through the steam-valve 

 at each stroke will be 2 w A L (1 + c). 



Let W = the water in cubic feet admitted per minute in the form of steam 

 to each cylinder, 



Let S = the number of cubic feet produced by a cubic foot of water at 

 the density which the steam has in the cylinders, 



Hence we shall have WS = 2wLA(l +c). 



And if P = the pressure of steam in the cylinder in lbs. per square foot, 

 we shall have -^ ^ ^ 



p= "^ " h 



2«LA(l+c) 



where a = 4347826 and 5 = 618. 



In the present case we have L = 1-5, A = 0"853. Let us take c = 0'05. 



The mean rate of evaporation per hour was 89'5 cubic feet of water. The 

 rate per minute would then be P-iQ cubic feet. Hence, since half the steam 

 is supplied to each cylinder, we shall have W = 0"745. 



The mean speed of the train being 30'93 miles per hour, the velocity in 

 feet per minute was 2722. To find the velocity of the piston, this must be 

 reduced in the ratio of the circumference of the driving wheel to twice the 

 length of the stroke. But the driving wheel being 5 feet in diameter, its cir- 

 cumference will be 15*7 feet; and since 2L = 3, the velocity of the piston 

 will be q 



2722 X -4-= 520; 

 15-7 



and the value of n will therefore be 

 Hence we obtain 



« = -5^ = ^= 173-3. 

 2L 3 



P = 0-74-5 X 4347826 ^^g ^ gg,^. 



346-6 X 1-5 X 0-853 x 1-05 



* See Lardner on the Steam-Engine, 7th edit., p. 514, 



