Prof. Barnard on a modification of the Ericsson Engine. 247 
Then, if « represent the area of the curve, 
du=t"'dx=tl*xa- dz 
fia rr74€. 
lt 
But, when z=/, u=o: Hence cirepnae b= - , 
_ At the end of stroke, z=1; Hence, 
mt _ egrs-t) 
nee OE CT AE 
which is to be multiplied into a, the area of the working piston. 
For the negative term, or that expressing the resistance of the 
air undergoing compression, make, x, the absciss, the supplement 
of the stroke, or the part of the stroke which remains at any 
time to be performed. Then the bulk of the partially compressed 
air, at that ome will be made up of the part amz, in cylinder 
C’C’, and a-(1—2), in cylinder C’C”; or will be a total of 
<(1+(mn—I)z). And the corresponding density will be found 
from this proposition : 
a . ee . VT tee mn 
5 (+ Gun — 1) x) an Lod Hea =0e 
The tension, ¢”, corresponding to d’, will be (1), 
sees aie 
t= 15( ix ae x3) = 15mn’ (I+ (mn—I)z) . 
And, as before, du=t’dr=15mnv(l+(mn—1)x) "dx 
15mn? =e 
Pe act 1+ (mn— z) +C. 
ie ‘as (1-7) (mn— —y( emia 
15mn?7. = 
anne man fe EEE RRO Be ISM eee pi y 
i ¢=0,.u=—0; or O= ro iteas =p* 
If z=1, the density is minimum ; and then 
15mn? 
ans Peres Semmens tein ea De fir? 
hee iar = 1ma=D 
15mn7 
mn' 1—['~7 
oF aejmt)™ 
which is to be multiplied by the difference of the surfaces P’ and 
15m nt* — 3— 
P’—ma-_a— a, giving, finally, P=—;— (mn 1_'-1)a, 
. The several _ then, which make up the mean pressure, 
will be as follows 
