248 Prof. Barnard on a modification of the Ericsson Engine. 
Positive pressure, before cut off = + alt 
lr-1—1 
“ “ce “ BE cme 
after =-+ali Sr 
alt 
Negative pressure, in supply cyl. = — ie 
1 
Balance, neg. pres. in cond. cyl. = ee (m?-Yn'-7—1!-7) 
Negative atmospheric pressure = —15a(1—m). 
Whence, 
1 alt l5ammr 
P=alt +ali7——— + . (nn '-7— I'-7)-150( 1m), 
which is the mean pressure during the stroke. By eliminating 
(= 15(" ai ) and reducing, we shall have, 
By y-1 
P=15al'-1mnr(n-1)+ 15al-ImIn— eS 
(m*-Yn\-7 — 19} Tat hoi 
But, as we have seen that » is dependent on J and m, this sym- 
bol also must be eliminated, in order that the equation may con- 
tain only 2a variables. 
T+0 tok 4 yet 
wm (G58) (2) 7 roan 8); an wae) 
O+6 m O86 
Substituting this value and reducing, the equation becomes 
9 an: g-1 2 2 —] 
i ue 2 i 
x 
fe fig oo: I mu? 
parse i? m 7 w™(n( 7) "ye Tera 
—i-y)? 1-7 
i? aiecayeactl 
= 15a 17 mal (ul? m9 Tyee as een eee en) 
+m—1] 
This equation is too complicated to admit of a — discus- 
sion. But, by making particular suppositions in regard to m and 
j, it may be, in some = Sn simplified. ‘Thus, putting 
m and i each =1, it beco 
Pte" ‘eh SF) 
Which, substituting the numerical values of the symbols, gives 
P=2:-55a=56400 Ibs. ; ; or, with ten revolutions and a six-foot 
stroke, a sah power o 
If we make m and / equal to each other, and pat a new sym- 
bol, as 4, to stand indifferently for either, the equation becomes, 
