38 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL. 



I Febrvarv, 



-To find the point of the stroke at which steam will he cut 



the eccentric arrives at 7, the port t e begins to open for tlic ad- 

 mission of steam beneatli tlie piston (see fig. 13), which has then 

 descended to r ; the arc c m .v heinff equal to the arc )' /r 7. ^^'hen 

 tlie eccentric lias readied tlie ]i<niit t", opjiosite to !, the port !> e 

 will be open to the extent of the lead A A', equal to c A, and the 

 piston will have completed its descent. 



Steam continues to enter the port be during the ascent of the 

 piston, until the eccentric reaches the point A', when the port h e 

 will be reclosed (fig. 13), the direction of the slide's motion being 

 downward, and the piston ha\ing ascended to /'. Exhaustion 

 ceases from above the piston when the eccentric reaches the point 

 t, the piston being then at «, and the slide again in the position 

 fig. 12. AVhen the eccentric reaches the point 71', ojiposite to », 

 exhaustion commences below the ])iston, the slide being tlien in 

 the position fig. 11, and the piston at 0'. Finally, — wlien the ec- 

 centric reaches the point f/', and the crank the point .v', opposite to 

 *, steam begins to enter the port erf for the return stroke, at the 

 commencement of which tlie ]iort cd will be open to the extent of 

 the lead c A ; the crank and eccentric occupying their original po- 

 sitions, ae and a i. 



It is here shown that four distinct circumstances result from the 

 use of a slide having lap on both sides of the port, with lead, 

 during a single stroke of the piston. These are — 



I<'irst : The cutting off the steam, for the purpose of exjiansion. 



Second: The cessation of exhaustion on the exhaustion side. 



Third: The commencement of exhaustion on the steam side. 



Fourth : The re-admission of steam for the return stroke. 



AVith regard to the first of these results, we found the steam 

 port frf closed, when the crank and eccentric had described the 

 equal arcs em, and idk. Now, erf, the steam port, is the versed 

 sine of rf/c; and A rf, the steam port minus the lead^ is the versed 

 sine of i rf. Hence, 



Rule V.- 

 off:- 



Divide the width of the steam port, and also that width minus 

 the lead, by half the slide's travel, and call the quotients versed 

 sines. ¥\nA tlieir corresponding arcs, and call them arc the first, 

 and arc the second, respectively. Then, if the sum of those arcs 

 be less than 90 degrees, multiply the versed sine of their sum by 

 half the stroke, in inches, and the product will be the distance of 

 the piston from the commencement of its stroke, when the steam is 

 cut off. 



If the sum of arcs the first and second exceed 90 degrees, sub- 

 tract it from 180 degrees; and the versed sine of the difference, 

 multiplied by half the stroke, equals the distance of the jiiston 

 from the end of its stroke, when the steam is cut off. 



Example 8. — The stroke of a piston is 60 inches ; the width of 

 steam port 3 inches ; lap on the steam side 2| inches ; lap on the 

 exhaust side ^th inch ; and lead ij inch : required the point of the 

 stroke at which steam will be cut off. 

 3 

 Here = -SUi = versed sine of 62" 58' (arc the first) ; 



3-5 



and =; -4545 = versed sine of 56° 57' (arc the second). 



Then G2°58' -1- 56° 57'= 119° 55'; and 180° - 119°55' = 60°5' = 

 arc of versed sine, -5012. -5012 x 30= 15-036 inches = distance 

 of the piston from the end of its stroke when the steam is cut off. 



Exhaustion was shown to cease, during the ascent of the piston, 

 when the eccentric had reached the point /, and the crank the 

 point ,r; the crank having described the arc dkx, equal to i'e< 

 described by the eccentric. 



Now i' e is equal to arc the second (Rule V.) ; and e t is equal to 

 90 degrees minus tt, or the arc of versed sine ef; and e/is half 

 the slide's travel minus the lap on the exhaust side. Hence, 



To find the point of the stroke at which exhaustion ceases : — 



Divide half the slide's travel, minus the exhaustion lap, by half 

 the travel, call the quotient versed sine, and add its corresponding 

 arc, calling it arc the third, to arc the second. The versed sine of 

 the difference between their sum and 180 degrees, multiplied by 

 half the stroke, equals the distance of the piston from the end of 

 its stroke when exhaustion ceases. 



Example 9. — The several proportions being as in the preceding 



example. 



Here 3 -}- 2-5 = 5-5 = half the slide's travel ; 



J 5-5 - -125 „ „ , , 



and — — = -9772 = versed sine of arc 88° 42'= (arc 



the third). 



Then 88° 42' + 56° 57' (arc the second) = 145° 39' ; and 180° - 

 145° 39' = 34° 21' = arc of versed sine, -1743. -1743 X 30 = 5-229 

 inches = the distance of the piston fiom the end of its stroke 

 when exhaustion ceases. 



Exhaustion was shown to commence from above the piston when 

 the crank and eccentric had described the equal arcs eA' 7;, and 

 i dn. 



Now i rf n is equal to 180 degrees minus n j' ; n i' is equal to n' i ; 

 and n'rf is ecjual to arc the third. Hence, 



To find the distance of the piston from the end of its stroke when 

 exhaustion commences : — 



Subtract arc the second from arc the third, and multiplv the 

 versed sine of their difference by half the stroke. The product 

 will be the distance required. 



Example 10. — The proportions being as in the two preceding 

 examples. 



Here 88° 42' - 56° 57' = 31° 45' = arc of versed sine, -1496; 

 and -1496 X 30 = 4-488 inches, the distance required. 



Steam was found to be re-admitted, for the return stroke, when 

 the piston had reached the point r in its descent, the crank and 

 eccentric having described the equal arcs ek' «, and i dq. 



Now, irf^ is equal to 180 degrees minus 9 T ; f being diametri- 

 cally opposed to i. And q f is equal to i q', the difference between 

 arcs the first and second. Hence, 



To find the distance of the piston from the end of its stroke when 

 steam is re-admitted for the return stroke : — 



Multiply the versed sine of the difference between arcs the first 

 and second by half the stroke, and the product will be the distance 

 required. 



Example 11. — The proportions being as before. 



Here 62° 58' - 56° 57' = 6° 1' = arc of versed sine -0055. 

 Then -0055 x 30 = "165 inches = the distance required. 



RiTLE VI. — To find the proportions of the steam lap and lead ; the 

 points of the stroke ivhere steam is cut off^ and re-admitted for the 

 return stroke, lieiny known : — 



When the steam is cut off before half-stroke, divide the portion 

 of the stroke performed by the piston, by half the stroke, and call 

 the quotient versed sine. Likewise, divide the distance of the 

 piston from the end of its stroke when steam is re-admitted for 

 the return stroke, by half the stroke, and call that quotient versed 

 sine. Find tlieir respective arcs, and also the versed sines of half 

 their sum and half their difference. The width of the steam port 

 in inches, divided by the versed sine of half their sum, equals half 

 the travel of the slide ; and half the travel, minus the width of 

 port, equals the lap. The difference of the two versed sines last 

 found, multiplied by half the travel of the slide, equals the lead. 



When the steam is to be cut off after half-stroke, divide the 

 distance of the piston from the end of its stroke by half the stroke ; 

 call the quotient \ersed sine, and subtract its corresponding arc 

 from 180 degrees. Divide the distance the piston has to move 

 when the steam is admitted for the return stroke, by half the 

 stroke ; call the quotient versed sine, and find its corresponding 

 arc. Then proceed with the two arcs thus found, as in the former 

 case. 



Example 12. — The stroke of a piston is 60 inches ; the width of 

 steam port 3 inches; distance of the piston from the end of its 

 stroke when steam is cut oft' 15-036 inches; and when steam is 

 admitted for the return stroke -165 inches : required the lap and 

 lead. 



Here 15-036 -^ 30 = -5012 = versed sine of arc 60° 5' ; 



and ] 80° — 60° 5' = 1 1 9" 55'. 



Then -165 -^ 30 = -0055 = versed sine of 6° 1'. 



119°55'-f 6-' 1' = 125° 56'; 119"55' - 6° 1' = 113°54'. 



125° 56' 



— ^: 62^ 58' ^ arc of versed sine -5454 ; 



— - := 56° 57' =: arc of versed sine -4545. 



3 -7- -5454 = 5-5 inches ^ half the slide's travel ; 



and 5-5 — 3 = 2-5 = lap. 



-5454 - -4545 = -0909 ; and -0909 X ^'^ = "5 inches = lead. 



To find the Lap and Lead by Construction. 



The stroke of the piston; width of steam port; and distances 

 of the piston from the end of its stroke when the steam is cut off, 

 and when it is re-admitted for the return stroke, being known ; 



