IS 



THE CIVIL ENGINEER AND ARCHITECT'S JOURNAL 



[Jaxuaby 



The Lap. 



A slide is s;ii<l to have hip wlien the width of its face is greater 

 tliiiiithat of the steam ports, the ports heing thereby overhipped, 

 «s ill fifl. 7. 



It is to he remarked that slides should have some degree of lap 

 on lioth the steam aiul exhaustion sides of the passage, because, 

 althouirh in theory an ajierture may he said to be completely closed 

 when covered by \i bar of similar width, yet, in the construction of 

 .-1 slide without' lap, we cannot insure such accuracy oi fit as to 

 ])reclude the possibility of steam entering or leaving both steam 

 ports at the same time. 



Lap on the steam side has the effect of cutting off the steam 

 from the cylinder, by closing the port before the coni|iletion of the 

 stroke, the remainder of the stroke being effected by the expansion 

 of the steam already admitted. 



Demnnstration. 



Case 3. When a Slide has Lap on the Steam side, 



WITHOUT Lead. 



Let a b and 6 c, diagram 4, represent the lap at both ends of the 

 slide ; and let a d and c e represent the two steam ports ; then d e 

 will re])resent tlie tra\ el of the slide, which, in this case, equals 

 twice the steam purt, plus twice the lap. 



Diagram 4. 



Supposing de also to represent the stroke of the piston, and that 

 the ])iston is on the top stroke, then hd and bf are the respective 

 IHisitionsof the crank and eccentric; for the slide, instead of occu- 

 pying its central position, when the piston is at the end of its 

 stroke (as in Case 1), must be set in advance of that position to 

 tlie extent of the lap, that steam may enter the cylinder when the 

 piston begins to move. (See fig. 5.) 



When the eccentric has advanced from / to c, the crank will 

 liave reached the point g ; the piston is therefore at a when the 

 port ce is fully open, the slide being then in the position fig. G. 



Again, — when the eccentric has reached the point/;, thejiortcf 

 will be re-closed (fig. .5), and i will be the position of the j)iston ; 

 therefore, the distance of tlie jiiston from the end of its stroke, 

 when the steam is cut off, is pri>))ortioned to the whole stroke, as 

 ieis to de. 



When the eccentric arrives at fr, the slide will oc<-u])y its central 

 position (fig. 7), and the piston will be at w, wliere exiuiustion 

 commences from above it ; but steam is not admitted below it, for 

 the return stroke, until the eccentric has reached the point n, 

 where the port o (/ begins to open, the position of the slide at that 

 moment being that shown in fig. 8. 



When the eccentric arrives at rf, the ])ort will be fully ojien, the 

 slide being then in its extreme position, fig. i) ; and it will be re- 

 closed when the eccentric arrives at (/, and the piston at p, where 

 the steam is cut off, the position of the slide being again that shown 

 in fig. 8. Again, — when the eccentric reaches tlie point r, exhaus- 

 tion ceases from above the piston, which is then at .v, and com- 

 mences from lielow it, the slide being then in its central position, 

 fig. 7, and moving downward. Finally, — the crank having arrived 

 at rf, and the eccentric at f, the piston will have completed its 

 ascent, and the slide will occupy the position fig. 5, as at starting. 



The steam was shown to be cut off when the piston had de- 

 scended frtmi d to ?, the crank having descrilied the arc d g H, and 

 the eccentric the arc/e/j. Now, di is the versed sine ui'dgii, and 

 ec is the versed sine of half _/>/*; and d g ii and felt are equal 

 arcs. Hence 



Rule III. — To find at what part of the stroke steam will be cut off 

 with a given ainoiiiit of lap : — 



Divide the width of steam jiort, by itself, plus the lap, and call 

 the quotient versed sine. Find its corresponding arc in degrees 

 and minutes, and call it arc the first. If arc the first be less than 

 45 degrees, multijily the versed sine of twice that arc 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 arc the fii'St exceed 45 degrees, multiply the versed sine of 

 the difference between double that arc and ISO degrees by half the 

 stroke, and the product v\ ill be the distance of the piston from the 

 end of its stroke when the steam is cut off. 



Rule IV. — To find the amount of lap necessary to cut off the steam 

 at any given part of the stroke : — 



If it be required to cut off the steam before half-stroke, divide 

 the distance the piston moves before steam is cut oft', by half the 

 stroke, and call the quotient versed sine. Find the arc of that 

 versed sine, and also the versed sine of half that arc. Divide the 

 difference between the versed sine last found and unity, by the 

 versed sine, and multiply the width of steam port by the quotient; 

 the product will be the lap. 



If it be required to cut off the steam at a point beyond half- 

 stroke, divide the distance of the piston from the end of its stroke, 

 when steam is cut off, by half the length of stroke ; call the ipio- 

 tient versed sine ; find its corresponding arc, and subtract it from 

 180 degrees. Find the versed sine of half the remainder, and sulv- 

 tract it from unity. Divide the remainder by the versed sine, and 

 multiply the width of the steam port by the quotient; the product 

 wUl be the lap. 



Example 3. — The stroke of a piston is 36 inches; width of steam 

 port I5 inch ; and lap 6 inches : required the point of the stroke 

 at which steam will be cut off. 



Here 1'5 -|- 6 ^ 7-5 ; and 1-5 -f- 7-5 = -2 = versed sine ; 



arc of versed sine '2 = 36° 52' (arc the first) ; 

 and 36° 52' X "-i — 73° 44' = arc of versed sine, -7198. 

 Then -7198 X 18 = 12-95 inches = distance of the piston from 

 the commencement of its stroke when the steam is cut off. 



Example 4. — The stroke of a piston is 36 inches ; width of steam 

 port l.i inch ; and extent of la|) If inch : required the point of the 

 stroke at which steam is cut off. 



Here 1-5 -(- 1-25 = 2-75; and 1-5 -h 2-75 =: -5454 = versed sine 

 of arc 62° 58' (arc the first). 



Then 62° 58' X 2 = 125° 56'; and 180° - 125° 56' = 54° 4' = arc 

 of versed sine, '4131 ; '4131 x 18 ^ 7*43 inches ^ distance of 

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



Example 5. — The stroke of a piston is 36 inches; width of steam 

 port 1"5 inches ; and distance of the piston from the commence- 

 ment of its stroke, when the steam is cut oft', 12'95 inches: required 

 the lap. 



Here 12-95 -f- 18 == "7198 = versed sine of arc 73° 44'; 

 73° 4. t' -^ 2 = 36° 52' = arc of versed sine '2. 

 Then 1 - -2 = -8 ; and -8-1-2 = 4; l-5x 4=6 inches = lap. 



