end of its travel, shown in Fig. 4, pp being equal to the exhaust ports b (figures 3, 4, 5, 6), and s will be the position of the piston in its down-stroke. It will be seen that during the time the piston is travelling from c to c', the bottom port p'p has been fully open, to allow the steam to escape from the under-side of the piston. When the crank arrives at w, the lower port is shut, and the slide is in the position shown in Fig. 6.

RULES AND EXAMPLES FOR THE LAP AND LEAD OF THE SLIDE.

If the lead, lap, lengths of the stroke of the slide and piston be given, to find where the steam is cut off:

Rule.-Add the lap and half the lead of the slide together, divide the sum by half the stroke of the slide, multiply the square of this quotient by the length of the stroke of the piston, subtract this product from the length of the stroke, and the remainder will be the distance travelled by the piston before the steam is cut off. Taking the dimensions the same as in the last problem, observing that they are all taken in inches,

= 17 inches.

And 40

17 23 inches the piston has moved over before the steam was cut off.

Given the length of the stroke of the piston, the distance moved by the piston before the steam is cut off, the length of the stroke of the slide and the lead: to find the lap or cover, subtract the distance moved by the piston before the steam is cut off, from the whole stroke of the piston; divide the remainder by the length of the stroke of the piston, and take the square root of the quotient; multiply this root by half the length of the stroke of the

slide, from this product subtract half the lead, and the remainder will be the lap required.

Example.-The length of the stroke of an engine is 40 inches, the steam is cut off at 23 inches, the travel or stroke of the valve 6 inches, and the lead of an inch: find the lap.

Now, 4023= 17 inches, the distance the piston has to travel after the steam is cut off.

then

17

40

= √·425 = ·6519; this, multiplied by half

the travel of the slide, 6519 × 3 = 1.9557,

and 1.9557-11-8557 inch, the lap required.

To find at what part of the Stroke the steam will be cut off, when the breadth of the Steam Port, the Lap, and Lead are given.

Divide the breadth of the steam port by half the travel of the slide, and the quotient is the versed sine of an arc; subtract the lead from the breadth of the port, and divide the remainder also by the travel of the slide, and the quotient is the versed sine of another arc; from the table of versed sines find the corresponding arcs. If the sum of these arcs be less than 90°, 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 the stroke when the steam is cut off. If the sum of the two arcs exceed 90°, subtract that sum from 180°, and the versed sine of the difference multiplied by half the stroke is equal to the distance of the piston from the end of the stroke when the steam is cut off.

The stroke of a piston, 60 inches; the width of the steam port, 3 inches; lap on the steam side, 2 inches; lap on the exhaust side,inch; and lead inch: required the point of the stroke at which steam will be cut off.

= ·5454 = versed sine of 62° 58′ (arc the

= ·4545 = versed sine of 56° 57′ (arc the

Then 62° 58′+ 56° 57′ = 119° 55′; and 180° 119° 55' = 60°5′ arc of versed sine, 5012. 5012 × 30 = 15.036 inches, distance of the piston from the end of the stroke when the steam is cut off.

To find at what part of the Stroke the exhaustion ceases.

Subtract the lap on the exhausting side from half the travel of the slide, and divide the remainder by half that travel, and the quotient is the versed sine of an arc; from the table find the corresponding arc, and add it to the second arc in the preceding rule; the versed sine of the difference between their sum and 180°, multiplied by half the stroke, gives the distance of the piston from the end of the stroke when the exhaustion ceases.

Example.-3 + 2·5 = 5·5 = half the slide's travel; 5.5.125 5.5

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 × 30 = 5·229 inches = the distance of the piston from the end of its stroke when exhaustion ceases.

Find the distance of the Piston from the end of the Stroke when exhaustion commences.

Subtract the second arc from the third, and multiply the versed sine of their difference by half the stroke; the product will be the distance required.

Using the values in the two preceding examples,

88° 42′ 56° 57′ 31° 45, the versed sine of which is •1496, and 1496 x 304-488 inches, the required dis

tance.

Find the distance of the Piston from the end of the Stroke when the steam is admitted for the Return Stroke.

Multiply the versed sine of the difference of the first and second arcs by half the length of the stroke, and the product will be the distance required.

The first arc is 62° 58′, and the second is 56° 57'.

62° 58' 56° 57′ = 6o 1', the versed sine of which is ⚫0055; and 0055 × 30·165 inch.

Rule. To find the proportions of the steam-lap and lead, the points of the stroke where steam is cut off, and re-admitted for the return stroke being 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 their 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 versed 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 off, 15·036 inches, and when steam is admitted for the return stroke 165 inch; required the lap and lead.

versed sine of arc 60o 5';

and 180° 60° 5' 119° 55'.

= 0055 versed sine 6o 1'.

15.036

Here

=5012

30

3

•5454

5.5 inches = half the slide's travel;

and 5.5 3 = 2.5 lap.

•5454-4545·0909; and 0909 × 5·5=·5 inches lead.

A Table of Multipliers to find the Lap and Lead, when the Steam is to be cut off at to ths of the Stroke. The lap must be equal to the width of the steam-port multiplied by Col. 1.

The lead must be equal to the width of the steam-port multiplied by Col. 2.