tiply the difference thus found by the quotient of the height, g c, divided by the versed sine, a g, and the product will be the curved surface required.

Fig. 44.

2

g

EXAMPLE.-The sine, a d, of half the arc of the base of an ungula is 5, the diameter of the cylinder is 10, and the height of the ungula 10 inches; what is the curved surface?

5×10=50=sine of half the arc by the diameter.

Length of arc, the versed sine and radius leing equal, under rule, page 78 15.708.

=

Again, as the versed sine and the radius are equal, the cosine is 0. Hence, when the cosine is 0, the product is 0. 50-0=50=the difference of the product before obtained and the product of the arc and the cosine.

50 × 10÷5=50×2=100=the difference multiplied by the height divided by the versed sine, which is the surface required.

Ex. 2. The sine of half the arc of the base of an ungula is 12 inches, the versed sine is 9, the diameter of the cylinder is 25, and the height of the ungula is 18; what is its curved surface? 12×25=300=product of sine and diameter. Arc of base of ungula, by rule, p. 77, the versed sine being 9, is 32.14795.

Then,

32.14795 × 12.5-9=112.51782, and 300-112.51782-187.48218, which, multiplied by 18÷9 =374,96436 inches, Ans.

4. When the Section passes through the Base of the Cylinder, and the Versed Sine exceeds the Sine, a g, Fig. 45.

RULE.-Multiply the sine of half the arc of the base by the diameter of the cylinder, and to this product add the product of the arc and the excess of the versed sine over the sine of the base.

Multiply the sum thus found by the quotient of the height divided by the versed sine, and the product will be the curved surface required.

EXAMPLE.—The sine, a d, of half the arc of an ungula is 12 inches, the versed sine, a g, is 16, the height, c g, 16, and the diameter of the cylinder 25 inches; what is the curved surface?

12×25=300=sine of half the arc by the diameter of the cyl

inder.

Length of arc of base, by rule, p. 74=arc of d b f—circumference of base 46.392.

Then 46.392 × 16—12.5=162.372=product of arc and the excess of the versed sine over the sine.

300+162.372=462.372=the sum of the above products. 16÷16=1=quotient of height divided by the versed sine. 462.372×1=462.372 inches=the sum of the products and the height divided by the versed sine the curved surface required.

=

Ex. 2. The sine of half the arc of the base of an ungula is O, the diameter of the cylinder is 10 inches, and the height of the ungula is 20 inches; what is its curved surface?

NOTE. The sine of the arc being 0, the versed sine is equal to the diameter (10), and the sine of the base is 10÷÷2=5.

0×10=0=product of sine of half the arc and diameter of the cylinder.

0+(31.416 (length of arc) x 10~5)=157.08=the sum of the product above obtained and the product of the arc and the excess of the versed sine over the sine.

157.08×20÷10=314.16=the above sum the height÷the versed sine the result required.

5. When the Section passes obliquely through both Ends of the Cylinder, a b c d, Fig. 46.

RULE.-Conceive the section to be continued till it meets the side of the cylinder produced; then, as the difference of the versed sines of the arcs of the two ends of the ungula is to the versed sine of the arc of the less end, so is the height of the cylinder to the part of the side produced.

Find the surface of each of the ungulas thus found by the rules 3 and 4, and their difference will be the curved surface required.

EXAMPLE.-The versed sines a e, d o, and sines i k, g r, of

the arcs of the two ends of an ungula, Fig. 46, are respectively 5 and 2.5, and 5 and 4.25; the height of the ungula within the cylinder, cut from one having 10 inches diameter, is 5 inches; what is the height of the ungula produced beyond the cylinder? 5 2.5 2.5, and 2.5: 2.5 :: 5:5=height of ungula produced beyond the cylinder.

Ex. 2. The versed sines of the base and arc of an ungula cut from a cylinder of 6 inches diameter are 6 and 2 inches, and its height within the cylinder is 4 inches; what is the distance it extends above or beyond the cylinder?

Lune.

Ans. 2 inches.

Definition. The space between the intersecting arcs of two eccentric circles.

To ascertain the Area of a Lune, Fig. 47.

RULE. Find the areas of the two segments from which the lune is formed, and their difference will be the area required. Or, s-s=a, when s and s′ represents the areas of the segments. Fig. 47.

d

EXAMPLE.-The length of the chord a c is 20, the height e d is 3, and e b 2 inches; what is the area of the lune? By Rule 2, p. 76, the diameters of the circles of which the lune is formed are thus found:

Then, by rule for the areas of segments of a circle, page 87, the area of a d c is 70.5577 in.

their difference 43.3939 in., the area of the lune required. Ex. 2. The chord of a lune is 40, and the heights of the segments 10 and 4 inches; what is its content?

Ans. 173.5752 inches. Ex. 3. The chord of a lune is 6 feet 8 inches, and the heights of the arcs 1.666 feet and 8 inches; what is its area? Ans. 694.2996 inches.

Ex. 4. The chord of a lune is 86.6024 inches, and the heights of the segments 25 and 15 inches; what is its area? Ans. 653.3551.

Centre of Gravity. On a line connecting the centres of gravity of the two arcs at a point proportionate to the respective areas of the arcs.

NOTE. If semicircles be described on the three sides of a rightangled triangle as diameters, two lunes will be formed, and their united areas will be equal to that of the triangle.

Cycloid.

Definition. A curve generated by the revolution of a circle on a plane.

To ascertain the Area of a Cycloid, Fig. 48. RULE.-Multiply the area of the generating circle a b c by 3, and the product will give the area required.