Ex. 2. Required the folidity of the parabolic conoid, whose

altitude is 21, and the diameter of its base 12.

Ans. 1187.5248.

Ex. 3. Required the solidity of a paraboloid, whose height is 30, and base diameter 40. Ans. 18849.6. N. B. The paraboloid conoid is =+ its circumscribing cylin

der.

PROBLEM XXV.

To find the folidity of the fruftum of a paraboloid.

RULE.

Multiply the sum of the squares of the diameters of the two ends by 7854, and this product by the altitude for the folidity.

EXAMPLE. I.

Required the content of the fruftum of a paraboloid, the greatest diameter being 30, the least 24, and the altitude 18.

Ii

30*=900
24=576

1476

.7854

5904

7380

11808

10332

1159.2504

9

10433.2536

Ex. 2. Required the solidity of the frustum of a paraboloid, the diameter of the greater end being 60, of the less 48, and length 18. Ans. 41733.0144.

Ex. 3. Required the solidity of the frustum of a parabolic conoid, whose diameters are 58 and 30, and the height 36.

PROBLEM XXVI.

Ans. 60281.0208.

To find the folidity of a parabolic spindle.

RULE.

8

Multiply the square of the middle diameter by .7854, and the product by the length, and fr, the last product, will be the folidity.

Note. The parabolic spindle is equal to the circumfcribing cylinder.

EXAMPLE I.

Required the solidity of a parabolic spindle, whose length is 18, and middle diameter 6.

Ex. 2. Required the folidity of a parabolic spindle, whose Ans. 1675.52.

length is 40, and middle diameter 10.

Ex. 3. Required the solidity of a parabolic spindle, whose

length is 100, and middle diameter 10.

PROBLEM XXVII.

Anf. 16755.2.

To find the folidity of the middle fruftum of a parabolic fpindle.

RULE.

Add into one fum 8 times the square of the greatest diameter, 3 times the square of the least diameter, and 4 times the product of the two diameters; multiply the sum by the length of the fruftum, and the product again by .05236 for the folidity.

EXAMPLE I.

Required the solidity of the middle frustum of a parabolic spindle, the length being 20, the greatest diameter 16, and the leaft 12.

162=256×8=2048 12=144×3= 432 16X12 =192X4= 768

3248

20

64960

.05236

389760

194880

129920

324800

3401.3056c Ans.

Ex. 2. Required the solidity of the frustum of a parabolic spindle, whose length is 10, the diameters being 8 and 6.

Anf. 425.1632.

Ex. 3. Required the solidity of the middle frustum of a parabolic spindle, whose length is 30 feet, and diameter 16 and Anf. 8243-5584

20.

SURVEYING,

SURVEYING

URVEYING of Land is confidered to have been the primitive part of Geometry, and consists of three principal parts, viz, The taking of the dimensions, and making the necessary obfervations on the ground; the laying down the same in a map or drawing on paper or vellum ;-and the finding the content or area thereof.

The instruments commonly used in surveying of land, are, the gunter's chain, a case of instruments, a set of plotting scales, the theodolite, and plain table.

The gunter's chain, whether Scots or English, is divided into 100 links. The English chain is 66 feet, and the Scots 74; consequently a link of the English chain is 7.92 inches, and that of the Scots 8.88 inches: likewise the English chain is divided into 4 poles or perches, each 16+ feet, and the Scots chain into 4 falls, each 18+ feet. 10 square chains are I acre, either Scots or English ; and 4 Scots acres are nearly equal to 5 English miles.

A