PROBLEM III.

To find the number of gallons or bushels contained in a body of a cylindrical form.

RULE. Multiply the square of the diameter by the height of the cylinder, and divide the product by 294.12, the quotient will be the number of wine gallons; if you divide by 359.05 the quotient will be the number of ale gallons; and if you divide by 2738, the quotient will be the number of bushels. NOTE. These divisors are found by dividing 231, 282, and 2150.4 by

.7854.

EXAMPLE. Required the number of wine gallons contained in the cylinder AFHD (See the fig. of Prob. VIII. of Mensuration) the diameter AD of its base being 26 inches, and length HD 18 inches? The diameter 26 multiplied by itself gives 676, this multiplied by the length 18, gives the solidity 12168, which divided by 294.12, gives the answer 41 wine gallons.

PROBLEM IV.

To find the number of gallons or bushels contained in a body of the form of a pyramid or cone. (See figures of Problem X. of Mensuration.)

RULE. Multiply the area of the base of the pyramid or cone by onethird of its perpendicular height; the product divided by 231 will give the answer in wine gallons; if divided by 282, the quotient will be the number of beer gallons; and if divided by 2150.4, the quotient will be the number of bushels.

EXAMPLE. Required the number of beer gallons contained in a pyramid DEFGK (See fig. Prob. X. Example I.) whose base is a square EFGK. a side of which, as EF, is equal to 30 inches, and the perpendicular height of the pyramid is 60 inches? The square of 30 is the area of the base 900, this multiplied by one-third of the altitude 20, gives the solidity 18000, which divided by 282, gives the answer in beer gallons 63.8.

PROBLEM V.

To find the number of gallons or bushels contained in a body of the form of a frustrum of a cone. (See the figure below.)

RULE. Multiply the top and bottom diameters together, and to the product add one-third of the square of the difference of the same diameters; multiply this sum by the perpendicular height, and divide the product by 294.12 for wine gallons, by 359.05 for ale gallons, and by 2738 for bushels.

EXAMPLE. Given the diameter DC of the bottom of a frustrum of a cone 36 inches, the top diameter AB=27 inches, and the perpendicular height, FE 50 inches. Required the content in wine gallons? The product of the two diameters 36 and 27 is 972; their difference is 9, which squared and divided by 3 gives 27; this added to 972 gives 999, which D inultiplied by the height 50 gives the solidity

A

B

IF

E

C

49950; this divided by 294.12 quotes the content in wine gallons 169.

PROBLEM VI.

To gauge a cask.

To gauge a cask, you must measure the head diameters FA, DC, and take the mean of them when they differ; measure also the diameter EB at the bung, (taking the measure within the cask); then measure the length of the cask, making due allowance for the thickness of the heads. Having these dimensions, you may calculate the content in gallons or bushels, by the following rule.

A

RULE. Take the difference between the head and bung diameters, multiply this by .62 and add the product to the head diameter, the sum will be the mean diameter; multiply the square of this by the length of the cask, and divide the product by 294.12 for wine gallons, by 359.05 for beer gallons, and by 2738 for bushels.

The quantity .62 is generally used by gaugers in finding the mean diameter of a cask; but if the staves are nearly straight. it would be more accurate to take 55 or less;* if, on the contrary, the cask is full on the quarter, it would be best to take .64 or .65.

EXAMPLE. Given the bung diameter EB=34.5 inches, the head diameter FA=DC=30.7 inches, and the length 59.5 inches; required the number of wine gallons this cask will hold? The difference of the two diameters 34.5, and 30.7, is 3.8; this multiplied by .62 gives 2.4 nearly, to be added to the head diameter 30.7 to obtain the mean diameter 33.1. The square of 33.1 is 1095.61, this multiplied by the length 59.3 gives the solidity 64969.673, which, divided by 294.12, quotes the content in wine gallons 220.8. To gauge a cask by means of the line of numbers on Gunter's Scale, or on the calipers used by gaugers.

Make marks on the scale at the points 17.15, 18.95, and 52.33, which numbers are the square roots of 294.12, 359.05, and 2738, respectively. A brass pin is generally fixed on the calipers at each of these points, which are called the gauge points. Having prepared the scale in this manner, you may calculate the number of gallons or bushels by the following

RULE. Extend from 1 towards the left hand to .62, (or less if the staves be nearly straight) that extent will reach from the difference between the head and bung diameters, to a number to the left hand, which added to the head diameter will give the mean diameter; then put one foot of the compasses upon the gauge point-which is 17.15 for wine gallons, 18.95 for ale gallons, and 52.33 for bushels-and extend the other to the mean diameter; this extent turned over twice the same way, from the length of the cask, will give the number of gallons or bushels respectively.

In the preceding example the extent from 1 to .62 will reach from 3.8 to 2.4 nearly, which added to 30.7 gives the mean diameter 33.1.

Then the extent from the gauge point 17.15 to 33.1, turned over twice from the length 59.3, will reach to 220.8 wine gallons.

If you had used the gauge point 18.95, the answer would have been in ale gallons; and if you had used 52.33, the answer would have been in bushels.

SURVEYING.

LAND is generally measured by a chain of 66 feet in length, divided into 100 equal parts called links, each link being 7.92 inches.

A pole or rod is 16 feet, or 25 links, in length; hence a square pole contains 2724 square feet, or 625 square links.

An acre of land is equal to 160 square poles, and therefore contains 43560 square feet, or 100,000 square links.

To find the number of square poles in any piece of land, you may take the dimensions of it in feet, and find the area in square feet, as in the preceding Problems; divide this area by 43560, the quotient will be the number of acres: or by 272.25, and the quotient will be the number of square poles. If the dimensions be taken in links, and the area be found in square links, you may obtain the number of acres by dividing by 100000 (that is, by crossing off

* In the example to Problem V. preceding (which may be esteemed as the half of a hogshead with staves perfectly straight) the multiplier is only 51. For this multiplied by 9, the difference between AB and DC, produces 4.59 or 4.6 nearly, which, added to 27, and the sum 31.6 squared, multiplied by 20 and divided by 204.12 quotes 169 gallons nearly.

SURVEYING.

the five right hand figures;) and the number of square poles may be obtained by dividing by 625.

PROBLEM I.

To find the number of acres and poles in a piece of land in the form of a rectangular parallelogram.

RULE. Multiply the base by the perpendicular height, and divide by 625, if the dimensions were taken in links, but by 272.25, if they were taken in feet; the quotient will be the number of poles, which, divided by 160, will give the number of acres.

EXAMPLE I. Suppose the base DB (see the figure of Ex. I. Prob. I. of Mensuration) of the rectangular parallelogram ACBD is 60 feet, and the perpendicular BC 25 feet; required the area in poles?

The product of the base 60 by the perpendicular 25 gives the content 1500 square feet, and by dividing by 272,25, we obtain the answer in square poles 5.5.

PROBLEM II.

To find the number of acres and poles in a piece of land in the form of an ob lique-angular parallelogram. (See the figure of Prob. I. Ex. IV. of Mensuration.)

RULE. This area may be found in exactly the same manner as in the preceding Problem, by multiplying the base AD by the perpendicular height BE, and dividing by 625, when the dimensions are taken in links, but by 272.25 when taken in feet; the quotient will be the answer in poles, which divided by 160 will give the answer in acres.

EXAMPLE. Suppose the base AD is 632 links, and the perpendicular BE 326 links; required the number of poles?

Multiply the base 632 links by the perpendicular 326 links, the product 206032 divided by 625, gives the answer in poles 329.

PROBLEM III.

To find the number of acres and poles in a piece of land of a triangular form.

RULE. Multiply the base by the perpendicular height, and divide the product by 1250 when the dimensions are given in links, but by 544.5 when they are given in feet; the quotient will be the answer in poles.

NOTE. Instead of dividing by 1250, you may multiply by 8, and cross off the four right hand figures.

EXAMPLE. Given the base AC (see fig. of Prob. II. of Mensuration) equal to 300 feet, and the perpendicular BD 150 feet; required the area in poles? Multiply the base 300 by the perpendicular 150, the product 45000, divided by 544.5, quotes the answer in poles 82.6.

PROBLEM IV.

To find the number of acres and poles in a piece of land of any irregular right-lined figure.

RULE. Find the area as in Problem III. of Mensuration, by drawing diagonals, and reducing the figure to triangles: the base of each triangle being multiplied by the perpendicular, (or by the sum of the perpendiculars falling on it) and the sum of all these products divided by 1250 when the dimensions are given in links, but by 544.5 when in feet, will give the area of the figure in poles.

EXAMPLE. Suppose that the piece of land is of the same form as the figure in Prob. III. of Mensuration, and that EB-22 feet, EC=33 feet, AF=13 feet, BG 14 feet, and DH=12 feet: it is required to find the area in poles? The product of EB 22 feet, by AF 13 feet, gives double the triangle EAB 236 square feet; and the diagonal EC 33 feet, multiplied by the sum of the perpendiculars BG, DH, 26 feet, gives double the figure BCDE, 858 square feet; the sum of this and 286, divided by 544.5 gives the area 2,1 or 2 poles.

To find the content of a field by the Table of difference of Latitude and De

parture.

This method is simple and much more accurate than by projection, the boundaries being straight lines, whose bearings and lengths are known. The rule for making these calculations is as follows.

RULE.

1. Begin at the western point of the field, as at the point A in the figure Prob. III. of Mensuration, for a point of departure; and mark down in succession the bearings and lengths of the boundary lines AB, BC, &c. as courses and distances in a traverse table. Find the corresponding differences of latitude and departure by Table I. or II. (or by logarithms) and enter them in their respective columns N. S. E. W. as in the adjoined Table.

2. Find the departures or meridian distances of the points B, C, &c. from the point A, by adding the departures when east, but subtracting when west, and mark them respectively against the bearings, in the column of meridian distance.

distance which stands on N. 429 35 W 15.1 11.1

the same line and that immediately above it. Thus on the second line I put 52.1 which is equal to the

549.

sum of 16.1 and 36.0. On the third line 66.236.0+30.2, &c. 4. Multiply the numbers in the column M by the differences of latitude in the same horizontal line, and place the product in the column of areas marked north or south, according as the difference of latitude is north or south. Thus in the first number in the column M is 16.1 which multiplied by the corresponding difference latitude 10.1 N. produces the north area 162.61. The second value of M, 52.1, multiplied by the second difference of latitude 2.1 S. produces the south area 109.41. The third values 66.2 and 19.1 S. produces the south area 1264.42. The fourth difference of latitude is 0, which multiplied by the fourth meridian distance 40.4 produces 0 for the corresponding area, which is the case whenever the bearing is east or west, &c.

5. Add up all the north and all the south areas; half their difference will be the area of the field in square measures of the same name as those made use of in measuring the lines, whether feet, links, or chains, &c. Thus the sum of all the north areas is 275.23, the south 1373.83; their difference is 1098, half of which is 549 square feet the area of the given field.

It may be observed that the bearings and lengths of the boundary lines in this example are not exactly the same as those in Problem III. of Mensuration, which is the reason of the difference in the area above calculated and that found in Problem III. by dividing the field into triangles.

If it had been thought necessary, the differences of latitude and departure might have been taken to one decimal place farther, by entering the table with ten times the length 19, 20, &c. and taking one-tenth of the corres ponding differences of latitude and departure.

In the above calculations we have supposed the survey to have been made with accuracy, in which case the sums of the differences of latitude in the columns N. S. ought to be equal to each other, also the sums of the depart ures in the columns E. W. This is the case in the above example where

the sum of the Diff. of Lat. is 21.2, and the sum of the departure 36.0: but it more frequently happens that the numbers do not agree; in which case the work ought to be carefully examined, and if no mistake can be found, and the error is great, the place ought to be surveyed again; but if the error be small, it ought to be apportioned among all the differences of latitude and departure, in such manner as to produce the required correction with the least possible changes in the given numbers. The method of doing this was explained by me in the fourth number of the Analyst, in answer to a prize question of Professor Patterson, and is as follows. Find the error in latitude, or the difference between the sums of southing and northing, also the sum of the boundary lines, AB, BC, &c. Then say, as this sum is to the error in latitude, so is the length of any particular boundary to the correction of the corresponding difference of latitude, additive if in the column whose sum is the least, otherwise subtractive. The corrections of the departure are found by the same rule, except changing diff. of lat. into departure. Thus in the adjoined example, the sum of the boundary lines is 161.6, the error of latitude is 0.10 and of departure .08,

and the corrections of the diff. of lat. and departure are found by the following proportions:

The first correction of lat. .02 is to be added to the first latitude, 28.28, because it is in the column whose sum 57.41 is less than the other 57.51, so that the first corrected diff. of lat. is 28.30. The second is the difference between 21.65, and the second correction .02, because 21.65 is in the greatest column, the corrected value is therefore 21.63. The third is found in the same manner to be 35.86-.02-35.84. The fourth corrected difference of latitude is simply the fourth correction .02 placed in the column N, because the sum in that column, 57.41 is the least and the fourth diff. of latitude in the original table is 0. The fifth is the sum of 29.13 and the fifth correction 0.02, making 29.15. These are placed in their proper columns in the corrected values. In a similar manner the first departure is equal to the sum of 28.28 and the first correction 0.02, which is equal to 28.30. The second is the difference between 12.50 and the second correction .01, making 12.49; and so as for the others, taking the sum when the departure is in the column whose sum is the least, which in the present case is the east, and the difference when in the other column. In the traverse table thus corrected, the sum of the differences of latitude is 57.47 in both columns, and the sum of the departures 42.08. Having corrected the values of this traverse table, you must find the meridian distances, the column M, the north and south areas, &c. as in the former example.

*The boundary lines in this example are so nearly of an equal length that the correction of the difference of latitude (taken to the nearest decimal) is 0.02 for each of them, but in general, they will be different. The table of difference of latitude and departure may be made use of in finding these corrections, thus: seek in the table till the first term 161.8 (or 162) is found in the distance column to cor respond to the second term .10 (or 10) in the departure column, thus opposite the third term 40, 25, 36, dc. will be the sought corrections, as is evident.