To gauge a cask by means of the line of numbers on Gunter's Scale, or that on the Callipers used by gaugers.

Make marks on the scale at the points 17.15, 18.95, and 52.33, which are the square roots of 294.12, 359.05, and 2738, respectively. A brass pin is generally fixed on the callipers 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 :

RULE. Extend from 1 towards the left hand to 0.62 (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 is to be added to the head diameter to get 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 0.62 will reach from 3.8 to 2.4 nearly, which, being added to 30.7, gives the mean diameter 33.1; then the extent from the gauge point 17.15 to 33.1, being turned over twice from the length 59.3, will reach to 220.8 wine gallons.

If we use the gauge point 18.95, the answer will be in ale gallons; and if we use 52.33, the answer will be in bushels.

14

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 163 feet, or 25 links, in length. Hence a square pole contains 2721 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; then divide this area by 43560, and 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 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 be taken in links, or by 272.25 if they be taken in feet; the quotient will be the number of poles. Dividing this by 160, we get the number of acres.

EXAMPLE. Suppose the base BC (see the figure of Ex. I. Prob. I. of Mensuration) of the rectangular parallelogram ABCD is 60 feet, and the perpendicular AB 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 it by 272.25, we obtain the answer in square poles 5.5, nearly.

PROBLEM II.

To find the number of acres and poles in a piece of land in the form of an oblique-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, or by 272.25 when taken in feet; the quotient will be the answer in poles, which, being 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 329.7.

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, or 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

EXAMPLE. Given the base AC (see figure of Problem 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, gives 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 a piece of land is of the same form as the figure in Prob. III. of Mensuration, and that BE22 feet, CE=33 feet, AF 13 feet, BG = 14 feet, and DH 12 feet; it is required to find the area in poles.

The product of BE 22 feet, by AF 13 feet, gives double the triangle ABE 286 square feet; and the diagonal CE 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, being 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 Departure. 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.

3. Place in the first line of the column M the first meridian distance 16.1, and, in the following lines, the sum of the meridian distance which stands on the same line and that immediately above it. Thus on the second line, I put 52.1, which is equal to the sum of 16.1 and 36.0. On the third line, 66.2: 36.0+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, being 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. produce the south area 1264.42. The fourth difference of

latitude is 0, which, being multiplied by the fourth meridian distance 40.4, produces 0 for the corresponding area, as 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 275.83, that of 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 between the area above calculated and that found in Problem III. by dividing the field into triangles.

If it be necessary, the differences of latitude and departure may be taken to one decimal place farther, by entering the table with ten times the length 19, 20, &c., and taking one tenth of the corresponding 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. must be equal to each other; also the sums of the departures in the columns E. W, This is the case in the above example, where the sum of the differences of latitude is 21.2, and the sum of the departures 36.0: but it most frequently happens that the numbers do not agree; in which case the work must be carefully examined, and if no mistake be found, and the error be great, the place must 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 difference of latitude 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 0,08;

161.6 57.41 57.51 42.02 42.10 0.10 0.08 57.47 57.47 42.08 42.08

and the corrections of the difference of latitude and departure are found by the following proportions:

The first correction of latitude 0.02 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

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 161.6 (or 162) is found in the distance column to correspond to the second term 0.10 (or 10) in the departure column; thus opposite the third term 40,

corrected difference of latitude is 28.30. The second is the difference between 21.65 and the second correction 0.02, 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 -0.02 35.84. The fourth corrected difference of latitude is simply the fourth correction 0.02 placed in the column N, because the sum in that column, 57.41, is the least, and the fourth difference 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 0.01, 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.

In projecting a survey of this kind, where there is a small error, you must plot off as usual the boundary lines AB, BC, CD, &c., and it will be found that the termination of the last line AE will not fall exactly in the point A, but will be at a point near it, which we shall call a. To correct this error, you must draw through the points B, C, D, &c., lines parallel to aA, in the direction from a to A, of such lengths as to be to Aa, as the distances of those points respectively from A (measured on the boundary ABCD, &c.) are to the whole length of the boundary line; through these points draw the corrected lines terminating on A.

The Manner of Surveying Coasts and Harbors.

From what has been said in the preceding problems, the intelligent reader will readily perceive the method of surveying a coast or harbor. But as this is an important subject, we shall enter more fully into an explanation of the different methods which may be used.

To take a draught of a coast in sailing along shore.

Having brought the ship to a convenient place, from which the principal points of the coast or bay may be seen, either cast anchor, if it is convenient, or lie-to as steady as possible; or, if the coast is too shoal, let the observations and measures be taken in a boat. Then, while the vessel is stationary, take, with an azimuth compass, the bearings, in degrees, of such points of the coast as form the most material projections or hollows. Write down these bearings, and make a rough sketch of the coast, observing carefully to mark the points, whose bearings are taken, with letters or numbers, for the sake of reference.

Then let the ship or boat run in a direct line (which must be very carefully measured by the log, or otherwise) one, two, or three miles, until she comes to another situation, from which the same points, before observed, can be seen again with quite different bearings. Then let the vessel lie steady, as at the former station, and observe again the bearings of the same points, and make a rough sketch of the coast. This sketch may be made more accurately while the vessel is running the base line.

To describe the chart from these observations, you must, in some convenient part of a sheet of paper, draw the magnetic meridian, and lay off the several bearings taken at the first station, marking them with their proper letters or numbers. Lay down also the bearings taken from the second station. Draw a line to represent the ship's run both in length and course, and from that end of the line expressing the first station, draw lines parallel to the respective bearings taken from that end; also from the other end draw lines parallel to the bearings taken at that end, and note the intersection of each pair of lines directed to the same point; and through these intersections draw by hand a curved line, observing to wave it in and out as near as can be like the trending of the coast itself. Then mark off the variation of the compass from the north end of the magnetic meridian, towards the right hand if it be west, or towards the left hand if it be east, and draw the true meridian through that point and the centre of the circle.

*In taking the bearings, if the vessel has much motion, the mean of several observations should be taken.