any two Stations take Bearings to the Object, and their Interfection will determine the Place where you are to infert it, in the Manner that the Tower is fet out in the Figure, from the Interfections taken at the first and fecond Stations of the above FieldBook.

A Protraction of this will render all plain, on which lay off your Off-fets and Interfections, and proceed to find the Content by any of the Methods in Section the 4th.

The foregoing Field-Book may be otherwise kept thus:

How to caft up Offfets by the Pen,

Plate XI. Fig. 2.

12-1f=2f, If-1e=fe, 1e-1d=ed.

Then Id x da ida, by Prob. 6, Page 183, anded x da+eb-adeb, by Lemma, Page 226, alfofe x eb+fc=befc, and 2f fccf2; the Sum of all which will be 1abc21; the Area contained between the ftationary Line 12, and the Mearing, 1 abc 2.

In the fame Manner you may find the Area of 2ihg2 of ik 3i, as well as what is without and with infide of the Stationary Line 71.

If therefore the left Hand Off-fets exceed the right Hand ones, it is plain, the Excess must be added to the Area within the stationary Lines, but if the right Hand Off-fets exceed the left Hand ones, the Difference must be deducted from the faid Area; if the Ground be kept on the right Hand as we have all along fuppofed; or in Words thus,

To find the Contents of Off-fets.

1. From the Distance Line, take the Distance to the preceding Off-fet, and from that Distance of the one preceding it, &c. in Four-Pole Chains; fo will you have the respective Distances from Off-set to Off-fet, but in a retrograde Order.

2. Multiply the last of these Remainders by the first Off-fet, the next by the Sum of the

first

first and second, the next by half the Sum of the fecond and third, the next by half the Sum of the third and fourth, &c. The Sum of these will be the Area produced by the Off-fets.

Thus, in the foregoing Field-Book, the first stationary Line is 22C. 12L. or 11C. 12L. of 4 Pole Chains. See the Figure.

C. L.

From 11.12=1,2

C. L.
6.50=1f

C. L.

3.90=1e

Take 6.50=1f

3.90=1e

2.25=1d

4.62=2f 2.60=ef

1.65= ed

C. L.

id=2.25×32L. half the firft Off-set=

7200

ed=1.65x1C.26L. the Sum of the 1st & 2d 2.0790

1.7094

ef=2.60x1C.32L. the Sum of 2d & 3d= 3.4320 2f=4.62×37L. half the last Off-set = Content of left Off-fets on the first Dift. in Square Four-pole Chains

7.9404

In like Manner the reft are performed. The Sum of the left Hand Off-fet will be 14.0856 And the Sum of the right Hand ones

3.6825

Excefs of left Hand Off-fets in Squ. 4PoleC. 10.4031

Acres 1.04031

.16124 4

Perch 6.4496

Excess of left Hand Off-fets above the right Hand ones, IA. OR. 6P. to be added to the Area within the ftationary Lines.

OF

OF INTERSECTIONS.

How to find the Area of a Piece of Ground by InterSections only, when all the Angles of the Field can be feen from any two Stations on the outfide of the Ground.

Plate XII. Fig. 1.

L

ET ABCDEFGA be a Field, H and I two Places on the out Side of it, from whence an Object at every Angle of the Field may be feen.

Take the Bearing and Distance between H and I, and fet that at the Head of your Field-Book, as in the annexed one. Fix your Inftrument at H, from whence take the Bearings of the feveral angular Points, A, B, C, D, &c. as they are here represented by the Lines HA, HB, HC, HD, &. Again fix your Inftrument at I, and take Bearings to the fame angular Points, reprefented by the Lines IA, IB, IC, ID, &c. and let the firft Bearings be entered in the fecond Column, and the second Bearings in the third Column of your Field-Book♣ Then it is plain that the Points of Interfection, made from the Bearings in the fecond and third Columns of every Line, will be the angular Points of the Field or the Points A, B, C, D, &c. which Points being joined by right Lines, will give the Plan ABCDEFGA required.

Bea.

Bea. 180 Dif. 28C. of the Sta. H and I.

The fame may be done from any two Stations within-fide of the Land, from whence all the Angles of the Field can be feen.

This Method will be found useful in Cafe the ftationary Distances from any Cause prove inacceffible, or fhould it be required to be done by one Party, when the other whofe Poffeffion it is, refufes to admit you to go on the Land.

To find the Content of a Field by Calculation, which was taken by Interfection.

In the Triangle AIH, the Angles AHI, AIH, and the Bafe HI being known, the Perpendicular Aa, and the Segments, of the Bafe Ha, Ai may be obtained by Trigonometry: And in the fame Manner all the other Perpendiculars Bb, Cc, Dd, Ee, Ff, Gg, and the feveral Segments at b, c, d, e, f, and g If therefore the feveral Perpendiculars be fuppofed to be drawn in the Scheme (which are here omitted to prevent Confufion arifing from a Multiplicity of Lines) it is plain that if from bBCDEeb,