CHAPTER III.

LAND SURVEYING BY THE CHAIN ONLY. THIS method of surveying has long been adopted by experienced surveyors; who have found it, in general, more accurate and expeditious, as well as better adapted to laying down extensive surveys, especially where no serious obstructions from woodlands, water, buildings, &c., exist; the use of the cross, in this method, being entirely excluded by some surveyors, and by others only used for secondary purposes, as for taking occasionally long offsets, or for squaring of lines obstructed by buildings, water, &c. Instead of the cross some use the Optical Square for these purposes; which will be hereafter described; while some erect perpendiculars with the chain only, as shall be shewn in the following Preliminary Problems.

The fundamental lines of surveys of this kind usually form a large triangle, or several triangles, abutting from one common base, which ought, if possible, to extend throughout the whole length of the survey. The sides of the triangle, or triangles, must run as near as possible to the external and internal fences of the estate, or district, to be surveyed; the sides of each triangle being connected by one or more lines, running anywhere within the triangle, to determine the accuracy of the work. These lines are called proof or tie-lines; and where the estate to be surveyed contains a great number of inclosures, the proof-lines may almost always be found available in determining the positions of some of these inclosures. Where a great number of lines run within the main triangle, they are called secondary lines, and are usually numbered for the sake of reference. Some surveyors number the stations, or extremities of the lines; but the former method is here recommended. In small surveys, for preliminary instructions, the numbering of the lines is unnecessary, the stations being referred to by the letters of the alphabet, as already done in Chap. II.

PRELIMINARY PROBLEMS.

PROBLEM I.

TO ERECT A PERPENDICULAR WITH THE CHAIN.

Let a B be a chain-line, and A B the extended chain. It is required to erect a perpendicular to a B at B. Fix the end of the chain to the ground with an arrow at B; fix also the 80th link of the chain, reckoning from B, at m, 40 links from B; 80 links of the chain now lying slack between B and m. Take

hold of the 30th link of the chain from B, and extend it till it take the position Bnm, the portions Bn, mn of the chain being pulled tight; then' shall Bn be perpen

dicular to the chain-line a B, and may be extended to any length required.

This method of erecting a perpendicular, though not so expeditious as that by the cross or optical square, is quite sufficient for those surveyors, who scarcely once require a perpendicular in their operations for weeks together; thus avoiding the inconvenience of daily carrying a cross, or other such like instrument for this purpose.

PROBLEM II.

TO MEASURE A LINE IMPEDED BY AN OBJECT NOT OBSTRUCTING THE SIGHT.

Let A B be a chain-line, the direct measurement of which is prevented by the unforeseen obstruction of the pond

P. Measure An till it reach to, or near to, the edge of the pond, as to n, and fasten the ends of the chain to the ground with arrows at m and n, the

A

P

B

m n

P

distance m n being made half a chain or 50 links. Take hold of the middle of the chain, and extend it firmly, till its two halves rest in the positions mo, on; thus making an equilateral triangle mno, each side of which is 50 links. In the direction mo, measure to nearly opposite the middle of the pond, as to q. Again, make p q equal 50 links, fasten the ends of the chain at and q, and extend its middle point to r, as before. In the direction qr, measure to s, till qs be equal to mq. Then s will be in the line AB, and mst will be equal to mq or q s which being added to Am will give the distance A s. Offsets being taken to the margin of the pond, during the measure

Р

* Since the parts Bn, nm of the chain are together 80 links, of which Bn is 30, the remainder n m is therefore 50; also Bm was made 40; whence 402302= 502, that is m B2 + Bn2 = m n2, therefore by Euc. I. 48, Bn is perpendicular to Bm, or m Bn is a right angle.

Because the triangles mno, pqr, are both equilateral, the angles at m and 7 are each 60° or one-third a right angle; whence by Theorem IV. the angle at s is also 60°; therefore all angles of the triangle, and consequently its sides, are equal, that is, m q = q s = m s.

ment of the line mq, qs, and proper notes of the operation made in the field-book, the measurement from s to B may be continued.

PROBLEM III.

TO MEASURE A LINE IMPEDED BY AN OBJECT OBSTRUCTING THE SIGHT, AS A BUILDING.

Let A B be a chain-line, the measurement of which is prevented by the building B. At m, four or five chains from the building, take a perpendicular mn, of such a length that the line ns may clear the building

At or near the building take another perpendicular pq, exactly equal to mn, (these perpendiculars ought to be measured with

the chain or a tape-line, if longer than the offset-staff,) and poles being put up, correctly vertical, at n and q, measure qsu in the direction ng of the poles, taking offsets to the building till it be cleared at s. Now on the line qu, at the distance su, at or about equal to mp, erect the perpendiculars s r, ut, each exactly equal to mn or pq, fixing poles, correctly vertical, at r and t. These poles are evidently in the true direction AB, and the measurement of the line may now be continued from r to B, after adding the distance qs (which is equal to pr) to Ap.

If the building, or other object, only protrude a few links over the line, the prependiculars mn, pq, sr, &c., may be erected by the offset-staff, as nearly correct as can be judged by the eye, and the results will be sufficiently accurate.

NOTE 1. When an object, as a pond or pit, not obstructing the sight, protrudes only a short distance over the line (see last figure); it will be sufficient to erect only the two equal perpendiculars pq, rs near its margin, with the offset-staff, as correctly as can be judged with the eye, and the distance q s, being measured, and added to Ap, will give the distance A r.

NOTE 2. Some unskilful surveyors square off the line, as they call it, when it is obstructed by a building, or other object, that impedes the sight, in the following manner. On arriving at or near the obstruction, as at p, a perpendicular p q is erected to Ap, another qs is erected to pq; a third sr, equal to pq, is erected to qs; and lastly a fourth perpendicular rB to rs. Though this last perpendicular is theoretically in the line A B, the student will at once perceive that when so many as four perpendiculars are taken, one upon another, at a very short distance from one another, that slight inaccuracies in the observations, as well as in the perpendicularity of the poles, placed at p, q, s, will have an almost unavoidable tendency to derange the accuracy of the work; since a small error, made at the beginning, multiplies as the operation proceeds. But by the method given in this Problem, the two perpendiculars on each side of the obstruction, are placed so far apart, that a slight deviation in the perpendicularity

of the poles cannot materially affect the accuracy of the work, while a slight error in erecting the perpendiculars, provided their lengths be made exactly equal, will not affect the work in the slightest degree. If, therefore, ordinary care be taken, the chance of error is almost impossible.

PROBLEM IV.

TO FIND THE WIDTH OF A RIVER, WHICH IS TOO WIDE TO BE REACHED ACROSS BY THE CHAIN.

Let A B be the chain line crossing a river, situated between o and p, a mark being fixed at p, on A B lay off on, nm, each equal 50 links, and with the ends of the chain successively fixed at o, n, and at n, m, lay down the equilateral triangles o qn, nrm, as in Problem II., poles being fixed at n, q, and r. In the two directions p q, nr, fix a pole at s, and measure the distance rs accurately with a tape line to one-eighth of a link. Then by the similar triangles srq, qop, we shall have rs : ġr :: oq : op.

But qroq = o n = 50 links, therefore,

B

P

n

m

Whence the distance op becomes known.

2

For those who do not understand a rule, when symbolically expressed, we give, in words at length, the following.

RULE. Divide 2500 by the distance rs, and the quotient will be the breadth of the river, or the distance op, which must be added to Ao to give the distance Apî.

EXAMPLES.

1. Required the breadth of a river by this method, when r s measures 15 links.

2. When r s measures 133 links, required the breadth of the

TRIANGULAR FIELDS.

When a triangular field, or piece of ground of that shape, is to be surveyed, set up poles or marks at each corner, and measure each side, leaving marks in at least two of the lines, and

entering their distances in the field-book; then measure the distance between the two marks for a proof-line :—or, one mark only may be left in one of the lines, which may be connected with its opposite angle for a proof-line.

EXAMPLES.

1. Required the construction and area of a field from the following dimensions.

When the triangle A B C is constructed, the proof-line mn will be found to measure 384 links, shewing that there has been no error in the work: also the perpendicular CD will be found to be 770 links; whence the area of the triangle = 1338 x 770 ÷ 2 = : 5.15130 =5a. Or. 24p. the area.

NOTE. If the proof line measured from the plan, does not exactly, or very nearly, agree with that measured in the field, some error has been made, and the work must be repeated.

TO FIND THE AREA OF A TRIANGLE FROM THE THREE SIDES. RULE. From half the sum of the three sides subtract each side severally and reserve the three remainders; multiply the half sum continually by the three reminders, and the square root of the product will be the area.

NOTE. By this rule the area of a triangle may be found without laying it down, or finding the perpendicular.

Adopting the preceding example, we have by the rule 1338 +852 + 1244

2

= 1717

half sum of the three sides.

Then 1717 - 1338 379 = 1st remainder; 1717 - 852 = 865 = 2nd remainder; 1717 1244 = 473

=

3rd remainder;

whence (1717 × 379 × 865 × 473) =6·15992 = 5a. Or. 25}p.