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Font size 133. Measure the SIDES op the field, and the
DIAGONALS by which it is divided into triangles.

By measuring the sides (Pig. 32.)
AB, BC, CD, DE, EA,
and the diagonals CA and CE, we have the three sides of
each of the triangles into which the whole figure is divided.
They may therefore be constructed, (Trig. 172.) and their
areas calculated. (Mens. 10.)

134. To measure an angle with the chainy set oS equal
distances on the two lines which ioclude the angle, as AB,
AC, (Fig. 39.) and measure the distance from B to C. There
will then be given the three sides of the isosceles triangle
ABC, to find the angle at A by construction or calculation.

The chain may be thus substituted for the compass, in
surveying a field by going round it, according to the method
explained in the preceding section ; or by measuring from
one or two stations, as in problems I and II.

PROBLEM IV.

To survey an irregular boundary by means of offsets^

136. Run a straight line in any convenient di-
rection, AND MEASURE THE PERPENDICULAR DISTANCE
OF EACH ANGULAR POINT OF THE BOUNDARY FROM THIS
LINE.

The irregular field (Fig. 40.) may be surveyed, by taking
the bearing and length of each of the four lines AE, EP, FI,
lA, and measuring the perpendicular distances BB', CC,
DD', GG', HH', KK'. These prpendiculars are called
offsets. It is necessary to note in a field book the parts

V

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into which Ihe line that is measured is divided by the oÂŁ&eld|
as iji tte following example. (See Fig. 40.)

Offsets on

the left.

Courses and Distanceig.

Offsets on the right.

Chains.

AE N. 85Â° E. 12.74 ch.

^

BB'

2.18

AB' 3.25

CC'

2.18

B'C 2.13

DD'

1.23

CD' 1.12
D'E 6.24

EPS. 24Â° E. 7.23

FI N. 87^ W. 13.34

GG'

2.86

FG' 3.84

HH'

1.48

G'H' 2.22

H'l 7.28

â€˘

lA N. 26Â° W. 6,32

IK'

WK.I 2.94

K'A

As the offsets are perpendicular to the lines surveyed, the
little spaces ABB', BB'CC, CC^DD^ <fcc. are either right an-
gled triangles, parcdlelograms, or trapezoids. To find the
contents of the field, calculate in the first place the area be-
tween the lin^s surveyed, as the trapezium AEFIA, (Fig. 40.)
and then add the spaces between the offsets, if they tail with-
in the boundary line ; or subtract them, if they fall without,
asAIK.

When any part of a side of a field is inaccessible^ equal
offsets may be made at each end, and a line run parallel tp
the boundary.

PROBLEM Y,

To measure the distance between any two points on the suT"
face of the earthy hy means of a series of triangles extend-
ing from one to the other.

136. Measure a side of one of the triangles for
A BASE LINE, take the BEARING of this or some
other side, and measure the angles in each of the
triangles.

If it be required to find the distemce between the two
points A and I, (Fig. 41.) so situated that the measure cai^
not be taken in a d&ect line frc^n one to the other; 1^ a 80^

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7g smtvfinNo.

ries of triangles be arranged in such a mimner between thenii
that one side shall be common to the first and second, as BC,
to the second and third as CD, to the third and fourth, &c.
Then measure the length of BC for a base line, take the bear-
ing of the side AB, and measure the aisles of each of the
triangles.

These data are sufficient to determine the length and bear-
ing of each of the sides, and the distance and bearing of
I from A. For in the two first triangles ABC and BCD, the
anÂŁ[les are ^ven and the side BC, to. find the other sides.
When CD is found, there are given, in the third triangle
CDE, one side and the angles, to find the other side. In die
same manner, the calculation may be carried ^om one trian-
gle to another, *till all the ^ides are found.

The bearings of the sides, that is, the angles which they
make with the meridian, may be determined from the bear-
infi[ of the first side, and the angles in the several triangles.
Thus if NS be parallel to AM, the Angle BAP, or its equal
ABN subtracted from ABD leaves NBD ; and this taken from
180 degrees leaves SBD.

From the bearing and length of AB may be found the
southing AP, and the easting PB. In the same manner are
found the several southings PP', P'P", P"P'", F^'M. The
sum of the southings is the line AM. And if the distance is
so smcdl, that the several meridians may be considered paral-
lel, the difference between the sum of the eastings and the
Â«uia of the westings, is the perpendicular IM. We have
then, in the right angled triangje AMI^ the sides AM and MI^
to find the distance and bearing of I from A,

137. Th s problem is introduced here for the purpose of
giving the general outlines of those important operations
which have been carried on of late years, with such admira-
ble precision, under the name of Trigonometrical Surveying.

Any explanation of the subject, however, which can -be
made in this part of the course, must be very imperfect. In
the demonstration of the problem, the several triangles are
supposed to be in the same plane, and the distances of the
meridians so small, that they may be considered parallel. But
in paractice, the ground upon which the measurement is to be
made is very irregular. The stations selected for the angu-
lar points of the triangles, are such elevated parts of the counr
tiy as ar^ visible to a consUerable distance. They should

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BUUyiEKiNCL 79

be so situated, that a signal staff, tower ^ orbther conq>iciious
object in any one of the angles, may be seen from th^ other
two angles in the same triangle. It will rarely be the case
that any two of the triangles will be in the same plane, or
any one of them paxallel to the horizon. Reductions will
therefore be necessary to bring them to a common leveL
But even this level is not a plane. In the cases in whidi this
kind of surveying is commonly practised, the measurement
is carried over an extent of country of many miles. The
several points, when reduced to the same distance from. the
center of the earth, are to be considered as belon^in^ to a
spherical surface. To make the calculations then, if the line
to be measured is of any considerable extent, and if nice ex-
actness is required, a knowledge of Spherical Trigonometry
is necessary.

138. The decided superiority of this method of surveying,
in point of accuracy, over all others wliich have hitherto
been tried, particularly where the extent of ground is great,
is owing partly to the fact that almost all the quantities meas-
ured are angles ; and partly to this, that for the single line
which it is necessary to measure, the ground may he chosen^
any where in the vicinity of the system of triangles. It
would be next to impossible to determine the precise horizon-
tal distance between two points, by carrying a chain over an
irregular surface. But in the trigonometrical measurements
which are made upon a great scale, there can generally be
found somewhere in the country surveyed, a level plane, a
heath, or a body of ice on a river or lake, of sufficient extent
for a ha^e. This is the only line which it is absolutely ne-
cessary to measure. It is usual, however, to measure a sec-
ond, which is called a line of verification. If the length of
the base BC (Pig. 41.) and the angles be given, all the other
lines in the figure may be found by trigonometrical calcula-
tion. But if GH be also measured, it will serve to detect any
error which may have been committed, either in taking the
angles, or in computing the sides, of ihe series of triangles
between BC and GH.

139.. In measuring these lines, rods of copper or platina
have been used in France, and glass tubes or steel chcdns ia
England. The results have in many instances been extreme-
ly exact. A base was measured, on Hounslow Heath, by
General Roy, with glass rods. Several years after, it was re-

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QQ 80B.VJCIIIMI.

meuiired bf Cobnel Mndge, vs^h a steel chain of very nice
Gonstrucdon. The difference in &e two measurements was
less than three inches in more than five miles. Two parties
measured a base in Peru of 6272 toises, or more than seven
mUes ; and the dififerenee in their resutts did not exceed two
inches.

Exact as these measurements are^ the exquisite con-
struction of the instruments which have been used for taking
the angles^ has given to that part of the process a still high^
degree of Defection. The amount of the errors in the an-

gles of each of the triangles, measured by Ramsden's Theodo-
te, did not exceed three seconds. In the great surveys in
France, the angles were taken with nearly t& same correct-
ness.

140. One of the most important applications of trigono-
metrical surveying, is in measuring arcs of the meridian, or
of parallels of latitude, particularly the former. This is ne-
cessary in determining the figure of the earth, a very essen-
tial problem in Geography and Astronomy. A degree of
zeal has been displayecj on this subject, proportioned to its
practical importance. Arcs of the meridian have been meas-
ured at great expense, in England, France, Lapland, Peru,
&c. Men of distinguished science have engaged ia the un-
dertaking.

A meridian line has been measured, under the direction of
General Roy and Colonel Mudge, from the Isle of Wight, to
Clifton in the north of England, a distance of about 200
miles. Several years were occupied in this survey. An-
other arc passing near Paris, has been carried quite through
France, and even across a part of Spain to Barcelona. In
measuring this, several distinguished mathematicians and as-
tronomers were engaged for a number of years. These two
arcs have been connected by a system of triangles runninff
across the English Channel, the particular object of which
was to feterimne the eyact difference of longitude between
the observatories of Greenwich and Paris. Besides the me-
ridian arcs, other lines intersecting them in various directions,
have been measured both in England and France. With
these, the most remarkable objects over the face pf the coun-
try have been so connected, that the geography of the vari-
ous parts of the two kingdoms is settled, with a precision
which could not be expected fcom any other method.

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flURVSYINO. Q^

141. Hie exactness of the surveys will be seen from a
comparison of the lines of verijicaiion as actually measured,
with the lengths of the same hnes as determined by calcula-
tion. These would be aflfected by the amount of all the er-
rors in measuring the base lines, in taking the angles, in com-
puting the sides of the triangles, and in making the necessary
reductions for the irregularities of surface. A base of veri-
fication measured on Iu>mney Marsh in England, was found
to di^r but about ttoo feet from the length of the same line
as deduced from a series of triangles extending more than 60
miles. A base of verification connected with the meridian
passing through France, was found not to differ one foot from
the result of a calculation which depended on the measure-
ment of a base 400 miles distant. A line of verification of
more than 7 miles, on Salisbury Plain, differed scarcely an
inch from the length as computed from a system of triangles
extending to a base on Hounslow Heath.*

Â« See Note K.
H

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g2 WURVBVINO.

SECTION m.

LAYING OUT AND DIVIDING LANDS.

Art. 142. To those who are famiUor with the principles
of geometry, it will be unnecessary to give particular rules,
for all the various methods of dividing and laying out lands.
The following problems may serve as a specimen of the man-
ner in which the business may be conducted in practice.

t>ROBLÂŁM t.

T\> lay out a given number of acres in the form qf a square.

143. Reduce the number of acres to Square rods

OR CHAINS, AND EXTRACT THE SQUARE ROOT. This wiU

give one side of the required field. (Mens. 7.)

Ex. 1. What is the side of a square piece of land contain-
ing 124 J acres ? Ans. 141 rods.

2. What is the side of a square field which contains 68f
acres?

PROBLEM IL

To lay out a field in the form of a parallelogram, when
one side and the contents are given.

144. Divide the number of square rods or chains
BY the length of THE GIVEN SIDE. The quotieut will
be a side perpendicular to the given side. (Mens. 7.)

JBx. What is the width of a piece of land which is 280
rods long, and which contains 77 acres?

Ans. 44 rods.

Ck>r. As a triangle is half a parallelogram of the same
base and height, a field may be laid out in the form of a tri-
angle whose area and base are given, by dividing twice the
area by the base. The quotient will be the perpendicidar
fix>m the opposite angle. (Mens. 8.

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SURYSTINCk 83

PROBLEM III.

To lay mU a piece of land in the form of a parallelogram, the
length of which shall be to the breadthin a given ratio.

146. As the length of the parallelo|raiD, to its breadth ;
So is the area, to the area oi a square of the same

The side of the square may then be found by problem I,
and the length of the parallelogram by problem II.

If BCNM (Fig. 42.) is a square in the right parallel o^am
ABCD, or in the oblique parallelogram ABC'D', it is evident
that AB is to MB or its equal BC, as the area of the parallel-^
ogram to that of the square.

Ex. If the length of a parallelogram is to its breadth as
7 to 3, and the contents are 52^ acres, what is the length and

PROBLEM IV.

TTie area of a parallelogram being given, to lap it out in
such a form, that the length shall exceed the breadth by a

given DIFFERENCE.

146. Let x^BG the breadth of the parallelogram ABCD

(Fiff. 42.) and the side of the square BCNM.

d^AM the difference between the length and

aÂ»the area of the paralleloffram.
Then a'^{x+d)xx^x^ +dx. (Mens. 4.)
Reducing this equation, we have
' "^a+id^â€”id^-x.

That is, to the area of the parallelogram, add one fourth
of the square of the difference between the length and the
breadth, and from the square root of the sum, subtract half
the difference of the sides ; the remainder will be the breadth
of the paridlelogram.

Ex. If four acres of land be laid out in the form of a par-
allelogram, the difference of whose sides is 12 rods^ what is

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84

PROBLEM V.

To lay out a triangle whose area and angles are given.

147. Calculate the area of any supposed triangle

WHICH HAS the SAME ANGLES^ ThEN

As THE AREA OP THE ASSUMED TRIANGLE,
To THE AREA OF THAT AVHICH IS REQUIRED ;
So IS THE SaUARE OF ANT SIDE OF THE FORMER,
To THE SaUARE OF THE CORRESPONDING SIDE OF THK
LATTER.

If the triangles B^CC and BCA (Pig. 43.) have equal an-
gles, they are similar figures, imd therefore their areas are
as the squares of their like sides, for instance, as A^Â« : ^'Â«.
(Euc. 19. 6.) The stjuare of CO' being found, extracting
the square root will give the line itself

To lay out a triangle of which one side and the area are
given, divide twice the area by the given side ; the quotient
will be the length of a perpendicular on this side firom tfie
opposite angle. (Mens. 8.) Thus twice the area of ABC
(Pig. 45.) divided by the side AB, gives the length of the per-
pendicular CP.

148. This problem furnishes the means of cutting ofl^ or
laying out, a given quantity of land in various forms.

Thus, from the triangle ABC, (Fi^. 43.) a smaller triangle
of a given area may be cut off, by a hue parallel to AB. The
line "CC being found by the problem, the point t)' will be
given, from which the parallel line is to be drawn.

149. If the directions of the lines AE and BD, (Pig. 44.)
and the length and direction of AB be given ; and if it be
required to lay off a given area, by a line parallel to AB ; let
the lines AE and BD be continued to C. The angles of the
triangle ABC with the side AB being given, the area may be
found. Prom this subtracting the area of the given bape-
&id, the remainder will be the area of the triangle DCE ;
from which may be found, as before, the point E through
which the parallel is to be drawn.

If the trapezoid is to be laid off on the ofther side of AB,
its area must be added to ABC, to give the triangle D'CE'.

150. If a piece of land is to be laid off from AB, (Pig.
45.) by a line in a given direction as DE, not paralM to AB \$

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suRVKviNa 85

let AC parallel to DE be drawn through one end of AB-
The required trapezium consists of two parts, the triangfle
ABC, and the trapezoid AGED. As the angles and one side
of the former are given, its area may be found. Subtracting-
this from the given area, we have the area of the trapezoid,
from which the distance AD may be found by the preceding
article.

151. If a given area is to be laid off from AB, (Fig. 46.)
by a line proceeding from a given point D ; first lay off the
trapezoid ABCD. . If this be too small, add the triangle
DCE ; but if the trapezoid be too large, subtract the triangle
DOE'.

PROBLEM VI.

To divide the area of a triangle into parts having given ra-
tios to each other, by lines drawn from one of the angles
to the opposite base,

152. Divide the base in the same proportion as

THE PARTS REQ,UIRE:D.

If the triangle ABC (Fig. 47.) be divided by the lines CH
and CD ; the small triangles, having the same height, lure to^
each other as the bases BH, DH, and AD. (Euc. 1. 6.)

PROBLEM VII.

To divide an irregular piece of land into any two given

parts.

153. Run a line at a venture, near to the true division line
required, and find the area of one of the parts. If this be
too large or too small, add or subtract, by the preceding artin
cles, a triangle, a trapezoid, or a trapezium, as the case may
require.

A field may sometimes be conveniently divided by redu-
cing it to a triangle, as in Art. 125, (Fig. 35.) and then di-
vidmg the triangle by problem VI.

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gQ UVBLLDIW.

SECTION IT,

LEVELLING.

Art. 164 It is frequently necessary to ascertain how much
one spot of ground is higher than another. The practica-
bility of supplying a town with water from a neighboring
fountain, wiil depend on the comparative elevation of the
two {)laces above a common level. The direction of the cur-
rent in a canal will be determined by the height of the seve-
ralparts with respect to each other.

The art of levellinff has a primary reference to the level
surface of water. The surface of the ocean, a lake, or a
river, is said to be level when it is at rest. If the fluid parts
of the earth were perfectly spherical^ every point in a level
surface would be at the same distance from the center. The
difference in the heights of two places above the ocean would
be the ^ame, as the mfference in their distances from the cen-
ter of the earth. It is well known that the earth, though
nearly spherical, is not perfectly so. It is not necessary, how-
ever, that the difference between its true figure and that of a
sphere should be brought into account, in the comparatively
small distances to which the art of levelling is commonly ap-
plied. But it is important to distinguish between the true
and the apparent level.

165. The TRUE LEVEL is a curve which either coincides
with, or is parallel to, the surface of water at rest.

The apparent level is a straight line which is a
TANGENT to the tTUC level, at the point where the observation

Thus if ED (Pig. 48.) be the surface of the ocean, and AB
a concentric curve, B is on a true level with A. But if AT
be a tangent to AB, at the point A, the apparent level as ob-
served at A, passes through T.

166. When levelling instruments are used, the level is de-
termined either by a fluid or a plumb-line. The surface of
the former is parallel to the horizon. The latter is pergeur

4'

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LBVSLUNO. g7

dicular. One of the most convenient instrninents for the
purpose is the spirit level. A ^lass tube is nearly filled with
spirit, a small space being left for a bubble of air. The tube
is so formed, that when it is horizontal, the air bubble will be
in the middle between the two ends. To the glass is attach*
ed an index with sight vanes ; and sometimes a small tele-
scope, for viewing a distant object distinctly. The surveyor
should also be provided with a pair of levelling rods^ which
are to be set up perpendicularly, at convenient distances, for
the purpose of measuring the height from the surface of the
ground to the horizontal line which passes through the spirit
fevel.

If strict accuracy is aimed at, the spirit level should be in
the middle between ^the two rods. Considering D'ED^D as
the sphericcd surface of the earth, and B'AB"B as a concen-
tric curve; a horizontal line passing through A is a tangent
to this curve. If therefore AT' and AT" are equal, the
points T' and T" are equally distant from the level of the
ocean. But if the two rods are at T and T', while the spirit
level is at A, the height TD is greater than T'D'. The dif-
ference however will be trifling, if the distance of the stations
T and T' be small.

157. With these simple instruments, the spirit level and
the rods, the comparative heights of any two places can be
ascertained by a series of observations, without measuring
their distance, and however irregular may be the ground be-
tween them. But when one of the stations is visible from
the other and their distance is known ; the difference of their
heights may be found by a single observation, provided al-
lowance be made for atmospheric refraction, and for the dif-
ference between the true and the apparent level.

PROBLEM I.

To find the difference in the height of ttoo places by levd^
ling rods.

168. Set up the levelling rods perpendicular to tfie horir
zouj and at equal distances from the spirit level; observe
the points where the line of ievel strikes the rods before and
behind, and measure the heights of these points above the
ground; level in the same manner from the second statim ,

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Og LBVELLINCk

to the thirdj from the third to the fourth, ^c The differ-
ence between the sum of the heights at the bcuJc stations^
mnd at the forward stations, wiU be the difference between
the height of the first station and the last

If the descent from H to H" (Fig. 49.) be required, let the
spirit level be placed at A, equally distant from the stations
H and H' ; observe where the line of level BF cuts the rods
which are at H and H^ and measure the heights BH and
FH'. The difference is evidently the descent from the first
station to the second. In the same manner by placing the
spirit level at A^ ihe descent from the second station to the
third may be found. The back heights, as observed at A
and A', are BH, and B'H' ; the forward heights are FH' and
F'H":

Now FH'â€” BH=the descent from H to H^
And F'H''â€” B'H' -the descent from H' to Wf ,
(FH' +F'H'0-(BH +B'HO=the whole descent fromH to H''.

159. It is to be observed, that this method gives the true
level, and not the apparent level. The lines BF and B'F'
are not parallel to each other; but one is parallel to a tangent
to the horizon at N^ the other to a tangent at N'. So that the
points B and F are equally distant from the horizon, as are
also the points B' and F'. The spirit level may be placed at
unequal distances from the two station rods, if a correction
is made for the difference between the true and the apparent
level by problem II.

. 160. If the stations are numerous, it will be expedient to
placd the back and the forward heights in separate columns
m a table, as in the following example.

Back heights.

Fore heights.

Feet. In.

Feet. Id.

1st. Observation

3 7

2 8

2. Â«

2 &

3 1

3. Â«

6 3

6 7

4. Â«

4 2

3 2

6. <Â«

6 9

4 10

-

22 2
19 4

19 4

DUSS^reuce

2 10

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LEVELUNQ.

If the sum of the forward heights is less than the sum of
the back heights, it is evident that the last station must be
higher than the first.

PROBLEM II.

161. To find the difference between the true and the ap-
parent levelffor any given distance.

If C (Fig. 12.) be the center oi the earth considered as a
sphere, AB a portion of its surface, and T a point on an ap-
parent level with A ; then BT is the diflference between the
tliie and the apparent level, for the distance AT.
Let 2BC =D, the diameter of the earth,

AT = rf, the distance of T, in a right line from A,

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