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the angle at the centre, subtended by the arc of 10 chains with a radius of 60 chains.

Or, 9:4248 ) 5400 ( 572':95 = 9° 32':95 = A OD, the angle at the centre.

That is, divide the length of the arc for the quadrant by 10, and by the quotient divide 5400, the number of minutes in the quadrant, which will give the number of minutes in the angle at the centre A OD, and this divided by 60, gives the angle in degrees and minutes.

As another example, let the length of the arc still be 10 chains, but the radius 25 chains.

3:1416 x 12.5 = 39:27 = length of arc in quadrant. And,

3.927 ) 5400 ( 22° 55' = angle A OD at centre. Let us return now to the radius of 60 chains. The first thing we shall require will be the length of a tangent, (not a trigonometrical tangent) but a line so called, because it touches the circle without cutting it, as before observed; this line A C, see Fig. 83, we require in order to set off from C, and at right angles to A C, the length of the offset CD, which will give a point on the curve.

In our first demonstrations, A OD has been called the complement of the angle DOB; we may reverse this, and call À O D the angle, and DO B the complement.

It has been shown that A C= cosine of DO B.

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90°-9° 32.95 = 80° 27':05 ;
AOD=9° 32-95, and DOB= 80° 27:05
Nat. Cos. of 80° 27' = -1659;
-1659 x 60 = 9.954 = AC;
Nat. Vers. 9° 32.95= .01386;
01386 x 60 = 0.8316=CD;
Nat. Tang. 9° 32-95 = 1682;
•1682 x 60 = 10092 = A G;
Nat. Secant 9° 32.95 = 1.01405;
1:01405 x 60 = 60-843 = 0 G;

60:813 – 60 (Radius) = 1843 = G D. To find A c, and cd;

Half the angle A OD = 4° 46' 30"
90°-4° 46' 30" = 85° 13' 30" = Complement of Angle;

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OFFSETS. PRACTICE OF RANGING CURVES.

137

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Ac= Cosine of 85° 13' 30" x 60

cd=Versine of 4° 46' 30" x 60 We shall by this means find A c=4.995

and c d=02082, or very nearly 21 links.

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With regard to the lengths of the intermediate offsets, observe that, as an approximation, the lengths of these will be at the consecutive chains as the squares of the times; that is,

if at 1 chain = 1 then at 2

2 x 2 x1 = 4 at 3

33x1=9 at 4

4 x 4 x 1 = 16, &c., &c. and at 10 10 x 10 x1 = 100. In the case now in hand, it is at 10 chains = 0·8316; then at 1 chain it will be 100th part of 0·8316 = 0·008316 at 2 chains it will be 4 times this

= 0.0332
at 3
9 9

= 0.0748
&c. &c. &c. &c.
The ten first chains are now set out.

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To carry on the course of the curve, make the angle TGT9° 32':95 2 = 19° 5'-90, and repeat along G E and E F the same operations we have been carrying out along A G. Having reached F, make the angle T F L=TG T, that is = 19° 5'-90.

On reaching K, after passing H, the length of the arc measures 4:40. To reach the tangent K L, we must multiply our offset at 1 chain = 0·008316 by (++ x tit) = 19:36, which will give for offset 0:1607 . . . which if the curve has been correctly set out, will be the exact measurement from the tangent to the point K, supposed previously fixed.

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In practice the fixing of A and K is the first thing to be done, in order that we may have the exact points at which the curve joins the tangents A T, KT. To ascertain this, range out as correctly as possible A T and K T from the plan, measure the angle K TX, one half of which, as already shown, will be equal to one of the angles at the centre, and which has in our former demonstrations been called the angle A OD.

From the tables find the natural tangent of half the angle KTX, and multiply it by the radius, which will give the length of T A and KT, A and K being the points at which the curve is to commence and terminate.

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At the same time that the natural tangent is taken from the tables, take out also the natural secant of the angle at the centre, multiply it by the radius, and then subtract the radius ; the remainder will be equal to T S, which in the Figs. 81, 82, 83, and 84, we have hitherto called G D. Ex. Angle KTX = 42° 24. 42° 24'

= 21° 12" = one of the two angles at the centre.

2 Nat. Tangent 21° 12 = -3879 x 60= 23.274 = A T or TK. Nat. Secant 21° 12 = 1.07258 x 60 = 64:3548-60=4:3548=T S.

But 180° -42° 24' = 137° 36' + 2 = 68° 48'

If then we make either of the angles A TS, K TS=68° 48', and make TS= 4:3548, we shall have the point S on the curve.

To find the Length of the Curve.—Refer to the proper table for the lengths of arcs, take out the tabular numbers referring to the degrees and minutes subtended by the angles at the centre, and multiply them by the given radius, which will give the length of curve. Ex: Angle 42° 24'; Tabular number for. 42° = 73303

24' = .00698

:

74001

60

Length of curve = 44:40060. Another method to ascertain the length of the curve will be found further on.

In Setting out by Secants.-If we choose to set out, as, for instance, in the case of a long curve, more points than the central one S by means of secants before proceeding with any other method, we may take half the tabular tangent already found, and ascertain by this in the tables the corresponding angle, and consequently the tabular secant, Ex: In Fig. 86 the tabular tangent for A T= 3879 as above; 3879

= '19395 = nat. tangent 10° 58' 30",

2 nat. secant 10° 58' 30" = 1:01863 x 60 = 61.1178-60= 1.1178 = GD; and,

90° -10° 58' 30" = 79° 1' 30" = AGD

X

RANGING CURVES WITH THE THÉODOLITE.

139

The method above explained we will call setting out by tangents and secants, in order to distinguish it from the method to be given next, and which is almost entirely performed by instrumental observation, and consists in setting off with a theodolite the angles T A a, TAb, TAc, &c., Fig. 87, at the same time that the chainage A a, ab, a c, &c., is carried forward, A a, Ab, A c, &c., subtending so many angles at the centre.

In setting out by this method we have first to ascertain the points A and k, at which the curve commences and terminates; this is obtained by measuring the angle XTL, Fig. 85, explained above. We thus ascertain the lengths of the tangents TA and Tk, and these being measured off from T give the points A and k.

Then divide the constant 1718-9 by the radius of the curve to obtain the angle T A a, and call this angle A.

Then TA b= 2 A

TAC=3A

TA d = 4 A, &c., &c. Ex: Let the angle X Tk, Fig. 85, be equal to 38° 30', equal to twice the angle at the centre, which we have called A OD in the former figures ; and let the radius = 60 chains.

Nat. tangent 19° 15 = •34921

34921 x 60= 20.953=T A or Tk. This distance measured off on the tangents gives the points A and k for commencement and termination of the curve.

1718:9-60 = 28.65 ET A a= tangential angle, Fig. 87. 28'-65 x 2 = 57:30 =TAB

28:65 x 3 = 1° 25.95 = T AC, &c., &c. Length of Curve.-Observe that half the angle X Tk (Fig. 85) = the angle A OD, Figures 81, 82, and 83, is also equal to the angle of deflection G A D in those Figures, and T Ak, Figures 85 and 87.

Divide half the angle x Tk or the angle of deflection in minutes by the tangential angle, and the product will be the length of the curve. Ex:

38° 30' +2=1155', and

28'65)1155(40:314 = length of curve in chains. Before removing the instrument from T set out the angle ATS, which is equal 90°-19° 15'.

Nat. secant 19° 15 = 1.05922;
1.05922 x 60 = 63-553-60=3553=TS.

.

O

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We thus obtain the point S at the centre of the curve, by the means of which we have check on reaching that point.

To Range the Curve by the Tangential Angle.—Plant the theodolite at A, with the zero on T, set off the tangential angle T A a = 28':65, and measure off 1 chain from A to a; the intersection of the visual line with the end of the chain fixes the point a; for the second chain set the vernier of the instrument at twice T A a = 57':30, and measure another chain from a to b, when the intersection will fix the point b, and so on.

Having thus set out a certain number of chains, say to d, move the instrument with the vernier fixed at the angle TAI to the point d. There fix the instrument with the vernier on A, and then move round to zero, which will place the visual line on the tangent d T'; reverse the telescope, which will give the tangent produced to T3, and set out the angle T3d e = 28'65, T3df = 57' 30, &c., as before.

Repeat this operation till we get to the end of the curve.

Generally 10 chains at each plant will be sufficient ;* when after reaching i, i k is a fraction of a chain, then the angle i dk will be a corresponding fraction of the tangential angle; thus if i k = 040, the id k = 11'46.

Given Two Points on the Ground to Pass a Curve through them when each is visible from the other.— Measure the length A k, Figures 85, 86, 87, and the angle T Ak, and divide the angle in minutes by the length, which will give the tangential angle. Possibly the difference between the length of the chord and that of the arc may be sufficient to prevent the last stump set out from reaching k properly; but we shall then have the correct length of the arc, by which again dividing the angle of deflection we shall obtain the correct tangential angle.

To find the Radius of the Curve thus set out divide the constant 1718-9 by the tangential angle, and the quotient will give the radius.

Therefore if from A we can obtain sight of k and the distance A k, we ascertain the angle of deflection, the tangential angle, the angle at the centre subtended by half the arc, and also the radius. On reaching the point k we have merely to fix the direction of the straight tangent at that point by the method already explained. Giren Two Points on the Ground not visible from each

* Unless the verniers should read a fraction of a minute.

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