radius MA. By geometry, the angle AMELCB = 15° 52'.

By logarithms we have:

As external secant 15° 52' LCB

8.597789
1.698970

ex. sec. 15° 52′ *039603

CASE 2D.

To find the tangent A C, or CB; or point of curve.

By logarithms:

As R.

By natural tangents:

1262 x (natural tangent 15° 52′ = '26546)= 388 feet

CA CB.
=

Having located a curve connecting two tangents, it is required to move the middle of the curve any given distance, either towards or from the vertex.

Let the angle TLG 36° whole amount of curvature; the

arc A B C 1200 feet; the radius A N = CN= 1910 feet, and I B

10 feet.

It is required to find the radii HM and EO.

We have by logarithms:

External secant 18° half of 36° = ANL

=

1910+ 183 2093 = MH = radius of a 2° 44' curve;
and 1910-183= = 1727 OE radius of a 3° 19' curve.

By natural tangents:

=

=

183 x (natural tangent 18° 32429) = 594 = HA=AE

PROPOSITION VII. FIG. 6.

It is required to locate a tangent from an inaccessible point on a

curve.

Let A B C be the given curve with a R. of 1637 feet curving 3° 30′ per 100 feet; C the inaccessible point. Assume a point B, if convenient, at a given distance, say 300 feet, from C. Throw off a tangent, and measure, at right angles therefrom, B E external

Fig.6.

H

G

secant of arc BC; then to find by logarithms the distance BÈ, we

By natural external secants:

=

=

1637 × (nat. ex. sec. 10° 30′ = 017030) 27.88. Measure the line B E the instrument over E, complement of 10° 30'. required.

=

27 88 feet at right angles to BH. Set and turn off the angle BEC 79° 30′ = ECF will be the direction of the tangent

CASE 2D.

Suppose there be no convenient accessible point between A and C, produce the curve to D, measure the external secant DF as before, place the instrument at F, and turn off the angle DFC. This will give the direction of the tangent FC as before.

CASE 3D.

Should the lines AI and IC be more practicable for operating

than the curve A B C, calculate and produce the tangent from A to I, the vertex of the curve AB C, and turn off the angle KIF= A O C, and make I CAI, as calculated.

CASE 4TH.

Again, should the last method be found impracticable, and the chord AD clear from obstructions, measure the chord AD, and turn off tangent from D.

Suppose angle KAD 25°, then we have 1687 × (nat. sine 25° = •42262) 2 = AD = 1384 feet.

Note.-The arc A B C D contains 50° curvature.

PROPOSITION VIII. FIG. 7.

It is required to find a curve which will connect two lines without producing the tangents to an intersection.

The principle involved in this diagram affords an easy mode of solving a very interesting geographical problem. Suppose AE is a mountain near the sea or a very extensive level. Measure with an instrument for taking vertical angles the depression or "dip of the horizon KEB BOA; then external secant KEBx radius of earth- AEheight of mountain.

Let the line be either a curve LA, HA, or a tangent D A, as the case may be. Suppose it impracticable, by reason of buildings or other obstructions, to produce the tangent to a vertex x.

At A lay off with the instrument a right angle to tangent, and produce it till it meets FB produced in E. Measure this distance, and the angle AEB; then its complement AOB will be the amount of curvature required to curve on to the tangent B F. Suppose the angle A E B = 65°, then A O B = 25°. Let A E be = 120 feet, then we have by logarithms:

And 11608 x (tangent 25° = 46631) EB = 541.28 feet.

=

Then will be 25° of curvature ÷ 4° 56′ the rate of curvature, give the length of curve between the two given points A and B = 506.2 feet.

PROPOSITION IX. FIG. 8.

To draw a tangent to two curves already located.

Let the curve CRAGH, of 2000 feet radius, be located from tangent CK; and let ESBD be a curve of 2605 feet radius, located from tangent E F. We are required to find the points A and B having a tangent common to both.

Suppose to be the point in the first curve, and S the point in the second. There being obstructions in the way, we will run the zigzag line RLPS, making RL tangent to R, and PS tangent to S.

Suppose R L Q = 20°

and TPS = 15°;

let R L = 1100 feet, L P = 1300, and PS = 1400.

Assume radius NR as a meridian; that is, suppose NR to be due north. Then will RL be due west, LP south 70° west, PS south 85° west, and radius SM north 5° west. These artificial courses, then, will show the relative bearings, with which we obtain the following traverse: