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Having determined the natural slope from observing that of the adjacent hills-and if no such hills are near, it is to be determined from the nature of the soil,

Let A C= required slope, making angle n degrees with the perpendicular A E; then CE tangent of angle n to radius A E.

Let s = secant of the angle C A E; then A C and angle n degrees. See fig. 42.

secant to radius A E

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When there is no slope, A C coincides with A E, and S

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Example. What dimensions must be given to the transverse profile (or section) of a canal, whose banks are to have 40° slope, and which is to conduct a quantity of water Q, of 75 cubic feet, with a mean velocity of 3 feet per minute?— Weisbach's Mechanics, vol. 1, p. 444.

40°, consequently <CA E 50°, and

Here we have the < D C A the sectional area of figure C A B D

a = 25 feet.

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1.919694 divided into 25, gives 13,022868,

the square root of which X depth A E

and tangent 1,191754 if multiplied by 3,609 × 3,609 area of the

3,60873,609 nearly,

triangles ACE + B F D

15,522309, which taken from 25, will leave

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which is the least surface with the given slopes, and containing the given

area = 25 feet.

The results here found are the same as those found by Weisbach's formula, which appears to me to be too abstruse.

145. From the above, the following equations are deduced:

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146. Hence it appears that the best form of Conduits are as follows:

Circular, when it is always filled.

Rectangular, that whose depth is half its breadth.

Triangular, when the triangle is equilateral.

Parabolic, when the depth of water is variable and conduit covered, and in accordance with section 133.

Rectilineal, when opened, and in accordance with section 144.

For the velocity and discharge through conduits, also for the laying out of canals, and calculating the necessary excavation and embankment, see Sequel.

147. TABLE, SHOWING THE VALUE OF THE HEIGHT A E

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X,

where a area of the given section, hav

tangent of the angle D B F,

ing given slopes, and such that the area a is inclosed by the least surface or perimeter in contact. S secant and n = or complement of the angle of repose (see fig. 42).

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Slopes for the sides of canals, in very compact soils, have 1 base to 1 perpendicular; but generally they are 2 base to 1 perpendicular, as in the Illinois and Michigan Canal.

Sea banks, along sea shores, have slopes whose base is 5 to 1 perpendicular for the height of ordinary tides; base 4 to 1 perpendicular for that part between ordinary and spring tides; and slopes 3 to 1 for the upper part. By this means the surface next the sea is made hollow, so as to offer the least resistance to the waves of the sea. The lower part is faced with gravel. The centre, or that part between ordinary and spring tides, is faced with stone. The upper part, called the swash bank, is faced with clay, having to sustain but that part of the waves which dashes over the spring tide line. (See Embankments.)

RIGHT ANGLED TRIANGLES.

148. Let the given angle be C A B' (fig. 9). Let A B and A C = c, be the given parts in the right angled triangle 149. Radius = A B' A C.

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cos. <AC D.

150. Sine <CA B'= C B = cosine of the complement = 151. Cos. <CABAB = sine of the comp. of < C A B <ACB.

sine

cot. <

152. Tangent < C A B' — B T = cot. of its complement cot. HAC.

153. Cotangent C A B HK: HAC.

tan. of its complement — tan.

tan. <

154. Secant < C A B' — A T = <HA C.

cosec. of its complement cosec.

155. Cosecant < C A B
sec. of its comp.
156. Versed sine < C A B' = B B'.

A K

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157. Coversed sine <C A B' = HI= versed sine of its complement. 158. Chord << CA B' C B' twice the sine of the <CA B'. 158a. Complement of an angle is what it wants of being 90°.

1586. Supplement of an angle is what it wants of being 180°.

158c. Arithmetical complement is the log. sine of an angle taken from 10, or begin at left hand and subtract from 9 each figure but the last, which take from 10.

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159. Let ACB (fig. 9) represent a right angled triangle, in which A B c, BC a, and A C = b, and A, B, C, the given angles.

a, and A C

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And the sides can be found as follows:

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And by having the < A = 53° 8′ ... the < C = 36° 52′.

Otherwise,

36° 52′ Log. sine 9,7781186

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171a. Let the side B C — a = 384, and the angle C — 36° 52′ be given to find c, b, and the angle A.

90° 36° 52/ = <A = 53° 8,

and a tan. C = c, that is 384 × 0,7499 = A B = 288 nearly.

1716. Let the sides be given to find the angles A and C.

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In like manner the angle C may be found.

These examples are sufficient to enable the surveyor to find the sides and angles.

The calculations may be performed by logarithms as follows:

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