Example 8.-Section 6.

192

722 feet, which is four times the focal distance.

= .0678 feet, the offset at 7 feet.

722

EXAMPLES.

Section 1.

1. Find the offsets, at the end of every two chains, to 10 chains from the tangent to a railway curve whose radius is 184 chains.

Section 2.

2. The distance from the vertex to the focus of a parabola is 7 feet 6 inches; find the deflection from a distance of 18 feet 9 inches along the tangent.

3. The beam of a first rate is 42 feet, and cambered 9 inches in the centre; what are the offsets at the distance along the tangent of 5, 10, and 15 feet, considering the curve as a parabola?

Section 3.

4. From the extremity of the major axis of an ellipse whose diameters are 18 and 14 feet, a distance of 5 feet 6 inches is measured; what is the deflection to the curve at this point?

5. The radius of the earth is 3979 miles; find the offset at the distance of 5 miles, supposing the earth to be a sphere.

Section 4.

6. The equatorial diameter of the earth is 7924.873 miles, and the polar diameter is 7898.634 miles; find the offsets from a tangent to the meridian at the equator, at the distances of 4, 16, and 20 miles.

Section 5.

7. The height of the hull and mast of a vessel is 100 feet; to what distance must it sail before it disappears below the horizon? Find the same when the vessel is 200 feet.

Section 6.

8. The beam of a first-rate is 38 feet, and cambered 6 inches in the middle; what are the offsets at the distances along the tangent of 7, 12, and 14 feet, considering the curve a parabola from the middle?

PROBLEM 7.

Given the base and height of the segment of a circle, to find the length of its arc AD B.

1st. When the radius

and degrees in the arc are

given.

B

Then, 180 degrees in ADB:: 3.1416 × radius to the length of the arc AD B.

2nd. When the base (A B), and height or versed sine (CD) are given.

Find the radius by Problem 4.

Divide (CD) the height by (A C) the base, and opposite to this quotient in Table (A) there will be found one-fourth of the degrees in the arc ADB.

Find the length corresponding to these degrees by case the first.

Approximate Rule (usually called Huygens' Theorem).

From eight times the chord of half the arc, subtract the chord of the whole arc, and one third of the remainder will be the length of the arc, nearly.

1. Find the length of the arc, when the base is 30 feet, and height 3 feet.

Therefore the degrees of the arc of the segment are

4 (11° 18′ 35′′) = 45° 14′ 20′′ = 45.239°.

.. 180 45.239 :: 3.1416 x 39: 30.793 feet, the length of the arc.

The approximate rule by Huygens' Theorem, gives 30.879 feet.

Section 1.

1. Find the length of the arc, when the base and height are 40 and 4 feet.

Section 1-(continued).

2. Find the length of the arc, when the base and height are 35 and 5 feet.

Section 2.

3. The bridge at Wearmouth has a rise of 30 feet, and a span of 240 feet; find the length of the intrados.

4. Vauxhall-bridge has a rise of 11 feet 6 inches, and a span of of 78 feet; find the length of the intrados.

5. Southwark-bridge has a rise of 24 feet, and a span of 240 feet; find the length of the intrados.

Section 3.

6. The Dunkeld-bridge crosses the Tay with a span of 90 feet, and a rise of 30 feet; find the length of the intrados.

7. The bridge of Augsburg, on the Lech, has a span of 114 feet with 10 feet 6 inches rise; find the length of the intrados.

Section 4.

8. The bridge of Freysingen, on the Isar, has a span of 153 feet, with a rise of 11 feet 6 inches; find the length of the intrados.

9. The bridge of Sharding, on the Rolt, has a span of 194 feet, with a rise of 18 feet 8 inches; find the length of the intrados.

PROBLEM 8.

Given the radius, length of helix at the extremity of the screw blade, the length of the screw; to find its pitch.

The line A E is the axis of the screw, ABDE is the screw blade, B D is the helix, A B is the radius, and D C AE is the length of the screw.