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Plan No. III.
Estimate of the expense of raising the valley fifty feet, lowering the summit at Foster's Booth twenty-seven feet, and reducing the slopes from 1 in 16, 17, and 18, to 1 in 30, on the east side of the valley; and from 1 in 14 and 15 to 1 in 30, on the west side of the valley.
Plan No. IV.
Estimate of the expense of raising the valley forty feet, cutting twenty-seven feet from the summit at Foster's Booth, and reducing the inclinations from 1 in 14 and 15 to 1 in 30 on the west side of Lovell's Hill.
To 104,671 cubic yards of earth-work, at
To 12,440 cubic yards from back cutting,
To 63,931 cubic yards from back cutting, at Id.
To 2,611 cubic yards of earth from back
cutting, at 6d. To three acres of land to be purchased for
supplying earth for embankment, at 100/. To twelve acres of land to be purchased
for the site of embankments, and for
the cutting at Foster's Booth and Lovell's
Hill, at 100/.
To field-gates, &c. -
- 5,233 11 0
Plan No. V.
Estimate for the expense of raising the valley forty feet, cutting twenty-seven feet from the summit at Foster's Booth, and fourteen from the summit at Stowe HilL
Note B. Page 67.
The resistance produced by collision is seldom a constant retarding force; loose stones, or hard substances, are sometimes met with, and will give a sudden check to the horses, according to the height of the obstacle: the momentum thus destroyed is often very considerable.
The power required to draw a wheel over a stone or other obstruction may be thus determined: — Suppose A B D to represent a carriage wheel fifty-two inches in diameter, the axis 2' 5 inches in diameter, the weight of the wheel 200 lbs., and the load on the axle 300 lbs. Let a stone or other obstacle four inches high be represented as at S; the power necessary to be applied to the axle to draw the wheel over the stone is thus found:—Suppose P the power which is just sufficient to keep the wheel balanced, or in equilibrio, when acting from the centre C in the direction C P. The force acting against this power is gravity, and is equal to the weight of the wheel and load on the axle, acting from the centre C in the direction C B. These forces act together against the point D in the direction C D. Gravity acts in the direction C B with the energy or length of lever D B, and the power acts in the direction C P, with the leverage B C; and the equation of equilibrium will be W DB P x CB. In this equation, C B the radius of the wheel diminished by the height of the obstacle, and BP equals -/DC—BO; hence the
Wxv/DCi-BCs . , power P= ——•: in the present example,