If D=the diameter of the surface of the liquor, d=the diameter of the nearest end,

m the middle diameter, and

l=the distance between the surface and the nearest end; The ullage in inches=(D2+d2+2m2)׆l×.7854.

Ex. If the diameter at the surface of the liquor, in a standing cask, be 32 inches, the diameter of the nearest end 24, the middle diameter 29, and the distance between the surface of the liquor and the nearest end 12; what is the ullage? Ans. 27 ale gallons, or 333 wine gallons.

PROBLEM VIII.

To calculate the ullage of a LYING cask.

131. Divide the distance from the bung to the surface of the liquor, by the whole bung diameter, find the quotient in the column of heights or versed sines in a table of circular segments, take out the corresponding segment, and multiply it by the whole capacity of the cask, and the product by 11 for the part which is empty.

If the cask be not half full, divide the depth of the liquor by the whole bung diameter, take out the segment, multiply, &c., for the contents of the part which is full.

Ex. If the whole capacity of a lying cask be 41 ale gallons, or 493 wine gallons, the bung diameter 24 inches and the distance from the bung to the surface of the liquor 6 inches; what is the ullage?

Ans. 7 ale gallons, or 9 wine gallons.

96

NOTES.

NOTE A. p. 16.

ONE of the earliest approximations to the ratio of the circumference of a circle to its diameter, was that of Archimedes. He demonstrated that the ratio of the perimeter of a regular inscribed polygon of 96 sides, to the diameter of the circle, is greater than 34: 1; and that the ratio of the perimeter of a circumscribed polygon of 192 sides, to the diameter, is less than 34: 1, that is, than 22: 7.

Metius gave the ratio of 355: 113, which is more accurate than any other expressed in small numbers. This was confirmed by Vieta, who by inscribed and circumscribed polygons of 393216 sides, carried the approximation to ten places of figures, viz.

3.141592653.

Van Ceulen of Leyden afterwards extended it, by the laborious process of repeated bisections of an arc, to 36 places. This calculation was deemed of so much consequence at the time, that the numbers are said to have been put upon his tomb.

But since the invention of fluxions, methods much more expeditious have been devised, for approximating to the required ratio. These principally consist in finding the sum of a series, in which the length of an arc is expressed in terms of its tangent.

If t=the tangent of an arc, the radius being 1,

This series is in itself very simple. Nothing more is necessary to make it answer the purpose in practice, than that the arc be small, so as to render the series sufficiently converging, and that the tangent be expressed in such simple. numbers, as can easily be raised to the several powers. The given series will be expressed in the most simple numbers, when the arc is 45°, whose tangent is equal to radius. If the radius be 1,

The arc of 45°-1-+-+- &c. And this multiplied by 8 gives the length of the whole circumference.

But a series in which the tangent is smaller, though it be less simple than this, is to be preferred, for the rapidity with which it converges. As the tangent of 30°, if the radius be 1,

And this muitiplied into 12 will give the whole circumfer

ence.

This was the series used by Dr. Halley. By this also, Mr. Abraham Sharp of Yorkshire computed the circumference to 72 places of figures, Mr. John Machin, Professor of Astronomy in Gresham College, to 100 places, and M. De Lagny to 128 places. Several expedients have been devised, by Machin, Euler, Dr. Hutton, and others, to reduce the labor of summing the terms of the series. See Euler's Analysis of Infinites, Hutton's Mensuration, Appendix to Maseres on the Negative Sign, and Lond. Phil. Trans. for 1776. For a demonstration that the diameter and the circumference of a circle are incommensurable, see Legendre's Geometry, Note iv.

The circumference of a circle whose diameter is 1, is 3.1415926535, 8979323846, 2643383279,

5028841971, 6939937510, 5820974944,

5923078164, 0628620899, 8628034825, 3421170679, 8214808651, 3272306647, 0938446+or 7-.

X.2821 the side of an equal square.

The circumf.x.2251=the side of an inscribed square. X.2756 the side of an ins'bed eq'lat. triang.

=

x1.128 the diameter of an equal circle. x3.545 the circumf. of an equal circle.

The side of a sq.X1.414-the diam. of the circums. circle. X4.443 the cir. of the circumsc. circle.

NOTE C. p. 19.

The following approximating rule may be used for finding the arc of a circle.

1. The arc of a circle is nearly equal to of the difference between the chord of the whole arc, and 8 times the chord of half the arc

2. If h=the height of an arc, and d=the diameter of the circle;

5. If s=the sine of an arc, and r=the radius of the circle;

NOTE D. p. 23.

To expedite the calculation of the areas of circular segments, a table is provided, which contains the areas of segments in a circle whose diameter is 1. See the table at the end of the book, in which the diameter is supposed to be divided into 1000 equal parts. By this may be found the areas of segments of other circles. For the heights of similar segments of different circles are as the diameters. If then the height of any given segment be divided by the diameter of the circle, the quotient will be the height of a similar segment in a circle whose d'ameter is 1 The area of the latter is found in the table; ar.d from the properties of similar figures, the two segments are to each other, as the squares of the diameters of the circles. We have then the following .ule:

To find the area of a circular SEGMENT by the TABLE.

Divide the height of the segment by the diameter of the circle; look for the quotient in the column of heights in the table; take out the corresponding number in the column of areas; and multiply it by the square of the diameter.

It is to be observed, that the figures in each of the columns in the table are decimals.

If accuracy is required, and the quotient of the height divided by the diameter, is between two numbers in the column of heights; allowance may be made for a proportional part of the difference of the corresponding numbers in the column of areas; in the same manner, as in taking out logarithms.

Segments greater than a semicircle are not contained in the table. If the area of such a segment is required, as ABD (Fig. 9.), find the area of the segment ABO, and subtract this from the area of the whole circle.

Or,

Divide the height of the given segment by the diameter, subtract the quotient from 1, find the remainder in the column of heights, subtract the corresponding area from .7854, and multiply this remainder by the square of the diameter.