If this area be cut out carefully and placed in one pan of a balance, while an ellipse of the same size, and cut out of the same material, be placed in the other pan, the two will be found to

balance.

FIG. 131.-Area of an ellipse obtained experimentally.

Another method of obtaining the area is to draw the ellipse on squared paper. Count the enclosed squares, also estimate as accurately as possible the total area of the squares partially cut by the profile of the figure. By adding the two together, the area can be found.

EXERCISES. LXVI.

1. The greater and lesser diameters of an elliptical manhole door are 2 feet 9 inches and 2 feet 6 inches; find its area.

2. What is the area of an elliptical door whose dimensions are 12 inches by 9 inches?

3. The greater diameter of an elliptical funnel is 4 feet 6 inches, and the lesser diameter 4 feet; what space will it occupy on the deck, and how many square feet of iron does it contain if its height be

16 feet?

4. Find the difference in area of two fields, the first in form of a circle whose radius=575 yards, and the other in form of an ellipse whose axes are 1000 yards and 1250 yards.

5. The semi-axes of an ellipse are 12 and 7 inches respectively. A circle is described on the minor axis. What is the area of that part of the ellipse without the circle.

6. The diameter of a piston is 25 inches, the piston-rod 43 inches; find the difference in the total pressure on both sides of the piston if the pressure of the steam be 160 lbs. per square inch.

7. The external and internal diameters of a hollow steel shaft are 16 and 8 inches respectively. Find the area of the cross-section.

8. If the pressure of steam is 100 lbs. per square inch, find the difference in the effective pressure on the two sides of the piston (Fig.127).

Summary.

Area of a Rectangle=base × altitude=length × breadth. Area of a Parallelogram = base x altitude = adjacent sides into the sine of included angle.

half the product of

Area of a Triangle = base × altitude; or, half the product of two sides by the sine of included angle; or, from half the sum of the three sides subtract each side separately. Multiply the half sum and the three remainders together and find the square root of the product.

Area of a Rhombus half the product of the two diagonals.

Area of a Trapezium = Multiply half the sum of the parallel sides by the perpendicular distance between them.

Circumference of a Circle Tx (diameter) = 2 (radius), where π denotes 3.1416 or approximately.

22 7'

π

Area of a Circle =π (radius)2 = (diameter)2.

Area of Sector of a Circle=2, where is the angle in radians subtended at the centre.

Area of an Annulus =π (R2 – r2), or subtract the square of the smaller radius from the square of the greater and multiply the difference by π.

Area of an Ellipse=π × ab=product of its semi-axes multiplied by π.

The length of each perpendicular from the base AB to the point where the vertical cuts the curve is carefully measured, and all the ordinates added together, the sum so obtained is divided by the number of ordinates. In this way the mean ordinate may be obtained.

A very convenient method of adding the ordinates is to mark

235

CHAPTER XVI.

SIMPSON'S RULE.

Areas of an irregular Figure. To find the area of figure ABDG, of which one of the boundaries is curved.

The usual way of estimating approximately the area of a curve is to divide the base AB (Fig. 132) into a number of equal parts, and to erect ordinates at the centre of each part, as shown.

2.46

2:57

them on a slip of paper, adding one to the end of the other until the total length is determined. The mean ordinate gives the mean height h approximately.

The approximation depends upon the number of ordinates taken. The approximation more closely approaches the actual value the greater the number of ordinates used.

The product of the mean ordinate and the base is the area required. For comparatively small diagrams, ten strips are usually taken. This number is sufficiently large to give a fair average, and, moreover, dividing by 10 can be effected by merely shifting the decimal point.

Thus, in estimating approximately the mean height of any diagram (Fig 133), the base GF is divided into a number of equal parts, usually 10. The height of the ordinate raised at the centre of each part is measured, the mean of these is approximately the mean height of the curve.

The length GF may correspond, on a reduced scale, to the travel of the piston in a cylinder, and the ordinates of the curve represent to a known scale the pressure per sq. inch of the steam in the cylinder at the various points of the stroke.

Hence the mean height (Fig. 201) indicates the mean pressure P of the steam in pounds per sq. inch throughout the stroke (the stroke being the term applied to the distance moved through by the piston in moving from its extreme position at one end of the cylinder to a corresponding position at the other end).

If A denote the area of the piston in square inches, then the total force is P x A, and the work done by this force in acting through a length of stroke L is PxAxL. If N denote the number of strokes per minute, the work done per minute PALN.

=

But the unit of power used by engineers is 33000 ft. lbs. per minute, and called a Horse-power.

PxLxAx N

Hence horse-power of the engine= 33000

The Trapezoidal Rule is another method to obtain the mean ordinate h, having divided the base into a number of equal parts, say 6 (Fig. 132), then :

h=={ } (h2+hy)+h2+h3+hq+hg+hq}.

"Divide the base into any number of equal parts; add half the

sum of the end ordinates to the sum of all the others. Multiply the result by the common interval s to obtain the area: divide by the number of ordinates into which the figure is divided to obtain the mean ordinate. Mean ordinate × length: = area of figure.

It will perhaps be found very instructive to calculate by means of this rule the areas in the following examples, and compare the results obtained with the more accurate values as found by the application of Simpson's rules.

Simpson's Rule.-Simpson's rule provides a means of obtaining the area of any curvilinear area, ABGD (Fig. 132), more accurately than by the method just described.

The base AB is divided into a number of equal parts (in this case 6). This ensures that the number of ordinates is an odd number, i.e. 3, 5, 7, 9, etc. Denoting as before the lengths of the ordinates AD, pm, sr, etc., by

h1, h2, h3, hy.

...

Then if s denotes the common distance or space between the ordinates, we have

Area of ABGD={h1 +hy+4(h2+hq+hc)+2(hg+h5)}

3

where A denotes the sum of the first and last ordinate,

. Add together the extreme ordinates, four times the sum of the even ordinates, and twice the sum of the odd ordinates (omitting the first and last). Multiply the result by one-third the common interval between two consecutive ordinates.

The end ordinates at A and B may either one or both be zero, the curve commencing from the line AB (Fig. 132), but the rule still applies. In certain cases the figure may be bounded by two curved lines DG and EF, and the two straight lines DE and FG, the ordinates would then correspond to mn, rq, etc.; if the two curved boundaries are symmetrical about a line AB, it is only necessary to give the half-ordinates pm, rs, etc., each of these may be doubled, or, by substituting the values of the half-ordinates in Simpson's rule half the area is obtained. This, when doubled, is the area required.