so as when open to let the water rise the 8 feet in two minutes. How many gallons per second must the open sluice let through? A cubic foot is 6.23 gallons. 27. A bed of coal 14 feet thick is inclined at 23° to the surface. Calculate the number of tons of coal that lie under an acre (4,840 square yards) of surface. A ton of coal occupies 28 cubic feet. (The 14 feet is to be regarded as a measurement at right angles to the surface of the coal bed.) 28. An arch in the form of an arc of a circle 40 feet in diameter crosses a stream 30 feet wide. Calculate, to the nearest inch, how much the arch rises in the centre, and check your calculation by a careful drawing in which 1 inch represents 10 feet. 29. Draw a circle with centre 0 and radius 8 cm. Make at three angles of 50°. Take 5 mm. on your dividers and measure the arcs these angles stand on by stepping the dividers along them. Calculate for each the ratio arc radius circumference How Step round the circumference and calculate many degrees must an angle contain to make the arc it stands on equal to the radius ? 30. M is a fixed point in a fixed line KL. A circle of radius 1 inch rolls along KL, and an unlimited straight line MA, passing through M, always touches it. Show the locus of the point of contact. Take M about the middle of the page, and use tracing-paper, drawing the circle and the tangent on it, the circle to be about the middle of the tracing-paper. 31. The height of the Peak of Teneriffe is 12,350 feet; the depression of the horizon from its summit is approximately 2°. What is the diameter of the earth? 32. Two wheels, 3 feet 6 inches and 6 inches in diameter, are fixed in the same plane with their centres 3 feet apart. Find, in any way you please, the length of belt required to go round them. 33. Draw to a scale of 1 inch to 50 yards the roads in front of Buckingham Palace from the dimensions given in the sketch (Fig. 13). The lengths are in yards. The middle line of the Mall goes through the centre of the memorial enclosure and bisects the front of the palace enclosure at right angles. The middle line of two other roads goes through the centre of the memorial enclosure, and is parallel to the front of the palace. 34. Draw a circle of radius 3 inches, and take any point P at a distance of 5 inches from the centre of the circle. Mark four points on the circle at distances 5, 6, 7, 8 inches respectively from P, and join them to P by straight lines. Then make a table like the following and fill it in. What conclusion can you draw from your results? What is the length of the tangent drawn from P to the circle? Draw the tangent. Prove, for the case in which the cutting line passes through the centre of the circle, that, if from a point outside a circle two straight lines are drawn, one of which cuts the circle and the other touches it, the rectangle contained by the whole line which cuts the circle and the part outside the circle, is equal to the square on the line which touches it. = = 35. Two straight lines AC and BD cut one another at K so that LAKB 53°, AK = 1.2", BK = 0·7′′, CK 1.6′′ and DK=2.4". A, B, C, D are the angular points of a quadrilateral. Measure the angles of the quadrilateral. Show from your measurements that it is impossible to describe a circle about the figure, and prove the theorem by which you justify your conclusion. How may the same conclusion be arrived at, from a consideration of the given dimensions of AK, BK, CK and DK? 36. From a point Coutside a circle two straight lines CAB and CT are drawn, CAB passing through the centre and cutting the circle in A and B, CT touching the circle at T. Prove that the square on CT is equal to the rectangle CA. CB, and express this relation algebraically, denoting CA by h, the radius of the circle by r, and the length of the tangent by t. Imagine your figure to represent a section through the centre of the earth, supposed to be a sphere of 4,000 miles radius, and C the top of a cliff. Show that, if the height of the cliff is k feet, the distance of the horizon from C is approximately 3 miles. CHAPTER XIV LOGARITHMS = 1. IN discussing the Slide Rule and other methods of speedy calculation (chap. XII) we considered two quantities x and y so related that y 1.01. How far would that discussion be altered if instead of 1.01 we used a different number, if for instance we used 1.001 ?—In this case also we should have a series of indices and corresponding powers. For the powers running from 1 to 10 the series of indices would be different, there would be more entries in our table. The graph made with 1·001 as base would serve the same end in just the same way as the graph made from 1.01. The curve would be the same kind of curve, and would differ only in the graduation of the index axis. 2. Calculate 1.0011000, 1.0012000, 1·0013000,... far enough to give a rough estimate of the number of entries in our table. The binomial series for 1·0011000 is in which we get the fourth term from the third by multi 0.997, ... we take them in the forms 1−0·002, 1 – 0·003, .... The series is 1+1+0-499+0·166+0·041+0·008+0·001, each term calculated to three places; the remaining terms do not affect the third decimal place. So the value of 1.0011000 is 2.715. By calculating 2.7152 and 2-7153 we find 10012000= 7.37 and 1·0013000 20.01. So that there would = be between 2000 and 3000 entries in our table. = = 3. We know that 1.01231 10, so that in using the table based on 1.01 we had frequently to add or subtract 231. It may be shown that 1.0012304 10, so that with a table based on 1.001, it would be necessary frequently to add or subtract 2,304. Can you suggest any means of simplifying this part of the work? Can you suggest another base that would be simpler to work with than 1.001 ?—If we could find a number k such that 1000 10, we should have to add or subtract 1,000, an easier operation than with 231 or 2,304. = 4. Expand (1+x)1000 in a series, and by substituting different values of x find a value that will make (1+x)1000 equal to 10.-The series is 1+1000x + 1000 × 999 1000 × 999 × 998 2 x2+ 2x3 We have just found that when x = 0·001 the value 2.7, so that this value is too small. 0.003, we find 1.0021000 7.4 and 1.0031000 x must lie between 0·002 and 0.003. = = 23+.... You could in the same way try 0·0021, 0·0022, 0·0023 as values of x, but the work is not especially instructive, and you may take it on trust that 1·00231000 10 pretty exactly. In what other ways can you express this relation ?—1.0023 multiplied by itself 1,000 times makes 10; or 1.0023 is the thousandth root of 10. 5. The following is an extract from the table calculated on 1.0023 as base: |