Walker's Fair book insert


Walker's Fair Book History Topics Index
Version for printing

In the archive of the Madras College, St Andrews, there is a jotter of a schoolboy who studied at the College in 1852. On the front of the book is 'FAIR BOOK, JAMES WALKER'. After one page which is joined to the cover of the book, there is an insert of many pages bound separately. It is clear that this is an insert rather than part of the original book which has come loose. It is not clear, however, whether these pages represent material covered before, after, or at the same time as the material in the rest of the Fair Book. Part of the difficulty is that the material in the Fair Book itself does not always advance in difficulty, and sometimes after quite hard problems, simple ones of the same type will appear. Also the material in the Insert contains problems similar to some in the rest of the book.


The first page of the insert contains three problems, the first being:
What is the area of a rhomboid, of which the base is 640 links, the slope end 436 links, and the inclined angle 47°20'.

Although no formula is stated, clearly the rule used is area = bc sin A. Again six-figure logs are used. The first point to note is that it is necessary to interpolate in finding the antilog. The antilog of 0.313136 is required. The nearest value from the tables is 0.311966 with antilog 2.051. The difference 0.313136-0.311966 gives 170. Adding a 0 gives 1700 which is then divided by 212 to obtain 80 to give the corrected antilog of 0.313136 as 2.05180.

The problems on this page all give solutions in square links. This is then divided by 100,000 to get acres. The decimal fraction is then multiplied by 4 to obtain roods (4 roods to 1 acre). The decimal fraction is then multiplied by 40 to obtain square poles (40 sq poles to a rood). Then the decimal fraction is multiplied by 301/4 to obtain square yards (301/4 sq yds to a sq pole). We give an example of this using the second problem on this page:

What is the area of a triangle, of which the base is 1254 links, and the perpendicular height 847 links.

Although no formula is stated, clearly the rule used is area = 1/2 base.height. After the calculation of 847 times 627 with six-figure logs, requiring an interpolation when computing the antilog, the result obtained is 531068. [Note. If he had noticed that 7 time 7 is 49, then he would have realised that the log calculation (even with six figures) had given the result 531068 instead of 531069.]

Divide by 100000 to get 5 acres. Multiply .31068 by 4 to get 1.24272, so 2 roods. Multiple .24272 by 40 to get 9.70880 so 9 poles [actually square poles but this distinction is seldom made]. Finally multiply .70880 by 30 1/4 to obtain 21.44120 to get 21.44 yds [again square yards and yards are not distinguished.]

Answer is 5 a 1 r 9 p 21.44 yds.

The third problem on the page is:

Two of the sides of a triangle are 621 and 842 links respectively, and the included angle 27°19'; what is its area.

Although no formula is stated, clearly the rule used is area = 1/2 bc sin A.

The next three pages contain further area problems.

If two of the sides of a triangle are 1580 and 1228 links, and the included angle 27°19'; what is its area.

What is the area of a triangle, of which the three sides are 628, 760, and 456 links.

Although no formula is stated, clearly the rule used is Heron's formula:
area = √(s(s - a)(s - b)(s - c)) where a, b, c are the sides and s = (a + b + c)/2.
Six-figure logs are used. Answer is worked out as 1 a 1 r 28 p 26.6 yds.

What is the area of a triangle, of which the three sides are 428, 614, and 517 feet.

Rule used as in the previous question but now answer is converted to sq yds and sq ft.

What is the area of a trapezoid, the parallel sides of which are 20 and 30 yds and the perpendicular distance between them 18 yds.

The first solution given to this problem is incorrect but then the correct solution is given using the rule area = (a + b)h/2 where a, b are the lengths of the parallel sides and h is the perpendicular distance between them.
The calculation (30+20)9 is worked out as 50 times 9 using six-figure logs! The correct answer of 450 sq yds is obtained [This time sq yds is correctly written.]

When the four sides of a trapezium inscribed in a circle are 600, 550, 750, and 400 links, what is the area of the trapezium.

Although no formula is stated, clearly the rule used is area = √((s - a)(s - b)(s - c)(s - d)) where a, b, c, d are the sides and s = (a + b + c + d)/2. The answer is given in acres, roods, sq poles and sq yds.

When the four sides of a trapezium inscribed in a circle are 800, 610, 508, and 420 feet; what is the area of the trapezium.

Answer is left in sq ft.

What is the area of the trapezium ABCD the diagonal BD being 896 and the perpendiculars AM, CN 437, 351 links.

The perpendiculars are clearly to the diagonal BD. The area is calculated as (437 + 351)/2 times 896 using six figure logs and converted to acres, roods, sq poles and sq yds.

What is the area of the trapezium ABCD the side AB being 484, BC = 438, CD = 632, DA = 470, and the diagonal AC = 686.

[We have corrected this problem which is incorrectly gives the diagonal as AB.] The solution just computes the areas of the two triangles ABC and ADC.

What is the area of the trapezium ABCD the side AB being 244, BC = 274, CD = 324, DA = 366, and the diagonal AC = 437.

The next page contains only the statement of the problem and a sketch of the area of the Firth of Forth to which it refers.


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Let M denote Nelson's Monument on the Calton Hill of Edinburgh, L the tower at the end of the quay at Leith, I the island of Inchkeith, P the pier at Pettycur, and B the school at Burntisland: also let PB = 10110 feet, the angles PBL = 29°56', IBL = 40°27', LBM = 8°24', MPB = 86°52', LBM = 4°35' and IPL = 37°52'. Find the distances of Pettycur and the school of Burntisland from Inchkeith from Leith, from Nelson's Monument respectively, and the distance of Leith from Inchkeith and Nelson's Monument.

The method of solution here is to solve triangles, one after the other, using b = a sin B/sin A, all done with six-figure logs.

How far could the Baltic fleet be seen from the Sidlaw hills being 2000 feet high.


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The formula used is d = √((2r + h)h) where h is the height above sea level and r is the radius of the Earth. The diameter of the Earth 2r is taken to be 41777360 ft. A modern value, for comparison, would be 41851509 ft (at least that is the equatorial value). The distance to the Baltic fleet is worked out in miles (by subtracting the log of 5280), giving 54.74 miles.

What is the distance of the visible horizon when the height of the eye is 6 feet.

Method is as the previous question, but the answer left in feet as 15832.

What is the distance of the visible horizon when the height of the eye is 81 feet.

The answer is given as 11.30 miles.

What is the difference between the true and apparent level at the distance of 2 miles.

If I see the flash of a cannon fired from a ship in distress at sea and hear the report 25 sec after.

The velocity of sound is taken as 114.2 ft per sec. (Today the value is 108.7 ft per sec.) The answer is converted to miles (by subtracting the log of 5280) to get 5.4072 miles. The fractional part is then multiplied by 8 to get furlongs, then the fractional part by 220 to get yds. Interestingly, Walker knows his 22 times table and is able to write the answer straight down. Finally the fractional part is multiplied by 3 to get feet. Answer give: 5 miles 3 furlongs 56 yds 2 ft.

Between seeing a flash of lightning and hearing the thunder 61/6 secs. What distance was the cloud.

He multiplies 114.3 by 61/6 without using logs, then uses six-figure logs to convert the resulting answer into 1.333 miles.

In what time after a gun is fired may the report be heard at the distance of 8 miles.

5280 is multiplied by 8 before logs are used to divide by 114.2. The answer is given as 36.98 secs.


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Dr Derham standing upon the bank of the Thames opposite to Woolwich observed that the echo of a single sound was reflected back from the houses in 3 secs. What is the breadth of the river at that point.

He computes the time taken for sound to travel 1.5 secs.

The next page contains a problem, a method of solving problems of this type, and a numerical example. It is the first time that a method of solution is given, all the previous problems being solved by use of a rule (there is no evidence that the rule is thought of in terms of a formula, rather it seems to be thought of as a procedure which has been learnt by heart).

Given the Area and Plan of a Survey, to find the Scale to which the Plan was laid down, should the Scale be unknown.

Construct another Plan, exactly similar to the given one, to any Scale you please; find its area:_ Then say as this Area is to the given Area of the Survey, so is the square of the Scale to which this Plan was laid down, to the square of the Scale required; the Square root of which will be the square of the Scale.

The Area of a Survey is 48 Acres 3 Roods 10 Poles and the area of a similar Plan constructed to a Scale of 3 Chains, on 12 Poles to the inch is:_ 12 Acres 0 Roods 32 1/2 Poles; required the scale to which the Plan was constructed.

First both areas are converted to Poles, that of the Survey being 7810 Poles and that of the similar Plan is 1952 1/2 Poles. He then calculates x such that

1952 1/2 : 7810 : : 9 : x.

He notes that the 9 is the square of 3 Chains per inch. The x here is my invention - it is not used in the solution where he has calculated that 36 goes in the space where I have the x. He writes "Then "36 = 6 Chains per Inch. Hence the Plan was laid down to a Scale of 6 Chains, on 24 Poles to 1 Inch."

The next three pages contain sketches of three dimensional solids. The are 12 on each page, in four rows of three all drawn in a little square box. The first few are labelled 'cube',' rectangle on parallelepiped', 'obtuse parallelepiped', 'upright triangular prism', 'oblique triangular prism', 'oblique pentagon', and 'oblique hexagon'.

Find the area of a circle of which the circumference is 5280.

The assumption here is that the circumference is given in feet. The standard method used here is to square the circumference and multiply by .07957747 (which is the reciprocal of 4π). However, in this first example of this type Walker claims to write down the log of 5280 but actually (correctly) writes down twice the log (so he has correctly squared the circumference). It would be reasonable to assume that he is capable of doubling the log in his head, but the argument against this is that he has written out the multiplication by 2 in the next two examples. It is also interesting that Walker is inconsistent in handling powers of 10. Here he uses log .07957747 = 8.900791 but in the next two problems uses log .07957747 = (bar 2).900791. He correctly adjusts when adding, however, to get the right answer whichever form is used.

Find the area of a circle of which the circumference is 24856 miles.

Again he squares the circumference and multiply by .07957747 (which is the reciprocal of 4π). The answer is given in square miles 49164612.


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Find the area of a circle of which the circumference is 5702 links.

Again he squares the circumference and multiply by .07957747 (which is the reciprocal of 4π). The answer is then converted to 25 acres 3 roods 19 poles 19.5 yds.


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Find the circumference of a circle of which the area is 3097600 sq yds.

First let us note that in fact with a slip of the pen Walker actually wrote 'Find the area of a circle of which the area is 3097600 sq yds.'. However, it was clearly intended as we have written it and this is the problem that is solved. The method used claims to be to divide the area by .7854 (which is π/4), then take the square root. In fact Walker uses the log of π/4 correctly to six places rather than the log of .7854.

The diameter of two concentric circles are 500 and 470 what is the area of the ring.

Walker squares 500 and 470 (in his head) and subtracts. Then takes the six figure log and adds the log of π/4. He obtains the answer 22855.1. (No units are given).

The next page is headed 'Practical Geometry'. The first construction is 'To find the centre of a line geometrically'. He writes 'by taking any distance greater than the centre, then drawing arcs bisecting one another and where they meet is the centre'. The illustrated method is correctly drawn. Although there is no description on the other constructions, they are to draw a perpendicular from a point on a line, to bisect an angle, to construct a right angle (as an angle in a semicircle).

The hypotenuse of a right angled triangle is 463 and perpendicular height 256.

Various errors are made in the solution. The triangle is wrongly labelled in that 463 is put on the base, not the hypotenuse. The other error is that 256 becomes 252 halfway through the sum. First Walker writes

463 : Sine 90 : : 256 : Sine A
Sine 90 : 463 : : B : BC

The first is used (with six figure logs) to give A = 32°58 (this answer is in fact wrong since log 252 is used instead of log 256 in the calculation). Angle A is subtracted from 90 to obtain the angle B. 57°2 is obtained. The side AB is then calculated from

Sine 32°5 : 252 : : Sine 57°2 : AB

to obtain 388.5. Of course there is an error in the data used in this last calculation.

[Note: Both Sine A and S A are used for the sine, but never sin A. In fact Walker uses log Sine A = log 1010sin A in his calculations. Also Pythagoras never seems to be used, perhaps because it is not 'log friendly'.]

The hypotenuse of a right angles triangle is 520, and the base 340 find the perpendicular and angles.

Walker uses the same method as the previous question to obtain A = 40°49, C = 49°11 and AB = 393.5.

There then follows a problem which is not stated but the solution clearly indicates that the problem was to solve a right angled triangle where the two shorter sides are 500 and 480. Solution given using methods as above is correct hypotenuse is 693.1 and the angles are 43°50 and 46°10.

The hypotenuse of a right angled triangle is 325, and the angle adjacent to the base 42°36 find the other acute angle and sides.

First calculates the other angle as 47°24 then uses the same type of proportional equations as before to calculate the sides:

Sine 90 : 325 : : Sine 42°36 : AB
Sine 90 : 325 : : Sine 47°24 : CB.

Walker uses log Sine A = log 1010 sin A in his calculations.

Now follow three more problems involving right angled triangles which are not stated but can be seen as examples where an angle and a non-hypotenuse is given. Again he uses log Sine A = log 1010sin A in his calculations.

Then follow seven problems on solving triangles which are not right angled. They are not stated but involve various given sides and angles. The first involves two sides and the included angle and is solved with the sine rule as above. The remaining six are all examples of triangles where all three sides are given. The method used to find the angles is

cos A/2 = √(s(s - a)/bc) where s = (a + b + c)/2.
Walker finds s and then s - a. He appears to have tables of log (1/n) since he writes these down without any calculation. As before powers of 10 are inserted in a way that is understood, so log (1/61) = 8.214670.


Let us look at the first of these problems. The sides of the triangle are given as 45, 53 and 61. Walker computes

    61
    53
    45
   ---
 2|159
   ---
s=  79.5
    45
    ---
s-a=34.5

log 79.5 = 1.900367
log 34.5 = 1.537819
log(1/61)= 8.214670
log(1/53)= 8.275724
        2|19.928580
          ---------
22°56      9.964290
    2
-----
45°52

This gives the angle between the two longer sides.

In the last four problems of this type, after using this means to find one of the angles, Walker then uses the sine rule to compute one of the other two angles.

What is the area of the back park of Madras College from the following survey.


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The back yard is divided by a diagonal from the top left to the bottom right. The left side of the park is composed of five straight lines and at each point where two join the survey has found the perpendicular distance to the diagonal. The sides of all the resulting triangles of quadrilaterals have been measured so that basically the whole back yard has been triangulated with triangles with known base and height. The area is then easily calculated.

What is the area of the field ABCDEF, AB being = 640, Aa = 400, FA = 260, Ab = 1020, bC = 200, AD = 1220, Dc =250, cC = 190 and DB = 1130 links.

Here ABCDEF is a six sided polygon with the lines AD and BD drawn in, and
a is the foot of the perpendicular from F to AD
b is the foot of the perpendicular from E to AD
c is the foot of the perpendicular from C to BD

Triangle ABD is computed from the formula √(s(s - a)(s - b)(s - c)) (of course these are different a, b, c. Other triangles calculated from area = base.height/2 and the final answer is converted into areas, roods, poles.

What is the area of the lower part of the back park of the Madras College from the following survey.

Again area is calculated from the formula √(s(s - a)(s - b)(s - c)) for one of the triangles and area = base.height/2 for the others. The final answer is converted into areas, roods, poles. In one place 35 has been copied as 53 but the correct value of 35 has been used in the calculation.

What is the surface of a cone its axis being 16 ft and the diameter of its base 2 ft 10 in.

Walker uses the formula πdm/2 + πd2/4 where m is the slant height and d the diameter of the base. He doesn't write the formula down - he never does. He first writes the diameter of the base as 34 inches. He then computes 34 times 34 times π/4 (appearing to look up the log of π/4 directly). He then divides by 144 to obtain 6.305 sq ft. Next he computes 34 times π and divides by 24 (12 to convert to feet, and the 2 from the formula). He then multiplies the answer by 16 to obtain 71.20. Adding the two parts gives 77.505.

What is the solidity of a pentagonal pyramid each side of its base being 3 feet, and its perpendicular height 20 feet.

He first computes the area of the base by squaring the length of the side of the regular pentagon and multiplying by 1.7204774. He then multiplies by the perpendicular height of 20 and divides by 3. He obtains 103.2.

What is the surface of the frustum of a cone the diameter pf the greatest end 9 and the lesser 5. Slant 16 ft 9 in.

Walker computes 9π + 5π by separately computing each part by logs then adding. He doesn't seem to think of calculating 14π. He makes an error when adding the two parts of the sum 28.27 + 15.70 obtaining 53.97 instead of 43.97. He also makes a copying error writing 6 instead of 9 for the diameter of the largest end but he has the log of 9 so it is only a copying error. The error in addition, however, persists to give an incorrect answer. He converts 16 ft 9 in to 16.75 ft and multiples 53.97 by this, ten divides by 2. He obtains 451.6 instead of the correct 368.35.

This is the end of the separately bound inserted pages. You can return to the original Fair Book using THIS LINK.

Article by: J J O'Connor and E F Robertson


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