James Walker's Fair Book of 1852


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Andrew Bell made his fortune in India. He founded Madras College in 1832 in St Andrews, the school being based on the Madras System which involved schoolboys tutoring their fellow pupils. Before Dr Bell founded and built Madras College there had been several schools in St Andrews initially intended for boys destined for the Church. There had been for a few hundred years before Dr Bell established Madras College, some educational establishments serving both sexes, providing a rather basic level of instruction. What Dr Bell did was to use his wealth to transform those existing schools into a single school well fitted to meet the needs of its day and to be run on the monitorial system which he had pioneered. The College provided an outstanding education and many families moved to St Andrews so that their children could benefit from this education. The school grew rapidly after its foundation and by 1838 there were 800 pupils at the College, by 1845 there were around 900 pupils, and by 1860 there were well over 1000 pupils. The school had many boarders and there were houses around the town for them run by masters at the school. When the school was founded there was no headmaster - the senior masters of each department were largely independent and received fees from the pupils.


In the archive of the school there is a jotter of about 80 pages of a schoolboy who studied at the College in 1852. On the front of the book is 'FAIR BOOK, JAMES WALKER'. It is unclear whether the material in this book was put there immediately after the rough working had been done elsewhere or whether this book was only created some time later. The first alternative seems the most probable.

We have transcribed the work done in the book and included some explanatory notes. Much of the interest in this volume (and almost certainly the reason it has been preserved) is in the art-work which accompanies the mathematics. Some of this is relevant to the problems being described, but the most impressive consists of some military sketches -- probably copied from engravings of the Crimean War, which was going on at the time.

You can see a separate page of the Crimean sketches at THIS LINK

One signature of the book has become separated from the rest. You can see this at THIS LINK

The first page is headed 'Tonnage of ships'.


Click it to see a larger version

If the length of the keel of tonnage be 100 feet and the extreme breadth of the ship 35 feet. Required tonnage by common rule.

Although no 'common rule' is specified, the solution tells us that the rule gives the tonnage as x . y3/2/ 94 where x is the keel in feet and y is the extreme breadth. The calculation is done using six-figure logarithms. We have inserted 'log' but the text does not.

log 100   = 2
log  35   = 1.544068
log  17.5 = 1.243038
--------------------
            4.787106
log  94   = 1.973128
---------------------
651.5     = 2.813978

Although no words appear in the solution, we see that the tonnage, found from log tables, is 651.5.

On the second page (back of the first page) there are two similar problems.


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If the length of the keel of tonnage be 80 feet and the extreme breadth of the ship 27 feet. Required tonnage by common rule.

If the length of the keel of tonnage be 96 feet and the extreme breadth of the ship 33 feet. Required tonnage by common rule.

The first is done in the same way as the above working, the second has been left unsolved with the values entered by the log tables not looked up, e.g. 96 = 1. etc.

At this point there is a fairly large insert of pages which are bound together. The content of these is given separately at THIS LINK.

It is possible that some pages are missing from the original book at this point but, given the content, it can't be more than a page or two. The problems about ships now change into problems about cattle but no heading to indicate this appears. Given the headings throughout the rest of the book, it is likely that a couple of pages have come loose containing the section heading. Note the common notation 5 / 10 for 5 shillings and 10 pence. This does not represent a division!

What is the value of an ox measuring 7 ft 3 in in girth and 5 ft 4 in in length at 5 / 10 a stone, reckoning the offal 1/3 the value of the four quarters.

[Walker should have written "fore quarter" rather than "four quarter".]

He uses a rule squaring the girth in feet, multiplying by the length in feet, then multiplying the answer by 5/21 to obtain the weight in stones. Multiplying by 5/10 (5 shillings and ten pence) converted to 5.833 shillings, he obtains 389.0 to which he adds a third to obtain 518.6 shillings. The then converts this to 25..18/6 by dividing by 20.
Ans. £25..18/6

What is the dead weight of the four quarters of an ox measuring 5 ft 7 1/2 in in length and 9 ft 3 1/2 in in girth, adding 1/12 part to the weight found by the rule as an allowance for the beasts extraordinary fatness, also living weight.

Walker seems confused in his answer. In the calculation he assumes that 9 ft 3 1/2 in is 9.35 ft (which it certainly isn't) and also 5 ft 7 1/2 in as 5.75 ft (which of course it isn't). He calculates as for the previous problem multiplying 9.35 by 9.35 by 5.75 by 5/21. This gives 119.6 to which he adds a 21st part (which I assume should be a 12th part). He then takes his answer of 125.3 and multiplies by 200/121.

The next page is headed Land Measuring.

Let AB or AC be 100 links, and BC 136 links, what is the angle BAC.

Walker takes a half of 136, then uses the sine rule to obtain half the angle at A. As in all his calculations, he uses log sine A = log 1010 sin A in his calculations. He should get 42°50 but (by a copying error) gets 45°50. He then doubles this to get the correct answer of 85°40 (confirming that the above was only a copying error and that he had it correct in his rough working).

There follow three similar problems which Walker does correctly in exactly the same way.

Let AB or AC be 100 links, and BC 63.5 links, what is the angle BAC.

Ans. 36°54'

Let AB or AC be 50 links, and BC 43.5 links, what is the angle BAC.

Ans. 51°34'

Let AB or AC be 50 links, and BC 68.75 links, what is the angle BAC.

Ans. 86°51'

Let AB be 100 links, and the perpendicular or tangent BC 71.5. Required angle.

Finds AB + BC = 171.5 and AB - BC = 28.5. He then writes
171.5 : 28.5 : : 45 : Tang x
and solves (using six figure logs) to obtain x = 9°26.

He then adds and subtracts this from 45 to obtain 54°26 and 35°34.
He next uses the sine rule Sine 9°26 : 171.5 : : Sine 45 : x
to obtain (using six figure logs with log Sine A = log 1010sin A) to obtain 122.9.
Then he writes
122.9 : 90 : : 71.5 : A
and solves to obtain 35°34.

Ans. 35°34'.

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

Heron's formula is used: 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 to calculate the area as 143000 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).

Ans. 4 ac 1 r 28 p 24 yds.

Two further problems are done using Heron's formula in the same way:

What is the area of a triangular field the three sides of which are 1270, 952 and 718 links.

Ans. 3 ac 1 r 21 p 13 yds.

What is the area of a triangular field, 3 sides, 816, 1048 and 1270 links.

Ans. 4 ac 1 r 1 p 13 yds.

What is the area of a triangular field the base and perpendicular of which are 2136 and 636 links.

Here Walker produces a strange method of solution. He seems to know that the area of a triangle is the same as that of a right-angled triangle on the same base and with the same height. He then finds the third side of the triangle using Pythagoras's theorem, and computes the area using Heron's formula. The answer is correct but an extremely lengthy and difficult method to obtain the solution. It is worth noting that some (but not all) of the solutions have a tick to the left of the question. This question has a tick. This may only indicate that the answer is right, or it may mean that Walker's Master approved the method - who knows.

Ans. 6 ac. 3 r 6 p 21 yds

What is the area of a triangular field, the base and perpendicular of which are 1890 and 470 links.

This time Walker uses the obvious method of computing the area as 1/2 base times height. He converts to obtain:
Ans. 4 ac 1 r 3 p 11 yds.

What is the area of a triangular field ABC the base AB being 823 links, the angle A = 37°10' and B = 64°52'.

Walker draws the figure wrongly in that he labels the side BC to be 823 rather than AB. He calculates the third angle of the triangle to be 78°58. This is an arithmetic error, it should be 77°58. He computes the side AC using the sine rule - finds 506.5. He then multiplies 506.5 by 416.5 (perhaps 1/2 times 823 wrongly done?) Again for some reason he chooses to work this out long-hand rather than use logs. He seems to take this to be his answer - so he forgets to multiply by sin 64°52.

If AB = 1232 links, BC = 283, CD = 1059, DA = 282 and BD = 1170; what is the area.

The solution is carried out with two applications of Heron's formula to the triangles ABD and BCD. Areas are converted to acres, roods, poles, square yards for each separate triangle before adding. Of course Walker very seldom bothers to write sq yds but writes yds. This does, on rare occasions, lead him to make errors.

If AB = 700 links, BC = 416, CD = 669, DA = 325 and AC = 793 links. Find the area.

Done in exactly the same way as the previous problem with the two triangles ABC and ACD.

What is the area of a rectangular field, its length being 1156 links and its breadth 948 links.

Takes the length times breadth and converts to acres, roods, poles, square yards.

What is the area and rent of a ridge of grass, the length being 965 links, the breadth 23 links at one end and 21 links at the other, at 10 guineas per acre.

Walker computes 965 times 22, then converts to acres. Although it would seem sensible to calculate in acres, he converts to acres, rooks, poles, yds and then a proportionality sum to arrive at £2..4/6 which is close but not right.

What is the content of a field in the form of a trapezoid whose parallel sides are 1260 and 984 links and their perpendicular distance 567 links.

Calculates half of 1260+984 times 567, then converts to acres, roods, poles, square yards.

Ans. 6 ac 1 r 17 p 22 yds. (He doesn't round 22.99 yds to 23 yds)

What is the area of a field in the form of a trapezoid, its parallel sides being 1051 and 850 links, and their perpendicular distance 436 links.

Method as previous problem.
Ans. 4 ac 0 r 23 p 1 yd.

If AB = 834, BC = 673, CD = 635, DA = 539 links and the angle A = 87°20: Required area of field.

Walker first solves triangle ABD using the formula tan(B - C)/2 = (b - c)/(b + c) tan(180 - A)/2. First he omputes AB + AD and AB - AD, then multiplies tan((180 - A)/2 by (AB - AD)/( AB + AD). Finds inverse tan. Adds and subtracts this from (180 - A)/2 to get the other two angles of ABD. Then uses sine rule to compute BD.

Once BD is found to be 971.6, he then uses Heron's formula on the two triangles ABD and BCD, converts the area to acres, rooks, poles, yds for each and adds.

Ans. 4 ac 1 r 18 p 7 yds.

What is the area of a four sided field its diagonals being 1000 and 850 links and the angle at their intersection 65°25'.

Sine 65°25' × 500 × 850 = area. Calculates in sq links, then converts to:

Ans. 3 ac 3 r 18 p 10 yds.

What is the area of a four sided field, its diagonals being 940 and 898 links, and the angle at their intersection 49°15'.

Method as previous problem.
Ans. 3 ac 0 r 31 p 17 yds.

If the diagonal AC = 848 links, the angles BAC = 30°56; CAD = 69°10; BCA = 62°15; and ACD = 41°12; what is the are of the field?

Walker computes ADC = 69°38, then uses the sine rule to find DC = 845.4. He then finds the area of ACD using area = 1/2 bc sin A. Similarly he computes the area of triangle ABC. Both are converted to acres, rooks, poles, yds and he then adds.

Ans. 3 ac 3 r 39 p 25 yds.

Suppose that from one corner, a measurer cannot see all the other corners of a field, but takes his observations from a point of rising ground at A, and that its angles are as follows; BAC = 105°; CAD = 59°30; DAE = 129° and EAB = 66°30'; and the lines AB = 480, AC = 550, AD = 665, AE = 730. What is its area?

Walker finds the area of each of the four triangles at A in turn using area = 1/2 bc sin A. He converts to acres, rooks, poles, yds and he then adds.

Ans. 6 ac 1 r 0 p 23 yds.

There follow a number of problems which involve fields which are drawn with given measurements (all lengths of lines). Areas are computed by summing up the areas of parts. Heron's formula is used for triangles of known sides. These contain up to 18 measured lines. After these rather lengthy sums, we go back to simpler examples.

Find the area of a triangular field, the three sides being 840, 460, 1120 links.

Calculation using Heron's formula.
Ans. 3 ac 3 r 28 p 29 yds.

Find the area of a triangular field the 3 sides being 630, 720, 940 links.

Calculation using Heron's formula.
Ans. 2 ac 1 r 2 p 16 yds.

What is the area of a four sided field of which the sides are 640, 520, 760, and 930 and diagonal 1170.

The two triangles are each solved using Heron's formula, He converts to acres, rooks, poles, yds and he then adds.
Ans. 4 ac 1 r 32 p.

What is the area of a four sided field the diagonal being 1460 links and perpendiculars 460 and 340 links.

The perpendiculars are from the other two vertices to the diagonal. The area of the two triangles are computed from 1/2 base times height.
Ans. 5 ac 3 r 14 p.

What length of a rectangular field, of which the breadth is 500 links, will make 1 acre, 2 roods, 30 poles.

Walker converts the area to square links, then divides by 500 by long division (not using logs).
Ans. 337 1/2 links.

What length of a ridge 21 links broad will make 8 poles.

Converts 8 poles to 5000 sq links by multiplying by 625, then divides by 21 to get 2382/21 links.

What breadth of a rectangular field 690 links long will make 3 roods 27 poles.

3 r 27 p is converted to square links, obtaining 91875. Divides by 690 (by long division) to obtain 133 with remainder 105. Finally reduces 105/690 = 21/138 = 7/46.
Ans. 133 7/46 links.

If 2 ac 20 p were to be cut off from the triangle ABC, which contains 4 ac, parallel to AC, the length of AB being 1125 links, in what point of AB must the line of division begin.

By subtracting he works out that the triangle remaining will have area 1 ac 3 r 30 p. He coverts this to 187500 square links. Now he uses the fact that the ratio of the squares of the lengths of the sides of the triangles will be the ratio of the areas to compute the length as 770.23 links. He uses six figure logs obtaining the answer correctly to 2 decimal places.

The next problem looks interesting but only the working is given, the statement of the problem being missing.

Divide a common of 244 ac 3 r 30 p among A, B, C, and D, whose estates, on which their claims are founded, are respectively £500, 400, 150, 100 a year; the quality of each being 20/18/15/12.

Walker divides 500, 450, 150, 100 by 20, 18, 15, 12 respectively. He then sums the answers 25, 25, 10, 8 1/3 to obtain 68 1/3. He then computes

68 1/3 : 25 : : 244.3.30 : A

by first multiplying 681/3 : 25 by 3 to obtain 205 : 75 then dividing by 5 to obtain 41 : 15. He converts 244 ac 3 r 30 p to 39190 poles and finds 15/41 of that using long multiplication and division. Converts back to acres to get

Ans. A = 89 ac 2 r 17 p 33/41. Similar calculations for C and D give 35 ac 3 r 15 p 5/41 and 28 ac 3 r 19 p 11/41respectively.

Next there is a section headed Conic Sections.

If an absciss of 9 corresponds to an ordinate of 12, what is the ordinate of which the absciss is 25.

√9 : √16 : : 12 : x
3 : 4 : : 12 : x
1 : 4 : : 4 : x

Ans. 16.

If the ordinate is 6 and the parameter 4 what is the absciss.

4 : 6 : : 6 : x
2 : 3 : : 6 : x
1 : 3 : : 3 : x

Ans. 9.

If an absciss is 16 and its ordinate 12 what is the parameter.

Walker multiplies 16 by 4, the answer by 2, then divides by 3 to get 422/3.
Ans. 422/3.

What is the area of a parabola, its absciss being 150 and the ordinate perpendicular to the absciss 450 links.

What is the area of a parabola its base being 600 links and its height 250 links.

Walker uses the fact that the area is 2/3 base times height. Computes using long multiplication and division that area is 1 acre.

If the base of a parabolic segment is 500 links and its absciss 100 makes an angle of 55° with it; what is the area of the segment.

Walker computes (using six figure logs) the height as 100 sin 55° = 81-91. The area is then 2/3 base times height which is computed without logs as 27303.33 (in fact 27305.06 with the error coming from taking the height to 2 decimal places). Converting to acres gives

Ans. 1 r 3 p 20 yds.

If the 2 parallel ends of a zone of a parabola be 10 and 6, and the part of the absciss perpendicular to and connecting the middle of those ends be 4; what will be the area of the zone.

He writes (10 + 62 divide 16) 4 × 2/3 = Ans. He computes using long division that
Ans. = 322/3.

If the two parallel ends of a zone of a parabola are 200 and 180 and the part of the absciss connecting the middle of those ends are 120 and makes an angle of 50°; required area of zone.

No solution given.

The length of a base line within a field curvilinear on the other side is 315 links and 11 equidistant ordinates erected thereon measure 70, 86, 96, 104, 109, 110, 108, 105, 99, 90 and 85 links, respectively, what is the area of the space between the base line and the curvilinear side of the field.

The average ordinate is calculated by adding the 11 ordinates to obtain 1062, then dividing by 11 to get 96.54. This is then multiplied by 315 using six figure logs.

Ans. 30410. (There is no conversion to acres.)

When the transverse axis is 210, the conjugate 180, and the two abscisses are 168 and 42, what is the ordinate.

An ellipse is drawn. The calculation is via 210 : 150 : : √(168 x 42) : x giving:

Ans. 60. (Note that the square root is done long hand without the use of logs.)

When the transverse axis is 180, the conjugate 60 and the two abscisses 144 and 36, what is the ordinate.

An ellipse is drawn. The calculation is via 180 : 60 : : √(144 x 36) : x giving:

Ans. 24. Note that the square root is done long hand without the use of logs or without the obvious 12 × 6.

What is the abscisses of the ordinate 20, the axes being 70 and 50.

An ellipse is drawn. The calculation is via
50 : 70 : : √(252 - 202) : x giving
50 : 70 : : 15 : x. Note that Walker recognises √225 = 15. This gives x = 21. He then calculates 35 ± 21 = 14 or 56.

Ans. 14 or 56.

What are the abscisses of the ordinate 12, the axes being 90 and 30.

30 : 90 : : √(152- 122): x so x = 27. But 45 ± 27 = 72 or 18.

Ans. 72 or 18.

When the abscisses to an ordinate of 20 are 56 and 14, and the transverse axis is 70, what is the conjugate axis.

An ellipse is drawn. √(14 x 56) : 20 : : 70 : x
Ans. 50.

When the abscisses to an ordinate of 24 are 36 and 144 what is the conjugate axis.

Transverse axis = 144 + 36 = 180.
144 × 36 : 242: : 1802: x2. Calculation via six figure logs to obtain:
Ans. 60.

When an ordinate and its less absciss are 18 and 12 respectively and the conjugate is 45, what is the transverse.

Walker computes 22.5 + √(22.52 - 182) = 36. He doesn't use logs but uses long multiplication and square roots. He the calculates
182 : 45 × 12 : : 36 : x
and obtains:

Ans. 60.

When an ordinate and its greater absciss are 248 and 144 respectively and the conjugate is 60, what is the transverse.

No solution is given.

Walker now computes the circumference and area of ellipses. For the circumference he uses the approximate expression π/2 × √(a2 + b2) where a, b are the lengths of the major and minor axes. This approximation is exact for a circle, but poor for ellipses with a high eccentricity. The true value of the circumference depends on an elliptic integral and was certainly not available in schools in the 1850s!
The area is (exactly) π/4 × a b

Find the circumference of an ellipse of which the axes are 140 and 120.

Walker computes (1402 + 1202)/2 = 17000. He then takes the square root (long-hand) to obtain 130.38. He then multiplies by 3.1416 (long-hand) to obtain 409.6.
Ans. 409.6.

Find the circumference of an ellipse of which the axes are 210 and 180.

Method as previous example.
Ans. 614.1.
The actual answer should be 614.4 and Walker's error comes from the fact that he calculates the square root of 38250 as 195.5 ending the calculation too soon since the correct answer should be 195.576.

Find the circumference of an ellipse the axes of which are 360 and 480.

Method as previous two questions. Gives:
Ans. 1332.855216.
This, of course, is far more figures than one could expect. The correct answer is 1332.865. Walker only computes the square root correct to two decimal places which is where the error arises.

Find the circumference of an ellipse of which the axes are 840 and 612.

Method as previous three questions. Gives:
Ans. 2308.4.
Correct answer is 2308.7. Again error comes from stopping the square root computation after one decimal place without realising that the next figure would be 9.

Find the area of an ellipse, of which the axes are 480 and 600 links.

Computes 480 x 600 x .7854 = 226195.2 (note π/4 = .7854). Then converts to acres.
Ans. 2 ac 1 r 1 p 27 yds.

Find the area of an ellipse, of which the axes are 210 and 180.

Computes 210 x 180 x .7854 =296881.2 (note π/4 = .7854).
Ans. 296881.2

Find the area of an ellipse, of which the axes are 140 and 120.

Method as previous problem.
Ans. 13194.7.

Find the area of an ellipse, of which the axes are 360 and 480 links.

Method as problem two back.
Ans. 1 ac 1 r 17 p 4 yds.

What is the ordinate conjugate to the transverse axis 15, the less absciss being 5 and its ordinate 6.

The calculation 15 + 5 = 20 is not written down.
√(20 x 5) : 6 : : 15 : x
10 : 6 : : 15 : x
2 : 6 : : 3 : x
1 : 3 : : 3 : x
Ans. 9.

What is the conjugate to the transverse axis 609, the less absciss being 116, and its ordinate 280.

The calculation 609 + 116 = 725 is not written down.
√(725 x 116) : 280 : : 609 : x
Long-hand multiplication and square root to get 290 : 280 : : 609 : x
Then (280 x 609)/290 is computed long-hand (not even the 0s are cancelled).
Ans. 588.

The conjugate being 160, the less absciss 25, and the ordinate 60, what is the transverse axis.

Computes √(602+ 802)+ 80 = 180.
602 : 25 x 160 : : 180 : x
Ans. 200.

END

Next section in headed: Mensuration of Solids.

What is the solidity of a parabolic conoid, of which the height is 20 and the diameter of its base 35.

The parabolic conoid is drawn. The calculation 352 × 20 × .3927 is carried out long-hand. (Note that .3927 = π/8)
Ans. 9621.150.

What is the solidity of a parabolic conoid of which the height is 28, and the diameter of its base 49.

The parabolic conoid is drawn. Method as previous problem.
Ans. 26400.43.

What is the solidity of a parabolic conoid of which the height is 36, and the diameter of its base 48.

The parabolic conoid is drawn. Method as previous problem.
Ans. 32572.108.

What is the content of the frustum of a paraboloid, the diameter of its ends being 40 and 32, and its height 12.

Calculation (402+ 322)12x .3927 = 12365.337. All calculations done long-hand.
Ans. 12365.337.

What is the content of the frustum of a paraboloid, the diameter of its ends being 20 and 16, and its height 6.

As previous problem. (202 + 162) 6 × .3927 =
But calculation incomplete and left as 3936 × .3927.

The length of a cask composed of two equal frustums of a paraboloid is 45 inches: what is its content in imperial gallons, the bung diameter being 40 and the head diameter 20 inches.

The cask is drawn but no calculations or solution are given.

What is the solidity of a parabolic spindle of which the length is 30, and the greatest diameter 12.

Calculation 122 × 30 × .7854 × 8/15 is carried out.
Ans. 1809.559.

What is the solidity of a parabolic spindle of which the length is 20, and the greatest diameter 8.

As previous problem: calculation 82 × 20 × .7854 × 8/15 is carried out.
Ans. 563.165.

What is the solidity of a parabolic spindle of which the length is 50, and the greatest diameter 20.

As previous problem: calculation 202 × 50 × .7854 × 8/15 is carried out.
Ans. 8377.58.

What is the solidity of the middle frustum of parabolic spindle its length being 25 in, greater diameter 20, and less diameter 15.

Calculation carried out is
(202 × 8 + 152 × 3 + 20 × 15 × 4) 25 = 126875
Then 126875 × .05236 = 6643.175 (note .05236 = π/60)
Ans. 6643.175

The length of a cask in the form of the middle frustum of a parabolic spindle is 46 inches, its bung diameter 31, and head diameter 24 in. Required content in gallons.

Calculation carried out is
(312 × 8 + 242 × 3 + 31 × 24 × 4) 46 = 570032
Then 570032 × .05236 = 29846.875 (note .05236 = π/60)
Ans. 126.99 imp. gallons.

The length of a cask in the form of the middle frustum of a parabolic spindle is 38 inches, its bung diameter 35.5, and head diameter 32 in. Required content in imperial gallons.

Calculation carried out is
(35.52× 8 + 322 × 3 + 35.5 × 32 × 4)38 = 543020
Then 543020 × .05236 = 28432.52720 (note .05236 = π/60)
Ans. 107.64 imp. gallons.

Walker then calculates the volumes and surface areas of various Platonic solids. He presumably has tables which allow him to scale up from the measurements of solids of unit side.

Find the surface and solidity of a tetrahedron of which the side is 2 ft.

Walker's first attempt is to add log 4 to √3 = 1.732050. He seems to realise his error for he then multiplies 1.732050 by 4 to obtain 6.928200.
Ans. 6.928.

To compute the volume Walker computes
23 × 0.11785113 (note 0.11785113 = 1/(3√8))
Ans. .9428.

Find the surface and solidity of a hexahedron, of which the side is 6 ft.

Walker computes 62 × 6 = 216 and 63 × 1 = 216.
Ans. 216 and 216.

Find the surface and solidity of an octahedron, of which the side is 8 ft.

Walker computes 82 × 3.4641016 = 221.7025024
Ans. 221.702.
He doesn't compute the volume.

Find the surface and solidity of a dodecahedron, of which the side is 12 ft.

Walker computes 122 × 20.6457788 = 2972.9921472 and 123 × 7.66311896 = 13241.86956288.
Ans. 2972.992 and 13241.869.

Find the surface and solidity of an icosahedron, of which the side is 20 ft.

Walker computes 202 × 8.660254 = 3461.01600 and 203 × 2.18169499 = 17453.55992000.
Ans. 3461.016 and 17453.5.

Find the surface and solidity of an icosahedron, of which the side is 30 ft.

Walker computes 302 × 8.660254 = 7794.228600 and 303 × 2.18169499 = 58905.76473000.
Ans. 7794.228 and 58905.764.

What freight, at the rate of 2 / 6 per barrel bulk, of 5 cubic feet, should be charged for a package of the lower part of which, the height, breadth, and depth, respectively measure 3 feet 6 inches, 4 ft 3 in and 1 ft 9 in; and the upper part, the height, breadth, and depth, respectively measures 4 ft 3 in, 4 ft 2 in, and 1 ft 3 ins.

Walker computes the volume of the lower and upper parts of the package working in feet and inches. However it appears incorrect. 3 ft 6 in times 4 ft 3 in is 14.875 sq ft or 14 sq ft 126 sq in. Walker writes this as 14..10..6 giving no units. Now if we multiply by 1 ft 9 in we get 26.03125 cubic ft or 26 cu ft 54 cu in. Walker writes this as 26..0..6..6 giving no units. After this strange calculation he converts to cubic in, obtaining the wrong answer (although it is not too far out). He divides by 5 then computes a proportion sum since 1728 cu in cost 30 pence.

Ans. 24 / 13/4. (The correct answer is 24 / 1.)

An irregular solid was put into a cubical vessel of which the side was 24 in; the vessel was then filled with water, and when the solid was taken out, the water descended 10 inches, how many solid feet did the irregular solid contain.

Walker computes 24 × 24 × 14 = 8064 and 24 × 24 × 24 = 13824. He subtracts to obtain 5760. (Of course he should have calculated 24 × 24 × 10 which is much easier!). He converts to cubic feet by dividing by 1728.
Ans. 31/3 cubic feet.

Find the content of a block of freestone 15 ft in length; the lower part being a parallelepiped of which the end is 8 feet in breadth and 6 ft in depth, and the upper part a triangular prism, of which the height is 3 ft.

Walker uses six figure logs to do the simple multiplications. He obtains 900 cubic feet.

Find the content of a block of freestone of which the dimensions taken in different places are as follows; the lengths 13 ft 5 in and 12 ft 7 in, breadths 5 ft 10 in, 5 ft 7 in and 5 ft 1 in, and depths 4 ft 9 in, 4 ft 7 in and 4 ft 2 in.

Walker computes the average length, breadth and depth which are 13..0, 5..6 and 4..6 respectively. He then makes the same strange error in multiplying as earlier. For example multiplying 13.0 by 5..6 should give 71.5 sq ft or 71 sq ft 72 sq in. Walker multiplies 13.0 by 5 to obtain 65.0 then 13.0 by 0.6 to obtain 6..6 and adds to obtain 71..6. His error therefore is in thinking that 1/2 sq ft is 6 sq in. He obtains

Ans. 321..9 cubic feet. (The answer should be 321.75 cu ft or 321..1298 cubic feet.)

Find the expense at 4d per cubic yard of baring or uncovering a rock in a limestone quarry; the north end of the excavation measuring to the top 72 feet, the south end 49 feet, the east end 63 feet the west end 57 feet, and the diagonal from NE corner to SW corner 70 ft, the mean depth of 6 equidistant sections being 2 feet, 5 ft, 9 ft, 11 ft, 13 ft and 17 ft respectively.

Uses Heron's formula to compute the area of the two triangles, finds the average depth, and thus the volume.
Ans. £19..13/1.

How many cubic yards have been dug out of part of a quarry, the areas of 5 perpendicular sections taken at right angles across the line of excavation at the common distance of 9 feet from one another being respectively 75, 210, 379, 712 and 924 square feet.

Walker writes

Here A = 75 + 924 = 999; 4B = 4 (210 + 712) = 3688; and 2C = 2 × 379 = 758.
Therefore (A + 4B + 2C) × D/3 = (999 + 3688 + 758) × 3 ÷ 27.

This he computes to be 605 and gives
Ans. 605 cubic feet. (It should be cubic yards!)

Let Fig. 56 represent a longitudinal section of a piece of ground over which a railway is to be made the line AB being at the bottom of the cutting: let the breadth of the road be 30 ft, the ratio of the slopes 1 1/2 to 1; the depths of the cross sections bg = 22 ft, ch = 14 ft, di = 16 ft, ek = 12 ft and fl = 28 ft, and their distances Ag = 360 ft, gh = 300 ft, hi = 180 ft, ik = 252 ft, hl = 342 ft and lB = 294 ft. How many cubic yards of earth work are there in the cutting.


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The reference to Fig. 56 must be from a book which contains this, and presumably other problems. The diagram is drawn. No solution is worked out.

Find the content of a block of freestone of which the dimensions taken in different places are as follows: the lengths 16 feet 5 inches and 14 feet 7 inches; breadths 7 feet 10 inches and 7 feet 7 inches and 7 feet 1 inch, and depths 6 feet 9 inches, 6 feet 7 inches and 6 feet 2 inches.

The average length, breadth and depth is calculated. The product is then taken. Walker computes the product in feet and inches in his usual (incorrect?) way.
Ans. 755 ft 7 in 6 pts.

Find the area of a square cistern, of which the side is 96 inches, or its content at 1 inch deep, in imperial bushels.

Walker uses six figure logs to compute the volume in cubic inches, then subtracts 3.346 from the answer (so dividing by 2218.192) to get:
Ans. 4.154 bushels.

Find the area of a rectangular vessel 144 inches long and 84 inches broad, or its content at 1 in deep in imperial gallons and bushels.

Standard calculation with six figure logs. Divides volume in cubic inches by 277.274 to get 43.62 gall and divides by 2218.192 to get 5.453 bushels. Note that the volume is computed from scratch for each calculation.

Find the area of a couch frame 130 inches long, and 80 in broad, or its content at 1 in deep, in imperial bushels.

Same method as before.
Ans. 4.658 bushels.

Find the area of a rhombus, of which the length is 60 in and the perp. height 52 1/2 in or its content at 1 in deep and this of hot soup.

Standard calculation with six figure logs. Divides volume in cubic inches by 277.274 to get 11.36 gall and divides by 26.76 to get 117.7 lbs. Note that the volume is computed from scratch for each calculation.

Find the area of a triangle, of which the base is 50 feet and the perpendicular 20 in or its content at 1 inch deep, in imperial gallons, and lbs of green starch.

The calculation is done as if the question gave the base of the triangle as 50 inches. Presumably this was intended and the misprint is in the question. Note that the volume is computed from scratch for each calculation.

Find the area of a trapezium, of which the diagonal is 78 in and the perpendiculars falling down from the opposite angles 23 and 151/2 inches, or its content at 1 inch deep, in imperial gallons and lbs tallow.

Note that the volume is computed from scratch for each calculation.
Ans. 5.415 gall and 47.81 lbs.

Find the area of a trapezoid, of which the parallel sides are 143 and 121 inches, and their perpendicular distance 10 inches, or its content at 1 inch deep, in imperial gallons and bushels.

Note that the volume is computed from scratch for each calculation.
Ans. 38.08 gall and 4.760 bush.

Find the area of a circular tun, of which the diameter is 60 inches or its content at 1 inch deep, in imperial gallons.

A tun is a cask for beer or wine. Walker computes 60 × 60 (using six figure logs) then divides by 353.036. Now this is 277.274 times 4/π, where 277.274 is what must be used as a divisor to convert cubic inches to gallons.
Ans. 10.19 gallons.

Find the area of a circle, of which the diameter is 48 inches or its content at i inch deep, in imperial gallons.

As in the previous problem, Walker squares the diameter and divides by 353.036. All done with six figure logs.

Find the area of an ellipse, of which the diameters are 66 and 50 inches or its content at i inch deep, in imperial gallons.

Similar to the previous problem, Walker takes the product of the diameters and divides by 353.036. All done with six figure logs.

Find the area of a regular heptagon, of which the side is 80 inches or its content at i inch deep, in imperial gallons and bushels.

Using six figure logs, Walker squares the length of the side and multiplies by 3.6339124. He carries out this calculation twice, the first time converting to gallons by dividing by 277.274, the second times converting to bushels by dividing by 2218.192.
Ans. 83.83 gall and 10.48 bush.

Find the area of a regular octagon, of which the side is 100 inches or its content at i inch deep, in imperial gallons and bushels.

Using six figure logs, Walker squares the length of the side and multiplies by 4.8284271. He carries out this calculation twice, the first time converting to gallons by dividing by 277.274, the second times converting to bushels by dividing by 2218.192.
Ans. 174.1 gall and 21.76 bush.

Find the area in imperial gallons of a curvilinear vessel, of which the traverse diameter is 148 in and 13 perp. ordinates of which the common distance is 15 in are as follows: 82.4, 96.5, 112, 119.2, 121.3, 122.4, 123, 122.6, 121.2, 118.9, 112, 96.3, 82.4, with a small segment of 2 in high at each end.

The imperial gallon is not a measure on area! Walker computes A = 82.4 + 82.4, B = 96.5 + 119.2 + 122.4 + 122.6 + 118.9 + 96.3, C = 112 + 121.3 + 123 + 121.2 + 112. He then calculates A + 4B + 2C = 4047.4. He also calculates D = 15/3 = 5. Next he calculates 96.5 × 2 = 193, multiplies again by 2 to obtain 386, divides by 3 to get 128.6 which he adds to 4047.4 to obtain 4176. He multiplies this by 5 to get 20880.00, finally dividing by 353.036 to convert to gallons.
Ans. 59 gallons.

The side of a cubical vessel is 30 inches, find its content in imperial gallons and bushels.

Using six-figure logs he cubes 30 and divides by 277.274 to get 97.4 gall and divides by 2218.192 to get 12.17 bushels. Note that the volume is computed from scratch for each calculation.

Find the content of a vessel in the form of a parallelepiped, of which the length is 80 inches, the breadth 24 inches, and the depth 25 inches in imp gallons and bushels.

Uses six-figure logs. Method as previous question with the volume computed from scratch for each calculation.

How many imperial gallons of wort will a back contain, of which the length is 110 in, the breadth 90 in, and the depth 10 in.

Standard method.
Ans. 357.1 gallons.

Find the content of a cylindrical vessel, of which the diameter is 40 inches, and depth 60 inches, in imperial gallons, and lbs of white soft soap.

Using six-figure logs, he computes 48 × 48 × 60 and divides by 30.609 to obtain lbs. Again he computes 48 × 48 × 60 and divides by 353.03604 to obtain 391.6 gallons.

Find the content of a vessel in the form of a triangular prism, of which the depth is 80 inches, one of the sides of the base 60 inches, and the perpendicular on that side from the opposite angle 15 inches, in imp. bush., lbs of hot soap and raw starch.

Using six-figure logs, he calculates the volume in cubic inches three times. On the first occasion he divides by 2218.192 to obtain bushels, on the second occasion he divides by 26.76 to obtain the lbs of hot soap, and on the third occasion he divides by 40.3 to obtain the lbs of raw starch.
Ans. 16.22 bush., 1345 lbs, 593.3 lbs.

Find the content of a vessel in the form of a regular hexagonal prism, of which the depth is 32 in, and a side of the base 40 inches, in imperial gallons and bushels.

Using six-figure logs, he twice calculates 40 × 40 × 32. First time he divides by 106.773 to obtain 479.7 gall. On the second occasion he divides by 853.782 to obtain 59.96 bush.

Find the content in imperial gallons of a vessel in the form of an elliptical cylindroid (that is, of a vessel of which the two ends are equal, similar, and similarly situate ellipses, and the sides perpendicular to them) the depth of the vessel being 50 in, the transverse axis of the base 154 in, and conjugate axis 54.96 in.

Using six-figure logs, he calculates 154 × 54.96 × 50 and divides by 353.035 to obtain 1198 gallons.

Find the content of a conical vessel, of which the perpendicular depth is 60 inches, and the diameter of the mouth 48 inches, in imperial gallons.

Using six-figure logs, he calculates 48 × 48 × 60 and divides by 3, and then by 353.035 to obtain 130.2 gallons.

Find the content of a vessel, in the form of a regular pentagonal pyramid, of which the depth is 90 in, and the side of the base 60 in, in imperial bushels.

Using six-figure logs, he calculates 60 × 60 × 90 and divides by 3, and then by 1289.288 to obtain 83.76 bushels.

Find the content of a sphere of which the diameter is 90 in, in imperial gallons and bushels.

Using six-figure logs, he twice calculates 90 × 90 × 90. First time he divides by 529.544 to obtain 1376 gallons. On the second occasion he divides by 4236.434 to obtain 172 bushels.

Find the content of a vessel in the form of a frustum, of which the bottom diameter is 55 in, the top diameter 20 in, depth 64 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 273.3 gallons.

Find the content of a vessel in the form of a frustum of a square pyramid, of which the depth is 48 in, and the sides of the bases and ends 21 in and 39 in respectively, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 160.5 gallons.

Find the content of a frustum of a rectangular pyramid, of which the depth is 60 in, sides of one base 54 and 24 in, the other 32 and 18 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 197.3 gallons.

Find the content of a vessel in the form of a frustum of a square pyramid, of which the depth is 50 in, and the sides of the bases and ends 20 in and 40 in respectively, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 168.1 gallons.

Find the content of a vessel in the form of a frustum of a square pyramid, of which the depth is 60 in, and the sides of the bases and ends 30 in and 50 in respectively, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 132.6 gallons.

Find the content of a vessel in the form of a frustum of a square pyramid, of which the depth is 80 in, and the sides of the bases and ends 70 in and 90 in respectively, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 1856 imp. gallons.

Find the content of a vessel in the form of a frustum of a rectangular pyramid, of which the depth is 40 in, sides of one base 30 and 20 in, of the other 20 and 10 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 55.1 imp. gallons.

Find the content of a vessel in the form of a frustum of a rectangular pyramid, of which the depth is 80 in, sides of one base 40 and 30 in, of the other 30 and 20 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 221.2 imp. gallons.

Find the content of a vessel in the form of a frustum of a rectangular pyramid, of which the depth is 180 in, sides of one base 100 and 60 in, of the other 80 and 40 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 2939 imp. gallons.

Find the content of a vessel in the form of a frustum of an elliptical cone, the diameter of the bottom being 40 and 35 in, those at the top respectively parallel to these 20 and 17 in, and the depth 36 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 82.59 imp. gallons.

Find the content of a vessel in the form of a frustum of an elliptical cone, the diameter of the bottom being 50 and 40 in, those at the top respectively parallel to these 30 and 27 in, and the depth 50 in, in imperial gallons.

Standard calculation with six-figure logs to get:
Ans. 190.4 imp. gallons.

Find the content of a copper pan in the form of a segment of a sphere, the diameter of the mouth being 90 in, that at the depth of 71/2 in 65.38 in, and the depth of the pan 15 in, in imp gall.

Walker copies down the log of 45 incorrectly. He writes the log as 1.693213 instead of 1.653213.

He computes 45 × 45 × 3 = 6002 (which of course is incorrect because of the miscopying of the log). He squares 15 and adds to this, multiplies by 15, then multiplies by .5236 to obtain the volume in cubic inches. Finally he divides by 277.274 to obtain the answer 176.4 imp gall. [If he had not made the copying error he would have got 178.45 imp gall.]

Find the content of a vessel in the form of a pyramid, of which the depth is 48 in, the length and breadth at the top 64 and 36 in, and the length and breadth of the bottom 48 and 27 in, in imp gall and bushels.

Walker computes, using six figure logs:

64 × 36 = 2304 and 48 × 27 = 1290 (this should be 1296 - probably a copying error from tables). He then multiplies 2304 by 1290 and takes the square root to get 1728. This answer is correct since in the calculation he uses the log of 48 × 27 which was right. He calculates 2304 + 1290 + 1728, multiplies by 48, then divides by 3. Finally he divides by 277.274 to obtain the answer 307.1 imp gall. [If he had not made the copying error he would have got 307.45 imp gall.] He does not calculate the volume in imp bushels.

Find the content of a vessel in the form of a pyramid, of which the depth is 68 in, the length and breadth at the top 84 and 56 in, and the length and breadth of the bottom 68 and 57 in respectively, in imp gall and bushels.

Calculation as in the previous problem. Again he does not calculate the volume in imp bushels.
Ans. 1050 imp gall.

Find the content of a vessel in the form of a pyramid, of which the depth is 68 in, the length and breadth at the top 84 and 56 in, and the length and breadth of the bottom 68 and 57 in respectively, in imp gall and bushels.

Calculation as in the previous two problems. Again he does not calculate the volume in imp bushels.
Ans. 1778 imp gall.

Find the content in imperial gallons, and tabulate it for every inch, of a round tun, of which the following are the dimensions in inches.

Parts of   Depth from     Cross        Mean
  depth	    the mouth   diameters    diameters
   10           5     35.75 .... 36.   35.875
   10          15     38.   .... 37.25 37.625
   10          25     39.25 .... 39.5  39.375
   10          35     40.75 .... 41.5  41.125
	    drip -- By measure 10 gallons

Walker computes, using six figure logs:

(39.375 × 39.375) / 353.036 × 10 = 43.91
(41.125 × 41.125) / 353.036 × 10 = 47.9
(35.875 × 35.875) / 353.036 × 10 = 36.45
(37.625 × 37.625) / 353.036 × 10 = 40.1
Then adds
43.91 + 47.9 + 36.45 + 40.1 = 168.36
168.36 + 10 = 178.36 (The 10 here is the drip)
Ans. 178.36 gall.

Find the content in imperial gallon of a copper, of which the following are the dimensions in inches.

Parts of  Distances from   Cross        Mean
  depth	    the mouth   diameters    diameters
   13         6.5     97.5 .... 98.1 97.8
   10          18     95.8 .... 96.4 96.1
   10          28     94.3 .... 93.9 94.1
   10          38     93.2 .... 93.  93.1
   43    Crown by measure 38 gallons

Walker computes, using six figure logs:

(97.8 × 97.8) / 353.036 × 13 = 352.2
(96.1 × 96.1) / 353.036 × 10 = 261.5
(94.1 × 94.1) / 353.036 × 10 = 261.5
(93.1 × 93.1) / 353.036 × 10 = 261.5
Then adds
352.2 + 261.5 + 261.5 + 261.5 = 1110
1110 + 38 = 1148 (The 38 here is the crown)
Ans. 1148 gall.

Find the content in imperial gallon of a still, of which the following are the dimensions in inches, computing the upper 9 in as the frustum of a sphere.


Parts of depth Cross Mean from collar diameters diameters 9 27 .... 27 27 56.3 .... 56.2 56.25 9 59.1 .... 60.3 59.7 9 63.8 .... 64.1 63.95 9 64.2 .... 64.5 64.35 10.6 62.1 .... 62.5 62.3 46.6 To cover the crown 30 gallons

Similar calculation to the previous two problems.

From this point on Walker draws fine sketches.

Find the content of a cask, of which the bung diameter is 31 inches, the head diameter 24 inches, and the length 32 1/2 inches and the perpendicular distance mn being 8 inches.

Calculation carried out with six figure logs.

24 × 24 + 31 × 31 × 2 = 2498.
(2498 × 32.5)/1059.108 = 76.66
Ans. 76.66 gallons

Find the content of a rum cask, of which the bung diameter is 31.7 inches, the head diameter 26.8 inches, and the length 32.7 inches and the perpendicular distance mn being 58 inches.

Calculation carried out with six figure logs.

26.8 × 26.8 + 31.7 × 31.7 × 2 = 2727.4 (misprint for 2727.2, the log is correct)
(2727.4 × 32.7)/1059.108 = 84.21
Ans. 84.21 gallons


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Find the content of a pipe of Madeira wine, of which the bung diameter is 29 inches, the head diameter 21.2 inches, and the length 47.4 inches and the perpendicular mn 9 3/4 inches.

Calculation carried out with six figure logs.

21.2 × 21.2 + 29 × 29 × 2 = 2131.4
(29 × 29 - 21.2 × 21.2) × 2/5 = 156.6
2131.4 - 156.6 = 1974.8
(1974.8 × 47.4)/1059.108 = 88.39
Ans. 88.39 gallons

Find the content of a pipe of Spanish wine, of which the bung diameter is 31 inches, the head diameter 24 inches, and the length 46 inches and the perpendicular mn 8 3/4 inches.

Calculation carried out with six figure logs.

24 × 24 + 31 × 31 × 2 = 2498
(2498 × 46)/1059.108 = 108.5
Ans. 108.5 gallons

Find the content of a cask, of which the bung diameter is 29 inches, the head diameter 23 inches, and the length 36 inches and the perpendicular mn 9 inches.

Calculation carried out with six figure logs.

23 × 23 + 29 × 29 × 2 = 2211
(2211 × 36)/1059.108 = 75.15
Ans. 75.15 gallons


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Find the content of a cask, of which the bung diameter is 31 inches, the head diameter 23 inches, and the length 40 inches and the perpendicular mn 1.4 inches.

Calculation carried out with six figure logs.

31 × 31 + 23 × 23 + 29.2 × 29.2 ×4 = 4900
(4900 × 40)/2118.217 = 92.53
Ans. 92.53 gallons

Find the mean diameter, and thence the content in imperial gallons of a cask, of which the bung diameter is 31 inches, the head diameter 23 inches, and the length 50 inches and the perpendicular mn 1.4 inches.

Calculation carried out with six figure logs.
Walker divides 23 by 31 and obtains .7419.
He then calculates 8 × .613 (which he labels Table area) to obtain 4.904.
To 4.904 he adds 23 to obtain 27.904 (which he labels Mean Diam.)
(27.904 × 27.904 × 50)/353.036 = 110.2
Ans. 110.2 gallons


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Find the mean diameter, and thence the content in imperial gallons of a spherical cask, of which the bung diameter is 31 inches, the head diameter 27 inches, and the length 301/2 inches.

Calculation carried out with six figure logs.
Walker divides 27 by 31 and obtains .879. [Note. He incorrectly looks up log 31 writing 1.491212 instead of 1.491362]
He then calculates 4 × .682 to obtain 2.728.
To 2.728 he adds 27 to obtain 29.728
(29.728 x 29.728 x 30.5)/353.036 = 76.35
Ans. 76.35 gallons

Find the mean diameter, and thence the content in imperial gallons of a spherical hogshead, of which the bung diameter is 29 inches, the head diameter 23 inches, and the length 27 inches.

Calculation carried out with six figure logs.
Walker divides 23 by 29 and obtains .7931.
He then calculates 6 × .691 to obtain 4.146.
To 4.146 he adds 23 to obtain 27.146
(27.146 × 27.146 x 27)/353.036 = 56.35
Ans. 56.35 gallons

Find the mean diameter, and thence the content in imperial gallons of a pipe of wine, of the variety of which the bung diameter is 28.8 inches, the head diameter 22.3 inches, and the length 47 inches.

Calculation carried out with six figure logs.
Walker divides 22.3 by 28.8 and obtains .7743.
He then calculates 6.5 × .677 to obtain 4.4.
To 4.4 he adds 22.3 to obtain 26.7
(26.7 x 26.7 × 47)/353.036 = 94.91
Ans. 94.91 gallons

Find the content of a cask, of which the bung diameter is 21 inches the head diameter 18 inches, and the length 30 inches.

Walker writes
21 = .8328
18 = .3059
Adds to get 1.1387 which he multiplies (using logs) by 30 to obtain 34.16.
Ans. 34.16 gallons.

Find the content of a cask, of which the bung diameter is 32 inches the head diameter 25.3 inches, and the length 47 inches.

Walker writes
32 = 1.9337
25.3 = 0.6044
Adds to get 2.5381 which he multiplies (using logs) by 47 to obtain 119.2.
Ans. 119.2 gallons.

Find the ullage of a lying aulm, of which the length is 24 inches the bung diameter 22 inches, the head diameter 19 inches, and the depth of liquor 12 inches.

Calculation carried out with six figure logs.
Walker divides 19 by 22 and obtains .8636.
He then calculates 3 × .606 to obtain 1.818.
To 1.818 he adds 19 to obtain 20.818 (which he labels mean diam.)
(20.818 x 20.818 x 24)/353.036 = 29.46.

He then computes 5 × 12 = 60 from which he subtracts 11 (which he labels 1/2 of 22 = bung diam.). He multiplies the resulting 49 by the 29.46 and divides the result by 88 (which he labels bung diam. 22 × 4 = 88) to obtain 16.4.
Ans. 16.4 gallons.

What is the ullage of a lying hogshead, of which the length is 27 inches, the bung diameter 29 inches, the head diameter 23 inches, and the depth of liquor 10 inches.

Calculation carried out with six figure logs.
Walker divides 23 by 29 and obtains .7931.
He then calculates 6 × .610 (he writes .79 = .610) to obtain 3.66.
To 3.66 he adds 23 to obtain 26.66.
(26.66 × 26.66 × 27)/353.036 = 54.35.

He then computes 10 × 5 = 50 from which he subtracts 14.5 (which he labels 1/2 of 29). He multiplies the resulting 35.5 by the 54.35 and divides the result by 116 (which he labels 29 × 4) to obtain 16.63.
Ans. 16.63 gallons.

What is the ullage of a standing hogshead, of which the length is 29.1 inches, the bung diameter 28.5 inches, the head diameter 24 inches, and the depth of liquor 18 inches.

Calculation carried out with six figure logs.
24 × 24 + 28.5 × 28.5 × 2 = 2200.5.
(2200.5 × 29.1)/1059.108 = 60.46

He then computes 18 × 11 = 198.0 from which he subtracts 14.55 (which is 1/2 × 29.1). He multiplies the resulting 183.45 by the 60.46 and divides the result by 291 to obtain 38.24. [Note. The 60.46 comes from the antilog of 1.781474 but when he uses it again in the next calculation he clearly looks it up in the tables and gets it wrong, writing 1.782902.]
Ans. 38.24 gallons


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What is the ullage of a standing butt, of which the length is 39.5 inches, the bung diameter 33.8 inches, the head diameter 29.8 inches, and the depth of liquor 30 inches.

Calculation carried out with six figure logs.
29.8 × 29.8 + 33.8 × 33.8 × 2 = 3172.
(3172 × 39.5)/1059.108 = 118.3

He then computes 30 × 11 = 330 from which he subtracts 19.75 (which is 1/2 × 39.5). He multiplies the resulting 310.45 by the 118.3 and divides the result by 395 to obtain 92.97.
Ans. 92.97 gallons.

There is now a sequence of problems involving artillery


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What is the first velocity of a 10 inch shell, weighing 90 lbs, when fired with a charge of 4 lbs of powder.

Calculation carried out with six figure logs.
Walker computes √(8/90) × 1600 = 477.
[Here 8 = 2 × 4, the weight of powder in lbs]
or
√90 : √8 : : 1600 : x
9.5 : 2.9 : : 1600
1.9 : 2.9 : : 320
1.9 x = 928
x = 477
Clearly he doesn't carry out this last calculation otherwise the approximations he has made would lead to a fair error. He would get x = 488.
Ans. 477 feet per second.


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What is the first velocity of an 8 inch shell, weighing 48 lbs, when fired with a charge of 2 lbs of powder.

Calculation carried out with six figure logs.
Walker computes √(4/48) × 1600 = 462.
Ans. 462 feet per second.

A 51/2 inch shell, weighing 16 lbs, is fired with 1 lbs of powder, with what velocity is it discharged.

Calculation carried out with six figure logs.
Walker computes √(2/16) × 1600 = 565.6.
Ans. 565.6 feet per second.

A 42/3 inch shell, which weighs 8 lbs, is fired with 1/2 lbs of powder, with what velocity is it discharged.

Calculation carried out with six figure logs.
Walker computes √(1/8) × 1600 = 565.6.
Ans. 565.6 feet per second.

The diameter of a 3 lb iron ball is 2.8 inches, what is its terminal velocity.

Calculation carried out with six figure logs.
Walker computes √2.8 × 175.5 = 293.6.
Ans. 293.6 feet.

What is the terminal velocity of a 9 lb ball, its diameter being 4.04 inches.

Calculation carried out with six figure logs.
Walker computes √4.04 × 175.5 = 357.2.
Ans. 357.2 feet.

What is the terminal velocity of a 42 lb ball, its diameter being 6.75 inches.

Calculation carried out with six figure logs.
Walker computes √6.75 × 175.5 = 455.9.
Ans. 455.9 feet.

What is the terminal velocity of a 13 inch shell, its diameter being 12.8 inches.

Calculation carried out with six figure logs.
Walker computes √12.8 × 147.3 = 527.
Ans. 527 feet.

What is the terminal velocity of a 8 inch shell, its diameter being 7.9 inches.

Calculation carried out with six figure logs.
Walker computes √7.9 × 147.3 = 414.
Ans. 414 feet.


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From what height must a body fall to acquire a velocity of 2000 ft per second.

Calculation carried out with six figure logs.
Walker computes (2000 × 2000)/64 = 62500
Ans. 62500 feet.

From what height must a body fall to acquire a velocity of 1600 ft per second.

Calculation carried out with six figure logs.
Walker computes (1600 × 1600)/64 = 40000
Ans. 40000 feet.

From what height must a body fall to acquire a velocity of 294 ft per second.

Calculation carried out with six figure logs.
Walker computes (294 × 294)/64 = 1350
Ans. 1350 feet.

What is the greatest range of a 42 lb iron ball, when discharged with a velocity of 2000 feet per second, and the elevation necessary for producing that range, the diameter of the ball being 6.75 inches.

Calculation carried out with six figure logs.
Walker computes √6.75 × 175.5 = 455.9.
(455.9 × 455.9)/ 64 = 3248.
2000/455.9 = 4.386 = 33°30.
He then writes 33°30 = 3.1031 which he multiplies by 3248 to obtain 10086.
He then writes 32°45 = 3.2968 which he multiplies by 3246 to obtain 10708.
(10708 - 10086)/2 = 311.
10086 + 311 = 10397
Ans. 33°30 elevation and 10397 feet (he appears to write 10.397 feet)


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What is the greatest range of a 13 inch shell and ball, when discharged with a velocity of 2000 feet per second, and the elevation to produce that range, the diameter of the shell being 12.8 inches.

Calculation carried out with six figure logs.
Walker computes √12.8 × 147.3 = 527.
(527 × 527)/ 64 = 4340.
2000/527 = 3.795 = 34°49.
He then writes 34°49 = 2.7631 which he multiplies by 4340 to obtain 11992.
Ans. 11992 feet and 34°49.

What is the greatest range of a 10 inch shell, of which the diameter is 9.84 inches, when discharged with a velocity of 1700 ft per sec and the elevation necessary.

    
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Calculation carried out with six figure logs.
Walker computes √9.84 × 147.3 = 462.
(462 × 462)/ 64 = 3336.
1700/462 = 3.679 = 35°10.
He then writes 35°10 = 2.6749 which he multiplies by 3336 to obtain 892.3.
Ans. 892 feet and 35°10.

The range of a shell at an elevation of 45° was found to be 3500 feet, at what elevation must the piece be set, to strike an object at the distance of 2920 feet, with the same charge of powder.

Walker writes
[log] Sine 90 = 10.000000 (he works here with log 1010 sin)
[log] 2920 = 3.465383
Adds to get 13.465383 from which he subtracts log 3500 = 3.544068 which gives 9.921313. He then writes 9.921313 = 56°32' (again log Sine 56°32 = 9.921313) which he divides by 2 to obtain 28°16'.
Ans. 28°16'.

Find the charge of powder necessary to fire a 13 inch shell weighing 196 lbs with a velocity of 485 ft per sec.

Walker writes
Log 485 = 2.685742
485 = 2.685742
485 = 2.685742
7.663740
5120000= 6.709270
9 lbs = 0.954470
Ans. 9 lbs.

This must be a simple miscopying. He should have computed
(485 × 485 × 196)/5120000 = 9.

Find the charge of powder necessary to fire a 10 inch shell weighing 90 lbs with a velocity of 500 ft per sec.

Calculation carried out with six figure logs.
(500 × 500 × 90)/5120000 = 4.394.
Ans. 4 lbs 6.3 oz.


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If, with a charge of 4 lbs of powder, a shell range 2000 feet, how far will its range when the charge is 5 lbs, the elevation being in both cases the same.

Walker computes, using six figure logs

4 : 5 : : 2000 : x.
Ans. 2500 feet.


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At what time will a shell range 4000 feet, when discharged at an elevation of 45°.

Walker writes
R : Tang 45° : : 4000 : x
Writing the log of Tang 45° and Sine 90° as 10, i.e. the log of 1010tan 45° and 1010sin 90°, he computes
(4000 × Tang 45°)/Sine 90°, takes the square root, then divides by 4 to obtain 15.80.
Ans. 15.80 secs.


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A shell discharged at an elevation of 40° ranges 3000 ft. required its time of flight.

Walker writes
R : Tang 40° : : 4000 : x
Writing the log of Tang 40° as 9.923813 and of Sine 90° as 10, i.e. the log of 1010tan 45° and of 1010sin 90°, he computes
(3000 × Tang 40°)/Sine 90°, takes the square root, then divides by 4 to obtain 12.54.
Ans. 12.54 secs.

A shell when discharged at an elevation of 35° ranges 1000 feet what is its greatest height.

Walker writes
R : Tang 35° : : 1000 : x
Writing the log of Tang 35° as 9.845227 (he writes 9.849227 in error but this is a copying error only since adding 3 gives 12.845227) and of Sine 90° as 10, i.e. the log of 1010tan 35° and of 1010sin 90°, he computes
(1000 × Tang 35°)/Sine 90° = 700.
Ans. 700 feet.


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With what impetus, velocity, and charge of powder must a 13 inch shell be discharged at an elevation of 32°12' to strike an object at the distance of 3250 feet.

Walker doubles 32°12 and halves 3250 writing
Sine 64°25 : rad : : 1625 : x
As before he uses six figure logs with log Sine as the log of 1010sin etc in the calculations.
(1625 x Sine 90°)/Sine 64°25 = 1802 feet.
1802/64 = 340 velocity
(340 × 340 × 196)/5120000 = 4.403 (this is not very accurate since Walker writes the log of 340 as 2.530418 when it should be 2.531479.
Ans. 4 lbs 6 oz charge.


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How far will a shot range on a plane which ascends 8°15 and on another which descends 8°15; the impetus being 3000 feet, and the elevation of the piece 32°20' the elevation above the plane in the first case 24°15, in the second 40°45'.


How much powder will throw a 13 inch shell 4244 feet on an inclined plane which ascends 8°15', the elevation of the mortar being 32°30.


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In what time will a shell strike a plane which was 10°, when discharging with an impetus of 2304 feet, at an elevation of 45°.


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At what elevation must a mortar be pointed to range 2662 feet on a plane which ascends 10°, the impetus being 2000 feet.


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Find the weight of an iron ball, of which the diameter is 6.41 inches.

Walker writes

43 : 6.413 : : 9 : x

and then, using six figure logs, computes x as
Ans. 37.03 lbs.

Find the weight of an iron ball, of which the diameter is 7.018 inches.

Walker writes

43 : 7.0183 : : 9 : x

and the, using six figure logs, computes x as
Ans. 48.6 lbs.


Find the diameter of an 18 lb iron ball.

Using the results of the previous problem, d3 should be 18 × (43/9). Now 43/9 = 7.1111111... which, interestingly, Walker writes as 7.1111/9. Using six figure logs he multiplies 7.1111/9 by 18 and takes the cube root to obtain
Ans. 5.039 inches.

Find the diameter of an 36 lb iron ball.

As in the previous problem, Walker computes 36 × 7.111 1/9 and takes the cube root to obtain
Ans. 6.344 inches.

Find the diameter of an 42 lb iron ball.

As in the previous problem, Walker computes 42 × 7.111 1/9 and takes the cube root to obtain
Ans. 6.684 inches. [Note that he writes 7.1191/2 instead of 7.1111/9 but since he has the correct log it must be a copying error.]

This picture shows the Serbian General Omer Pasha entering the town of Eupatoria which he captured from the Russians for the Turks in February 1855


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What is the weight of an iron shell, the external and internal diameter being 11.1 and 8 inches.

Walker writes

(11.13 - 83) x 9 ÷ 64 = Ans

Then he computes (3 log 11.1 - 3 log 8 + log 9) which is clearly wrong. He multiplies the log answer he gets by 2, writes down the log of 64 and ends in confusion leaving the answer blank.

What is the weight of an iron shell, the external and internal diameter being 9.8 and 7 inches.

Walker writes

(9.83 - 73) × 9 ÷ 64 = Ans

Then he computes (3 log 9.8 - 3 log 7 + log 9) which is clearly wrong. He writes down the log of 64 and ends in confusion leaving the answer blank.


How much powder will fire a shell of which the internal diameter in 8 inches.

Walker computes, using logs

83/57.3 = 8.93 lbs.
Ans. 8.93 lbs.

How much powder will fill a shell, of which the internal diameter in 9 inches.

Walker computes, using logs

93/57.3 = 12.72 lbs.
Ans. 12.72 lbs.

Find the diameter of a shell which will hold 6 lbs of powder.

Walker computes, using logs
6 × 57.3 then takes the cube root.
Ans. 7.005 lbs. [A misprint for 7.005 inches.]


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Find the diameter of a shell which will hold 9 lbs of powder.

As for the previous problem.
Ans. 8.019 inches.

Find the diameter of a shell which will hold 13 lbs of powder.

As for the previous problems.
Ans. 9.064 inches.

A box is 20 inches long, 14 inches wide and 12 inches deep how much powder will it hold.

Walker computes, using logs
20 × 14 × 12 ÷ 30 = 112.
Ans. 112 lbs.

How much powder will fill a cubical box of which the side is 18 inches.

Walker computes, using logs
18 × 18 × 18 ÷ 30 = 194.4.
Ans. 194.4 lbs.

Find the number of balls in a triangular pile, of which each side of the base contains 40 balls.

1/6 (40 × 41 × 42)
Walker computes, using logs
40 × 41 × 42 ÷ 6 = 11480.
Ans. 11480 balls.

Find the number of balls in a triangular pile, of which each side of the base contains 20 balls.

As previous problem.
Ans. 1540 balls.

Find the number of balls in a triangular pile, of which each side of the base contains 15 balls.

As previous problem.
Ans. 680 balls.

How many balls are there in a square pile of 20 rows.

Walker computes, using logs
20 × 21 × 41 ÷ 6 = 2870.
Ans. 2870 balls.

How many balls are there in a square pile of 25 rows.

As previous problem.
Ans. 5520 balls.


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Find the number of balls in a rectangular pile, the length and breadth of the base row being 50 and 30 respectively.

Walker computes, using logs
30 × 31 × 121 ÷ 6 = 18755.
Ans. 18755 balls.

Find the number of balls in an incomplete triangular pile, one side of the bottom course being 50, and the top course being 25.

Walker computes, using logs
50 × 51 × 52 ÷ 6 = 23100.
Then
25 × 26 × 24 ÷ 6 = 2600.0.
23100-2600 = 19500
Ans. 19500 balls.

Find the number of balls in an incomplete triangular pile, one side of the bottom course being 30, and the top course being 10.

Walker computes, using logs
30 × 31 × 61 ÷ 6 = 9455.
Then
9 × 10 × 19 ÷ 6 = 285.

Ans. 9170 balls.

That finishes the excursion into artillery!

The base of a right-angled triangle is 300 and the sum of the other sides 1000, what are these sides.

Interestingly Walker is now writing a little more about the problems. Here he writes
Let x = the hypotenuse
1000-x = other side
x2 = 3002 + (1000 - x)2
x2 = 9000 + 10000000 - 2000 x + x2
2000 x = 1090000
2 x = 1040
x = 545 hypotenuse
1000 - x = 455 other side.

The paving of a triangular court, at 4/- a yard; came to £26..13/4, the length of the three sides was 80 feet, find the sum of the other two sides.

Let x = perpendicular
(80 x /2)4 = (5331/3)9
160 x = 4800
x = 30
x2= 302 + 402
x2=
900 + 1600
x2= 2500
x = 50
2x = 100
Ans. 100 feet.


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The area of a right-angled triangle is 2 roods 16 poles, and the hypotenuse 500 links, what are the legs.

First he converts 2 roods 16 poles to 96 poles, then computes 625 x 96 = 60000 to obtain the area in square links. He then writes

Let x = one leg
Then √(5002 - x2) = other leg
x/2√(5002 - x2) = 60000

He then squares the equation obtaining a quadratic in x2 which he solves by completing the square. This gives x2 = 160000 or 90000 so x = 400 or 300.
Ans. 400 and 300 links.


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If from a right-angled triangle, of which the base is 18 feet, and the perpendicular 24 feet, a triangle of which the area is 54 square feet be cut off by a line parallel to the perpendicular, what will be the sides of the latter.

Walker computes 18 × 12 = 216 using six figure logs. He then writes
216 : 54 : : 18 : x
which he solves to obtain x = 9.
Again he solves 2 : 1 : : 24 : x to obtain x = 12.
He then uses Pythagoras to compute the hypotenuse √(122 + 92) = 15.
Ans. Base = 9 feet, Perp. = 12 feet, Hyp. = 15 feet.

If from a triangle, of which the three sides are 13, 14, 15, a triangular area of 24 was cut off by a line parallel to the longest side, what will be the length of the sides of the triangle containing that area.

He uses Heron's formula to compute the area as 84. He then writes
√84 : √24 : : 13 : x
calculating x = 7.
√84 : √24 : : 14 : x
calculating x = 7.48.
√84 : √24 : : 15 : x
calculating x = 8.
Ans. 7, 7.48 and 8.


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The segments of the base of a triangle made by a straight line bisecting the vertical angle are 72 and 58, and the sum of the other sides 270, what is the area of the triangle.

(72 + 58) : 270 : : 72 : x
giving x = 1497/13.
(72 + 58) : 270 : : 58 : x
giving x = 1206/13.
He now has the three sides as 130, 1497/13, and 1206/13. Using Heron's formula he obtains 7496.
Ans. 7496 area.

A roof which measures 30 feet 8 in by 16 feet 6 inches, is to be covered with lead, at 8 lbs per square foot, find the expense of the lead @ 41/6 a cwt.

Walker computes, using six figure logs

30.66 2/3 × 16.5 × 8 ÷ 112.
He then multiplies by 41.5 and divides by 20 to obtain 74.98 (pounds sterling) which he then calculates is
Ans. £74..19..7.

If the expense of paving a semicircular plot at 3 / 6 a yard amounts to £19..1/9 1/2 what was the diameter of the circle.

£19..1/91/2 is converted to 4581.5 (pence).
He the computes, using six figure logs
4581.5 × 2 × 9 ÷ 42
He divides the answer by .7854 (which is π/4) and takes the square root to obtain 50.
Ans. 50 feet.

A triangle of which the three sides are 140, 180, 82 and is inscribed in a circle, what is the diameter of the circle.

We have corrected the question. Walker actually missed out the length of the third side!
First he finds the area by Heron to be 4036. Then
4036 ÷ 70 = 57.66 perp.
He then divides 8200 by 57.66 to obtain 142.2
Ans. Diameter = 142.2

A circular pond occupies an imperial acre, find the perimeter of the circumscribed square.

Walker computes 43560 ÷ .7854 (which is π/4). He then takes the square root, multiplies by 4 and obtains 942.
Ans. 942 feet.

A perpendicular drawn from one of the angles of an equilateral triangle to the opposite side measures 12 feet, find the length of a side of the triangle.

Sine 60° : Sine 90° : : 12 : x
Using (as usual log Sine A = log 1010sin A) he calculates 13.85.
Ans. 13.85.

A straight line 330 links long, drawn from the right angle of a right-angled triangle, divides the hypotenuse into two segments which respectively measure 217 and 480.12 links, what is the area of the triangle.

Walker writes
267 : Sine 45° : : 330 : Sine C
After calculating the angle to be 60°55 he adds 45°, subtracting the answer from 180° to get BDC = 74°5
Sine 45° : 267 : : Sine 74°5 : BC
giving 363.1.
Sine 45° : 480.12 : : 105°55 : AB
He computes Sine 74°5 × 480.12 ÷ Sine 45°.
Multiplies the answer by 363.1, divides by 2, then divides by 100000 to obtain 1.185. He converts to get
Ans. 1 ac 0 r 28 p.

A field in the form of an equilateral triangle contains half an acre, what must be the length of the tether fixed to a horses nose to allow him to graze exactly half of it.

2 : Sine 56°18 : : 2.4 : Sine BEF
Calculates 86°49
Sine 56°18 : 2 : : Sine 36°53 : FE
Calculates 1.442
Sine 86°49 : 1.442 : : Sine 6°32 : AE
Calculates 10.92. (writes 10:92)
Sine B : AC : : Sine BAC : BC
Calculates 19.12.
Ans. C = 30°30, AE = 10.92 (writes 10:92), BC = 19.12.


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In order to ascertain the height of an inaccessible object CD, I selected two stations, A and B, on a level with the object's base, but in a different direction, I found the horizontal angle DAB 87°5; the horizontal angle ABD 53°15; the vertical angle CAD 47°30 and the distance AB 283 feet; what was the height of the object, my eye being 5 feet from the ground.

Walker writes

180° - (53°15 + 87°5) = 39°40 = ADB
90° - 47°30 = ACD
Sine 39°40 : 283 : : 53°15 : AD
Sine 42°30 : AD : : 47°30 : AC

Calculating with six figure logs he obtains 387.6 to which he adds 5 to obtain 392.6.
Ans. 392.6 feet.


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Lord Melville's statue in St Andrews Square Edinburgh is 16 ft high and stands on the top of a column 136 feet high; at what distance from the base of the column in the same horizontal plane will the statue appear under the greatest possible vertical angle, what will that angle be, and at what distaance may the statue be viewed under an angle of .

Only a beautiful drawing of a statue (not, as it happens, Lord Melville's in St Andrews Square) is given. No working or answer.

A piece of cable 3 feet in length and 9 inches in girt weighs 22 lbs, what will the cable weigh per fathom of which the girt is 12 inches.

3 : 8 : : 22 : x
9 : 12
27 : 96 : : 22 : x
9 : 32 : : 22 : x
9 x = 704
x = 78 2/9
Ans. 78 2/9 lbs.


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If 20 foot of railing weigh half a ton, when the base of the bars are 11/4 inches square, what will 50 feet come to at 31/2 per lb, when the bars are 7/8 inch square.

20 : 50 : : 1120 : x
100 : 49 : :
2000 : 2450 : : 1120 : x
4 : 49 : : 112 : x
x = 1372
Walker then multiplies by 3 1/2 and converts the answer, which is in pence, into pounds shillings and pence
Ans. £20..0..2

If 20 grain of gold gild a globe which weighs 512 ounces, how many grains will gild a globe of the same kind of wood that weighs 1331 ounces.

Let x = diameter of small globe.
Let y = diameter of large globe.
3.1416 x2 = 20
x = √20/3.1416
512 : 1331 : : x3 : y3
512 : 1331 : : (20/3.1416)(3/2): y3 [This doesn't seem right. He is now using x = √(20/3.1416).]
From this he calculates y (using logs), squares it and multiplies by 3.1416 to obtain 37.8
Ans. 37.8 grains.


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Find the expense of gilding a ball 6 feet in diameter, at 31/2 per square inch.

Walker writes
D2 × 3.1416 × 144 × 3.5 = Ans

He computes this using six figure logs with D = 6> he divides by 240 to obtain the answer in pounds.
Ans. 237..10/-

The length of the keel of a ship of 250 tons is 72 feet, find the tonnage of another ship of the same form, of which the keel is 81 feet long.

723 : 81 3 : : 250 x.
Then x is computed using logs.
Ans. 355.9 tons.

If a cistern which is 7 feet long 4 feet 6 inches broad and 3 feet deep were proportionally enlarged in all its dimensions, so as to be made to hold 3 times its present content, what would then be its dimensions.

1 : 3 : : 73: x3
calculated with logs as 10.09
1 : 3 : : 4.53: x3
calculated with logs as 6.49
1 : 3 : : 33: x3
calculated with logs as 14.37
Ans. length 10.09, breadth 6.49. depth 14.37.

If a cubic foot of brass were drawn out into wire 1/40 of an inch in diameter, what would be the length of the wire, supposing no loss in the metal.

(1/40)2 × .7854 x = 1728
Walker calculates x (using logs) then converts to miles by dividing by 63660.
Ans. 55 miles 4 f 10.5 yds.

The National Debt amounts to £840,000,000; find the side of a cube, equal in value to that sum, 11 ounces of pure gold being worth £46..14..6.

Walker converts to shillings then writes
934.5 : 16800000000 : : 11 : x
Walker calculates x (using logs) then multiplies by 480, divides by 437.5, then divides by 19640. Finally he takes the cube root to obtain 22.25.
Ans. 22.25 inches.

A gentleman has a bowling green 300 feet in length, and 200 feet in breadth which he wishes to be raised one foot higher by means of the earth to be dug out of a ditch by which he intends to surround it; to what depth must be ditch be dug, if the breadth be everywhere 8 feet.

300 × 200 = content in sq. ft
length breadth
300 × 8 = 2400 × 2 = 4800 content of ditch
200 × 8 = 1600 × 2 = 3200 content of ditch

He then divides by 300 × 200 by 8000 to obtain 7.32
Ans. 7.32 ft depth to which it must be dug.

My coppersmith agreed to make me a flat bottomed kettle, to contain 13.88 imperial gallons, the depth being 12 inches, and the top and bottom diameters in the ratio of 5 to 3; find the diameters.

Walker writes
Let D = x inches
d = 3x/5
Contents of a Frustum = (A + a +√A a) 4/3
13.88 × 277.274 = (x2 + 9 x2/25 + 3 x2/5).7854 × 4

Calculates x as 25 inches

5 : 3 : : 25 : x
calculates x = 15 inches.
Ans. Greater = 25 inches. Less = 15 inches.

Find the depth of a tub in the form of a conic frustum, which contains 25 imperial gallons, and which the bottom and top diameters are 20 inches and 10 inches respectively.

Walker writes
25 × 277.274 = Content in Cubic Inches
which he calculates as 6930
He then computes 102 × .7854 = 78.54
and
202× .7854 = 314.16.
He then takes the square root of the product 78.54 × 314.16 to obtain 157.08
At this point the solution terminates and no answer is given.

Find the depth of a tub in the form of a conic frustum, of which the greater diameter is 60 inches, the diagonal 66 inches, and the length of the staves 30 in.

The solution to this question again terminates and no answer is given.
Walker writes
60 AC = 8.221849
66 AC = 8.180456
He then adds the log of (60 + 66 + 30)/2 =78.
He then adds the log of 78 - 30 =48.
He divides the resulting log by 2 to obtain 9.987820. From this he gets the angle 13°30'23" (log of the cosine)
He multiples by 2 to get 27°0'46"
30 : 27°0'46 : : 66 : Sine
Obtains 87°44
He adds 27°0'46 to this, and subtracts from 180 to obtain 65°15'14"
R : 30 : : Sine 87°44 = 9.999660
30 = 1.477121
29.97 = 1.476781
Terminates here with no answer is given.

In the oblique angled triangle ABC, let the side BC = 532, AB + AC = 637 and the angle BAC = 40°30, to find the others.

Standard calculations using the sine rule.
Ans. AC = 402.1, AB = 234.8, C = 24°16

Given the base 428, the vertical angle 49°16, and the sum of the other two sides 918; find the rest.

Calculations with the sine rule but no answer is given except during the working.

The base of a plane triangle is 384 feet, and the other sides 288 and 192, find the length of the perpendicular upon the base, and the length of the segments of the base made by the line bisecting the vertical angle.

Heron's rule and the sine rule are used. No answer is given except during the working.

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


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