James Walker 1852年的公平簿James Walker's Fair Book of 1852

Andrew Bell在印度发家致富。他于1832年在圣安德鲁斯创办了Madras College，该校基于Madras System，即由学生辅导同学。在Bell博士创办并建立Madras College之前，圣安德鲁斯已有几所学校，最初是为 destined for the Church 的男孩设立的。在Bell博士建立Madras College之前的几百年里，已有一些教育机构同时服务于两性，提供相当基础的教育。Bell博士所做的，是用他的财富将那些现有学校转变为一所非常适合满足当时需求、并按他开创的导生制运行的单一学校。该学院提供了卓越的教育，许多家庭搬到圣安德鲁斯，以便他们的孩子能从这种教育中受益。学校在创立后迅速发展，到1838年学院有800名学生，到1845年约有900名学生，到1860年则远超1000名学生。学校有许多寄宿生，镇上周围有由学校教师管理的寄宿公寓。学校创立时没有校长——各部门的高级教师基本上独立，并从学生那里收取费用。

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在学校档案中，有一本约80页的笔记本，属于1852年在该学院学习的一名学生。书的封面上写着“FAIR BOOK, JAMES WALKER”。不清楚这本书中的材料是在别处完成粗略计算后立即写上去的，还是这本书只是后来某个时候才制作的。第一种可能性似乎最大。

我们已经转录了书中所做的工作，并附上了一些解释性注释。这一卷中的许多兴趣（几乎可以肯定也是它被保存下来的原因）在于伴随数学的艺术作品。其中一些与所描述的问题相关，但最令人印象深刻的是一些军事素描——可能是从克里米亚战争的版画中临摹的，当时这场战争正在进行。

你可以在THIS LINK看到克里米亚素描的单独页面。

这本书的一个印张已与其余部分分离。你可以在THIS LINK看到这一点。

第一页的标题是“Tonnage of ships”。

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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.

虽然没有指定“common rule”，但解答告诉我们，该规则给出的吨位为$x . y^{3/2}/ 94$，其中$x$是龙骨长度（英尺），$y$是最大宽度。计算使用六位对数进行。我们插入了“log”，但原文没有。

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

尽管解答中没有出现文字，但我们看到，由对数表求得的吨位是651.5。

在第二页（第一页的背面）有两个类似的问题。

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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.

第一个按与上述运算相同的方式完成，第二个尚未解出，只填入了数值而未查对数表，例如96 = 1.等。

此处有一处相当大的插页，这些页面装订在一起。其内容另见THIS LINK。

原书在此处可能缺了一些页，但根据内容判断，缺失不会超过一两页。关于船的问题现在变成了关于牛的问题，但没有出现任何标题来表明这一变化。鉴于全书其余部分都有标题，很可能有几页脱落了，其中包含该节的标题。注意5先令10便士的常见记法5 / 10。这并不表示除法！

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 $\large\frac{1}{3}\normalsize$ the value of the four quarters.

[Walker本应写“fore quarter”而不是“four quarter”。]

他使用一条规则：将周长（以英尺为单位）平方，乘以长度（以英尺为单位），再将结果乘以5/21，得到以英石为单位的重量。乘以5/10（5先令10便士）换算为5.833先令，他得到389.0，再加上其三分之一得到518.6先令。然后除以20将其转换为25..18/6。
答：£25..18/6

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

Walker在答案中似乎有些混乱。在计算中，他假设9英尺3 1/2英寸等于9.35英尺（这当然不对），并且5英尺7 1/2英寸等于5.75英尺（这当然也不对）。他按照前一个问题的方法，将9.35乘以9.35，再乘以5.75，再乘以5/21。这得到119.6，然后他加上二十一分之一（我猜应该是十二分之一）。接着他将得到的125.3乘以200/121。

下一页的标题是土地测量。

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

Walker取136的一半，然后使用正弦定理求出A处角度的一半。如同他所有的计算一样，他在计算中使用了$\log \sin A = \log 10^{10} \sin A$。他应该得到42°50，但由于抄写错误得到了45°50。然后他将此加倍，得到正确答案85°40（这证实了上述只是抄写错误，他在草稿中已经算对了）。

接下来是三个类似的问题，Walker以完全相同的方式正确解答了它们。

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

答：36°54'

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

答：51°34'

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

答：86°51'

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

求出$AB + BC = 171.5$和$AB - BC = 28.5$。然后他写出
171.5 : 28.5 : : 45 : tan $x$
并解出（使用六位对数表）得到$x$ = 9°26。

然后他将此与45相加和相减，得到54°26和35°34。
接着他使用正弦定理 sin 9°26 : 171.5 : : sin 45 : x
（使用六位对数表和$\log \sin A = \log 10^{10}\sin A$）得到122.9。
然后他写出
122.9 : 90 : : 71.5 : A
并解出得到35°34。

答：35°34'。

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

使用海伦公式：面积 = $√(s(s - a)(s - b)(s - c))$，其中$a, b, c$为边长，$s = (a + b + c) / 2$。使用六位对数表计算面积得143000平方链。然后除以100,000得到英亩数。再将小数部分乘以4得到路德（4路德等于1英亩）。再将小数部分乘以40得到平方杆（40平方杆等于1路德）。然后将小数部分乘以$30\large\frac{1}{4}\normalsize$得到平方码（$30\large\frac{1}{4}\normalsize$平方码等于1平方杆）。

答：4英亩1路德28平方杆24平方码。

另外两个问题用同样的方法通过海伦公式解决：

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

答：3英亩1路德21平方杆13平方码。

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

答：4英亩1路德1平方杆13平方码。

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

这里Walker给出了一种奇怪的解法。他似乎知道三角形的面积与同底等高的直角三角形的面积相同。然后他用毕达哥拉斯定理求出三角形的第三边，并用海伦公式计算面积。答案是正确的，但这种方法极其冗长且难以求得解答。值得注意的是，有些（但不是全部）解答在题目左侧有一个勾号。这道题有一个勾号。这可能只表示答案正确，也可能意味着Walker的老师认可了这个方法——谁知道呢。

答：6英亩3路德6平方杆21平方码

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

这次Walker使用了显而易见的方法，将面积计算为$\large\frac{1}{2}\normalsize$底乘以高。他转换后得到：
答：4英亩1路德3平方杆11平方码。

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画图有误，他将边$BC$标为823而非$AB$。他计算出三角形的第三个角为78°58。这是一个算术错误，应为77°58。他使用正弦定理计算边$AC$，得到506.5。然后他将506.5乘以416.5（或许是错误地计算了1/2乘以823？）同样出于某种原因，他选择用长乘法而非对数来计算。他似乎将此作为答案，因此忘记乘以sin 64°52。

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

解法是对三角形ABD和BCD两次应用海伦公式。在相加之前，每个三角形的面积都转换为英亩、路德、平方杆、平方码。当然，Walker很少费心写sq yds，而是写yds。这偶尔会导致他犯错。

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

以与前一题完全相同的方式完成，涉及两个三角形ABC和ACD。

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

取长度乘以宽度，并转换为英亩、路德、平方杆、平方码。

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计算965乘以22，然后换算为英亩。尽管以英亩计算似乎更合理，但他先换算为英亩、路德、杆、码，再通过比例求和得出£2..4/6，这个结果接近但不正确。

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.

计算1260+984的一半乘以567，然后换算为英亩、路德、杆、平方码。

答案：6英亩1路德17杆22平方码。（他没有将22.99平方码四舍五入为23平方码）

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.

方法同前一题。
答案：4英亩0路德23杆1平方码。

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

Walker首先使用公式$\tan(B - C)/2 = (b - c)/(b + c) \tan(180 - A)/2$解三角形ABD。他先计算$AB + AD$和$AB - AD$，然后将$\tan((180 - A)/2$乘以$(AB - AD)/( AB + AD)$。求出反正切。将此结果与$(180 - A)/2$相加和相减，得到$ABD$的另外两个角。然后使用正弦定理计算$BD$。

一旦求出$BD$为971.6，他便对两个三角形$ABD$和$BCD$使用海伦公式，将面积分别换算为英亩、路德、杆、码并相加。

答：4英亩1路德18杆7码。

一块四边形的田地，其对角线为1000和850链，交角为65°25'，面积是多少？

sin 65°25' × 500 × 850 = 面积。以平方链计算，然后换算为：

答：3英亩3路德18杆10码。

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

方法同前一题。
答：3英亩0路德31杆17码。

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计算$ADC$ = 69°38，然后用正弦定理求出$DC$ = 845.4。接着他用面积 = $1/2 bc \sin A$求出$ACD$的面积。同样地，他计算三角形$ABC$的面积。两者都换算为英亩、路德、杆、码，然后相加。

答：3英亩3路德39杆25码。

假设测量员从一个角看不到田地的所有其他角，而是从A处的一个高地取观测值，且其各角如下：BAC = 105°；CAD = 59°30；DAE = 129°；EAB = 66°30'；且直线AB = 480，AC = 550，AD = 665，AE = 730。它的面积是多少？

Walker依次用面积 = $1/2 bc \sin A$求出$A$处四个三角形各自的面积。他换算为英亩、路德、杆、码，然后相加。

答案：6英亩1路德0平方杆23平方码。

接下来是一些涉及田地的题目，这些田地按给定的测量数据（所有线段的长度）绘制。面积通过将各部分的面积相加来计算。对于已知三边的三角形，使用海伦公式。这些题目包含多达18条测量线段。在这些相当冗长的求和之后，我们回到较简单的例子。

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

使用海伦公式计算。
答：3英亩3路德28杆29平方码。

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

使用海伦公式计算。
答：2英亩1路德2杆16平方码。

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

两个三角形均使用海伦公式求解，他转换为英亩、路德、杆、码，然后相加。
答：4英亩1路德32杆。

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

垂线是从另外两个顶点到对角线。两个三角形的面积由$\large\frac{1}{2}\normalsize$底乘高计算。
答：5英亩3路德14杆。

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

Walker将面积转换为平方链，然后用长除法除以500（不使用对数）。
答：337 1/2链。

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

将8杆乘以625换算为5000平方链，然后除以21得到$238\large\frac{2}{21}\normalsize$链。

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

将3路德27平方杆换算为平方链，得到91875。用长除法除以690，得到商133余105。最后约简105/690 = 21/138 = 7/46。
答案：133又7/46链。

如果要从三角形ABC中平行于AC截去2英亩20平方杆，而该三角形面积为4英亩，AB边长为1125链，那么分割线应从AB上的哪一点开始？

通过相减，他算出剩下的三角形面积为1英亩3路德30杆。他把它换算成187500平方链。现在他利用三角形边长的平方之比等于面积之比这一事实，算出边长为770.23链。他使用六位对数，得到的答案精确到小数点后2位。

下一道题看起来很有意思，但只给出了演算过程，题目本身缺失。

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分别用500、450、150、100除以20、18、15、12。然后他将所得结果25、25、10、8又1/3相加，得到68又1/3。接着他计算

68又1/3 : 25 : : 244.3.30 : A

先把$68\large\frac{1}{3}\normalsize : 25$乘以3得到205 : 75，再除以5得到41 : 15。他把244 ac 3 r 30 p换算成39190杆，并用长乘法和长除法求出其15/41。再换算回英亩得到

答。$A$ = 89 ac 2 r 17 p 33/41。对$C$和$D$的类似计算分别给出35 ac 3 r 15 p 5/41和28 ac 3 r 19 p 11/41。

接下来有一节标题为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

答：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

答：9。

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

Walker将16乘以4，所得结果再乘以2，然后除以3得到$42\large\frac{2}{3}\normalsize$。
答：$42\large\frac{2}{3}\normalsize$。

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利用面积等于$\large\frac{2}{3}\normalsize$底乘以高这一事实。通过长乘法和除法计算得出面积为1英亩。

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（使用六位对数）计算出高度为100 sin 55° = 81-91。然后面积是底乘以高的2/3，不用对数计算得到27303.33（实际上应为27305.06，误差来源于将高度取到两位小数）。转换为英亩得到

答：1路德3平方杆20平方码。

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.

他写下（$10 + 6^{2}$ ÷ $16) 4 \times \large\frac{2}{3}\normalsize$ = 答。他用长除法计算得出
答 = $32\large\frac{2}{3}\normalsize$。

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.

未给出解法。

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.

平均纵坐标的计算方法是：将11个纵坐标相加得到1062，然后除以11得到96.54。接着用六位对数表将其乘以315。

答案：30410。（没有换算成英亩。）

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

画出一个椭圆。计算通过210 : 150 : : √(168 x 42) : $x$得出：

答案：60。（注意平方根是用长除法手工计算的，没有使用对数。）

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

画出一个椭圆。计算通过 180 : 60 : : √(144 x 36) : $x$ 给出：

答：24。注意平方根是用长除法手算的，没有使用对数，也没有用显然的 12 × 6。

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

画出一个椭圆。计算通过
50 : 70 : : √(252 - 202 ) : $x$ 给出
50 : 70 : : 15 : x。注意 Walker 认识到 √225 = 15。这给出 $x$ = 21。然后他计算 35 ± 21 = 14 或 56。

答：14 或 56。

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

30 : 90 : : √(152 - 122) : x，所以 x = 27。但 45 ± 27 = 72 或 18。

答：72 或 18。

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

画一个椭圆。√(14 x 56) : 20 : : 70 : $x$
答：50。

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

横轴 = 144 + 36 = 180。
$144 \times 36 : 24^{2}: : 180^{2}: x^{2}$。通过六位对数计算得到：
答：60。

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

Walker计算$22.5 + √(22.5^{2} - 18^{2}) = 36$。他不用对数，而是用长乘法和平方根。然后他计算
$18^{2} : 45 12 : : 36 : x$
并得到：

答：60。

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

未给出解法。

Walker现在计算椭圆的周长和面积。对于周长，他使用approximate表达式$\pi/2 \times √(a^{2} + b^{2})$，其中$a, b$是长轴和短轴的长度。这个近似值对于圆是精确的，但对于高偏心率的椭圆则很差。周长的真实值取决于一个椭圆积分，这在1850年代的学校里肯定没有！
面积（精确地）是$\pi/4 \times a b$

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

Walker计算$(140^{2} + 120^{2})/2 = 17000$。然后他（用长除法）取平方根得到130.38。接着他（用长乘法）乘以3.1416得到409.6。
答：409.6。

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

方法同前一例。
答：614.1。
实际答案应为614.4，Walker的错误在于他将38250的平方根算作195.5，过早结束了计算，因为正确答案应为195.576。

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

方法同前两题。得：
答：1332.855216。
这当然比人们所能期望的位数多得多。正确答案是1332.865。Walker只把平方根计算到两位小数，错误即由此产生。

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

方法同前三题。得：
答：2308.4。
正确答案是2308.7。错误同样来自把平方根计算到一位小数后就停止，而没有意识到下一位数字会是9。

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

计算480 x 600 x .7854 = 226195.2（注意π/4 = .7854）。然后换算为英亩。
答：2英亩1路德1平方杆27平方码。

求轴长为210和180的椭圆的面积。

计算210 x 180 x .7854 = 296881.2（注意π/4 = .7854）。
答：296881.2

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

方法同前一题。
答：13194.7。

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

方法同前两题。
答：1英亩1路德17平方杆4平方码。

与横轴15共轭的纵坐标是多少，较小截距为5，其纵坐标为6。

计算15 + 5 = 20未写出。
√(20 x 5) : 6 : : 15 : $x$
10 : 6 : : 15 : $x$
2 : 6 : : 3 : $x$
1 : 3 : : 3 : $x$
答：9。

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

计算609 + 116 = 725未写出。
√(725 x 116) : 280 : : 609 : $x$
用长乘法与开平方得290 : 280 : : 609 : $x$
然后长除法计算(280 x 609)/290（甚至没有约去0）。
答：588。

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

计算$√(60^{2}+ 80^{2)}+ 80 = 180$。
$60^{2} : 25 x 160 : : 180 : x$
答：200。

结束

下一节标题为：立体求积。

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

画出抛物面体。计算$35^{2} \times 20 \times .3927$用长除法进行。（注意$.3927 = \large\frac{1}{8}\normalsize \pi$）
答：9621.150。

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

画出抛物面体。方法同前一题。
答：26400.43。

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

画出抛物面体。方法同前一题。
答：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.

计算$(40^{2}+ 32^{2)12}x .3927 = 12365.337$。所有计算均用长除法完成。
答：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.

同前一题。$(20^{2} + 16^{2}) 6 \times .3927 =$
但计算未完成，留作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.

桶已画出，但没有给出计算或解答。

一个长为30、最大直径为12的抛物面纺锤体的体积是多少。

计算$12^{2} \times 30 \times .7854 \times \large\frac{8}{15}\normalsize$已进行。
答：1809.559。

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

同前一题：计算$8^{2} \times 20 \times .7854 \times \large\frac{8}{15}\normalsize$已进行。
答：563.165。

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

同前一题：计算$20^{2} \times 50 \times .7854 \times \large\frac{8}{15}\normalsize$已进行。
答：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.

所进行的计算为
$(20^{2} \times 8 + 15^{2} \times 3 + 20 \times 15 \times 4) 25 = 126875$
然后 126875 × .05236 = 6643.175（注：.05236 = π/60）
答：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.

计算过程为
$(31^{2} \times 8 + 24^{2} \times 3 + 31 \times 24 \times 4) 46 = 570032$
然后 $570032 \times .05236 = 29846.875 (note .05236 = \large\frac{1}{60}\normalsize \pi)$
答：126.99 英制加仑。

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.

所进行的计算是
$(35.5^{2}\times 8 + 32^{2} \times 3 + 35.5 \times 32 \times 4)38 = 543020$
然后$543020 \times .05236 = 28432.52720$（注意$.05236 = \large\frac{1}{60}\normalsize \pi$）
答：107.64英制加仑。

Walker随后计算了各种正多面体的体积和表面积。他大概有表，可以让他从单位边长的立体的测量值按比例放大。

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

Walker 的第一次尝试是把 log 4 加到 √3 = 1.732050 上。他似乎意识到了自己的错误，因为他随后把 1.732050 乘以 4，得到 6.928200。
答：6.928。

为了计算体积，Walker 计算了
$2^{3} \times 0.11785113$（注 0.11785113 = 1/(3√8)）
答：.9428。

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

Walker 计算了 $6^{2} \times 6 = 216 and 6^{3} \times 1 = 216$。
答：216 和 216。

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

Walker计算$8^{2} \times 3.4641016 = 221.7025024$
答：221.702。
他没有计算体积。

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

Walker 计算 $12^{2} \times 20.6457788 = 2972.9921472$ 和 $12^{3} \times 7.66311896 = 13241.86956288$。
答：2972.992 和 13241.869。

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

Walker 计算 $20^{2} \times 8.660254 = 3461.01600$ 和 $20^{3} \times 2.18169499 = 17453.55992000$。
答：3461.016 和 17453.5。

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

Walker 计算 $30^{2} \times 8.660254 = 7794.228600$ 和 $30^{3} \times 2.18169499 = 58905.76473000$。
答：7794.228 和 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 用英尺和英寸计算包裹下部和上部的体积。然而这似乎不正确。3 英尺 6 英寸乘以 4 英尺 3 英寸是 14.875 平方英尺，即 14 平方英尺 126 平方英寸。Walker 将其写作 14..10..6，未给出单位。现在如果我们乘以 1 英尺 9 英寸，得到 26.03125 立方英尺，即 26 立方英尺 54 立方英寸。Walker 将其写作 26..0..6..6，未给出单位。在这个奇怪的计算之后，他转换为立方英寸，得到了错误的答案（尽管相差不太远）。他除以 5，然后计算比例和，因为 1728 立方英寸价值 30 便士。

答：$24 / 1\large\frac{3}{4}\normalsize$。（正确答案是 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 计算 $24 \times 24 \times 14 = 8064$ 和 $24 \times 24 \times 24 = 13824$。他相减得到 5760。（当然他本应计算 $24 \times 24 \times 10$，那要容易得多！）他除以 1728 转换为立方英尺。
答：$3\large\frac{1}{3}\normalsize$ 立方英尺。

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用六位对数表进行简单的乘法运算。他得到900立方英尺。

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计算平均长度、宽度和深度，分别为13..0、5..6和4..6。然后他在乘法中犯了与之前相同的奇怪错误。例如，将13.0乘以5..6应得到71.5平方英尺或71平方英尺72平方英寸。Walker将13.0乘以5得到65.0，然后将13.0乘以0.6得到6..6，相加得到71..6。因此他的错误在于认为1/2平方英尺是6平方英寸。他得到

答案：321..9立方英尺。（答案应为321.75立方英尺或321..1298立方英尺。）

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.

使用Heron公式计算两个三角形的面积，求出平均深度，从而得到体积。
答案：£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写道

这里$A$ = 75 + 924 = 999；4$B$ = 4 (210 + 712) = 3688；以及2$C$ = 2 × 379 = 758。
因此$(A + 4B + 2C) \times D/3$= (999 + 3688 + 758) × 3 ÷ 27。

他计算出这个值为605，并给出
答：605立方英尺。（应该是立方码！）

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.

插图：d36.JPG ↗
点击查看大图

提到⟦E数字⟧图56，必定出自一本包含此图、且很可能还包含其他问题的书。图中画出了图形。没有给出解答。

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.

计算平均长度、宽度和深度。然后取乘积。Walker用他惯常的（不正确的？）方法以英尺和英寸计算乘积。
答：755英尺7英寸6分。

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使用六位对数计算以立方英寸为单位的体积，然后从答案中减去3.346（即除以2218.192）得到：
答：4.154蒲式耳。

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.

使用六位对数的标准计算。将以立方英寸为单位的体积除以277.274得到43.62加仑，除以2218.192得到5.453蒲式耳。注意每次计算都从头计算体积。

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

与之前相同的方法。
答：4.658蒲式耳。

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.

用六位对数进行标准计算。将体积（立方英寸）除以277.274得到11.36加仑，除以26.76得到117.7磅。注意每次计算体积都是从零开始算的。

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.

计算时假定题目给出的三角形底边为50英寸。想必这是有意为之，而题目中有印刷错误。注意每次计算体积都是从零开始算的。

Find the area of a trapezium, of which the diagonal is 78 in and the perpendiculars falling down from the opposite angles 23 and $15\large\frac{1}{2}\normalsize$ inches, or its content at 1 inch deep, in imperial gallons and lbs tallow.

注意每次计算体积都是从零开始算的。
答案：5.415加仑和47.81磅。

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.

注意每次计算体积都是从零开始算的。
答案：38.08加仑和4.760蒲式耳。

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

大桶是装啤酒或葡萄酒的桶。Walker用六位对数计算60 × 60，然后除以353.036。而353.036等于277.274乘以4/π，其中277.274是将立方英寸换算为加仑时必须使用的除数。
答案：10.19加仑。

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

与上一题一样，Walker 将直径平方并除以 353.036。全部用六位对数完成。

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.

与上一题类似，Walker 取直径的乘积并除以 353.036。全部用六位对数完成。

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.

使用六位对数，Walker 将边长平方并乘以 3.6339124。他将此计算进行两次，第一次除以 277.274 换算为加仑，第二次除以 2218.192 换算为蒲式耳。
答：83.83 加仑和 10.48 蒲式耳。

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.

使用六位对数，Walker 将边长平方并乘以 4.8284271。他将此计算进行两次，第一次除以 277.274 换算为加仑，第二次除以 2218.192 换算为蒲式耳。
答：174.1 加仑和 21.76 蒲式耳。

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.

英制加仑不是面积上的度量！Walker 计算 $A = 82.4 + 82.4, B = 96.5 + 119.2 + 122.4 + 122.6 + 118.9 + 96.3, C = 112 + 121.3 + 123 + 121.2 + 112.$ 然后他计算 $A + 4B + 2C = 4047.4$。他还计算 $D = 15/3 = 5$。接着他计算 96.5 × 2 = 193，再乘以 2 得到 386，除以 3 得到 128.6，将其加到 4047.4 得到 4176。他将其乘以 5 得到 20880.00，最后除以 353.036 换算为加仑。
答：59 加仑。

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

使用六位对数，他将 30 立方并除以 277.274 得到 97.4 加仑，除以 2218.192 得到 12.17 蒲式耳。注意每次计算都从头计算体积。

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.

使用六位对数。方法与前一题相同，每次计算都从头计算体积。

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.

标准方法。
答案：357.1加仑。

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.

使用六位对数，他计算48 × 48 × 60并除以30.609得到磅数。他再次计算48 × 48 × 60并除以353.03604得到391.6加仑。

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.

使用六位对数，他三次计算以立方英寸为单位的体积。第一次他除以2218.192得到蒲式耳，第二次他除以26.76得到热皂的磅数，第三次他除以40.3得到生淀粉的磅数。
答案：16.22蒲式耳，1345磅，593.3磅。

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.

使用六位对数，他两次计算40 × 40 × 32。第一次他除以106.773得到479.7加仑。第二次他除以853.782得到59.96蒲式耳。

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.

使用六位对数，他计算154 × 54.96 × 50并除以353.035得到1198加仑。

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.

他用六位对数计算 48 × 48 × 60，除以 3，再除以 353.035，得到 130.2 加仑。

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.

他用六位对数计算 60 × 60 × 90，除以 3，再除以 1289.288，得到 83.76 蒲式耳。

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

他用六位对数两次计算 90 × 90 × 90。第一次他除以 529.544，得到 1376 加仑。第二次他除以 4236.434，得到 172 蒲式耳。

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.

用六位对数进行标准计算，得到：
答：273.3 加仑。

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.

用六位对数进行标准计算，得到：
答：160.5 加仑。

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.

用六位对数进行标准计算，得到：
答：197.3 加仑。

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.

用六位对数进行标准计算，得到：
答：168.1加仑。

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.

用六位对数进行标准计算，得到：
答：132.6加仑。

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.

用六位对数进行标准计算，得到：
答：1856英制加仑。

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.

用六位对数进行标准计算，得到：
答：55.1英制加仑。

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.

用六位对数进行标准计算，得到：
答：221.2英制加仑。

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.

用六位对数进行标准计算，得到：
答：2939英制加仑。

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.

用六位对数进行标准计算，得到：
答：82.59 英制加仑。

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.

用六位对数进行标准计算，得到：
答：190.4 英制加仑。

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 $7\large\frac{1}{2}\normalsize$ in 65.38 in, and the depth of the pan 15 in, in imp gall.

Walker 把 45 的对数抄错了。他写成了 1.693213，而不是 1.653213。

他计算 45 × 45 × 3 = 6002（由于对数抄错，这当然是错的）。他把 15 平方后加到上面，乘以 15，再乘以 .5236，得到以立方英寸为单位的体积。最后他除以 277.274，得到答案 176.4 英制加仑。[如果他没有抄错，他会得到 178.45 英制加仑。]

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 用六位对数计算：

64 × 36 = 2304 和 48 × 27 = 1290（这应该是 1296——可能是从表中抄写时的错误）。然后他把 2304 乘以 1290，再取平方根，得到 1728。这个答案是正确的，因为他在计算中使用了 48 × 27 的对数，而那是正确的。他计算 2304 + 1290 + 1728，乘以 48，再除以 3。最后他除以 277.274，得到答案 307.1 英制加仑。[如果他没有抄错，他会得到 307.45 英制加仑。] 他没有计算以英制蒲式耳为单位的体积。

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.

计算如前一个问题。他仍未以英制蒲式耳计算体积。
答：1050英制加仑。

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.

计算如前两个问题。他仍未以英制蒲式耳计算体积。
答：1778英制加仑。

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 用六位对数计算：

(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
然后相加
43.91 + 47.9 + 36.45 + 40.1 = 168.36
168.36 + 10 = 178.36（这里的10是滴漏）
答：178.36加仑。

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 用六位对数计算：

(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
然后相加
352.2 + 261.5 + 261.5 + 261.5 = 1110
1110 + 38 = 1148（这里的38是皇冠）
答：1148加仑。

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

与前两题类似的计算。

从这一点起，Walker绘制了精细的草图。

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.

使用六位对数表进行计算。

24 × 24 + 31 × 31 × 2 = 2498。
(2498 × 32.5)/1059.108 = 76.66
答案：76.66加仑

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.

使用六位对数表进行计算。

26.8 × 26.8 + 31.7 × 31.7 × 2 = 2727.4（2727.2之误植，对数正确）
(2727.4 × 32.7)/1059.108 = 84.21
答：84.21加仑

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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.

使用六位对数表进行计算。

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
答案：88.39加仑

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.

使用六位对数表进行计算。

24 × 24 + 31 × 31 × 2 = 2498
(2498 × 46)/1059.108 = 108.5
答：108.5 加仑

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.

使用六位对数表进行计算。

23 × 23 + 29 × 29 × 2 = 2211
(2211 × 36)/1059.108 = 75.15
答：75.15 加仑

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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.

使用六位对数表进行计算。

31 × 31 + 23 × 23 + 29.2 × 29.2 ×4 = 4900
(4900 × 40)/2118.217 = 92.53
答：92.53 加仑

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.

用六位对数进行计算。
Walker 将 23 除以 31，得到 .7419。
然后他计算 8 × .613（他标记为 Table area），得到 4.904。
他在 4.904 上加 23，得到 27.904（他标记为Mean Diam.）
(27.904 × 27.904 × 50)/353.036 = 110.2
答：110.2 加仑

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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 $30\large\frac{1}{2}\normalsize$ inches.

用六位对数进行计算。
Walker 将 27 除以 31，得到 .879。[注：他查 log 31 时出错，写成 1.491212，而不是 1.491362]
然后他计算 4 × .682，得到 2.728。
他在 2.728 上加上 27，得到 29.728
(29.728 x 29.728 x 30.5)/353.036 = 76.35
答：76.35 加仑

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.

用六位对数进行计算。
Walker 将 23 除以 29，得到 .7931。
然后他计算 6 × .691，得到 4.146。
他在 4.146 上加上 23，得到 27.146
(27.146 × 27.146 x 27)/353.036 = 56.35
答：56.35 加仑

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.

用六位对数进行计算。
Walker将22.3除以28.8，得到.7743。
然后他计算6.5 × .677，得到4.4。
他将4.4加上22.3，得到26.7
(26.7 x 26.7 × 47)/353.036 = 94.91
答：94.91加仑

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写下
21 = .8328
18 = .3059
相加得到1.1387，他用对数乘以30，得到34.16。
答：34.16加仑。

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写下
32 = 1.9337
25.3 = 0.6044
相加得到2.5381，他用对数乘以47，得到119.2。
答：119.2加仑。

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.

用六位对数进行计算。
Walker将19除以22，得到.8636。
然后他计算3 × .606，得到1.818。
他将1.818加上19，得到20.818（他标记为平均直径）
(20.818 x 20.818 x 24)/353.036 = 29.46。

然后他计算5 × 12 = 60，从中减去11（他标记为22的1/2 = 桶孔直径）。他将得到的49乘以29.46，并将结果除以88（他标记为桶孔直径22 × 4 = 88），得到16.4。
答：16.4加仑。

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.

用六位对数进行计算。
Walker将23除以29，得到.7931。
然后他计算6 × .610（他写下.79 = .610），得到3.66。
他将3.66加上23，得到26.66。
(26.66 × 26.66 × 27)/353.036 = 54.35。

然后他计算 10 × 5 = 50，从中减去 14.5（他将其标记为 29 的 1/2）。他将所得的 35.5 乘以 54.35，并将结果除以 116（他将其标记为 29 × 4），得到 16.63。
答：16.63 加仑。

一个直立大桶的缺量是多少？其长度为 29.1 英寸，桶腹直径为 28.5 英寸，桶端直径为 24 英寸，液体深度为 18 英寸。

使用六位对数计算。
24 × 24 + 28.5 × 28.5 × 2 = 2200.5。
(2200.5 × 29.1)/1059.108 = 60.46

然后他计算 18 × 11 = 198.0，从中减去 14.55（即 1/2 × 29.1）。他将所得的 183.45 乘以 60.46，并将结果除以 291，得到 38.24。[注：60.46 来自 1.781474 的反对数，但当他下一次计算再次使用它时，他显然在表中查找它并弄错了，写成了 1.782902。]
答：38.24 加仑

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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.

使用六位对数计算。
29.8 × 29.8 + 33.8 × 33.8 × 2 = 3172。
(3172 × 39.5)/1059.108 = 118.3

然后他计算 30 × 11 = 330，从中减去 19.75（即 1/2 × 39.5）。他将所得的 310.45 乘以 118.3，并将结果除以 395，得到 92.97。
答：92.97 加仑。

现在有一系列涉及火炮的问题

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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.

用六位对数进行计算。
Walker 计算 √(8/90) × 1600 = 477。
[这里 8 = 2 × 4，即火药的重量（磅）]
或
√90 : √8 : : 1600 :$x$
9.5 : 2.9 : : 1600
1.9 : 2.9 : : 320
1.9 $x$ = 928
$x$ = 477
显然他没有进行最后这一步计算，否则他所做的近似会导致相当大的误差。他会得到$x$ = 488。
答：每秒 477 英尺。

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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.

用六位对数表进行的计算。
Walker计算√(4/48) × 1600 = 462。
答案：每秒462英尺。

A $5\large\frac{1}{2}\normalsize$ inch shell, weighing 16 lbs, is fired with 1 lbs of powder, with what velocity is it discharged。

用六位对数表进行的计算。
Walker计算√(2/16) × 1600 = 565.6。
答案：每秒565.6英尺。

A $4\large\frac{2}{3}\normalsize$ inch shell, which weighs 8 lbs, is fired with 1/2 lbs of powder, with what velocity is it discharged.

用六位对数表进行的计算。
Walker计算√(1/8) × 1600 = 565.6。
答案：每秒565.6英尺。

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

用六位对数进行计算。
Walker计算√2.8 × 175.5 = 293.6。
答：293.6英尺。

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

用六位对数进行计算。
Walker计算√4.04 × 175.5 = 357.2。
答：357.2英尺。

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

用六位对数进行计算。
Walker计算√6.75 × 175.5 = 455.9。
答：455.9英尺。

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

用六位对数表进行的计算。
Walker计算√12.8 × 147.3 = 527。
答：527英尺。

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

用六位对数进行计算。
Walker 计算 √7.9 × 147.3 = 414。
答：414 英尺。

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

用六位对数表进行的计算。
Walker计算(2000 × 2000)/64 = 62500
答：62500英尺。

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

用六位对数表进行的计算。
Walker计算(1600 × 1600)/64 = 40000
答：40000英尺。

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

用六位对数表进行的计算。
Walker计算(294 × 294)/64 = 1350
答：1350英尺。

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.

用六位对数进行计算。
Walker 计算 √6.75 × 175.5 = 455.9。
(455.9 × 455.9)/ 64 = 3248。
2000/455.9 = 4.386 = 33°30。
然后他写下 33°30 = 3.1031，将其乘以 3248，得到 10086。
然后他写下 32°45 = 3.2968，将其乘以 3246，得到 10708。
(10708 - 10086)/2 = 311。
10086 + 311 = 10397
答：仰角 33°30，10397 英尺（他似乎写成了 10.397 英尺）

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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.

用六位对数表进行的计算。
Walker计算√12.8 × 147.3 = 527。
(527 × 527)/ 64 = 4340。
2000/527 = 3.795 = 34°49。
然后他写下34°49 = 2.7631，将其乘以4340得到11992。
答：11992英尺和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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用六位对数表进行计算。
Walker计算√9.84 × 147.3 = 462。
(462 × 462)/ 64 = 3336。
1700/462 = 3.679 = 35°10。
然后他写下35°10 = 2.6749，再乘以3336得到892.3。
答：892英尺和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写道
[对数] sin 90 = 10.000000（他在这里使用$\log 10^{10} \sin$）
[对数] 2920 = 3.465383
相加得到13.465383，从中减去对数3500 = 3.544068，得到9.921313。然后他写下9.921313 = 56°32'（同样对数sin 56°32 = 9.921313），将其除以2得到28°16'。
答：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写道
Log 485 = 2.685742
485 = 2.685742
485 = 2.685742
7.663740
5120000= 6.709270
9 lbs = 0.954470
Ans. 9 lbs.

这一定只是简单的抄写错误。他本应计算
(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.

用六位对数表进行计算。
(500 × 500 × 90)/5120000 = 4.394。
答：4磅6.3盎司。

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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使用六位对数计算

4 : 5 : : 2000 : $x$。
答：2500英尺。

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

Walker写道
$R$ : tan 45° : : 4000 : $x$
将tan 45°和sin 90°的对数写作10，即$10^{10}\tan 45°$和$10^{10}\sin 90°$的对数，他计算出
($4000 \times \tan 45°)/\sin 90°$，取平方根，然后除以4得到15.80。
答：15.80秒。

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

Walker写道
$R$ : tan 40° : : 4000 : $x$
将tan 40°的对数写作9.923813，sin 90°的对数写作10，即$10^{10}\tan 45°$和$10^{10}\sin 90°$的对数，他计算出
($3000 \times \tan 40°)/\sin 90°$，取平方根，然后除以4得到12.54。
答：12.54秒。

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

Walker写道
$R$ : tan 35° : : 1000 : $x$
将tan 35°的对数写作9.845227（他误写为9.849227，但这仅是抄写错误，因为加上3得到12.845227），并将sin 90°的对数写作10，即$10^{10}\tan 35°$和$10^{10}\sin 90°$的对数，他计算出
(1000 × tan 35°)/sin 90° = 700。
答：700英尺。

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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将32°12加倍，将3250减半，写作
sin 64°25 : rad : : 1625 : x
如前所述，他在计算中使用六位对数表，其中log sin作为$10^{10}\sin$的对数等。
($1625 \times \sin 90°)/\sin 64°25$= 1802英尺。)
1802/64 = 340速度
(340 × 340 × 196)/5120000 = 4.403（这不太准确，因为Walker将340的对数写作2.530418，而它应为2.531479。
答：4磅6盎司装药。

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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写道

$4^{3} : 6.41^{3} : : 9 : x$

然后，使用六位对数，计算出$x$为
答：37.03磅。

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

Walker写道

$4^{3} : 7.018^{3} : : 9 : x$

然后，使用六位对数，计算出$x$为
答案：48.6磅。

Find the diameter of an 18 lb iron ball.

利用前一题的结果，$d^{3}$应为18 × $(4^{3}/9)$。现在$4^{3}/9 = 7.1111111...$，有趣的是，Walker将其写作$7.111\large\frac{1}{9}\normalsize$。使用六位对数，他将$7.111\large\frac{1}{9}\normalsize$乘以18并取立方根，得到
答案：5.039英寸。

Find the diameter of an 36 lb iron ball.

如前一题一样，Walker计算36 × 7.111 1/9并取立方根，得到
答案：6.344英寸。

Find the diameter of an 42 lb iron ball.

如前一题一样，Walker计算42 × 7.111 1/9并取立方根，得到
答案：6.684英寸。[注意他写的是$7.119\large\frac{1}{2}\normalsize$而不是$7.111\large\frac{1}{9}\normalsize$，但既然他有正确的对数，这一定是抄写错误。]

这张图片展示了塞尔维亚将军奥马尔帕夏进入叶夫帕托里亚镇，他在1855年2月为土耳其人从俄罗斯人手中夺取了该镇。

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

Walker写道

$(11.1^{3} - 8^{3}) x 9 \div 64$ = Ans

然后他计算了(3 log 11.1 - 3 log 8 + log 9)，这显然是错误的。他将得到的对数答案乘以2，写下64的对数，最终陷入混乱，留下答案空白。

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

Walker写道

$(9.8^{3} - 7^{3}) \times 9 \div 64$ = Ans

然后他计算了(3 log 9.8 - 3 log 7 + log 9)，这显然是错误的。他写下64的对数，最终陷入混乱，留下答案空白。

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

Walker使用对数计算

$8^{3}/57.3 = 8.93$磅。
答案 8.93磅。

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

Walker使用对数计算

$9^{3}/57.3 = 12.72$磅。
答：12.72磅。

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

Walker用对数计算，
6 × 57.3，然后取立方根。
答：7.005磅。[7.005英寸的印刷错误。]

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

同前一题。
答：8.019英寸。

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

同前几题。
答：9.064英寸。

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

Walker利用对数计算
20 × 14 × 12 ÷ 30 = 112。
答：112磅。

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

Walker利用对数计算
18 × 18 × 18 ÷ 30 = 194.4。
答：194.4磅。

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用对数计算
40 × 41 × 42 ÷ 6 = 11480。
答：11480个球。

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

同前一题。
答：1540个球。

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

同前一题。
答：680个球。

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

Walker用对数计算
20 × 21 × 41 ÷ 6 = 2870。
答：2870个球。

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

同前一题。
答：5520个球。

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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用对数计算
30 × 31 × 121 ÷ 6 = 18755。
答：18755个球。

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使用对数计算
50 × 51 × 52 ÷ 6 = 23100。
然后
25 × 26 × 24 ÷ 6 = 2600.0。
23100-2600 = 19500
答案：19500个球。

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使用对数计算
30 × 31 × 61 ÷ 6 = 9455。
然后
9 × 10 × 19 ÷ 6 = 285。

答案：9170个球。

这就结束了炮术的探讨！

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

有趣的是，Walker现在对问题写得稍微多了一些。他写道
设$x$ = 斜边
1000-x = 另一边
$x^{2} = 300^{2} + (1000 - x)^{2}$
$x^{2} = 9000 + 10000000 - 2000 x + x^{2}$
$2000 x = 1090000$
$2 x = 1040$
$x = 545$ 斜边
$1000 - x = 455$ 另一边。

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.

设$x$ = 垂线
$(80 \times /2)4 = (533\large\frac{1}{3}\normalsize )9$
$160 x = 4800$
$x = 30$
$x^{2} = 30^{2} + 40^{2}$
$x^{2} = 900 + 1600$
$x^{2} = 2500$
$x = 50$
$2x = 100$
答：100英尺。

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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.

他先把2路德16杆换算成96杆，然后计算625 × 96 = 60000，得到以平方链为单位的面积。他接着写道

设$x$ = 一条直角边
则$√(500^{2} - x^{2})$ = 另一条直角边
$x/2√(500^{2} - x^{2}) = 60000$

然后他将方程平方，得到一个关于$x^{2}$的二次方程，并通过配方法求解。由此得到$x^{2}$ = 160000或90000，所以$x$ = 400或300。
答：400和300链。

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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用六位对数计算18 × 12 = 216。他接着写道
216 : 54 : : 18 : $x$
解此式得到$x$ = 9。
他又解2 : 1 : : 24 : $x$，得到$x$ = 12。
然后他用毕达哥拉斯计算斜边$√(12^{2} + 9^{2}) = 15$。
答：底 = 9英尺，垂直边 = 12英尺，斜边 = 15英尺。

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.

他用Heron公式计算面积为84。然后他写出
√84 : √24 : : 13 : $x$
计算得$x$ = 7。
√84 : √24 : : 14 : $x$
计算得$x$ = 7.48。
√84 : √24 : : 15 : $x$
计算得$x$ = 8。
答：7、7.48和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 : $x$
得到$x = 149\large\frac{7}{13}\normalsize$。
(72 + 58) : 270 : : 58 : $x$
得到$x = 120\large\frac{6}{13}\normalsize$。
他现在得到三条边为$130, 149\large\frac{7}{13}\normalsize$和$120\large\frac{6}{13}\normalsize$。用Heron公式，他得到7496。
答：面积7496。

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使用六位对数计算

30.66 2/3 × 16.5 × 8 ÷ 112。
然后他乘以41.5并除以20，得到74.98（英镑），接着他计算出这是
答：£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/$9\large\frac{1}{2}\normalsize$ 转换为4581.5（便士）。
然后他使用六位对数计算
4581.5 × 2 × 9 ÷ 42
他将答案除以.7854（即π/4）并取平方根，得到50。
答：50英尺。

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

我们已更正了问题。Walker实际上漏掉了第三条边的长度！
首先他用Heron公式求得面积为4036。然后
4036 ÷ 70 = 57.66 垂线。
然后他将8200除以57.66，得到142.2
答：直径 = 142.2

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

Walker计算43560 ÷ .7854（即π/4）。然后他取平方根，乘以4，得到942。
答：942英尺。

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.

sin 60° : sin 90° : : 12 : $x$
使用（如通常的$\log \sin A = \log 10^{10}\sin A$）他计算出13.85。
答：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写道
267 : sin 45° : : 330 : sin C
计算出角度为60°55后，他加上45°，再从180°中减去所得结果，得到$BDC$ = 74°5
sin 45° : 267 : : sin 74°5 : $BC$
得出363.1。
sin 45° : 480.12 : : 105°55 : $AB$
他计算$\sin 74°5 \times 480.12 \div \sin 45°$。
将结果乘以363.1，除以2，再除以100000，得到1.185。他换算得到
答：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 : sin 56°18 : : 2.4 : sin $BEF$
计算得86°49
sin 56°18 : 2 : : sin 36°53 : $FE$
计算得1.442
sin 86°49 : 1.442 : : sin 6°32 : $AE$
计算得10.92。（写作10:92）
sin $B : AC$ : : sin $BAC : BC$
计算得19.12。
答：C = 30°30，AE = 10.92（写作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写道

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

用六位对数计算，他得到387.6，再加上5得到392.6。
答：392.6英尺。

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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 3°.

只给出了一幅精美的雕像图（实际上并非圣安德鲁广场的梅尔维尔勋爵雕像）。没有计算过程或答案。

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
答：78 2/9磅。

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If 20 foot of railing weigh half a ton, when the base of the bars are $1\large\frac{1}{4}\normalsize$ inches square, what will 50 feet come to at $3\large\frac{1}{2}\normalsize$ per lb, when the bars are $\large\frac{7}{8}\normalsize$ inch square.

20 : 50 : : 1120 : $x$
100 : 49 : :
2000 : 2450 : : 1120 : $x$
4 : 49 : : 112 : $x$
$x$ = 1372
Walker 然后乘以 $3 \large\frac{1}{2}\normalsize$ 并将以便士为单位的答案转换为英镑、先令和便士
答：£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.

设$x$ = 小球体的直径。
设$y$ = 大球体的直径。
$3.1416 x^{2} = 20$
$x$= √20/3.1416
$512 : 1331 : : x^{3} : y^{3}$
$512 : 1331 : : (20/3.1416)^{(3/2)}: y^{3}$ [这似乎不对。他现在使用$x$ = √(20/3.1416)。]
由此他计算y（使用对数），将其平方并乘以3.1416得到37.8
答：37.8格令。

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Find the expense of gilding a ball 6 feet in diameter, at $3\large\frac{1}{2}\normalsize$ per square inch.

Walker 写道
$D^{2} \times 3.1416 \times 144 \times 3.5$ = 答

他使用六位对数计算，其中 $D$ = 6 > 他除以 240 得到以英镑为单位的答案。
答：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.

$72^{3} : 81 ^{3} : : 250 x$。
然后使用对数计算 x。
答：355.9 吨。

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.

$1 : 3 : : 7^{3}: x^{3}$
用对数计算得10.09
$1 : 3 : : 4.5^{3}: x^{3}$
用对数计算得6.49
$1 : 3 : : 3^{3}: x^{3}$
用对数计算得14.37
答：长10.09，宽6.49，深14.37。

If a cubic foot of brass were drawn out into wire $\large\frac{1}{40}\normalsize$ of an inch in diameter, what would be the length of the wire, supposing no loss in the metal.

$(1/40)^{2} \times .7854 x = 1728$
Walker计算$x$（使用对数），然后除以63660换算成英里。
答：55英里4弗隆10.5码。

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换算成先令，然后写出
934.5 : 16800000000 : : 11 : $x$
Walker计算$x$（使用对数），然后乘以480，除以437.5，再除以19640。最后他取立方根得到22.25。
答：22.25英寸。

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 = 面积（平方英尺）
长 宽
300 × 8 = 2400 × 2 = 4800 沟的容积
200 × 8 = 1600 × 2 = 3200 沟的容积

然后他用300 × 200除以8000得到7.32
答：必须挖到7.32英尺深。

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写道
设$D = x$英寸
$d = 3x/5$
截锥体积 = $(A + a +√A a) \large\frac{4}{3}\normalsize$
13.88 × 277.274 = $(x^{2} + 9 x^{2}/25 + 3 x^{2}/5)$.7854 × 4

将$x$计算为25英寸

5 : 3 : : 25 : $x$
计算出$x$ = 15英寸。
答：较大者 = 25英寸。较小者 = 15英寸。

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写道
25 × 277.274 = 体积（立方英寸）
他将其计算为6930
然后他计算$10^{2} \times .7854 = 78.54$
和
$20^{2}\times .7854 = 314.16$。
接着他取乘积78.54 × 314.16的平方根，得到157.08
此时解答终止，未给出答案。

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.

此题的解答同样终止，未给出答案。
Walker写道
60 $AC$ = 8.221849
66 $AC$ = 8.180456
然后他加上(60 + 66 + 30)/2 = 78的对数。
然后他加上78 - 30 = 48的对数。
他将所得对数除以2，得到9.987820。由此他得到角度13°30'23"（余弦的对数）
他乘以2得到27°0'46"
30 : 27°0'46 : : 66 : sin
得到87°44
他将27°0'46加上此值，并从180中减去，得到65°15'14"
$R$ : 30 : : sin 87°44 = 9.999660
30 = 1.477121
29.97 = 1.476781
在此终止，未给出答案。

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

使用正弦定理的标准计算。
答：$AC$ = 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.

使用正弦定理进行计算，但除计算过程中外未给出答案。

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.

使用了Heron规则和正弦规则。除了在计算过程中，没有给出答案。

作者：J J O'Connor 和 E F Robertson
最后更新：2009年2月