8
“By Plain and Practical Rules”: Mathematics
at Work
A GREAT DEAL OF WRITING ABOUT MATHEMATICS IS ABOUT MATHEMATICS in use: applied mathematics, as the modern term has it. Naturally quite a lot of that material is aimed squarely at specialists—whether professionals or amateurs in the field concerned—and assumes a good deal of background knowledge about the subject to which mathematics is being applied. Much makes no sense without specific instruments or tables in hand (a tendency was for books to be tailored to the use of specific instruments and tables produced by the same author or a crony), and many deal in extreme detail with highly specific technical procedures, never intended to be interesting to the uninitiated.
Frustratingly, this means that a good many of the most frequent kinds of mathematics writing are beyond the “popular” scope of this book. The design of fortifications, for instance, was one of the most common places where mathematics was written about in the early modern period, yet I have failed to find an example which is really accessible to the general reader. On the other hand, the mathematics of hobbies, like the geometry of kite design, the computation of the strength of home brew, or the calculation of cricket averages, is seldom, if ever, mathematical enough to engage the nonenthusiast.
So this chapter samples some of the more accessible reaches of “mathematics at work,” and shows how the professional uses of mathematics changed between the sixteenth and the twentieth centuries. The first four selections survey some of the breadth of what traditionally was called “practical mathematics,” in military and navigational uses, and the measurement of both the large and the (relatively) small. The final four extracts are from technical works aimed at beginners from the nineteenth and twentieth centuries and discuss such things as steam engines, plumbing, and typesetting, showing a shift in the subject matter to which mathematics could be applied.
The extracts also give a strong sense of the enduring tension between specificity and generalization: between the single-use rule of thumb and the wide-ranging tools of mathematical modeling provided by algebra. We can compare this with the different styles of arithmetic teaching we saw in Chapter 2, but different writers made different decisions, and the pendulum of fashion was always in motion.
High Marshal and Camp Master
Leonard Digges was, like Robert Recorde whom we met in Chapter 1, one of the earliest writers to publish on mathematics in English. His Tectonicon (1556) covered measuring, calculation, and the use of instruments and was reprinted many times, right down to 1692. This passage is from his posthumous Stratioticos, which presented arithmetic and algebra tailored to the needs of the military man.
Leonard Digges (c. 1515–c. 1559), An Arithmeticall Militare Treatise, named Stratioticos: Compendiously teaching the Science of Numbers, as vvell in Fractions as Integers, and so much of the Rules and Æquations Algebraicall and Arte of Numbers Cossicall, as are requisite for the Profession of a Soldiour. Together with the Moderne Militare Discipline, Offices, Lawes and Duties in euery wel gouerned Campe and Armie to be obserued: Long since attempted by Leonard Digges Gentleman, Augmented, digested, and lately finished, by Thomas Digges, his Sonne 
. (London, 1579), pp. 57–58, 53–55.
Certain Questions touching the Office of the High Marshal and Camp Master
Although the form of Camps may be altered according to the diversity of Situations, in respect of Rivers, or woods that do adjoin thereto, yet for lodging both Horsemen and Footmen commodiously, readily and without confusion, there is none better than the Square. It behooveth therefore those Officers to understand first what quantity of ground sufficeth for the lodging and encamping of some certain Regiment of Horsemen and Footmen. Which known, by the Rules ensuing he shall be able to extend the same to all numbers, and to know readily upon the view of any ground what number it is able to receive both of Footmen and Horse, and accordingly to give order to inferior Officers in what sort they shall proceed to divide their ground for every Regiment.
Question
I find by experience that 1000 Horsemen will demand as much ground to encamp on as 10,000 Footmen; I find also by experience that 550 paces Square of ground will suffice to receive either of them in one main Squadron Camp. But my desire is to divide 
of them into 3 Regiments, and to lodge them 
in 3 square Camps. I demand how many paces Square of those Camps must be.
For the number sought I set down 1x.° That, squared, maketh 1x2 of 
paces, the 
content of one of the 3 Camps. Therefore shall the three Camps be 3x2 of paces. But those three were contained in 302,500 paces, for so much is the content of the Camp of 550 paces Square. Behold therefore your Equation:
The Equation
3x2 = 302,500
which reduced makes
The value of 
is 317 paces, and so much Square ought every one of the three Foot Camps and Horse Camps to be, for the convenient receipt of such Regiments.
Question concerning the office of the Sergeant Major
The high Marshal commandeth that the Army shall be divided into 3 like square Battalions, every Battalion to be Armed in the front with 7 Ranks of Pikes, and that these battalions making up one front upon the Enemy be 
on either side with a sleeve of Pikes of 5 in a rank. This order being prescribed, he delivereth the Sergeant Major 18,000 Soldiers’ Pikes and short weapons, commanding that the Squadrons be made as great as possibly may be of those men. It is demanded how many in every rank of the Battalions, and in what sort the Sergeant Major shall shift his weapons; how many Pikes for the sleeves, and how many of short weapons and Pikes in every Battalion 
.
Figure 8.1. Three Battalions and two Sleeves, with seven ranks of Pikes.
To resolve this Question, first I put the Marshal’s order in the figure (8.1), A, B, C representing the 3 Battalions armed in the front with 7 ranks of Pikes, D and E the Sleeves, 5 in a Rank.
Now because I know not GH, the 
of the Battalion or length of the Sleeves, I suppose that 
1x, the which increased by 5 maketh 5x for one Sleeve. That, doubled, maketh 10x for the 2 Sleeves. Then square x; so have I x2, the quantity of one Battalion, and so consequently 3x2 for the 3 Battalions. Which, adjoined to 10x, the two Sleeves, maketh 3x2 + 10x for Battalions and Sleeves.
But that should be 18,000, for so many men were delivered to make the Battalions. Behold therefore the Equation: 3x2 + 10x = 18,000, which reduced maketh 1x2 + 
x = 6000, and so consequently 1x2 = 6000 − x. The Value of this Root is found to be 75 and certain fractions, which always in these Military Questions may be omitted.
I conclude, therefore, every Battalion must have 75 in a Rank, and that, multiplied by 5, maketh 375, the number of Pikes in either Sleeve. And because the Battalions are armed in the front with 7 Ranks of Pikes, I multiply 7 in 75: so have I 525, the number of Pikes in every Squadron. And that 
Deducted from 5625, the Square of 75, 
5100, the number of short weapons in every Squadron. Thus standeth every demand resolved as followeth:
Pikes in either Sleeve: 257
Pikes in every Squadron: 525
Short weapons in every Squadron: 5625
The number of Ranks in every Squadron: 75.
Note
1x: Digges actually used “cossic” notation, in which the unknown, its square, its square root, and so on, were each denoted by a single symbol. Here modern notation has been substituted.
The Practical Gauger
One use of mathematics that was quite prominent in the seventeenth century was “gauging”: working out the volumes of vessels containing liquids. The purpose was usually to tax the contents. Here we present a remarkably purple commendatory poem to William Hunt, a master of the art, together with part of his treatise, showing the reader how to compute the volume of liquid in various barrels. It is characteristic of this genre of writing to include some rather fantastic solid shapes as exercises, and I have excised from what follows the discussion of the baffling “cylindroid”: “When the Bases are both Elliptical, but unequal, and Disproportional, or Inverted, or if one Base be a Circle, and the other an Ellipsis . . . .”
William Hunt (fl. 1673–1739), A guide for the practical gauger with a compendium of decimal arithmetick. Shewing briefly I. Many plain and easie ways how to gauge brewers tuns, coppers, backs, etc. also the mash-tun, either in whole, or gradually from inch to inch, with divers new tables for facilitating the work. II. The gauging of any wine, brandy, ale or oyl-cask, either in whole, or in part, with the construction and use of two tables of area’s of circles, and Sybant Hantz his table of area’s of segments of a circle. III. The mensuration of all manner of superficies, as board, glass, pavement, wainscot, tiling, floors, roofs, etc. also brick-work, timber and stone. Added as an appendix to the former work. Collected and published principally for the service of the farmers of his Majesties revenue of excise. By William Hunt, student in the mathematicks. (London, 1673). Commendatory poem by Henry Coley, “Philom
,”. A8r–v.
William Hunt, The Gaugers Magazine wherein the Foundation of his Art Is briefly Explain’d and Illustrated with such Figures, As may render the Whole intelligible to a mean Capacity. By William Hunt, Gauger. (London, 1687), pp. 223–227, 230–232.
Upon The Author And his Work
Let Criticks Carp against thy Work and thee,
Whilst I commend it, since thou art so free
Thus to discover what thou know’st in ART,
Both in the Theory and the Practick part
Of Gauging, which is now become so pure
A Ne plus ultra° may be put I’m sure.
How much of late years has been writ hereon?
And yet room left for thee to Build upon;
Go on! Brave Soul, still search, thy Active Pate
Is always Busied; thou regard’st not Fate,
But Aim’st at Common, more than private good,
Which shows thy veins are filled with Generous blood.
To such as in strange paths are led aside,
Thou here hast sent a Welcome pleasant Guide
For to Direct them; let them view each page
With understanding, and Attain to Gauge.
All Vessels, though of various forms they be,
Round, Square or Oval, ’tis all one to thee,
Tuns, Backs and Coolers, Coppers, and the rest,
Cum multis aliis,° work as you like best.
Here’s all variety you can desire,
Return our Author thanks, and then Aspire
To understand each Problem in his Guide,
Which he with Diligence hath often tried.
Words will not Reach his Worth, Hyperbolize
I can’t, nor sound his Fame up to the skies;
Yet this I’ll say, and still add to his praise,
That he alone in this deserves the Bays.°
Problem 118. Definition
A Prism is a Solid Figure contained under several Planes, two of which, being opposite, are called the Bases, and are equal, parallel, and alike situated. But the rest of the Planes are Parallelograms, in which a 
line may be everywhere applied from one Base to another, (which 
may be a Triangle, Quadrangle, Pentagon, or any other plane Surface).
Theorem
Multiply the Area of the Base 
by the Altitude ; the Product will be the Content.
Problem 121. Definition
A Pyramid is a Solid Figure contained under diverse Planes, set upon one which is called the Base, from whence it decreaseth equally, less and less, till it end in a Point at the Top or Vertex. Also, in 
of these Planes, a line may be everywhere applied from the Base to the Vertex.
Theorem
Multiply the Area of the Base (Whether it be Triangular, Quadrangular, Pentagonal, etc.) by one third part of the Altitude. The Product will be the Solid Content.
we cut a pyramid into two parts using a cut parallel to the base, two solid figures will remain. One will be another pyramid; the other will be what is called a frustum of a 
Problem 122
Given AH and CI, the Sides at the Base
of the Frustum of a 
Pyramid, and GP the Altitude, to find the Solid Content. (See Figure 8.2.)
Theorem
As AK (the Semi-difference of AH and CI) is to CK (= CP, the Frustum’s Altitude), so is AP (= half AH the Side of the greater Base) to ZP (the Altitude of the Pyramid AZB).°
From ZP subduct GP; the Remainder is ZG, the Altitude of the Pyramid CZD.
Lastly, from AZB, the whole Pyramid, subduct the Lesser Pyramid CZD; the Remainder will be the Content of the Frustum ABC D.
Now to reduce the former Theorems to Practice
You must take notice that Brewers’ Tuns, though of various Forms, are generally comprehended under these three, viz. Square, Round, and Elliptical, their Sides being supposed to be straight from Top to Bottom, and their Bases parallel.
Figure 8.2. Given AH and CI, the Sides at the Bases of the Frustum of a square-based Pyramid, and GP the Altitude, to find the Solid Content.
I. Of Square Tuns
1. When the Bases are both Square, and Equal, the Tun is a Prism and Gauged by Theorem 118.
2. When the Bases are both Rectangular Parallelograms, equal and alike situated, the Tun is a Prism.
3. When the Bases are both Square, but unequal, the Tun is the Frustum of a Pyramid, and Gauged by Theorem 122.
4. When the Bases are both Rectangular Parallelograms, proportional and alike situated, but unequal, the Tun is the Frustum of a Pyramid.
II. Of Round Tuns
1. When the Bases are both Circular, and equal, the Tun is a Cylinder and Gauged by Theorem 118.
2. When the Bases are both Circular, but unequal, the Tun is the Frustum of a Cone, and Gauged by Theorem 122.
III. Of Elliptical Tuns
1. When the Bases are both Elliptical, equal, and alike situated, the Tun is a Prism, and Gauged by Theorem 118.
2. When the Bases are both Elliptical, proportional, and alike situated, but unequal, the Tun is the Frustum of an Elliptical Cone, and Gauged by Theorem 122.
Notes
Ne plus ultra: literally, “no further”; so, a boundary, an outer limit.
Cum multis aliis: with many others.
the Bays: the victor’s crown of laurels.
As AK . . .: There is some mistake here. It is evident from the diagram that K is supposed to be the foot of a line through C perpendicular to the base of the frustum. The length AK is 
, or 
, which is 
. Hunt’s argument is then that APZ and AKC are similar triangles, so that the length ZP can be deduced. From there it is straightforward to calculate the volumes of the pyramids CZD and AZB, and, subtracting one from the other, to find the volume of the frustum.
Geodæsia
Surveying was one of the oldest uses of mathematics (“geometry,” after all, means “measuring the earth”), and it flourished in the early modern period as a practical art which anyone could learn. New concerns with the production of accurate maps for navigation and in the colonies gave the subject a new urgency.
Figure 8.3. A tower whose height you would know. (Love, p. 181. © The Bodleian Libraries, University of Oxford. F 2.36 Linc.)
John Love described himself as a “philomath” (“lover of,” or perhaps “friend to” mathematics), and he presumably had some professional experience as a surveyor; he does not seem to have written anything else, but this book was hugely successful, remaining in print for over a century and later appearing in special editions for the American surveyors he had envisaged on his original title page.
John Love, Geodæsia: or, the Art of Surveying and Measuring of Land, Made Easie. Shewing, By Plain and Practical Rules, How to Survey, Protract, Cast up, Reduce or Divide any Piece of Land whatsoever; with New Tables for the ease of the Surveyor in Reducing the Measures of Land. Moreover, A more Facile and Sure Way of Surveying by the Chain, than has hitherto been Taught. As Also, How to Lay-out New Lands in America, or elsewhere: And how to make a Perfect Map of a River’s Mouth or Harbour; with several other Things never yet Publish’d in our Language. (London, 1688), pp. 180–183.
How to take the Height of a Tower, Steeple, Tree, or any such thing
Let AB be a Tower whose Height you would know (see Figure 8.3).
First, at any convenient distance, as at C, place your Semi-circle, or what other Instrument you judge more fit for the taking an Angle of Altitude, as a large Quadrant, or the like, and there observe the Angle ACB. Measure next the distance between your Instrument and the foot of the Tower, viz. the line Cd, which let be 25 Yards. Then have you all the Angles given (admitting the Angle of the Tower makes with the Ground, viz., d, to be the Right Angle), and the Base, to find the Perpendicular AB; which you may do, as you were taught in Trigonometry. For if you take 58 from 90, there remains 32 for the Angle at A. Then say,
the Sine of the Angle A, 32, is to the Base Cd, 25, so is the Sine of the Angle C, 58, to the Height of the Tower, AB, or rather Ad.
So 0.530 : 25 = 0.848 : Ad
and Ad = 0.848 × 25/0.530
= 40 
°
To this 40 Yards you must add the height of your Instrument from the Ground . In this way of taking Heights, the Ground ought to be very level, or you may make great Mistakes. Also the Tower or Tree should stand perpendicular, or else you must measure to such a place where a Perpendicular would fall if let down. As, AB is not a Perpendicular, but Ad; therefore measure the Distance Cd for your Base.
This you may plainly understand by the foregoing Figure, for 
, standing at C, you were to take the Height of the Tower and Steeple to E. The Angle ECB is the same as the Angle at ACB, and if you measure only CB or Cd, you will make the Height FE the same as dA, which by the Figure you plainly perceive to be a great Error. Therefore, to take the Height FE, you should measure from C to F.
How to take the Height of a Tower, etc., when you cannot come nigh the Foot thereof
In the foregoing Figure, let AB be the Tower, and suppose CB to be a Moat, or some other hindrance, that you cannot come higher than C to take the Height. Therefore at C plant your Instrument, and take (as before) the Angle ACB, 58 degrees. Then go backwards any convenient distance, as to G, 
there also take the Angle AG B, 38 degrees. This done, subtract 58 from 180, so have you 122 degrees, the Angle ACG. Then, 122 and 38 being taken from 180, 
remains 20 for the Angle GAC. The Distance GC, measured, is 26. Now by Trigonometry, say,
the Sine of the Angle A, 20, is to the Distance GC, 26, so is the Sine of the Angle G, 38, to the Line AC.
So 0.342 : 26 = 0.616 : AC,
and AC = 0.616 × 26/0.342
= 46.8.
Again,
As the Sine of the right angle, B, is to the Line AC, 46.8, so is the Sine of the Angle C, 58, to the Height of the Tower Ad.
So 1 : 46.8 = 0.848 : Ad,
and Ad = 0.848 × 46.8
= 39.7 °
Notes
As the Sine . . .: Love’s procedure is slightly more complex, using the logarithms of the quantities rather than the quantities themselves.
39.7 Yards: Love rounds at two different stages in the working, to produce, a little dubiously, the same result—40 yards—as in the previous calculation.
Plain Sailing
Navigation was one of the most common uses—in print—for mathematics, certainly in the seventeenth and eighteenth centuries. We find works on the “longitude problem,” of course, and books describing the use of new and more advanced instruments and tables; but there was also a basic literature on the application of spherical (or plane) geometry to course finding. So routine, indeed, was this material, that many authors were content to “borrow” one another’s examples and even text, perpetuating what were often not particularly lucid ways of explaining matters.
Patoun, more original than most, was a Fellow of the Royal Society, although he does not seem to have published anything in its Transactions; and although it was something of a success and went to at least eight editions, this seems to have been his only book.
Archibald Patoun (dates unknown), A Complete Treatise of Practical Navigation Demonstrated from it’s First Principles: With all the Necessary Tables. To which are added, the useful Theorems of Mensuration, Surveying, and Gauging; with their Application to Practice. (London: 6th edition, 1762), pp. 160–163.
Of Plain Sailing
This Method of Sailing supposes the Earth to be a Plane, and the Meridians parallel to one another, and likewise the Parallels of Latitude at equal Distance from one another, as they really are upon the Globe. Though this Method be in itself evidently false, yet in a short Run, and especially near the Equator, an Account of the Ship’s Way may be kept by it tolerably well.
The Angle formed by the Meridian and Rhumb,° that a Ship sails upon, is called the Ship’s Course. Thus, if a Ship sails on the NNE Rhumb, then her Course will be 22°, 30′. And so of others.
The distance between two Places lying on the same Parallel (counted in Miles of the Equator), or the Distance of one Place from the Meridian of another (counted as above) on the Parallel passing over that Place, is called Meridional Distance, which in Plain Sailing goes under the Name of “Departure.”
Let A denote a certain Point on the Earth’s Surface, AC its Meridian, and AD the Parallel of Latitude passing through it (see Figure 8.4). And suppose a Ship to sail from A on the NNE Rhumb ’til she arrive at B. And through B, draw the Meridian, BD (which, according to the Principles of Plain Sailing, must be parallel to CA), and the Parallel of Latitude, BC. Then the Length of AB, viz., how far the Ship has sailed upon the NNE Rhumb, is called her Distance. AC or BD will be her Difference of Latitude, or Northing, CB will be her Departure, or Easting, and the Angle CAB will be the Course.
Hence it is plain that the Distance sailed will always be greater than either the Difference of Latitude or 
Departure, it being the Hypotenuse of a right-angled Triangle, whereof the other two are the Legs, 
the Ship sails either on a Meridian or a Parallel of Latitude. For if the Ship sails on a Meridian, then it is plain that her Distance will be just equal to her Difference of Latitude, and she will have no Departure; but if she sail on a Parallel, then her Distance will be the same with her Departure, and she will have no Difference of Latitude. It is evident also, from the Scheme, that if the Course be less than 4 Points, or 45 Degrees, its Complement, viz., the other Oblique Angle, will be greater than 45 Degrees, and so the Difference of Latitude will be less than the Departure. And, lastly, if the Course be just 4 Points, the Difference of Latitude will be equal to the Departure.
Figure 8.4. Suppose a Ship to sail from A on the NNE Rhumb ’til she arrive at B.
Since the Distance, Difference of Latitude, and Departure form a right-angled Triangle, in which the Oblique Angle opposite to the Departure is the Course, and the other its Complement, therefore having any two of these given, we can find the rest. And hence arise the Cases of Plain Sailing, which are as follow.
Case 1
Course and Distance given, to find Difference of Latitude and Departure.
Example
Suppose a Ship sails from the Latitude of 30°, 25′ North, NNE, 32 Miles. Required: the Difference of Latitude and Departure, and the Latitude come to.
the Sine of 90° is to the Distance AC, 32, so is the Sine of the Course, A: 22°, 30′, to the Departure BC.
So 1 : 32 = 0.383 : BC,
and BC = 0.383 × 32
= 12.25
So the Ship has made 12.25 Miles of Departure Easterly, or has got so far to the Eastward of her Meridian. Then, for the Difference of Latitude, or Northing, the Ship has made, we have:
the Sine of 90° is to the Distance AC, 32, so is the Cosine of the Course, A: 22°, 30′, to the Difference of Latitude, AB.
So 1 : 32 = 0.924 : AB,
and AB = 0.924 × 32
= 29.56.
So the Ship has differed her Latitude, or made of Northing, 29.57 Minutes.
And since her former Latitude was North, and her Difference of Latitude also North, therefore to the Latitude sailed from, 30°, 25′N, add the Difference of Latitude, 29.57′, and the Sum is the Latitude come to: 30°, 54.57′N.
Notes
Rhumb: a path on the earth’s surface at a constant compass direction.
29.56: Patoun makes it 29.57, due presumably to a rounding error.
High-Pressure Engines
This Pocket Companion has much in common with the practical manuals of a much earlier era, and William Templeton, an engineer, remained in print on subjects like “The locomotive engine popularly explained” until the early twentieth century. He was the author of mathematical tables “for practical men,” but like many of his predecessors he also aimed his work at “scientific gentlemen.”
At the same time, this section on the mathematics of steam engines bears witness to the fact that the nineteenth century saw any number of new technological uses for mathematics, in a development which would ultimately drive the nonspecialist away from much of this type of practical mathematics.
William Templeton (fl. 1833–1852), The Millwright & Engineer’s Pocket Companion; comprising decimal arithmetic, tables of square and cube roots, practical geometry, mensuration, strength of materials, mechanic powers, water wheels, pumps and pumping engines, steam engines, tables of specific gravities, etc. (London: 2nd edition, 1833), pp. 142–144.
The effective power obtained by means of a high-pressure engine is nearly two-thirds of the force of the steam, one-third being expended in friction, etc. Hence, multiply the cylinder’s area in inches by the force of the steam in pounds, and by the velocity of the piston in feet per minute; deduct 
of the product, and divide the remainder by 33,000. The quotient will be the force of the engine, expressed in horsepower.
Example
Required: the power of an engine, the cylinder being 8 inches diameter, and stroke 2 feet, the engine making 50 revolutions per minute, and the weight upon the safety valve equal 
30lbs. per square inch.
8 inches diameter = 50.2656 inches area, and
50 revolutions × 4 feet = 200 feet velocity.
Or 
, multiply 49,500 by the number of horsepower required; divide the product by the force of the steam in pounds, multiplied by the velocity of the piston in feet per minute, and the quotient will be the area of the cylinder.
Example
Required: the diameter of a cylinder for an engine of 12 horsepower, 
working pressure 35lbs. per square inch, length of stroke 2 feet 6 inches, and making 45 revolutions per minute.
To find an Equivalent Force of the Steam, when the Engine is working expansively
Rule 1
Divide the length of the stroke in inches by the distance (also in inches) that the piston moves before the steam is shut off. 
Divide the pressure on the boiler in pounds by this quotient. 
2
Add 1 to the 
logarithm of the number of times 
which the steam is expanded, and multiply the logarithm by the number of pounds to which the steam is expanded, and the product is the uniform force of the steam acting throughout the whole stroke.
Example
Let the steam in the boiler of a high-pressure engine equal 45lbs. per inch, the length of stroke 4 feet, and the steam to be shut off after the piston has moved 16 inches. Required, an equivalent force of the steam in the cylinder.
4 feet = 48 inches, and 48 ÷ 16 = 3.
Then 45 ÷ 3 = 15lbs.
and 1 + 1.0986123 = 2.0986123. 
× 15 = 31.4791845lbs., 
uniform force of the steam.
The Strength of Materials
Sections on the strength of materials and the rules for scaling constructions up or down had long appeared in mathematical manuals. Lucius D. Gould, architect, had some new materials to work with in the nineteenth century. Gould himself is an obscure figure; this volume, in various versions, seems to have been his only book.
Like Templeton’s account of high-pressure engines in the previous extract, this passage illustrates the resurgence in the nineteenth century of rules of thumb rather than general principles in works of this type, reflecting the increasing complexity of the underlying mathematical models compared with their eighteenth-century counterparts.
Lucius D. Gould (dates unknown), The American House Carpenters- and Joiners- Assistant: Being a New and Easy System of Lines, Founded on Geometrical Principles. For Cutting every Description of Joints. And for Framing the most Difficult Roof. To which is added A Complete Treatise on Mathematical Instruments. Also, Mensuration, Tables of the Weights and Cohesive Strength of the Several Materials used in the Construction of Buildings, etc. (New York, 1853), pp. 125–127.
1. To find the Breadth of a uniform Cast-Iron Beam, to bear a given weight in the middle
Rule
Multiply the length between the supports in feet, by the weight to be supported in pounds, and divide the product by 850 times the square of the depth in inches; the quotient will be the required breadth in inches.
2. To find the Depth of a uniform Cast-Iron Beam, to bear a given weight in the middle
Rule 1
Multiply the length of bearing in feet, by the weight to be supported in pounds, and divide this product by 850 times the breadth in inches; and the square root of the quotient will be the required depth in inches.
When no particular breadth or depth is determined by the nature of the situation for which the beam is intended, it will be found sometimes convenient to assign some proportion; as, for example, let the breadth be the nth part of the depth, n representing any number at will. Then the rule will be as follows:
Rule 2
Multiply n times the length in feet, by the weight in pounds; divide this product by 850, and the cube root of the quotient will be the depth required; the breadth will be the nth part of the depth.
Note
It may be remarked here that the rules are the same for inclined as for horizontal beams, when the horizontal distance between the supports is taken for the length of bearing.
Example
In a situation where the flexure of a beam is not a material defect, it is required to support a load which cannot exceed 33,600 pounds, or 15 tons, in the middle of a cast iron beam, the distance of the supports being 20 feet, and making the breadth a fourth part of the depth. In this case,
The cube root of 3162.35 is nearly 14.68 inches, the depth required; the breadth is
14.68 ÷ 4 = 3.87 inches.
In practice, therefore, whole numbers should be used, and the beam be made 15 inches in depth, and 4 inches in breadth.
3. To find the Breadth of a uniform Cast-Iron Beam, when the load is not in the middle between the supports
Rule
Multiply the distance between the load and the nearest support in feet, by the distance between the load and the furthest support in feet; and four times this product, divided by the whole length between the supports, will give the effective leverage of the load in feet; this quotient being used instead of the length, in any of the foregoing Rules, will give the breadth and depth.
Example
Instead of placing 15 tons in the middle of a beam, as in the last Example, let it be placed at 5 feet from one end, and 15 feet from the other.
which is the number to be employed instead of the whole length, as in the last example; that is
And the cube root of 2372 is nearly 13.34 inches, the depth of the required beam; and
13.34 ÷ 4 = 3.33 inches
gives the breadth, or the beam is nearly 13
inches by 3 inches. In the foregoing Example it was 15 inches by 4 inches.
4. To substitute Beams of Wrought Iron, of Oak, or of Yellow Fir, instead of having Beams of Cast Iron, in either of the foregoing problems
Rule
Instead of using 850 as a divisor, when the beam is of wrought iron, employ 952; when it is of oak, 212; and when it is of yellow fir, 255.
5. To find the Breadth of a uniform Cast-Iron Beam, when the load is distributed over the length of a Beam, supported at both ends
Rule
When the load is uniformly distributed over the length of a beam, it supports double the weight that it would do if the load were laid on the middle; therefore the divisor, in the preceding examples, is changed from 850 to 1700.
Example
In a situation where an arch cannot be used for want of abutments,° it is necessary to leave an opening 15 feet wide, in an 18-inch brick wall. Required: the depth of two cast-iron beams to support the walls over the openings, each beam to be 2 inches thick, and the height of the wall intended to rest on the beam being 30 feet.
The wall contains
30 × 15 × 1 = 675 cubic feet;
and as a cubic foot of brickwork weighs about 100 lbs., the weight of the wall will be about 67,500 lbs.; and half this weight, or 33,750 lbs., will be the load upon one of the beams. Since the breadth is supposed to be given, the depth will be found by problem 2, if 1700 be used as the constant divisor; thus,
The square root of 149 is 12
nearly; therefore each beam should be 12inches deep, and 2 inches in thickness. This operation gives the actual strength necessary to support the wall, but double the weight is usually taken in practice to allow for accidents. In this manner the strength for brestsummers,° lintels, and the like, may be determined.
Notes
abutments: supports which resist outward thrust.
brestsummer: “a ‘summer’ or beam extending horizontally over a large opening, and sustaining the whole superstructure of wall, etc.; e.g. the beam over a shop-front, the lower beam of the front of a gallery, and the like.” (OED.)
Plumbing and Hydraulics
Authors named William H. Dooley published works on industrial education, clothing style, and shoemaking in the early twentieth century, but it is not quite clear whether any of them was the same person as the author of this volume of “vocational mathematics.” Firmly in the tradition of practical mathematics addressed to practical readers stretching right back to, say, the works on “gauging” of the seventeenth century, it is concerned with mathematical procedures for solving specific practical problems in various areas. Dooley follows the nineteenth-century tendency to give practical recipes with little in the way of proof. Algebra is used only sparingly, even when it would clarify the relationships between some of the problems and reduce the amount of material needing to be rote-learned.
For a different take on air pressure and barometers, see the extracts from The Ladies’ Diary in Chapter 3.
William H. Dooley (dates unknown), revised by A. Ritchie-Scott, Vocational Mathematics (London, 1920), pp. 169–172, 174–177.
Atmospheric pressure
Atmospheric pressure is often expressed as a certain number of “atmospheres.” The pressure of one “atmosphere” is the weight of a column of air, one square inch in area.
At sea level the average pressure of the atmosphere is approximately 15 pounds per square inch.
The pressure of the air is measured by an instrument called a barometer. The barometer consists of a glass tube, about 31 inches long, which has been entirely filled with mercury (thus removing all air from the tube) and inverted in a vessel of mercury.
The space at the top of the column of mercury varies as the air pressure on the surface of the mercury in the vessel increases or decreases. The pressure is read from a graduated scale which indicates the distance from the surface of the mercury in the vessel to the top of the mercury column in the tube.
Questions
1. Four atmospheres would mean how many pounds?
2. Give in pounds the following pressures: 1 atmosphere; atmosphere; 
atmosphere.
3. If the air, on the average, will support a column of mercury 30 inches high with a base of 1 square inch, what is the pressure of the air? (One cubic foot of mercury weighs 849 pounds.)
Water pressure
When water is stored in a tank, it exerts pressure against the sides, whether the sides are vertical, oblique, or horizontal. The force is exerted perpendicularly to the surface on which it acts. In other words, every pound of water in a tank, at a height above the point where the water is to be used, possesses a certain amount of energy due to its position.
It is often necessary to estimate the energy in the tank at the top of a house or in the reservoir of a town or city, so as to secure the needed water pressure for use in case of fire. In such problems one must know the perpendicular height from the water level in the reservoir to the point of discharge. This perpendicular height is called the head.
Pressure per Square Inch
To find the pressure per square inch exerted by a column of water, multiply the head of water in feet by 0.434. The result will be the pressure in pounds.
Head
To find the head of water in feet, if the pressure (weight) per square inch is known, multiply the pressure by 2.31.
Lateral Pressure
To find the total lateral (sideways) pressure of water upon the sides of a tank, multiply the area of the submerged side, in inches, by the pressure due to one half the depth.
Thickness of Pipe
To find the thickness of a lead pipe necessary for a given head of water, multiply the head in feet by the size of the pipe required, expressed as a decimal, and divide this result by 750. The quotient will represent the thickness required in hundredths of an inch.
Velocity of Water
Velocity through Pipes
To calculate the velocity of water flowing through a horizontal straight pipe of given length and diameter, the head of water above the centre of the pipe being known, multiply the head of water in feet by 2500 and divide the result by the length of the pipe in feet multiplied by 13.9, divided by the inner diameter of the pipe in inches. The square root of the quotient gives the velocity in feet per second.
Let
Then
Head and velocity
To find the head which will produce a given velocity of water through a pipe of a given diameter and length, multiply the square of the velocity, expressed in feet per second, by the length of pipe, multiplied by the quotient obtained by dividing 13.9 by the diameter of the pipe in inches. Divide this result by 2500, and the final quotient will give the head in feet.
Power
Power is the time rate of doing work. The unit of power is the horsepower (H.P.), which represents 33,000 foot pounds a minute or 550 foot pounds a second. A foot pound is the amount of work necessary to lift a pound through a distance of one foot
Raising water
To find the power necessary to raise water to any given height, multiply the product of cubic feet required per minute and the number of feet through which it is to be lifted by 62.3 and divide this product by 33,000. This will give the nominal horsepower required. If the amount of water required per minute is in gallons, the multiplier should be 8.3 instead of 62.3.
Water Power
When water flows from one level to another, it exerts a certain amount of energy, which is the capacity for doing work. This energy may be utilized by such means as the water wheel, the turbine, and the hydraulic ram.
Friction, which must be considered when one speaks of water power, is the resistance which a substance encounters when moving through or over another substance. The amount of friction depends upon the pressure between the surfaces in contact.
When work is done, a part of the energy which is expended is apparently lost. In the case of water this is due to the friction. All the energy which the water has cannot be used to advantage, and efficiency is the ratio of the useful work done by the water to the total work done by it.
Efficiency
To find the amount of useful work done when a pump lifts or forces water to a higher level, multiply the weight of the water in pounds by the height in feet through which it is raised.
Since friction must be taken into consideration, the total work done by the pump will be greater than the useful work. To find the amount of the total work, multiply the amount of the useful work by the reciprocal of the efficiency of the pump.
Automobiles and Printing
This account of workshop mathematics is another thoroughly practical manual, by a team which was also responsible for the often reprinted Mathematics for Technical and Vocational Schools (1922). The latter has been translated into Spanish, and in a revised form was reprinted as recently as 2002. The extracts given here show how both old and new technologies required mathematical knowledge and procedures for those who worked with them.
Samuel Slade and Louis Margolis (dates unknown), Essentials of Shop Mathematics (New York, 1941), pp. 62–63, 103–105, 123–133.
Automobile mechanics
Work and power
When we lift a weight of one pound a height of one foot, we do one foot-pound of work. If we exert a pull of 50 pounds in dragging a load a distance of 10 feet, we do 10 × 50 or 500 foot-pounds of work.
A foot-pound is the unit of work. It is the work done in raising one pound a height of one foot, or it is the pressure of one pound exerted over a distance of one foot in any direction.
A freight elevator lifting 2,000 pounds a height of 40 feet does 80,000 foot-pounds of work whether it takes 2 minutes or 4 minutes to do it. But an elevator that would lift the load in 2 minutes would have twice the power of one that would take 4 minutes to do the same amount of work.
Power is the rate at which work is done, and is measured in foot-pounds per minute. Thus, if an elevator did 80,000 foot-pounds of work in 2 minutes, its power would be
or 40,000 foot-pounds per minute. If it took 4 minutes to do it, its power would be 
or 20,000 foot-pounds per minute.
The unit of power is the horsepower and is equal to 33,000 foot-pounds per minute.
Example
An engine does 120,000 foot-pounds of work in 3 minutes. Compute the horsepower.
Solution
Horsepower = 
Explanation
Since it takes 3 minutes to do 120,000 foot-pounds of work, in 1 minute it will do 
of 120,000 or 40,000 foot-pounds of work.
Since 33,000 foot-pounds = 1 H.P., we divide by 33,000.
Problems
1. A man carries a load of 50 pounds up a flight of stairs 10 
high. Compute the work done in foot-pounds.
2. Compute the horsepower required for an elevator if it is to lift 2,400 pounds a height of 120 feet in 1
minutes.
3. Find the horsepower of an engine that pumps 30 cubic feet of water per minute from a depth of 320 . Water weighs 62.5 pounds per cubic foot.
4. What must be the horsepower of an engine to lift a girder weighing 7 tons a height of 40 in 5 minutes?
5. A man in pushing a wheelbarrow a distance of 450 exerts a pressure of 45 pounds. How much work does he do?
Printing
The point system
The standard unit of measurement used by printers is the pica. There are six picas to the inch. The pica is divided into twelve parts or points.
Therefore there are 6 × 12, or 72, points to the inch.
The pica is actually slightly less than one-sixth of an inch. The difference, however, is so small that for all practical purposes, it may be disregarded. The pica measures 0.16608 of an inch. Reduce one-sixth of an inch to a five-place decimal fraction and compare it with the pica.
Measuring types
The size of type is stated as the number of points in the height of the body of the piece of type. Thus, a piece of 10-point type is a piece of type which measures 10 points; a pica type is a piece of type which measures 12 points, or 
. Knowing the size of the type and the length of the page, we can compute the number of lines in the page or the length of the page required for a given number of lines.
Leaded type
In order to separate the lines and make the page easier to read, it is customary to insert strips of type metal between the lines. These are called leads. They are usually two points thick, though one-point leads and leads of other sizes are also used. This of course reduces the number of lines of type that can be printed on the page; or, if the same number of lines of type is required, the length of the page must be increased.
Measuring composed type
Another unit of measurement used in printing is the em. The em is a square of the type size used; that is, the width is the same as the height. For example, a 10-point em is a square 10 points on a side. The number of ems in a line depends on the size of the type.
The area of the printed page is expressed in ems. This is computed by multiplying the number of ems in the width of the page by the number of ems in the length of the page.
Space required
A common method for computing the amount of space or the number of pages of a stated size required for setting a given amount of copy is to use an average figure of 3 ems per word. The number of words in the copy is readily estimated by counting the number of words on one page of the typewritten or handwritten copy and multiplying by the number of pages.
Example 1
A manuscript consists of 42 typewritten pages and the average number of words on a page is 280. How many ems will the copy make?
Solution
Example 2
How many pages 23 picas by 40 picas will be required to set the job in Example 1 if 10-point type solid 
is used?
Solution
