The Calcard

Linear scales – addition and measurement

Nature only knows counting numbers. That’s why they’re called “Natural Numbers”. If you have five sticks, you can count them – 1….2….3….4….5 sticks. If you break one you don’t have 4 sticks and two half sticks – you have 6 sticks. Nature doesn’t understand fractions. There is no zero in nature. If you don’t have any sticks to count, there are no sticks. There is no such thing in nature as zero sticks – there is no empty set.

It’s easy enough to see how natural numbers work. 1 is the symbol and “one” is the name of the number of items in the smallest group of anything. If you add another item to the group, the symbol used to designate the number of items is 2 and the name of that number is “two”. In the same way we have 3 “three”, 4 “four”, 5 “five”, 6 “six”, 7 “seven”, 8 “eight”, 9 “nine”, and * “ten”, and
if you’re counting with fingers, that’s as far as you can go. That would have been the most primitive form of counting – using small, convenient objects, such as body parts, to count other objects like sheep. This process is called enumeration, the numbers still being firmly a part of the physical world.

Later, people came up with names for the numbers and they were originally the names of the objects used for counting – the body parts – and we still call them “digits” (fingers) today. This process is called numeration.

When people started using numbers, a step had been taken that separated numbers from things being counted. Numbers became “things in their
own right”. Numbers were abstracted by the process of numeration.

In order to remember large numbers – the number of sheep in everybody’s flocks in the neighborhood or the number of young men of military age, people needed some way of writing the numbers and so humanity developed numeration systems. Numeration systems are for counting – not doing arithmetic. The most common numeration system from ancient times still in use, is the tally system. As you count, you make a mark on paper. When you get to five items, you make a mark across the four marks you’ve already made. When you get to ten items, you make a circle out to the side, and you keep toting up marks, slashes and circles.

Such a system is awkward and, in time, the Indian people came up with the idea of a placeholder that allowed the development of our current place system. The Arabic people translated the Indian works into Latin and we had zero. The natural numbers and zero make up the whole number system. A number system is a further abstraction that allows arithmetic computations.

You can know everything about whole numbers by starting with 3 symbols and 5 ideas.

‘ze’ zero: an individual constant
‘N’ ‘is a whole number’: a one-place predicate
‘sc’ ‘the successor of’ or ‘the whole number following’: a one-place functor

1) Zero is a number
2) The successor of a number is a number
3) Numbers with the same successors are identical
4) Zero is not the successor of any number
5) Every number is F if the property F (F being any particular property you might think of) satisfies the two conditions: 1) zero is F; and 2) if any individual is F, then so is its successor. (Principle of Mathematical Induction)

These “axioms” (an axiom is a principle that is simply taken for granted because you have to start somewhere), were developed by a man named Peano. Starting with them, you can define and explain natural and whole numbers, counting, addition, and multiplication.

Addition is easy enough to demonstrate. Pour a few beads into one tray. Pour some more into another tray. Count the beads in each tray separately, then pour all the beads from the two trays into one tray and count them. The final count is the sum of the numbers of beads from the two trays added together.

That’s what addition is. If you have two counts and you want to know how many things there are in all, you don’t have to count them all over again. You can just add the two counts to get the total.

Furthermore, all those mysterious properties of natural numbers are rendered very mundane when you remember that you’re just counting things.

The symmetrical property of addition states that:
a + b = b + a = c

In other words, it doesn’t matter in what order you add two numbers, the result will still be the same.

The association property states that:
a + (b + c) = (a + b) + c = d

It doesn’t matter, again, what order you add three numbers (parentheses are used to group two or more numbers to be worked with first), the result will be the same.

Zero is the additive identity:
a + 0 = 0 + a = a

You can add zero to any number and the result will be that number. The addition of zero to any number doesn’t change the count.

Most people take place values for granted but it was one of the most important inventions of humanity.

Consider:

Before the Hindi invented the place value system, people simply had certain symbols for certain quantities. The Roman numeral system was pretty typical. I stood for 1. V stood for 5. X stood for 10. L stood for 50. C stood for 100. D stood for 500. M stood for 1000. If you placed a smaller number to the right of a number, the two are added. If the smaller number is to the left, it is subtracted from the larger number.

So, 1563 would be represented by MDLXIII and 2719 would be represented by MMDCCXIX. To add the two, you would group like symbols:
M MM D D CC L X X IX III

The two Ds are the same as another M. The IX and one of the Is make an X. Then you end up with: MMMMCCLXXXII.

Now imagine that you can’t think “M is the same
as 1000″ because, without the place value system, there is no “1000”; there’s only “M”.

Now, you should have an idea of how difficult “simple addition” was without the place value system.

You can’t add things that are different. You can’t add apples and oranges, although you can add fruit. Place values tell you what you are adding.

In our system of numbers, the first place value at the right is 10⁰(ones). The next place value is 10¹(tens). The next place value is 10²(hundreds). and so forth. You can’t add tens and hundreds – they’re different things. You can add one ten to ten tens (which is a hundred) to get eleven tens.. That’s why you line up place values when you add numbers with multiple digits – so you will be adding the same things.

When you work with fractions, the denominator specifies the place value. Therefore, you can add halves or quarters but you can’t add halves to quarters unless you first translate the halves as quarters. In algebra, you can only add like terms, again, because you can’t add things that are different. The different terms are the different place values.

And zero is always the place keeper for a place value that has no members. A number like 3020 means that there are no ones or hundreds, but if you didn’t include a symbol to represent the empty places, the numbers wouldn’t line up properly in an addition problem. For instance, if you had to add 35, 750, and 3, you wouldn’t line up the numbers like:
35
750
3__

The
result, 1400, would be incorrect. You have to add like items so you would need to group like place values like:

 35
 750
   3

and the correct sum would be 788.

There are some special techniques for adding quickly and accurately.

When adding a group of numbers, you can usually group successive digits to form larger numbers.

For instance in the following problem,

375
205
136
113

when adding the ones column you could think, “5 and 5 makes 10. 6 and 3 makes 9. 9 and 10 makes 19. Carry 1.” In the tens column, “7 and 3 makes 10 and 1 and the 1 I carried makes 12. Carry 1.” For the hundreds column you can quickly see that 3 and 2 make 5, and the three 1s makes 8. So the sum is 829.

It’s actually fairly easy to learn to add two and even three columns at a time. Of course, the more columns you add at a time, the quicker you will be able to add.

Most people can get used to adding up multiple columns in large addition problems so that for:

instead of thinking, 5 and 5 is 10, and 9 is 19, and 6 is 25 to add up the first column, you might think 25 and 5 is 30, and 36 is 66, and 19 is
85. Or you might even see 305 + 25 to be an easy problem and say “330 + 136 is 466 and 219 is 685.” That would take care of a lot of time consuming calculations.

You can also use Finger Math (as discussed on earlier Excursion pages) to add each column.

There have been whole books written on special tricks that can be used to speed up addition and to increase accuracy, but that’s overkill. There are a few tips that should take care of just about any errors that might creep into your work.

It’s a good idea to keep as much control over the internal workings of a large problem as you can, so, if accuracy is important, take the time to record intermediate results. In long addition problems, write down the partial sums. That will reduce the chances of making errors, and it will make it easier to see errors if they occur. Recording partial sums is particularly useful when the numbers you’re adding are scattered throughout a document, for example, on multiple ledger pages.

Let’s say that the numbers 1237, 34219, 65, and 7532 are on three different pages. Page through once and add up the units places in the numbers:

then page through again and add up the tens places:

 23
 13

then the hundreds places:

  23
  13
  9

then the thousands:

Finally, add in the ten thousands places:

and add them together to get 43053.

One of the best checks of accuracy is to estimate the answer to see if you’re even in the right ball park. Round each number to one significant digit and add them to get an estimate. For instance, for the last example 8000 and 60 is 8060. Add 30000 to get 38060. add another 1000 to get 39060. Estimate that the answer will be somewhere in the area of 40000.

Another easy and effective method is to sum the problem in both directions. Start at the top and go down the numbers. If the sum is the same as that that you get if you start at the bottom and go up, your result is probably correct.

Another useful check is a procedure called “casting out nines”. A peculiar and entertaining property of numbers in our decimal number system is that, if you add the digits of a number together, and keep adding digits until you come up with a single digit, that will be the remainder that is left if you divide the original number by 9.

Look at the result of the last problem. If you add the digits across, you will get, 4 + 3 + 0 + 5 + 3 = 15. Repeat the process to get 1 + 5 =
6. If you divide 43053 by 9 you will get 4783 with a remainder of 6 which was your sum of digits.

If you repeat this process for each of the numbers being added in a long addition problem and add the remainders together, the result should equal the result of adding the digits of the sum, if the sum is correct. For example:

 1237   1 + 2 + 3 + 7 = 13,            1 + 3 = 4
 34219   3 + 4 + 2 + 1 + 9 = 19, 1 + 9 = 10, 1 + 0 = 1
     65    6 + 5 = 11                                1 + 1 = 2
  7532   7 + 5 + 3 + 2 = 17                 1 + 7 = 8 
                                                   4 + 1 + 2 + 8 = 15
 43053 4 + 3 + 0 + 5 + 3 = 15           1 + 5 = 6

If the sum of the remainders of the numbers is the same as the remainder of the sum, it doesn’t guarantee that the sum is right. All you have to do to have an incorrect sum without changing any of the remainders is to switch two digits in any of the numbers being added. But if the sum of the remainders of the numbers is different from the remainder of the sum, you know that there is an error.

You can add any number of natural numbers together and the result will always be a natural number. Mathematicians say that the natural number system (and the whole number system, for that matter) is closed under the operation of addition. In other words, you can’t get outside the natural number system by adding natural numbers. The same can be said of multiplication. As we’ll see later, that’s not necessarily true of other operations.

Now, in the construction of a slide rule (like the Calcard) you will have to construct several kinds of scales. The simplest is the linear scales. A linear scale is simple because the line is divided into a number of equal length intervals and they’re numbered according to the whole numbers, starting at zero. Each interval may then be divided into a number of equal distances to create fractional divisions, and so on. You will find that linear scales are useful in two ways – they allow you to measure lengths and they allow you to add.
Before we can begin to create linear scales such as length scales for rulers, we’ll have to explore one more topic. Early on, the Greeks asked a question about geometry, “What can you do with just an unmarked straightedge and a compass.”

Why a straightedge and compass?”

Well go outside into nature and look around and see what shapes you can put a name to. You may understand that when a ball is thrown, it follows a parabola, and that when a slack line is stretched out horizontally, it forms a caternary, but what is really going to catch your attention are the horizon, the sun, a line hanging down toward the earth. What you will notice first are lines and circles. Straightedges draw straight lines; and compasses draw circles. And you will find that you can do a lot with straight lines and circles.

For instance, you can draw perfect right angles to make either right triangles or perfectly bisected lines. You can also bisect angles (divide them into two equal angles). you can make regular hexagons and you can either inscribe or exscribe circles on triangles so that all the vertices (angle points) of the triangle lie on the circle or one point on each leg of the triangle lies on the circle. Lines and circles might be considered the most primitive shapes – the natural numbers of the shape kingdom.

What you will need to do to construct (drawing with only a straightedge and compass is called geometric construction) a linear scale is divide a straight line into a number of equal intervals, but before we get into that, it’s interesting to note that there are some things that are impossible to construct using only an unmarked straight edge and compass.

There are actually two ways that geometric constructions are useful. In the most obvious case, they allow you to draw accurate geometric drawings. They can also be used as a form of geometric calculator to calculate special numbers. For instance, you can take two lines of known length and construct a line that is the same length as the sum of the lengths of the two lines. In that way, you are actually adding two numbers (the lengths of the two lines) to find their sum (the length of the constructed line.) The process is as follows:

Given two lines A and B of length a and b respectively.

Copy line A by setting the point of a compass at one end and adjusting it until the drawing arm touches the other end. Set the point of the compass in a blank space and strike off an arc the distance of a.

Using a straightedge, draw a line from the point of the compass to the arc. That line is the same length as line A.

Now set the point of the compass at one end of line B and adjust it so that the drawing arm touches the other end.

Set the point of the compass at the right end of the copied line and mark off an arc to the right of the line.

Continue the copied line from it’s current end point, using a straightedge, to the new arc. The resulting straight line is the same length as the sum of the lengths of the two shorter lines.

You can also calculate the length of the square root of two by constructing a right triangle (a triangle with one angle that measures 90°.A 90° angle is the kind of angle that you see at each corner of a square) with the two legs that meet the right angel being a unit length. When you draw the third leg, it will be a length of the square root of two. This follows from Pythagoras’ theorem that states that the length of the square of the longer leg of a right triangle (the hypotenuse), is equal to the sum of the squares of the lengths of the other two legs.

The process, briefly, is to draw a line of 2 units length. Bisect it with a perpendicular bisector (you do that by drawing two circles of same size using a compass so that the centers are on the ends of the line and that they intersect above and below the line. if you draw a line between the two points where the two circles meet, you will have a straight line through the middle point of the first line so that four equal (right) angles are formed). Measure the length of one half of the first line using he compass and transfer that length to the bisector – place the point of the compass at the intersection and mark off the unit length above the original line. You will have perpendicular lines of equal length. All you have to do then is connect the top end of the upright line with the left end of the horizontal line to have a right triangle with two legs of length 1 (the horizontal leg of the triangle and the upright) and a longer leg whose length is the square root of two.

So, believe it or not, there are several things that you cannot do with just an unmarked straightedge and a compass. Some classical problems that are impossible under those conditions are:

Trisect any arbitrary angle: You can, in fact, trisect certain angles but not all.

Square a circle, that is construct a square that has the same area as a particular circle.

Double a cube, that is construct a cube that has twice the volume of a given cube.

Calculate the cubed root of a number; that is, construct a line that has a length equal to the cubed root of the length of a given line.

These problems were known to have no known solution in ancient Greece and it was proven that they have no solution, period, only in the 19th Century.

But I am not particularly interested here in what can be done with just a straightedge and a compass – I want to know what can be done with primitive equipment and with the addition of two very accessible pieces of equipment, even these “impossible” tasks can be accomplished.

The first is the string, which can be used to measure the lengths of curves (such as the circumference of a circle) and to draw curves, and an neusis, which is simply a straightedge which can be marked.

As an example, a neusis can be used to trisect any angle. Given an angle that is greater than 180°,

1. Draw a circle using one leg of the angle as a radius and the vertex of the angle as the center of the circle. Extend the base of the semicircle out to either side.

2. Mark the neusis so that a segment from the end to the mark is the length of the radius of the circle

3. Slide the neusis along the drawing so that mark rests on the line that extends beyond the base of the semicircle, and the edge rests where the slanted leg of the angle intersects the circle.

3. Mark the other place where the neusis intersects the semicircle on the circumference of the semicircle.

4. Draw a line from that point on the semicircle to the vertex of the angle. The angle containing that line and the base of the semicircle is one third the original angle.

The final drawing tool needed is a drawing frame. The ideal drawing frame is a corkboard with a large number of tacks and pins (the kind with little beads for heads).. The cord can then be used to draw various curves. For instance, an ellipse can be drawn by placing two pins where you want the foci, looping the cord loosely around the pins, and pulling the cord tight with a pencil point. Drawing around a curve, keeping the cord tight with the pencil point, draws an ellipse.

With those tools, you can accurately draw just about anything. Specifically, we can draw a couple of scales on our Calcard that will allow us to do addition. Let’s take a look at the kind of scales we need first.

In order to add two numbers using line lengths, all you do is take a line of one length and extend it by adding on a length equal to the length of the other line.. You could do that with two rulers if one ruler scale was backward. Therefore, adding 5 + 2 would look like this:

                            0           2
                             |     |     |     |     |    
    |     |     |     |     |     |     |     |     |     |
    0                        5           7

So, we need two scales divided into 10 equal lengths with tick marks labeled 0 to 10.

The way the Calcard will be constructed, there will be two types of index cards. Some of them will be laminated together in accordion fashion. Others will fit between the folds and will slide back and forth along the fold. Let’s call the folds of the accordion “fixed” cards, and the others “slider” cards. The slider cards will have scales along their edges that will line up next to the scales on the edges of the fixed cards; therefore, the fixed cards need to be a little shorter than the slider cards by just enough to let the scale at the edge of the fixed card be fully visible. Be sure that when you cut the strip off the fixed cards, you use a constant width. Of course, cut the strip off the long edge.

There are a few tips that will help you to make accurate constructions:

Don’t use a scissors to cut paper – few can cut a straight line with scissors. Use a sharp pointed knife (preferably an X-Acto type knife) with a straightedge or other guide.

Use a sharp pencil and fine tip pen.

Be as precise as you can. Small errors can easily be multiplied quite a lot. Double check your work. Don’t repeat constructions – make a standard and copy it to work from.

Don’t do your construction work on your final copy. Copy your line on a workpage and do you construction there, then transfer the end product back to the position you want it to have.

Work in pencil. Overdraw the final construction in ink.

Work in a well lighted area.

Locate points with register marks. It’s much easier to hit an intersection than it is to hit a point. Tick marks, targets, cross hairs are worth drawing.

When you draw a line from one point to another, line up your straightedge or compass, place your pencil on each point to make sure that it’s on mark, then draw.

Now, to draw the first linear scale (linear because the intervals will be equal in length) on the first fixed Calcard segment, take an index card and shorten it by cutting off a strip along the long edge by your constant length. Then copy the edge onto a worksheet. Just lay the card down on the worksheet and draw a line using the edge of the card, remembering to draw the line around the ends of the card to provide tick marks for the ends. You may want to make the line a little shorter than the width of the index card so you can label the scale to the side. If you do, though, make sure that you keep this length as a standard. Now you have to divide the line into 10 equal segments on the worksheet. At this point you don’t care what the lengths are – just that they’re equal.

A common construction technique is the bisection of a line (marking a point on a line that exactly divides it into two equal length), but you can’t use that here because, after you mark the midpoint of your line, you would have to divide the resulting lengths into 5 equal parts each. You can’t use bisection to divide a line into an odd number of equal lengths.

The solution uses a geometric property of a line that cuts down through two parallel lines. The key insight is to realize that, if two lines are parallel and are divided into the same equal intervals, then lines drawn from one line to the other at the end points of the intervals will also be equally spaced, and any line cut by these parallel equally spaced lines will also be divided into equal intervals. If the intermediate line doesn’t have the same slope and is not parallel to the two original lines, then the divisions will not be the same size as the divisions on those two lines, but they will all be equal to the other divisions on the intermediate line. So the procedure is as follows:

1. From one end of the line you want to divide into n equal intervals, draw a line at an acute angle. The particular size of the new line or the angle doesn’t matter. Just choose the new line to be long enough so that you can conveniently divide it into n (in this case, 10) equal intervals.

2. From the other end of the line you want to divide, draw a line downward at the same angle. You copy the angle like this:

a. From the vertex of your first angle, using the compass, draw an arc that cuts both legs of the angle.

b. From the other end of the line you want to divide, draw the same radius arc so that it cuts the line and extends downward.

c. Go back to your angle and, from the point where the first arc intersects the first line, measure the distance to the other leg of the angle by drawing an arc that intersects the point where the first arc crosses the upper leg of that angle.

d. Transfer that distance down to the second arc. In other words, without changing the setting of the compass, place the point of the compass at the place where the line you want to divide intersects the second arc and then draw an arc that intersects the arc that passes downward from the line.

e. Draw a line from the end of the line you want to divide down through where the two arcs intersect. Make sure the line is at least as long as the upper line.

These two new lines will be parallel because they leave the original line at the same angle.

3. Using the compass, divide the upper line into 10 equal lengths.

4. Using the same setting of the compass, divide the lower line into the same equal intervals.

5. Starting at one angle vertex and the interval tick furthest from the other angle on the other parallel line, draw 10 lines, using the straightedge, from the interval endpoints on one parallel line to the interval endpoints on the other parallel line.

6. These 10 vertical lines will cut the original line into 10 exactly equal intervals. You can then transfer the interval tick marks back to the line on the Calcard.

Now you have to perform the same operation on one of the intervals. Although the process is the same, the horizontal lines from the two parallels will be much closer together and you will have to be much more precise. You may also have to extend the line you are dividing past it’s endpoints but be careful you don’t loose track of where the angles have to be and where the horizontals must be drawn. When you have an interval subdivided, you can transfer those tick marks to the Calcard. It may help to make the primary divisions from the first construction longest, every fifth division of the primary intervals a little shorter, and the other subdivisions shortest.

Number the primary intervals from left to right from 0 to 10.

You want two linear scales, at least; one on the fixed card and one on the slider, so that, when the slider is placed under the fixed card, the scales will line up – 0 to 0, 1 to 1, 2 to 2, etc. Hold onto an unnumbered template linear scale because there are other linear scales that you will find useful in the future. For instance, if you would like to go ahead and place it on the slider, you will want an inverse linear scale, the same markings as the ones you’ve already done but numbered in reverse.

Label the linear scales L. On a slide rule, “L” stands for “Logarithm” for reasons we’ll get around to, but on the Calcard, “L” stands for “Linear”. If you graph each primary division to it’s distance from one end of the scale, you’d have a straight line. Label the inverse scale LI.

Now, since we’re constructing, you can go ahead and construct a ruler. You have your standard. Index cards are standard size. Since my cards are 4″ x 6″, I only had to divide the 6″ long edge into 6 equal intervals, and then divide each interval into 16 subdivisions to have a standard half foot ruler.

Then, since an inch is 2.54 centimeters, A 6 ” line can be divided into 15 1/4 equal intervals to form a 15 1/4 centimeter ruler.

Hey, all that on one fixed card and a slider. Now let’s see how to add with it. The 0 and 10 marks on the L scales are called the left and right indexes respectively. Just like the rulers mentioned above, if you line up the left index on the top L scale with the first number on the bottom scale, and then find the second number on the top L scale, just below it will be the sum. So you can add up any numbers as long as the sum isn’t more than 10. You’re not impressed?! Okay, let’s see if we can do a little better.

If the sum is off the scale on the right (the sum is more than 10), you can use the right index. Place the right index of the top scale over the first number on the bottom scale and read the number directly under the second number on the top scale. You have to reinterpret the bottom scale as numbers from 10 to 20. Why does that work? Well, if you had three L scales, you could set them up like this:

 

             0               8   10

             | | | | | | | | | | | 

   | | | | | | | | | | | | | | | | | | |

   0                   0     3   5

                       10   (13)

Then if you count 5 marks across the bottom left L scale, start there and count 8 more marks on the top scale, you will have counted 13 marks, which is, of course, the number just below the 8 on the top L scale. So now you can add sums to 20. Still not impressed? Well, okay……

I think you can see that the secondary marks can be read as fractions. So, if you set the left index of the top L scale over 1.5 on the bottom scale and read the number over 2.6, you will see 4.1. You can scale the scale up by a factor of 1. In other words, instead of reading the marks as 1, 2, 3, 4…. you can read them as 10, 20, 30, 40….. In that case, in the last example, you would have been adding 15 + 26 to get 41. In the same way, you can reinterpret the scales as running from 0 to 1000 or 0 to 10,000, as long as you scale both scales by the same factor and you can be satisfied with only 2 significant figures (for most real world applications, 2 significant figures is fine).

You don’t usually see two L scales on a slide rule for two reasons:
1) Rocket scientists don’t need a slide rule to add, they can do that in their head.
2) The real reason an L scale is on the slide rule is to read logarithms to the base 10 off the C and D scales , and to do that, you only need one L scale. (we’ll get to that later.)

The reason I have two L scales on my Calcard are:
1) I’m not a rocket scientist and I want something I can add with.
2) Adding with L scales provides a good preliminary explanation for how you can multiply with C and D scales (when we get to that…..later).

Subtraction

Now, once you understand addition, subtraction is easy. It can be thought of in two ways. Subtraction is the opposite ot reverse of addition – it undoes addition. And subtraction is counting backward.

Looking at the bead example, if you have 5 beads in a group and you throw in two more, you will have 7 beads in all. That’s addirion. If you remove the two beads, you will only have the original 5 – 7 minus 2 is 5 – that’s subtraction.

If A plus B equals a sum, you can undo the process by taking B out of the sum to leave A. Since subtraction undoes addition, it’s called the inverse function of addition.

On a number line, subtraction is counting backward. Counting to the right is addition; counting to the left is subtraction.

So, to add 5 and 2 you would count 5 marks to the right (that will bring you to the mark labeled “5”), then count two more marks to the right (that will bring you to “7”, which is the sum of 2 and 5).

To subtract 2 from 7 you would count 7 marks to the right and, from that point, count 2 marks to the left. That will bring you to “5”. When you subtract 2 from 7 the remainder (the result of a subtraction is called a remainder or difference) is 5.

The number line instance gives us a way to subtract using the Calcard. That’s where the inverse linear scale comes in. Since the LI scale counts to the left, it’s a natural for subtraction. Using the example above, to subtract 2 from 7, set the LI scale on the slider over the L scale on the fixed card. Move the 0 index on the LI card over the 7 on the L card. What’s under the 2 on the LI card? 5 on the L card.

What you have done is count forward 7 marks on the L card and used the LI card to count back 2 marks.

You can do the same thing with two L scales. Set the L scale on the slider over the L scale on the fixed card. Set the 2 on the slider over the 7 on the fixed L scale and then count back to 0. Under the 0 on the slider wil lbe th 5 on the fixed scale.

Now, that brings us to a problem. What if we subtract 7 from 5. You count 5 marks on the L scale and then set the 0 on the LI scale over it. When you start counting back on the LI scale you get to 5 and run out of L scale.

There is nothing less than 0 in nature (or in the natural numbers). If you subtract 5 pebbles from 5 pebbles, you are left with no pebbles. If you subtract 7 pebbles from 5 pebbles, you have precisely the same result – no pebbles. But with humans, you begin to have another situation – you owe me two pebbles, and you begin to need a numerical way to represent debt, so we invented the unnatural negative number. If we have a numberline with a zero at the midpoint and having numbers from 1 on going to the right of the 0 and numbers from 1 on going to the left of the zero, you would be able to deal with negative numbers, Just place a minus (-) before all the numbers to the left of the zero. Then, to subtract 7 from 5, you could count 5 marks to the right on the number line and then start counting back to the left (subtracting is counting back) for 7 marks. You will end up on the mark labeled “-2”.

Now there are convolutions you can use with Calcard to subtract numbers from smaller numbers to get negative numbers but it’s much easier to remember two simple rules.

Notice that, when you subtract 5 from 7 you get 2. When you subtract 7 from 5 you get -2. In both cases, you get the same number but in one case the number is negative. The number itself (2) is called the absolute value. 2,whether it’s to the left or the right of 0 is a distance of 2 marks from 0. Absolute value can be thought of as distance from zero.

The two simple rules are these:

If you subtract a number from a larger number, the remainder is positve.

If you subtract a number from a smaller number, the result is negative.

So, if you want to subtract 7 from 5 on the Calcard, go ahead and subtract 5 from 7 and place a – before the result.

Keep in mind one more thing: -7 is smaller than -5. Numbers to the left of a number on a number line are always smaller than that number regardless of whether they’re positive or negative.

Now, let’s summarize:

Addition is counting

You can only add things that are alike.

Subtraction is the reverse of addition; or, subtraction is counting back.

You can only subtract things that are alike.

If you subtract a number from a larger number the result is positive.

If you subtract a number from a smaller number, the result is negative.

Multiplication of whole numbersNow, we want to be able to multiply using our calculator. Where addition is a little useful, being able to multiply would be would be of considerable use. Trying to do large multiplication problems in your head is painful. But you can’t multiply directly using a linear scale and what you have to do to get a product is long and tedious and would be too prone to error. Still we have a ways to go before we can teach our calculator how to multiply efficiently.

Let’s look at just exactly what multiplication is and let’s stick to the numbers we have available, the whole numbers. If addition is counting and subtraction is the reverse of addition (counting backward), then multiplication is repeated addition.

Think about multiplying 5 times 7. You are, in essence counting 5 seven times. Visualize a cup with 5 beads in it. How many beads would you have if you had 7 such cups, each with 5 beads in it? If you add all the beads together, you would have 35 beads in all and you will have multiplied 5 times 7.

In such a problem, the size of the group of items that is being counted repeatedly (in this case 5) is called the multiplicand. The number of times it is counted (in this case 7) is called the multiplier. And the result of the repeated counting (35) is called the product.

Traditionally, you learn all the products of multiplication up to 12 times 12 so that multiplying a single digit by a single digit is pretty much automatic. There are a lot of methods of multiplying numbers larger than 10 – two are common in Western culture – but they are all different forms of one procedure.

Multiply each place value in the multiplicand by each place value in the multiplier and add the results. Notice that in multiplicantion, you can combine apples and oranges. Let’s look at a traditional long multiplication problem to see how that works (also look at the essay on this site – The Right Way to Multiply – for some exposure to the other popular form of multiplication, lattice multiplication.

To multiply 25 by 72, we set the problem up as:

25
72

Then we multiply 25 times 2 and 25 times 70 and add the two results. 25 times 2 is 50 and 25 times 7 is 175 making 25 times 70 1750 (by the way, this problem is easy to do in your head if you think of 25 as a quarter and think of, for instance, how much money 7 quarters is.). If you add the two partial products together, the result is 1800.

The way we usually write that is:

25
72
50
175
1800

Notice that we ignored the zero at the right end of 70 and the partial product 1750 and, instead, staggered the second partial product. We would do the same with the hundreds place, the thousands place, and so on, if they were present in the multiplier.

There is an interestig geometric “construction” that will allow you to multiply (if you think about it in terms of lattice multiplication, I think you’ll see why it works.)

Draw a series of vertical parallel lines for the multiplicand. In the case above, you would draw 2 lines to the left, some separation, and 5 lines to the right. Next, intersecting those line, draw horizontal parallel lines for the multiplier – 7 to the top and 2 to the bottom.

Now count the number of intersections – 14 at the upper left, 35 at the upper right, 4 at the lower left, and 10 at the lower right. Add the diagonal numbers at the upper right and lower left together to get 39. Set the three numbers you now have up as follows and add:

14
39
10
1800

This will work for any multidigit multiplication problem. It gets more complicated as you have more digits and there’s the special problem of what to do with zeros, but its fairly easy to figure those complications out and I’ll leave that to you for your own entertainment.

But we seem no closer in finding a way to easily do multiplication with our homemade calculater. As it stands, we would have to add 25 over and over 72 times to get the relatively easy product 1800. That is rediculously tedious.

Before we can untangle the problem we will have to take some detours through division, fractions, and exponents (repeated multiplication) but, in the end, we will have an easy solution.

DivisionAs subtraction is the inverse of addition, division is the inverse of multiplication.. Just as subtraction “undoes” addition, division “undoes” multiplication. Whereas addition is counting and subtraction is counting backwards, multiplication is repeated addirion and division is repeated subtraction. Multiplication is adding equal sized groups together; diviision is separating into equal sized clusters. Whereas subtraction caused problems with the number system, division will also create new complcations. An example:

If you took 150 beads and began dropping them into 5 cups arranged in a circle one at a lime until you ran out of beads, you would end up with 30 beads in each cup. This is the opposite – the undoing of – taking 5 cups with 30 beads in each and adding all the beads together, which would be multiplication. Approached in a different way, you could take 150 beads ad separate out (subtract) 5 into their own pile. Then male another pile of 5, and continue until you only have little piles of 5 beads each. You would end up with 30 little piles of beads.

We say that 150 divided into 5, or 5 goes into (or as Jethro Clampit would have it “guzinta”)150 gives us 30. In this instance, the dividend is the number being divided – 150, the divisor is the of groups it’s being divided into – 4, and the result – 30 – is called the quorient. We write it:

150/5 = 30

5÷150 = 30

30
5)150

Notice that I chose the problem rather conveniently. Try the same problem but begin with 149 beads. The last cup, the last pile, will not have the same number of beads. You end up with 4 cups with with 30 beads and one with 29 or you end up with 29 piles of 5 beads and one with 4.

We say that 5 doesn’t go into 149 evenly, or 150 divided by 5 is 29 with 4 left over – 4 what? – 4 parts. Again, if the dividend is 149 and the divisor is 5 then the quotient is 29 with a remainder of 4.

Let’s take a brief detour to emphasize that division, indeed, undoes multiplication.

If 30 x 5 = 150

then 150/5 = 30

Adding 30 groups of 5 beads, you end up with 150 beads. If you start separating out the same groups of 5 beads, you will end up where you started – with 30 groups of 5 beads.

Now, when we started using subtraction, we suddenly had to invent a number that did not represent anything in reality – a negative number – but that new kind of “unnatural” number ended up being useful in it’s own right, as a way to represent a deficit.

Now the problem with division is that, more often than not, you will end up with a result that requires two whole numbers to describe – and, really, what does those “remainders” represent in reality, anyway?

Let’s look at a problem that is a little more modest. Let’s divide 6 by 2 using a paper strip. Take a paper strip and mark it off into 6 equal parts. To divide the strip into two equal parts, you would tear it across right at the middle. Both parts ould have 3 marked-odd sections each. 6/2 = 3

Now what if you had a strip of paper marked off into 5 equal parts. What happens when you divide that by tearing it in half? You would have two strips with 2 sections and part of another. 5/2 is 2 and a remainder of 1 part – what we would call one half.

Notice that we have started using a new, but familiar language – “parts”, “sections”, “halves”. We’re talking fractions.

But keep in mind that “fraction” is not a natural concept. In nature, if you break a stick in two, you don’t end up with two halves of a stick – you just end up with two smaller sticks. Fractions are human concepts; they don’t exist outside of people’s heads.

But this is useful. Consider how difficult it would have been for the Romans to have divided 150 by 5. That is, how do you go about dividing CL by V? We now have a way to do that.

Now, to pin a name on the animal, this new set of numbers – the whole numbers along with these new fractions – is called “the rational numbers” – not because they are particularly smart (“rational”) but because fractions are also ratios. 5/2 is also the ratio “5 to 2”.

Do you see now what the remainder represented when we tried to divide 149 by 5 is? We ended up with a smaller group of 4, instead of 5. The remainder was 4 extra parts left over. If you invision strips of paper, you’ll recognize that those parts are fifths. 149/5 = 29 4/5.

What is 1 divided by 2? it’s one half – 1/2. What is 2 divided by 5? It’s two fifths – 2/5.

Now, since we are now faced with coming to terms with this new kind of number – we will be running into it over and over again – let us understand just exactly what it is that we faced with.

When adding with the L scales of the CalCard, we found it useful to be able to scale up – to be able to look at marks on the linear scales as ones, or tens, or hundeds, and so on. Fractions allow us to scale down and thereby work with ever smaller and smaller quantities. Larger scales are decided by the place value taken as the unit. Remember that place values – the value that tells us what sort of digit we are working with at any particular time – for whole numbes are powers of ten. The 9 in 9 and the 9 in 90 are two different kinds of 9. The 9 in 9 are 9 ones. The 9 in 90 are 9 tens.

But what are the place values when you are scaling down from one? Well, by analogy. it looks at though, in a fraction, it is the bottom number. Whereas the 9 in 9 and the 9 in 90 are two different kinds of 9s, the 1 in 1/2 and the 1 in 1/4 are two different kinds of 1s. The first is 1 half and the second is one quarter. By the way, in mathematical language the number on top in a fraction is called the numerator and the number on the bottom is called the denominator.

If you divide one of any particular thing – a stick, a board, a dollar, an interval on the L scale of the CalCard, or just an abstract 1 – into four parts and you take one of those parts, that’s 1/4 – one quarter. If you take 3 of the parts, that’s 3/4 – three quarters.

Here we run into an inconvenience. With whole numbers, we can line up place values and make one number out of them that we can handle as one number. 2,431 actually means two thousands, 4 hundreds, 3 tens, and 1 one. What do we do with a creature such as 5 , one half, 3 thirty seconds, and 14 sixty sevenths? 5 1/2 3/32 14/67 is just confusing. There are, we will see, ways of dealing with such numbers but it’s tedious and we don’t usually have time to fiddle around with the complexity and the headache. The problem is that halves, thirty-secondths, and sixty-sevenths don’t seem to have a lot in common. What whole number place values have in common is readily apparent. They are all powers of ten. This suggests a ready alternative. Use fractional powers of ten.

Instead of using just any number as the denominator of a fraction, we can use a power of ten and, in that way, represent the result of any devision to any speciific level of precision. Use a symbol (.) to separate the whole place values from the fractional place values and you have what we call “decimal fractions”. Therefore, 1.43424 is understood to mean 1 one, 4 tenths, 3 hundredths, 4 thousandths, 2 ten-thousandths, and 4 hunderd-thousandths.

When you’re dealing with non-metric measurements like those used in the United States, you still have to work with cumbesome stacked numbers like 3 sixteenths of an inch and quater cups, and we’ll explore ways of handling them later For now we will only deal with decimal fractions because results of division will almost always be in terms of decimal fractions (we’ll call them “decimals” from now on.)

Keep in mind our goal, we’re trying to teach our Calcard how to calculate, so we will need a way to teach it to multiply and divide – we’re not even close yet, but we’re getting there.

Let’s look at how we divide.

I know of no way of dividing drectly. There are two practical method of division and both of those involve guessing, checking the guess, and then adjusting the answer accordingly. The most straight forward way of doing this is tedious; therefore, it’s often used as a division algorithm for computers as they need simple procedures (computers are not very smarrt) and they don’t mind tedium.

Since division is the inverse of multiplication, and you can multiply directly, you guess at what the quotient of the two numbers would be and then multiply the guess by the divisor to see how close you were. If the guess was too large, use a smaller number and try again. If it was too small, use a larger number and try again. By repeating that process and makng shrewd guesses, you should be able to get more and more accurate answers.

You should realize that most quotients are going to be fractions. For instance, when dividing by seven, every seventh larger dividend is going to divide evenly. All the others will yield fractional quotients. Some of the fractional quotients are going to have few decimal places but many will just keep right on going and, with some, you will never be able to find the perfectly accurate quotient by dividing. For instance, try dividing 10 by 3 and see what happens.

Now, let’s use the simple method to divide 263 by 35. Since, if you round both numbers to one significant digit you will get 300 and 30, 10 might be a good initial guess. The product would be 10 x 30 = 300. That’s way too large, so lets decrease the dividend to somewhere between 5 and 10 (The quotient seems to be closer to 10 than to 1.)

If we choose 7, then 7 x 35 = 245, which is too small, so let’s go with 8. 8 x 35 – 280, which is too large, so we need something between 7 and 8. Let’s go for the middle and try 7.5. (Notice that, if we’d just tried half way between 5 and 10 to start with, we would have settled on 7.5 instead of 7 and saved two steps. It’s always most effective to divide search intervals by half at each step.)

35 x 7.5 = 262.5. That.s too small so we need to increase our guess. The answer will be berween 7.5 and 7.6 so let’s half our search interval and try 7.55. Continuing in this way we have:

35 x 7.55 = 264.25…T0o big…..try 7.52.
35 x 7.52 = 263.2…..Too big…..try 7.51.
35 x 7.51 = 262.85…Too small…..try 7.515.
35 x 7.515 = 263.025…..Too big…..Try 7.512
35 x7.512 = 262.920……Too small…..Try 7.514.
35 x 7.514 = 262.99…..That’s pretty close. But how much further will we have to go before the process ends.

That’s the problem with this method; you can never tell until the process actually ends – and the process may nver end.

The second method, long division, is the one used by most of us when dividing with pencil and paper. It allows you to estimate the quotient a piece at a time so that you’re always working with easy estimates. This is how it works:

35[263.

By playing around with single digit quotients, it’s easy to see that 7 is the largest that will divide 263 by 35, so that’s our first digit. 35 x 7 = 245. The remainde is 18. Bring down the next digit, which is a zero and we have 180, so:

7
35[243
245
180

Now we figure out how many times 35 will go into the remainder 180. The answer is 5, so that’s our next digit. Put in a decimal point and we have:

7.5
35[243
245
180
175
50

Continuing this process of dividing the remainder by the divisor and coming up with a new remainder and bringing down the next digit in the dividend for two more steps, we get:

7.514
35[243
245
180
175
50
35
150
140
10

The remainder is 10, bring down the next digit and we have 100. 35 goes into 100 2 times so the answer is going to round off to 7.514, which is what we got with the other method (with a lot more effort.) If we keep dividing, we come up with 7.5142857 with a remainder of 5 and the process is still not complete. 7.514 is a really good estimate and we can cut our losses there with no pain.

The advantage of the second method (in addition to relative ease and speed) is that you don’t have to guess at each step. Each step requires an easy division problem that yield, at most, one more decimal.

Neither method will tell you how much further you will have to go before the process stops, but you will usually know how accurate you want your answer to be, so you will know when you can stop, satisfied with the precision or your answer.

Well, we now know how long dvision works and why it works, but I don’t see how that will bring us any closer to teaching Calcard how to multiply or divide.

We need some other tool that we haven’t been introduced to yet. There are two other basiv arithmetic operations we haven’t met yet and they may provide a key. If they don’t work, we have a problem. We’ll be out of ready options.

Addition is counting. Subtraction is the reverse of addition. Multiplication is repeated counting and diviision is the reverse of multiplication. When you repeat multiplication, you call it finding a power or expoentiation and there will be a reverse operation for that.

But that’s scary. Subtraction caused us to go beyond what we were familiar with in the natural numbers, and division was a lot more diffcult than subtraction. What in the world is going to happen now?

Here’s a teaser – your fears are well founded. The inverse of exponentiation is going to complicate the heck out of everything – it’s going to give us a really weird (almost incomprehensible) new number and it’s going to turn out to be really difficult, but it’s also going to give us a way to teach Calcard how to multiply and divide. In fact, it will allow Calcard to perform all the arithmetic operations including exponentiation and it;s inverse. But we have another small bit of business to clear up before we go on. How go we work with fractions in general and with decimals in particular?

FractionsIt’s easy to work with fractions if you keep in mind the basic rules of arithmetic operations. Let’s review them:

Addition is counting.
Subtraction is the reverse of adding.
You cannot add or subtract different kinds of things.
Multiplication is repeated addition.
Division is the reverse of multiplication.
To multiply or divide different kinds of things, you multiply or divide each kind of thing in the first term by each kind of thing in the second term and then add the results together.
To add numbers with different signs, sutract them and take the sign of the lrger number.
To add numbers with the same sign, add them and give the result the common sign.
To multiply of divide numbers with different signs, go ahead and multiply or divide and make the result negative.
To multiply or divide two numbers with the same sign, go ahead and do the multiplication or division and give the result a positive sign.

For fractions, multiplication and division is easier than addition and subtraction. Remember that the denominator names the place value in a fraxtion, so there is only one place value in each fraction to multiply. You simply multuply the numerators of two fractions, multiply the denominators, and stack the product of the numerators over the product of the denominators. For instance:

2/3 x 3/4 = (2 x 3)/(3 x 4) = 6/12

Since you can multiply or divide the numeraor and demoninator of a fraction by the same number without changing it’s value – that’s equivalent to multiplying or dividing by one since a fraction with the same numerator and denominator is equal to one.) this fraction can be simplified by dividing buth numbers by 6 to get 1/2.

Addition and subtraction are a bit more involved and it’s best from this point to consider subtraction as the addition of a negative and a positive number. You can add two fractions with the same denominator by simply adding the numerators and placing them over the common denominator. You can’t add two different knds of things, and fractions with different denominators are different kinds of things; but there is a way to convert one kind of fraction into another. You simply multiply the numerator and denominator by the same number. The simplest way to do that with two different kinds of fractions is to muleiply the numerator and deominator of each fraction by the denominator of the other fraction, add the two products of the numerators and place the result over the, now common, deominator. That will look like:

a/b + c/d = ((a x d) + (c x b))/(b x d)

For instance:

2/3 + 3/4 = ((2 x 4) + (3 x 3))/(3 x 4) =
(8 + 9)/12 = 17/12

That can be cleaned up some by realizing that 12/12 = 1 so that the result is 1 5/12.

There are often several ways of adding a particular pair of fractions to get the same result. For instance,you can add 1/2 and 1/4 by going through the whole rigamarole:

1/2 + 1/4 = ((1 x 4) + (1 x 2))/(2 x 4) =
(4 + 2)/8 = 6/8

and then clean up the result by dividing the numerator and the denominator by 2 to get 3/4, or you could simply realize that 1/2 is the same as 2/4 and add 2/4 + 1/4 to get 3/4.

Division, in the case of fractions, is quite literally the inverse of multiplication. To divide two fractions, you simply invert (flip upside-down) the divisor and multiply. For example:

3/4 ÷ 1/2 = 3/4 x 2/1 = 6/4 = 3/2 = 1 1/2

Working with decimals is easy; the difficult part is figuring out where to place the decimal point in the result, but there are easy rules and it’s always a good idea to estimate the answer and make sure the result is close.

To add decimals, simply line up the numbers so that the decimal points are in line. Therefore, for 1.065 +32.4 + 0.014, stack them like:

1.0650
32.4000
0.0014
33.4664

To multiply two decimals, say, 1.065 x 32.4, line them up flush to the right, like:

1.065
32.4

and multiply as usual, you get:

1.065
32.4
345060

the way you place the decimal place is count the number of decimal places in the numbers being multiplied – 1 in 32.4 and 3 in 1.065, 4 in all – and that’s the number of decimal places that should be in the answer:

34.5060

Since the number has to be close to 1 x 32 = 32, you can easily see that the answer has the correct number of decimal places.

When dividing, move the decimal place in the divisor to make it be a whole number and move the decimal place in the dividend the same number of places in the same direction. The decimal place in the result should be placed above the decimal place in the dividend. For instance:

21
15.3{325.4
306
194
153
41

Moving the decimal place to the right one place:

21.
153.{3254.
306
194
153
41

you would have the answer 21 with a remainder of 4.1. As a check, you can see that 15 will go into 300 20 times, so the answer has to be close to 20. 21 in, so you can feel confident that you have the decimal place in the right place.

Now we can move on to exponents.

ExponentsPowers are actually pretty simple. They are just repeated multiplication. 2 x 2 x 2 x 2 x 2 can be written as 2⁵ which is read “2 to the 5th power”. I know of no “long exponentiation” method like there is a long multiplication method. To find a power, you simply keep multiplying. It gets tedious after a while.

Fortunately, most applications of powers involve small powers. Area and volume calculations involve powers up to 4th powers – rarely beyond. Growth and decay functions usually involve small exponents. You run into them a lot in economics since money accounts generally grow (you hope) or decay. You also see them in ecology since you are often interested in population growth and decay.

The major exception to small exponents is place values. In our number system, as numbers get larger, left-most digits are larger and larger powers of 10. That’s still not difficult because, to get larger powers of 10, you just add zeros. If you’re working with computers, you may have to deal with place values that are powers of 2, 8, or 12, but less so now. When computers first came out, the only programming language was machine code, which was simply the binary code in which everything was stored. Codes were byte (8 bnary digit (bit)) or double byte (16 bit) numbers which meant that the programmer had to be familiar with powers of two up to 16 – they just memorized them. Later, assembly languages made things a little easier but the code was often represented as octal and hexadecimal numbers which used powers of 8 and 12, respectively. Again, programmers just memorized the place values.

Can exponents be negative or fractional? Well, yes they can, and that’s where it gets complicated. To see easily what happens with negative powers, consider what happens as you come to progressively smaller and smaller place values in our decimal system of numbers. You drop off zeros:

10³ = 1000
10² = 100
10¹ = 10 (any number to the first power is itself)
Intuitively (and we’ll see why later)
10⁰ = 1
Any number to the zeroth power is 1.
The next step carries us into the negative exxponents:
10^-1 = 1/10 = 0.1
10^<-2 = 1/10² = 1/100 = 0.01
10^-3 = 1/10³ = 1/1000 = 0.001
and so on

So a negative power is a fraction with one as the numerator and the denominator equal to the number to the same power, except positive.
5^-3 = 1/5³
75^-32 = 1/75³²

And, what does fractional exponents involve? To understand that, we’re going to have to consider what hapens when we multiply and divide exponents (because a fractional exponent is one exponent divided by another) and that’s when things really get weird.

But, first, let me make very clear what I mean by “factor”. If you have a series of terms multiplied together, each term is a factor.

Now, let’s look a little closer at the consequences of exponeniation. Specifically, let’s look at a string of twos multplied together:

2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

If we partition the series of factors into two groups, we can see that:

(2 x 2 x 2 x 2 x 2) x (2 x 2) = 2⁵ x 2 ²

Well, since the exponent of the result of the first expression (7) equals the sum of the exponents of the result of the second expression (2 + 5) you can see that:

2⁷ = 2^{5 + 2} = 2⁵ x 2²

So, as long as all the factors are the same, you can multiply two series of factors by adding their exponents.

Let’s stack a couple of repetaed products of 2 and see what happens.

2 x 2 x 2 x 2 x 2
2 x 2

Here, you are justified in canceling out any factor that appears both in the numerator and the denominator because, for instance, 2/2 = 1 and multiplying by one doesn’t change the products, so:

2 x 2 x 2 x 2 x 2
2 x 2

I think you can see that:

2⁵
2²

is the same as

2^{5 – 2} = 2³

And, so, as long as the factors are the same, you can divide two exponentiated terms by subtracting their exponents.

Now, I think you see that we have a way of showing why:

a¹ = a

and

a⁰ = 1

In the first case:

a^2-1 = a¹

is the same as

a x a
a

One a cancels out in the numerator and denominator leaving only a.

In the second case:

a^{1 – 1} = a⁰

is the same as

a
a

a/a = 1

What about a negative exponent?

a^{0 – 1} = a^-1

is the same as

a⁰/a¹

which is the same as

1/a

Now, it’s evident that a^{b x 1/b} = a¹ = a, so there is an inverse of powers just like there is an inverse of addition (subtraction) and multiplication (division). We give it the name “taking a root” and it’s the fractional exponent.

For instance, if you want to undo a square, you take the square root. the square root of a is the number that when multiplied by itself gives a as the product.

The square root of 4 is 2 because 2 x 2 = 4. The square root of 9 is 3 because 3 x 3 = 9. This extends to cube roots and fourth roots and so on.

Now we get into the complexities. There is a procedure for finding a square root that’s similar to long division. There are also “long division” methods and other methods for higher roots but the larger the root index gets, the more complicated the procedures get, and it’s usually just as easy to use an iterative method like the alternative method to long division. Let’s see how these procedures work.

In the case of the square root, you begin the iterative method by making an educated guess. Then you check the guess (by squaring it) and adjust your guess up or down according to whether the square is too large or too small. Let’s take the square root of 79.5 as an example.

The largest whole number with a square less than 79.5 is 8, so we can start there.

8² = 64

That’s too small so we adjust up and try 8.5.

8.5² = 72.25

Still too small so we try 8.75.

8.75² = 76.5625

That’s still too small, so let’s try 8.9.

8.9 x 8.9 = 79.21

Much closer, but let’s see how accurate we can get. Try 8.95.

8.95 x 8.95 = 80.1025

That over shot our target so let’s back up some. Try 8.93.

8.93² = 79.7449

Too large; try 8.92.

8,92² = 79.5664

If you’re not wore out yet from multiplying big numbers, you can continue (like I did) and you’ll see that:

8.915² = 79.477225
8.916² = 79.495056
8.9165² = 79.50397225

That is really close and I’m happy with it. As in division, you can never know how far away you are from the end of the iterative process and it’s possible that the process will never end.

The long division method for square roots is worth knowing. I’ll let you look up the procedures for higher roots – I’d never use them. We can use the same 79.5 and make it accurate to the same 4 decimal paces. You start by writing it like:

____
\|79.50 00 00 00

There will be a digit in the answer for each pair of digits in the square. Noticed that I’ve grouped the square into pairs of digits. If there had been an odd number of digits, sat 35351,I would have set the first digit apart like – 3 53 51.

Make a guess for the square rootof the first pair of digits. Again, 8 is the largest whole number that yields a square less than 78.5. Put that over the first pair of digits.

8
\|79.50 00 00 00

Place the square of 8 under the 79 and subtract it from 79.

8
\|79.50 00 00 00
64
1550

I dropped the next two digits down to the remainder. Notice also that I indented the whle thing over some. We’ll be doing some work over on the left side.

Next we place the square of the number on top (8) to the right of the first remainder, leaving room at the right for one digit. We’re going to be multiplying our divisor by the new digit and we want to place one more digit on the divisor which will bring it closest to the remainder. That digit will also be the next digit in the answer.

8.9
\|79.50 00 00 00
64
169|1550

Now, we multiply the divisor (168) by the new digit (9) and place it under the remainder. Subtract the product from the remainder to get the new remainder.

8.9
\|79.50 00 00 00
64
169|1550
1521
29 00

That will be the process we use over and over until we reach the accuracy we desire – place the new divisor to the left of the remainder, leaving room for one digit, figuring out which digit will bring it closest to the remainder when multiplied by the new digit, doing the multiplicarion and placing that digit into the approximate answer, and subtracting the product from the old remainder to get the new remainder. Finally, we bring down the next two digits from the number we’re finding the square root of.

8.91
\|79.50 00 00 00
64
169|1550
1521
1781|29 00
1781
111900

Two more times will get us the four decimal places in the answer we want.

8.9162
\|79.50 00 00 00
64
169|1550
1521
1781|29 00
1781
17826|111900
106956
178322| 484400
356644
127756

Our answer is 8.9172, and if you square that you will get 79.49862244. This is a more accurate answer than what we got with the iterative method. Of course, we will rarely require a square root to that level of accuracy so either the iterarive method or the long division method should serve well and, of the two, if you’re programming, you’ll want the iterative method.

To find higher roots, you can use the same iterative method. If the root is a multiple of 2 or 3, the solution is a lot simpler. For instance, to find the sixth root of 191102976, you can find the square root, which is an even 13824, and then find the cube root of that, which is 24. Alternatively, you could find the cube root and then the square – he order doesn’t matter.

We must now indulge in a little algebra in order to explore some of the consequences of exponents and deal with a small issue pertaining to the Calcard.

Algebra is a game in which you have numbers you know and numbers you don’t know and the objective is to find out what the “mystery” numbers are. The basic rule is that you can do any arithmetic operation to both sides of an equation – add, subtract, mutiply, divide the same quantity, square, find the square root, etc. – and it won’t change the unknowns.

Equations contain known quantities and unknown quantities. The unknown quantities are usually represented by letters. Equations are like balances that you can place the same weights on both pans and you won’t change the balance. The procedure of algebra requires that you keep the balance of both sides of the equation. A common strategy is to try to isolate the unknown on one side of the equation and move everything else to the other side. For instance, what is the value of x in:

5x – 32 = 3x + 9

You can get rid of the -32 on the left side by adding 32 to both sides so that the equation becomes:

5x = 3x + 41

To get all the x’s on the left side, we simply subtract 3x from both sides (remember, you can only add or subtract like things. x’s are like.)

2x = 41

But we’re not lookng for 2x, we’re looking for x, so we must divide both sides by 2 to get:

x = 20.5

Let me show you a trick with the square root of 2.

All the kinds of numbers we have run into so far have been such that they could be represented by fractions that have whole numbers in both numerators and denominatiors, the whole numbers having no common factor between them – that’s why we called them “rational numbers”. (All rational numbers can be reduced to simplest terms such that the numerator and denominator have no common factor.)

Lets try to represent the square root of two that way – let’s say as:

a/b = 2^1/2

We can square both sides, keep the balance of the equation, and come up with:

a²/b² = 2

Solving for a, we can isolate a² to the left side of the equation.

a² =2b²

We still don’t know what a is and we can’t get rid of that b to find out, but we do know that a must be an even number because 2 is a factor (if 2 is a factor of a², then it’s a factor of a.) Therefore, we can represent a by 2m where m is an integer and, squaring both sides:

a² = 4m²

Since a² = 2b²:

4m² = 2b²

and, dividing both sides by 2:

2m² = b²

Now we know that b is also an even number (has a factor of 2.)

Okay – that’s a problem. Can you spot it?

We assumed that the square root of 2 could be represented by a fraction, a/b, in which the numerator and denominator were whole numbers that do not have common factors. If a and b are both even, they share the factor 2.

What does that mean? It means that the square root of 2 cannot be represented as a fraction with two whole numbers without common factors and that it is not a rational number. If you try to determine the square root of 2 you’ll find out that you never reach the end – the decimal fraction never terminates – and unlike, say, 1/3, the decimal fraction never repeats. This is a new kind of number. Since it’s not a rational number, we call it an irrational number

So why do we even need 2^1/2?

There is a well known law of geometry called the Pythagorean Theorem that can be proved in many (many!) ways , and it’s fun to prove, so you might try it. It states that, if you have a right triangle (one with one angle measuring 90°) that has two short legs, a and b, and one long leg, c, that:

a² + b² = c²

That means that, if you know the length of two legs, you can always find the other. Pythagoras’ theory finds many applications in surveying and geometry.

When you’re working with areas (areas of carpet, areas of wall paper, etc.) the units you use is the unit square – a square with each side measuring 1 unit – so, of course, the unit square is one of the most important entities in geometry and it behooves us to know as much as we can about it.
One thing we should know is the length of the dagonal – the line drawn from opposing corners of the unit square. Well, what is the length of the diagonal of the unit square?

The sides of the unit square and the diagonal form a right triangle with the diagonal as the long side, each side is 1 unit, so the diagonal is the square root of 1² + 1², or the square root of 2. Actually, the square root of two turns up all over mathematics so it’s good to know how to approximate it (since we can’t determine it exactly).

What’s more, 2^1/2 isn’t the only irrational number, Several irrational square roots are important as are two of the most important constants of mathematics, pi and e. So we should at least nail down the positions of the square roots of 2 and 3, pi and e on the scales of Calcard so we won’t have to figure out the values every time we want to use them.

The square roots of 2 and 3 are easy now that you know the way to find square roots and you only need them to two decimal places (you won’t be able to locate a point on a scale with any more precision than that.

Alternately, for the square root of 2, you can draw (accurately) a unit square and measure the diagonal.

I find that:

2^1/2 = 1.42
3^1/2 = 1.73

These are approximations, but they’re pretty close and usable for most purposes.

The value of pi is somewhat more difficult and the value of e is a bear. Let’s look at them.

Pi arises when workng with circles. The area of a circle is pi times the radius (distance from the center to the circle) squared. The circumference (distance around the circle) is equal to pi times the diameter (distance from one side of the circle to the other through the center). Pi is devined fromantiquity as the ratio of the circumference of the circle to the diameter.

Pi has been calculated with accuracies of who knows how many digits (they know it never ends – somebody just has way too much time on their hands) and there are interminable tables of digits of pi. These tables are usually used as sources of random numbers. You start at any place at the table and read horizontally, vertically, diagonally, any direction you wish and read off the number of digits you want. Since pi’s digits don’t repeat systematically, they might as well be random.

The geometry of the circle gives us ways of conveniently calculating an approximation of pi. I have calculated pi fairly accurately by cutting out a circle of analyical aluminum foil (which has a very consistent thickness and density). measuring the thickness and diameter, dissolving it in acid and titrating the resulting solution to determine how much aluminum chloride was formed and then figuring back to determine the volume and area of the aluminum disc and then the value of pi.

You could similarly weight a disc of analytical aluminum foil with known density and radius and calculate the area from the weight and density. That really requires laboratory accuracy and we’re not in an advanced laboratory, so we need something a little more available.

I drew a circle 32 units in diameter on a sheet of graph paper and counted the squares, hoping that those partial squares on the rim would average out. I counted 778 squares. Since the radius is 16 units and the square of the radius is 256, that puts pi at 778/256, or 3.04. Well, if you remember any constant from math class, it will be pi = 3.1415, so 3.04 isn’t nearly good enough for the Calcard. We need it accurate to 2 decimal places.

My next experiment was much better. After dispatching a cylindrical can full of creme-filled tubular wafer cookies (chocolate, I think they were), I wound a thread 10 time around the can (10 times to provide replication and increase the accuracy) and measured the length of the thread.
It was 102 inches long making the circumference of the can 102/10 or 10.2 inches around. The diameter of the can, measured by a caliper (inexpnsive plastic calipers that fit quite well into an Excursions kit are available), was 3.25 inches. Since pi equals the circumference divided by the diameter, the result was 3.1453. Rounded off to 3.14, that’s a number I could be happy with and a survivalistic method for approximating pi that I could keep.

Now e was a problem. It’s important because it’s the base of natural logarithms which we will soon become quite intimate with as we continue the construction of the Calcard, but there is no easy, memorable way to approximate e. It’s usually calculated using a long, complicated series of algebraic fractions. We need something that can be remembered without too much trouble.

A little calculus, now. If you draw a graph of an equation, the slope of the line or curve of that equation tangent to a point on that line or curve is the derivative of that equation at that point. There are ways of calculating derivatives that I won’t go into here. It’s not that hard and there are sites all over the Internet to explain the process quite well. But it doesn’t help us with the Calcard. The point is that, if you graph the equation y = e^x and check the slope of any point along the resulting curve, you will find that it is e^x. That simpifies a lot of calculations with exponential quantities, and that’s why e is so important.

I found that e is the limit, as n gets larger and larger (increases to infinity) of (1 + 1/n)ⁿ. I had no idea how far I could take that (as n gets larger, (1 + 1/n)ⁿ gets more and more difficult to calculate) or how accurate the result would be.

My patience held out to calculate n = 1/4, 1/2, 1, 2, 3, 4, 5, and 10. The results were (to two decimal places):

(1 + 1/.25)^1/4 = 1.5
(1 + 1/.5)^1/2 = 1.73
(1 + 1)¹ = 2
(1 + 1/2)² – 2.25
(1 + 1/3)³ = 2.35
(1 + 1/4)⁴ = 2.44
(1 + 1/5)⁵ = 2.49
(1 + 1/10)¹⁰ = 2.59

Then, I took a sheet of graph paper and marked off the units along the bottom long edge starting at the left end, 4, 8, 12, 16, and so on all the way to the right end. Up thee left edge from the bottom, I labled the units 0, .1, .2, .3, and so on to the top. That gave me an x-axis (or, n-axis) from 0 to 164, and a y-axis from 0 to 3.0. I hoped that would give me enough room to get a reasonably good idea of the shape of the graph. Then I graphed the points I had. No surprise, I had the beginning of an exponential curve. But it was only the beginning. I couldn’t tell where the asymptote would be (an asymptote is a point (or value) that a graph (or value of an equation) approaches but never reaches)

I knew the curve would be symmetric around some point so I tried to estimate about where I would be able to fold the paper so the curve would overlay itself. My best guess was at n = 8 so I folded the chart at a 45° angle to the x axis so the fold would run through the n=8 point and I traced the rest of the curve through the sheet. Opening it back up, I continued the graph by tracing back to the graph side of the sheet. I had an exponential curve all the way out to n = 108.

Next I drew some tangent lines along the upper part of the curve to see about where they were pointing and estimated that they would flatten out around 2.72. Looking up e, I found to my surprise, that e = 2.71828…

So, if you can remember (1 + 1/n)ⁿ, you can , with a huge amount of work and patience, figure out a good approximation for e.

Now, I mentioned logarithms – what is a logarithm?

Logarithms and Log ScalesMost people who study logarithms in introductory algebra courses usually end up hating the enigmatic things. Regardless of their obvious usefulness in complex calculations, most introductory level stuents never get a grasp on what they are, so they remain nebulous and hard to grasp mentally. They remain nebulous because the course never raises the one pivotal statement that makes everything fall into place – the part of the jigsaw puzzle that falls into place and makes everyone gasp, “Oh, that’s what it is.” And that statement is:

Logarithms are exponents.

Isn’t that simple?

2⁶ = 64, is the same as saying:
6 is the base-2 logarithm of 64.

10⁴=10000, is the same as saying:
4 is the base-10 logarithm of 10000.

In general, if:

a^b = c

then:

Log_a c = b

Or, to put it another way:

The logarithm to the base a of c is equal to b.

It’s easiest for me to parse this out by remembering that the base-10 logarithm of a power of 10 is the number of zeros in the number. And since a logarithm is an exponent, logarithms act exactly like exponents.

Look at the following pairs of statements:

10^a x 10^b = 10^{a + b}
Log₁₀(c x d) = Log₁₀c + Log₁₀d

10^a/10^b = 10^{a – b}
Log₁₀(c/d) = Log₁₀c – Log₁₀d

(10^a)^b = 10^{a x b}
Log₁₀(c^d) = d x Log₁₀c

In the last case, d can also be negative or fractional.

I think you can see from this that you can multiply two numbers by adding their logarithms or divide two numbers by subtracting their logarithms. You can also calculate just about any exponent by multiplying the logarithm by the value of the exponent, whether positive, negative, or fraction.

For instance, you could find the sixth root of a number by finding it’s logarithm and dividing iit by 6, and then finding what number the resulting logarithm represents.

Now, if a is the logarithm of b for any base, then b is called the antilogarithm of a.

So now we have a way to multiply, divide and find exponential values using a sliding scale. All we need are two scales of logarithmic values labeled, not with the logarithmic values but with the antilogs.. That way, you don’t even have to worry about translating to and from logarithms.

In order to create the scales, we’ll need a way to figure logarithms. There are published tables of logarithms to rediculous precision but I’m assuming that we don’t have one of those available when we build the scales – we’re doing this “from scratch”.

We will need two logarithmic scales to base 10 labled C and D (because that’s the labels traditionally given logarithmic scales on slide rules.) We’ll also need two inverse logarithmic scales labeled CI and DI (remember the LI scales). It will also be useful to have scales that “fold” at e labeled CF and DF. It will also be nice to have the same scales for natural logarithms.

A scale consisting of two half-sized logarithm scales labeled B will allow us to calculate squares and square roots directly and one having three smaller logarithm scales labeled K will allow us to calculate cubes and cube roots.

All these scales should be drawn corresponding to the L scales. That will allow us to read logarithms directly from our slide rule. From there, if we can figure out the general rule for constructing scales for special functions such as trigonometric functions and hyperbolic functions, we will know all there is to know about slide rules and, indeed, analog calculating machines in general since they all work using logarithms.

Before we get back to building, though, there is one more small issue I want to cover concerning the consequences of exponents.

ComplexitiesI want to play a little more of the algebra game now. We saw how easy it is to solve equations without exponents – we just keep the two sides of the equation balanced while isolating the “mystery” number on one side (the “mystery” number is called a “variable”, by the way.)

Let’s see what happens when we introduce an exponent, say a squared variable. Let’s try 4x² – 4x – 15 =0

We can move the -15 to the other side of the equation:

4x² – 4x = 15

but, then, what do we do with the two fours and how do we get the x² to turn into an x so that all we have on the left side of the equation is x?

In this case, it turns out to be rather simple but I have to tell you how to multiply algebraic equations first. To multiply any two things composed of different components, you multiply each componet in the first thing be each component of the second thing and add them. For instance, when you mutiply 32 and 14 (which are both composed of both ones and tens) you multiply each digit in 23 by each digit in 14 and add all the products together. Therefore:

4 x 2 = 8
4 X 30 = 120
10 x 2 = 20
10 X 30 = 300
23 X 14 = 8 + 120 + 20 + 300 = 448

So, to multiply (3x + 2) by (5x – 7) you would multiply everything in (3x +2) by each thing in (5x – 7) and add the products all together:

3x x 5x = 15x²
2 x 5x = 10x
3x x -7 = -21x
2 x -7 = -14

Add them together to get:

15x² + 10x – 21x -14

You can simplify that by adding the like terms to obtain:

15x² – 11x – 14

Once you get used to it, you can run the process in reverse to find the factors of some equations with squared variable (they’re called “quadratic equations”.) For instance, we can factor 4x² – 4x – 15 =0.

Two numbers that will multiply to give 4x² are 2x and 2x, so let’s try them in the first positions of our two factors:

(2z + __) x (2x + __) = 4x² – 4x – 15

Now we need to fill the blanks with numbers that, when multiplied, will give -15 and when multiplied by 2x and after adding the products will give -4x. A little trial and error will assure us that the two numbers should be 3 and -5, so the factors we’re looking for are:

(2x + 3) x (2x – 5)

To make this statement equal to 0 and, therefore, solve for x, we only have to make each factor equal to x (because any number multipied by 0 equals 0).

To make (2x + 3) = 0, x must be -3/2.
To make (2x – 5) = 0, x must be 5/2.

If you place either -3/2 or 5/2 back into the original equation for x, you’ll see that either work. Quadratic equations always have two solutions (the solutions may be equal).

Now, what if the equation was:

4x² + 15 =0

You’ll find that you can’t factor this equation in the same way . You could move the 15 to the right side of the equation and then take the square root of both sides:

4x² = -15
2x = -15^1/2

Wait, how can you have the square root of a negative number. Any number multiplied by itself, whether it’s positive or negative, is going to give a positive square. But one solution to this equation is most certainly (-15)^1/2/2.

This is the problem that exponents cause. With many equations, you have to resort to square roots of negative numbers to solve them. So mathematicians invented square roots that have no associated squares called complex numbers (an example is 5 + 3 x (-2)^1/2)and the square root of the negative number is called an imaginary number.

Mathematicians have continued to run into such problems and have continued to invent new kinds of numbers to get them out of the problem, but this is the last new kind of number I will discuss.

It’s time to teach Calcard how to multiply and divide.

Logarithmic scalesSo now we know what we need to do to make Calcard be able to multiply and divide. We need scales marked at the positions corresponding to logarithmic values corresponding to the L scale, but marked with the antilog values. Specifically, we’ll need the logarithmic values for 1, 1.1, 1.2, 1.3, 1.4, 1.5, 1/6, 1.7, 1.8, 1.9. 2, 3, 4, 5, 6, 7, 8, 9, and 10.

Starting with the common logarithms (logarithms to base 10) to form the C and D scales, we already know the logarithms of 1 (0, because anything to the 0 power is 1), 10 (1, because 10 to the first power is 10) and it’s a good idea to remember the common logarithm for 2 which is 0.020103. How do we calculate the other logarithms?

The bad news is that there is no way to calculate logarithms that 99.99% of the population would be able to remember. The good news is that there are tricks. Since the logarithm of a number is the sum of the logarithms of the factors of that number, we can immediately calculate the logarithms of the powers of 2 – 4 and 8.

Log(4) = Log(2) + Log(2) = 0.60206
Log(7) = Log(2) + Log(2) + Log(2) = 0.90309

We only need accuracy to a couple of decimals, so we should be able to graph this fnction and read the other logarithms from the graph. I used a graph with the numbers along the x-axis in two-tenths from 1 to 9.4 (that’s all I could fit on the long edge of my sheet)., and the logarithms on the y-axis from 0 to 1 with every 3 marks being a tenth. Reading off the other values along the resulting curve I obtained the following values (the numbers in parentheses are actual logarithms I looked up in a table of logarithms):

Log(1) = 0 (0.00)
Log(1.1) = 0.05 (0.04)
Log(1.2) = 0.08 (0.08)
Log(1.3) = 0.14 (0.11)
Log(1.4) = 0.17 (0.17)
Log(1.5) = 0.18 (0.19)
Log(1.6) = 0.21 (0.20)
Log(1.7) = 0.25 (0.23)
Log(1.8) = 0.27 (0.25)
Log(1.9) = 0.28 (0.28)
Log(2) = 0.30 (0.30)
Log(3) = 0.46 (0.48)
Log(4) = 0.60 (0.60)
Log(5) = 0.71 (0.70)
Log(6) = 0.78 (0.78)
Log(7) = 0.82 (0.84)
Log(8) = 0.90 (0.90)
Log(9) = 0.95 (0.95)
Log(10) = 1 (1.00)

I’m frankly surprised at how close these figures are. There’s only one figure off by more than two-hundredths and I will bet I notice that and can correct it when I start building the C scale.

You can easily determine other logarithmic values in this range. for instance, You can easily determine the logrithmic value of 2.2 since you know those of 1.1 and 2. The logarithm of 2.2 is the sum of the logarithms for 1.1 and 2. Therefor, you can find the logarithms for most of the tenths between 1 and 10. The rest, you can estimate once you have the calculated values placed on the scale.

I only need to construct one log scale – the others are copies of or modified copies of the same scale.

Before we get into constructing this scale, let’s get how we will assmeble the end product firmly in our minds. There will actually be a couple of designs yeu can use (or more if you’re really imaginitive.) You can put the stable cards together in an accordion shape and slide the slider cards between them, or you can use a star design and connect all the stable cards together along one edge. If you use the former design, you can probably get away with placing scales on each of the long edges of the stable cards, front and back, for 4 scales per stable card. If you use the other scheme, you will only be able to place scales along one edge, the one that won’t be joined to the other cards. Take your pick. I’m using the fanfold design although the star design may be more portable.

Now, the log scales.

Work from a sheet of paper to design the scales. Use a pencil to costruct the template. The scales must be the same length as the L scales you have already constructed. The two log scales will be labeled C and D, because that is the traditional labels for the log scales on slide rules. The A scale is for squares and square roots. B isn’t used.

1. Use the edge of your L scale to draw a line the length of the card (and alittle around the corners for registration). Place end marks on the paper scale at 0 and 10 on the L scale. You can label these 1 and 10 (because the logarthm of 1 is 0 and that of 10 is one – this scale will be scaled up one factor from the L scale.

2. Find the logarithm of 2 (0.3 – since you’re scaling up, call it 3) on the L scale and put a corresponding mark on the paper scale. Label it 2 (Not .3 or 3. You don’t want the C and D scales labeled with the logarithms but with the antilogarithms!).

3. Continue for each of the logarithmic values determined above and any other you determine. Remember to label the marks with the antilogarithmic values.

4. once you get all the other values in place, you can estimate where the logarithms of the square roots of 2 and 3, pi, and e should go.

5. The marks on the logarithmic scale should get closer and closer to each other as to go from the left end of the scale to the right and they should do so smoothly. You may need to do some clean up work on the scale to make this happen (because of error in the estimation process.)

When you have the paper template finished, you can use it to transfer a log scale to a stable card edge (remember that the stable cards have to be a little shorter than the slider cards) and a slider card edge. Label the log scale on the stable card as C and the one on the slider card as D.

Now, we’re ready to try the log scales out..

To review: muliplying two numbers is the same as adding their logarithms and dividing two numbers is the same as subtracting their logarithms. So, you multiply using two logarithmic scales in exactly the same way that you would add using two linear scales.

To multiply 2 x 3 I place the 1 on on my D scale over the 2 on my C scale, then I look under the 3 on my D scale to see that 2 x 3 = 6. Keeping the two cards togethre in the sme configuration I can also look under the other numbers on the D scale, up to 5, to see that:

2 x 2 = 4
2 x 4 = 8, and
2 x 5 = 10

None of the answers from the Calcard are off by more than 0.1. I’m okay with that.for right now. To determine the accuracy of the other half of the scales, I can divide. I place 2 on the D scale over 1 on the C scale and look across the numbers on the D scale. They should all orient over their halves on the C scale. Sure enough:

3/2 = 1.5
4/2 = 2
5/2 = 2.5
6/2 = 3
7/2 = 3.5
8/2 = 4
9/2 = 4.5
and 10/2 = 5

Playing around in this fashion convinces me that the whole nuubers and their halves are portrayed farly accurately on the log scales. Now I try some of the tenth marks (the actual calculated values are given in parentheses):

2.4 x 3.2 = 7.5 (7.68)
5.7 x 8.3 = 48 (47.31)
3.1 x 5.6 = 16.3 (17.36)

Now, I’m sorta surprised at even this accuracy – I expected the results to be off more because the extra steps required in calculating the fractional logarithms all introduced error – round off error, if nothing else. But the first figure above is off by more than 0.1 and that is troublesome.

Nevertheless, there is a way of refining the scales and I’m going to take that expedient. Since the unit and half markings are about where they should be, I will redraw new scales using the old ones and include only the unit and half markings. Then I will use those to figure out where the other tenth markings should go by mutlipying and dividing with the new cards. Thus, if I place 1 on the new D scale over 5 on the C scale, 1.1 on the D scale is over where 6.1 should be on the C scale. I will place a mark there and call it 6.1.

That’s going to take a while so I’ll get back to you as soon as I have that done.

Okay, let’s uses the same problems above to see how my new and improved C and D scales perform.

2.4 x 3.2 = 7.5 (7.68)
5.7 x 8.3 = 47.2 (47.31)
3.1 x 5.6 = 17 (17.36)

Oh yes. That is quite acceptible.

Now, let’s review the ways the C and D scales can be used:

1) The direct method is to set the left index (1) of the D scale on the slider over the first number to be multipliied, and looking under the second number on the D scale to get the answer on the C scale. In that way, you add lgarithmic lengths to get the sum of two logarithms.

2) If the result is off the scale (In other words, if you try multiplying 5.7 x 8.3, the result is going to be somewhere to the right of the right end of the C scale), you can set the right index of the D scale (10) over one of the two numbers to be multiplied on the C scale, and then looking under the other number on the D scale to see the result on the C scale. Remember that the result is off the scale using the direct method simply because the result is larger than 10 and place the decimal point accordingly.

What you are doing, in effect, by placing 10 over one of the number to be multiplied, is scaling that nimber down by an order of 10, In the case above, if you place the right index of the D scale over 5.7 to perform the multiplication, you are, in effect, multiiplying 8.3 by 0.57 to get 4.72. You must remeber to multiply the result by 10 to get the correct result.

There are other ways to get off-the-scale results using other scales. One such set of scales is the double-decade A and B scales. These scales are identical, the A on the stable card and the B on a slider. They consist of two logarithmic scales, half length and set end to end. I constructed two for my Calcard using the following steps (I eliminated the necessity of much of the cosntructions used in creating the linear scales by using grid paper (quad paper or quadrille paper) to determine the positions of the parallels):

1. Using the C or D scale from the Calcard, copy it onto a line of a sheet of grid paper. Draw a straight line the same length as the log scale elsewhere.
2. To get the size of a half length line, bisect the straight line by setting the point of a compass at one end and drawing an arc across the line (the arc must be further than half the length of the line from the end where the point of the compass is). keeping the setting of the compass, transfer the point to the other end of the line and strike off an arc so that it intsersects the other arc in two places. Draw a line between the two places where the arcs meet. The point where that line meets the original straight line bisects it into two lines of equal length. Use this bsection to determine the length of the next line by setting the compass to one end and the midpoint.
3. Transfer the half length down, setting the point of the compass at the right end of the copy of the log scale and strike an arc so as to cross a parallel line under it.
4. Draw another copy of the log scale begining at where the arc crosses the parallel and drawing it back to the right. Make sure that it’s oriented in the smake direction as the first log scale and not reversed.
5. Draw parallels from each mark along the first log scale to the matching mrk on the secod log scale.
6. Where these parallel lines cross the half length line, place the tick marks and label them with the same values on the full sized log scales.
7. Using the C or D scale on Calcard, determine where the left index of the A and B scales should be on an edge of a stable and a slider card. Starting at this point, copy the half legth log scale to the edge of the card. At the right index of that line, start a new half length log scale.The two hal length scales should fit onto the edge and the left and right indexes should coincide with the left and right indexes of the C or D scale set against it.

What you end up with are two double logarithm scales.

They function exactly like the C and D scales. If you set the left index of the B scale of the sliider card over 5.7 on the A scale on the stable card, and look under 8.3 on the B scale you will be between 4.5 and 5, the result of 5.7 x 4.5.

Notice that the unit marks on the A and B scales tend to crowd together a lot worse than on the C and D scales, so for accuracy it is best to use the C and D scales for multiplication.

The A and B scales are intended for another purpose. Set the A (B) scale against The D (C) scale so that their indices are aligned. Look across the numbers on the logarithm scale to the numbers on the double logarithm scale:

2 – 4
3 – 9
5 – 25
8 – 64
2.5 – 6.2 (should be 6.25)

Yes, the numbers n the A scale are the corresponding squares of the numbers on the D scale, and, consequently, the numbers on the D scale are the square roots of the corresponding numbers on the A scale. This is the real utility of the A and B scales.

It’s fairly obvious how this works. The positions on the A scale are half the distance from the left index than the positions on the D scale with the same lables. The A scale multiplied the logariths on the D scale by two, which is the same thing as squaring the antiogarithms.

Notice that, since you can multilyby adding diistances on the log scales, you can also divide by subtracting differences.

For instance, to divide 4 by 2, set the 4 on the D scale over the 2 on the C scale and look above the left index on the C scale. To divide 3,7 by 1.2. set 3.7 on the D scale over 1.2 on the C scale and look over the left index of the C scale. I get 3.1. The acrual quotient is approximately 3.08, so it’s close enough for government work.

If we can find squares and square roots by uing double log scales, we can find cubes and cube roots by using a triple log scale and that is exactly what the K scale is. I will construct one of those next in the same way that I did the double log scales, using a third length instead of a half length line in the construction.

Since the reduction to get a third size scale proceeds in exactly the same way as for the A and B scales, I won’t repeat that. The construction for the length of the third sized scale is something else, though.

There are actually several ways to trisect a line and you can look them all up on the Internet if you wish. I chose a proceedure that requires three circles and two lines. All of them require lots of room so you need to get a sheet of poster paper to do the construction. Transfer the result to a scratch card or a strip cut off one of the sliders.

1. Draw a lne the length of a scale in the center of the poster paper.

2. Set the point of a compass at one end of the line and the drawing point at the other end and draw a circle the radius of the line. Keeping the compass setting, move the point to the other end of the line and draw another circle of the same radius with the other end as the center.

3. Keeping the compass at the same setting, move the point down to where the two circles intersect below the line and draw the third circle from there. You should notice that the line forms a chord across the top of this last circle.

4. Now, using a straightedge, draw a straight line from the right end of the original line down through the point where the upper two circles intersect to the lower edge of the bottom circle. This will be a diameter of the lower cicle that passes through its center and the right end of the scale-length line.

5. This last line will likely require a long straightedge. Draw a line from the lower end of the last lne (where it intersects the lower edge of the lower circle) up to the top point where the two upper circles intersect.

6. This last line crosses the scale-length line a third of its distance from the left end. That distance should fit three times exactly onto the scale-length line (It worked for me.)

Using a scratch piece of card, transfer this length down between two parallel copies of a C or D scale (as you did for the A/B scales to complete the reduction. You will probably only be able to locate the major units and half units – tenths are going to be so crowded as to be very impractical for this project. You may be abe to get them between the 1 and 3 matks if you’re very good.

Once you have a third size log scale, transfer it three times onto an edge of a stable card of Calcard.

I’m sorta surprised at the accuracy of my results. I set the K scale over my C scale and I see that the cube of 2 is 8 (which is right on the mark), 3 is about 27 (again, correct), 4 is about 62 (correct), 5 is about 135 (it’s actually 125), 6 is about 226 (216), 7 is close to 350 (343), 8 is near 500 (512), 9 is a little more than 700 (729), and 10 is 1000 (It would be surprising if this were not correct since the left and right ends were set from the left and fight ends of the C scale.).

Next, I will construct a pair of folded scales, CF and DF. These are useful in two instances. If you know that a result is going to be close to the right end of a C or D scale but you don’t know if it will be less than 10 or off the scale, you can use a folded scale to multiply. Also, a log scale that begins at pi is convenient for all those engineering and calculus formulas that involve pi. The CF scale will be on a slider directly beneath the C scale and the DF scale will be on a stable card. Further, the DF scale should be on the stable card facing the stable card with the D scale.

Use an edge of a scratch card to construct the CF scale. Mark off the left and right ends using the D scale. Locate the place of pi on the D scale (you might want to place a mark in red there and label it with a Greek “pi” for continued reference.). Label the left end of the scratch CF scale “pi”, set it to the mark for “pi” on the D scale and copy the marks from pi on the D scale to the right end starting at the left end of the CF scale. The 10 mark on the D scale will end up somewhere in the middle of the CF scale. Lable it 1 on the CF scale, move the 1 on the D scale over that point and continue transferring marks from the D scale to the CF scale. The right end mark on the CF scale should coincide with the pi mark on the D scale. You can label it “pi” also.

You can now use this scratch scale to transfer marks to a line just below the C scale on the slider card and to an edge of a full sized, stable card. A multiplication using the folded scales require both forlded scales in addition to the C and D scales. Let’s multiply 2 x 8 to demonstrate how it’s done.

Set the 2 on the D scale over the left index of the C scale. Now locate the 8 on the CF scale and look straight below on the 8 to the DF scale. It should be close to 16. On my Calcard, it’s more like 15.5, so there’s a little error.

Next, it will be useful to have some inverted scales, which are simply the C, D, and DF scales backward. They will be called the CI, DI, and DIF scales. They are also used to simplify calculations that go off the scale using C and D.

It’s easy eough to produce an inverted scale; simply use the strips cut off the shorter slifer cards to transfer “backward versions of C, D, and DF scales to blank edges of slider, stable and stable cards, respecively. Label them CI, DI, and DIF.

The primary characterostic of the inverted scales is that they display the inverse values of values on the C, D, and DF scales. For instance,if you place the DI scale over the D scale, lining up the end ponts, the inverses line up with each other. If the DI scale runs from 10 on the left to 1 on the right, an the C scale runs from 0.1 o the right to 1 on the left, then 0.5 is directly beneath the 2 on the DI scale, 0.35 (should be 0.33) beneath 3, 0.2 beneath 5, 0.29 is directly beneath 3.5, and so forth

You can also use the inverse scales to perform multiplications that go off scale when using the C and D scales together. For instance, if you try to mulpiply 7 by 2 using the C and D scales, the answer is off scale to the right. If, instead, you line up the 7 on the DI scale above the 2 on the C scale, you can read the result on the C scale beneath the left index of the DI scale.

Scales can also be constructed for calculating the values of special functions such as logarithmic and hyperbolic functions. Lets try a scale for sines and cosines – only one new scale is required since sines and cosines are inverses of each other. Since sines and cosines range between 0 and 1, all we have to do is make a scale on a stable card which has the values of angles dierctly opposite sine (and cosine) values on the C scale. (We could use the L scale instead of the C scale but, using the C scale allows us to perform multiplications with the results of our calculations.) We can place the sines in back ink and the cosines in red ink.

Now, the problem is to calculate the sines and cosines without a calculator. That can be done geometrically, but we need to review a little about trigonometry.

Trigonometry and trigonometric scales

Now days, trigonometric functions tend to be defined in terms of wrapping functions – that is, what happens to angles and arcs, and such as a radius rotates around the center of a circle, but until very recently, they were defined in term of the ratios of the lengths of sides of right triangles. The six most common trigonometric functions (there are actually more, but the others are rarely used and are normally ignored) are defined as follows.

A right triangle is defined as a three sided figure (meaning that there are also three angles) in which one angle is a right triangle. If you select either of the other angles (call it’s measure in degrees x), you can define the longest leg of the triangle as the hypotenuse (h), the other leg connected to the selected angle as the adjacent side (a), and the remaining leg as the opposite side (o). The trigonometric functions are the values of the following ratios:

Sine x = sin x = o/h
Cosine x = cos x = a/h
Tangent x = tan x = o/a
Cotangent x = cot x = a/o
Secant x – sec x = h/a
Cosecant x = csc x = h/o

Well, it’s pretty clear that a geometrical way of calculating these functions is immediately available. Simply draw the triangle, measure the sides, and do the division. When working with sines and cosines, there is even a very convenient fact that we only have to determine values between 0° and 90°.

Visualize what happens as the hypotenuse of a right triangle sweeps from the adjacent side up to vertical. First it lies along the adjacent side and has the same length and the opposite side doesn’t exist, so the sine is 0/h or 0 and the cosine is h/h or 1. As it swings up the sine gets larger and the cosine gets smaller until at 90°, the hypotenuse is the same length as the opposite side and the sine is 1 and the cosine is 0. As the hypotenuse continues to rotate around the selected angle, the values cycle. That means that we only need values on our scale for sines and cosines between 0 and 1 and we can include both on the same scale in different colors.

The problem is, without some way to measure angles, how do we draw acceptably precise angles – preferably of angles between 0° and 90° in increments of 10°? It’s easy enough to bisect an angle and, if you use certain primitive, nonmetric equipment other than a straightedge and compass, you can even trisect an angle. But how do you dissect an angle?

Here’s my plan of attack.

First, draw a 90° angle. I’m using graph paper, so I can draw the angle along the gridlines. I’ll be wanting to make a protractor for use when I start surveying, so I want an area the size of an index card. To get the most area, I draw the base line along one long side of the graph paper and position the perpendicular about halfway from the top to the bottom of the page.

Without graph paper, it’s easy enough to draw a perpendicular. Just strike off two arcs of a circle using a compass on the line. From the arcs, use the compass to draw two circles of the same diameter to intersect above and below the line. Where the two circles intersect, draw a straight line to the center of the circle on the baseline. It will intersect the base line at a 90° angle.

2. Bisect the 90° angle to form two 45° angles. To bisect an angle, place the point of the compass at the vertex of the angle. Strike off an arc on each side of the angle equal distances from the vertex. Placing the point of the compass at each arc in turn, strike off two more arcs to intersect between the sides of the angle. Draw a line from the intersection of the two arcs to the vertex of the angle. This line will neatly divide the angle in half.

3. Trisect the lower 45° angle to form a 15° and a 30° angle. To trisect an angle, you need a marked straight edge – a neusis. The marked distance (from one end to the mark) should be the radius of a circle drawn with the 90° angle at it’s center. Make the circle the same size or larger than the semicircle of your protractor on an index card will be. I use one of the paper strips cut from the edge of one of the stable index cards as a neusis.

Place the end of the neusis on the baseline outside the circle and the mark where the 45° angle intersects the circle. The other point where the neusis and the circle meet is where the 15° angle crosses the circle. Mark it and draw a line from there to the center of the circle, which is also the vertex of the 90° angle.

You’re going to find that this method doesn’t work for angles less than 45°. There’s a more involed method that does work. First, you’re going to need a longer baseline – I used a second sheet of graph paper to extend the baseline to the left. You can use a piece of tape or line the two sheets up against a frame.

Next, you’re going to need a longer neusis – a foot ruler or long strip that you can mark with the radius of the circle.

Here’s how you do it. Anchor the ruler at the point where the 45° angle (and later, the 15° angle) crosses the circle. At the other end of the neusis, where your marks are, place the farthest mark on the baseline and move the neusis, keeping one end on the first mark on the baseline and the other end on the pace where the angle crosses the circle, until the other mark is on the circle opposite the angle. Draw a line along the strip from the baseline across the circe. The angle where the baseline meets the neusis is one third of the angle you’re trisecting.

Next, transfer the new angle so that it’s vertex is at the center of your circle. The process is described above.

4. Now trisect the 15° angle in the same way to get a 5° angle.

5. Set the compass to the distance marked off on the circle of the 5° angle and make 5° marks from the baseline up to the perpendicular around the circle. You should end up with 18 marks from the baseline to the perpendicular. The third mark (including the baseline) should coincide with the 15° angle. The 9th should coincide with the 45° angle. And the 19th should coincide with the 90° angle.

I suggest that you make 15° marks first from the vaseline to the 90° angle and use the 45° and 90° angles to correct slight inaccuracies in your drafting. Then make the 5° marks using the 15° marks to correct your work.

6. Draw lines from the center of the circle to each of the 5° marks on the circle.

7. Once you have the circle marked off with 5° marks from the base line to the perpendicular, you can make the protractor. Cut a semicircle out of index card, giving yourself 1/2″ margins with the diameter of the semicircle along one long edge of the card (this should be a full sized slide card). If you have a compass that will take a pencil and an X-Acto knife that will fit the pencil holder, cutting out the semicircle will be easy. Otherwise, just draw the semicircle with a compass and be precise with your cutting.

Place the card over your graduated 90° angle with the baseline along the base of the cut out semicircle so that the vertex of the 90° angle is at the center of the base of the cut out semicircle. Use the lines of the 5° angles to mark off the 5° graduations on the protractor. The 5° marks should be close enough together that you can “eyeball” 4 equally spaced marks between them with fair accuracy to give you 1 ° graduations. Make short marks for the 1° graduations, slightly longer marks for 5° and longer marks for the 10° graduations. The 90° mark should be especially long. Also mark the center point on the base line of the protractor.

Now, turn the protractor 90° on the graduated circle to mark it from 90° to 180°.

The protractor will be rather flimsy but when it’s laminated, it will be sturdier.

8. Now to calculate the sines and cosines. Just calculate them af the 5° angles between 0° (the baseline) and 90° Set up a table of lengths of the hypotenuse, adjacent side, and opposite side for each 5° increment from 0° to 90°. Leave a column for the sine value and the cosine value.

9 For each 5° mark on the circle, drop a perpendicular down to the base line An index card would be useful for that since the corners make 90° angles. Measure each side of the resulting triangle and enter the lengths into your table.

10. For each triangle, divide the length of the opposite side by the length of the hypotenuse to get the sine and the length of the adjacent side by the hypotenuse to get the cosine. If you wish, since we will later be making scales for the other trigonometric functions, you can go ahead and calculate the other functions using the ratios given above. Calculate each function to two decimal places.

Now we can construct the Sine/Cosine (SC) scale.

Here are my values for the trigoometric functions between 0° and 90°

				Sin	Cos	Tan	Csc	Sec	Cot
	Hyp	Opp	Adj	o/h	a/h	o/a	h/o	h/a	a/o
0°	95	0	95	0.00	1.00	0.00	∞	1.00	∞
5°	95	8	94	0.08	0.99	0.08	12.00	1.00	12.00
10°	95	17	93	0.18	0.98	0.19	5.60	1.00	5.50
15°	95	25	92	0.26	0.96	0.27	3.80	1.00	3.70
20°	95	34	89	0.36	0.94	0.38	2.80	1.10	2.70
25°	95	40	86	0.42	0.90	0.42	2.40	1.10	2.15
30°	95	47	82	0.49	0.86	0.51	2.00	1.10	1.70
35°	95	54	78	0.57	0.82	0.69	1.80	1.20	1.40
40°	95	61	73	0.64	0.77	0.84	1.60	1.30	1.20
45°	95	67	67	0.70	0.70	1.00	1.40	1.40	1.00
50°	95	73	61	0.77	0.64	1.20	1.30	1.60	0.84
55°	95	78	54	0.82	0.57	1.40	1.20	1.80	0.69
60°	95	82	47	0.86	0.49	1.70	1.10	2.00	0.51
65°	95	86	40	0.90	0.42	2.15	1.10	2.40	0.42
70°	95	89	34	0.94	0.34	2.70	1.10	2.80	0.38
75°	95	92	25	0.97	0.26	3.70	1.00	3.80	0.27
80°	95	93	17	0.98	0.18	5.50	1.00	5.60	0.19
85°	95	94	8	0.99	0.08	12.00	1.00	12.00	0.08
90°	95	95	0	1.00	0.00	∞	1.00	∞	0.00

If you check a table of sines and cosines, you will see that these values are quite satisfactory.

Now the S scale is easy to construct. It should go on the slider card with the C scale (I now have a C, CI, and S scale on the slider card). Draw a straight line for the scale across the card and begin marking off tick marks using the C scale as a guide. The Sine values are found on the C scale and the degree values are marked off on the S scale. Since the left index on the C scale represents 0.1, the leftmost mark on the S scale will be (approximately) 5° since the sine of 5° in our table is 0.08. Continue to the right finding the Sine values on the C scale and marking the corresponding degree values on the S scale. If you want to be terribly accurate the Sine of 6° is .1045, so you can mark the leftmost index of the S scale 6°.

I jammed the long edge of my slider card against the lower edge of my frame (which, for me is an old, rolling TV cart I picked up at a yard sale.) I used a plastic triangle to line up the values on the C scale with the places I should make tick marks on the S scale. Another index card would work just as well.

Notice that the S scale is actually a log arcsine scale since the values are arcsines corresponding to Sine values on a logarithmic C scale. Because of that, you can use the C, D, and S scales together to calculate values such as kSinx.

Since the S scale is the reverse of the Cosine scale, you can place the two scales on the same scale by marking the arccosine values above the arcsine values in red. Therefore, above the black arcsine 5° you can write a red 85° as the arccosine. Above the black 10°, place a red 80° and so one until you place a red 0° over the back 90° on the right side of the scale. The right side will be rather crowded.

When you finish placing the 5° and 10° intervals on the S scale, you can eyeball the 1° intervals. Remember that the intervals are logarithmic. That doesn’t make much of a difference at the crowded right side of the scale but it does on the left side.. For the 5° to the 46° intervals I marked each off on the edge of a sheet of graph paper and found 5° marks on another section of the S scale that gave me about 5 intervals in that space and used those lengths to give me 1° marks. Then I transferred them back to the S scale.

In other words, I marked the distance between the 5° and the 10° marks onto the edge of the graph paper and then I slid that interval down the S scale until I found 5 5° marks about the same distance across, then I marked those positions back onto the graph paper and then transferred those marks back to the S scale between the 5° and 10° marks.

At the right side of the scale, you can get away with simply placing 4 marks between the 5° ticks at about equidistant.

To use the finished S scale, find the value of the angle (in degrees) on the S scale and look straight across to the C scale to find the Sine value.

You will rarely run into situations where you will have to use small angles less than 5° in survivalistic work but if you absolutely must have them, I would suggest calculating the angles (you’re going to have to draw some really big triangles to be that accurate) and then mark the angle values in green on the S scale between the 5° and 10° marks, taking the C scale in that region to be 0.01 to 0.1.

It strikes me that, for my surveying project, the only other scale I will need is the tangent scale, so I decide to finish the Calcard with that, and I place it on the slider with the D scale.

The T scale presents a bit of a problem. Angles from 0° to 90° have tangents ranging from 0 to infinity. Most of the values range between 0 and 100. My solution is to first use the D scale as a scale running from 0.1 to 1.0. That, according to my table takes care of the angles from 10° to 45° so I construct that part of the scale just as I did the S scale and I mark the degree marks in black.

Then I take the D scale to be between 0.01 and 0.1 and I place the 5° mark under 0.8 on the D scale in red. Next I take the D scale to be from 1 to 10 and place the marks for 50° to 80° in blue. Finally, I take the D scale to be from 10 to 100 and place the 85° mark on the T scale in green.

Finally, I place a little legend below the T scale indicating the color codes and the ranges for the D scale for each and the construction of the scales for the Calcard is finished.

I do have a few more things to do before I have a finished instrument. I want to make a card with some tables that will be useful for surveying and such – some important constants and conversions and some trigonometric diagrams. I also want to construct some way to sight in a straight line from one scale to another and I want to make a way to hang a plumb line accurately off the protractor to sights angles while surveying. Then, of course, I will have to laminate the whole thing.

I used clear packing tape to laminate the cards and a swizzle stick (the little plastic sticks for stirring coffee and such) as a sight for the protractor. I’ve also thrown in some big, flat paper clips to use as sliders so that numbers of the cords can be lined up accurately.

I have pictures of the finished product right here;