Spiral Slide Rule

Posted on December 17, 2019 by Lin McMullin

As I wrote last week, I found an old spiral slide rule last summer. It is about the size of a rolling pin and in fact has a handle like a rolling pin’s at the bottom. The device consists of a short wide cylinder that slides around, and up or down on a longer thin cylinder.

: Figure 1

: Figure 2

The short wide cylinder has a spiral common (base 10) logarithm scale starting at the top at the 100 mark after the words “slide rule” (see Figure 3). The scale runs around the cylinder 50 times ending precisely under the starting mark. By my measurement the scale is about 511 inches or 42.6 feet long. (1.30 meters). The scale is marked for 4 digits reading with a 5th digit that can be reasonably estimated. (By way of comparison, the common 10-inch slide rule scale discussed last week allows for 2 digits reading with the third digit estimated.) These are the mantissas of the common logarithms from 1 at the zero point (since log (1) = 0) to 1.0 (log (10) = 1) at the lower end.

The thin cylinder is marked with several formulas and other information including a table of natural sines from 0 to 45 degrees, from which you can have the value of any trig function if you’re clever enough. This cylinder is not used for calculations; it is there to allow the wider cylinder to move.

There are also two pointers. The shorter one is attached to the bottom and fixed. The cylinder is moved into position for this pointer. The longer pointer is attached to the thin cylinder and can be moved to the position needed – up, down left or right. Both the top end and the bottom end of the long pointer may be used. The pointers are made to slide past each other if necessary. If the long pointer covers the number needed the other side of it may be used instead (just don’t switch back-and-forth in the same computation).

Here is how it works. For the multiplication problem 15.115 x 439.65.

For the moment we ignore the decimal points.

: Figure 3: Setting top marker at 1 and the first factor 1.5115.

: Figure 4: Setting the second factor 4.3965 and reading the product 6.645

The top, “T” shaped, pointer is moved to the start value after “Slide rule.”
The bottom pointer is first set at 15115 (the 151 is marked, the next 1 is the first mark following 151 and the 5 is estimated. See Figure 3 (Click to enlarge). The distance between the two measured almost 9 times around the cylinder is log (1.5115)
Next the cylinder is moved without disturbing the pointers so that the top pointer is at 4.3965 and again estimating the last digit. Figure 4 upper long pointer.
The product is at the fixed pointer: 6.645 Figure 4 lower pointer.
Finally, we put the decimal in the proper place. The product is 6645.
The full value is 6645.30975 by calculator. So the answer is correct to 4 digits, good enough for most practical work.

By moving the top pointer to log (4.3965) and using the pointers to add to it log (1.5115) we have performed the calculation log (1.5115) + log (4.3965) = log (1.5115. x 4.3965) = log (6.645)

To divide the procedure is reversed. $\frac{{6645}}{{439.65}}$

Set the fixed pointer to the dividend and move the top pointer to one of the divisors. (Figure 4)
Without moving the pointers, move the cylinder so the top pointer is at 1.
The quotient is at the fixed pointer (Figure 3)
Adjust the decimal point for the quotient: 15.115.

If the cylinder is moved so that the pointer is off the bottom of the cylinder, the bottom pointer is used instead of the top. (This is the reason it is directly below the top pointer.)

If this seems like a lot of trouble, it is. But remember, a working computer was not available until near the end of World War II and filled a room. Electronic calculators were not available until around 1970. Computations before then were done by hand or with logarithms.

When I was in college in the early 1960s, I worked for an engineer on my summer vacations. My boss had and occasionally used a large table of logarithms. Large, as in a whole book! As I recall, it was good without interpolating for at least 6 digits accuracy. I used a large desktop mechanical calculator that had a hand crank to do calculations. Hence the term “crank out the answer.”

As for teaching: In those old days before about 1970, you spent 3 to 4 weeks in Algebra 2 teaching students how to use logarithm table and compute with logarithms. I gave that up when the students started using calculators to do the adding and subtracting of their logarithms.

The one advantage of the spiral slide rule is that it doesn’t need batteries!

Happy Holidays!

Inverses

Posted on October 2, 2018 by Lin McMullin

This series of posts reviews and expands what students know from pre-calculus about inverses. This leads to finding the derivative of exponential functions, a^x, and the definition of e, from which comes the definition of the natural logarithm.

Inverses Graphically and Numerically

The Range of the Inverse

The Calculus of Inverses

The Derivatives of Exponential Functions and the Definition of e and This pair of posts shows how to find the derivative of an exponential function, how and why e is chosen to help this differentiation.

Logarithms Inverses are used to define the natural logarithm function as the inverse of e^x. This follow naturally from the work on inverses. However, integration is involved and this is best saved until later. I will mention it then.

Two new post coming soon:

My Favorite Function

Posted on July 31, 2013 by Lin McMullin

My favorite function is $f\left( x \right)=\sin \left( \ln \left( x \right) \right)$ . I like to ask folks how many zeros this function has on the interval $0<x\le 1$ .

Most folks will get their calculator out and graph the function on the interval

Two zeros: one at 1 and the other about 0.05 more or less.

So then I suggest they look at $0<x\le 0.1$ . This is the left 10% of the first window.

Sure enough there is the zero near 0.05 but there is another near 0

So another window $0<x\le 0.01$

Pretty soon they get the idea. Every time we stretch out the graph, there are more roots.

What is going on? The first thing is that this is not a question to be answered on a graphing calculator, the nice graphs notwithstanding.

So try to solve it by hand. Since $\sin \left( x \right)=0$ for $x=k\pi$ where k is an integer, we need to see when $\ln \left( x \right)=k\pi$ . That will be when $\displaystyle x={{e}^{k\pi }}$ . And since our domain is proper fractions it must be that $k\le 0$ . So the zeros are infinite in number, namely $\displaystyle x={{e}^{0}},{{e}^{-\pi }},{{e}^{-2\pi }},{{e}^{-3\pi }},\cdots$ . Which answers the original question but raises others.

Why can’t we see the zeros on the graph?

This is not a calculator glitch; in fact computers can do no better. Each root is the next largest root divided by $\displaystyle {{e}^{\pi }}\approx 23$ . So each root is about $\displaystyle \tfrac{1}{23}$ of the larger next root.

The calculator screen is made up of pixels. The number you choose for xmin is the center of the column of pixels; the number you choose for xmax is the center of the right-most column of pixels. The distance between the two ends is divided and assigned evenly to the centers of the other columns of pixels. The y-coordinates of the pixels are calculated the same way. The calculator evaluates the function at each pixel value and turns on the pixel in that column closest to (rarely at) the function’s value. A lot can go on between the pixels and the graphing calculator and its operator will not see what is happening there.

In this example, all the missing roots are between the first and second pixels on the left! When you change xmax to see the left 10% of the screen you see one and every now and then two roots, but the rest are still between the two pixels on the left.

Would a wider screen help? Perhaps a little, but not much.

Here’s a good exercise for a class: Suppose you could print the graph on a paper 1 mile (5280 feet) wide with the root at x = 1 on the right edge. Where would the next several roots be?

$\displaystyle {{e}^{-\pi }}$ is 228.169 feet from the left edge
$\displaystyle {{e}^{-2\pi }}$ is 9.860 feet from the left edge
$\displaystyle {{e}^{-3\pi }}$ is 0.426 feet or 5.113 inches from the left edge
$\displaystyle {{e}^{-4\pi }}$ is 0.221 inches from the left edge (less than ¼ inch)
And all the remaining roots are within 0.00955 inches from the left edge.

If the paper stretched from the earth to the sun you could see a few more. At 93,000,000 miles, the zero at $\displaystyle {{e}^{-10\pi }}$ is about 0.134 inches from the edge.

So why do I like this problem?

Look at all the math we did.

We learned that graphing is not always the path to the answer.
We learned how calculators choose the points they graph, and which they miss.
We practiced how to solve a trig equation.
We practiced how to solve a natural logarithm equation.
We consider the actual size of the negative powers of e and saw how they got exponentially smaller.
We did a practical problem in scaling to illustrate how fast the numbers diminish.

Why do I like this function?

What’s not to like?

Update (February 7, 2015) Chip Rollinson made this cool Geogebra applet to illustrate My Favorite Function. Use the slider on the screen and notice the x-axis scale as it changes. Thank you, Chip.

Logarithms

Posted on February 6, 2013 by Lin McMullin

Students first encounter logarithms in Algebra 2 or their last pre-calculus course. They are usually defined as exponents and their properties “proven” by reference to the corresponding properties of exponents. This is good enough at the time, but in calculus we are ready to define logarithms properly. The result is the definition in terms of a function defined by an integral.

In fact, the definition flows naturally from the properties of inverse functions (no pun intended, because the flow is quite natural). But is this a definition? Or a proof? The solution of certain differential equations discussed in my last post may be used to derive the logarithm function.

Let’s begin by pretending we know nothing about logarithms, and investigate the inverse of e^x. (See The Derivative of Exponential Functions)

This function e^x contains points of the form $\left( X,{{e}^{X}} \right)$ so its inverse will contain points of the form $\left( {{e}^{X}},X \right)$ , which, since we like the first coordinate of functions to be x, we may also call $\left( x,\ln \left( x \right) \right)$ , where ln(x) will be the name of the inverse of e^x. Remember, at the moment ln(x) is just a notation for the inverse of e^x, we do not know anything about logarithms (yet).

So $\left( {{e}^{X}},X \right)$ = $\left( x,\ln \left( x \right) \right)$ and $X\ne x$ . For example, (0, 1) is a point on e^X, so (1, 0) is a point on the ln(x) function, and so ln(1) = 0.

In a previous post we defined the number e and the function e^x in such a way that $\displaystyle \frac{d}{dx}{{e}^{x}}={{e}^{x}}$ . Now, at $\left( X,{{e}^{X}} \right)$ the derivative is e^X, so at the corresponding point on its inverse, $\left( {{e}^{X}},X \right)$ , the derivative is the reciprocal of the derivative of e^X which is $\frac{1}{{{e}^{X}}}=\frac{1}{x}$ . That is

$\displaystyle \frac{d}{dx}\ln \left( x \right)=\frac{1}{x}$

We can use this idea to define ln(x) as a function defined by an integral. Solving the differential equation by the method suggested in a recent post we get:

$\displaystyle \ln \left( x \right)=\ln \left( a \right)+\int_{a}^{x}{\frac{1}{t}dx}$

We can pick any positive number for a and a convenient one is the one we already found a = 1 where ln(1) = 0, so

$\displaystyle \ln \left( x \right)=\int_{1}^{x}{\frac{1}{t}dx},\quad x>0$

This then is the “official definition” of the natural logarithm function. The domain is x > 0 (the range of e^x) and the range is all real numbers (the domain of e^x).

Graphically, ln(x) is the area between the graph of $\displaystyle \frac{1}{t}$ and the t-axis between 1 and x. If $0<x<1$ , then $\ln \left( x \right)<0$ and if $x>1$ , $\ln \left( x \right)>0$ .

From this definition we can prove all the familiar properties of logarithms $\left( \ln \left( ab \right)=\ln \left( a \right)+\ln \left( b \right)\text{ etc}\text{.} \right)$ and show that this is a function of the kind we called “logarithm” all along. This is in all the textbooks so I will not go into this here.

One of these properties tells us that for $a>0\text{ and }a\ne 1$ , $\displaystyle {{\log }_{a}}\left( x \right)=\frac{\ln \left( x \right)}{\ln \left( a \right)}$ and since ln(a) is a constant

$\displaystyle {{\log }_{a}}\left( x \right)=\frac{1}{\ln \left( a \right)}\int_{1}^{x}{\frac{1}{t}dt}$

Finally, you will see this antiderivative formula: $\displaystyle \int_{{}}^{{}}{\frac{1}{x}dx=\ln \left| x \right|+C}$ . The absolute value sign is to remind you that the argument of the logarithm function must be positive, since in some situations x itself may be negative.

Revised slightly 8-28-2018

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Tag Archives: logarithm

Spiral Slide Rule

Inverses

My Favorite Function

Logarithms

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