Formula for escalator steps

Here's my solution to the escalator problem in my previous post.

Per minute, I climb n steps and the escalator moves along by s steps. So relative to the bottom of the escalator I am travelling at the rate of n + s steps per minute.

If there are p 'visible steps' on the escalator and I take t minutes to get to the top, then t = p/(n + s).

But in t minutes I actually climb nt of the moving steps. This is the variable I call m, the actual number of steps that I take in getting to the top.

So m = nt = np/(n + s).

So below is my formula!

m = np/(n + s).

I must admit I never expected it to be as complicated as this.

To understand this formula, think of it as saying that the proportion of visible steps I climb is the ratio of my speed (n) to the sum of my speed and the speed of the escalator (n + s).

For example, if the escalator has 30 'visible steps' and is moving at 90 steps per minute, my speed is 120 steps per minute, then the proportion of the 30 steps I will tread on will be 120/210, or 4/7 of them, which is about 17.

If you plot m  against n, from n = 0, you get a curve that starts at the origin (m = 0) and tends to p as  n tends to infinity.

When n = s, notice that m = s/2, which makes sense intuitively: if I climb at the same rate as the escalator is moving I will cover half the steps before I get to the top.

Note also that as n increases the gain in my step count decreases. For example, with a 30-step escalator, and a speed of 90 steps rising per minute ...

When n = 0, m = 0

When n = 60, m = 12

When n = 120, m = 17

When n = 180, m = 20

When n = 240, m = 22

When n = 300, m = 23

It's intriguing to think about how the graph behaves for values of n less than zero down to n = –s. These values of n correspond to walking down the escalator when it is going up. Provided you walk down slower than the escalator is going up, then you will get to the top! The number of steps you take, m, will turn out to be negative because you are walking down the escalator.  If you walk down as fast as the escalator is going up (n = –s) then the formula involves division by zero, which is impossible: you never get to the top. In fact, you never get anywhere.

You can have even more fun now, by thinking about how the formula works for climbing up an escalator going down!

How many steps on an escalator?

One of the joys (or burdens?) of being fascinated by mathematics is that you see interesting mathematics problems all around you in your everyday life!

I have recently acquired one of those little devices you attach to your belt that counts the number of steps you make in a day. The result of this is that you get obsessional about maximising your number of steps in every situation. There is even a temptation to mince along with very short steps rather than my usual manly stride, just to increase the step-count!

Today, I had to go up an escalator in a shopping centre. I never stand still on an escalator, unless it is totally blocked with other people. So, I found myself thinking about how many steps might I manage going up this escalator. It seemed intuitive to me that the faster I went the more steps I would get in before I reached the top. After all, if my speed was zero (i.e. standing still on one step) then my step-count for the escalator ride would also be zero.

Intuitively I imagined that there would be a linear relationship here. So, if I went twice as fast I would get in twice as many steps before I reached the top. But, wait a minute! That can't be true. Because I could never get more than the total number of visible steps on the escalator, however fast I went. In fact, to get in a number of steps equal to the number of visible steps on the escalator I would have to travel at an infinite speed!

This was getting so interesting, I forgot to do my shopping and just walked home puzzling it out.

It's a really nice mathematical problem. I managed to calculate, for example, that ...

• if there are 30 visible steps on the escalator (i.e. the number of steps you would see if the escalator were stationary), and
• the steps are disappearing at the top at the rate of 90 steps per minute, and
• I go up the escalator at the rate of 120 steps per minute (pretty fast),

then I will add 17 steps to my step-counter.

If I went twice as fast up this escalator, doing 240 steps per minute, I calculated that I would increase my step-count to 21 or possibly 22 steps. And at half the speed, i.e. 60 steps per minute, I would do 12 steps before reaching the top of the escalator.

I'll leave this as a challenge for readers ... can you produce a formula for the number of steps I add to my step-counter (m) if there are p visible steps; the steps are disappearing at the top at the rate of s steps per minute; and I go up at the rate of n steps per minute?

m  = ....?

And what would the graph look like, if you plotted m against n?

Solution in my next post.

Average age

As I get older I find that I pay more attention to the ages of the people who appear each day in the obituary pages of The Times! Now I find that I have started to calculate the average (mean) age at which they died – and I am encouraged if it is greater than my age (it usually is!).

But, I have an interesting mathematical observation, related to the issue of rounding errors, to offer you.

Take this example. Four people are listed in the obituaries, dying at the ages of 69, 73, 76 and 80. What is the average (mean) age at which they died?

Now, 69 + 73 + 76 + 80 = 298. Divide this by 4 and we get 74.5 years.

But, this is NOT the average age of these four individuals.

Remember that when we say that someone died at the age of 69 this means that they could have been as little as one day short of their 70th birthday. Different conventions for rounding up or rounding down are used in various contexts. We always round down to the year below when we give someone's age in years.

So, the best estimate for the average age of the four individuals in this example would be the mean of 69.5, 73.5, 76.5 and 80.5 years. And that gives the mean age as 75 years.

At my age, let me assure you, that extra half a year is quite significant!

Calculations: foundations of mathematics?

The myth that the mastery of the processes of written calculations, particularly long multiplication and division, is fundamental to doing mathematics continues to be perpetrated by people with political influence and control of our school curriculum. This myth was exposed for me recently in the experience of helping one of my grandsons prepare for his A-level mathematics examination.

We worked together through loads of questions from past examination papers – pure maths, mechanics and statistics. I made the following observations.

Not once in doing A-level mathematics was he required to do a written calculation, since he always had a calculator to hand. His calculator skills were stunning and showed real mathematical understanding, in terms of processing the steps of a complex calculation in the appropriate order, in selecting the correct function keys and handling brackets, and doing this with speed and accuracy.

More important than written calculation skills were the ability to interpret the calculator answer and checking whether it looked reasonable. Additionally, with a little encouragement from me, he improved markedly in using mental strategies for calculations that could be done more efficiently that way than by resorting to the calculator.

But, I repeat, not once did he use a formal written calculation procedure. Yet, there he was doing advanced level mathematics! If he had had to take his eye off the structure of the problem to do a written calculation it is very likely that he would have lost his grasp on where he was going.

For centuries mathematicians have devised ways of avoiding or reducing the demand of written calculations, simply because they get in the way of the real mathematics and effective problem-solving, and take up too much of your precious time. So, we had Napier's bones, and logarithm tables and slide rules, and so on. Now we have modern technology, so please let's give younger children the chance to use it and start doing real mathematics.

Haylock and Cockburn

My apologies to any readers of this blog for a long period of silence. Anne Cockburn and I have been busy working on a fifth edition of our Book, Understanding Mathematics for Young Children. We finished this last week and have sent off the 'manuscript' to the Sage Publications.

The word 'Manuscript', which is still used in publishing circles, means, of course, 'written by hand', which is strangely archaic, given that no pens and no paper were involved in either the writing process or the submission of the new edition to the publisher.

We hope to see the new edition on the shelves (another anachronism, since most of the sales will be online!) in the first half of next year.

We have had to rework the book to ensure that it is consistent with the language and content of the new primary mathematics curriculum in England. This has introduced some new material for Year 2 children, such as fractions. The new curriculum has meant a general shift of content down from Year 3 to Year 2 (and likewise from Year 4 to Year 3). So, to avoid our book getting even longer, we have had to make a decision to reduce the age range covered. So, it is now described as 'a guide for teachers of children aged 3–7 years' (rather than 3–8). This is a better fit for teacher training courses anyway, since it is now clearly aimed at Early Years Foundation Stage and Key Stage 1.

So, with that done, I hope to get back to writing the occasional blog again! Watch this space.

Roman numerals

Here are three questions about Roman numerals that genuinely puzzle me! 

First, why, in the twenty-first century, do we persist in using Roman numerals in particular contexts, such as the hours on clock or watch faces, and dates on buildings or at the end of a movie? 

Second, in such a technological age, can anyone really justify the inclusion of Roman numerals in the statutory English primary mathematics curriculum and in the associated national assessment of mathematics? 

Third, there’s something odd I’ve noticed recently about Roman numerals on clock and watch faces. It is important for teachers to be aware of this, because there are limited contexts for assessment items in Key Stage 2 national mathematics tests in this country, so clock and watch faces with Roman numerals turn up often. 

In an early form of Roman numeration, the numbers we call ‘four’ and ‘nine’ would be represented by IIII and VIIII. A later development was to represent these more concisely as IV and IX, the convention being that when a letter representing a smaller value is written in front of another letter, then the value is to be subtracted, not added. So, XC would represent 90 (100 subtract 10). So, here’s what we have noticed: in most cases where Roman numerals are used on a clock or watch face, the four is written using the early system (IIII) and the nine is written using the later system (IX). Check this out and see if we are right. The question that puzzles me is, simply, why?

The Haylock Bookmark

When I am in the middle of reading a book I need a bookmark that not only helps me to find the page I am on, but also exactly where on the page I should start to continue my reading! I offer readers my design for an effective bookmark for this purpose and a simple example of mathematics applied to another area of the curriculum!

First measure the height of the page of the book (h mm) and the height of the smaller of the top and bottom margins for the text on the page (x mm). Cut a rectangular piece of firm card so that it has a length that is equal to h/2 + x. The width can be whatever you wish, say, about a third of the width of the book.


For example, for a standard paperback novel I find that h = 198 and x = 19. So we need the bookmark to be 99 + 19 = 118 mm long.

Then on one side of the card draw a double-headed arrow x mm from the top (i.e. 19 mm for our standard paperback). The diagram shows a bookmark made for this standard paperback.

That's it! Now, when you have finished your reading session place the bookmark in the fold of the book on the opposite page to the one you are on, with the arrow showing and pointing to your place on the page. This might involve rotating the bookmark, of course, depending whether you are on the top half or the bottom half of the page. Then just close the book. When you open it next time just make sure you have the side of the card with the arrow visible, and remember that the arrow tells you where on the opposite page you should start reading again!

With this system any starting point on either the even-numbered page (verso) or the odd-numbered page (recto) can be marked, without the bookmark sticking out of the closed book (which I don't like).

The diagrams below show the bookmark placed inside the book to mark (a) a point half-way down the left-hand page; and (b) a point near the bottom of the right-hand page.

 (a)

(b)

Well, it works for me. And it is a good application of simple measurement and spatial reasoning!