This page is for those curious minds that took up the challenge I presented in Chapter 8 of ** “Fractonia”**. If you don’t have any idea what I am referring to, you should go back and read Chapter 8 again before you look at the rest of this page.

If you took up the challenge, then you know that in order to answer two of the questions I asked, you need to do some calculations. You can use your very own Rotaluclac for quick results, or you can do the calculations on a piece of paper.

### How many yonkinots fit into one second?

To determine the number of seconds that fit into one yonkinot, we will start by gathering all the clues that Merc gave us in ** “Fractonia” **(Chapter 8).

Merc related a yonkinot to an Earth second by saying that the number of seconds in the period of eleven days, thirteen hours, forty-six minutes and forty seconds would be same as the number of yonkinots that fit into one second. So, to find the number of yonkinots that fit into one Earth second, we must calculate the number of seconds in this period (from now on referred to as the “stated period”) of time: eleven days, thirteen hours, forty-six minutes and forty seconds.

Let’s start by stating what we know about Earth time:

- There are 60 seconds in a minute.
- There are 60 minutes in an hour.
- There are 24 hours in a day.

Since we need to determine the number of seconds in the stated period of a little more than eleven and a half days, let’s convert each element in the stated period to seconds using what we know about Earth time units.

- Forty seconds is 40 seconds (no surprise there!).
- Forty-six minutes is equivalent to 46 x 60 = 2760 seconds.
- Thirteen hours is equivalent to 13 x 60 = 780 minutes. Since there are sixty seconds in a minute, it follows that 780 minutes is equivalent to 780 x 60 = 46800 seconds.
- Eleven days is equivalent to 11 x 24 = 264 hours. Each hour has sixty minutes, so 264 hours is equivalent to 264 x 60 = 15840 minutes. To convert these minutes to seconds, we multiply by sixty: 15840 x 60 = 950400 seconds.

We have now calculated all the seconds in the minutes, the hours, and the days that form the stated period. Have we answered the question? Almost, but no – not yet. We have to add all these seconds to get the total number of seconds in the stated period, so let’s do that now:

- Total seconds in the stated period = 40 + 2760 + 46800 + 950400 = 1000000 seconds (one million seconds)

Let’s go back to question to see if we answered it. Did we? *How many yonkinots fit into one second?* According to Merc and the clue he gave us, the number of yonkinots that fit into one Earth second is the same as the number of seconds in the stated period. How many seconds are in the stated period? We calculated one million seconds in the stated period. So how many yonkinots fit into one second? The final answer is:

**ONE MILLION yonkinots**

Now let’s answer the other question that required a calculation.

### How many seconds fit into one yonkinot?

Well, we already know that the second is a LOT bigger than a yonkinot. In fact, our previous calculations showed that the second is a million times bigger than the yonkinot, because one million yonkinots fit into one second. If you already know how to write this relationship as a fraction, go ahead and check your answer further down the page. If you are unsure of how to write a number that represents this relationship, take a deep breath and let’s figure it out.

Before we start any calculations, we should agree that our answer has to be smaller than ONE second. How do we know this?

Think of the **yonkinots** as colorful marbles, and the **second** as a big bucket. Our previous calculations showed that a million yonkinots fitted into one second. Imagine a bucket big enough to hold one million marbles. That’s a big bucket, right?

Keeping with the analogy of the bucket of marbles, we can conclude that the question is asking how many buckets (*i.e. seconds*) can fit inside one marble (*i.e. yonkinot*). Obviously a whole second cannot fit into a yonkinot. The second is just too big, and the yonkinot is just too small for this kind of squeezing to happen. But what about part of a second? Could a part of a second be matched to a yonkinot?

Imagine cutting up the bucket into tiny pieces. Imagine that all the bucket fragments are the same size. How big would the piece of bucket be if it could be matched to the tiny marble? If we can answer that question, we can also say how many seconds fit into one yonkinot.

Since a million marbles (yonkinots) fit in the bucket (second), we should cut the bucket into one million equally sized pieces. That way, each bucket piece would match up with one marble. The mathematical operation equivalent to “cutting or dividing up” is DIVISION. Division can be presented by the symbol /.

Let’s take the process of dividing up the bucket (second) so that one bucket piece is associated with each marble, and write this process in the language of mathematics. It would look like this:

(One BUCKET)/(One million MARBLES) = (One second)/(One million yonkinots)

Now let’s rewrite this equation with numbers:

1/1000000 = 0.000001 seconds in each yonkinot

The answer is a small fraction and can be written as **1/1000000** or **0.000001** (decimal fraction). That’s a small fragment of a second!

*Here’s something to think about (if you haven’t already given this some thought):*

While a million yonkinots pass in Fractonian time, only one second of time passes here on Earth. Do you think Matthew returned from Fractonia a lot older than he when he left for Fractonia? Was he missed while he was away? Would Fractonians be easy to spot if they spent half their lifetime here on Earth? Would we even notice if a Fractonian translated into our world for a visit lasting a very long one hundred yonkinots? It makes you think, doesn’t it?