Author Topic: No84_How many different registrations can I create? (Read 2218 times)

Peter Anderson · « **on:** February 26, 2018, 03:51:54 PM »

On any given pipe organ, just how many different registrations can I form from the number of stops that I have available?

In Peters Pearls No 81 - Basic Organ Stops Explained and specifically in Reply #12, we noted (sorry about the pun) that even with a very few number of stops, you can create a surprisingly high number of different combinations. This is especially true as the number of stops, in that instance, increases.

To read that article, click this link to open it in a new window:

http://www.ar-group.org/smforum/index.php?topic=2976.0

It is a long time since I did Maths at school, but there is a formula to show how many different combinations that can be created from a set number (usually designated n) of items to choose from. I cannot recall, either what it was, or how to generate it.

So let’s call in the mathematicians to give us the formula, please?

Now Hugh is/was the Maths teacher …… so what is the formula, please? This is really bugging me now.

Peter

Peter Anderson · « **Reply #1 on:** February 26, 2018, 05:22:52 PM »

I think that I have just reasoned how to get the formula.

Let there be n stops and the organist can choose any number of them up to and including all n stops.

Obviously, they can either choose each stop to be on or off.

So they get two options for dealing with each stop.

To summarise, there are n stops and they have 2 options for each one, namely either on or off.

In total they will have 2 X 2 X 2 X 2 X 2 X……………………n times options.

So there will be 2ⁿ total combinations of stops. This means 2 multiplied by itself n times.

However, this also happens to include the option of them selecting no stops at all, which for our purposes is meaningless.

So we have to remove that possibility.
So, simply take 1 from every result.

The formula then for determining the total number of possible combinations on an organ, which has n stops I reckon is 2n - 1

Wow! I feel better now that I have solved that! Been so long since I did practical Maths for A and S level.

Peter

Peter Anderson · « **Reply #2 on:** February 26, 2018, 05:50:49 PM »

Now take the organ in the parish Church at Odiham, which you can view by opening a new window here:

http://www.ar-group.org/smforum/index.php?topic=3025.0uu

Now using the formula we worked out we can do some simple calculations, as follows.

It has 10 stops on the Upper Manual, which gives us 1,023 possible combinations.

It has 8 stops on the Lower Manual, which gives us 255 combinations.

It has 4 stops on the Pedals, which gives us 15 combinations.

If we multiply these together, this organ can give us a total of 3,912,975 different registrations from the basic set of 22 stops!

If you set up each of these different registrations one after the other and took just one minute over each, including playing a chord or two for every one, it would take you a fraction over 65,216 hours! That is in excess of 2,717 days, without a single seconds break! That equates to nearly 7¹/₂ years.

People see this type of organ with a comparatively small number of stops and think of the instrument as small!

How wrong can they be?

Your AR has far more stops (voices) available to you than this organ in Odiham!

Peter

AR-Group

News: