A good craft for our office. It is about a **perpetual calendar**, of those that we change the day.

There are 2 models of simple perpetual calendars that we can build:

- With cubes or wooden dice
- With 3 rings of different diameters.

## Perpetual calendar with wooden cubes

It is clear that we only need wood, the cutting tools, a little glue and paint. But here we see an exploded view with the amount of blocks we need.

Two cubes, for the days, a support to hold everything and three bars for the months.

It's very simple, but if you don't want to think about anything, here are the measurements. We work with 2 cm thick wood.

- the rear 10,5 cm long and 9 high
- the bottom 10,5 cm by 5
- 2 pieces of 7 by 9
- 2 cubes of 5 cm side
- 3 pieces of 10 and a section of 2 cm

When it comes to numbering and writing the months ... The months are very simple, 3 bars with 4 sides each make 12 months. So

- the first will take January, February, March and April
- the second May, June, July and August
- the third September, October, November and December

The numbers is a bit more complicated, if you like riddles, try to guess which numbers you would put on each of the faces of the dice to be able to make the combinations of the 31 days of a month.

If there are more number combinations are welcome :)

you are one of those who like not to get too hot your head point ...

- the first die will carry 0,1,2,3,5,6
- the second given 0,1,2,4,7,8.

With this you can make all the necessary combinations. Notice that we do not have the nine, so we will use the 6 turning it over.

I have adapted the measurements of the pieces because those of the source do not fit.

Scrap Book Crazy font

## Calendar with concentric rings

It is a model that I have seen in a etsy shop. Doing it is not very complicated, and if we want we can use other simpler materials than wood. From cardboard or painted cardboard to 3D printed

These images give us an idea of how we should build it. I do not leave measurements, but if there is someone interested I can do some calculations.

The inner ring must be divided into 7 parts, the middle ring in 12 and the outer ring in 31

Do you know of any other type of perpetual calendar?

Great, I went in just because I thought three cubes would be needed, and to see how they were distributed. I think the solution is simple about the distribution (now that I have seen this and have corrected my errors). In each die there must be a zero, a one and a 2, and then we distribute the other numbers as we want, because it does not matter.

EASY TURN THE NO. 6 AND ECHO

How do you mark the numbers? Some template

I think the 9 replaces one of the 0, because in the calendar there is no combination 00

You won't be able to do 30, because 0 and 3 must be on different cubes!

Hello, for easier; You turn the 6 upside down and it gives you a 9 :) I hope it works for you ...

How did you paint the numbers and letters? Any advice, idea or recommendation?

Great! Any ideas / advice on how to paint the numbers and letters?

It can be done in many ways, but if you want it to be perfect, it will be a good idea to make a template.

Print the numbers you like to the right size and cut them out leaving the hole of the number on the sheet and paint the hole ;-)

Or my friend, ask for more combinations of numbers. He has. São 57.600 alternatives: a combination of 6 elements 3 to 3, twice (6 or 9) results in 240 possibilities that, raised to or squared (two cubes), result in 57.600 options - see minha page http://www.ghiorzi.org/quebraca.htm.

As discriminated on minha page, there are 40 (forty) probabilities of pulling two two cubes, to allow the formation of any day of the month, from 01 to 31.

We have 40 probabilities of pulling two two cubes, to allow training for two days a month, from 01 to 31, as discriminated on each page http://www.ghiorzi.org/quebraca.htm:

012345-012678 012345-012789 012346-012578 012347-012568 012347-012589

012348-012567 012348-012579 012349-012578 012356-012478 012357-012468

012357-012489 012358-012467 012358-012479 012359-012478 012367-012458

012368-012457 012378-012456 012378-012459 012379-012458 012389-012457

012456-012378 012457-012368 012457-012389 012458-012367 012458-012379

012459-012378 012467-012358 012468-012357 012478-012356 012478-012359

012479-012358 012489-012357 012567-012348 012568-012347 012578-012346

012578-012349 012579-012348 012589-012347 012678-012345 012789-012345

We have 40 probabilities of pulling two two cubes, to allow training for two days a month, from 01 to 31, as discriminated on each page http://www.ghiorzi.org/quebraca.htm:

012345-012678 012345-012789 012346-012578 012347-012568 012347-012589

012348-012567 012348-012579 012349-012578 012356-012478 012357-012468

012357-012489 012358-012467 012358-012479 012359-012478 012367-012458

012368-012457 012378-012456 012378-012459 012379-012458 012389-012457

012456-012378 012457-012368 012457-012389 012458-012367 012458-012379

012459-012378 012467-012358 012468-012357 012478-012356 012478-012359

012479-012358 012489-012357 012567-012348 012568-012347 012578-012346

012578-012349 012579-012348 012589-012347 012678-012345 012789-012345

Strictly speaking, there are 20 probabilities, because they are repeated twice by two. For example, a first option (012345-012678) reveals itself identical to the penultimate one (012678-012345), barely reversed at two cubes position.

There are 20 probabilities of numbering two cubes, using a combination of two algarisms 0, 1 and 2 (fixed) and 3, 4, 5, 7, 8 and 6 or 9.

You two cubes ficariam assim (as seen in minha page http://www.ghiorzi.org/quebraca.htm):

012345-012678 012345-012789 012346-012578 012347-012568 012347-012589

012348-012567 012348-012579 012349-012578 012356-012478 012357-012468

012357-012489 012358-012467 012358-012479 012359-012478 012367-012458

012368-012457 012378-012456 012378-012459 012379-012458 012389-012457

There are 20 combinations of the figures 0, 1 and 2 (fixed) and 3, 4, 5, 7, 8 and 6 or 9. The two cubes would look like this:

012345-012678 012345-012789 012346-012578 012347-012568 012347-012589

012348-012567 012348-012579 012349-012578 012356-012478 012357-012468

012357-012489 012358-012467 012358-012479 012359-012478 012367-012458

012368-012457 012378-012456 012378-012459 012379-012458 012389-012457

See on my page http://www.ghiorzi.org/quebraca.htm

Friend Nacho

Much confusion has already been made with the definition of the figures that each of the two cubes must contain. I think the friend should delete all comments and replace them with the definitive information:

«There are only 20 possible combinations of the figures 0 to 9. The two cubes would look like this:

012345-012678 012345-012789 012346-012578 012347-012568 012347-012589

012348-012567 012348-012579 012349-012578 012356-012478 012357-012468

012357-012489 012358-012467 012358-012479 012359-012478 012367-012458

012368-012457 012378-012456 012378-012459 012379-012458 012389-012457.»

How to do it was very helpful to me? a creative calendar. I saw the steps to follow and I was super clear, no doubt. Thank you.