clock face tarot arrangement
Posted: 07 Sep 2009, 06:54
this is just a bit of fun with classification. i have been reading some feng shui books and noticed that in certain diagrams the twelve earthly branches are arranged in a circle, and the four points which are at 90 degrees to one another are grouped for various purposes. that led me to thinking about Tarot, since the numbers from 1-12 sum to 78.
so if we take the twelve numbers on a clock face, and then pair them as opposites, and group two pairs together that are at right angles to each other, we have the following groups:
12 + 6 / 9 + 3; total is 30
1 + 7 / 10 + 4; total is 22
2 + 8 / 11 + 5; total is 26
the middle group has 22 members, which immediately suggests the 22 trumps, (or 21 trumps and the excuse).
the two remaining groups have 56 members, to equal the minor cards. of these, the group that equals 26 suggests the 16 court cards and one whole suit, while the group that equals 30 would be the three remaining suits.
it's an interesting exercise to put various cards in these groups. starting with the trumps, i think the magician is number 1, while the 7 'eschatological' cards (XV - XXI) are number 7. this leaves the four Papi to be the number 4, with the Fool and the 9 Allegories of Life (V-XIV) to be the number 10, (this is the Fool as everyman, going through the trials of life).
for the group of 26, I think these could be one suit of ten and 16 court cards, but there is no easy division of them. the number 2 might be the ace and ten of disks, while the number 8 would be the eight other disks cards. the 11 and 5 would be court cards, but again no easy division here. perhaps the four pages/princesses, along with a special zodiacal card (if you assign the zodiac to the courts) could be the number 5, with the remaining court cards as the 11.
the last group of 30 is also not simple. i would say the three remaining aces as the number 3, opposite the 3's, 6' and 9's from each remaining suit as the number 9, then you could have the six remaining Swords cards as the number 6, and the 12 remaining Wands and Cups cards as the 12.
so you would have:
1 = Trump I, the Magician
2 = Ace & Ten of Disks
3 = Aces of Wands, Cups, Swords
4 = Trumps II-V
5 = Four princesses and Knight of Disks(?)
6 = 2,4,5,7,8,10 of Swords
7 = Trumps XV-XXI
8 = 2-9 of Disks
9 = 3,6,9 of Wands, Cups, & Swords
10 = Trumps VI-XIV plus the Fool
11 = All court cards except those in (5)
12 = 2,4,5,7,8,10 of Wands & Cups
Number totals from cards:
1 = 1
2 = 11
3 = 3
4 = 14
5 = 0
6 = 36
7 = 126
8 = 44
9 = 54
10 = 90
11 = 0
12 = 72
obviously, other schemes are possible.
so how would you arrange the 78 cards?
so if we take the twelve numbers on a clock face, and then pair them as opposites, and group two pairs together that are at right angles to each other, we have the following groups:
12 + 6 / 9 + 3; total is 30
1 + 7 / 10 + 4; total is 22
2 + 8 / 11 + 5; total is 26
the middle group has 22 members, which immediately suggests the 22 trumps, (or 21 trumps and the excuse).
the two remaining groups have 56 members, to equal the minor cards. of these, the group that equals 26 suggests the 16 court cards and one whole suit, while the group that equals 30 would be the three remaining suits.
it's an interesting exercise to put various cards in these groups. starting with the trumps, i think the magician is number 1, while the 7 'eschatological' cards (XV - XXI) are number 7. this leaves the four Papi to be the number 4, with the Fool and the 9 Allegories of Life (V-XIV) to be the number 10, (this is the Fool as everyman, going through the trials of life).
for the group of 26, I think these could be one suit of ten and 16 court cards, but there is no easy division of them. the number 2 might be the ace and ten of disks, while the number 8 would be the eight other disks cards. the 11 and 5 would be court cards, but again no easy division here. perhaps the four pages/princesses, along with a special zodiacal card (if you assign the zodiac to the courts) could be the number 5, with the remaining court cards as the 11.
the last group of 30 is also not simple. i would say the three remaining aces as the number 3, opposite the 3's, 6' and 9's from each remaining suit as the number 9, then you could have the six remaining Swords cards as the number 6, and the 12 remaining Wands and Cups cards as the 12.
so you would have:
1 = Trump I, the Magician
2 = Ace & Ten of Disks
3 = Aces of Wands, Cups, Swords
4 = Trumps II-V
5 = Four princesses and Knight of Disks(?)
6 = 2,4,5,7,8,10 of Swords
7 = Trumps XV-XXI
8 = 2-9 of Disks
9 = 3,6,9 of Wands, Cups, & Swords
10 = Trumps VI-XIV plus the Fool
11 = All court cards except those in (5)
12 = 2,4,5,7,8,10 of Wands & Cups
Number totals from cards:
1 = 1
2 = 11
3 = 3
4 = 14
5 = 0
6 = 36
7 = 126
8 = 44
9 = 54
10 = 90
11 = 0
12 = 72
obviously, other schemes are possible.
so how would you arrange the 78 cards?