## Pratesi 2014 on Bologna 1477, cards & triumphs

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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I proposed you could print 26 cards on half a standard (17.5 x 22.5, English historical "medium" size) sheet of paper, by using a 3 x 9 layout. A 3x9 layout fits 27 cards, so one card worth of paper is wasted if you only wanted 26. This layout with 7 columns upright and one turned sideways:
26 on half sheet.png Viewed 4306 times 8 KiB
improves on that very slightly, wasting 5/8 of a card's worth of paper instead of a full card. The cards are 3% wider: they are 2 inches (50.8 mm) wide. A modern bridge card is 2 1/4 inches (57.15 mm) wide. These cards are taller than modern cards.

I have no evidence either layout was ever used.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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I want to revisit this topic after a two year lapse, because re-reading the document in question, I think I have found another approach to it that might be useful, in the sense of getting closer to a useful conclusion.

Orioli's 1907 article is at http://emeroteca.braidense.it/eva/sfogl ... olo=276349. At the very end, in document 2, p. 119, at the bottom of the page, here is the relevant passage, isolated by Huck at viewtopic.php?p=18216#p18216:
Item che el prefato ser Roberto sia obligato dare e pagare al prefato maestro Pietro o a suo figliolo in suo nome soldi diexedotto de quattrini per ogni centovinticinque para de carte, o vero triumphi para tanto manco de centovinticinque para, quanto gette el numero de le carte che ha più li iochi de li triumphi da quilli de le carte.

(Item that the aforesaid Mr. Roberto is obligated to give and pay to the aforesaid Master Pietro or his son in his name eighteen soldi of money for each 125 packs of cards, or true triumphs sufficiently less than for 125 packs, in so far as the number of the cards is more of Triumphs than of cards.)
The problem is, from this information, perhaps supplemented by other things known, can it be deduced how many cards there are in a single pack of ordinary cards and how many in a deck of triumphs, or at least their ratio?

First, what does it mean? It seems to me, looking at it without remembering Pratesi's analysis, that it is saying that he will pay 18 soldi for expenses of 125 packs of ordinary cards or the equivalent number of triumph packs, a lesser number because the number of cards in each pack is more. Pratesi infers, I think correctly, that the number 125 is chosen because with that number there will be no cards left over, but precisely enough for such-and-such many Triumph decks.

It is odd that the number of Triumph decks is not specified. Pratesi suggests that it is because everyone knows how many more cards are in a triumph deck than in a regular deck. That seems to me not quite enough, since what is required is how many triumph packs equal how many regular packs. Some numbers will be easier than others for getting to the answer.

Pratesi then interprets the contract as calling for a combination of ordinary packs plus Triumph packs whose total number of cards will be the same as in 125 packs of ordinary cards. He finds that a 70 card triumph deck will be the most logical, with triumphs to regular suit cards in a ratio of 5 to 4. I objected that with his formula numerous combinations with numerous numbers of cards per triumph deck were possible, so that one could conclude nothing.

However, I did not take the additional step of questioning his interpretation of the contract. In fact it seems to me now, looking at it afresh, that the contract says nothing about providing mixes of Triumph packs and regular packs, or having to provide enough cards for 125 packs of regular cards. In fact, an earlier section called for 250 packs. The implication is that the 18 soldi are for the expense of making 125 packs of ordinary cards or an unspecified but somhow equivalent number of Triumph packs, a lesser number since Triumph decks have more cards. It is the same rate per card, whether triumph packs or regular packs, since both take approximately the same amount of paper, glue, paint, and so on. But 125 packs of regular cards has an equal number of cards as an unspecified number of triumph packs.

If we could determine what the possibilities are for two groups of packs, with no cards left over, we would at least know the ratio of the number in one to the number in the other, and perhaps even precisely how many cards each had, or at least what was possible, given certain reasonable assumptions to narrow the range.

There are only a certain number of possibilities for the number of cards in an ordinary pack. We know from Bernardino's sermon that packs in his day, 1423, consisted of four suits, each of ten number cards and four courts. We know that by the second quarter of the 16th century the regular pack in Bologna consisted of ten cards per suit, 40 cards total, and 62 in the tarocchini, which simply adds 22 to the same 40. We know that 48 card regular packs were common in Germany. We know that at some point before 1700, a reduced pack in France had 36 cards (see Wikipedia on piquet). We know that 52 card packs were common in Italy, at least later. It is possible that Marziano's game had 11 cards per suit (10 plus the King), so perhaps another possibility. This is not a great number of alternatives to investigate.

So for 125 packs we have:
56 x 125 = 7000.
52x125 = 6500.
48x125 = 6000
44x125 = 5500.
40x125 = 5000
36x125 = 4500.
The problem now is, what numbers will divide the numbers on the right above into an equal number of packs with an equal but higher number of cards, with the same four suits but some additional cards? Essentially this is a matter of looking at these numbers' factors, i.e., the numbers that divide evenly into the total, to see which will yield numbers reasonably possible for a triumph deck.

What are the possibilities on that end? Well, 78 is probably as high as we should go, or perhaps 80, like the CY. I do not know what the lower limit would be. The PMB has 14 triumphs and 16 court cards done in the same style and paints. The number cards have pure gold paint, as do 6 triumphs in a different style. That would give 30 as the low end (14+16 and no number cards). But we know that however many cards it is, it is more than what is in the regular pack. So we have 6 lower limits, from 36 to 56, depending on the size of the regular pack under consideration. We also know that whatever number it is, for triumph packs, it will be 4x + y, where x and y are integers with x = number of cards per suit in triumph decks and y=number of triumphs. How far up and down should we go in finding the values for y? I would think no more than 26 (which would be the maximum for the CY, assuming 3 theologicals and Prudence) and no less than 8. Also, for this purpose any card other than trumps (i.e. the Fool) not in a regular deck counts as a triumph, i.e. part of y in the formula. I am not, at least at this point, assuming that the number of cards in a Triumph deck are the same as those of a regular deck plus the special cards, the Triumphs. The Triumph deck may have been reduced while the other not, even if we know of no actual case of this kind. The sad fact is that we have no regular Italian decks from the 15th century, only German. If they are any indication of what was true in Italy, regular decks were reduced and triumph decks not. So we have to consider both possibilities.

Let us go one by one with the different factors, starting with the factors of 7000 (the number of cards in 125 regular packs of 56 cards each). Our object is to find out how many triumphs there would be in the corresponding triumph pack. From https://www.gcflcm.com/factors-of-7000 we learn that there aren't many that are relevant:
28 × 250 = 7000
35 × 200 = 7000
40 × 175 = 7000
50 × 140 = 7000
56 × 125 = 7000
70 × 100 = 7000
100 × 70 = 7000
of which the first five are too small and last too large. Now it is a matter, for the value of 70 cards per triumph pack, of varying the value of x (number of regular cards per suit) and solving the equation 70=4x + y for y.

If x is 14, 70x100 fits our condition of 70=4x+y if (and only if) y is also 14. If x is 13, y will be 18. If x is 12, y will be 22. If x is 11, y will be 26. We can stop there.

Part of the object of this exercise is to find combinations where there will be 22 triumphs. Another part is to find combinations that make sense. For the moment, let us go on.

For 52 card regular decks, https://www.gcflcm.com/factors-of-6500
52 × 125 = 6500
65 × 100 = 6500
100 × 65 = 6500.
52 is too small (not more than the size of the corresponding regular deck) and 100 too big (for a Triumph deck). 65 = 4x + y is fulfilled if x is 14 and y is 9, if x is 13 and y is 13, if x is 12 and y is 17, if x is 11 and y is 21, if x is 10 and y is 25.

Now for 48 card regular decks, https://www.gcflcm.com/factors-of-6000:
50 × 120 = 6000
60 × 100 = 6000
75 × 80 = 6000
80 × 75 = 6000
50 only allows for 2 triumphs, not enough. 60 = 4x + y is satisfied if x=13 and y=8, x=12 and y=12, if x=11 and y=16, if x=10 and y=20, if x=9 and y=24. 75 = 4x + y if x=14 and y=19, x=13 and y=23. 80 = 4x + y if x=14 and y=24.

Now for 44 card decks. The factors are, https://www.gcflcm.com/factors-of-5500:
50 × 110 = 5500
55 × 100 = 5500
For 50 = 4x + y is satisfied if x is 10 and y is 10, and if x is 9 and y is 14; for 55, if x is 11 and y is 11, if x is 10, y is 15; if x is 9, y is 19.

For 40 card regular decks, https://www.gcflcm.com/factors-of-5000, there is only one factor within range:
50 × 100 = 5000, where if x=10, y=10, if x=9, y=14. (In this case y = 22 if x = 7.)

For 36 card decks, https://www.gcflcm.com/factors-of-4500
45 × 100 = 4500
50 × 90 = 4500
60 × 75 = 4500
75 × 60 = 4500
90 × 50 = 4500
if 45 = 4x + y, x=9, y=9. If 50 = 4x + y, then x=9, y=19, or x=10, y=10. If 60, then x=9, y=24.

The only solution for 22 is on the assumption that the regular decks are 14 cards per suit and the triumph decks are 12 cards per suit. The problem is that we know of no such instances where the regular deck has fewer suit cards than the triumph deck. The solutions where there is the same number of cards per regular suit in both regular decks and triumph decks are all where the number of triumphs is the same as the number of cards in the regular suits. This gives a ratio of 5:4 of triumph decks in these cases to regular decks, corresponding to Pratesi's conclusion (and with reasoning much in common with his). It also gives a nice round 100 triumph decks as the equivalent of 125 regular decks. For x=y=14, this of course is the 5x14 theory. And 70 cards, which was also Pratesi's suggestion. I think now the same conclusion as his is reached in a more rigorous manner.

But the 5:4 ratio applies to the 4x12+22 deck as well; both it and the 5x14 bear a 5:4 ratio to regular decks of 4x14. Moreover, both alternatives are possible configurations for the 1457 triumph decks of Ferrara as well, of 70 cards each.

In favor of 5x14, however, it can be said that it is odd and unprecedented for the regular deck to have a smaller number of cards in the regular suits than the regular deck of the same place and time. It is true that the Bolognese tarocchi was a shortened deck, but that was when the regular deck shortened as well, due to the spread there of the Spanish game of Primera, which used a 40 card deck, according to Dummett a game not documented in northern Italy until the early 16th century.

Moreover, in the 5x14 deck it is not only that the deck is in a 5:4 ratio to the regular deck, but it is easier to see that ratio with that deck than with the 4x12+22, for someone who may be unsure of their arithmetic. If you have 5 regular decks, divide one of them into fourths - the same as the number of cards in a suit - and add one suit to each of the other decks, you get 4 triumph decks. Likewise 125 to 100 is the same 5 to 4; everyone knows that 25 is a fourth of a hundred. If the suits are all the same size, the equation is so obvious that it is not worth putting the ratio in the contract.

As usual, the result remains inconclusive, with considerations on both sides. But I think the 1477 Bologna contract does give some additional considerations on the side of 5x14.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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deleted, double posting

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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I wrote in the discussion ....
The text, which is focussed by Franco Pratesi and translated by Michael S. Howard ...
Item che el prefato ser Roberto sia obligato dare e pagare al prefato maestro Pietro o a suo figliolo in suo nome soldi diexedotto de quattrini per ogni centovinticinque para de carte, o vero triumphi para tanto manco de centovinticinque para, quanto gette el numero de le carte che ha più li iochi de li triumphi da quilli de le carte.

(Item that the aforesaid Mr. Roberto is obligated to give and pay to the aforesaid Master Pietro or his son in his name eighteen soldi of money for each 125 packs of cards, or true triumphs sufficiently less than for 125 packs, in so far as the number of the cards is more of Triumphs than of cards.)
... is marked by me with red signs.

I get problems with this text. I don't know, what 18 soldi in Bologna precisely means in comparison to the Florentine money, but if I assume, that it is somehow in the range of 18 soldi in Florence, then my own logic about the playing card prices strikes. 18 soldi for 125 decks is too far below the prices, which the silk dealers paid for the decks of Niccolo di Calvello, who made the cheapest decks in Florence in the 1440s/50s. How shall this be possible?
It's clear, that 100 and 125 would be a good choice for decks with 14 trumps and 56 common cards. The problem are the 18 soldi and what they are good for. Possibly the 18 soldi are the price for the printing color. 18 soldi for 100 triumph decks or 125 normal decks with 56 cards, it's hard to imagine, that so much decks had only a value of 18 soldi.
Or. Mr. Roberti might be the organizing painter, who paid for color and paper and the two working printers Master Pietro and/or his son.

100 x 56 are 5600 cards, possibly - if the decks were small and cheap - 2 blocks for each deck, which would make 250 prints for 125 decks, which possibly could be done in 1-3 days. If 10 prints are possible in one hour, then it would be 25 hours. Are 18 soldi enough for a work, which wasn't considered very complicated?

"Income and working time of a Fencing Master in Bologna" in ...https://bop.unibe.ch › apd › article › download PDF von A Battistini · 2016 ....
.... has 5 Bologna Lira (= 100 Soldi) for a notary in a month, which should be about 24-25 in a week ... and 60 Lira for a Preacher Bishop.
So 18 Soldi paid for a 2 men cooperation for 3 days would be a common price, I think, perhaps a little too much. Perhaps it included the work for painting the cards, who knows.

But all this doesn't change the situation, that we simply don't understand the document completely. Looking through the earlier text I note, that we had proceeded with the situation more than I remember now. But it stays in the situation, that we don't get a sure argument ...

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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The 18 soldi is fairly clearly for expenses other than paper and cardboard, i.e., glue, paint, brushes, etc. Here is what Orioli says about it (emphasis his), p. 113:
. . . Master Pietro was also required not to allow his son or other of his [people] to work or sell cards for others, except for Blanchelli, and neither to help nor give advice to other people about such profession, nor teach it to others; he instead promised that for the following eighteen months he would be entirely dedicated to preparing [ordinary] cards [carte] and Triumphs [trionfi] on behalf of Blanchelli, who, in turn, must supply paper and the necessary cardboard to make “said cards or true Triumphs [dicte carte o vero triumphi].” Besides the payment [or salary: mercede] agreed upon, Blanchelli also had to add eighteen soldi for expenses for every 120 decks of [ordinary] cards, or as many corresponding decks of Triumphs [mazzi di trionfi], keeping in mind the greater number of pieces needed to form a deck, since “decks of the Triumphs [iochi di li trionfi] have more cards than those of [ordinary] cards[/i]."

(Si obbligava pure detto maestro Pietro a non permettere che suo figlio od alcuno altro de’ suoi lavorasse o vendesse carte per altri, eccetto che per il Blanchelli, nè che aiutasse o consigliasse altri intorno a detto mestiere nè molto meno lo insegnasse ad altri; prometteva invece che per lo spazio di diciotto mesi continui si sarebbe dedicato a preparare carte e trionfi unicamente per conto del Blanchelli; il quale a sua volta, doveva fornire la carta e i cartoni necessari per fare "dicte carte o vero triumphi” Oltre la mercede convenuta doveva anche il Blanchelli aggiungere soldi diciotto a titolo di spese, ogni centoventi mazzi di carte o per altrettanti di mazzi di trionfi corrispondenti, tenendo però conto del maggior numero di pezzi che occorrevano per formarne un mazzo, poichè "ha più iochi de li triumphi da quelli de le carte.”)
The 18 soldi are in addition to the "payment agreed upon," which presumably is the money for the workers' time [mercede currently means wage, salary]. Orioli unaccountably uses the word "centoventi" (120) instead of "centoventicinque" (125) in describing how many packs the 18 soldi are for. It seems to me that this is a mistake on his part (also to Andrea Vitali, who is my go-to guy on such matters).

"Roberto" is Roberto di Blanchelli, the person receiving the cards wholesale for him to sell retail. Piero Bonozzi is a "maestro" (teacher? master of a workshop?) in Bologna, member or leader on the council of elders ("mezziere di anziani)", who is agreeing to require his son or employee ("figliolo") to observe certain conditions in producing the cards for Blanchelli. This information is again Orioli p. 113.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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mikeh wrote: 16 Sep 2022, 02:55 The 18 soldi are in addition to the "payment agreed upon," which presumably is the money for the workers' time [mercede currently means wage, salary]. Orioli unaccountably uses the word "centoventi" (120) instead of "centoventicinque" (125) in describing how many packs the 18 soldi are for. It seems to me that this is a mistake on his part (also to Andrea Vitali, who is my go-to guy on such matters).
No, MUST NOT be a mistake. If there was a rule, that the number relation between triumph cards and normal cards is 1:4, then from 600 decks 120 are made as triumphs and 4x120=480 are made as normal cards. Likely triumph cards were more expensive in the selling prize (higher than 5:4 in the relation to the normal cards). Let's assume the double or a relation of 3:2 cause of "Sonderanfertigung" (special production). As far I remember, the production numbers between Minchiate decks and normal cards deck had about a similar ratio in the 18th century production in Florence.

Indeed: this passage gives some strength to the idea, that the triumph decks were really 70 card decks.
Although 1477 looks a little bit late for 5x14 decks. Likely Florence had already a preference for the number 20/21 or 40/41.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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Interesting discussion. Here's something you might want to consider: from the cardmaker's point of view, the number of woodblocks required for the decks might have been more important in determining the expenses than the number of cards in the decks. Having a couple of blank spaces on a woodblock probably didn't make much difference to the cardmaker's expenses when printing a sheet from that woodblock. So perhaps you really need to be thinking about how many woodblocks were required for trionfi decks, versus how many were required for regular decks. This significantly expands the possibilities for numerical equivalence for your 125 decks.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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Likely it were cheap Trionfi decks, I would assume. Small cards.
56 cards and 48 cards likely had common solutions with 28 and 24 cards. 14 trumps would make 2x14 at one block. Smaller blocks with 2x7 possibly would have more careful results, perhaps used for the trumps.
The 28 cards solution would have 1200 prints for 600 decks of the two blocks with common cards. And 60 prints with a double set of 28 triumph cards. One could offer 2 different Trionfi decks with 14 cards on one block for the different taste of the customers.

The 24 card solution wouldn't work for the production of two different-number decks. 24 cards were good for a system realized with the Rosenwald Tarocchi structure with Minchiate cards. 4x24 = 96.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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Huck wrote,
No, MUST NOT be a mistake. If there was a rule, that the number relation between triumph cards and normal cards is 1:4, then from 600 decks 120 are made as triumphs and 4x120=480 are made as normal cards. Likely triumph cards were more expensive in the selling prize (higher than 5:4 in the relation to the normal cards).
I do not understand this. The 120 or 125 applies only to ordinary decks. Where do you get 120 for triumph decks? And why do combinations of triumph decks and ordinary decks enter in? I do not follow your reasoning.

It seems to me more likely that Orioli copied the document correctly than that his paraphrase is correct. But only a trip to the archives could settle the matter. That is something Andrea is not fond of doing: you have to make an appointment for a specific time, and the traffic is unpredictable, usually more or less terrible from Faenza. It would be almost as easy to fly from Koln. Both Andrea and Franco go with 125, although neither has checked with the archives. I haven't examined the possible configurations if the number of ordinary decks is 120. They will surely be different, and I think more numerous, because 120 has more factors than 125.

On Nathaniel's point, the contract says nothing about molds/woodblocks, so I assume they didn't enter into consideration, as far as the ratio of ordinary decks to triumph decks. It is only the difference in the number of cards that is mentioned. However, the number of molds might be important as far as the overall cost. To make decks with 48 suit cards plus 22 triumphs as well as decks with 56 suit cards would seem to involve more molds than 56 and 70. If triumph decks are 4x48+22, you need 24 + 24 + 22 as well as 28 + 28 (or just an extra one for the 8 extra suit cards, perhaps repeated twice more, for another 24), so 4 or 5 molds overall). For decks of 4x14 and 5x14 you just need 2 for the suits, 28 + 28, and a 4x7 for the triumphs (two sets on one mold) so 3 molds overall.

I don't think you want blank spaces in a mold, because that wastes paper when the mold is used on a blank sheet.

### Re: Pratesi 2014 on Bologna 1477, cards & triumphs

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There was a 125 and from this developed a 100 in the mind of Franco cause of the idea, that there is a relation similar to 56 and 70 connected to the condition, that there should be 14 trumps.

If then in another document to the same problem a 120 is noted between 125 and 100, then you have 5 between 120 and 125 and 20 between 120 and 100, which means, that again the relation 1:4 and 4:5, now as 5:20 and 20:25, appears. That is too much for an accident, and so one should look out for a reason, which explains it.

125-120-100-0 .... 5x5x5 ... 5-25-125

Naturally there might be a humble number error, but my calculation has much more elegance.

************

From another perspective I remember, that the "dozen" = "12 decks" was important in the playing card notes of the silk dealers in Florence. 120 decks and 18 soldi makes 10 dozen decks totally and 1 dozen decks gets 1.5 soldi = 1 soldi and 6 denari.
And likely 8 dozen decks were normal cards (56 cards) and 2 dozen decks were Trionfi decks (70 cards).