Yerushalmi Kilayim 5:1 ירושלמי כלאים ה׳:א׳

MISHNAH: A dried-up vineyard, if it contains ten producing vines per bet seah and they are planted regularly¹If the surviving vines are equally spaced in one direction., is called vineyard. A poor vineyard, one that is planted irregularly, if in it one can ascertain three parallel to two²If somewhere in the vineyard one can identify a minimal vineyard following the House of Hillel. is a vineyard, otherwise it is not a vineyard. Rebbi Meïr said, since it has the appearance of vineyards, it is a vineyard.

HALAKHAH: Rebbi Joḥanan said: A bald spot in a vineyard is like a³Not that it is the same thing but that the same rules apply, and one may sow only at a distance of 16 cubits from the nearest vine. Clearing the ground means tearing out the dead vines. vineyard that dried up. One clears the ground of a bald spot in a vineyard in its interior, one clears the ground of a dried-up vineyard on all sides.

Rebbi Zeïra showed to the colleagues: Nine rows of seven each⁵He explains the Mishnah by an example. We have a vineyard whose vines are planted 8 cubits apart from one another. This follows from the discussion later (Notes 17,18), but it is understood that one wants to construct a vineyard in which the 16-cubit rules forbid one to sow with a minimum number (10) of vines standing. Hence, the vines must be spaced as far apart as possible and still form a vineyard. In modern language, we make a model of that vineyard by drawing an (x,y)-coordinate system in the plane and indicating the vines by the points with integer coordinates, where in each row (“woof”) x = 1, 2, …, 7, and the rows are numbered by (“warp”) y = 1, 2, …, 9. In all, we have 63 vines. Now, we cross out alternate rows and columns, i. e., we cross out all points with coordinates x = 2, 4, 6 and y = 2, 4, 6, 8. There remain 5 y-coordinates (1,3,5,7,9) for each of the remaining 4 x-coordinates (1,3,5,7). Hence, there remain 5×4 = 20 vines.. Take away one next⁶Greek παρά. to another for the warp and one next to another for the woof. There remain 20 vines. Take away two from each side in one direction, one from each side in the other direction⁷On the long sides, one removes the two vines at the corners and the two adjacent ones, leaving only those in positions (1,5) and (7,5). On the short sides, one removes one additional vine each, leaving those in positions (5,1) and (5,9). That means that for x = 1, one has only y = 5, for x = 3, y =3,5,7, for x = 5, y = 1,3,5,7,9, for x = 7, y = 5; the total of standing vines is 1 + 3 + 5 + 1 = 10.
FIGURE 5-1 HERE. There remain 10 vines. That is what we have stated: “Ten vines per bet seah”. It turns out that there are four times two parallel two and one protrudes as tail⁸From any angle one looks at it, one can always find a minimal vineyard contained in the standing pattern. For example, the standing vines {(1,5) (3,5) (5,5) (3,7) (5,7)} form such a vineyard, as do {(1,5) (3,5) (5,5) (3,3) (5,3)} or {(5,9) (5,7) (5,5) (3,7) (3,5)} or {(5,1) (5,3) (5,5) (3,3) (3,5)} or even {(1,5) (3,7) (5,9) (3,5) (5,7)}. In all, one may identify 8 minimal vineyards. Hence, from all angles, the pattern describes a valid vineyard. (One certainly cannot simply create one large vineyard, three times three vines plus one additional, since the maximum area covered by such a vineyard is only 1532 square cubits.).

One assumes that⁹כילו is a contracted form of כְּאִילּוּ. another [vine] was planted there, for all four sides¹⁰Since R. Zeïra explains the Mishnah according to R. Joḥanan, one has to follow the latter’s opinion in Halakhah 4:6 that the area covered by a minimal vineyard is not the trapezoid defined by the segments containing, respectively, two and three vines, but the smallest rectangle containing the trapezoid, adding an imaginary third vine in the second row. In our case, one cannot actually have these vines added, because that would immediately generate three rows of three vines each, which define a large vineyard for which the rule of tail does not apply. Then the question arises of how to compute the area of the remaining vineyard. This will be dealt with in the next paragraph where it is shown that one cannot have exactly a bet seah(2500 square cubits) as required by the Mishnah unless one reduces the distances between the vines.. One cannot actually plant on any of the four sides because it would form a tail and there is no tail for a tail. This implies that there is no tail for a large vineyard, which gives a simple answer to the question of Rebbi Yose ben Zabida in the name of Rebbi Joḥanan: one considers as if another [vine] was planted there¹¹In Halakhah 4:6, R. Yose bar Zemina in the name of R. Joḥanan had a statement, about which R. Jonah asked whether minimal vineyards can be combined to form a large vineyard. The answer given here is that one does not add the imaginary vines to form a theoretical large vineyard..

It turns out to be 64 by 48¹³One has to take the stems of the vines included in the eight cubits distance from one to the next, otherwise one could sow a row of vegetables between the rows of vines. Then the row of originally nine vines has a total length of 64 cubits and that of seven vines one of 48 cubits (Note 5). Since 64 × 48 = 3072, the area is 572 square cubits larger than the 2500 square cubits of a bet seah. Since we have removed the corners, we have to remove some of the area; the problem is to do it in a uniform way to end up with 2500 square cubits. {According to Samuel in Halakhah 4:6, the kite-shaped quadrilateral defined by the extremal vines in R. Zeïra’s design has an area of only 1536 square cubits. Hence, Samuel’s opinion may be disregarded.}. Take two from here and put them there, that turns out to be 48 by 48¹⁴If, for example, one moves the vine at position (5,1) to (1,3), (7,3) to (1,7), and completes the rectangle, one has an area that falls 196 square cubits short of the required 2500.. Take out one in between, one in between, and make a bald spot 16 by 16 that can be tended by eight cubits¹⁵The distance between two vines now is 16 cubits since every second vine was removed. Even according to him who in the preceding Mishnah treated a dried-up vineyard more severely than a newly planted one, one could think of removing two squares of 16 by 16 in the interior of the vineyard. But that would leave one with 3072 - 2×256 = 2560 square cubits, an area that is still too large.. How do you treat it¹⁶Before continuing to compute, we have to settle the question whether the bet seah is the area defined by the standing vines or whether it includes the strip of 4 cubits all around the vineyard for tending the vineyard.? Itself and the area to tend it within a bet seah, or without the area to tend it? If you take it with the area to tend it within the bet seah, then 348 are missing¹⁷This number is difficult to explain. All commentators change the numerical value to agree with their computation. All we can do is understand the problem. The basis of the argument is the computation underlying the next sentence. There it is established that the area of the vineyard in R. Zeïra’s example is cross-shaped (fig. 2), formed from the rectangle of sides 64 and 48 by excising corner squares of area 16×16, of area 2048 square cubits.
FIGURE 5-2 HERE
The strip of width 4 around it, not counting the corners, has area (64+48) ×2×4 = 896 square cubits. Of these four are concave corners that are counted twice, hence, one has to remove 4 times 16, for a total area of 832 square cubits. It is questionable whether one has to fill out the 90° circular sectors around the virtual vines but one has to add two 90° circular sectors of radius 4 at the two vines on the border of the vineyard. Hence, the total area in question here is A = 2048 + 832 + 8π = 2880 + 8π. Even if one uses the approximation π ≅ 3, A = 2500 + 404 and not A = 2500 + 348 as asserted. It is impossible to account for the missing 56 square cubits.
On the other hand, the number 452 in the next sentence is certainly correct and both mss. have 348 here; emendation is impossible. All we can do is discover a possible way of computing the area, keeping in mind that in computations of this kind one suppresses fractions at the end. It seems that here the boundary strip is only three cubits wide except near where real vines are standing. Let h be the width of the boundary strip. Then the computation of the area of the boundary strip is done in five steps.Since the perimeter of the figure is 224 (all measures in cubits), the strip along the sides except for the convex corners is A1 = 224h-4h2 since the convex corners are counted twice in the first term.On the top and bottom edges in fig. 2 are two real vines, both of which have to be enclosed in a circular domain of radius 4. Letq = (16-h²)^1/2
and
α = arctan h/q.
Then the center angle of the sector outside the strip is
φ = π - 2α;
the area to be added is
A₂ = 32φ - 2hq.
FIGURE 5-3 HEREAt the two vines in position (5,1) and (5,5), one adds two semicircles of radius 4 minus two isosceles right triangles of sides h, forA₃ = 32π - 2h².
FIGURE 5-4 HEREAt the four concave corners with real vines, one has to add quarter disks and subtract the parts covered by the strip wide h. The half-angle of the sector outside the strip is β = π/4 - α. This yieldsA₄ = 128β - 64sin α (cos α - sin α).. If you take it without the area to tend it, there is an excess of 452¹⁸FIGURE 5-5 HEREFinally, at the six convex corners occupied by virtual vines, one adds quarter discs of radius h, or A5 = 3πh2. The total area is the sum of these five. The entire computation is within the level of difficulty of Euclid’s Data. The next problem is the determination of π. Mishnaiot Eruvin 3:1, Ahilut 12:6 state that π = 3. Maimonides writes that they adopted this raw approximation since in the opinion of Aristotle, the straight and the curved are original opposites and it is impossible to express curved quantities precisely in terms of linear measures. (In the framework of Archimedean mathematics, the falsity of this assertion was shown only in the Seventeenth Century by John Wallis.) Since we tend to disregard fractions in Talmudic computations, it seems reasonable to take π = 3. We do not have access to the table of chords used by the Talmudists to compute the contribution of A4 unless we want to take Ptolemy’s table (Ptolemäus, Handbuch der Astronomie, transl. K. Manitius, ed. O. Neugebauer. Leipzig 1963, pp. 37–40). This means all our computations here are only an approximate reconstruction. If we take modern values for the trigonometric functions in step 4 the total area is 800 for h = 2.94. (For h = 3, the deviation of the modern value from Ptolemy’s is negligible, about 0.001, much less than the precision of our computation.) It is reasonable to assume that in talmudic approximation, this means h = 3. [Better values of π known from Antiquity, 3 ⅛ or 3 ⅐, yield smaller values of h which have no reasonable explanation.]According to R. Joḥanan, we have to add to the standing vines the ones that complete the minimal vines, i. e., the vines at positions (1,3), (1,7), (3,1), (3,9), (7,3), (7,7). This means that from the original 64×48, at each corner we excise a square of edge length 16. The remaining area is 3072 - 4×16×16 = 3072 - 1024 = 2048 = 2500 - 452 cubits.. You might say, they are separated as single ones¹⁹Since the scheme of R. Zeïra does not work out to produce 10 vines on exactly 2500 square cubits, we have to conclude that the 10 vines are isolated in the field. Then, each vine has 250 square cubits around it, which means that it is situated in the center of a square of edge √250 = 15.81 cubits. This length is also the distance of two vines (or better, the distance of the centers of any two vines) from one another, and this is just below the 16 cubits that would diqualify the field from being a “vineyard”.. You might say, they are closer together²⁰Than 15.811 cubits.: we have stated only “10 per bet seah;” you could not state that.

“²¹Tosephta Kilaim 3:4; the text is almost identical with the one given here, including the queer use of the
singular in the first two statements that make sense only if applied to at least three vines. The later statement, commented on by R. Yose, is from a source other than the Tosephta. The trunk is aligned but the top is not aligned, that is a vineyard. The top is aligned but the trunk is not aligned, that is not a vineyard. If they were thin, but when they grow thick they will be aligned, that is a vineyard. How does one know that they are aligned? One brings a rope and stretches it.” Some Tannaïm state: “From within²²This Tanna requires that the centers of the trunks be aligned. That obviously is not directly measurable; hence, one has to measure from the outside and determine that the centers are within one hand-breadth of the streched rope. The second version, that one puts the rope along the outsides of the vines, is that of the Tosephta..” But some Tannaïm state: “From without.” Rebbi Jonah said, he who says “from within,” within a hand-breadth, but he who says “from without” requires that it touches. “If the outer row is not aligned in its trunks, one reserves for it the space to tend it²³Since the vines in this case do not form a vineyard and do not belong to the vineyard, they are treated as single vines and have only six hand-breadths reserved for tending. The language of the Tosephta is: “If one [vine] deviates to the inside and one to the outside (i. e., the straight segments connecting the vines form a zig-zag line), that is not a vineyard.” and one may sow the rest.” Rebbi Yose says, if one sows within six hand-breadths all is forbidden; outside of six hand-breadths single vines are permitted but a vineyard is forbidden²⁴Since a vineyard needs four cubits for tending..

What means “a poor vineyard?²⁵The one mentioned in the Mishnah. It needs a large tending area since the imaginary vines also belong to the tending area.” Rebbi Jonah said, it is poor in vines but rich in tending areas.