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CT 09, pl. 08-13, BM 085194 (P274707)
Mathematical, Old Babylonian, Sippar · c. 1900 BCE–1600 BCE
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| Designation | CT 09, pl. 08-13, BM 085194 |
|---|---|
| ID | P274707 |
| Museum no. | BM 085194 |
| Collection | British Museum, London, UK |
| Language | Sumerian |
| Period | Old Babylonian · c. 1900 BCE–1600 BCE |
| Genre | Mathematical |
| Find spot | Sippar |
| Coordinates | 33.0592, 44.2538 |
Transliteration (ATF)
From Open Richly Annotated Cuneiform Corpus (ORACC) · CC BY-SA 3.0 (ORACC default); per-file JSON license fields say: This data is released under the CC0 license (133618)
Machine draft. The right-hand column is machine English from ixmon/atf-to-english (atf-en-t5base-F, 2026-10-17), generated line by line from the transliteration. It is unreviewed and is not a scholarly translation.
| Line | ATF | Machine English (draft) |
|---|---|---|
| o i | ||
| 1 | a-ra-am-mu-um 1 NINDA DAGAL KI.TA 30 NINDA mu-hi 4 SUKUD | Aramum: 1 rod, the lower width, 30 rods, the height 4 (rods). |
| 2 | i-na UR₂ SAHAR-HI.A 1.30 DAGAL KI.TA 1 NINDA DAGAL AN.NA 6 SUKUD | From the surface of the earth, the lower width is 1;30 rods, the upper width 1 rod, the height 6 rods. |
| 3 | i-na pa-ni a-bu-li-im SAHAR-HI.A EN.NAM a-na 1 LU₂ | What is the earth for 1 man? |
| 4 | US₂ pu-lu-uk ZA.E 1.30 u₃ 1 UL.GAR 2.30 ta-mar | You: sum 1;30 and 1. You will see 2;30. |
| 5 | 1/2 2.30 he-pe₂ 1.15 GAR.RA 1 u₃ 30 UL.GAR 1.30 ta-mar | Break 1/2 of 2;30. Put down 1;15. Sum 1 and 30. You will see 0;01 30. |
| 6 | 1/2 1.30 he-pe₂ 45 GAR.RA 1.15 u₃ 45 UL.GAR 2 ta-mar | Break 1/2 of 1;30. Put down 45. You will see 1;15 and 45. |
| 7 | 1/2-šu 1 ta-mar GAR.RA 6 SUKUD u₃ 4 SUKUD UR₂ SAHAR-HI.A | You will see 1/2 of 1. Put down 0;06, the height, and 4, the height of the surface. |
| 8 | UL.GAR 10 ta-mar 1/2-šu he-pe₂ 5 ta-mar 10 US₂ a-na 5 i-ši | Solve the reciprocal of 0;10. Break off 1/2 of it. You will see 5. Multiply 10 00, the length, by 5. |
| 9 | 50 ta-mar SAHAR-HI.A IGI 10 EŠ₂.GAR₃ DU₈.A 6 ta-mar 50 SAHAR-HI.A | You will see 50. Solve the reciprocal of 0;10, the dividing line. [You will see] 6. The volume is 50. |
| 10 | a-na 6 i-ši 5 ta-mar ERIN₂-MEŠ IGI ERIN₂-MEŠ DU₈.A 12 ta-mar | Multiply by 6. You will see 0;05. Look at the troops. You will observe 12 (groups). |
| 11 | 12 a-na 10 US₂ i-ši 2 ta-mar 2 1 LU₂ i-ṣa-ba-at | Multiply 0;12 by 10, the length. You will see 2. The man will weigh out 1 man. |
| 12 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 13 | ki-ku₈-ru-⸢um⸣ 30 US₂ 20 SAG i-na 10-TA.AM₃ im-ta-har | The length is 30 (rods), the width 20 (cubits). It squares itself off ten times. |
| 14 | i-na ki-im-ṣi₂-im 3 KUŠ₃ BUR₃ SAHAR-HI.A <EN.NAM> a-na 1 LU₂ | At the same time, the depth is 3 cubits, the volume for 1 man. |
| 15 | US₂ pu-lu-uk ZA.E 10 NIGIN 1.40 ta-mar 1.40 a-na | You: square 0;10, the dividing line. You will see 1;40. Multiply [1;40] by [0;20]. |
| 16 | 3 KUŠ₃ BUR₃ i-ši 5 ta-mar SAHAR-HI.A ki-im-ṣi₂-im | The depth is 3 cubits. You will see 0;05. The volume is the same. |
| 17 | NIGIN₂.NA SAHAR-HI.A <US₂> a-mur 20 <US₂> a-na 10 SAG <i-ši> 3.20 ta-mar | You will see the volume. Multiply 0;20 by 10, the width, and you will see 3;20. |
| 18 | 3.20 a-na 3 KUŠ₃ BUR₃ i-ši 10 ta-mar 10 SAHAR-HI.A US₂ | Multiply 3;20 by 3 cubits, the depth. You will see 0;10. The volume is 10. |
| 19 | 10 SAHAR-HI.A 5 u₃ 5 SAHAR-HI.A UL.GAR 20 ta-mar | You will see 0;20. |
| 20 | SAHAR-HI.A IGI 10 EŠ₂.GAR₃ DU₈.A 6 ta-mar 20 a-na 6 i-ši | Solve the reciprocal of 10. You will see 6. Multiply 20 by 6. |
| 21 | 2 ta-mar 2 US₂ 1 LU₂ i-ṣa-ba-at | You will see 2. The length is 2. The man will weigh out 1 man. |
| 22 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 23 | ⸢ki-ku₈⸣-ru-um 30 US₂ 20 SAG 3 KUŠ₃ BUR₃ i-na 1/2 | 30 lengths 20 widths, the depth 3 cubits. |
| 24 | ⸢U₄⸣ <a-na> 1 LU₂⸣ US₂ pu-lu-uk ZA.E 10 NIGIN 1.40 ta-mar | You: square 0;10, the dividing line. You will see 1;40. |
| 25 | 1.40 a-na 3 KUŠ₃] BUR₃ i-ši 5 ta-mar SAHAR-HI.A NIGIN₂.NA SAHAR-HI.A | Multiply 1;40 by 3 cubits, the depth. You will see 0;05. Total volume. |
| 26 | ki-im-ṣi₂-im ša-ni-im a-mur 10 NIGIN 1.40 a-na 3 KUŠ₃ | Look at the reciprocal of 0;10, the dividing line. 1;40 by 3 cubits |
| 27 | i-ši 5 ta-mar SAHAR-HI.A ša-ni-im 20 US₂ a-na | You will see 0;05. That earth, 20 lengths to |
| 28 | 10 SAG i-ši 3.20 ta-mar 3.20 a-na 3 KUŠ₃ BUR₃ i-ši | Multiply 0;10, the width, by 3;20, the depth. |
| 29 | 3.20 a-na 3 KUŠ₃ BUR₃ i-ši 10 ta-mar SAHAR-HI.A NIGIN₂.NA i-na 1/2 | Multiply 3;20 by 3 cubits, the depth. You will see 0;10. The volume is enclosed. In 1/2 |
| 30 | <U₄> a-na 1 LU₂ US₂ pu-lu-uk 1/2 5 SAHAR-HI.A he-pe₂ 2.30 ta-mar | Multiply <day> by 1 man, the length. Break in half 0;05, the volume. You will see 2;30. |
| 31 | IGI 10 EŠ₂.GAR₃ DU₈.A 6 ta-mar 2.30 a-na 6 i-ši 15 ta-mar | Solve the reciprocal of 0;10. You will see 6. Multiply 2;30 by 6 00. [You will see] 15. |
| 32 | 15 ERIN₂-MEŠ IGI 15 DU₈.A 4 ta-mar 10 a-na 4 | Solve the reciprocal of 0;15. You will see 4. Multiply 10 by 4. |
| 33 | i-ši 40 ta-mar 40 a-na 1 LU₂ US₂ ta-pa-la-ak | Multiply 0;40 by 1 man, the length. |
| 34 | ⸢NIGIN₂.NA⸣ 10 EŠ₂.GAR₃ a-mur 1/2 10 he-pe₂ 5 ta-mar 40 a-na 5 i-ši | You will see 0;10. Solve the reciprocal of 1/2 00. Break off half of 10. [You will see 5]. Multiply 40 by 5. It squares itself. |
| 35 | 3.20 ta-mar 3.20 a-na 3 KUŠ₃ BUR₃ i-ši 10 ta-mar EŠ₂.GAR₃ | Multiply 3;20 by 3 cubits, the depth. You will see 0;10. The procedure. |
| 36 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 37 | IRI{ki} 1 ŠU GUR₂ ak-pu-up | City: 1 talent of kiln-fired bricks |
| 38 | 5-TA.AM₃ it-te-ṣi₂-ma | it amounted to 5 each, and |
| 39 | hi-ri-tam ab-ni 6 GAM | I have strewn a censer. 6 lines. |
| 40 | 1.07.30 SAHAR-HI.A BA.ZI | 0;01 07 30, the volume, was removed. |
| 41 | 5-TA.AM₃ <it-te-ṣi> 10 SAG hi-ri-tim | It amounted to 5 each. 0;10, the width of a circle. |
| 42 | E ab-ni E šu-u₂ | “Hand of a cow, Ḫābû-šumu-iddin |
| 43 | i-na 1 KUŠ₃ 1 KUŠ₃ ŠA₃.GAL | (measured) by the 1 cubit, the width |
| 44 | sa₃-sum₂ mu-hu-um u₃ DAGAL EN.NAM u₃ GUR₂ E EN.NAM | What are the width and breadth, and how much is the ...? |
| 45 | ZA.E i-nu-ma 1 ŠU GUR₂ DAL EN.NAM IGI 3-GAL₂ 1 ŠU GUR₂ BA.ZI | You: when 1 talent of kukru is wide, what is the reciprocal of 3? It is reduced to 1 talent. |
| 46 | 20 ta-mar 20 DAL 5 di-ik-ša-am TAB.BA 10 ta-mar | You will see 0;20. The square-side is [0;05]. The depth is 10. You (ll see) 10. |
| 47 | 10 a-na 20 DAL DAH.HA 30 ta-mar DAL šu-li-iš | Multiply 0;10 by 20, the width. You will see 30, the length. |
| 48 | 1.30 ta-mar 1.30 GUR₂ ša hi-ri-tim | You will see 1;30. 0;01 30, the dividing line of a square. |
| 49 | NIGIN₂.NA 1.30 NIGIN 2.15 ta-mar 2.15 a-na 5 GUR₂ | Take away 1;30. You will see 2;15. Multiply [2;15] by 0;05. |
| 50 | i-ši 11.15 ta-mar KI 11.15 a-na 6 GAM i-ši 1.07.30 SAHAR-HI.A> | Multiply 0;11 15 by 6. The volume is 1 07 30. |
| o ii | ||
| 1 | IGI 10 EŠ₂.GAR₃ DU₈.A 6 ta-mar 6 a-na 1.07.30 SAHAR-HI.A | Solve the reciprocal of 0;10. You will see 6. Multiply 6 by 1 07 30, the volume. |
| 2 | i-ši 6.45 ta-mar 6.45 ERIN₂-MEŠ IGI 6.45 DU₈.A | You will see 0;06 45. Solve the reciprocal of 6 45. |
| 3 | 8.53.20 ta-mar 8.53.20 a-na 1.30 GUR₂ i-ši | You will see 0;08 53 20. Multiply 8 53 20 by 1;30, the depth. |
| 4 | 13.20 ta-mar i-na hi-ri-tim uš-ta-pa-la-ak | You will see 0;13 20. It goes around in the street. |
| 5 | NIGIN₂.NA 10 EŠ₂.GAR₃ a-mur IGI 1.30 GUR₂ DU₈.A 40 ta-mar 40 a-na 13.20 | Return. Look at 0;10, the work rate. Solve the reciprocal of 1;30, the dividing line. You will see 40. Multiply 40 by 13;20. |
| 6 | i-ši 8.53.20 ta-mar 8.53.20 a-na 1.07.30 SAHAR-HI.A | Multiply 0;08 53 20 by 1 07 30, the volume. |
| 7 | i-ši 10 ta-mar EŠ₂.GAR₃ NIGIN₂.NA E a-mur 5 ŠA₃.GAL TAB.BA 10 ta-mar | You will see 0;10. Find the reciprocal of a neširšu-commentary. |
| 8 | 10 TAB.BA 20 ta-mar 20 a-na 1.07.30 i-ši 22.30 | You will see 10. Multiply 0;20 by 1 07 30. It is 22 30. |
| 9 | ta-mar EN.NAM a-na 22.30 DAH.HA IB₂.SI₈(TAG₄) li-pu-<ul> | What should I add to 22;30? Let him build a square-side. |
| 10 | u₃ ša DAH.HA IB₂.SI₈ li-pu-ul 5.03.45 DAH.HA | And subtract what you added from the square-side. Add 5 03 45. |
| 11 | 27.33.45 ta-mar 5.15 IB₂.SI₈ sa₃-sum₂ | You will see 27 33 45. The square-side is 5 15. |
| 12 | 5.03.45 EN.NAM IB₂.SI₈ 2.15 IB₂.SI₈ mu-hu | What does 5 03 45 square? The square-side is 2 15 00. |
| 13 | IGI 2 ŠU GAM E DU₈.A 30 ta-mar 30 a-na 1.07.30 <i-ši> | Solve the reciprocal of 2. You will see 0;30. Multiply 30 by 1 07 30. |
| 14 | 33.45 ta-mar NIGIN₂.NA sa₃-sum₂ u₃ mu-<<hu>>-ha-<am> | You will see 0;33 45. The coils of the colon are squared and you will hear it. |
| 15 | UL.GAR 7.30 ta-mar 1/2 7.30 he-pe₂ 3.45 ta-<mar> | Solve the reciprocal of 0;07 30. Break 1/2 of 7;30. You will see 3;45. |
| 16 | IGI 3.45 DU₈.A 16 ta-mar 16 a-na 33.45 | Solve the reciprocal of 3 45. You will see 16. Multiply 16 by 33 45. |
| 17 | i-ši 9 ta-mar 9 SUKUD ša i-ki-im | You will see 9. The height is 9. |
| 18 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 19 | BAD₃ 1 ŠU US₂ 20 mu-hu 1 sa₃-sum₂ 6 DAGAL SAHAR-HI.A <EN.NAM> | A wall. The length is 1. The width is 0;20. The depth is 1. <What is> the volume? |
| 20 | a-na 1 LU₂ US₂ pu-lu-uk ZA.E 30 u₃ 1 UL.GAR | You: find the reciprocal of 0;30 and 1. |
| 21 | 1.30 ta-mar 1/2 1.30 he-pe₂ 45 ta-mar 45 | You will see 1;30. Break off 1/2 of 1 30, you will see 0;45. |
| 22 | a-na 6 SUKUD i-ši 4.30 ta-mar SAHAR-HI.A IGI 10 EŠ₂.⸢GAR₃⸣ | Multiply by 6, the height. You will see 4;30. The volume is 0;10. |
| 23 | DU₈.A 6 ta-mar 4.30 a-na 6 i-ši 27 ta-mar | Solve the reciprocal of 6. You will see 0;06. Multiply 4;30 by 6 00. [You will see] 27. |
| 24 | ERIN₂-MEŠ IGI 27 ERIN₂-MEŠ DU₈.A 2.13.20 ta-mar 2.13.20 | Solve the reciprocal of 27 people. You will see 2;13 20. |
| 25 | a-na 1 US₂ i-ši 2.13.20 ta-mar 1 LU₂ i-ṣa-ba-at | Multiply by 1, the length. You will see 2;13 20. 1 man will die. |
| 26 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 27 | {giš}DIB.DIB ip-te-e-ma 1/2 SILA₃ {giš}DIB.DIB | He opened the door and 1/2 qû of door jamb |
| 28 | IGI 4-GAL₂ 10 ⸢ŠU.SI⸣{na} a-na i-si-iq-tim | a 4th of a cubit, 10 fingers of ēšiqtu-stone for a circle. |
| 29 | u₂-ul i-⸢sa⸣-an-ni-<iq> qa₂-qa₂-rum UGU qa₂-qa₂-rum | He will not raise a claim against the other side. |
| 30 | EN.NAM SUKUD ZA.E IGI 1 40 SUKUD {giš}DIB.DIB DU₈.A | What is the height? You: solve the reciprocal of 1 0;40, the height of a steppe. |
| 31 | 36 ta-mar 36 a-na 30 i-ši 18 ta-mar | You will see 36. Multiply 36 by 30. [You will see] 18. |
| 32 | 18 a-na 2.30 i-ši 45 ta-mar KI UGU KI DIRI | Multiply 18 by 2;30. You will see 0;45. The area exceeds the area. |
| 33 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 34 | {giš}⸢DIB.DIB⸣ ip-te-e-ma 1/2 SILA₃ 1.45 qa₂-qa₂-rum | He opened the door and 1/2 qû 1 45 qaqarum |
| 35 | UGU ⸢qa₂-qar⸣ <DIRI> 1 SILA₃ a-na i-si-iq-tim | <Add> 1 qû for the isiqtu-festival. |
| 36 | EN.NAM ⸢BA.ZI⸣ ZA.E IGI 30 1 SILA₃ {giš}DIB.DIB DU₈.A | What is the height? You: solve the reciprocal of 0;30. You will see 1 qû of a steppe. |
| 37 | 2 ta-mar 2 a-na 40 i-ši 3.20 ta-mar | You will see 2. Multiply 2 by 0;40. [You will see] 3;20. |
| 38 | 3.20 a-na 45 i-ši 2.30 ta-mar 1 SILA₃ | Multiply 3;20 by 45. You will see 2;30. |
| 39 | a-na i-si-iq-tim u₂-ul i-sa₃-an-ni-iq | He shall not raise a claim against the other. |
| 40 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 41 | {giš}DIB.DIB 3.20 ip-te-e 1 SILA₃ he-pi {giš}DIB.DIB | A wedge. The opening of the wedge is 3;20. One qû of broken bricks. |
| 42 | 44.26.40 IGI 9-GAL₂ 2/3 10 ŠU.SI 1 SILA₃ | 44 26 40 is the reciprocal of 9 2/3 |
| 43 | a-na i-si-iq-tim u₂-ul i-sa₃-an-ni-iq | He shall not raise a claim against the other. |
| 44 | qa₂-qa₂-rum UGU qa₂-qa₂-rum EN.NAM DIRI | What is the area on top of the area? |
| 45 | IGI 1.40 DU₈.A 36 ta-mar 36 a-na 3.20 | Solve the reciprocal of 1;40. You will see 36. Multiply 36 by 3;20. |
| 46 | i-ši 2 ta-mar 2 a-na 44.26.40 i-ši | You will see 2. Multiply 2 by 44 26 40. |
| 47 | 1.28.53.20 qa₂-qa₂-rum UGU qa₂-qa₂-rum | 1 28 53 20, the volume over the volume. |
| 48 | ki-a-am ne-pe₂-šum | Thus the procedure |
| o iii | ||
| 1 | a-aš-ba-tum 20 im-ta-har A.ŠA₃ EN.NAM | The square-side is 0;20. What is the area? |
| 2 | ZA.E 20 NIGIN 6.40 ta-mar IGI 4-GAL₂ 6.40 | You: square 0;20. You will see 6;40. The reciprocal of 4 is 6;30. |
| 3 | BA.ZI 3.20 ta-mar 2(IKU){gana₂} A.ŠA₃ | You will see 3;20. The area is 2 iku. |
| 4 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 5 | 1 IM.LA₂ GUR₂ EN.NAM ZA.E 4 u₃ 3 BAL GAR.RA | If the north wind roars: what are yours? You put down 4 and 3. |
| 6 | 4 a-na 3 i-ši 12 ta-mar IGI 18 IM.LA₂ <DU₈.A> | Multiply 0;04 by 3. You will see 12. Solve the reciprocal of 18 00, the height. |
| 7 | 3.20 ta-mar 3.20 a-na 12 i-ši | You will see 3 20. Multiply 3 20 by 12. |
| 8 | 40 ta-mar 40 GUR₂ | You will see 0;40. |
| 9 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 10 | 40 GUR₂ IM(KI).LA₂-BI <EN.NAM ZA.E 4 u₃ 3 BAL GAR.RA> 4 a-na 3 i-ši 12 ta-mar | You: put down 4 and 3. Multiply 4 by 3. You will see 12. |
| 11 | IGI 12 DU₈.A 5 ta-mar 18 IM.<LA₂>-BI | Solve the reciprocal of 12. You will see 0;05. 18 is its inverse. |
| 12 | a-na 5 i-ši 1.30 ta-mar 1.30 a-na 40 GUR₂ i-ši 1 ta-mar | Multiply by 0;05. You will see 1;30. [Multiply 1];30 by 40 kurru. [You will see] 1. |
| 13 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 14 | 30 IM.LA₂-BI GUR₂ EN.NAM ZA.E 4 a-na 3 i-ši | If the Moon reaches its normal length: what are yours? Multiply 4 by 3. |
| 15 | 12 ta-mar IGI 12 DU₈.A 3.20 ta-mar 3.20 a-na 12 | Solve the reciprocal of 12. You will see 3;20. Multiply 0;03 20 by 12. |
| 16 | i-ši 40 ta-mar 40 a-na 30 i-ši 20 ta-mar GUR₂ | Multiply 0;40 by 30. You will see 20. Return. |
| 17 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 18 | 20 GUR₂ IM.LA₂ EN.NAM ZA.E 4 a-na 3 i-ši | What are 20 kurru of bricks? You: multiply 4 by 3. |
| 19 | 12 ta-mar IGI 12 DU₈.A 5 ta-mar | You will see 12. Solve the reciprocal of 12. [You will see] 0;05. |
| 20 | 18 IM.LA₂ a-na 5 i-ši 1.30 <ta-mar> | Multiply 0;18, the brickage, by 5. You will see 1;30. |
| 21 | 1.30 a-na 20 GUR₂ i-ši 30 ta-mar 30 IM.LA₂ | Multiply 1;30 by 20 00 00. You will see 0;30. |
| 22 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 23 | GI.SA 4 GUR₂ KI.TA(BI) 2 GUR₂ AN.NA 6 SUKUD | a reed bundle of 4 kurru; its lower side is 2 kor of lead, 6 sukurru. |
| 24 | SAHAR-HI.A EN.NAM u₃ DAL SAHAR-HI.A AN.NA u₃ SAHAR-HI.A | What are the volume and the dilapidation of the volume of lead and the volume? |
| 25 | KI.TA(BI) ZA.⸢E 4 NIGIN 16 ta-mar 16 a-na | The lower part is elevated. The coils of the colon are 4 in number. You will see 16. Multiply 16 by 1. |
| 26 | 5 GUR₂ i-ši 1.20 ta-mar 2 NIGIN 4 ta-mar | Multiply 0;05, the depth, by 1;20. You will see 2. The coils of the colon are 4 in number. |
| 27 | 4 a-na 5 GUR₂ i-ši 20 ta-mar 1.20 u₃ 20 UL.GAR | Multiply 4 by 0;05, the reeds. You will see 20. Combine 1;20 and 20. |
| 28 | 1.40 ta-mar 1/2 1.40 he-pe₂ 50 ta-mar 50 a-na 6 SUKUD | You will see 1;40. Break off 1/2 0;01 40. [You will see] 50. Multiply 50 by 6. The height is 6. |
| 29 | i-ši 5 ta-mar SAHAR-HI.A GI.SA-HI.A | You will see 0;05. The volume of the reeds. |
| 30 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 31 | GI.⸢SA⸣ 4 sa₃-sum₂ 3 mu-hu 6 SUKUD 5 ⸢SAHAR⸣-[HI].⸢A⸣ | A reed fence. The square-side is 4, the height 3. The height is 6, the height 5. |
| 32 | 3 KUŠ₃ e-li DAL u₃ SAHAR-HI.A EN.NAM ZA.[E] | What are the width and volume? |
| 33 | 4 sa₃-sum₂ UGU 2 mu-hu EN.NAM i-tir | The square-side is 4. What is the reciprocal of 2? |
| 34 | 2 DIRI IGI 6 SUKUD DU₈.A 20 ta-mar 10 [a-na 2 i]-ši | Solve the reciprocal of 0;02, the height. You will see 20. Multiply 10 by 2. |
| 35 | 20 ta-mar 20 a-na 3 ša te-lu-u₂ i-ši | You will see 20. Multiply 0;20 by 3. It comes up. |
| 36 | 1 ta-mar 1 i-na 4 sa₃-sum₂ BA.⸢ZI⸣ | You will see 1. Subtract 1 from 4. |
| 37 | 3 DAL ki-a-am ne-pe₂-šum | 3 DAL is the procedure. |
| 38 | SIG₄.AL.UR₃.RA PU₂ 3.20 US₂ 2.30 SAG AN].⸢NA⸣ | A well. The length is 3;20, the width 2;30, the height of a ... |
| 39 | 1.40 SAG KI.TA a-na PU₂ ⸢SIG₄⸣ [EN.NAM ša-ak]-na-at | 1;40, the lower front, to the well of the mudbrick house; the governor is appointed. |
| 40 | ZA.E 2.30 SAG AN.NA UGU 1.40 SAG KI.TA] | You: the upper width is 2;30, the lower width 1;40. |
| 41 | DIRI 50 DIRI IGI 50 DU₈.⸢A⸣ 1.12 ta-mar] | Multiply 50 by 0;50. You will see 1;12. |
| 42 | 1.12 a-na 1.40 i-ši 2 ⸢ta⸣-[mar 2 ⸢NIGIN⸣ 4 ta-mar | Multiply 0;01 12 by 1;40. You will see 2. Sum 2 and a total of 4. |
| 43 | 3.20 US₂ a-na 4 i-ši 13.20 ta]-mar DAL GAL | Multiply 3;20, the length, by 4. You will see 13;20. The large square-side is 0;00 03 20. |
| 44 | 13.20 A.RA₂ 3 TAB.BA 40 ⸢ta⸣-[mar 40 GUR₂ [NIGIN₂.NA] | 13;20 times 3 is 0;40, the diagonal of 40. Total. |
| 45 | ti-ib-ka-at iš-te-[en EN.NAM IGI 1.40 SAG KI].TA DU₈.A | What is the reciprocal of 1;40, the lower width? |
| 46 | 36 ta-mar 36 [a-na 40 GUR₂ i-ši 24 ta-mar] | You will see 36. Multiply 36 by 0;40, the šangû-square. [You will see] 24. |
| 47 | 24 ti-ib-[ka-at iš-te-en (...) NIGIN₂.NA] | 24 tibkātu-vessels, this one ... |
| 48 | GUR₂ qe₂-er-be₂-tum [EN.NAM (...) 3.20 US₂] | The length is 3;20, the width 0;30. |
| 49 | SIG₄ TAB.BA 6.40 a-na 13.20 DAH.HA 20 ta-mar] | Add 6;40 to 13;20. You will see 20. |
| 50 | 20 A.RA₂ 3 TAB.BA 1 ta-mar 1 ⸢GUR₂⸣ [qe₂-er-be₂-tum] | You will see 0;20 times 3. |
| 51 | ki-a-am ne-⸢pe₂-šum⸣ | thus the ritual procedures |
| r i | ||
| 1 | BAD₃ 36 SUKUD 1/2 [NINDA 3 KUŠ₃ mu-hu-⸢um⸣ | The city wall is 36 cubits, the width 1/2 rod and 3 cub[its]. |
| 2 | ma-la sa₃-⸢sum₂ e-le-nu⸣-um ur-dam | as much as the sa-esum of upper-class is to be added, |
| 3 | ša ur-dam sa₃-sum₂ DAL u₃ SAHAR-HI.A EN.NAM | What are the length, width, and volume? |
| 4 | ZA.E i-na 1 KUŠ₃ 50 ŠA₃.GAL 50 TAB.BA 1.40 ta-mar | You: with 1 cubit, 0;50, the depth. You will see 1;40. |
| 5 | 1.40 a-na 36 SUKUD i-ši 1 ta-mar GAR.RA 45 | Multiply 1;40 by 36, the height. You will see 1. Put down 0;45. |
| 6 | NIGIN 33.45 ta-mar 1 u₃ 33.45 UL.GAR-ma | You will see 0;33 45. Combine 1, so that you see 3;35 45. |
| 7 | 1.33.45 ta-mar EN.NAM IB₂.SI₈ 1.15 IB₂.SI₈ | You will see 1;33 45. What is the square-side? It is 1;15. |
| 8 | 1.15 sa₃-sum₂ 1.15 a-na 12 BAL SUKUD i-ši | Multiply 1;15 by 12, the diagonal of the height. |
| 9 | 15 ta-mar 25 ŠA₃.GAL GAR.RA NIGIN₂.NA 1.15 sa₃-sum₂ | You will see 0;15. The constant of a square is 25. Solve the reciprocal of 1;15, the square-side. |
| 10 | UGU 45 EN.NAM DIRI 30 DIRI 25 ŠA₃.GAL | What is the difference between 45 and 30? The square-side is 25. |
| 11 | a-na 30 DIRI i-ši 12.30 ta-mar 12.30 a-na 45 ⸢mu-hi⸣ | Multiply by 0;30, the width. You will see 12;30. Add 12 30 to 45, the height. |
| 12 | DAH.HA-ma 57.30 ta-mar DAL NIGIN₂.NA SAHAR-HI.A a-⸢mur⸣ | Add (it) and you will see 0;57 30. The diameter is a circle. You will gather earth. |
| 13 | 57.30 DAL u₃ 45 mu-hu UL.GAR 1.42.30 | 57;30 the long side and 45 the short side are 1 42;30. |
| 14 | ta-mar 1/2 1.42.30 he-pe₂ 51.15 ta-mar 51.15 | You will see 1/2 0;01 42 30. Break (it) down and you will see 51 15. |
| 15 | a-na 15 ša tu-ur-dam <i-ši> 12.48.45 ta-mar SAHAR-HI.A AN.[NA] | Return. You will see 0;12 48 45. The volume of the lead. |
| 16 | NIGIN₂.NA SAHAR-HI.A KI.<TA> a-mur 57.30 u₃ 15 UL.GAR 2.12.30 <ta-mar> | You will see the lower volume. Sum 57;30 and 0;15, the constant. <You will see> 2;12 30. |
| 17 | 1/2 2.12.30 he-pe₂ 1.06.40 ta-mar 1.06.40 a-na 21 SUKUD | Break 1/2 of 2 12;30. You will see 1 06;40. Multiply 1 06 40 by 21. |
| 18 | i-ši 23.11.15 ta-mar SAHAR-HI.A KI.<TA> | You will see 0;23 11 15. The lower volume. |
| 19 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 20 | BAD₃ SAHAR-HI.A 36 36 SUKUD ⸢i-na⸣ 1 KUŠ₃ 50 ŠA₃.GAL | The volume of a wall is 36 00. The height of the eši-ina-e23 is 36 cubits. The square-side is 50. |
| 21 | mu-hu u₃ sa₃-sum₂ <EN.NAM> ZA.E IGI 36 SAHAR-HI.A DU₈.A | You: find the reciprocal of 36, the volume. |
| 22 | 1.40 ta-mar 1.40 a-na 36 i-ši 1 ta-mar 1(2) GAR.RA | Multiply 1;40 by 36. You will see 1. Put down 0;01. |
| 23 | 25 ŠA₃.GAL a-na 36 i-ši 15 ta-mar 15 i-na 1 BA.⸢ZI⸣ | Multiply 0;25, the volume, by 36. You will see 15. Subtract 15 from 1. |
| 24 | 45 ta-mar mu-hu 15 a-na 1 DAH.HA 1.15 ta-mar sa₃-sum₂ | You will see 0;45. Add 15 to 1, you will see 1;15. Square (it). |
| 25 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 26 | SIG₄.AL.UR₃.RA PU₂ 3.20 US₂ 50 SAG AN.NA UGU <SAG> KI.<TA> 50 | a well. The length is 3;20, the width 0;50, the capital of the lead, over the lower width is 50. |
| 27 | DIRI SAG AN.NA u₃ SAG KI.<TA> EN.NAM ZA.E | What are the difference between the capital of the lead and the lower capital? |
| 28 | IGI 40 20 BA.ZI 13.20 ta-mar IGI 3.20 DU₈.A 18 ta-mar | Solve the reciprocal of 0;40. You will see 13;20. Multiply 3;20 by 18 00 00. |
| 29 | 18 a-na 13.20 i-ši 4 ta-mar 1/2 4 he-pe₂ 2 ta-mar | Multiply 0;18 by 13;20. You will see 4. Break off 1/2 of 4 and you will see 2. |
| 30 | 50 DIRI a-na 2 i-ši 1.40 ta-mar 1.40 a-na BAL SAG AN.<NA> | Multiply 0;50, the width, by 2. You will see 1;40. [Add 1];40 to the width of the top of the tin. |
| 31 | u₃ SAG KI.<TA> GAR.RA 50 a-na 1.40 DAH.HA 2.30 ta-mar | Add 0;50 to 1;40. You will see 2;30. |
| 32 | 2.30 SAG AN.NA ki-a-am ne-pe₂-šum | The width is 2;30, the procedure. |
| 33 | 1 GUR₂ 2 NINDA ur-dam DAL EN.NAM ZA.E | You multiply 1 kurru 2 rods. What is the diameter? |
| 34 | 2 NIGIN 4 ta-mar 4 i-na 20 | You will see 2. Multiply 4 by 0;20. |
| 35 | DAL BA.ZI 16 ta-mar 20 DAL NIGIN 6.40 ⸢ta⸣-[mar] | You will see 16. The width is 0;20. The total is 6;40. |
| 36 | 16 NIGIN 4.16 ta-mar 4.16 i-na 6.40 BA.ZI | Total: 4 16. You will see 4 16. Subtract 4 16. From 6 40. |
| 37 | 2.24 ta-mar 2.24 EN.NAM IB₂.SI₈ 12 IB₂.SI₈ | You will see 2 24. What are 2 24? The square-side is 12. The square side is 12. |
| 38 | DAL ki-a-am ne-pe₂-šum | DAL, thus the procedure |
| 39 | šum-ma GUR₂ 1 ak-pu-up | If kurru (= 100 sila3) is 1 I put down |
| 40 | 12 DAL ša ur-dam <EN.NAM> ZA.E 20 DAL NIGIN | You: find the reciprocal of 0;12, the length. |
| 41 | 6.40 ta-mar 12 NIGIN 2.24 i-na 6.40 BA.ZI | You will see 0;06 40. The total is 12. Subtract 2;24 from 6;40. |
| 42 | 4.16 ta-mar 16 EN.NAM IB₂.SI₈ 4 IB₂.SI₈ 1/2 4 he-pe₂ | You will see 0;04 16. What are the square-sides? Break 4 in 1/2. |
| 43 | 2 ta-mar 2 ša tu-ur-dam ne-pe₂-šum | You will see 2. Return. The procedure. |
| 44 | {giš}BA:RI.GA 4 DAL 1 ŠE GUR₂ u GAM EN.NAM | A scepter. Its length is 4 (rods), its depth 1 (rod). What are its depth and its weight? |
| 45 | ZA.E DAL NIGIN 16 ta-mar IGI 16 DU₈.A | You: square 0;16. You will see 16. Solve the reciprocal of 16. |
| 46 | 3.45 ta-mar 3.45 GAM ne-pe₂-šum | You will see 0;03 45. The procedure. |
| 47 | {tug₂}ṣu₂-ba-tum 48 US₂ i-na U₄ 1-KAM 20 im-ha-aṣ | a šubatu-garment, 48 lengths on the 1st day, 20 was broken. |
| 48 | i-na ki-ma u₄-mi e-pa-ra-<as> ZA.E IGI 20 DU₈.A | You: solve the reciprocal of 0;20. |
| 49 | 3 ta-mar 48 a-na 3 i-ši 2.24 ta-mar | Multiply 0;48 by 3. You will see 2;24. |
| 50 | ITI 4-KAM U₄ 24-KAM i-pa-ra-as | He will perform (the service) on the 4th day and the 24th day. |
| r ii | ||
| 1 | {giš}BAN še-a-am ma-li-a-at i-na ŠA₃-⸢BI⸣ 1 SILA₃ ŠE] BA.ZI 2 <DAL> | A seah of barley equals one-third of a šapik-container; 1 qû is the price of one-half. |
| 2 | GAM EN.NAM u₂-ri-id-ma lu-u₂ 1 SILA₃ [...] ⸢ZA⸣.E 2.30 GAM | What is the reciprocal of 0;00 02 30? |
| 3 | a-na 10 ṣi-lip IGI 10 DU₈.A 6 ta-mar 2.30 a-na 6 i-ši | Multiply by 0;10. You will see 6. Multipy 2;30 by 6. |
| 4 | 15 ta-mar 10 ŠU.SI 1/2 10 ŠU.SI GAM 1 ⸢SILA₃⸣ lu-mur 2 DAL NIGIN | You will see 0;15. The ‘square-side’ is 10 fingers 1/2. It is 1 finger. Let me see 1 of a šalâ-vessel. The width is 2. |
| 5 | 4 ta-mar 15 GAM a-na 4 i-ši 1 ta-mar 1 SILA₃ ŠE | Multiply 0;15 by 4. You will see 1. 1 qû is the work rate. |
| 6 | ki-a-am ne-pe₂-šum | Thus the procedure |
| 7 | i-na SAHAR-HI.A 1.30{gana₂} IRI{ki} na-ki-⸢ir⸣ {[d}AMAR].⸢UTU⸣ a-ṣa-ba-at | From the earth, 1 30 00 square meters: the city Nakī-ērir, Marduk, I shall call. |
| 8 | 6 UR₂ SAHAR-HI.A u₂-ki-⸢in 8 [a]-⸢na⸣ BAD₃ la sa-na-qam₂ | 6 layers of earth. 8 layers of wall not to be strewn. |
| 9 | 36 zi-⸢iq⸣-[pi] ⸢SAHAR⸣-HI.A ki ma-ṣi₂ US₂ | 36 ‘cutting-of-the-throat’ reed bundles, length by volume. |
| 10 | lu-uk-⸢bu⸣-[us₂ IRI{ki]} lu aṣ-ba-at u₃ US₂ EGIR | May I be able to enter the city. Further, the length is further. |
| 11 | hur{⸢hu⸣}[ru EN.NAM ZA.E IGI] 6 UR₂ SAHAR-HI.A DU₈.A 10 ta-mar 10 a-na | What are the heights? You: solve the reciprocal of 6. You will see 0;10, the volume. Multiply 10 by 1. |
| 12 | 1.30 SAHAR-HI.A i-ši] 15 ta-mar IGI 8 DU₈.A 7.30 ta-mar | Solve the reciprocal of 8. You will see 7;30. |
| 13 | 7.30 a-na 15 i]-⸢ši⸣ 1.52.30 ta-mar 1.52.30 TAB.BA | Multiply 0;07 30 by 15. You will see 1;52 30. The width is 1;22 30. |
| 14 | 3.45 ta-mar] 3.45 a-na 36 i-ši 2.15 ta-mar 1.52.30 | Multiply 3;45 by 36. You will see 2;15. 1 52;30 |
| 15 | [NIGIN 3.30.56.15 ta-mar 2.15 i-na 3.30.56.15 | [Total: 3 30 56 15]. You will see 2;15. [Multiply] by 3;30 56 15. |
| 16 | [BA.ZI 1.15.56.15 EN.NAM IB₂.SI₈ 1.07.30 ta-mar | What is the square-side of 1 15 56 15? You will see 1 07 30. |
| 17 | 1.07.30 i-na] 1.52.30 BA.ZI 45 ta-mar SUKUD BAD₃ | You will see 0;45, the height of the city wall. |
| 18 | 1/2 45 he-pe₂] 22.30 ta⸣-mar IGI 22.30 DU₈.A 2.40 | Break off 1/2 of 45. You will see 22;30. Solve the reciprocal of 22 30. 2;40. |
| 19 | [ta-mar 15 a-na] 2.40 i-ši 40 US₂ NIGIN₂.NA 1.30 ⸢SAHAR-HI⸣.A a-mur 22.30 | Multiply 0;15 by 2;40, the length. A total of 1;30 esaharḫu-plants. Look at 22;30. |
| 20 | [...] a-na 40 US₂ i-ši 15 ta-mar 15 a-na 6 i-ši | Multiply ... by 40, the length. You will see 0;15. [Multiply] 15 by 6. |
| 21 | 1.30 ta-mar 1.30 SAHAR-HI.A ne-pe₂-šum | You will see 0;01 30. The volume is 1;30. The procedure |
| 22 | i-na SAHAR-HI.A 1.30 IRI{ki} na-ki-ir {d}AMAR.UTU a-ṣa-ba-at | From the earth, 1 30 cities are destroyed. Marduk is seized. |
| 23 | iš-tu UR₂ SAHAR-HI.A 32 US₂ a-na pa-ni-ia al-lik | From the surface of the earth I marched 32 lengths in front of me. |
| 24 | 36 SUKUD SAHAR-HI.A US₂ EN.NAM lu-uk-bu-us₂ IRI{ki} | What are the length and volume? How much should I build a city? |
| 25 | lu aṣ-ba-at US₂ ša pa-ni hur{hu}ru EN.NAM ZA.E | The length is two-thirds of the face of the mountain. What are you? |
| 26 | IGI 32 DU₈.A 1.52.30 ta-mar 1.52.30 a-na 36 SUKUD | Solve the reciprocal of 32. You will see 1;52 30. Multiply 1;22 30 by 36. |
| 27 | i-ši 7.30 ta-mar IGI 6 UR₂ SAHAR-HI.A DU₈.A 10 ta-mar | You will see 0;07 30. Solve the reciprocal of 6 00, the surface of the earth, so that you will see 10. |
| 28 | 1.30 SAHAR-HI.A a-na 10 i-ši 15 ta-mar 15 a-na 2 TAB.BA | Multiply 1;30, the volume, by 10. You will see 0;15. [Multiply 15 by 2]. |
| 29 | 30 ta-mar 30 a-na 1.07.30 i-ši 33.45 ta-mar | Multiply 0;30 by 1 07 30. You will see 33 45. |
| 30 | 33.45 EN.NAM IB₂.SI₈ 45 IB₂.SI₈ 45 SUKUD BAD₃ | What is the square-side of 0;33 45? It is a square-sided square. The height of the circle is [0;45]. |
| 31 | 45 SUKUD BAD₃ UGU 36 SUKUD SAHAR-HI.A EN.NAM DIRI 9 DIRI | What are the height, width, and volume? The square-side is 0;45. What is the volume, and how much is 9? |
| 32 | IGI 1.07.30 DU₈.A 53.20 ta-mar 53.20 a-na 9 i-ši | Solve the reciprocal of 1 07;30. You will see 0;53 20. Multiply [0;03 20] by 9. |
| 33 | 8 ta-mar 8 US₂ a-na pa-ni-ka₃ ta-ka-ba-as | You will see 8. You will fold 8 lengths in front of you. |
| 34 | [i]-⸢na⸣ 4 AB.SIN₂ 30 im-qu₂-ut i-na 1/2 NINDA 1 SILA₃ ŠE | Four furrows, thirty fell from a rod, one qû is the work rate |
| 35 | ⸢e-si-ip⸣ [...] ⸢US₂⸣ 1(BUR₃){gana₂} A.ŠA₃{im} ŠE.GUR EN.NAM ZA.E | E-sip-e23 ... E-UŠ-E23 1 bur3 area: what is the debt of the king? |
| 36 | IGI 4 AB.SIN₂ ⸢DU₈.A⸣ 15 ⸢ta-mar 15 a-na 30 NINDA i-ši | The reciprocal of 4 furrows is 15. You will see 15. Multiply 15 by 30 rods. |
| 37 | 7.30 ta-mar IGI 7.30 DU₈.A 8 ta-mar 8 a-na 30 | You will see 0;07 30. Solve the reciprocal of 7;30. [You will see 8]. Multiply 8 by 30. |
| 38 | A.ŠA₃{im} i-ši 4 ta-mar US₂ 1(BUR₃){gana₂} A.ŠA₃ 8 DANNA-ma | You will see 4. The length is 1 bur. The area is 8 double-hours. |
| 39 | IGI 30 A.ŠA₃ DU₈.A 2 ta-mar 4 a-na 2 i-ši 8 | Solve the reciprocal of 30. You will see 2. Multiply 4 by 2. It gives 8. |
| 40 | ta-mar 8 i-na 1(DIŠ).3(U).6(AŠ) ŠE.GUR i-ba-ši-i | You will see 8. What is the dividing line of 1/6 (shekel?). |
| 41 | hi-ri-tum 10-TA.AM₃ mu-hu 18 SUKUD i-na 1 KUŠ₃ 1 ŠA₃.GAL | The width is 0;10 each. The height is 18 cubits. |
| 42 | sa₃-sum₂ u₃ SAHAR-HI.A <EN.NAM> ZA.E 5 u₃ 5 UL.GAR 10 ta-mar | You: sum 5 and 5. You will see 0;10. |
| 43 | 10 ⸢a⸣-na 18 SUKUD i-ši 3 ta-mar 3 i-na 10 BA.ZI 7 | Multiply 0;10 by 18. The height is 18. You will see 3. Subtract 3 from 10. |
| 44 | [ta-mar 7 sa₃-sum₂ NIGIN₂.NA sa₃-sum₂ u₃ mu-hu UL.GAR 17 ta-mar | You will see 7. Put down 0;07, the dividing line, the square-side, and the perimeter. You will observe 17. |
| 45 | 1/2 17 he-pe₂] 8.30 ta-mar NIGIN 1.12.15 ta-mar | Break off 1/2 of 17. You will see 8 30. Total: 1 12 15. |
| 46 | 1.12.15 GAR].⸢RA⸣ IGI 4(2-)GAL₂ DIRI ša mu-hu UGU | The reciprocal of 4 double-hours is 0;01 12 15. It is incumbent on |
| 47 | sa₃-sum₂ ⸢DIRI BA.ZI⸣ 45 a-na 1.12.15 DAH.HA-ma | Multiply 0;45 by 1;12 15, so that you see 4. |
| 48 | 1.13 ta-mar 18 a-na 1.13 i-ši 22.30 ta-⸢mar⸣ | You will see 1 13. Multiply 18 by 1 13 00. You will observe 22 30. |
| 49 | 2(EŠE₃).1(IKU).1/2(IKU){gana₂} SAHAR-HI.A ki-<a-am> ne-pe₂-šum | 2 e1 1/2 ikû of earth is the procedure |
| r iii | ||
| 1 | U₄.GAN₂.SAKAR 1 ŠU GUR₂ 50 DAL A.ŠA₃ <EN.NAM> | The month of Kislimu, the 1st of which followed the 30th of the preceding month, sunset to moonset: 1°. The length is 0;50. What is the field? |
| 2 | ZA.E 1 ŠU GUR₂ UGU 50 EN.NAM DIRI(SI) | You: multiply 1 by 0;50. What is the difference? |
| 3 | 10 DIRI 50 a-na 10 DIRI i-ši 8.20 | Multiply 0;50 by 10 00 00. The square-side is 8;20. |
| 4 | ta-mar 10 DAL NIGIN 1.40 ta-mar 1.40 | You will see 0;10, the diagonal. You will [see] 1;40. [1;40] |
| 5 | [i]-⸢na⸣ 8.20 BA.ZI 7.30 ta-mar | You will see 0;07 30. |
| 6 | ⸢U₄⸣.[GAN₂].⸢SAKAR⸣ A.ŠA₃ ne-<pe₂>-šum | E23.GAN.E22SAKAR₂ E23 means “hand-of-a-man’. |
| 7 | ⸢šum⸣-ma {giš}[MA₂] 1 SAR SIG₄ i-na-aš-ši-i | If a boat of 1 sar of bricks is to be brought |
| 8 | ⸢še⸣-a-am EN.⸢NAM i⸣-na-aš-ši ZA.E 41.40 SAHAR-HI.A | You, when you multiply 0;41 40, the volume. |
| 9 | 1 SIG₄-HI.A 41.40 a-na 5 GAN₂ i-ši 3.28.20 | 1 of bricks is 0;41 40. Multiply by 5 00 00, the surface, by 3 28 20. |
| 10 | SAHAR-HI.A 3.1/3 SILA₃ 8.1/3 GIN₂ <ŠE> 1 SIG₄ | The volume is 3 1/3 qû, 8 1/3 shekels. |
| 11 | 3.28.20 a-na 12 ŠU.SI i-ši 41.40 ta-mar | Multiply 3 28 20 by 12 fingers. You will see 0;41 40. |
| 12 | 8(AŠ).1(GEŠ₂).4(U) ŠE.GUR ta-mar ne-pe₂-šum | You will see 840 šE.GUR. The procedure |
| 13 | ⸢šum⸣-ma pa-ru-um 1 NINDA im-ta-[har ...] | If a pāru-container received 1 rod |
| 14 | [a]-na 1/2 pa-ri-im 2 PI [...] | For 1/2 pānum, 2 PI ... |
| 15 | x BA.ZI iš-tu x [...] | ... ... |
| 16 | x x x x [...] | |
| 17 | a-na mu-hi-šu [...] | against him |
| 18 | a-na SAG SAG [...] | for the head of the head ... |
| 19 | i-nu-ma 30 [...] | When the moon ... |
| 20 | le-qe₂ 10 [...] x | Take 10 ... |
| 21 | 50 ta-mar 50 [...] x | You will see 0;50. |
| 22 | a-na še-pi₂-šu [...]-im | to the ... of ... |
| 23 | a-na 50 1/2 DAH.[HA ... ta]-mar 1.05 1/2 | Add 1/2 to 50. You will see 1 05 1/2. |
| 24 | iš-tu SAG [...] x a-na še-pi₂-ia | From the beginning ... to my head |
| 25 | ⸢ki-a-am⸣ [ne]-pe₂-šum | the ritual of E-am-e-23 |
| 26 | [...] x x AN.TA 2 | ... upper 2 |
| 27 | [... uṭ]-⸢ṭe₄⸣-ti-ša EN.NAM ZA.E 40 | What is its uṭeti function? You: 0;40 |
| 28 | [a-na 1.12 IGI.GUB i-ši 48 ta-mar 48 | Multiply by 1;12, the constant. You will see 48. |
| 29 | [...]-nu-um uṭ-ṭe₄-ti-ša | ...-num, its ... |
| 30 | ⸢ki⸣-a-am ne-pe₂-šum | The ritual of Ea |
| 31 | 30 US₂ 10 A.ŠA₃ SAG EN.NAM ZA.[E 1/2 30 he-pe₂] | What are the length and width? You: break 1/2 of 30. |
| 32 | 15 ⸢ta⸣-[mar] IGI 15 DU₈.A 4 ⸢ta⸣-mar | 0;15, the reciprocal of 15. You will see 4. |
| 33 | 10 a-na 4 [i-ši 40 ta-mar 40 a-na 30 | Multiply 0;10 by 4 00. You will see [0;40]. Multipl[ay] 1 00 by 30. |
| 34 | US₂ i-[ši 20 ta]-⸢mar⸣ SAG | The length is 0;20. You will see the width. |
| 35 | pi₂-sa-nu-um 3 [ŠU.SI GUR₂] ⸢ak⸣-pu-pa(UL-)am | Basket-of-tablets: 3 fingers, xxx |
| 36 | A ma-ah-ru-tum ki-⸢ma⸣ [...]-ša-di-im | son of Aḫi-rūtum, the ... of the netherworld |
| 37 | wa(ŠI-)ar-ku-tum i-na ka-[...] ⸢ga⸣-am-ru | “It is an arrow in the mouth of a lion |
| 38 | SUKUD u₂-ri-di-im EN.NAM ZA.⸢E⸣ 3 NIGIN 9 ⸢ta⸣-mar | What is the height of a circle? |
| 39 | 9 a-na 5 GUR₂ i-ši 45 ta-⸢mar⸣ [IGI 45 DU₈].⸢A⸣ | Multiply 9 by 5 00 00. You will see 0;45. [Return the reciprocal of 0] 45. |
| 40 | 1.20 ta-mar IGI 6.40 a-na 9 i-ši [...] | You will see 1;20. Multiply the reciprocal of 6;40 by 9. |
| 41 | 12 NINDA SUKUD u₂-ri-id(TE) ne-pe₂-⸢šum⸣ | 12 nindanu, the height of a rope, ritual procedure |
| 42 | 1 SILA₃ pi₂-<sa-nu-um> 10 ŠU.SI am-ṣu₂-ur US₂ EN.⸢NAM⸣ | 1 qû of pisanum-flour, 10 fingers of amšūr-gold, the length of the en-priestess. |
| 43 | al-<<li>>-lik ZA.E 10 NIGIN 1.40 ta-mar 1.40 | You: square 0;10. You will see 1;40. |
| 44 | a-na 5 GUR₂ i-ši 8.20 ta-mar IGI 8.20 DU₈.A | Multiply by 5 00 00. You will see 8 20. Solve the reciprocal of 8;20. |
| 45 | 7.12 ta-mar IGI 6.40 a-na 7.12 i-ši | You will see 7 12. Multiply the reciprocal of 6 40 by 7 12. |
| 46 | 1.04.48 US₂ ta-mar ne-pe₂-šum | You will see 1 04 48, the length. The procedure. |
| 47 | ŠU.NIGIN₂ 35 ki-bi-is ŠID | A total of 35 kibris, harvested. |
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