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CT 44, 39 (P356054)
Mathematical, Old Babylonian, Sippar-Yahrurum · c. 1900 BCE–1600 BCE
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| Designation | CT 44, 39 |
|---|---|
| ID | P356054 |
| Museum no. | BM 080209 |
| Collection | British Museum, London, UK |
| Language | Akkadian |
| Period | Old Babylonian · c. 1900 BCE–1600 BCE |
| Genre | Mathematical |
| Find spot | Sippar-Yahrurum (mod. Tell Abu Habbah) |
| Coordinates | 33.0605, 44.2552 |
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 | ||
| 1 | [šum-ma n im]-ta-har A.ŠA₃ mi-nu-um | If ... has received, what is the field? |
| 2 | [šum-ma n] ⸢im⸣-ta-har A.ŠA₃ mi-nu-um | If n (shekels of) emtahar-plant, what is the field? |
| 3 | [šum-ma] 20 im-ta-har DAL mi-nu-um | If the sun faces the sky, what is the dividing line? |
| 4 | [šum]-ma 10 im-ta-har di-ik-šum mi-nu-um | If the square-side is 0;10, what is the thickness? |
| 5 | šum-ma 8.20 A.ŠA₃ GUR₂ mi-nu-um | If the area is 8;20, what is the dividing line? |
| 6 | šum-ma 2.13.20 A.ŠA₃ GUR₂ mi-nu-um | If the field is 2 13 20, what is the dividing line? |
| 7 | šum-ma 3.28.20 A.ŠA₃ GUR₂ mi-nu-um | If the field is 3 28 20, what is the dividing line? |
| 8 | šum-ma 5 A.ŠA₃ GUR₂ mi-nu-um | If the field is 5 00, what is the dividing line? |
| 9 | a-na A.ŠA₃ GUR₂ 1/2 US₂ DAH 8.25 | Add 1/2 length to the area of a field: 8 25. |
| 10 | i-na A.ŠA₃ GUR₂ 1/2 US₂ <BA.ZI> 8.15 | From the field, ... 1/2 length is 0;08 15 |
| 11 | a-na A.ŠA₃ GUR₂ 1 US₂ DAH 8.30 | Add 1 length to the area of a field: 8 30. |
| 12 | i-na A.ŠA₃ GUR₂ 1 US₂ BA.ZI 8.10 | Multiply 1 length by 0;00 08 10 |
| 13 | a-na A.ŠA₃ GUR₂ 1.1/3 US₂ DAH 8.33.20 | Add 1 1/3 lengths to the area of a field: 8 33 20. |
| 14 | i-na A.ŠA₃ GUR₂ 1.1/3 US₂ BA.ZI 8.06.40 | Multiply 1 1/3 lengths by 0;00 08 06 40. |
| 15 | a-na A.ŠA₃ GUR₂ 1.1/2 US₂ DAH 8.35 | Add 1 1/2 lengths to the area of a field: 8 35. |
| 16 | [i]-⸢na A⸣.ŠA₃ GUR₂ 1.1/2 US₂ BA.ZI 8.05 | When you multiply by the volume of a reed, you multiply 1 1/2 lengths by 8 05. |
| 17 | [a-na A].ŠA₃ GUR₂ 1.2/3 US₂ DAH 8.36.40 | Add 1 2/3 lengths to the area of a field: 8 36 40. |
| r | ||
| 1 | [i-na A.ŠA₃] ⸢GUR₂⸣ 1.2/3 US₂ BA.ZI 8.03.20 | From a field of eškur-tree 1 2/3 lengths is the constant of 8 03 20. |
| 2 | [a-na A.ŠA₃] GUR₂ 1 IGI-4-GAL₂ US₂ DAH 8.32.30 | Add 1 1/4 lengths to the area of a field: 8 32 30. |
| 3 | i-na A.[ŠA₃ GUR₂ 1 IGI-4-⸢GAL₂⸣ US₂ BA.ZI 8.07.30 | Take away 1 1/4 lengths from the area of a square: 8 07 30. |
| 4 | a-na A.⸢ŠA₃⸣ [GUR₂ 1 IGI]-5-GAL₂ US₂ [DAH 8.32 | [To add] 1 00, the length, to the depth. 8 32 |
| 5 | i-na A.⸢ŠA₃⸣ [GUR₂ 1 IGI-5-GAL₂ US₂ ⸢BA.ZI⸣ 8.08 | From the depth of the reed bed 1 5/8 lengths is 8 08 |
| 6 | A.ŠA₃ 2 GUR₂ UL.GAR 41.40 GUR₂ UGU GUR₂ 10 DIRI | Field: 2 00 00, completed. 41 40 00 is the dividing line of 10 00. |
| 7 | A.ŠA₃ 2 GUR₂ UL.GAR 3.28.20 GUR₂ UGU GUR₂ 10 DIRI | Solve the reciprocal of 2 00, the square-side. 3 28 20 00 squares 10 00 00. |
| 8 | A.ŠA₃ 2 GUR₂ UL.GAR 5.39.40 GUR₂ UGU GUR₂ 10 DIRI | A field is 2 00 00. Solve the reciprocal of 0;05 39 40. Add 10 to the 00 10 00, the length. |
| 9 | A.ŠA₃ 2 GUR₂ UL.GAR 8.28.20 GUR₂ UGU GUR₂ 10 DIRI | Solve the reciprocal of 2 00, the square-side. 8 28 20 00 squares 10 00 00. |
| 10 | A.ŠA₃ GUR₂ DAL GUR₂ u₃ si₂-hi-ir-ti GUR₂ UL.GAR-ma; 8.32.20 | Solve the reciprocal of 0;00 08 32 20 so that it gives 8 32 20. |
| 11 | A.ŠA₃ GUR₂ DAL GUR₂ u₃ si₂-hi-ir-ti GUR₂ UL.GAR-ma 1 | Add the dividing line of the šūzû-mark and the square-side (of the circle) so that (it is) 1. |
| 12 | [A].⸢ŠA₃⸣ [GUR₂] ⸢DAL⸣ GUR₂ ⸢u₃⸣ si₂-hi-ir-ti GUR₂ UL.GAR-ma 1.55 | Add the šû-song, the kurru of the ēmurru-house (and) the reed bundles, so that 1;55. |
| 13 | [A.ŠA₃ GUR₂ DAL] GUR₂ u₃ si₂-hi-ir-ti GUR₂ UL.GAR-ma 3.06.40 | Solve the reciprocal of 0;03 06 40 so that it gives 3 06 40. |
| 14 | 3.06.40 15.33.20 40 | [3 06 40] 15 33 20 40 |
| 15 | [...] 26.40 | ... 26;40 |
| 16 | [... 16 4 3.20 40 | ... 16 4 3 20 40 |
| 17 | 1.55 9.35 33.40 4.40 | 1 55: 9 35 33 40: 4 40 |
Sources & licences
- Cuneiform Digital Library Initiative · CDLI: reuse of catalogue/transliteration data with citation of CDLI; images restricted (not fetched) · Cuneiform Digital Library Initiative (CDLI), https://cdli.earth/ ; cite CDLI and the P-number
- 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) · ORACC project editions, https://oracc.museum.upenn.edu/ ; cite the individual project (see texts.extra_json.project)
Cite the original edition and translator. Page generated by Unread Atlas from the public corpus export of 2026-10-18.