page 68

68 / 102 Russian, pre-1918 confidence 0.95 not checked by a human

Provenance

transcript transcript: 2026-09-29T23:26:40+00:00 machine-generated
normalization normalization: 2026-10-02T18:15:39+00:00 machine-generated

Full TPME chain

[
  {
    "_schema": {
      "id": "https://doi.org/10.5281/zenodo.21287085",
      "name": "AI4LAM TPME",
      "version": "1.0"
    },
    "conventions": [
      "diplomatic",
      "https://github.com/selic/bookolot/blob/main/docs/transcription-guide.md"
    ],
    "features": {
      "bbox_space": "deskewed page, see deskew step",
      "line_bbox": true,
      "line_confidence": true,
      "word_bbox": false
    },
    "file_format": "ALTO 4",
    "human_review_level": "machine-generated",
    "media_id": "http://www.moldavica.bnrm.md/biblielmo?l=ro\u0026a=d\u0026c=carti\u0026d=JD3364",
    "media_title": "Priblijonnoe iscislenie",
    "modification_date": "2026-09-29T23:26:40+00:00",
    "originating_media_file": "68.tif",
    "processing": [
      {
        "application_name": "kraken blla",
        "application_parameters": {
          "device": "cuda",
          "model": "blla.mlmodel (встроенная)"
        },
        "application_provider": "Benjamin Kiessling",
        "application_repo": "https://github.com/mittagessen/kraken",
        "application_type": "line segmentation",
        "application_version": "7.1.1"
      },
      {
        "application_name": "kraken",
        "application_parameters": {
          "device": "cuda"
        },
        "application_provider": "Benjamin Kiessling",
        "application_repo": "https://github.com/mittagessen/kraken",
        "application_type": "OCR",
        "application_version": "7.1.1",
        "inference_model": "ppocrv6_medium (sha256:15313b51ace64cbf)"
      }
    ],
    "provider": "Bookolot",
    "transcript_checksum": "sha256:c49af2f6a60fa1a3116520c4197872559c4589015dc0a61af59a5851829ecd03",
    "transcript_id": "bookolot:JD3364:68:ocr-v1",
    "transcript_language": [
      "ru-petr1708"
    ],
    "type": "transcript"
  },
  {
    "_schema": {
      "id": "https://doi.org/10.5281/zenodo.21287085",
      "name": "AI4LAM TPME",
      "version": "1.0"
    },
    "characteristics_note": "For reading and search only; quote the parent transcript.",
    "human_review_level": "machine-generated",
    "media_id": "http://www.moldavica.bnrm.md/biblielmo?l=ro\u0026a=d\u0026c=carti\u0026d=JD3364",
    "media_title": "Priblijonnoe iscislenie",
    "modification_date": "2026-10-02T18:15:39+00:00",
    "originating_media_file": "68.tif",
    "parent_transcript_id": "bookolot:JD3364:68:ocr-v1",
    "processing": [
      {
        "application_name": "bookolot.rag.normalize",
        "application_parameters": {
          "normalizer_version": 3,
          "script": "ru_prereform"
        },
        "application_provider": "Bookolot",
        "application_repo": "https://github.com/selic/bookolot",
        "application_type": "spelling normalization",
        "application_version": "fb8b81d-dirty"
      }
    ],
    "provider": "Bookolot",
    "transcript_checksum": "sha256:98dea7decc345b348ffb81a9d567c141650c6fa13d902fb0d65639f21268e441",
    "transcript_id": "bookolot:JD3364:68:modern-v3",
    "transcript_language": [
      "ru-t-ru-petr1708"
    ],
    "type": "normalization"
  }
]

confident worth a look check the scan Open the full scan

Text as printed modern spelling both

as printed − 1 + modern spelling − 5 +

0 -65 – 0.86
1 V37 = 6,0827+0,0001: 0.96
2 V 43 = 6,5574+-0,0001 0.91
3 V37+V 43=12,6401+0,00027 0.92
4 V12,64010000 — 3,5552 0.94
5 Примѣръ III. 0.95
6 у—3,5553—1-0,00018 0.94
7 Вычислить = 52378,4137+ V 73 съ точн. до-+0,1. 0.89
8 V 73 равенъ 8 съ дробью, поэтому целая часть 52378,4137+V 73 0.97
9 будетъ выражаться 5-ю цифрами, а цѣлая часть квадратнаго 0.99
10 корня изъ этого числа будетъ выражаться тремя цифрами. Весь 1.00
11 квадратный корень будетъ выражаться 4-мя цифрами. Подко- 0.98
12 ренную величину можно укоротить, начиная съ сотыхъ долей, 1.00
13 на три цифры, т. е. до десятковъ. Такъ какъ V73 менѣедвухъ 0.97
14 десятковъ, то мы можемъ его отбросить и тогда V52378,41 бу¬ 0.99
15 детъ приближеннымъ значеніемъ ꙁ съ требуемой точностью. 0.99
16 V 5.23.78,41 – 228,8 0.88
17 42 123 0.87
18 84 1.00
19 448 3978 0.90
20 3584 1.00
21 4568|39441 0.95
22 37544 1.00
23 =228,9-+-0,1. 0.76
24 Извлеченіе кубическаго корня. 1.00
25 54. Извлекаемъ кубич. корень изъ целаго часла 53784237 0.99
26 съ точи. до+1 но обыкновенному способу. 0.97
27 V 53-784-237 =377 0.89
28 27 0.99
29 27 26 784 0.90
30 23 653 0.99
31 4107 1.00
32 3 131 237 0.98
33 2 929 633 0.98
34 201 604 1.00
35 Найденное такимъ образомъ число 377 есть приближен¬ 1.00
36 ный кубическій корень изъ цѣлаго числа 53784237 съ точи. 0.98