page 102

102 / 102 Russian, pre-1918 confidence 0.90 not checked by a human

Provenance

transcript transcript: 2026-09-29T23:43:13+00:00 machine-generated
normalization normalization: 2026-10-02T18:15:40+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": "2026-09-29T23:43:13+00:00",
    "originating_media_file": "102.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": "sha256:62df2c915f8960d62c395abaaf21a33943d385b664373a6e96dcb8bc708e43ea",
    "transcript_id": "bookolot:JD3364:102: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": "2026-10-02T18:15:40+00:00",
    "originating_media_file": "102.tif",
    "parent_transcript_id": "bookolot:JD3364:102: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": "sha256:0b4a3c73b5522391bfc7ff70ae04d57ba6c3d01a6f34fef7f7d2f51c39c0fdfd",
    "transcript_id": "bookolot:JD3364:102: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 − 2 + modern spelling − 3 +

0 –99 - 0.77
1 75. Приводимъ здѣсь примѣры на вычисленіе посредствомъ 0.99
2 логариемовъ, съ целью показать, какъ опредѣляется погрѣш¬ 0.95
3 ность въ результатв вычисленія, зависящая отъ неточностп 0.98
4 логариомовъ. Замѣтимъ, что, вычисляя алгебранческую формулу 0.98
5 посредствомъ логариемовъ, мы не можемъ предъявдять требо- 0.98
6 ванія вычпслнть искомую величину съ заранђе заданной точ¬ 0.96
7 ностью, а можемъ предъявить требованіе вычислить ее съ най¬ 0.98
8 болышею точностью, допускаемой неточностью тѣхъ или другихъ 0.99
9 логариомическихъ таблицъ. 0.97
10 Вычислить съ наӥбольшей точностью. 0.98
11 U=3 π ³, = 537,41863 0.82
12 logr=2,730313-+ 0.82
13 3logr=8,190939-+-3κ 0.78
14 π=0,497150-+-0,5 0.83
15 log 4 =0,602060 - 1-0,50 0.87
16 9,290149-+4 0.84
17 log 3 =0,477 121-+0,55 0.89
18 log U=8,813028-+-4,5+ 0.83
19 U= 650171000—+ 7000, 0.90
20 Вычислить съ найбольшей точностью катетъ с и углы 0.99
21 В и С прямоугольнаго треугольника, въ которомъ даны гиппо- 0.99
22 тенуза а и катетъ . 0.96
23 a=43,7842 0.89
24 b=27,4038 0.95
25 c=V(a+b)(a -b) sin B= 0.89
26 log(a –b) =1 214335-+» 0.82
27 log(a +b) =1,852407-+-0,5v 0.85
28 3,066742-+-1,53 0.87
29 logc= 1,533371-+-0,87 0.92
30 c== 34, 1484-1-0,00026 0.88
31 logb = 1,437811-+ε 0.87
32 loga= 1,641317-+0,56 0.94
33 logsin A= 9,796494-+1,5v 0.92
34 A= 38°44'50"+-2"ω 0.81
35 B= 41015'101-2"ω 0.84
36 a 0.98
37 a=43,7842 0.89
38 b= 27.4038 0.91
39 a – b = 16,3804 0.85
40 +=71,1880 0.92