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1 lbf·ft to N·m through the SP 811 factor table, the ½-13 UNC thread of Handbook H28 Table 2.8 and 12 AWG copper of Handbook 100 Table 5
| Tool | Engineering Constants and Tables |
| Source | Thompson, A., and Taylor, B. N. (2008). Guide for the Use of the International System of Units (SI). NIST Special Publication 811, Appendix B.8 (pound-force foot → newton meter); National Bureau of Standards (1969). Screw-Thread Standards for Federal Services, Part I. NBS Handbook H28, Table 2.8 and Appendix A5; National Bureau of Standards (1966). Copper Wire Tables. NBS Handbook 100, Table 5. |
| Level | Undergraduate |
Problem
Convert 1 pound-force foot to newton metres with the NIST SP 811 factor and show the unit algebra. Find the basic pitch diameter, the internal and external minor diameters, the lead angle, the area at the minor diameter and the tensile stress area of a ½-13 UNC thread from the H28 formulas and compare with Table 2.8; give the shear areas per inch of engagement at basic dimensions and the length of engagement that develops the full strength of the screw. Find the diameter, circular-mil area, resistance per 1,000 ft at 20 °C and mass per 1,000 ft of 12 AWG standard annealed copper by the gage law and compare with Handbook 100 Table 5.
Inputs to Enter
- Converter: value 1, from lbf·ft, to N·m
- Unified thread: D 0.500 in, n 13 threads per inch; metric thread d 10 mm, p 1.5 mm
- Wire: 12 AWG at 20 °C; search blank, all tables, 25 rows per table
Answers
- 1 lbf·ft = 1.355 818 N·m (SP 811 B.8 row 'pound-force foot (lbf · ft) → newton meter (N · m) × 1.355 818 E+00', rounded; the exact 0.453 592 37 × 9.806 65 × 0.3048 = 1.355 817 948); dimensions M L² T⁻², base units kg·m²·s⁻²
- ½-13 UNC: p = 0.076 923 in, E = D − 0.649 519 p = 0.4500 in, K_n = D − 1.082 532 p = 0.4167 in, K_s = D − 1.226 869 p = 0.4056 in, λ = atan[1/(nπE)] = 3° 7′, area at D − 2h_b = 0.1257 in², A_t = 0.7854 (D − 0.9743/n)² = 0.1419 in² (all as Table 2.8 prints them)
- Shear areas at basic dimensions: external 0.75 π K_n = 0.982 in² per inch of engagement, internal 0.875 π D = 1.374 in² per inch, equal-strength πE/2 = 0.707 in² per inch; L_e = 4A_t/(πE) = 0.401 in for equal materials
- M10 × 1.5: d₂ = 9.026 mm, D₁ = 8.376 mm, d₃ = 8.160 mm, A_s = (π/4)[(d₂ + d₃)/2]² = 58.0 mm² (the ISO 898-1 value)
- 12 AWG: d = 0.005 × 92^(24/39) = 80.81 mils = 2.053 mm, 6,530 circular mils (0.005 13 in²), R₂₀ = 10.371/6,530 × 1,000 = 1.588 Ω per 1,000 ft (Table 5: 1.59), 1.93 Ω at 75 °C (Table 6: 1.93), 19.77 lb per 1,000 ft (Table 5: 19.8)
Notes
The tool's default inputs are this case; press Submit. Every number is a definition or a printed table value: test_engineering_tables.php (2,741 checks) matches the 445 SP 811 factors to an independent parse of the 2008 PDF text layer, re-derives all 82 printed H28 rows and all 44 Handbook 100 rows from the formulas, and checks the default against the numpy script in References/Engineering-Tables/checks/. Two handbook misprints are kept as printed and flagged (H28 Table 2.10 lead angle of 1.375-18 UNEF, 0° 48′ for 0° 45′; Handbook 100 No. 19, 1200 circular mils for 1289).