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Landing-gear loads of a 2,400 lb Part 23 airplane: 23.473 descent velocity, the drop-test energy balance, the Appendix C conditions and the strut trunnion loads
| Tool | Landing Gear Loads |
| Source | Federal Aviation Administration (2016). Code of Federal Regulations, Title 14, Part 23, Sections 23.473, 23.479–23.509, 23.725 and Appendices C and D (2016 edition, Amendment 23-62). U.S. Government Publishing Office, Washington, DC; Milwitzky, B., and Cook, F. E. (1953). Analysis of Landing-Gear Behavior. NACA Report 1154 (Figure 11, strut effectiveness). |
| Level | Undergraduate |
Problem
A 2,400 lb normal-category airplane with 160 ft² of wing lands on two main gears whose oleo struts have a 6 in stroke at 80 % efficiency with 2 in of tire deflection at 47 % efficiency; the nose wheel is 57 in ahead of the c.g., the mains 8 in behind it, the c.g. 48 in above the ground and the track 100 in. Find the limit descent velocity of 14 CFR 23.473(d), the vertical ground reaction and the landing load factors from the energy balance with two-thirds wing lift, the drop-test quantities of 23.725, the Appendix C landing conditions and the braked-roll, nose-wheel and towing loads, and the largest axial load and bending moment at the trunnion of a vertical cantilever strut 20 in above the axle with a 7.5 in loaded tire radius.
Inputs to Enter
- Rule Part 23 (2016 ed.); W 2,400 lb, S 160 ft²; descent velocity and lift at the rule values; ultimate factor 1.5
- Strut stroke 6 in at η_s 0.80, tire deflection 2 in at η_t 0.47, two main units, rake 0°, unsprung weight 40 lb
- A 57 in, B 8 in, E 48 in, T 100 in, tail-down 12°; axle to trunnion 20 in, trail 0, loaded tire radius 7.5 in
Answers
- W/S = 15 lb/ft²: V = 4.4 · 15^(1/4) = 8.66 ft/s (23.473(d), within 7–10 ft/s); 23.725(a) drop height 3.6 √15 = 13.9 in, the same case since √(2 g h) = 4.39 (W/S)^(1/4)
- Energy balance ½ (W/g) V² + (W − L)(S_s + S_t) = F (η_s S_s + η_t S_t): 2,797 + 533 = 3,330 ft-lb = F (0.8 · 0.5 + 0.47 · 0.1667) ft → F = 6,961 lb on both mains, 3,481 lb per gear
- Load factors: n_g = F/W = 2.90, n = n_g + L/W = 3.57 (≥ 2.67 and ≥ 2.0 of 23.473(g)); ultimate 5.35; struts absorb 84 % of the energy; drop-test form W_e = 909 lb, n_j = 3.83, n = n_j W_e/W_M + L = 3.57; reserve energy at 1.2 V with lift = weight: 4,209 lb per gear
- Appendix C level landing: V = 3,481 lb, D = K n W/2 = 0.25 · 3.567 · 2,400/2 = 1,070 lb per main; side-load condition 1,596 lb vertical with 1,200 lb inboard / 792 lb outboard; static reactions 295 lb nose, 1,052 lb per main
- Braked roll three-point (n = 1.33, μ 0.8): mains 1,760 lb total (880 each) and nose 1,432 lb by V_m = n W A/(A + B + μ E); nose wheel 2.25 × static = 665 lb; towing 540 lb main, 720 lb auxiliary
- Trunnion: largest axial load 3,481 lb (level landing), largest bending moment 35,100 lb-in (braked roll on the mains alone, 1,277 lb drag × 27.5 in)
Notes
The tool's default inputs are this airplane; press Submit. The rule constants, the energy balance, the 23.725 effective-weight identity, the 25.493(e) braking formula, the trunnion resolution and the Milwitzky–Cook generalized impact (Report 1154 Figure 11 and the TN 2755 Figure 12 example, 2.50 g) are checked in test_landing_gear_loads.php against the filed rule text and an independent numpy/scipy computation (References/Landing-Gear/checks/gearloads_check.py). The brief's 7 ft/s is the rule's floor, not its value at this wing loading.