Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
小齿轮上的转矩
ω = 2π·2900/60 = 303.7 rad/s
T = P/ω = 11000/303.7 = 36.22 N·m
分度圆直径
d = m·N = 5·18 = 90 mm
切向力
Ft = 2T/d = 2·36.22/0.09 = 804.9 N
答案
Ft = 804.9 N
4.齿轮与齿轮系
答案: ND = 32
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
中间轴转速
nB = nC = 2880·20/40 = 1440 rpm
齿轮 D 的齿数
ND = C·nC/nout = 20·1440/900 = 32
答案
ND = 32
5.数据采集与采样
答案: M = 14.4 MB
数据速率:每个通道每秒提供 fs 个 N 位的采样,因此原始速率为 通道数 × fs × N bit/s。在内存中一个采样占用整数个字节(12 位采样存为 2 个字节),因此一段记录的内存为 通道数 × fs × 每个采样的字节数 × 时间。多路复用 ADC 依次转换各通道:每次转换(含建立时间)需要 t,所以 n 个通道转一轮需要 n·t,每个通道的采样频率为 1/(n·t)。