Fulling Mill Trip Hammers

What is modelled

The stream delivers power rho g Q H to the wheel, which turns at a rim speed set by the jet. Each cam lifts a hammer of mass m by h, costing m g h of shaft work divided by an assumed train efficiency, and the hammer then falls freely and delivers m g h at the cloth. If the stream cannot supply all the lifts at the free wheel speed, the wheel slows until power balances. If the cam period is too short for the hammer to finish falling, the next cam catches it early and the blow is weaker. The 3D page shows the wheel, camshaft and stamps; their numbers are computed here.

Equations

• Water power P = ρ g Q H η; free rim speed taken as a quarter of the jet speed sqrt(2 g H).
• Work per lift m g h / ηm with η_m = 0.8 (assumed).
• Wheel speed n = min(nfree, P / (hammers x cams x work)); cam period T = 1 / (cams n).
• Fall time t = sqrt(2h/g), impact speed sqrt(2 g h); effective drop heff = g ((1 - 0.35) T)2 / 2 when the fall does not fit.
• Shaft torque from virtual work τ = m g (dy/dθ) / ηm summed over hammers.

Checks run in the page

• A 1 microsecond free-fall integration against sqrt(2h/g) and sqrt(2gh).
• Integral of the torque curve against cams x hammers x work per lift.
• Counting the cam passes of 60 s of wheel rotation against the closed-form lift rate.
• Mean torque times angular speed against the shaft power.

Honest simplifications

A smooth cosine cam profile with a fixed 35 percent lifting share, rigid wood parts, a hammer treated as a point mass falling freely, no impact losses or rebound, assumed efficiencies, and a rim speed taken as a fraction of the jet speed. The figures are illustrative, not measurements of the historical mill.

• Press Dry season and watch the wheel slow and the blow rate drop because the stream cannot supply all the lifts.
• Press Too many cams, high lift: the fall no longer fits between cams and the blow energy collapses.
• Add hammers and see the shaft torque curve smooth out.