
Active Heave Compensation
Understand the control. Separate peak power from energy over the duty.
Two numbers. Two different design questions.
The existing 100 t inline-AHC example gives both. Peak actuator power describes a rate; battery energy accumulates over the stated ten-hour duty.
How fast must work be delivered?
The highest mechanical power in the stated cycle. It is not an installed supply rating or an average draw from the battery.
How much energy does the duty use?
The ideal net energy after the declared drive and regeneration efficiencies. It is not a specified battery capacity.
100 t payload in air · 1 m sinusoidal amplitude · 10 s period · 10 h / 3,600 cycles. Inline gas-supported architecture: R = 10, maximum stroke 4 m, γ = 1.4.
Efficiency convention: 90% battery-to-actuator drive efficiency; 50% of available negative mechanical work returned to the battery.
Drive draw minus recovered energy.
The original calculation integrates 130.2 MJ of positive work and 130.2 MJ of available return work over ten hours. Apply each efficiency to its own path.
130.2 MJ ÷ 0.90 ÷ 3.6
130.2 MJ × 0.50 ÷ 3.6
79.6 MJ over ten hours
Energy-path values are rounded to one decimal place. The full calculation and efficiency definitions remain below.
Read the sign of power before adding the work.
Red marks drive work. Navy marks mechanical work available for regeneration; only the declared fraction returns to the battery.
Original three-line worked summary
The energy budget, in three lines
- The case. 100 t payload in air, sinusoidal zeta equals 1 metre at a period T of 10 seconds; inline AHC with gas ratio R equals 10, maximum stroke S max equals 4 metres, and the adiabatic nitrogen exponent gamma equals 1.4. Here drive efficiency eta drive equals 0.90 is battery-to-actuator drive efficiency and regeneration efficiency eta regen equals 0.50 is the fraction of negative mechanical work returned to the battery.
- The physics. Over ten hours, positive drive work and available return work are each 130.2 MJ (36.16 kJ of each per cycle). Peak mechanical actuator power is 11.6 kW.
- The answer. battery energy equals drive work divided by drive efficiency, minus regeneration efficiency multiplied by return work; the result is 79.6 megajoules, equal to 22.1 kilowatt-hours.
This is an ideal screening calculation with an explicit gas and efficiency convention. Friction, hydraulic losses, real-gas behaviour and full pump/motor/battery maps refine it — that project model is part of a CONSTELLATION study.
This idealised case excludes friction, auxiliaries, real-gas behaviour and detailed pump, motor and battery maps. Project sizing also needs the actual duty and motion spectrum, losses, control demands, reserve margin and operating limits.
Read the full model and six original equations ↓