A C130 lands on a carrier. How much energy must be dissipated and where does it all go?

Question

Assumptions must be described and answers must be math justified. Extra credit for plausible obscurity.

c130 carrier landing

update: From the video the landing roll was 270ft at 85klb and 460ft at 121klb using upgraded antilock brakes and full reverse thrust.

Answer 1

The kinetic energy to dissipate is easy to calculate:

$$ E_\text{kin} = \frac{1}{2} m \times \text{GS}^2 $$

Here, the ground speed GS must be w.r.t. to the carrier deck, not the sea.

The landing weight of the C-130 must be between the Operating Empty Weight ($\sim 75 \, 000 \, \text{lbs}$) and the Maximum Landing Weight ($\sim 130 \, 000 \, \text{lbs}$). These tests were performed at $85 \, 000 \, \text{lbs}$ and $121 \, 000 \, \text{lbs}$. To figure out the landing speed, we can look at this graph for full flaps from the performance manual:

We will use the touchdown speeds from the graph for the calculation, which are defined as

The touchdown speeds presented in figures A8-1 and A8-2 are the initial speeds at the point of contact in landing. [...] The speeds represent 1.2 times power-off stall speeds for each flap setting.

Hypothetically, one could touch down at the stall speed, but it's probably more realistic to use the factor of 1.2 times stall speed here.

However, these speeds are airspeeds. To calculate the ground speed, we need to subtract the headwind component. The video says it was up to 50 kt, but one cannot always get that. Some comments under the video note that a modern aircraft carrier can always achieve 32 kt, when travelling into the wind.

Let's plug in a few numbers to see how much kinetic energy we get:

Weight	Airspeed	Headwind	Energy
75 000 lbs	81 kt (Stall)	50 kt	4.3 MJ
75 000 lbs	97 kt	50 kt	9.9 MJ
75 000 lbs	97 kt	32 kt	19.0 MJ
85 000 lbs	97 kt	50 kt	11.3 MJ
121 000 lbs	104 kt	50 kt	21.2 MJ
130 000 lbs	89 kt (Stall)	50 kt	12.0 MJ
130 000 lbs	107 kt	50 kt	25.4 MJ
130 000 lbs	107 kt	32 kt	43.9 MJ

^{(Airspeed is touchdown speed read from the graph, or stall speed without the safety factor of 1.2)}

While megajoules sounds like quite a lot, these numbers are surprisingly small, mostly due to the relatively slow ground speed after subtracting the headwind. For reference: an A380 RTO corresponds to about 1.7 GJ according to this answer.

Where did it all go?

Usually, most energy is dissipated by the wheel brakes, which should be more than capable of absorbing this energy (a single brake on the A380 can absorb 120 MJ). However, the C130 can also use reverse thrust during landing:

Reverse thrust is applied by moving the throttles from FLIGHT IDLE to GROUND IDLE, and then into REVERSE range in coordination with nosewheel steering.

^{(C130 Flight Manual - 2-55 - Normal Landing)}

This could further reduce the energy going into brakes, but since the video shows the C-130 stopping very quickly, there was likely not much time spent in reverse.