Mathematical question.
 

I'm aware of a story of a compassionate case needing to be sent home from S Georgia PDQ (some time ago, details of the case not really relevant).

The resident frigate is despatched from the Falklands area towards S Georgia at max speed (say 30 kt). Once within Lynx range (say 300nm), Lynx is launched to pick up pax, refuel and return to ship for transit to FI and onward flight home.

Let's say the helo can do 150kt, and (unrealistically) winds are still and there are no sea currents opposing or aiding the ship's progress.

What is the best plan for the ship? Continue on toward S Georgia for a period then await the return of the Lynx, or just do a U-ey and wait for the helo to return for 4-5 hours? Every mile toward S Georgia that the ship covers from the point of release of the helicopter takes 5x as long as the helo can do it in. But the ship also has to backtrack those same number of miles at one-fifth of the speed of the helo.

My rudimentary maths says the best plan is for the ship to stay where it is and twiddle its hypothetical thumbs while the Lynx goes and comes back. But it doesn't "feel" right. Surely there is some benefit to be had by reducing the return range the helo has to fly, even if only by 30nm.

What am I missing? Or is that it, the ship just stays put?

I defer to incredibly highly-paid Navigators, who enjoy doing sums. :)

But staying on station, or indeed creeping home whilst staying in Lynx max range, seems sensible.

No Maths Required is there?
 

As the Heli is faster than the ship it needs to be used for the maximum possible portion of the route length - 300nm in your example. So yes the dispatch and pick up point are the same assuming no wind / current.

Pete

I've just done a bit more thinking. If I've got my equation right, and calling the extra distance the ship sails on "x" (nm), then there IS indeed a benefit, but it is vanishingly small.

Time taken for ship to return to point of release is (x/30)+((300-x)/150). Boiling that down a bit, I get (2x+150)/75. Differentiating that gets me to 2/75ths of a nautical mile, or probably less than the turning circle of the ship!

Or have I stuffed up my algebra/calculus? It's been a while...

If refuelling is available at SG there is nothing to be gained by reducing the range further apart from any range reduction due to additional payload or, as you discounted, weather.

But ... what's the max [safe] range of the Lynx on its way back to Mother? Surely after launch the ship wants to be there, to maximise the helo's speed advantage? If that's at the start point, then orbit port/starboard. If not, adjust for wind/tide .. which lies outwith the original question.

A few years back a much more complex problem was posed in a recruiting advert.

IIRC it involved a F4 at Coningsby, an intruding Badger bound for Boulmer and a tanker out of Marham. A solution was a supersonic intertcept by the F4, a link with the tanker etc. While many of us came up with an effective answer there was one solution that effected the intercept even quicker.

If you got it right you were well on the way to being a fighter controller.

What was the solution? Intercept using the tanker ?

It sounds like the problem I was given back in 1971 at Biggin Hill. However, my vessel was a fishing boat and my helo was a speedboat.;)

You launch the fastest asset at its maximum range and recover it at the same. In still air and water that is the same place. Pointless the ship disappearing back west in advance if the Lynx is going to ditch fifty miles astern...
In real weather you need a whizz wheel to work it out.

The trains will pass each other at Crewe.

Much depends upon what is happening to the pax after getting on board the ship.....if he needs to be taken to MPA to catch an aircraft back to Blighty then perhaps the odd hour won't matter and safety/fuel planning takes over. No point rushing back if there is then a two day wait for the airbridge.

Shaft, IIRC it was using full burner that would achieve earliest intercept, vector tanker to an intercept point with F4 after its Badger intercept.

There was a twist for the F4 but can't remember the details.

It was one of those puzzles that you deliberately got wrong if you didn't want to be an aerospace battle manager. As a mere plane driver you executed the intercept as ordered and left the tanker control to the ABM. :)

R280, the basic plot seems reasonable,however,if the `package` is injured,rather just compassionate,then the ship should continue to SG,to give the injured best chance for earlier /better medical treatment on board.
Then,I would suggest that a `moving Crit Point`is considered if the Lynx should have to do a return to Mother in the event of a single engine failure and return at a lower speed,ie 90 kts,or continue to Grtvkn,in which case the ship will go there as well...!
Lots of things to consider...

Or the fully fuelled Lynx could take some drummed fuel and a Zenith Pump to top up at SG (assuming no fuel there) and then the ship could have done a 180 and be steaming FI-wards to take advantage of the (new, increased) PLE of the Lynx.

When one was working in HQBFFI (many years ago) someone had attempted to write a definitive answer to this question and put it in the "Contingency Planning" book - an answer which of course does not exist! (and it was Sea King and HIFR in them days)

There are too many variables, most of which have been stated above: weather, light, servicability, and the reason for the evacuation. In the end we just put a list of assets and capabilities in the Book, with the message - "don't rush - let the SMEs work it out on the day!"

and don't forget 2 POB out to SG and 3 POB for the return to mother. How does that impact on Lynx range?

Does a Lynx have different reserve requirements when flying to a land base vs a ship? If so, that could alter the equation.