Rescue Plane: CR4 Challenge (11/10/09)

I voted good answer, but "*" or "x" in lieu of "." for 2 by distance in the first calculation would have been clearer.

As a matter of interest, one technique used here in Australia is to let the life raft go with long ropes attached, with the view to having the ropes fall across the person being rescued. All this so they don't have to swim too far to get themselves rescued.

I suspect in the real world that less altitude and less speed would be preferred to that given in the problem, but I doubt that the preferred airspeed would be as low as 80 kph as one respondent suggested. At that speed in many aircraft you could easily finish up with even more people to rescue, along with their aeroplane.

GA from me. Best you can do on the data.

Ref the speculation part, as I noticed a comment; 80 km/h is a bit low (~43 kn)

Commonly used is this one

for the visibility (over-wing construction) and reasonable range in class.

Real cruise is ~107 kn = 198 km/h.

70 kn would be a reasonable 'slow flying' speed, = ~130 km/h.

http://www.risingup.com/

Cessna 172 C - Performance Data

Horsepower: 145	Gross Weight: 2250 lbs
Top Speed: 121 kts	Empty Weight: 1330 lbs
Cruise Speed: 114 kts	Fuel Capacity: 42 gal
Stall Speed (dirty): 43 kts	Range: 515 nm
Takeoff	Landing
Ground Roll: 900 ft	Ground Roll 600 ft
Over 50 ft obstacle: 1450 ft	Over 50 ft obstacle: 1200 ft
Rate Of Climb: 675 fpm
Ceiling: 14200 ft

Not that real world is in the mix here.

I probably misinterpreted - "boating" to imply that the accident was on a lake or inshore, and dropping a capsule to imply that the aircraft would not land (the planes used to illustrate the challenge certainly couldn't...).
Locally, lake/inshore rescue can allow something a bit more aerodynamically flexible than a seaplane - which is why i(in ignorance) I chose a low slow-flying speed.
(As I said, I'm no expert. But 500-m (!) also sounds high to me - probably high enough for rescue pilots to reduce somewhat below standard slow-flying speed immediately prior to capsule drop?)

Hi Fyz, there are a whole bunch of rules that centre on 1500 ft as a minimum over terrain/water. I guess this is why 500m. We used to play a game similar using toilet roles, but over the air field so you use 500ft or a 'bit lower'. It's bombardier controlled. Pilot just does vector altitude and air speed. But if this package had a chute and you'd' need about 300ft to allow full deployment before it hit them. Problem is they then tangle and drown in the chute. Better to - drop a floaty with a smoke marker - and call in the chopper. A floaty is a bit like thistledown. You might "accidentally" lose some altitude in the process.

I've a memory that many years ago a read this number as the minimum search height for military aircraft. My 'common sense' says two things - civilian aircraft might be allowed lower, and you would circle and return (possibly at a lower altitude and speed) for a drop. I can see the problem with a standard chute fowling everything - I had assumed some kind of release mechanism would have been designed to avoid this.
Thistledown sounds different from capsule - and good. But I suppose (in spite of first impressions) capsule doesn't carry any implications of mass or solidity.

1500ft AGL is the civilian rule - 'without clearance', meaning permission by the radio folk. Generally 500ft AGL is as low as they are comfortable with. Military is a whole different thing with minimums for the exercise or obstacle clearance (and "no hitting the civilians" rules, like no going under the ski lift cables - oops!).

Thanks. Doesn't rescue work generally get permissions? (Actually 500-ft is the height I had in the back of my mind for drops and sweeps).

See how easy it is to get "picky" about challenges?

A GA anyway. You're right, the realistic approach, at least here in the Colonies, is to drop a rescue swimmer. One hopes neither he nor any equipment hits the water in excess of 50 km/hr.

Thanks. That sounds sensible.

Ignoring wind resistance in this problem induces huge errors. The plane is flying at 198 km/hr which is 443 mph terminal velocity against sea level atmospheric resistance is between 160 and 120 mph for a free falling person depending mostly upon the frontal area he or she exposes to the wind.

Your capsule would decelerate along its horizontal velocity component at an alarming rate, falling short of its target by comfortable safety margin.

Unfortunately, the reality of the situation is that darn few "rescue capsules" are dropped from air planes. However, men have been dropping explosives on each other from planes for a long time. In that case falling short like this would strongly tend to kill your own "team mates". Not, what I would rate as a "good answer"!

I tried to comment on the limitations - which I imagine is why some others considered my limited answer worthwhile. (I didn't mention this at the time, but it's also possible to use an aerodynamic load with a glide angle that would give alpha greater than 48^O )

But please don't hide your light under a bushel - everyone watching this thread would love to see a better answer to the challenge as posed.

Hi Guest, How on Earth did you get 443 mph? = 712 km/hr. It's at 5000 not FL 50

Might I suggest that next time read the question and list what is defined and what is 'open'.

For instance; the capsule is not defined.

Would a polished Teflon coated teardrop of depleted uranium "decelerate along its horizontal velocity component at an alarming rate, falling short of its target by comfortable safety margin"?

Not, what I would rate as a "good answer"! calculation.

The answer without aerodynamic drag is A = 48° (time to fall 500m 10.09sec; Horizontal velocity constant at 55m/s)

BUT

Since the capsule is dropped from a plane which is experiencing lift (and drag) from the surrounding air. the released capsule will also experience aerodynamic lift and drag. If the pilot uses 48°, the capsule will fall short.

The drag on the capsule is a function of the capsule's shape and size, and of its instantaneous velocity, which is non-linear in magnitude and changing in direction. This leads to non-linear differential equations which require sophisticated solving techniques, such as Truncated series expansion; Predict and correct (e.g. Runge Kutta); or computer simulation.

Use a chopper and lower it down to him on a tether. Better still, lower the tether and pull him out.

Do you want to rescue or harm the person in accident still further? Without parachute, the capsule with horizontal speed of 198 kms/hr + vertical speed newly achieved due to gravity fall is certainly going to harm the person. Even though it doesn't hit the person, it will surely creat waves in water creating problems for a person already in problem.

Answer = 48° ignoring air resistance etc.

The capsule will have two velocity components, 1. horizontal at 198 km/h and 2. vertical at 9.81 m/s2 (g gravity).

Say the plane is at x horizontal km from the boat. Time taken to reach the boat horizontally = (198000/3600)*x sec

For the vertical componenet, the equation will be

s = ut + 1/2 g t^2, s (vertical dist.) = 500 m, u (initial vertical vel) = 0, g =9.81 m/s^2 , t (time) is the same as horizontal time = (19800/3600)*x

This gives x = 555.3 m i.e. the plane should be 555.3 m horizontally from the boat.

Angle A then is = tan-1 (555.3/500) or 47.99 or say 48°.

It should be approximately 45 degrees.

Then it will reach approximately the person, and save a person approximately

32 degree's

Hmmmm....if you really want to get it really close to the boating "victim" at that airspeed, why not just use an old WWII "Norton" bombsight and "pickle" the rescue raft???

"LETS SEE NOW" with out breaking out the calculator, we have an airplane traveling a little faster than 200 mph at a height of a little more than 1500 ft. above the water, and the pilot or his assistant is attempting to drop this capsule near but not on top of the person in distress.

Depending upon the size of said capsule, it's shape and the density of the air and, and possible wind speed and direction of said wind, "not any of which has been provided", and even more important any current speed and direction issues that have likewise "not been provided" "I offer a prayer to the poor bastard in the water", and suggest that the capsule should be dropped just a little later than is shown in the example as the capsule, once dropped is not going to continue to travel at the same speed as said airplane. Friction being what it is, "Don't You See"!

That is why those folks practice those kinds of drops. It doesn't do any good to make such an effort if said current is going to carry the device away from the distressed individual faster than they can swim. This is why a practice drop of a weighted flare/smoking device is advisable prior to the dropping of the capsule of aid.

TMF