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

Posted November 08, 2009 5:01 PM

This week's Challenge Question:

A rescue plane flies at 198 km/hr and at a constant elevation of 500 m trying to rescue a person involved in a boating accident. The pilot releases a rescue capsule so that it hits the water close to the victim. Determine the angle (A, in the figure) of the pilot's line of sight when the release is made.

As indicated in the given figure, we will put the origin of our coordinate system at the point of release of the capsule. Let's make the time of release as t = 0.

Angle A is given by

Because the capsule, once released, is a projectile, it follows that the horizontal and vertical velocities are independent. The horizontal velocity of the capsule is the same as that of the plane, and it is constant, or its initial velocity is equal to its velocity at any time t. Thus we can express the horizontal velocity by

Or,

Where tf is the time the capsule reaches its destination. The initial velocity is 198 km/h or 55 m/s.

We see now that to determine d we must calculate tf. This can be accomplished if we consider the vertical displacement. For a projectile released at time t = 0, the vertical displacement is given by

At t = tf the displacement is y = -500 m. With g = 9.8 m/s2, using the equation given above we get tf = 10.1 seconds. Then the horizontal distance, d, is calculated as

Finally the angle A is calculated

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#1

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

11/09/2009 1:33 PM

Ignoring air resistance etc, the capsule would take about (2.500/9.81)0.5, or 10.1 seconds to fall. 198 km/hr is 55-m/sec, so he would have to release the capsule 555-metres before reaching the target. So the angle A between the vertical and the intended landing point would be tan-1(555/500), or about 48O.

If this resembles the actual scenario, the term victim seems particularly apt - a capsule with a landing speed in excess of 113-m/s had best land at some distance from any prospective 'beneficiary'.

I'm no expert here, but I imagine that a more realistic scenario for a civilian rescue would have a plane flying at closer to 80-km/hr, and then use a parachute to slow the capsule. If the parachute opened 0.2 second after release and was designed for a terminal velocity of 10-m/s, I suppose that we'd be looking to release about 21-metres before coming over the intended landing spot - that would give an angle nearer 1.8O from vertical (so perhaps the required position would depend more on the wind).

If the rescue was military, maybe they would need to use such speeds - and indeed delay the opening of the parachute until the capsule was relatively close to the water (50-m??). The angle could then be relatively close to the 48O calculated above - perhaps even as large as 45O???

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#2

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

11/09/2009 1:46 PM

I agree, Fyz, the problem appears to belong in bomber pilot class rather than rescue pilot class.

Are extra points given for direct hit?

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#4

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

11/09/2009 3:16 PM

I'm certainly missing something here. Hopefully it's not merely a few words in the challenge - e.g. "a person about to become involved in a boating accident".

Recollecting my driving teacher's comments (don't ask when) on my early "motoring skills": doubtless he gets additional points for "little old ladies", the disabled, children, etc.

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#23

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

11/10/2009 8:49 AM

Well, sir, the only 'trickery' I can see in the question is the angle solution is from the vertical, which you caught. My natural inclination would have been to solve relative to the horizontal.

My driving instructor was, perhaps the younger brother to yours! I do recall that no extra points are given for felines or canines... I am certain there are members here who agree with that!

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#10

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

11/09/2009 11:24 PM

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.

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#11

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

11/10/2009 12:24 AM

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.

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#19

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

11/10/2009 7:21 AM

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?)

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#26

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

11/10/2009 5:11 PM

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.

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#27

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

11/10/2009 5:24 PM

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.

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#28

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

11/10/2009 10:13 PM

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!).

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#29

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

11/11/2009 6:17 AM

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).

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#30

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

11/11/2009 7:48 AM

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.

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#31

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

11/11/2009 8:14 AM

Thanks. That sounds sensible.

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#32

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

11/17/2009 11:33 AM

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"!

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#33

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

11/17/2009 11:52 AM

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 48O )

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.

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#34

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

11/17/2009 12:32 PM

I do notice the official answer is, indeed, a direct hit... no mention of extra points for this. What you suppose is the probability of the victim having a father named Arthur, Bill, or Charlie?

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#35

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

11/17/2009 1:25 PM

"I see the official answer matches your solution exactly" Sad that - you would have thought such a silly solution would have been stifled at birth.

But they are probably using a laser guidance system...

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#36

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

11/17/2009 1:45 PM

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.

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#3

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

11/09/2009 3:06 PM

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.

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#6

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

11/09/2009 3:29 PM

Unfortunately the advanced solving techniques are no use without input - including mass and aerodynamics of the capsule, downdraught from the rescue plane. Maybe the capsule has its own lift - and could be released even further back.

By the bye, I suspect a simple trajectory such as this should still be adequately soluble using high-school-standard integration methods.
Also (and I'd appreciate a comment from someone more directly involved), earlier in my so-called career I was roundly abused for describing RK2 (I think that SPICE describes this as "trapezoidal Newton Raphson") as "predict and correct"; I think the logic was that it was "predict, backtrack and repredict until backtrack is close enough", but even to this day I'm not certain about that.

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#5

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

11/09/2009 3:17 PM

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

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#7

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

11/09/2009 10:38 PM

Looks like the plans for the dam busters.

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#8

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

11/09/2009 11:19 PM

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.

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#16

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

11/10/2009 5:33 AM

Hi GS, did you see the quality of quizzes?

I tried to make it complicated (assuming the aircraft in tangential motion rather than a curved motion)- but the difference at earth radius- was only 2 cm (ie the height will be 500.02m - and that will be negated by the waves formed by a drowning boat (and man/persons )

Only way for complicating it is the usage of drag but unfortunately no information on the constants (k , CD or A )

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#18

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

11/10/2009 5:54 AM

Hello my friend Guest (Are you Mr. Guest)

I agree with you on this particular quiz. This was very simple school level physics problem. Thats why I participated in surcastic way. This was not a quiz at all. Earlier was OK. But many of the quizes are quite brain storming.

Once again I invite you to join as member with some name.

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#21

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

11/10/2009 8:32 AM

I am a member, only on guest missin (as I said in previous threads) I am not happy the way GAs and appreciations are being given. So rather than my own name/avatar where the names are attracting GAs, I am anonymous, so that the worthy GAs are only awarded.

And as I said if I see a post (where a deserving GA is not marked, I will just log in, mark GA and log out )

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#9

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

11/09/2009 11:22 PM

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°.

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#12

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

11/10/2009 1:46 AM

It should be approximately 45 degrees.

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#13

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

11/10/2009 1:59 AM

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

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#14

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

11/10/2009 3:15 AM

Yup.

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#15

### Re: Rescue Plane: C) R4 Challenge (11/10/09

11/10/2009 3:58 AM

It will be interesting to calculate how much electricity can be generated by two or three propellers mounted on the rescue capsule which must be having batteries to store the electricity, so that the person rescued will not be hit or thrown away by drastic drop and at the same time he can have the required electricity for light and or signals.

The reverse breaking generator can make the capsule land like a helicopter. The whole propeller, drive shaft and mountings can be in FRP with only motors and battery being heaviest parts.

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#17

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

11/10/2009 5:53 AM

32 degree's

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#20

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

11/10/2009 8:29 AM

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???

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#22

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

11/10/2009 8:43 AM

Oops. Perhaps the Norden bombsight?

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#24

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

11/10/2009 9:03 AM

A rescue plane flies at 198 km/hr and at a constant elevation of 500 m trying to rescue a person involved in a boating accident

The impact velocity would be .....? I don't know, but suggest 'bad' as an answer. 'spot-on' is not always the best solution.

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#25

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

11/10/2009 1:57 PM

"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

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