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Riddle me this Batman

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Okay, so we know the answer is forty-two, but what's the freaking question?
 
give it a shot...

An airplane releases a balloon that stays at that altitude over point A. It

Hang on a second. "It"? Which "it"? The balloon or the plane? If "it" is the balloon, then the problem is solvable. Only problem is the second plane is assumed to be kinda hovering there next to the balloon, travelling over the ground between points A and B at the same speed as the wind: 100 knots.

If I can assume that "it" is the balloon, then I can assume the second plane is hovering. And yes, I'm a still-wet-behind-the-ears private pilot.
 
OtterFO said:
The answer is 0, the rotation of the earth makes the 75 mi difference.
That would only apply in a vacuum. The Earth's rotation affects movement of the atmosphere through friction, and you can't just discount that.
 
Re: Re: Riddle me this Batman

enigma said:
I hesitate to speak for fear of confirming that I am in fact a fool. However, I have to take a guess. The balloon traveled 75 miles in 90 minutes, giving a balloon speed of 50 knots.

Assuming that the airmass is completely uniform, that every molecule of air at that certain altitude moved the exact same distance and direction during the duration of the experiment; I don't think that it should really make any difference how fast the airplanes are, only that they are flying at an equal speed. Nor do I think that the wind direction makes any difference versus the aircraft direction, since all three (two airplanes and one balloon) are in the same airmass.

regards,
enigma
YES!

Trophy prize goes to Enigma (ironic isn't it that Ed Nigma was the Riddler in the Batman movie Forever--just saw it on the layover) for answering the question correctly.

Headwind, tailwind, crosswind are all irrelevant. Since speed was the only criteria, and not velocity, direction has no bearing but only the magnitude.

Some assumptions were not spelled out --that the aircraft didn't traverse any weather fronts thus assuming a solid airmass, that speed translated directly to KIAS or what ever shows up your indicator (KCAS), and others were not necessary, i.e. height, color, etc., or even the speed of the aircraft themselves.

Years ago, when the Captain I was flying with posed this question to me, I started writing out D = r x t. I also drew a triangle showing points A and B and two longer sides representing the other two aircraft flight distances. My mistake right away was to reference this to the ground.

Needless to say, I could not determine the time or distance the second aircraft traveled, so I couldn't answer this one.

The Captain, who as a civilian (or as military guys say, has no formal training) was quite an aviator and his life and his wife's life (who is a United 767 Captain) revolve around flying. They have several airplanes (but no kids) and he actively teaches.

He told me one Air Force Engineer type wrote him a three page letter saying that this problem was unsolvable!

Needless to say, here's how it works:

Assuming a constant airmass, if you mark a point (deploy a balloon, relative wind speed = zero) and fly away from it for a give amount of time, and have a second leg flying a reverse course (+/- 180) at the same speed through the airmass (KIAS) back to the point, the time is the same, hence 45 +45 = 90 minutes.

The balloon having no speed in the airmass is stationary, but now the speed of the airmass is included with the additional information that the reference point on the ground has moved 75 miles (it doesn't matter which direction either, the airmass is totally oblivious to how it moves across the ground).

Now you can plug in D = r x t where 75nm = r x 1.5 hours and dividing 75 by 1.5 yields 50nm/hr.

Enigma has gotten all the assumptions correct, and his answer as well and so deserves a hearty well done in addition to his self-deflating humbleness in answering this query.

Aviation is a wonderful business; I wish I knew more about it.
 
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Question: why bother with exactly what the two airplanes were doing?
Super 80 said:
An airplane releases a balloon that stays at that altitude over point A. [A] second airplane...catches the balloon over point B on the ground. Points A and B are 75 nautical miles apart.
Isn't that all you need to know to answer this?
 
Typhoon,

Well, you need something to define the elapsed time. "What the two airplanes are doing" is what gives you that.

Realizing that 45 minutes outbound will, at the same speed, require the same 45 minutes inbound, is the missing part, along with ignoring the motion of the ground beneath the airmass containing the 2 aircraft and the balloon (at least until you've solved for elapsed time).

I wasn't particularly impressed with the riddle; I got the same answer with no great mental strain, and I am one of those Air Farce trained engineering types.

Maybe the time in the civillian world has had an effect!

Snoopy

P.S. Should we rehash the old "Let's Make a Deal" puzzler about opening one of the doors you didn't pick, then asking if you want to switch?
 
Blue. No yelloooooooooooooooooooooooooooooooow...
 
According to the creators of the waveform editing program Cool Edit Pro, Syntrillium Software, there is no significant difference in the characteristics of the African or European swallow.

These fans make this little tidbit appear as one of the "daily tips" that pop up when the program is opened on the desktop.

You just never know when a Python fan will show himself...


" 'ow did 'e know that????" :D
 
Was this a poll?

Seriously, I didn't see any information regarding how long it took the second airplane to fly back to the balloon. Without that piece of info it's unsolvable.
 
slide33 said:
I agree with Goggles. Speed being the only criteria, you should have given the second plane's speed.


"An airplane releases a balloon that stays at that altitude over point A. It then flies outbound for 45 minutes when it passes another airplane at the same altitude going the same speed but in the opposite direction. (instant turn) The second plane makes a bee line from that passing directly to the balloon. It catches the balloon over point B on the ground. Points A and B are 75 nautical miles apart.

What is the wind speed at that altitude?"


Enigma had to assume that plane 2 was traveling the same speed as plane 1. What if plane 2 was traveling twice as fast as plane 1? Without mentioning plane 2's speed, it's just a swiss cheese riddle. . . . and I'm an Air Force Engineer.

No such assumption was required. I just carefully read the question. It seemed pretty simple, but then I learned most of what I know about flying from Naval Aviators. Thanks Commander Muetzel.

regards,
enigma
 
GogglesPisano said:
Seriously, I didn't see any information regarding how long it took the second airplane to fly back to the balloon. Without that piece of info it's unsolvable.
Well that's the missing piece if you try to solve it like Slide33 suggests. I said originally that the second plane was going the same speed.

So if D = r x t and all I give you is t, but not the rate, the Distance is immaterial. (Assuming some speed will necessitate some distance however.)

Now in the second part, the second plane has the same r as the first.

The missing piece of the puzzle is the concept of airmass. What is constant is the distance from the balloon to the passing of the airplanes.

Take two pieces of paper on a surface you can make an impression. Tape the first piece of paper down. This will be the ground. Put the second piece of paper over it. This will be the airmass. Press a point on both pieces of paper. This is the instance the balloon is released.

Now draw a ray (a line starting at a point) outward from the point in any direction while sliding the top paper across the bottom in any direction. Stop at any time before you run out of paper. This is the first airplane.

Now draw a line back to the point (basically retracing your first line) while moving the paper at the same rate in the same direction as before. When you reach the dot, stop moving the paper and impress another dot on top of the first.

Remove the second sheet of paper. You will now have two impressions on the lower sheet (the ground). These two points are 75 nm apart in the riddle. This is the D in D = r x t.

The upper sheet will have a double line of a given distance. In the riddle, it takes 45 minutes to do. Since the second airplane covers the same distance at the same rate, it also takes 45 minutes. So the total time is now 1 hour, 30 minutes. This is the t for D = r x t.

Now you can plug in the values and solve for r, the wind speed, or the scale rate you moved the second piece of paper. r = D / t or r = 75 nm / 1.5 hours, so r = 50 nm/hr.

If you drew your exercise so it didn't have a pure head or tailwind and transcribed the double airmass lines to the ground paper, you'd end up with some kind of triangle. (With a pure headwind or tailwind, you'd just have unequal lines.) This is the ground track of the two airplanes as the resultant vector between their reciprical velocity and the airmass velocity.

However, different in the ground track, the airmass distance both aircraft travel is the same as shown by the second piece of paper. This is the lesson every new pilot learns in landing an airplane in a cross wind, but we tend to forget that as we just mechanically fly the airplane to touchdown.

So the t for the second airplane is 45 minutes. Enigma inferred that, he did not assume it.
 
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Not speed. Time. In other words, how much time did that 75 mile journey by the balloon take? Nowhere in the clue does it state that the second airplane arrived at the balloon 45 minutes after passing the first plane.

Is this what they call mental mastrubation? If it is, is it supposed to hurt?
 

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