Riddle me this Batman

[Typhoon1244]

Blue. No yelloooooooooooooooooooooooooooooooow...

 

[Timebuilder]

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????"

 

[TonyC]

" 'ow did 'e know that????"

We Kings must know such things.

 

[GogglesPisano]

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.

 

[Typhoon1244]

We Kings must know such things.

'ow did you know 'e's a king?

 

[slide33]

I'm retarded. I didn't read the question carefully enough.

 

[enigma]

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

 

[crjdxr]

We Kings must know such things.

Well I didn't vote for you.

 

[Super 80]

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.

 

[GogglesPisano]

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?

 

[Super 80]

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?

Last chance to help you conceptualize this.

The total time is 90 minutes, or 1.5 hours.

In the airmass, the first airplane flies out from a point (the balloon) for 45 minutes. This means some distance is covered. A second plane passes it going the same speed and travels the same distance back to the starting point in the airmass. So the second airplane takes 45 minutes to fly back to the starting point.

None of the three, the balloon, or the two airplanes are affected by the movement of the airmass over the ground. Each one of them relative to the other are in perfect symmetry. The balloon is stationary (in the airmass) and the two airplanes are on reciprocal headings at identical indicated speeds.

The ground track and ground speed are a totally different matter. That is controlled by movement of the airmass that all three are in. This is the measurement you get with points A and B on the ground as the stationary point moves over it.

With the distance being 75 nm and the total time being 1.5 hours you can now calculate the speed of the airmass or the windspeed at altitude.

 

[TonyC]

Last chance...

None of the three, the balloon, or the two airplanes are affected by the movement of the airmass over the ground.

If this property is fundamental to the solution of your riddle, then it must be so stated, since it is not a property of air in the nature in which we live.

The balloon is stationary (in the airmass) ...

This contradicts the first sentence of the riddle as you posed it:

Post which began this thread, originally posted by Super 80
An airplane releases a balloon that stays at that altitude over point A.

Either it remains over point A, as the first post SPECIFIED, or it moves perfectly with the airmass as your "solution" demands. It is improper for an unbiased observer, an honest scientist, to make such leaps of assumption.

 

[Super 80]

Re: Last chance...

quote:
--------------------------------------------------------------
Originally posted by Super 80
None of the three, the balloon, or the two airplanes are affected by the movement of the airmass over the ground.
--------------------------------------------------------------

If this property is fundamental to the solution of your riddle, then it must be so stated, since it is not a property of air in the nature in which we live.

No it is a principle of flying. The concept of airmass is the lesson here Tony. As a concept, a large block of air can be thought to move in unison. For a general locale, such as the runway environment, this is generally true. However we know this concept is rudimentary at best, but to show the difference between heading and airspeed versus ground track and ground speed it is a necessary component.

quote:
--------------------------------------------------------------
Originally posted by Super 80
The balloon is stationary (in the airmass) ...
--------------------------------------------------------------

This contradicts the first sentence of the riddle as you posed it:

Not really. A balloon has no lateral self-propulsion. If you ride in a hot air balloon, there is very little or no wind at all. The balloon is carried by the airmass current at that altitude in comparison to the ground.

quote:
--------------------------------------------------------------
Post which began this thread, originally posted by Super 80
An airplane releases a balloon that stays at that altitude over point A.
--------------------------------------------------------------

Either it remains over point A, as the first post SPECIFIED, or it moves perfectly with the airmass as your "solution" demands. It is improper for an unbiased observer, an honest scientist, to make such leaps of assumption.

The balloon stays at that altitude. It does not stay at that point.

The balloon is stationary relative to the airmass in which it is floating. The airmass is moving across the landscape, so the balloon is not stationary relative to the earth.

This was meant for fun, not to start an argument. A couple of people got it right, and ironically it was Enigma that answered the Batman riddle.

 

[TonyC]

Re: Re: Last chance...

No it is a principle of flying. The concept of airmass is the lesson here Tony. As a concept, a large block of air can be thought to move in unison. For a general locale, such as the runway environment, this is generally true. However we know this concept is rudimentary at best, but to show the difference between heading and airspeed versus ground track and ground speed it is a necessary component.

"principle..." "concept..." "generally true..." "concept is rudimentary at best..." you're making my point for me. It is not a property of our atmosphere, so we can NOT assume it to be true. It IS a "necessary component" of your riddle, so the assumption should have been stated.

It's kind of like the compressibility of air. Due to the incredible complexities of considering the compressibility of air in many aerodynamic calculations, we assume that air is incompressible when looking at most subsonic scenarios. The answers we get, then, are close, but lack a certain degree of accuracy owing to those assumptions. If we embark on solving problems, we must agree to certain assumptions, or make none. We can't just go about assuming anything that makes the math easier.

If you ride in a hot air balloon, there is very little or no wind at all. The balloon is carried by the airmass current at that altitude in comparison to the ground.

Again, you make the point for me. If the balloon moves at the exact velocity of the surrounding air, there can be NO wind -- not even "very little" -- relative to the passenger in the balloon. Such is not the case in our atmosphere. If motion of the balloon exactly matches the movement of the airmass in your riddle, that must be included in the assumptions.

This was meant for fun, not to start an argument. A couple of people got it right, and ironically it was Enigma that answered the Batman riddle.

Aww, come on. You can't seriously think that ANY question can be posed here without inciting an argument, do you? Enigma came up with YOUR answer because he made the assumptions you wanted him to make. The jury's still out on the RIGHT answer.

 

[bart]

'ow did you know 'e's a king?

'cause 'e 'asn't got sh_t all over 'im

and then there is watery tarts distributin' swords as a basis of government