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Riddle me this Batman
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Wankel7
It's a slippery slope...
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I think he was thinking that if after the 1st plane dropped the ballon it flew for 45 minutes. Then the 2nd plane past head on with the 1st plane then found the ballon. Maybe if those planes were going the same speed that would have been 45 minutes from the dropping. Then 45 mins getting back to it. Assuming no wind. But wait.....that is the question.....hum
Wankel
Wankel
Kingairrick
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siucavflight
Back from the forsaken
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Typhoon1244
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dmspilot00
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mattfish42
missed the ground
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mattfish42
missed the ground
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give it a shot...
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.
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.
Typhoon1244
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Super 80
Rube Goldberg device
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Re: Re: Riddle me this Batman
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.
YES!enigma said:
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.
Last edited:
Typhoon1244
Member in Good Standing
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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?
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?
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