Cue PH brainiacs - logic puzzle!
Discussion
The Airplane one is not logical
The plane has to move forwards before the conveyor can match it
In "balance" the plane will be stationary, which means that the conveyor will be stationary, so the engines will creep the plane forwards
The question is the point of reference
Is it a tree next to the conveyor belt runway or is it the conveyor itself.
If it is a tree, then the plane thavels at + 5 mph, the conveyor goes at -5 mph, the plane wheels travel as if it were at 10 mph
Plane accelerates, plane at 50 mph, conveyor - 50, wheels 100
Result normal take off, expensive runway, and wheel bearings last half the time they do normally (Dont try this in a German filled Concord)
The plane has to move forwards before the conveyor can match it
In "balance" the plane will be stationary, which means that the conveyor will be stationary, so the engines will creep the plane forwards
The question is the point of reference
Is it a tree next to the conveyor belt runway or is it the conveyor itself.
If it is a tree, then the plane thavels at + 5 mph, the conveyor goes at -5 mph, the plane wheels travel as if it were at 10 mph
Plane accelerates, plane at 50 mph, conveyor - 50, wheels 100
Result normal take off, expensive runway, and wheel bearings last half the time they do normally (Dont try this in a German filled Concord)
chickensoup said:
The Airplane one is not logical
The plane has to move forwards before the conveyor can match it
In "balance" the plane will be stationary, which means that the conveyor will be stationary, so the engines will creep the plane forwards
The question is the point of reference
Is it a tree next to the conveyor belt runway or is it the conveyor itself.
If it is a tree, then the plane thavels at + 5 mph, the conveyor goes at -5 mph, the plane wheels travel as if it were at 10 mph
Plane accelerates, plane at 50 mph, conveyor - 50, wheels 100
Result normal take off, expensive runway, and wheel bearings last half the time they do normally (Dont try this in a German filled Concord)
Wrong answer. Your plane is travelling forwards faster than the conveyor is traveling backwards, so it is not a solution to the problem.
If I state "What is the solution for x to the problem x^2 = 4 and x < 0", there's no point in going on and on about 2 being a valid answer because 2^2 = 4, it doesn't meet the conditions of the problem as stated so it is an invalid answer.
chickensoup said:
The Airplane one is not logical
The plane has to move forwards before the conveyor can match it
In "balance" the plane will be stationary, which means that the conveyor will be stationary, so the engines will creep the plane forwards
The question is the point of reference
Is it a tree next to the conveyor belt runway or is it the conveyor itself.
If it is a tree, then the plane thavels at + 5 mph, the conveyor goes at -5 mph, the plane wheels travel as if it were at 10 mph
Plane accelerates, plane at 50 mph, conveyor - 50, wheels 100
Result normal take off, expensive runway, and wheel bearings last half the time they do normally (Dont try this in a German filled Concord)
woah, you've got totally wrong end of the stick. in fact, i'm not sure what stick you've got hold of either.
the question is quite simple and I can't believe there are quite so many dumbasses on PH who just don't get it.
the plane is on the conveyor and for whatever speed the plane can taxi at, the conveyor moves in the opposite direction keeping the plane stationary.
now, just to make this really clear - its a magical conveyor, and the wheels on the plane are magical wheels with magical bearings. but essentially, the plane is not moving.
the question is - the (magical) conveyor is doing -200mph and the plane's (magical) wheels are doing +200mph, keeping the (magical) plane at 0mph in overall speed - WILL THE PLANE TAKE OFF?
and the answer is NO.
geesh
Size Nine Elm said:
Nope. As I said sometime earlier, if the conveyor is defined to move backwards at the same speed the plane moves forwards, I don't have to provide any reason whatsoever, except to state that it is a precondition of any proposed solution to the problem as stated that this is true.
Speed is relative and this may be why you're happy making assumptions that seem silly to me. Perhaps you interpret the reference to the plane speed as meaning the speed of the plane relative to the conveyor belt? That's the only way I can see to come to the conclusion that the plane remains stationary. If that's the interpretation you want to make then I agree the rest of your argument does follow. Must say I don't think that's a particularly sensible interpretation, but when did sensible ever matter when we're talking about this sort of nonesense?
Take a more sensible interpretation that the speed is measured relative to a fixed ground plane that the conveyor is mounted on, and it is no longer valid to assume that the plane can't move forwards.
GreenV8S said:
Speed is relative and this may be why you're happy making assumptions that seem silly to me. Perhaps you interpret the reference to the plane speed as meaning the speed of the plane relative to the conveyor belt? That's the only way I can see to come to the conclusion that the plane remains stationary. If that's the interpretation you want to make then I agree the rest of your argument does follow. Must say I don't think that's a particularly sensible interpretation, but when did sensible ever matter when we're talking about this sort of nonesense?
Ah, I see what you mean. If you mean the conveyor travels at -v, and the plane travels forward at v (i.e. 2v relative to the conveyor), then of course it can take off.
If it means the conveyor travels at -v, and the plane travels at v relative to the conveyor, then it is at rest.
Anyway, I've got this formula for racehorses, guarranteed to predict a winner...
Size Nine Elm said:
Anyway, I've got this formula for racehorses, guarranteed to predict a winner...
Oh no, they aren't on a bloomin conveyor belt too are they? I predict it's going to be a long race, ... unless they're winged horses.
Hmmm, you didn't get this formula from Vixpy did you? tinman0 said:You are wrong. That couldn't work. The only thing that "works" is the bearings in the wheels have siezed, the conveyer belt is actually glued using an unknown glue that can stick rubber to the belt so strong that it can withstand at least 40,000lbs of thrust, and neither is moving because the motor on the belt has enough torque to keep the plane from moving forward.
the question is - the (magical) conveyor is doing -200mph and the plane's (magical) wheels are doing +200mph, keeping the (magical) plane at 0mph in overall speed - WILL THE PLANE TAKE OFF?
and the answer is NO.
A few seconds later the wheels would sheer off the undercarriage and the plane would go shooting off down the conveyer.
If the wheel moves, then the plane goes forward.
Don't you get it?
Sheesh
J
Cue U turn coming up!
I have just read the original plane /conveyor question and (in spite of my previous posts) the plane would definitely take off. As they say in exams, always best to read the question!
I think we ought to take a vote on it - all those in favour of the plane taking off raise their hand!
I have just read the original plane /conveyor question and (in spite of my previous posts) the plane would definitely take off. As they say in exams, always best to read the question!
I think we ought to take a vote on it - all those in favour of the plane taking off raise their hand!
Took me a while to work out, but the answer given to the car/goat problem is wrong - you do not improve your odds by swapping.
At first, it does appear so. There appear to be eighteen strategies, e.g.:
Choose1, Car in 3, Stick = lose
Choose 1, Car in 3, Swap = win
etc.
and this suggests that three Stick options produce a win, while 6 swap options produce a win. However, the key factor is that the host knows which door to open after the frst round. This creates a situation where what appear to be different options are actually the same. There are several logical ways to approach the problem, and here's just one of them.
First, we have to remember that probability is the science of making predictions of the unknown, based on the facts known.
So let's assume that we know how the game is played.
This makes the issue of the host opening one losing door irrelevant. He will always choose a door that he knows is a loser. Right from the start, your chance of choosing the winning door is 50%. The door that he chooses is dependent on which of the three original doors you select:
The probability (Pr) of a particular scenario of doors open after the first 'round' is:
Pr (you chose Door 1 = 1/3) AND Pr(Car is behind Door 2 =1/3) AND Pr (host opens Door 2 = 1)
OR
Pr (You chose Door 1 = 1/3) AND Pr(Car is behind door 1 = 1/3) AND Pr(host opens Door 2 = 1/2)
= 1/9 +1/18 =3/18 = 1/6
Surprise, surprise, there are six potential layouts (Door 1, chosen, Door 2 open; Door 1, chosen, Door 3 open, etc.), three of which already make you a potential winner. So the odds of you being a potential winner after round 1 is 1/2 or 50%
Next let's assume you do not know the how the game is played.
You do not know that the host will definitely open a losing door. So the odds of you choosing the correct door in the first round 1/3 (9 scenarios, 3 win). If the game stopped here, the chances of you winning were 1/3.
The host now opens a door. It is a losing door. You now have more information, and a new decision to make, which makes everything previous irrelevant. For round 2, you have a simple decision - choose 1 door from 2: you have a 50% chance of winning.
Either way, the odds are 50% that you will choose the winning door.
At first, it does appear so. There appear to be eighteen strategies, e.g.:
Choose1, Car in 3, Stick = lose
Choose 1, Car in 3, Swap = win
etc.
and this suggests that three Stick options produce a win, while 6 swap options produce a win. However, the key factor is that the host knows which door to open after the frst round. This creates a situation where what appear to be different options are actually the same. There are several logical ways to approach the problem, and here's just one of them.
First, we have to remember that probability is the science of making predictions of the unknown, based on the facts known.
So let's assume that we know how the game is played.
This makes the issue of the host opening one losing door irrelevant. He will always choose a door that he knows is a loser. Right from the start, your chance of choosing the winning door is 50%. The door that he chooses is dependent on which of the three original doors you select:
The probability (Pr) of a particular scenario of doors open after the first 'round' is:
Pr (you chose Door 1 = 1/3) AND Pr(Car is behind Door 2 =1/3) AND Pr (host opens Door 2 = 1)
OR
Pr (You chose Door 1 = 1/3) AND Pr(Car is behind door 1 = 1/3) AND Pr(host opens Door 2 = 1/2)
= 1/9 +1/18 =3/18 = 1/6
Surprise, surprise, there are six potential layouts (Door 1, chosen, Door 2 open; Door 1, chosen, Door 3 open, etc.), three of which already make you a potential winner. So the odds of you being a potential winner after round 1 is 1/2 or 50%
Next let's assume you do not know the how the game is played.
You do not know that the host will definitely open a losing door. So the odds of you choosing the correct door in the first round 1/3 (9 scenarios, 3 win). If the game stopped here, the chances of you winning were 1/3.
The host now opens a door. It is a losing door. You now have more information, and a new decision to make, which makes everything previous irrelevant. For round 2, you have a simple decision - choose 1 door from 2: you have a 50% chance of winning.
Either way, the odds are 50% that you will choose the winning door.
HiRich said:
The probability (Pr) of a particular scenario of doors open after the first 'round' is:
Pr (you chose Door 1 = 1/3) AND Pr(Car is behind Door 2 =1/3) AND Pr (host opens Door 2 = 1)
OR
Pr (You chose Door 1 = 1/3) AND Pr(Car is behind door 1 = 1/3) AND Pr(host opens Door 2 = 1/2)
= 1/9 +1/18 =3/18 = 1/6
Surprise, surprise, there are six potential layouts (Door 1, chosen, Door 2 open; Door 1, chosen, Door 3 open, etc.), three of which already make you a potential winner. So the odds of you being a potential winner after round 1 is 1/2 or 50%
To be honest I'm having trouble following your logic, but I don't agree with your conclusions.
I think the actual situation is simple, but unintuitive.
You have a 1/3 probability of picking the right door at your first guess. Therefore a 2/3 probability that one of the other two doors is the right one (at this stage you can't tell which; each door has an equal 1/3 probability of being the right one).
It isn't explicitly stated, but I think it's implicit that the host will choose to open a wrong door. After they've done this, there is still a 1/3 chance that the door you originally picked is right, but now a 2/3 chance that the other door is right.
The reason it's 2/3 not 1/2 is that the host added some information by looking at the two doors you didn't pick, and telling you that one of them was the wrong one. Therefore the probability that the remaining one is the right one is the same as the probability that one of them is the right one.
GreenV8S said:
The probability (Pr) of a particular scenario of doors open after the first 'round' is:
Pr (you chose Door 1 = 1/3) AND Pr(Car is behind Door 2 =1/3) AND Pr (host opens Door 2 = 1)
OR
Pr (You chose Door 1 = 1/3) AND Pr(Car is behind door 1 = 1/3) AND Pr(host opens Door 2 = 1/2)
= 1/9 +1/18 =3/18 = 1/6
Surprise, surprise, there are six potential layouts (Door 1, chosen, Door 2 open; Door 1, chosen, Door 3 open, etc.), three of which already make you a potential winner. So the odds of you being a potential winner after round 1 is 1/2 or 50%
To be honest I'm having trouble following your logic, but I don't agree with your conclusions.
I think the actual situation is simple, but unintuitive.
You have a 1/3 probability of picking the right door at your first guess. Therefore a 2/3 probability that one of the other two doors is the right one (at this stage you can't tell which; each door has an equal 1/3 probability of being the right one).
It isn't explicitly stated, but I think it's implicit that the host will choose to open a wrong door. After they've done this, there is still a 1/3 chance that the door you originally picked is right, but now a 2/3 chance that the other door is right.
The reason it's 2/3 not 1/2 is that the host added some information by looking at the two doors you didn't pick, and telling you that one of them was the wrong one. Therefore the probability that the remaining one is the right one is the same as the probability that one of them is the right one.
i think the whole car goat thing is made up by statisticians so that they can feel smug and superior.
edited put to order my in right words the
>> Edited by tinman0 on Tuesday 13th December 12:58
The answer to the conveyor/jet question is that the plane takes off.
The planes wheels are simply there to support the plane and allow it to move on the ground. The wheels arent powered.
A car for example would stay still because the conveyor moving backwards would negate the traction and friction of the powered wheels against the ground in the opposite direction, just like a rolling road.
The jet on the other hand relies on its thrust against the air to push it forward. The jet moves backwards whilst it produces no thrust. Once it starts producing thrust it would need nominal time to counter and negate the backwards force present at that time BUT the action of the conveyor becomes irrelevant once the jet is thrusting. As the conveyor speeds up the jet wheels just turn faster BUT the jets motion would begin to move forwards as its thrust is against the air and eventually gain enough speed to produce lift and take off as normal
>> Edited by swilly on Tuesday 13th December 14:03
The planes wheels are simply there to support the plane and allow it to move on the ground. The wheels arent powered.
A car for example would stay still because the conveyor moving backwards would negate the traction and friction of the powered wheels against the ground in the opposite direction, just like a rolling road.
The jet on the other hand relies on its thrust against the air to push it forward. The jet moves backwards whilst it produces no thrust. Once it starts producing thrust it would need nominal time to counter and negate the backwards force present at that time BUT the action of the conveyor becomes irrelevant once the jet is thrusting. As the conveyor speeds up the jet wheels just turn faster BUT the jets motion would begin to move forwards as its thrust is against the air and eventually gain enough speed to produce lift and take off as normal
>> Edited by swilly on Tuesday 13th December 14:03
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