maths based riddles... See if you can answer them.
Discussion
I can think of an experiment that would give you an answer.
Take a model sailing yacht and a paddling pool. Set up fans to provide a constant wind, and observe the boat sailing.
Next, take the paddling pool and place it on a flat railway car, in front of a locomotive to provide an unimpeded airflow and drive the train at the same speed as the wind the fans were creating. Can the model boat still sail in the pool?
Take a model sailing yacht and a paddling pool. Set up fans to provide a constant wind, and observe the boat sailing.
Next, take the paddling pool and place it on a flat railway car, in front of a locomotive to provide an unimpeded airflow and drive the train at the same speed as the wind the fans were creating. Can the model boat still sail in the pool?
tank slapper said:
I can think of an experiment that would give you an answer.
Take a model sailing yacht and a paddling pool. Set up fans to provide a constant wind, and observe the boat sailing.
Next, take the paddling pool and place it on a flat railway car, in front of a locomotive to provide an unimpeded airflow and drive the train at the same speed as the wind the fans were creating. Can the model boat still sail in the pool?
But that's not the same at all - you are stilling missing the critical point: The wind is created ONLY by the boat itself - there is no external force acting on the boatTake a model sailing yacht and a paddling pool. Set up fans to provide a constant wind, and observe the boat sailing.
Next, take the paddling pool and place it on a flat railway car, in front of a locomotive to provide an unimpeded airflow and drive the train at the same speed as the wind the fans were creating. Can the model boat still sail in the pool?
The only way you could prove this (other than on a fast tide in a flat calm) would be with a remote controlled model yacht in moving water.
kVA said:
The wind is created ONLY by the boat itself - there is no external force acting on the boat
The wind is created by the movement of the water not the boat. As long as the water continues to flow, there is energy available to sail. If the water stops flowing, the energy input stops and the boat coasts to a halt.tank slapper said:
The wind is created by the movement of the water not the boat. As long as the water continues to flow, there is energy available to sail. If the water stops flowing, the energy input stops and the boat coasts to a halt.
No it is not - the wind is created by the hydrodynamic resistance of a free floating boat in the water... If something pushes against the boat (like the wind in the sails, for example), the boat will slow down or stop - and so will the apparent wind. The hull design and the angle of the keel relative to the flow alone will determine how much it slows down. However, critically, the water will keep flowing at 10 knots, but it cannot give any more energy to the boat once it is matched by the force acting on the boat in the opposite direction.Edited by kVA on Wednesday 14th September 15:37
This really does have to be my last post on this subject as I have to prepare for our Scally Rally effort - starting tomorrow!
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As I said previously, once you realise you are wrong - no need to apologise, just make a small donation here and all will be forgiven
http://www.justgiving.com/CanineCapersScallyRallyE...
Cheers...
kVA said:
No it is not - the wind is created by the hydrodynamic resistance of a free floating boat in the water... If something pushes against the boat (like the wind in the sails, for example), the boat will slow down or stop - and so will the apparent wind. The hull design and the angle of the keel relative to the flow alone will determine how much it slows down. However, critically, the water will keep flowing at 10 knots, but it cannot give any more energy to the boat once it is matched by the force acting on the boat in the opposite direction.
You are over thinking this. The question is simply checks whether you understand frames of reference. From your answers you appear not to.Strangely Brown said:
Google for "The Jeep Problem" if you're into reading maths papers.
There are some quite interesting variations on the problem, like optimising the time taken, or restricting the amount of fuel that can be put into the jeep, to using a fleet of jeeps that either do or don't have to return to base. I can see why it has kept people busy for such a long time.kVA said:
If you (and the University of the Internet) are correct, surely, all you would need to do is give a yacht a little shove to create an apparent wind and then it could sail all day, in a flat calm, with no water movement, sailing on it's own apparent wind, that it must create by sailing faster than the wind that started its movement! In fact, sailed well, it would get faster and faster as the day went on and would probably end up causing climate change in its Antipodes, or something 
This would also be known as perpetual motion - well done chaps!
Noone is saying that... this problem isn't the same as giving it a quick little shove then leaving it to its own devices. It's the same as giving it a constant push. If you stop pushing, the boat slows to a stop, because now there'd no longer be an exterior force pushing it through the air at 10 knots...This would also be known as perpetual motion - well done chaps!
Edited by kVA on Wednesday 14th September 14:55
Next time read other peoples' posts and make sure you understand them before responding and wasting my time saying the same thing over and over again.
kVA said:
This really does have to be my last post on this subject as I have to prepare for our Scally Rally effort - starting tomorrow!
As I said previously, once you realise you are wrong - no need to apologise, just make a small donation here and all will be forgiven
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If you care to quote Newton's 3rd law, I will explain to you how it works. I'm rather hoping that if you write it down, you will understand yourself, anyway. Hint: it does not mention "perpetual motion".As I said previously, once you realise you are wrong - no need to apologise, just make a small donation here and all will be forgiven
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The train is another red herring which is confusing you, however the situation is exactly analogous to the toy boat on the train, it doesn't matter if the water is pushed at 10 mph by a train, or by the water behind it (i.e.current).
If you want me to explain using sailing terms from your day skipper theory, I assume you understand "apparent wind", motion "through the water" and motion "over the ground".
Both boats start exactly stern to current, so doing 0 knots through the water, and 10 knots over the ground. One boat has a real 10 knot tailwind, so no apparent wind. The other boat has no real wind, so an apparent 10 knot headwind. Turn this boat 45 degrees to port. You now have a 10 knot apparent wind over the starboard bow, which is enough to sail. For simple explanation purposes, that 45 degree track through the water does have a component in the direction of the current , however small, which you add to the 10 knots to get the speed over the ground. Thus it will beat the first yacht which can only do 10 knots over the ground.
Yes that does mean it's making headway into an apparent wind, by tacking - that's fine, sailing boats do that every day.
In reality, the yacht cannot quite maintain the 45 degree track through the water. First, as it starts to move through the water, it will bring the apparent wind towards the bow, and will have to bear away a fraction more. Second, it will make some leeway. However, as long as it keeps its track through the water ahead of 90 degrees, it will win. And the fact that sailing boats make headway into apparent winds every day proves this can be done.
Edited by deeen on Wednesday 14th September 20:14
Lost_BMW said:
dibbly dobbler said:
Well done Blinky - that's what I've got the answer 
It could also be written as 3^^0 ( ie 3 to the power of 0) 3^^1, 3^^2, 3^^3 - surely not a coincidence...
The really brilliant thing would to be able to explain why this series works - like all the problems that triangular numbers, or Fibonacci series values help solve - as concisely and clearly as possible. Something my old boss (the best mathematician I've ever worked with) loved and was very good at doing. Isn't maths brilliant! It could also be written as 3^^0 ( ie 3 to the power of 0) 3^^1, 3^^2, 3^^3 - surely not a coincidence...
deeen said:
kVA said:
This really does have to be my last post on this subject as I have to prepare for our Scally Rally effort - starting tomorrow!
As I said previously, once you realise you are wrong - no need to apologise, just make a small donation here and all will be forgiven
http://www.justgiving.com/CanineCapersScallyRallyE...
Cheers...
If you care to quote Newton's 3rd law, I will explain to you how it works. I'm rather hoping that if you write it down, you will understand yourself, anyway. Hint: it does not mention "perpetual motion".As I said previously, once you realise you are wrong - no need to apologise, just make a small donation here and all will be forgiven
http://www.justgiving.com/CanineCapersScallyRallyE...
Cheers...
The train is another red herring which is confusing you, however the situation is exactly analogous to the toy boat on the train, it doesn't matter if the water is pushed at 10 mph by a train, or by the water behind it (i.e.current).
If you want me to explain using sailing terms from your day skipper theory, I assume you understand "apparent wind", motion "through the water" and motion "over the ground".
Both boats start exactly stern to current, so doing 0 knots through the water, and 10 knots over the ground. One boat has a real 10 knot tailwind, so no apparent wind. The other boat has no real wind, so an apparent 10 knot headwind. Turn this boat 45 degrees to port. You now have a 10 knot apparent wind over the starboard bow, which is enough to sail. For simple explanation purposes, that 45 degree track through the water does have a component in the direction of the current , however small, which you add to the 10 knots to get the speed over the ground. Thus it will beat the first yacht which can only do 10 knots over the ground.
Yes that does mean it's making headway into an apparent wind, by tacking - that's fine, sailing boats do that every day.
In reality, the yacht cannot quite maintain the 45 degree track through the water. First, as it starts to move through the water, it will bring the apparent wind towards the bow, and will have to bear away a fraction more. Second, it will make some leeway. However, as long as it keeps its track through the water ahead of 90 degrees, it will win. And the fact that sailing boats make headway into apparent winds every day proves this can be done.
Edited by deeen on Wednesday 14th September 20:14
Thanks for your patronising post, deeen: I do understand Newton's 3rd Law and I know it doesn't mention perpetual motion, but it does say that every action must have an equal and opposite reaction... All the well meaning folks on here that think I am wrong are conveniently forgetting the reaction of the 'apparent wind' on the boat's speed: If you exert a force on anything floating in a current, you will affect it's speed relative to the land (and the water). In this riddle, the apparent wind is always in the opposite direction to the current (by definition) and so will slow the boat down - reducing the apparent wind-speed in the process. You are also conveniently ignoring gravity!!! The heavier something is, the easier it is for some of the water to slip past it, rather than push it along at the same speed - especially something specifically designed to be smooth and slippery underneath, precisely so that water can easily slip past it and allow unimpeded forward motion: If the boat always moved at the same speed as the current, regardless of weight, logging in Canada would be a seriously dangerous occupation for those guys that stand on them in the fast flowing rivers!!!
I will stand by my opinion that boat B will win this race, as, unless boat A has zero aerodynamic drag (in which case it will be a draw), any other solution will contradict the basic laws of physics - it just cannot be done.
I have been away for a week (windsurfing as it happens - lots of apparent wind) so I may have the stomach for one more try.
(Even though it is astonishing that the multiple links to University Physics departments aren't enough to convince you.)
Since you are so up on your physics KvA could you show me on a vector diagram the SIGNIFICANTLY different forces acting on two boats in these situations.
One: middle of the sea - no tide or current - 10 knot breeze.
Two: no wind (relative to land) - current of 10 knots (relative to land).
I am not sure you have ever given your sailing credentials but your language appears to show that you know something about sailing.
Put yourself in the shoes of the sailors in the above situations.
There is simply no difference between the two.
The sails will feel 10 knots of breeze.
The hull is sitting in the water. The FRICTION of the hull's bulk sitting in the water is stopping it going backwards (relative to the water).
If it were frictionless then it would be shooting backwards at 10knots (relative to the water) in BOTH situations.
Please give me a force - point me to ANYTHING that the sailor could use to determine which of the two situations he was in?????
He can't. It's impossible.
(Even though it is astonishing that the multiple links to University Physics departments aren't enough to convince you.)
Since you are so up on your physics KvA could you show me on a vector diagram the SIGNIFICANTLY different forces acting on two boats in these situations.
One: middle of the sea - no tide or current - 10 knot breeze.
Two: no wind (relative to land) - current of 10 knots (relative to land).
I am not sure you have ever given your sailing credentials but your language appears to show that you know something about sailing.
Put yourself in the shoes of the sailors in the above situations.
There is simply no difference between the two.
The sails will feel 10 knots of breeze.
The hull is sitting in the water. The FRICTION of the hull's bulk sitting in the water is stopping it going backwards (relative to the water).
If it were frictionless then it would be shooting backwards at 10knots (relative to the water) in BOTH situations.
Please give me a force - point me to ANYTHING that the sailor could use to determine which of the two situations he was in?????
He can't. It's impossible.
kVA said:
If you (and the University of the Internet) are correct, surely, all you would need to do is give a yacht a little shove to create an apparent wind and then it could sail all day, in a flat calm, with no water movement.
This is where you have the misunderstanding as far as I can tell.The water (in the constant current situation) is CONSTANTLY pushing the boat through the air.
As I said above - if it were frictionless the boat would shoot backwards over the water as the air resistance met no opposing force.
However the boat has lots of opposing force from the hull, rudder and centreboard all stuck in the water creating plenty of drag.
Forget the apparent wind argument.
It makes the whole thing far more complicated than it needs to be.
The equal and opposite reaction that you are so desperately searching for is the force applied by the water on the hull.
It is there when you face a 10 knot breeze on still water (otherwise you would be shooting backwards at 10 knots).
AND it is there when you are pushed along at 10 knots by the current (otherwise you would be stationary).
I've been quiet on the boat front - mainly because it goes over my head (despite having sailed and windsurfed), and I'm not sure who it is that I agree with.
I can't explain the physics in the way that others are doing here, but I just can't see how boat A (still air, 10knot current) could be faster than boat B (10knot wind behind it, 10 knot current).
I understand that boat B is becalmed, and can travel at a velocity of 10 knots max - slightly less because the water would flow around the boat, and nothing moves at the same speed as the current, but can therefore 'top up' this energy with the slight wind differential (i.e. if current is 10, friction, inertia, resistance, gravity, whatever you want to call it means the boat would travel at 9; boat therefore feels wind behind it of 1 which can be used to bring the velocity back closer to 10).
I cannot understand how boat A can use the 'apparent headwind'. Yes, I know it is possible to sail into the wind and understand that, but as soon as the boat raises its sails, it will slow due to air resistance, therefore reducing the speed of this apparent headwind. This isn't useable wind, it is air resistance.
My frame of reference would be a sand yacht travelling down a hill, is the suggestion that the sand yacht faced with air resistance could zig zag down the hill and get to the bottom faster than another yacht that just went straight down and had the wind behind it (travelling at the same speed (or slightly faster) as the yacht)?
ETA - This isn't a troll. I genuinely don't understand this in the slightest.
I can't explain the physics in the way that others are doing here, but I just can't see how boat A (still air, 10knot current) could be faster than boat B (10knot wind behind it, 10 knot current).
I understand that boat B is becalmed, and can travel at a velocity of 10 knots max - slightly less because the water would flow around the boat, and nothing moves at the same speed as the current, but can therefore 'top up' this energy with the slight wind differential (i.e. if current is 10, friction, inertia, resistance, gravity, whatever you want to call it means the boat would travel at 9; boat therefore feels wind behind it of 1 which can be used to bring the velocity back closer to 10).
I cannot understand how boat A can use the 'apparent headwind'. Yes, I know it is possible to sail into the wind and understand that, but as soon as the boat raises its sails, it will slow due to air resistance, therefore reducing the speed of this apparent headwind. This isn't useable wind, it is air resistance.
My frame of reference would be a sand yacht travelling down a hill, is the suggestion that the sand yacht faced with air resistance could zig zag down the hill and get to the bottom faster than another yacht that just went straight down and had the wind behind it (travelling at the same speed (or slightly faster) as the yacht)?
ETA - This isn't a troll. I genuinely don't understand this in the slightest.
Edited by therealpigdog on Friday 23 September 11:31
therealpigdog said:
I understand that boat B is becalmed, and can travel at a velocity of 10 knots max - slightly less because the water would flow around the boat, and nothing moves at the same speed as the current...
This is the one essential element that eveeryone else seems NOT to understand...
ok, so just read this http://www.phys.unsw.edu.au/~jw/sailing.html and whilst I don't understand it, it has been written by people who probably know a lot more than me so whilst it seems illogical to me I'm going to just accept that these clever professor types might know what they are talking about.
as you were.
as you were.
walm said:
This is where you have the misunderstanding as far as I can tell.
The water (in the constant current situation) is CONSTANTLY pushing the boat through the air.
As I said above - if it were frictionless the boat would shoot backwards over the water as the air resistance met no opposing force.
However the boat has lots of opposing force from the hull, rudder and centreboard all stuck in the water creating plenty of drag.
Forget the apparent wind argument.
It makes the whole thing far more complicated than it needs to be.
The equal and opposite reaction that you are so desperately searching for is the force applied by the water on the hull.
It is there when you face a 10 knot breeze on still water (otherwise you would be shooting backwards at 10 knots).
AND it is there when you are pushed along at 10 knots by the current (otherwise you would be stationary).
Surely you understand that the heavier a boat is, the slower it goes - however it is being pushed? If you were being carried by the tide alone in a flat calm and a fat bloke was suddenly dropped onto your board, do you really believe that you would continue at exactly the same pace? The water (in the constant current situation) is CONSTANTLY pushing the boat through the air.
As I said above - if it were frictionless the boat would shoot backwards over the water as the air resistance met no opposing force.
However the boat has lots of opposing force from the hull, rudder and centreboard all stuck in the water creating plenty of drag.
Forget the apparent wind argument.
It makes the whole thing far more complicated than it needs to be.
The equal and opposite reaction that you are so desperately searching for is the force applied by the water on the hull.
It is there when you face a 10 knot breeze on still water (otherwise you would be shooting backwards at 10 knots).
AND it is there when you are pushed along at 10 knots by the current (otherwise you would be stationary).
The resistance of the hull against the moving water it will still slow down if you make it too heavy - the water will slip past the hull, rather than pushing it along at the same speed. Agreed?
Well, the 'apparent wind' is worse than the fat bloke, as it directly opposes the direction of motion (rather than being a component of force based on gravity, which is of course perpendicular to the direction of motion).
therealpigdog said:
I cannot understand how boat A can use the 'apparent headwind'. Yes, I know it is possible to sail into the wind and understand that, but as soon as the boat raises its sails, it will slow due to air resistance, therefore reducing the speed of this apparent headwind. This isn't useable wind, it is air resistance.
There is no difference between "useable wind" and "air resistance".They are the same thing.
Either there are a bunch of nitrogen and oxygen molecules brushing past your face or there aren't.
To be specific relative to the becalmed B here's what would happen:
A would be pushed backwards (relative to B) from the word go by the wind he feels.
(This is just like when you are sitting in a boat on a pond in the wind. You get pushed down-wind if you do nothing, unless you are anchored.)
A then hoists his sails and gets pushed backwards even faster as the wind is now pushing against more than just the spars. (Same on pond.)
But - and here is the part you guys are struggling with. A then bears away from the wind and sails upwind back to his starting point (so now level with B) and then past.
I think the really important point to realise is that the wind doesn't push him backwards at 10 knots.
It is FAR LESS than 10 knots. As a result there are still PLENTY of N and O molecules passing by him.
Those little guys bounce off his sails, pushing him say right, and then the resistance of the centreboard translates that right-hand movement forwards a little bit = upwind.
therealpigdog said:
My frame of reference would be a sand yacht travelling down a hill, is the suggestion that the sand yacht faced with air resistance could zig zag down the hill and get to the bottom faster than another yacht that just went straight down and had the wind behind it (travelling at the same speed (or slightly faster) as the yacht)?
The difference here is that you are DRAMATICALLY slowing the zig-zagging sand yacht by making him avoid the path of least resistance DOWNHILL.As soon as he veers away from straight down he will start to slow.
That doesn't happen for the boat. It can point any direction and the current will still push it 10 knots.
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