Can you solve the maths question for Singapore schoolkids that went viral?

[Declan]

27 minutes ago, quilp said:

Which part of the ship travels the furthest? 

I want to say the side furthest from the centre of the circle - like on a running track. 

But for that to be true and the ship not tear apart the inside would need travel fractionally faster than the outside and that seems irrational. 

Edited February 3, 2018 by Declan

[TheTeapot]

2 minutes ago, Declan said:

I want to say the side furthest from the centre of the circle - like on a running track. 

But for that to be true and the ship not tear apart the inside would need travel fractionally faster than the outside and that’s seems irrational. 

There would be a slight tilt to solve that paradox

[quilp]

20 minutes ago, Declan said:

I want to say the side furthest from the centre of the circle - like on a running track. 

But for that to be true and the ship not tear apart the inside would need travel fractionally faster than the outside and that seems irrational. 

You're on the right track... 

[quilp]

55 minutes ago, quilp said:

Which part of the ship travels the furthest? 

Ah, read it wrong. The boat's just travelling around in a circle. I was basing it on the vessel circumnavigating the globe. But if a boat was sailing a complete circumference, which part of the boat travels the furthest..? 

[wrighty]

6 hours ago, Chinahand said:

A ship which can sail at 2 x pi miles an hour is steaming around a 1 mile radius circle.  The sea in this area has a uniform current flowing from north to south at a speed of 0.5 x pi miles an hour.

How long does it take for the ship to sail around the circle?

I don't think the effect of the current just cancels out because if the current was 2 pi miles an hour the boat wouldn't be able to do the into the current leg.

Vectors ... what fun!

I do enjoy that this is a reasonable analogy of a work problem I'm doing at the moment ... the joy of maths! Any ideas ... anyone!?

I've solved it, but it ends in an integral that I don't think I can integrate!  Numerically my calculator comes up with 0.8 hours.  Is this right?

Edited to add - No it isn't! Calculator was in degree mode, should have been radians.  1.0498 hours is my revised answer.

Edited February 3, 2018 by wrighty
Correct my error!

[Uhtred]

1 hour ago, wrighty said:

I've solved it, but it ends in an integral that I don't think I can integrate!  Numerically my calculator comes up with 0.8 hours.  Is this right?

Edited to add - No it isn't! Calculator was in degree mode, should have been radians.  1.0498 hours is my revised answer.

Of course if this is the Ben My Chree the answer is “It hasn’t left port, it’s broken again”.

[quilp]

The top of the mast. 

[P.K.]

15 minutes ago, quilp said:

The top of the mast. 

In a flat calm I would say the stern quarter furthest from the axis.

[John Wright]

12 minutes ago, P.K. said:

In a flat calm I would say the stern quarter furthest from the axis.

For quilps question it’s the side of the ship furthest from the centre of the circle that travels slightly farther.

Heres the analogy that explains, pragmatically, why, what happens, and why there aren’t dire self destruct consequences.

Forget it’s a ship steaming in circles. Think of it as a clock face. Take the minute hand. Take a spot close to the centre, half way along the hand and at the tip. They each travel a different distance around the circle in an hour. The mast won’t travel any farther than any spot in the centre of the ship and it doesn’t depend upon the ship or clock being angled.

 Neither the ship, nor the minute hand will tear themselves apart.

Edited February 3, 2018 by John Wright

[P.K.]

So you agree with me John. But Mr Quilp must be aware of this so what has he got up his sleeve?

A very tall mast?

If you drop a perpendicular from the top of the mast to the centre of the earth then it's possible.

I think.

Eratosthenes measured the circumference of the earth circa 200 BC with a similar method.

Mind you, his accuracy was out by at least a couple of percent.....

Edited February 4, 2018 by P.K.
Typo

[wrighty]

8 hours ago, Chinahand said:

 

I don't think the effect of the current just cancels out because if the current was 2 pi miles an hour the boat wouldn't be able to do the into the current leg.

To be pedantic China, the current would have to be strictly greater than 2pi for it not to work. If the current were exactly 2pi the boat’s speed would be instantaneously zero at a single point on the circle but since there’s an acceleration it would move through this point and carry on. Akin to if you throw a ball straight up in the air its speed is zero at the apex, but it doesn’t stop 

Edited February 3, 2018 by wrighty
Grammar typo - in a pedantic post can’t have that!

[Bobbie Bobster]

1 hour ago, John Wright said:

Neither the ship, nor the minute hand will tear themselves apart.

Oh, my!

 

[Chinahand]

2 hours ago, wrighty said:

To be pedantic China, the current would have to be strictly greater than 2pi for it not to work. If the current were exactly 2pi the boat’s speed would be instantaneously zero at a single point on the circle but since there’s an acceleration it would move through this point and carry on. Akin to if you throw a ball straight up in the air its speed is zero at the apex, but it doesn’t stop 

Is that right? 

Imagine looking down on the situation with the ships circuit like it traversing a clock face with 12 o'clock due North.  The current is from uniformly from the north and the ship starts at 12 o'clock on the clock face.

The ship would rush down from 12 to 6 o'clock and then at 6 o'clock try to start sailing up from 6 back to 12, but even with all it's power pushing north it will never be able to move that way.

Wouldn't it just move off westerly on a tangent - all its power unsuccessfully trying to overcome the current and carried westerly by the residual momentum it had when its motors were still pushing it an infintesimal amount westward, an infintesimal distance before 6 o'clock

[quilp]

2 hours ago, P.K. said:

A very tall mast? 

Er, yes, a very tall mast. This is the kind of pointless factoid that Ken Cooil (science teacher) would delight in puzzling short-panted 11 year-olds with. Relating to Chay Blyth's circumnavigation (1971ish). Funny the things you remember...

[John Wright]

1 hour ago, Chinahand said:

Is that right? 

Imagine looking down on the situation with the ships circuit like it traversing a clock face with 12 o'clock due North.  The current is from uniformly from the north and the ship starts at 12 o'clock on the clock face.

The ship would rush down from 12 to 6 o'clock and then at 6 o'clock try to start sailing up from 6 back to 12, but even with all it's power pushing north it will never be able to move that way.

Wouldn't it just move off westerly on a tangent - all its power unsuccessfully trying to overcome the current and carried westerly by the residual momentum it had when its motors were still pushing it an infintesimal amount westward, an infintesimal distance before 6 o'clock

I’m no mathematician or expert in fluid mechanics, China. But doesn’t your analogy assume a drag coefficient of one at 9 o’clock? And that’s a a current speed equalling ship speed. In the question they aren’t equal.

Thats like a wall facing the current, head on.

The hull will have water piercing and dividing flow properties so the current is divided and diverted, so, unlike wrighty’s gravity analogy, the ship will never lose forward way or momentum.

As the drag coefficient will change with hull design, and the effect of the current will change with things like depth, don’t we just have to assume, because that’s what we’re told,  the ship has enough power to maintain the 2 pi speed constantly, with and against the current.

in other words isn’t it a trick question, you’re overthinking, and the current is irrelevant, it’s just the amount of power and fuel necessary to maintain the constant speed that changes en route. We are looking at speed, not momentum.

Best example I can give is in storms, where ships can be moving at less than wave speed, they don’t move at the speed of the current or waves if not underway, it’s an undesirable effect, potentially dangerous.

I think it was one of the factors of the wreck of Riverdance on Cleveleys beach  as it lost way in Lune Deep in the MAIB report.

If I’m correct, and I’ve indicated my limitations already, then with r = 1  2 pi and 2 pi r are identical. The circumference is 6.2832 Miles, so is the speed. Ergo it’s an hour.

Or am I guilty of a reductio ad absurdum?

Edited February 3, 2018 by John Wright