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Old 24-06-2010, 21:10   #31
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Bash -
Right you are. My fault for thinking knots instead of kilometers per hour. Hell, I'm a yank so I can't think in terms of kph anyway.
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Old 24-06-2010, 21:23   #32
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No, any headway will be enough, because the current is taking both boats towards the finish line at the same VMG. Beyond that, one boat is slowed by the wind, the other is making way into it.
Ahhhhh... yes....of course....

so the winning time would be 2 hours less boatspeed divided by 2 ish
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Old 26-06-2010, 16:44   #33
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Ok, it's been a couple of days and there are no more people answering, here is the web site of the answers that the engineers gave.
WD
CR4 - Blog Entry: Sailing: Newsletter Challenge (06/01/10)
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Old 03-07-2010, 05:14   #34
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You tease... why post the link to the brainteaser when the answer hasn't even been posted yet??

1 - the straight line boat: if a log floats at the same speed as the tide (5km/h), it will take 2 hours to travel 10km. I learnt that in high school physics. However, a boat will have windage due to its sail luffing (even if it is pointing into [apparent] wind), so it will be a little slower than the log.

2 - The tacking boat (me). If tacking at rightangles (45 deg to the wind) we can use pythagorus theory (c2 = a2 + b2)to calculate distance travelled. Therefore, if the hypotenuse is 10km, my tacks would be the total of 8km and 6km (14km).

If you are correct, and the windspeed generated by apparent wind is 2.7knots, on top of the 5km/h generated by tide, then my speed would be more. If on my dart 18, I would prob be doing about 2 knots (4.5ish km/h?) through the water, giving me a total speed of 9.5 km/h. So I would take just over 1 hour.

Therefore, I win.
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Old 03-07-2010, 05:54   #35
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Originally Posted by Aaza Dana View Post
Y
If on my dart 18, I would prob be doing about 2 knots (4.5ish km/h?) through the water, giving me a total speed of 9.5 km/h. So I would take just over 1 hour.

Therefore, I win.
Duh - my speed would be about 1.5 hours, since distance covered over ground when tacking would be 14km. But I still win.
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Old 03-07-2010, 07:27   #36
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please stop this could go on for ever,the cold beer is the winner....
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Old 08-07-2010, 06:41   #37
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Whats the tide doing?
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Old 08-07-2010, 10:00   #38
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The answer is posted now. I won. Of course. And my answer was better than the engineers' answer. Yeah.
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Old 08-07-2010, 10:07   #39
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How experienced and/or drunk is the crew on the tacking boat? If it's a novice crew of drunks, they might lose all speed on the tack, and end up way behind the straight boat.
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Old 08-07-2010, 17:50   #40
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The answer is posted now. I won. Of course. And my answer was better than the engineers' answer. Yeah.
Is it posted by a bunch of engineers?

One should listen to the stories my father in law has about the stuff he had to make work despite the engineers.

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Old 08-07-2010, 18:43   #41
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The only thing not considered is the set caused by the apparent. The set will be about 90 degrees to the heading. If the tacking boat has a heading anything different than directly at the line the vector of the set will be towards the start line reducing VMG.

This will affect the finish time but the tacking boat will still win. Any forward progress by the tacking boat is additive to the 2.7 kts current.

BTW 2.7 kts won't even fill my sails. I'd retire, get back to the dock and get stuck into the margaritas.
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Old 08-07-2010, 18:52   #42
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Engineers, that provided no math formulas for their conclusion....scary...
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Old 08-07-2010, 19:34   #43
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What a lame answer. Completely ignored the question "what was the winning time?"

In fairness to the engineers, a number of sailors on our forum missed it.

Brett
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Old 08-07-2010, 21:13   #44
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In mild defense of those engineers and sailors who didn't do or attempt the math

- Most engineers don't create new math. We use formula created by folks who like that sort of thing. We use the formula to make predictions, then we test for outcomes. Usually the outcome doesn't match the prediction because the formula is incomplete as all variables are not considered. Then we send the results back to the math guys to modify the model.

Some of the variables in this example have been discussed. Boat polars, current differences across the channel., how well the crew actually trims the sails and probably several others. Another to discuss is windage. If the drifting boat has a large frontal area there will be drag against the apparent and likely won't make 2.7 knots. Do both boats have sails up? How will the apparent affect the drifting boat? What is the sail area of the drifting boat? can the drifting boat drop the sails?

Basically I think the answer of time cannot be derived accurately with the given info. We can solve for the winning boat if crews and boats are "identical".

A couple of folks here did some math and are to be applauded. If I was getting paid for it I might do some math. I am here for fun...
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Old 09-07-2010, 18:34   #45
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I would like to strangle some "engineers"......Why do you insist on putting oil filters on with the open end down...(Ford Lehmans)?

Why do you insist on putting a sharp edge...directly opposite of the bolt/nut that usually freezes up....?

Have you ever repacked a stuffing box?
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