Puzzling me, but not a puzzle
Justin C
serious maths or trig/geometry always get the better of me. Hence my
position.
I'm taking part in an on-line version of the Volvo Ocean Race (for those
interested: <URL: http://www.volvooceanracegame.org/>). It's basically
an on-line sailing game. Boats sail at different speeds depending on
their angle to the wind, as well as depending on the wind speed. What
is puzzling me is, what is my best heading to optimise speed and close
on the finish (which is 1400 miles away).
Here's the scenario I have:
The finish line is exactly where the wind is coming from.
Boats cannot sail directly into the wind.
Boats sail faster the greater their angle to the wind.
I can manage
7.9 knots at 45 degrees
8.7 knots at 50 degrees
9.4 knots at 55 degrees
10.6 knots at 73 degrees
Increasing my angle to the wind reduces my speed until I'm actually
heading in the wrong direction completely and can use the spinnaker.
What I'm trying to figure out is at which angle to the wind is it better
for me to sail, bearing in ming the at the higher speed I'm not closing
on the finish that much. There has to be a 'best' angle for speed versus
distance toward the finish made good.
I'm sure there must be a formula for this too. I'd be happy to work it
out myself but I don't really know where to start... A sheet of graph
paper seems sensible, but it's got to be do-able with numbers rather
than much plotting.
A solution would be useful, but if someone can point me to somewhere
where I'd be able to learn it would be better.
Thank you for any help you can give with this.
Justin.
--
Justin C, by the sea.
Patrick Hamlyn
Simple trig will tell you the component of your velocity which is in the correct
direction. Absent any info on future wind shifts, this is all you need to know.
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
John Bailey
<justi...@purestblue.com> wrote:
>The finish line is exactly where the wind is coming from.
>Boats cannot sail directly into the wind.
>Boats sail faster the greater their angle to the wind.
>I can manage
> 7.9 knots at 45 degrees
> 8.7 knots at 50 degrees
> 9.4 knots at 55 degrees
> 10.6 knots at 73 degrees
>
>A solution would be useful, but if someone can point me to somewhere
>where I'd be able to learn it would be better.
If you polar plot the velocity of the boat at various angles to the
wind, your plot will be a cardiod--a very flattend heart shape. The
velocity is zero in the direction the wind is coming from. It grows
larger as the heading of the boat becomes at right angles to the wind
and then falls off again as the heading becomes away from the wind,
falling to the wind velocity at a heading directly from the wind.
Using this plot, the component of velocity in the direction you want
to go (in your case, directly into the wind) is the projection of the
velocity vector on to that axis. If you imagine a line which is just
tangent to the upper lobes of the heart, those two points of tangency
represent the direction and speed which has the maximum component in
the direction of the wind. If you turn toward the wind, the
component becomes slower; if you turn away from the wind, the
component becomes slower, although the boat will go faster.
The other post is correct, but, may only be a more exact definition of
the problem.
For more information a search of wiki/sailing is a good start.
humunculus
We'd also have to know how wide the finish line is and much time it
takes you to tack. If tacking loses no time at all, then sail 50
degrees to the wind and tack as often as possible. If taking takes
time, or if the finish line is excessively wide, then you want to tack
less often.
--riverman
Justin C
>
[snip]
> We'd also have to know how wide the finish line is and much time it
> takes you to tack. If tacking loses no time at all, then sail 50
> degrees to the wind and tack as often as possible. If taking takes
> time, or if the finish line is excessively wide, then you want to tack
> less often.
It's not that straight-forward. There is a course, and the wind
varies... considerably. What I was trying to figure out was the best
angle off the wind regardless of wind direction. Which I now have (at
least for one set of sails) with the help of the polar graph.
Now I've to extract that info for four other sales.
I don't know if they've put any time penalty in for tacking.
Where the course is over 1400 miles, and the speed of these boats
averages around the 10 to 12 knot mark we're going to be at this for
several days, I think frequent tacking could get a bit tedious!
Thank you for the suggestions.
Justin C
>
> If you polar plot the velocity of the boat at various angles to the
> wind, your plot will be a cardiod--a very flattend heart shape. The
> velocity is zero in the direction the wind is coming from. It grows
> larger as the heading of the boat becomes at right angles to the wind
> and then falls off again as the heading becomes away from the wind,
> falling to the wind velocity at a heading directly from the wind.
> Using this plot, the component of velocity in the direction you want
> to go (in your case, directly into the wind) is the projection of the
> velocity vector on to that axis. If you imagine a line which is just
> tangent to the upper lobes of the heart, those two points of tangency
> represent the direction and speed which has the maximum component in
> the direction of the wind. If you turn toward the wind, the
> component becomes slower; if you turn away from the wind, the
> component becomes slower, although the boat will go faster.
I've done this. It's quite time consuming though to extract the speed
for each degree! For those angles to the wind where I'm not sailing I've
skipped most of them, it's only where the data gets close to being right
that I've been precise... I was going to post my result, but, hey, there
might be a competitor here!
Thank you for this solution. For me it has been the easiest. I am going
to go away now and have a think about how to get the answer using
trigonometry - it's been twenty five years since I've done any, and to
say I'm rusty is an understatement!
Justin C
>
> Simple trig will tell you the component of your velocity which is in the correct
> direction. Absent any info on future wind shifts, this is all you need to know.
This may be simple to you, not so for me, I haven't done any for a
little over twenty five years!
I'm thinking aloud here, so please forgive my ramblings. Perhaps you, or
someone, can tell me where I'm going wrong, if at all.
OK. So, from the data I've extrated for the polar graph (see other post)
I have a range of speeds and angles off the wind. Below 21 degrees off
the wind I cannot sail. At 21 degrees I get the lowest measurement of
.5kt. There are then various, but ascending, speeds for the incrementing
angles up to 75 degrees where I am sailing at 11.4kts, beyond which I
have more measurements, but they are decreasing.
For the problem I set I need to calculate, for each possible
heading (spreadsheet time), the distance I will be from the finish line
to know which heading will give me the best closing speed.
OK. Sticking my thinking head on - and trying to think trig. My course
will be a triangle, the base of which will be the line linking start and
finish, I will have the angle between that line and the course I have
steered. I will have a length for the start finish line (course length),
and length for my sailed course (speed * time). The calculation I am
after is the length of the (currently missing) line - current position
to the finish line. So, I have the length of two sides, and the angle
between them.
Googling I come up with the "law of cosines". If the two sides that I
have are A and B, the side I don'e have is C, and the angle I have is X,
then:
hmmmm... this isn't easy with plain text!
C^2 = A^2 + B^2 - 2ABcosX
OK, I think I can now figure that out (if I've got the above right).
If anyone sees that I've got something please let me know!
Thank you to all who have help me reach this point.
CBFalconer
>
... snip ...
>
> I've done this. It's quite time consuming though to extract the
> speed for each degree! For those angles to the wind where I'm not
> sailing I've skipped most of them, it's only where the data gets
> close to being right that I've been precise... I was going to
> post my result, but, hey, there might be a competitor here!
All you need is a slip-stick with trig scales.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Patrick Hamlyn
Once you've worked out the 'optimum angle' for each sail, you need to work out
where the 'ley line' is for that sail.
From the left edge of the finish line (or from the next turning point), draw
back a line as if you were approaching that point from the left. From the right
do the same approaching from the other tack. If you ever sail outside of these
two lines you've lost time because you'll be forced to sail at a
less-than-optimal angle at some point subsequently (barring wind shifts which
might save you). At the beginning of the course those lines are way off and not
a worry. As you get closer they are nearby and you need to pay close attention.
Furthermore, the wind is going to shift. So if you sail close to the ley line at
any point and the wind shifts, you either get a lift which advantages you or you
get pushed and you lose 'height on the wind' and find yourself with more
'upwind' to make up.
So the further you sail from the 'centre line' of the course the bigger the punt
you are taking on the wind shifting in one direction and not the other.
In big one-on-one races if somebody finds himself at a 5% speed disadvantage he
will generally try to guess the next wind shift and sail in such a way that he
will get a huge lift it it comes off.
Of course the other guy frequently 'covers' this option by sailing in the same
direction (but not quite as far from the centre line as the other guy).
humunculus
OK, this is not as simple as you want it to be. If you are trying to
head directly into the wind, its a pretty straightforward calculation.
However, if the wind is at an angle to the direction that you want to
head, it changes the calculations. Of course, the winds are variable
over the length of the course you want to sail, so your angle to wind
has to be constantly adjusted. And on top of that, you have to take
into account that wind will be changing over time itself....so all
this comes together to make for a very complex set of factors.
However, for the very simplest one; you want to know how to adjust
your angle so that you have the greatest velocity directly into the
wind (assuming that the finish line is directly upwind from you), then
the answer is V*cosX where V is your ships actual velocity in the
direction its pointed, and X is the angle between your direction and
the wind.
Which ship are you? I'm "Irregardless"
--riverman
Justin C
> Justin C wrote:
>>
> ... snip ...
>>
>> I've done this. It's quite time consuming though to extract the
>> speed for each degree! For those angles to the wind where I'm not
>> sailing I've skipped most of them, it's only where the data gets
>> close to being right that I've been precise... I was going to
>> post my result, but, hey, there might be a competitor here!
>
> All you need is a slip-stick with trig scales.
I have no idea what either of those are, the first page of hits for
slip-stick on Google makes the mind boggle. I now know what trig scales
are, but can't see the need to make a purchase! Excel is my calculator -
it may be clunky, but it works.
Justin C
> Justin C <justi...@purestblue.com> wrote:
>>
>>It's not that straight-forward. There is a course, and the wind
>>varies... considerably. What I was trying to figure out was the best
>>angle off the wind regardless of wind direction. Which I now have (at
>>least for one set of sails) with the help of the polar graph.
>>
>>Now I've to extract that info for four other sales.
>>
>>I don't know if they've put any time penalty in for tacking.
>>
>>Where the course is over 1400 miles, and the speed of these boats
>>averages around the 10 to 12 knot mark we're going to be at this for
>>several days, I think frequent tacking could get a bit tedious!
>
> where the 'ley line' is for that sail.
>
> From the left edge of the finish line (or from the next turning point), draw
> back a line as if you were approaching that point from the left. From the right
> do the same approaching from the other tack. If you ever sail outside of these
> two lines you've lost time because you'll be forced to sail at a
> less-than-optimal angle at some point subsequently (barring wind shifts which
> might save you). At the beginning of the course those lines are way off and not
> a worry. As you get closer they are nearby and you need to pay close attention.
Ah, this is racing tactics. Being one of the ones who "does what he's told, when he's told" (in between the vomiting), such nuances have never been explained to me, but this is all very clear and obvious now you mention it.
> Furthermore, the wind is going to shift. So if you sail close to the ley line at
> any point and the wind shifts, you either get a lift which advantages you or you
> get pushed and you lose 'height on the wind' and find yourself with more
> 'upwind' to make up.
We do have a 12, 24, and 36 hour wind forecast, but everyone is seeing the same forecast, so it's not much of an advantage. Nor is it a certainty (AIUI they're programming real wind into the game). Also, the wind in the area might be doing shown as doing one thing, but in your little 400 yard circle it can be completely different. It wasn't until I started sailing that I learnt how odd the wind's behaviour can be.
> So the further you sail from the 'centre line' of the course the bigger the punt
> you are taking on the wind shifting in one direction and not the other.
>
> In big one-on-one races if somebody finds himself at a 5% speed disadvantage he
> will generally try to guess the next wind shift and sail in such a way that he
> will get a huge lift it it comes off.
>
> Of course the other guy frequently 'covers' this option by sailing in the same
> direction (but not quite as far from the centre line as the other guy).
This is all very interesting. Thank you for taking the time to post. It's definitely adding to my understanding of what goes on when racing - not that I do any more than round the cans on a Sunday morning out in the bay. Principles are the same, and the don't just apply to racing.
Good stuff, thanks again.
Eric Sosman
> In article <49482874...@yahoo.com>, CBFalconer wrote:
>>
>> All you need is a slip-stick with trig scales.
>
O tempora! O mores! O gimme anutha; make it a double.
(As oldfashioned as I'm feeling at the moment, I can't
help but suspect CBF is more oldfashioned still: By the
time I met the slipstick its hyphen had selfelided.)
--
Eric Sosman
eso...@ieee-dot-org.invalid
Justin C
> On Dec 17, 2:23 am, Justin C <justin.0...@purestblue.com> wrote:
[snip]
>> Googling I come up with the "law of cosines". If the two sides that I
>> have are A and B, the side I don'e have is C, and the angle I have is X,
>> then:
>>
>> hmmmm... this isn't easy with plain text!
>>
>> C^2 = A^2 + B^2 - 2ABcosX
>>
>> OK, I think I can now figure that out (if I've got the above right).
>>
>> If anyone sees that I've got something please let me know!
> OK, this is not as simple as you want it to be. If you are trying to
> head directly into the wind, its a pretty straightforward calculation.
> However, if the wind is at an angle to the direction that you want to
> head, it changes the calculations. Of course, the winds are variable
> over the length of the course you want to sail, so your angle to wind
> has to be constantly adjusted. And on top of that, you have to take
> into account that wind will be changing over time itself....so all
> this comes together to make for a very complex set of factors.
I didn't want to over-complicate the problem. The course I follow is up
to me and I have to make judgement calls based on... well, you know,
you're playing too. I just wanted to know how to go as fast as possible
in the right sort of direction without using excess speed going in the
wrong direction.
> However, for the very simplest one; you want to know how to adjust
> your angle so that you have the greatest velocity directly into the
> wind (assuming that the finish line is directly upwind from you), then
> the answer is V*cosX where V is your ships actual velocity in the
> direction its pointed, and X is the angle between your direction and
> the wind.
OK. I think I follow. Put in what I'm getting now, the output is a
target velocity, find the angle that gives that velocity, then I'm
making the best out of it?
> Which ship are you? I'm "Irregardless"
I'm purestblue. I was doing OK until the re-start, even then I was doing
OK for a while, now I'm right at the bottom of the field. I do think
some of the leaders are sailing into still air though (hope so!).
CBFalconer
> Justin C wrote:
>> CBFalconer wrote:
>>
>>> All you need is a slip-stick with trig scales.
>>
>> I have no idea what either of those are, [...]
>
> O tempora! O mores! O gimme anutha; make it a double.
>
> (As oldfashioned as I'm feeling at the moment, I can't
> help but suspect CBF is more oldfashioned still: By the
> time I met the slipstick its hyphen had self elided.)
I wonder where one buys them these days. They must still exist. A
6 incher fit into the shirt pocket, but a 12 or 18 incher required
a briefcase. In my day ten bucks got a pretty good one. Two bucks
was adequate for classes.
For Justin - a slip-stick is a slide rule. The basic one has log
scales, allowing easy multiplication and division.
humunculus
Not quite. You can't really determine your target (desired) velocity
beyond bigger is better. Your ship cannot head into the wind at an
angle less than 45 degrees (this is with real boats also), and most
boats can't even do that....about 50 degrees is the best they can do.
What you have to do in this game is determine where you want to be in
12 hours, and aim for that place. If that place is directly upwind (as
it seems to be for the start of this leg), then you can figure out how
much 'headway' you are making by that formula: VcosX where V is your
ships velocity in the direction its headed, and X is your angle off
the wind. What I am doing is tweaking the angle off the wind, looking
at the velocity, and quickly calculating what my velocity towards my
goal is. Of course, I also am taking account of the reality that an
oblique course is LONGER, so even though I'm approaching my goal
faster, I'm taking the long way around.
> > Which ship are you? I'm "Irregardless"
>
> I'm purestblue. I was doing OK until the re-start, even then I was doing
> OK for a while, now I'm right at the bottom of the field. I do think
> some of the leaders are sailing into still air though (hope so!).
>
Oh, they will be. I'm headed up the coast for those big winds that are
coming in 12-24 hours. We both lost lots of places on the leaderboard
because we are tacking across the field, and not gaining ground on the
finish line. But the folks who are racing each other rather than
looking at the weather will be doing 5-8 knots in about 15 hours,
while I expect to be doing 25, all of it toward the Molucca Straits.
Even if I have to go twice as far as them, I think its a better
strategy. Its that current I'm worried about....
--riverman
Justin C
>
[snip]
> For Justin - a slip-stick is a slide rule. The basic one has log
> scales, allowing easy multiplication and division.
Ah! I know what one of they is! I've even seen one. Have no idea how to
use one though - and being of 41 years I'm considered 'old skool' by
most of today's youngsters.
By the time I was at the point in my education where one would use a
slide-rule, log tables, etc. we were allowed to use pocket calculators
so that made most of those tools redundant. I've still got my Casio
fx82, too... And still use it if I've need to, for something that's
quicker than firing up Excel.
Eric Sosman
> Eric Sosman wrote:
>> Justin C wrote:
>>> CBFalconer wrote:
>>>
>>>> All you need is a slip-stick with trig scales.
>>> I have no idea what either of those are, [...]
>> O tempora! O mores! O gimme anutha; make it a double.
>>
>> (As oldfashioned as I'm feeling at the moment, I can't
>> help but suspect CBF is more oldfashioned still: By the
>> time I met the slipstick its hyphen had self elided.)
That's not quite what I wrote. The quoter has added
a character and there by deleted the joke. Toobad.
> I wonder where one buys them these days. They must still exist. A
> 6 incher fit into the shirt pocket, but a 12 or 18 incher required
> a briefcase. In my day ten bucks got a pretty good one. Two bucks
> was adequate for classes.
We were taught the theory and use of the slide rule in
school, in about the ninth or tenth grade -- I imagine the
topic was put off until we'd learned about logarithms so the
gadget wouldn't seem to work by magic. (And also so we'd
understand why you could get more accuracy out of a one-and-
change result than out of nine-and-change.)
> For Justin - a slip-stick is a slide rule. The basic one has log
> scales, allowing easy multiplication and division.
Most also had scales for squares and perhaps cubes (and
for extracting roots thereof), but the crucial point for the
O.P.'s purposes is that there were often some trigonometric
scales, sometimes printed on the reverse side of the slipstick.
A confounding difficulty with the O.P.'s problem is that
plain plane trigonometry won't suffice. He's aiming for a
point 1400 miles (2250 km) away, which is about one-eighteenth
the circumference of the Earth. Over distances of that scale
the flat-Earth fiction becomes an inadequate approximation,
and I think he'll need to resort to spherical trigonometry to
win his race. Spherical trig is a subject I never studied,
not with a slipstick, a slip-stick, or a slop jar.
--
Eric Sosman
eso...@ieee-dot-org.invalid
Justin C
>
> A confounding difficulty with the O.P.'s problem is that
> plain plane trigonometry won't suffice. He's aiming for a
> point 1400 miles (2250 km) away, which is about one-eighteenth
> the circumference of the Earth. Over distances of that scale
> the flat-Earth fiction becomes an inadequate approximation,
> and I think he'll need to resort to spherical trigonometry to
> win his race.
Win?! If only I could *dream* of being in with a chance! There are over
100,000 entrants, I'm currently a few places below 85,000th. There are a
lot of people sailing into dead air just now, and the forecast is
looking favourable for the route I have chosed. I hope to make
significant progress in the next 36 hours.
humunculus
SO far, so good. Last night the winds were steady and I could leave my
ship on a tack all night long, and I gained about 40,000 places on the
field and even 20,000 on you. Today is the day...I need to cut across
the N Indian current, and then I get to sit on a broad reach for the
next 36-48 hours all the way to Molucca while the leaderpack has to
beat upwind. IF I time it right, I will be tracking at a steady 60-
degree angle, which is the fastest angle, while the leaders are
tacking every few hours to gain ground. I think if I get across the
current without losing too much distance, I have a chance to actually
beat the leader pack to the strait....I predict I will be in the top
500 (out of 102,000) to get into the gap.
We'll see...
--riverman
Justin C
>
> SO far, so good. Last night the winds were steady and I could leave my
> ship on a tack all night long, and I gained about 40,000 places on the
> field and even 20,000 on you. Today is the day...I need to cut across
> the N Indian current, and then I get to sit on a broad reach for the
> next 36-48 hours all the way to Molucca while the leaderpack has to
> beat upwind. IF I time it right, I will be tracking at a steady 60-
> degree angle, which is the fastest angle, while the leaders are
> tacking every few hours to gain ground. I think if I get across the
> current without losing too much distance, I have a chance to actually
> beat the leader pack to the strait....I predict I will be in the top
> 500 (out of 102,000) to get into the gap.
>
> We'll see...
I too gained a lot of places overnight... which was nice! I should have
been working today, but by working all day in a spreadsheet I can make
it look like I am even when I'm not... So I've refined all sorts of
things, been looking lots of stuff up too.
The spreadsheet has improved, and I've found a 'great circle' Excel
formula. With that I'm able to put in my position and that of the next
mark, and better calculate my closing on it.
A bit more work and I hope to be able to feed it wind speed and
direction, my position, position of the next mark, and get best course
to steer out of it. Looking forward to the next leg!
humunculus
Well, I think you're on a good track. My plan is to beat the leaders
to the opening of the Molucca Strait, but unfortunately I will be
entering the strait on Sunday morning, local time, which is when I
have to get on an airplane for Thailand and I'll be incommunicado for
the next 2 weeks. So look forward to Irregardless to be at the head of
the pack in 36 hours, only to run straight on to the rocks.
Still, its a fun game.
--riverman
Justin C
>
> Well, I think you're on a good track. My plan is to beat the leaders
> to the opening of the Molucca Strait, but unfortunately I will be
> entering the strait on Sunday morning, local time, which is when I
> have to get on an airplane for Thailand and I'll be incommunicado for
> the next 2 weeks. So look forward to Irregardless to be at the head of
> the pack in 36 hours, only to run straight on to the rocks.
That's a shame. To put the work in and then have to abandon it. You
should be back, though, for the next but one leg.
Patrick Hamlyn
humunculus
wrote:
> Justin C <justin.0...@purestblue.com> wrote:
> >In article <541e67d9-4a82-4169-a927-241da83d0...@c36g2000prc.googlegroups.com>, humunculus wrote:
>
> >> Well, I think you're on a good track. My plan is to beat the leaders
> >> to the opening of the Molucca Strait, but unfortunately I will be
> >> entering the strait on Sunday morning, local time, which is when I
> >> have to get on an airplane for Thailand and I'll be incommunicado for
> >> the next 2 weeks. So look forward to Irregardless to be at the head of
> >> the pack in 36 hours, only to run straight on to the rocks.
>
> >That's a shame. To put the work in and then have to abandon it. You
> >should be back, though, for the next but one leg.
>
> Or he could hand over to someone...
> --
> Patrick Hamlyn posting from Perth, Western Australia
> Windsurfing capital of the Southern Hemisphere
True....interested in taking over my vessel, Patrick?
--riverman
Patrick Hamlyn
Sure. At least then, when the Irregardless tuns on to the rocks, it'll be
incompetence rather than plain dereliction :>
--
Patrick Hamlyn posting from Perth, Western Australia
Windsurfing capital of the Southern Hemisphere
Moderator: polyforms group (polyforms...@egroups.com)
humunculus
wrote:
>
> - Show quoted text -
LOL. Well, it seems a waste to just throw it all away. I'll send you
the username and password via your multipro.com address in about 12
hours when I head off for the airport. I can't resist the challenge of
getting into the top 1000...this is a VERY addicting game.
I'm not too deeply invested in the long term of this, so don't worry
if you pile up on Molucca or get hijacked by pirates or anything. I
can always create a new boat. :-)
--riverman
humunculus
> >True....interested in taking over my vessel, Patrick?
>
>
> incompetence rather than plain dereliction :>
Check your email. :-)
--riverman
Patrick Hamlyn
Tricky little user interface.
When I tried to set my course a little higher to avoid an inconvenient little
island it said something rude in French which sounded like 'not my problem dude'
and then I had to go to bed. In the morning it had managed to run aground, but
by now the server was responding again so I got it afloat again and rejoined
quite a few places down from where I would have been. Now it looks like those
boats in the lead are heavily into the light air and I would have benefitted in
a big way if I had managed to point a little more North like I was trying to.
Justin C
> When I tried to set my course a little higher to avoid an inconvenient little
> island it said something rude in French which sounded like 'not my problem dude'
> and then I had to go to bed. In the morning it had managed to run aground, but
> by now the server was responding again so I got it afloat again and rejoined
> quite a few places down from where I would have been. Now it looks like those
> boats in the lead are heavily into the light air and I would have benefitted in
> a big way if I had managed to point a little more North like I was trying to.
There is an little flag on the top right of the screen you can click to
change the language to English.
If you get that 'server not responding' type message, logout and log
back in again.
tomf...@googlemail.com
skills or time to write excel formulae, but hope it helps you!
i was then just seeing how you were getting on, to find that we were
only 3 miles apart, having run the same tactic!
bomboat
good luck,
tom
bristol uk
Justin C
> just trying to answer the same question... i dont have your maths
> skills or time to write excel formulae, but hope it helps you!
>
> i was then just seeing how you were getting on, to find that we were
> only 3 miles apart, having run the same tactic!
I'm in terrible shape at the moment, falling back through the field at a
great rate. I've got almost no wind, while a few miles off shore people
are steaming along. If I change course and head out it'll take me hours
to get where the wind is.... I'm screwed!
I've added you to my 'friends' list.... oooh! And there's been a wind
change!
Justin C
Tom,
I was opening a gap on you for a while there, then I ran in to low wind.
You seem to have a little more at this time (don't head inshore!). It's
hard to believe that there are 8000 boats between me and Irregardless.
Those are places that it's definitely possible to make up before the
finish of the leg, I just need to not sleep tonight or until about 11pm
tomorrow when I predict I'll finish.