Vespoli boats
TidewayUmpire
please assist?
Are they a viable option? e.g. Can damage be repaired easily in the
UK, do any UK manufacturers make riggers for them, whats the delivery
time for spares imported from the US, is it easy to do and are they
[the spares], like for like, any more expensive than other boat
manufacturers?
Thanks for any input
Christopher Anton
news:1171825482.2...@a75g2000cwd.googlegroups.com...
> Can anyone in the UK who has had any experience with Vespoli boats
> please assist?
>
> Are they a viable option?
Another important question, do they pass the FISA minimum guidance for the
safe practice of rowing?
Teaplant
I highly recommend Eric Sims for repairs - Vespoli or otherwise. The
silver lining around the ERB-RIP cloud is that he ought to be much
easier to get to now!
> do any UK manufacturers make riggers for them.
Any/all of them - all riggers should be made 'bespoke' for the boat
after careful measuring and there is usually nothing in particular to
be gained from buying a new boat rigged - saves a few quid perhaps but
do all boatbuilders measure and check every single boat before having
riggers made?
Carl does a very tidy rigger and you can get his details from his
occasional posts to rec.sport.ranting. Just don't mention that the
boat is a poor quality import from a nation with questionable
political situation and a favourable currency exchange rate.
Vespolis themselves always seem quite robust and therefore durable.
However, you don't seem to see many of them in the finals of the World
Champs.... I presume that we are about to see a few more of them about
at club level with the exchange rate as it is at present. Who knows?
Janousek (bullet proof club boats) to go bust next?
Michael Walker
One of our Vet eights used one on loan when they went to the Head of
The Charles a couple of years back. They liked it so much they bought
one and imported it. Still cheaper than buying new in the UK from
someone like Aylings or Jano. Do bear in mind - if you go down this
route - that because they're American, fittings etc are NOT metric!
marco...@gmail.com
other cheaper and more reliable options. If you buy a Vespoli V1 (the
top of the line boat) you spend as much as you spend for a Resolute,
but you end up getting half of the quality. Now that Resolute have a
better customer service and offer some more economical options they
are the better option in the market in my humble opinion.
Neither of the 2 boat builders at the moment are producing a FISA
standards compliant boat. I assume though that at least Resolute will
have no problem to fit their boats with sealed bulkhead.
If you are going for less expensive boats there Hudson and Pocock that
are good alternatives. I think you need to consider the skills of the
crew that is going o use the boat. If your crew is skilled shorter
boats like V1 and Resolute are better options than longer boat like
the M2 and Pocock. Also keep in mind the crew weight: Pocock and
Resolute build a real Midweight boat; V1 are not available yet on that
size so you would get a a cut down HW boat. Let me know if you have
any other questions.
Marco
Christopher Anton
<marco...@gmail.com> wrote in message
news:1171905951.2...@p10g2000cwp.googlegroups.com...
> Neither of the 2 boat builders at the moment are producing a FISA
> standards compliant boat.
In that case I wouldn't hesitate to stop it boating at a UK regatta because
it would not comply with the water safety code.
marco...@gmail.com
builder will consider to fix the problem in order to sell a boat. When
I asked Resolute they told me that they didn't think it would be a
problem. Unfortunately most of the market of the USA boat builder is
internal so they are really not that motivated in making changes that
will increase the weight of the boat (even if only by a little). Here
the race "the-lighter-the-better" is still going on.
M
On Feb 19, 12:50 pm, "Christopher Anton"
<c.an...@NOSPAM.blueyonder.co.uk> wrote:
> <marco.b...@gmail.com> wrote in message
Robin
> My suggestion is that if you are going to buy an American boat try for
> other cheaper and more reliable options. If you buy a Vespoli V1 (the
> top of the line boat) you spend as much as you spend for a Resolute,
> but you end up getting half of the quality. Now that Resolute have a
> better customer service and offer some more economical options they
> are the better option in the market in my humble opinion.
> Neither of the 2 boat builders at the moment are producing a FISA
> standards compliant boat.
I'm not normally a huge fan of Vespolis - but the V1 I coxed in
Seattle last summer had closed bulkheads at every seat AND hatches
between the tracks. I can't comment whether the space enclosed would
'support the crew if swamped' - but it looked at least as enclosed as
a Janousek does. Nice and stiff, and relatively responsive to
steering.
Regarding Pococks and Hudsons. Pococks VIIIs - although good solid
beasties (in particular the side-rigged E8) are not currently built
with enclosed buoyancy. The E8 (based on the late 80s Empacher shape)
sits up nicely for most crews and both the boats I've coxed were a
delight in head and sprint races - but until the underdeck situation
is sorted out, not ideal for the UK.
Hudsons are certainly capable of being built enclosed and were so
during the 2004 (Olympics) season (I think you have to ask them to do
it now) - but I would have some reservations about recommending them
for anything except technically competent crews, because the one I
have experience coxing was very stiff - but also unforgiving and very
twitchy. It either went very fast (in a straight line) - or would
trundle along with one or other side of the boat in the water each
stroke if you have people whose finishes aren't great or just don't
have the power to get it up to running speed. And they don't steer -
6 kilometres of head race wobbling along below running speed and
unable to get easily round even wide corners aren't my idea of fun.
Speaking as a cox, though, despite the problems with steering, they
have a superbly comfortable cox seat area - tonnes of room to stretch
out, and indentations into the inner skin giving space for hips.
Finally - does anyone have any further information as to what the
"alternative" range options are sold by Resolute? They had something
on their website about introducing two new ranges a couple of years
ago, but I was unable to find out any details such as prices,
construction / specification differences etc.
marco...@gmail.com
> On Feb 19, 5:25 pm, marco.b...@gmail.com wrote:
>
> > My suggestion is that if you are going to buy an American boat try for
> > other cheaper and more reliable options. If you buy a Vespoli V1 (the
> > top of the line boat) you spend as much as you spend for a Resolute,
> > but you end up getting half of the quality. Now that Resolute have a
> > better customer service and offer some more economical options they
> > are the better option in the market in my humble opinion.
> > Neither of the 2 boat builders at the moment are producing a FISA
> > standards compliant boat.
>
> I'm not normally a huge fan of Vespolis - but the V1 I coxed in
> Seattle last summer had closed bulkheads at every seat AND hatches
> between the tracks. I can't comment whether the space enclosed would
> 'support the crew if swamped' - but it looked at least as enclosed as
> a Janousek does. Nice and stiff, and relatively responsive to
> steering.
You are right on the V1 but they are not built with sealed under-seat
compartment as standard (unless they started this year) and the
stiffness of the boat is really a gamble. at my club they bought 2
last years a HW and a MW (that was actually a cut down HW). The
stiffness of the second wasn't really the same of the HW. Compared to
a Resolute they felt soft. I rowed and coached a MW entry level
Resolute for a couple of summers and I can tell you that was still
stiffer than the latest V1.
>
> Regarding Pococks and Hudsons. Pococks VIIIs - although good solid
> beasties (in particular the side-rigged E8) are not currently built
> with enclosed buoyancy.
>The E8 (based on the late 80s Empacher shape)
> sits up nicely for most crews and both the boats I've coxed were a
> delight in head and sprint races - but until the underdeck situation
> is sorted out, not ideal for the UK.
Actually the "entry level" E8 are nothing more of bath tubs with
riggers. Sturdy and stable but really heavy. Ideal for novices.
The Hypercarbon it's another kind of beast. I really like it and i
think the boat is really stiff and fast in the water and it doesn't
require as much proficiency to be rowed as other shorter shells.
Still they had up to last year any sealed compartment making it really
dangerous. As I mentioned already in this group, I had to fish out of
the Charles crew telling me how they were used to get swamped in
choppy water with their Pocock. The latest episode at the head of the
Charles with the Kids from the Beijing University that sank in the
borrowed Pocock just increase my concerns about the safety of that
boat.
>
> Hudsons are certainly capable of being built enclosed and were so
> during the 2004 (Olympics) season (I think you have to ask them to do
> it now) - but I would have some reservations about recommending them
> for anything except technically competent crews, because the one I
> have experience coxing was very stiff - but also unforgiving and very
> twitchy. It either went very fast (in a straight line) - or would
> trundle along with one or other side of the boat in the water each
> stroke if you have people whose finishes aren't great or just don't
> have the power to get it up to running speed. And they don't steer -
> 6 kilometres of head race wobbling along below running speed and
> unable to get easily round even wide corners aren't my idea of fun.
> Speaking as a cox, though, despite the problems with steering, they
> have a superbly comfortable cox seat area - tonnes of room to stretch
> out, and indentations into the inner skin giving space for hips.
>
> Finally - does anyone have any further information as to what the
> "alternative" range options are sold by Resolute? They had something
> on their website about introducing two new ranges a couple of years
> ago, but I was unable to find out any details such as prices,
> construction / specification differences etc.
I had the opportunity to evaluate the full line of Resolute, but I
don't have a latest prices list.
I think the entry level was good enough for most of the crew up to
national level although I had always preferred to give it to expert
crews since it's a little too responsive for novices. The web sites
now days doesn't show any entry level boat so I am wondering if they
decided to go for just the top of the line version.
Ciao,
Marco
Robin
> You are right on the V1 but they are not built with sealed under-seat
> compartment as standard (unless they started this year) and the
> stiffness of the boat is really a gamble. at my club they bought 2
> last years a HW and a MW (that was actually a cut down HW).
The V1 I experienced was either an 06 or an 05. It was definitely
sealed. Re stiffness - bear in mind that my normal VIIIs are 27 year
old english-built Carbocrafts (ie equivalent to the very earliest
Vespolis built on license when they were still called Carbocraft
USA)... so my experience of genuinely stiff boats is relatively
infrequent. I've never been in a Resolute on the water so can't
comment.
>
> > Regarding Pococks and Hudsons. Pococks VIIIs - ...
> >The E8 (based on the late 80s Empacher shape)
....
> Actually the "entry level" E8 are nothing more of bath tubs with
> riggers. Sturdy and stable but really heavy.
I think you're confusing the E8 with either the C8 (also side rigged,
double skin composite but based on the original Pocock Cedar shape
which is a bit longer, heavier, and more stable from memory), or even
their M8 learner shell which genuinely was a bathtub with riggers,
whether built with the earlier wood / composite construction or the
double-skin composite construction later on.
O.K - so the E8 design is quite conservative, being based on an 80s
Empacher shape, and they don't go all out to take them down to FISA
minimum; nor are they built with the Hypercarbon composite. But in my
experience of them they are competitive with all but the top of the
range racing models, and it doesn't take an Olympic crew to make them
fairly fly if you need them to. If they were built enclosed, and
moreover, sectioned at 4-seat, they would work pretty well in UK
conditions for clubs needing a hard-wearing "club"-level VIII, much as
Janousek build good honest boats here in the UK for that kind of
market.
> The Hypercarbon it's another kind of beast. I really like it and i
> think the boat is really stiff and fast ... but .. the latest episode at the head of the
> Charles with the Kids from the Beijing University that sank in the
> borrowed Pocock just increase my concerns about the safety of that
> boat.
>
True. But that goes for any boat which is built and sold unenclosed,
and ultimately, if we want the average safety of the vessels out there
in North America to improve, moves such as those by Mr. Kaschper to
enclose his boats as standard from last year should be applauded,
providing they enclose enough volume to adequately support a seated
crew when swamped. Having experienced a Men's Hwt crew getting
swamped by launch wake in an 06-model midweight Kaschper 4+ with the
enclosed bulkheads - and then being able to row back to the dock to
empty it out - they already seem to be there or thereabouts, or
certainly are a big step in the right direction.
>
>
Lawrence Edwards
news:1171905951.2...@p10g2000cwp.googlegroups.com...
> Neither of the 2 boat builders at the moment are producing a FISA
> standards compliant boat. I assume though that at least Resolute will
> have no problem to fit their boats with sealed bulkhead.
>
was fully bouyant, as were all of the other Resolutes I saw. They even had a
section in their brochure explaining that they have tested the boats with
over-weight crews full of water and they are perfectly rowable. I also saw a
US crew in a Resoltue 8 at Women's Henley last year, which was enclosed
under the seats. I've just looked on their website and it is not clear
whether the current designs have enclosed underseat areas, but I would be
surprised if they didn't after having spoken to the owner of the company
about this issue!
They are also very well made, with an excellent finish and very high quality
fittings.
It is the quality of the fittings and finish on the Chinese boats that I
would question. At the World Masters Row Show 2 years ago there were some
Swift boats, the basic hulls and bulkheads looked fine but the riggers,
seats, rails, and nuts and bolts looked as though they had been selected on
the basis of price not quality.
It would be a false economy to save a few grand on a new 8 only to find that
after a moderate amount of use everything started to break.
I apply the same logic to purchasing cars. I've tried many makes over the
years, and settled on Mazdas because they seem to be the only cars which are
both well made and good value for money. NB I have no connetion with either
Mazda or Resolute!
P.S. following the SARA ruling we have retrofitted bulkheads to our Empacher
8. It was not that difficult to do and has not affected the performance
enough to notice. Perhaps at Olympic or World Championship level it might
make a difference, but the quality of the crew is far more important than a
few extra kilos in an 8. It also means you can keep your kit dry....
Lawrence
Gripper
wrote:
yes, and here it is being overtaken by an 11 year old Sims,
http://clodaghphotos.fpic.co.uk/p38746876.html
Nice boat though!
KC
Two seat is doing a great demonstration of what Sully and Paul would
probably call "rowing it in".
-Kieran
Mike Sullivan
"KC" <kc_...@sonic.net> wrote in message
news:erik7d$kee$2...@prometheus.acsu.buffalo.edu...
> Gripper wrote:
>> On Feb 19, 1:10 pm, "Michael Walker" <webmas...@cantabsrowing.org.uk>
>> wrote:
>>> On Feb 18, 7:04 pm, "TidewayUmpire" <pen...@aol.com> wrote:
snip
> Two seat is doing a great demonstration of what Sully and Paul would
> probably call "rowing it in".
"but I have the biggest puddle!"
paul_v...@hotmail.com
wrote:
> "KC" <kc_n...@sonic.net> wrote in message
IOW, size matters? LOL
He'd probably taken his ViErgra, but still missed the timing.
Good lord! Do any of those Ports have two arms?
And here I was thinking it was only Starboards that screw up... [;o)
- Paul Smith
Kieran
And here I was thinking you were going to just ignore all my posts! Oh
wait, that was a reply to Mike not me. Hmmm...
-Kieran
paul_v...@hotmail.com
>
> - Show quoted text -
Are there any in particular that you would like me to give some
attention (respond) to?
You point me at a web resource that I could have sworn you knew I was
very familiar with (Ken Youngs pages), and then another that describes
the blade path of a boat underway (Kleshnev), and then declare that
the path is the same (with presumably the same forces in play)
regardless of the boat being underway or not. Now how can I respond
to something like that? If you don't see the large difference, being
far more expert in the area than my "ignorant" self, I'm having doubts
that I will be able to make much headway attempting an explanation.
The non-sequitor about the pocock integrated backstay "pointing down
at the water", that you came up with, instead of recognizing that it
was bracing against force in the direction of travel, which the vast
majority of force on the pin is being directed, seemed to indicate
that you did not want to seriously consider what I had been trying to
explain. I even offered a small thought experiment, which you did not
respond to, near as I could tell, though maybe I missed your response
in the fray.
- Paul Smith
KC
>> And here I was thinking you were going to just ignore all my posts! Oh
>> wait, that was a reply to Mike not me. Hmmm...
>>
>> -Kieran- Hide quoted text -
>>
>> - Show quoted text -
>
> Are there any in particular that you would like me to give some
> attention (respond) to?
>
> You point me at a web resource that I could have sworn you knew I was
> very familiar with (Ken Youngs pages), and then another that describes
> the blade path of a boat underway (Kleshnev), and then declare that
> the path is the same (with presumably the same forces in play)
> regardless of the boat being underway or not. Now how can I respond
> to something like that? If you don't see the large difference, being
> far more expert in the area than my "ignorant" self, I'm having doubts
> that I will be able to make much headway attempting an explanation.
Paul, that's just a cop-out on your part. At least *I* am trying to
explain this to you. If you don't understand the underlying reasons
behind your claims, then you shouldn't make those claims. If you see a
large difference in blade path due to boat speed, please explain it!
I'm happy to be shown where MY misunderstanding lies. You just have
made no effort at all to support your claim that "the vast majority of
force on the blade at the catch points in the direction of boat travel."
You have stated that claim in not so many words MANY times, and it is
just plain wrong. If you can prove otherwise, please do. If not, quit
claiming it.
> The non-sequitor about the pocock integrated backstay "pointing down
> at the water", that you came up with, instead of recognizing that it
> was bracing against force in the direction of travel, which the vast
> majority of force on the pin is being directed, seemed to indicate
> that you did not want to seriously consider what I had been trying to
> explain. I even offered a small thought experiment, which you did not
> respond to, near as I could tell, though maybe I missed your response
> in the fray.
>
> - Paul Smith
Let's tackle one thing at a time, shall we? Blade path first, the back
stay thing can wait.
(BTW, I had no idea you were "very familiar" with Ken Young's pages.
How was I to know that? Maybe you forgot that while I started on RSR in
1994/5, I took a long break from it for about two or three years, during
which time I think you started posting.)
The path that the blade takes, as illustrated by Young, Kleshnev and
others is determined by the arc of the oar about the pin, and the boat
traveling past the blade. It is not determined by the boat's speed
(small variations in the blade path shape do occur at different speeds,
but on a gross level the shape of the path is roughly the same at all
speeds). If you can explain to me how the path of the blade through the
water is significantly affected by the boat's speed (enough to account
for your claims of where the "majority" of force points), I'll gladly
hear it. One way the path will change: the amount of slip during
midstroke will be much less with lower forces applied to the oar handle
by the rower. So at paddle/light pressure, there is less slip. At full
pressure there is more slip. But the differences in the blade path
shape between full pressure at a 18 and full pressure at a 38, is
negligible despite the large increase in boat speed. Likewise a full
pressure stroke from a stand still will have a VERY similar path to that
of a full pressure stroke in the middle of a race.
If you can accept the above, then the next point is to understand that
the lift that the blade creates is determined by the direction of flow
about the blade. Since the blade path through the water is not affected
by speed, the direction of the force on the blade is the same at various
speeds, for a given oar angle. The lift force on the blade is by
definition perpendicular to the direction of flow about the blade, which
AGAIN is determined by the path of the blade THROUGH the water. The net
resultant force on the blade (lift and drag) is roughly perpendicular to
the blade cord, which for most oars is parallel to the oar shaft. So,
the net resultant force vector acting on the blade during the stroke is
always perpendicular to the oar shaft (or very close to it).
At the catch, given the angle of the oar, much of the force on the blade
points toward the hull. For a 30deg catch angle, 86.6% of the net force
points at the hull, and 50% of the net force points in the desired
propulsive direction. And in case anyone forgot, at 30deg, Pythagoras
gave us the relationship of sqrt(86.6^2 + 50^2) = 100%.
-Kieran
paul_v...@hotmail.com
> paul_v_sm...@hotmail.com wrote:
> > On Feb 22, 12:46 pm, Kieran <kc_n...@sonic.net> wrote:
> >> And here I was thinking you were going to just ignore all my posts! Oh
> >> wait, that was a reply to Mike not me. Hmmm...
>
> >> -Kieran- Hide quoted text -
>
> >> - Show quoted text -
>
> > Are there any in particular that you would like me to give some
> > attention (respond) to?
>
> > You point me at a web resource that I could have sworn you knew I was
> > very familiar with (Ken Youngs pages), and then another that describes
> > the blade path of a boat underway (Kleshnev), and then declare that
> > the path is the same (with presumably the same forces in play)
> > regardless of the boat being underway or not. Now how can I respond
> > to something like that? If you don't see the large difference, being
> > far more expert in the area than my "ignorant" self, I'm having doubts
> > that I will be able to make much headway attempting an explanation.
>
> Paul, that's just a cop-out on your part. At least *I* am trying to
> explain this to you. If you don't understand the underlying reasons
> behind your claims, then you shouldn't make those claims. If you see a
> large difference in blade path due to boat speed, please explain it!
Okay. From a dead stop, take a full power drive and the blade tip
will be extracted from the water futher from the finish line than it
began, this will not happen once the boat is up to racing speed, and
probably not even considerably prior to that. This means the blade is
making a much different path through the water under way than when
stationary and that path varies with the overall speed.
> I'm happy to be shown where MY misunderstanding lies. You just have
> made no effort at all to support your claim that "the vast majority of
> force on the blade at the catch points in the direction of boat travel."
No, I have tried, just to no avail. Plus, you should be able to
explain it much better than I anyhow. (My ignorance and all.)
> You have stated that claim in not so many words MANY times, and it is
> just plain wrong. If you can prove otherwise, please do. If not, quit
> claiming it.
No. And your record of telling me I "have it wrong", "backwards",
whatever, is none to good, IMO.
> > The non-sequitor about the pocock integrated backstay "pointing down
> > at the water", that you came up with, instead of recognizing that it
> > was bracing against force in the direction of travel, which the vast
> > majority of force on the pin is being directed, seemed to indicate
> > that you did not want to seriously consider what I had been trying to
> > explain. I even offered a small thought experiment, which you did not
> > respond to, near as I could tell, though maybe I missed your response
> > in the fray.
>
> > - Paul Smith
>
> Let's tackle one thing at a time, shall we? Blade path first, the back
> stay thing can wait.
>
> (BTW, I had no idea you were "very familiar" with Ken Young's pages.
> How was I to know that? Maybe you forgot that while I started on RSR in
> 1994/5, I took a long break from it for about two or three years, during
> which time I think you started posting.)
Darn, I was sure we'd discussed the Stroke Phase Guide of ErgMonitor,
which was derived directly from Ken Young's work, which I have
mentioned pretty much every time I discuss it. Maybe we didn't
discuss it, but it strains belief that it was not.
> The path that the blade takes, as illustrated by Young, Kleshnev and
> others is determined by the arc of the oar about the pin, and the boat
> traveling past the blade.
Please, make up your mind! That AND is all I was referring to. Now
are you going to claim I said something else and that you are right
and I was wrong? (Let's not go there again, okay?)
> It is not determined by the boat's speed
> (small variations in the blade path shape do occur at different speeds,
> but on a gross level the shape of the path is roughly the same at all
> speeds). If you can explain to me how the path of the blade through the
> water is significantly affected by the boat's speed (enough to account
> for your claims of where the "majority" of force points), I'll gladly
> hear it. One way the path will change: the amount of slip during
> midstroke will be much less with lower forces applied to the oar handle
> by the rower. So at paddle/light pressure, there is less slip. At full
> pressure there is more slip. But the differences in the blade path
> shape between full pressure at a 18 and full pressure at a 38, is
> negligible despite the large increase in boat speed. Likewise a full
> pressure stroke from a stand still will have a VERY similar path to that
> of a full pressure stroke in the middle of a race.
It seems we have a complete failure to communicate going on somewhere,
as that last sentence makes absolutely no sense at all. As discussed
in my reply above.
> If you can accept the above, then the next point is to understand that
> the lift that the blade creates is determined by the direction of flow
> about the blade.
Are you thinking that you are playing a game with me? Of course I
can not accept that, as it's clearly not acceptable. The lift is
result of circulation about the blade and acts in direct opposition to
the resisting forces of drag and inertia, which last time I checked
were parallel with the direction of travel.
BTW - There is almost no lift involved at the blade until the boat is
moving forward, and it's more in proportion to the actual system
speed. i.e. A quick recovery after the first stroke may have the boat
moving fast forward but since the system is still moving quite slow
the lift is still small. One of the reasons that we use abbreviated
strokes off the start, since at that point the blade drag is the
dominate resistive force in use.
> Since the blade path through the water is not affected
> by speed, the direction of the force on the blade is the same at various
> speeds, for a given oar angle. The lift force on the blade is by
> definition perpendicular to the direction of flow about the blade, which
> AGAIN is determined by the path of the blade THROUGH the water. The net
> resultant force on the blade (lift and drag) is roughly perpendicular to
> the blade cord, which for most oars is parallel to the oar shaft. So,
> the net resultant force vector acting on the blade during the stroke is
> always perpendicular to the oar shaft (or very close to it).
>
> At the catch, given the angle of the oar, much of the force on the blade
> points toward the hull. For a 30deg catch angle, 86.6% of the net force
> points at the hull, and 50% of the net force points in the desired
> propulsive direction. And in case anyone forgot, at 30deg, Pythagoras
> gave us the relationship of sqrt(86.6^2 + 50^2) = 100%.
>
> -Kieran- Hide quoted text -
>
> - Show quoted text -
Can't do much with that last bit, since it was relying on the
acceptance of earlier incorrect claims.
- Paul Smith
martin+x@y.z
> points toward the hull. For a 30deg catch angle, 86.6% of the net force
> points at the hull
Very easy to prove. Take the catch with a single blade. Keep the
other one feathered. Just take the first third of the drive in this
configuration. The boat will turn very nicely. Most of the force is
going sideways.
A while ago I initiated a thread about not being able to go straight.
I was told to watch my releases and catches. That solved the
problem. Any differentials during the catch will produce a turning
force. It's a pretty amazing sport. I appreciate more and more of it
as I learn. It requires one to develop a lot of skills.
-Martin
KC
Excellent, Martin. Thank you. That was definitely a simple "why didn't
I think of that?" explanation.
Paul if you're still unconvinced that the "pinch force" exists, I'll
respond to your other post. Meanwhile, Martin's example proves pretty
much everything I was trying to say. Note that this phenomenon happens
regardless of whether the boat is moving or not.
Now, to preempt a likely rebuttal. Given the off axis location of the
point of force application (the pin) even a force applied to only one
pin pointed directly forward would still turn the boat, but not nearly
as much as happens in Martin's example.
-Kieran
martin+x@y.z
> I think of that?" explanation.
Well, if you haven't studied basic Physics (and paid attention)
understanding force vector decomposition could be difficult. I deal
with having to communicate a lot of highly complex technical
information to semi or non-technical people in my business. Because
of this I've become quite adept at coming-up with simple analogies and
simplifications that work.
-Martin
paul_v...@hotmail.com
> marti...@y.z wrote:
> >> At the catch, given the angle of the oar, much of the force on the blade
> >> points toward the hull. For a 30deg catch angle, 86.6% of the net force
> >> points at the hull
>
> > Very easy to prove. Take the catch with a single blade. Keep the
> > other one feathered. Just take the first third of the drive in this
> > configuration. The boat will turn very nicely. Most of the force is
> > going sideways.
>
> > A while ago I initiated a thread about not being able to go straight.
> > I was told to watch my releases and catches. That solved the
> > problem. Any differentials during the catch will produce a turning
> > force. It's a pretty amazing sport. I appreciate more and more of it
> > as I learn. It requires one to develop a lot of skills.
>
> > -Martin
>
> Excellent, Martin. Thank you. That was definitely a simple "why didn't
> I think of that?" explanation.
What, you've never sculled the bow around when at a stake boat in
cross wind conditions? I find that difficult to believe, but I'll
take your word for it.
> Paul if you're still unconvinced that the "pinch force" exists, I'll
> respond to your other post. Meanwhile, Martin's example proves pretty
> much everything I was trying to say. Note that this phenomenon happens
> regardless of whether the boat is moving or not.
Still completely unconvinced.
But nice crossing of the eyes and dotting the tease. There is a
forest behind those trees.
You keep stating the extreme of "pinch force exists", as if I have
said somewhere that it 'does not exist', which I don't think I have,
but if I have, let's make it clear right now that I am saying that it
is completely insignificant (when the system is at speed), you claim
87% at 30deg catch angle, which may be just fine in the static
situation, but it is not the same, or even close, in the dynamic one.
BTW - If you continue to misrepresent what I have said (aka - making
false claims) and use that to argue against, I'll start asking about
your paper again, since apparently that's equally irritating to you.
For example, I do not ask you when you stopped beating your wife, only
to get the answer that you have not stopped. Which could equally be
interpretted as you are continuing to beat her, or you never started
in the first place. It's all up to the point of view of the person
doing the interpretting. Oops, there is one other answer that would
involve the admission that you had stopped, but I don't think lowly
enough of you to assume that an option.
> Now, to preempt a likely rebuttal. Given the off axis location of the
> point of force application (the pin) even a force applied to only one
> pin pointed directly forward would still turn the boat, but not nearly
> as much as happens in Martin's example.
>
> -Kieran- Hide quoted text -
>
> - Show quoted text -
Perhaps you should dig up Kleshnevs newsletter regarding the different
force profiles required to produce straight tracking in a 2- (I know
you are familiar with that boat.) The turning moment he describes is
between the center of mass of the system and the pins, not our to the
blades.
And still, there are vast differences in what is happening when the
system is underway or not. Sorry that I don't have the required
vocabulary or tools to get this across to you, but in the long run I
don't see it making much difference to either of us. Perhaps if we
ever get a chance to row together it will be easier to demonstrate
many of these things, but I suppose that's quite unlikely too.
- Paul Smith
Carl Douglas
The blade path has to be different, depending on whether there is a
standing start or what passes in rowing for steady state. The question
is; "How different?"
At the static start you do not normally reach so far forward because,
there being no forward motion, there can not yet be any component in the
intended direction of the boat to give you a resolved flow along the
blade. However, this being a transient flow situation, you do not get
stall either - that takes some time to become established.
A boat accelerates quite swiftly off the start, reaching ~60% (off my
cuff) of eventual cruise velocity by the end of that first stroke, and
its initial acceleration is greater than that. So you develop
significant boat velocity very soon after the catch & lift will start to
develop in that first part of the stroke. By the finish you will get
the normal development of lift on the blade, following the mid-stroke
stall phase (which may be somewhat extended, especially if you fail to
dig deep enough in the middle to prevent air entrainment, which builds
up rapidly &, because velocity is relatively low, has more time to do so.
Naturally you will can resolve forces acting around a single blade not
only in terms of lift & drag acting perpendicular to & along the blade,
but instead to reveal a side force perpendicular to the boat & a
propulsive force parallel to the boat, at both catch & finish. If you
liken the early & late parts of the blade action to those of a section
of a propeller, then that side force relates directly to the torque
applied to drive the propeller. As with aircraft, rowing has ways to
cancel the effect of the side force, leaving only the axial, propulsive
force - in boats by having a blade on the other side, working in mirror
fashion.
The side force, of itself, does no work because it (normally) moves
nothing. And there is no such thing as pinching the boat at the catch.
What you have is a range of effective loadings & angles of attack on
the blade, which are determined by the prevailing combination of boat
velocity & amount of forward reach, within which range there will be an
optimum for any given boat speed. You get the same situation in
close-hauled sailing (i.e. into the wind), where again the large side
force generated on the sail(s) (if you choose to resolve forces in that
sense) is exactly balanced by an equal but opposite side force acting on
the keel, rudder & immersed hull of the boat. Some sailors try to haul
in the rig too tight and point to close to the wind, some let out too
much mainsheet & sail too far off the wind, but the expert gets it just
right, somewhere between those 2 extremes. And, with sailing as with
rowing, the faster you go the closer the sailor can head into the wind &
the further it makes sense for the rower to reach for the catch.
Was that any help?
Carl
--
Carl Douglas Racing Shells -
Fine Small-Boats/AeRoWing low-drag Riggers/Advanced Accessories
Write: The Boathouse, Timsway, Chertsey Lane, Staines TW18 3JY, UK
Email: ca...@carldouglas.co.uk Tel: +44(0)1784-456344 Fax: -466550
URLs: www.carldouglas.co.uk (boats) & www.aerowing.co.uk (riggers)
KC
In contrast, I spend much of my time explaining highly complex technical
information to people who are REQUIRED to understand it (students) so
rather than skip the vector math with an analogy, I must cover it.
Although, often the real-world analogy helps in grasping the tough stuff.
-Kieran
KC
The reason that might happen (I've not noticed it but don't dispute that
it might be possible) is due to the extra slip that occurs during the
first stroke. The extra slip is almost all during the mid-stroke. The
first 1/3 of the first stroke has a blade path very similar to any other
stroke, and it is the first 1/3 of the stroke that we're talking about
here. The reason you get extra slip during the first stroke is because
the rowers are able to apply more force on the first stroke (maybe first
few) than other, higher joint velocity strokes. As I said before, the
more force you apply, the more slip you'll get during the mid-stroke.
> this will not happen once the boat is up to racing speed, and
The slip still occurs, just not as much, and the reason why is because
the rowers can't apply as much force as they can when the boat is moving
slower.
> probably not even considerably prior to that. This means the blade is
> making a much different path through the water under way than when
> stationary and that path varies with the overall speed.
Again, the main variation in the path shape is during the mid-stroke,
not during the high-lift portion(s) of the stroke.
>
>> I'm happy to be shown where MY misunderstanding lies. You just have
>> made no effort at all to support your claim that "the vast majority of
>> force on the blade at the catch points in the direction of boat travel."
>
> No, I have tried, just to no avail. Plus, you should be able to
> explain it much better than I anyhow. (My ignorance and all.)
Paul, I can't explain your point of view because I think it is wrong.
If you can't explain your beliefs, then find someone who can. Find and
share ANY web site, paper, book, picture or anything that shows or
explains that the "pinch force" doesn't exist, please. You're the ONLY
person I've come across who makes this claim. All diagrams in every
book or paper or website I've seen show blade force vectors at the catch
that point perpendicular to the oar shaft.
>> You have stated that claim in not so many words MANY times, and it is
>> just plain wrong. If you can prove otherwise, please do. If not, quit
>> claiming it.
>
> No. And your record of telling me I "have it wrong", "backwards",
> whatever, is none to good, IMO.
>
>>> The non-sequitor about the pocock integrated backstay "pointing down
>>> at the water", that you came up with, instead of recognizing that it
>>> was bracing against force in the direction of travel, which the vast
>>> majority of force on the pin is being directed, seemed to indicate
>>> that you did not want to seriously consider what I had been trying to
>>> explain. I even offered a small thought experiment, which you did not
>>> respond to, near as I could tell, though maybe I missed your response
>>> in the fray.
>>> - Paul Smith
>> Let's tackle one thing at a time, shall we? Blade path first, the back
>> stay thing can wait.
>>
>> (BTW, I had no idea you were "very familiar" with Ken Young's pages.
>> How was I to know that? Maybe you forgot that while I started on RSR in
>> 1994/5, I took a long break from it for about two or three years, during
>> which time I think you started posting.)
>
> Darn, I was sure we'd discussed the Stroke Phase Guide of ErgMonitor,
I don't even know what the Stroke Phase Guide is. I don't use EM much,
maybe I'll look at it next time. What's it do, and what's it based on?
> which was derived directly from Ken Young's work, which I have
> mentioned pretty much every time I discuss it. Maybe we didn't
> discuss it, but it strains belief that it was not.
>
>> The path that the blade takes, as illustrated by Young, Kleshnev and
>> others is determined by the arc of the oar about the pin, and the boat
>> traveling past the blade.
>
> Please, make up your mind! That AND is all I was referring to. Now
> are you going to claim I said something else and that you are right
> and I was wrong? (Let's not go there again, okay?)
The boat moves past the blade even on the first stroke. How does that
"AND" in my statement contradict anything I've said? Maybe I've
misunderstood how you think boat velocity affects lift on the blade.
But since you have never explained why or how you think this is so, I
can hardly be blamed for misunderstanding your thoughts. Please share them?
>> It is not determined by the boat's speed
>> (small variations in the blade path shape do occur at different speeds,
>> but on a gross level the shape of the path is roughly the same at all
>> speeds). If you can explain to me how the path of the blade through the
>> water is significantly affected by the boat's speed (enough to account
>> for your claims of where the "majority" of force points), I'll gladly
>> hear it. One way the path will change: the amount of slip during
>> midstroke will be much less with lower forces applied to the oar handle
>> by the rower. So at paddle/light pressure, there is less slip. At full
>> pressure there is more slip. But the differences in the blade path
>> shape between full pressure at a 18 and full pressure at a 38, is
>> negligible despite the large increase in boat speed. Likewise a full
>> pressure stroke from a stand still will have a VERY similar path to that
>> of a full pressure stroke in the middle of a race.
>
> It seems we have a complete failure to communicate going on somewhere,
> as that last sentence makes absolutely no sense at all. As discussed
> in my reply above.
It makes perfect sense. The shape of the blade path is very similar
regardless of boat speed. Yes, there is more slip mid-stroke, but the
general shape of the path is still similar, and in the first 1/3 of the
stroke, it is very similar.
>> If you can accept the above, then the next point is to understand that
>> the lift that the blade creates is determined by the direction of flow
>> about the blade.
>
> Are you thinking that you are playing a game with me? Of course I
Absolutely not.
> can not accept that, as it's clearly not acceptable. The lift is
> result of circulation about the blade and acts in direct opposition to
> the resisting forces of drag and inertia, which last time I checked
> were parallel with the direction of travel.
Where did you check? Where did you get this information that "lift is
result of circulation about the blade and acts in direct opposition to
the resisting forces of drag and inertia,"???? That is a false
statement. Look up the definition of lift, and show me where it says
that it acts in direct opposition to drag and inertia. Lift acts
PERPENDICULAR to drag, and "inertia" is not a force.
http://en.wikipedia.org/wiki/Inertia
So maybe you used the wrong term, what did you mean rather than
"inertia"? The force that acts in direct opposition to the
boat-system's inertia is DRAG. LIFT (the propulsive *component* of it
anyway) acts in the same direction as the boat-system's inertia.
http://en.wikipedia.org/wiki/Lift_%28force%29
>
> BTW - There is almost no lift involved at the blade until the boat is
> moving forward,
That is a chicken-or-egg statement, and not entirely true. Lift on the
blade is produced by the blade slicing through the water (traveling on
it's arc away from the boat), not by the boat moving forward, per se.
The boat must move forward for the blade to travel on its path though,
thus the chicken/egg situation.
> and it's more in proportion to the actual system speed.
Again, I wonder where you got this information. Lift is proportional to
the *square* of velocity of the flow about the lifting body (e.g. the
blade). For a given drive time and drive length (oar angle range), the
velocity of fluid about the blade is going to be virtually the same,
regardless of boat speed.
> i.e. A quick recovery after the first stroke may have the boat
> moving fast forward but since the system is still moving quite slow
> the lift is still small. One of the reasons that we use abbreviated
> strokes off the start, since at that point the blade drag is the
> dominate resistive force in use.
Actually, but using abbreviated strokes, we limit the blade to the
drag-dominant portion of the stroke, not the other way around, as you
suggest. The reason we use abbreviated strokes is to give us
effectively lower gearing while we get the mass of the system up to speed.
>
>> Since the blade path through the water is not affected
>> by speed, the direction of the force on the blade is the same at various
>> speeds, for a given oar angle. The lift force on the blade is by
>> definition perpendicular to the direction of flow about the blade, which
>> AGAIN is determined by the path of the blade THROUGH the water. The net
>> resultant force on the blade (lift and drag) is roughly perpendicular to
>> the blade cord, which for most oars is parallel to the oar shaft. So,
>> the net resultant force vector acting on the blade during the stroke is
>> always perpendicular to the oar shaft (or very close to it).
>>
>> At the catch, given the angle of the oar, much of the force on the blade
>> points toward the hull. For a 30deg catch angle, 86.6% of the net force
>> points at the hull, and 50% of the net force points in the desired
>> propulsive direction. And in case anyone forgot, at 30deg, Pythagoras
>> gave us the relationship of sqrt(86.6^2 + 50^2) = 100%.
>>
>> -Kieran- Hide quoted text -
>>
>> - Show quoted text -
>
> Can't do much with that last bit, since it was relying on the
> acceptance of earlier incorrect claims.
Still waiting for your evidence or references that show that your claims
are correct.
-Kieran
KC
> On Feb 23, 8:59 am, KC <kc_n...@sonic.net> wrote:
>> marti...@y.z wrote:
>>>> At the catch, given the angle of the oar, much of the force on the blade
>>>> points toward the hull. For a 30deg catch angle, 86.6% of the net force
>>>> points at the hull
>>> Very easy to prove. Take the catch with a single blade. Keep the
>>> other one feathered. Just take the first third of the drive in this
>>> configuration. The boat will turn very nicely. Most of the force is
>>> going sideways.
>>> A while ago I initiated a thread about not being able to go straight.
>>> I was told to watch my releases and catches. That solved the
>>> problem. Any differentials during the catch will produce a turning
>>> force. It's a pretty amazing sport. I appreciate more and more of it
>>> as I learn. It requires one to develop a lot of skills.
>>> -Martin
>> Excellent, Martin. Thank you. That was definitely a simple "why didn't
>> I think of that?" explanation.
>
> What, you've never sculled the bow around when at a stake boat in
> cross wind conditions? I find that difficult to believe, but I'll
> take your word for it.
What gave you that impression? I just didn't think of the analogy,
that's all. Doesn't mean I haven't done it.
>
>> Paul if you're still unconvinced that the "pinch force" exists, I'll
>> respond to your other post. Meanwhile, Martin's example proves pretty
>> much everything I was trying to say. Note that this phenomenon happens
>> regardless of whether the boat is moving or not.
>
> Still completely unconvinced.
> But nice crossing of the eyes and dotting the tease. There is a
> forest behind those trees.
>
> You keep stating the extreme of "pinch force exists", as if I have
> said somewhere that it 'does not exist', which I don't think I have,
> but if I have, let's make it clear right now that I am saying that it
> is completely insignificant (when the system is at speed), you claim
> 87% at 30deg catch angle, which may be just fine in the static
> situation, but it is not the same, or even close, in the dynamic one.
Fine, forgive me for stating the extreme of "exists". I *ALSO* used
statements like "you think the MAJORITY of force on the blade points
forward." (in the early portions of the drive). Using words like
"majorty" there acknowledges your claim that the side force is "so small
as to be insignificant" (which is STILL false.)
So quit your belly aching. I acknowledge your OFFICIAL, clear stance on
the subject. It's still incorrect.
> BTW - If you continue to misrepresent what I have said (aka - making
> false claims) and use that to argue against, I'll start asking about
> your paper again, since apparently that's equally irritating to you.
I submitted my preliminary work to, it was accepted by, and I'll be
presenting my work at, a regional conference of the American Society of
Biomechanics this spring. Happy?
Oh, and I have not misrepresented what you have said. Your claims about
the pinch/side/lateral forces ARE false claims.
> Perhaps you should dig up Kleshnevs newsletter regarding the different
> force profiles required to produce straight tracking in a 2- (I know
> you are familiar with that boat.) The turning moment he describes is
> between the center of mass of the system and the pins, not our to the
> blades.
Paul, here's some advice: when you find your self in a hole, stop digging.
That newsletter, as well as the portions of _Rowing Faster_ on the same
topic, only contradict your claims. And your statement that the forces
on the pins are some how separate from those at the blades just further
shows your lack of understanding of force mechanics and vectors.
I'm starting to see why you use the Sisyphean animation for your avitar
on the C2 forums: Because anyone who tries to communicate with you is
doomed to never make any progress, and just when they think they've got
somewhere with you, you kick them back down.
-Kieran
paul_v...@hotmail.com
Strike one.
> So quit your belly aching. I acknowledge your OFFICIAL, clear stance on
> the subject. It's still incorrect.
Strike two..
> > BTW - If you continue to misrepresent what I have said (aka - making
> > false claims) and use that to argue against, I'll start asking about
> > your paper again, since apparently that's equally irritating to you.
>
> I submitted my preliminary work to, it was accepted by, and I'll be
> presenting my work at, a regional conference of the American Society of
> Biomechanics this spring. Happy?
No, I turned off interest until I get to see it, as you requested.
All the best however.
> Oh, and I have not misrepresented what you have said. Your claims about
> the pinch/side/lateral forces ARE false claims.
Strike three, yer out!
> > Perhaps you should dig up Kleshnevs newsletter regarding the different
> > force profiles required to produce straight tracking in a 2- (I know
> > you are familiar with that boat.) The turning moment he describes is
> > between the center of mass of the system and the pins, not our to the
> > blades.
>
> Paul, here's some advice: when you find your self in a hole, stop digging.
Thank you for the advice, I'm sure it's well intentioned.
> That newsletter, as well as the portions of _Rowing Faster_ on the same
> topic, only contradict your claims. And your statement that the forces
> on the pins are some how separate from those at the blades just further
> shows your lack of understanding of force mechanics and vectors.
Perhaps, but it's not that complicated is it?
> I'm starting to see why you use the Sisyphean animation for your avitar
> on the C2 forums: Because anyone who tries to communicate with you is
> doomed to never make any progress, and just when they think they've got
> somewhere with you, you kick them back down.
>
> -Kieran
Wow, so I have a lurker/stalker on the C2 forum? Have you argued
anonymously with me there? There are a few with a similar demeaning
tone. Unmask!
It's more like the futile attempt to convince the unconvincible, or
would that be inconvincible?
SissyPuss Must Die! [;o)
Oops, now Carl has weighed in, but I'm at a loss to see where my grave
error has occured. Darn education, I really should have paid more
attention! [:o\
- Paul Smith
>
> > And still, there are vast differences in what is happening when the
> > system is underway or not. Sorry that I don't have the required
> > vocabulary or tools to get this across to you, but in the long run I
> > don't see it making much difference to either of us. Perhaps if we
> > ever get a chance to row together it will be easier to demonstrate
> > many of these things, but I suppose that's quite unlikely too.
>
> > - Paul Smith- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
KC
>
>
> The blade path has to be different, depending on whether there is a
> standing start or what passes in rowing for steady state. The question
> is; "How different?"
>
> At the static start you do not normally reach so far forward because,
> there being no forward motion, there can not yet be any component in the
> intended direction of the boat to give you a resolved flow along the
> blade. However, this being a transient flow situation, you do not get
> stall either - that takes some time to become established.
The confounding part of what you said Carl, is "the intended direction
of the boat". The boat must move for the blade to slide through the
water on it's lift-producing path, but it is not the motion of the boat
that is imperative to producing lift, it is the motion of the blade away
from the hull, through the water, that gives us flow over the blade.
Then of course there must also be some load applied, for a lifting body
in flow sans load produces no lift.
> A boat accelerates quite swiftly off the start, reaching ~60% (off my
> cuff) of eventual cruise velocity by the end of that first stroke, and
> its initial acceleration is greater than that. So you develop
> significant boat velocity very soon after the catch & lift will start to
> develop in that first part of the stroke. By the finish you will get
> the normal development of lift on the blade, following the mid-stroke
> stall phase (which may be somewhat extended, especially if you fail to
> dig deep enough in the middle to prevent air entrainment, which builds
> up rapidly &, because velocity is relatively low, has more time to do so.
>
> Naturally you will can resolve forces acting around a single blade not
> only in terms of lift & drag acting perpendicular to & along the blade,
> but instead to reveal a side force perpendicular to the boat & a
> propulsive force parallel to the boat, at both catch & finish. If you
The lift force is not actually perpendicular to the blade, it is
perpendicular to the free-stream flow. So if the angle of attack at
some point during the stroke is say, 7deg, then the lift vector will be
at an angle of 83 degrees to the blade cord, not 90deg.
However, the total resultant force acting on the blade (resultant of
drag and lift) is very nearly perpendicular to the cord of the blade,
and therefore also to the oar shaft.
> liken the early & late parts of the blade action to those of a section
> of a propeller, then that side force relates directly to the torque
> applied to drive the propeller. As with aircraft, rowing has ways to
> cancel the effect of the side force, leaving only the axial, propulsive
> force - in boats by having a blade on the other side, working in mirror
> fashion.
The side forces cancel out in that they (ideally) do no work. But they
do not disappear insofar as stresses on the boat parts are concerned.
Please correct me if I'm wrong here, Carl.
> The side force, of itself, does no work because it (normally) moves
> nothing. And there is no such thing as pinching the boat at the catch.
Whether you want to call it "pinching" or "two opposing forces that do
no work" the forces do exist, and since the boat parts are not perfectly
incompressible, some lateral compression must occur. If not, please
explain in more detail how this is so.
> What you have is a range of effective loadings & angles of attack on
> the blade, which are determined by the prevailing combination of boat
> velocity & amount of forward reach, within which range there will be an
> optimum for any given boat speed. You get the same situation in
> close-hauled sailing (i.e. into the wind), where again the large side
> force generated on the sail(s) (if you choose to resolve forces in that
> sense) is exactly balanced by an equal but opposite side force acting on
> the keel, rudder & immersed hull of the boat. Some sailors try to haul
I think you mean that the moments cancel out, so the boat doesn't pitch
over, but the forces are not necessarily equal and opposite (unless the
moment arms are equal in length).
> in the rig too tight and point to close to the wind, some let out too
> much mainsheet & sail too far off the wind, but the expert gets it just
> right, somewhere between those 2 extremes. And, with sailing as with
> rowing, the faster you go the closer the sailor can head into the wind &
> the further it makes sense for the rower to reach for the catch.
>
> Was that any help?
Maybe. Keep going, please. I'm not much of a sailor (the scouting
merit badge is about as far as I went) so that analogy is not doing a
lot for me.
Let's focus on the instant before the catch. Pause the frame right
where the squared blade is about to enter the water at the catch. Say
the angle of the shaft to the water flow is 30deg. So we have a blade
about to enter water that is flowing past it at an angle of 30deg. But
*this* angle of the flow to the blade is meaningless, for as soon as
the blade enters the water, it moves along the well described path
that's been linked to and mentioned previously. IF ANYTHING that 30deg
angle of flow to the blade right before the catch would produce NEGATIVE
lift and slow the boat (as is the case with "backing it in"). The flow
toward the blade that produces lift is due ONLY to the blade traveling
away from the boat and through the water, and the resultant "flow"
approaches the blade from the concave (face) side of the blade. This is
why I've been saying that it matters not how fast the flow of water
around the hull is (boat speed) - because that is not the flow that
causes lift forces on the blade. Indeed, the blade virtually never sees
that flow, for as soon as it enters the water, it is moving WITH *that*
flow (negating it) and the only resultant flow about the blade is due to
the blade's travel away from the hull.
-Kieran
-Kieran
KC
> Thank you for the advice, I'm sure it's well intentioned.
Very much so, and you're welcome.
>> That newsletter, as well as the portions of _Rowing Faster_ on the same
>> topic, only contradict your claims. And your statement that the forces
>> on the pins are some how separate from those at the blades just further
>> shows your lack of understanding of force mechanics and vectors.
>
> Perhaps, but it's not that complicated is it?
I don't think so.
>> I'm starting to see why you use the Sisyphean animation for your avitar
>> on the C2 forums: Because anyone who tries to communicate with you is
>> doomed to never make any progress, and just when they think they've got
>> somewhere with you, you kick them back down.
>>
>> -Kieran
>
> Wow, so I have a lurker/stalker on the C2 forum? Have you argued
> anonymously with me there? There are a few with a similar demeaning
> tone. Unmask!
Nope. If I've ever posted there (I have but it's been VERY seldom) it
was under my own name.
> It's more like the futile attempt to convince the unconvincible, or
> would that be inconvincible?
My point exactly... but YOU are the "unconvincible"[sic] ;^)
-Kieran
Carl Douglas
>
> Oops, now Carl has weighed in, but I'm at a loss to see where my grave
> error has occured. Darn education, I really should have paid more
> attention! [:o\
>
Paul - I was criticising no one, nor arguing any point.
On the contrary. By shining another light (I hope) on one of the
issues, I sought to facilitate discussions between you & Kieran.
It's not about any one person being right, but about teamwork -
collectively arriving at the right (or best available) answer & everyone
feeling a glow of satisfaction over the achievement, as when a crew wins
a race.
Cheers -
paul_v...@hotmail.com
Well then your point was exactly wrong. At least if it relies on ME
being inconvincible, as I clearly am convincible when being convinced
to a correct conclusion. i.e. I was convinced that: Sliding Riggers do
not produce faster times over 2000m race courses by any inherent
benefit contained in the system.
- Paul Smith
KC
Glad to hear it. Maybe this means then that some day, some one will
convince you that the net force on a blade is nearly always
perpendicular to the shaft at virtually all instances during the drive,
since that is a true statement.
-Kieran
KC
I'd like to also point out the irony in your last claim. "I am ...
convinceable [sic] when being convinced to a correct conclusion."
One can not be "convinced" to a conclusion they see as correct, for if
they see it as correct they need no convincing. So, that *you* see my
claims as incorrect should not affect your ability to be convinced,
unless you truly are "unconvinceable" (to continue to use our new non-word).
-KC
paul_v...@hotmail.com
> paul_v_sm...@hotmail.com wrote:
>
> > Oops, now Carl has weighed in, but I'm at a loss to see where my grave
> > error has occured. Darn education, I really should have paid more
> > attention! [:o\
>
> Paul - I was criticising no one, nor arguing any point.
Completely understood that part, but thank you for confirming my
meager understanding.
> On the contrary. By shining another light (I hope) on one of the
> issues, I sought to facilitate discussions between you & Kieran.
Yes indeed, I recall another instance of the "propeller" analogy,
along with a reference to how a screw works. i think I even posted a
link to a relevant kitchen utensil, since the blade is acting much
like a finite slice of the propeller at any given point throughout the
drive.
> It's not about any one person being right, but about teamwork -
> collectively arriving at the right (or best available) answer & everyone
> feeling a glow of satisfaction over the achievement, as when a crew wins
> a race.
I don't completely dismiss the teamwork aspect, though even within a
team there can be friendly adversarial relationships, it's good for
proper motivation. My University V8 didn't have to "get along"
outside the boat, just inside. Frankly, I feel as if I'd like Kieran
more than a couple of guys from that boat, and would probably get
along with him better if we happen to meet one day.
> Cheers -
> Carl
> --
> Carl Douglas Racing Shells -
> Fine Small-Boats/AeRoWing low-drag Riggers/Advanced Accessories
> Write: The Boathouse, Timsway, Chertsey Lane, Staines TW18 3JY, UK
> Email: c...@carldouglas.co.uk Tel: +44(0)1784-456344 Fax: -466550
> URLs: www.carldouglas.co.uk(boats) &www.aerowing.co.uk(riggers)
Cheers Coach! I'll work on the teamwork aspect.
- Paul Smith
PS - Seat arrived, installed and in testing now. Fixed studs would
have been considerably better for mounting, and having a bit of a
riser (much like a standard seat, but probably 30mm in width) would
put some needed clearance above the rollers. I'll get some pictures
up on the web for illustrative purposes. So far I have noticed that
bum placement (settling in) is much better than it was on the rather
cushioned stock Erg Seats.
paul_v...@hotmail.com
Oh, so now it's "nearly" is it?
How about this then: "The propulsive force at the Pin is nearly
always in the direction of hull travel, and that force is parallel to
the force acting at the Blade." (Note, this is exactly in line with
what I have been saying all along. Just a new bunch of words strung
together.)
> > -Kieran
>
> I'd like to also point out the irony in your last claim. "I am ...
> convinceable [sic] when being convinced to a correct conclusion."
>
> One can not be "convinced" to a conclusion they see as correct, for if
> they see it as correct they need no convincing. So, that *you* see my
> claims as incorrect should not affect your ability to be convinced,
> unless you truly are "unconvinceable" (to continue to use our new non-word).
>
> -KC- Hide quoted text -
>
> - Show quoted text -
Why in the heck do I try?
You can't even stick with the word (real word) that I was using.
I bothered to look it up, though one can only wonder why.
inconvincible
Main Entry: in·con·vinc·ible
Pronunciation: "in-k&n-'vin(t)-s&-b&l
Function: adjective
: incapable of being convinced
Though there doesn't seem to be the opposition term, "convincible".
But it communicates the required meaning well enough.
Yes, the irony is thick, but I think you miss it completely. At least
I am convinced so.
Socrates is rolling in his grave as this continues. My confidence in
reaching critical mass is still high, hopefully a sign of my
persistent optimism.
- Paul Smith
Carl Douglas
> Carl Douglas wrote:
>
>>
>>
>> The blade path has to be different, depending on whether there is a
>> standing start or what passes in rowing for steady state. The
>> question is; "How different?"
>>
>> At the static start you do not normally reach so far forward because,
>> there being no forward motion, there can not yet be any component in
>> the intended direction of the boat to give you a resolved flow along
>> the blade. However, this being a transient flow situation, you do not
>> get stall either - that takes some time to become established.
>
>
> The confounding part of what you said Carl, is "the intended direction
> of the boat". The boat must move for the blade to slide through the
> water on it's lift-producing path, but it is not the motion of the boat
> that is imperative to producing lift, it is the motion of the blade away
> from the hull, through the water, that gives us flow over the blade.
> Then of course there must also be some load applied, for a lifting body
> in flow sans load produces no lift.
You won't get flow _along_ the blade at a modest angle of attack, the
pre-requisite for generating lift on a foil, until you have forward
motion. You can do the triangle if velocities as well as I can.
Fortunately, you get rapidly increasing forward motion from the outset.
>
>> A boat accelerates quite swiftly off the start, reaching ~60% (off my
>> cuff) of eventual cruise velocity by the end of that first stroke, and
>> its initial acceleration is greater than that. So you develop
>> significant boat velocity very soon after the catch & lift will start
>> to develop in that first part of the stroke. By the finish you will
>> get the normal development of lift on the blade, following the
>> mid-stroke stall phase (which may be somewhat extended, especially if
>> you fail to dig deep enough in the middle to prevent air entrainment,
>> which builds up rapidly &, because velocity is relatively low, has
>> more time to do so.
>>
>> Naturally you will can resolve forces acting around a single blade not
>> only in terms of lift & drag acting perpendicular to & along the
>> blade, but instead to reveal a side force perpendicular to the boat &
>> a propulsive force parallel to the boat, at both catch & finish. If you
>
>
> The lift force is not actually perpendicular to the blade, it is
> perpendicular to the free-stream flow. So if the angle of attack at
> some point during the stroke is say, 7deg, then the lift vector will be
> at an angle of 83 degrees to the blade cord, not 90deg.
It is a convention that we talk about lift & drag forces as being
mutually perpendicular, for which I blame the father of Cartesian
coordinates. Lift forces operate perpendicular to every art of the
surface of a foil (can't be any other way in a fluid), & the true lift
vector is the sum of those local lift vectors (including the negative
ones on the pressure face of the foil & any negative ones on the lifting
face). Similarly with drag, which can only operate parallel to the
local surface. But we opt to combine those forces into a single vector
which we then resolve into lift, perpendicular to a chosen axis of the
foil (easily determined for a symmetrical foil) & drag parallel to the
same axis.
>
> However, the total resultant force acting on the blade (resultant of
> drag and lift) is very nearly perpendicular to the cord of the blade,
> and therefore also to the oar shaft.
Sure, provided that lift predominates over drag, but nothing is for nothing.
>
>> liken the early & late parts of the blade action to those of a section
>> of a propeller, then that side force relates directly to the torque
>> applied to drive the propeller. As with aircraft, rowing has ways to
>> cancel the effect of the side force, leaving only the axial,
>> propulsive force - in boats by having a blade on the other side,
>> working in mirror fashion.
>
>
> The side forces cancel out in that they (ideally) do no work. But they
> do not disappear insofar as stresses on the boat parts are concerned.
> Please correct me if I'm wrong here, Carl.
But I've never noticed a boat getting hot, nor do they change
perceptibly in shape through the stroke. In fact, it is hard to measure
stresses in a rigger because they produce relatively small amounts of
elastic deformation. Much easier to strain-gauge an oar.
>
>> The side force, of itself, does no work because it (normally) moves
>> nothing. And there is no such thing as pinching the boat at the catch.
>
>
> Whether you want to call it "pinching" or "two opposing forces that do
> no work" the forces do exist, and since the boat parts are not perfectly
> incompressible, some lateral compression must occur. If not, please
> explain in more detail how this is so.
As above, the movements are insignificant so the losses are similarly
insignificant.
>
>> What you have is a range of effective loadings & angles of attack on
>> the blade, which are determined by the prevailing combination of boat
>> velocity & amount of forward reach, within which range there will be
>> an optimum for any given boat speed. You get the same situation in
>> close-hauled sailing (i.e. into the wind), where again the large side
>> force generated on the sail(s) (if you choose to resolve forces in
>> that sense) is exactly balanced by an equal but opposite side force
>> acting on the keel, rudder & immersed hull of the boat. Some sailors
>> try to haul
>
>
> I think you mean that the moments cancel out, so the boat doesn't pitch
> over, but the forces are not necessarily equal and opposite (unless the
> moment arms are equal in length).
No, they don't. The moments are additive (in opposite directions above
& below the roll axis) so they work together to _increase_ the degree of
heel of the boat. That's why a monohull keel-boat always has a dense
lump of metal attached to the bottom of a keel as far below the surface
as possible to generate a righting moment as the boat heels. And why
advanced monohulls have canting keels, while dinghies & other
light-displacement sailboats have to stick lumps of humanity out as far
as possible to windward of the centreline, sometimes hiking out on
trapezes - all to generate the necesasry righting moment to keep the
airfoil (the sail) as nearly upright as possible.
>
>> in the rig too tight and point to close to the wind, some let out too
>> much mainsheet & sail too far off the wind, but the expert gets it
>> just right, somewhere between those 2 extremes. And, with sailing as
>> with rowing, the faster you go the closer the sailor can head into the
>> wind & the further it makes sense for the rower to reach for the catch.
>>
>> Was that any help?
>
>
> Maybe. Keep going, please. I'm not much of a sailor (the scouting
> merit badge is about as far as I went) so that analogy is not doing a
> lot for me.
OK. Consider a landyacht (so we can forget about keels). If you point
it head to wind all it feels is wind drag & no lift. Yet it has the
potential, with a good rig, of sailing into the wind at a speed over the
ground of up to 10x the actual wind speed.
Now point the craft at 10 deg to the wind. Sheet in the sail at say 5
deg (you can't exactly, because the sail is a flexible thing, but no
matter). Now you will be getting lift from the sail at 5 deg ahead of
abeam, & you can calculate from that the effective propulsive force felt
by the craft as sin(5) x the actual lift force. You may also be able to
compute the total wind drag on rig & the rest of the craft. If this
multiplied by sin(85) does not exceed the propulsive force, you still
won't be sailing.
Now point another 5 degrees off the wind, keeping the sail similarly
trimmed. By doing this you have doubled your lift force, but hardly
altered your drag forces (you're working in the drag bucket part of the
drag vs angle of attack curve, where drag hardly alters with AOA). You
now have twice as much propulsive force along the new heading and
hopefully that now exceeds the drag forces, so you're sailing!.
OK, you ain't sailing on your intended heading (dead into the wind), but
your velocity made good in that direction is cos(80) of you speed over
the ground, which ain't too far off 100%.
Now ease off another 5 deg on your course. Lift is up by 50%, drag is
still quite low but beginning to creep up, so you go faster still
(sailing speed is all about the ratio of lift to drag), but you are now
only getting cos(75) of your actual speed in your intended direction, so
there's a trade-off to be made. At some point you will find you are
failing to make ground quite so fast in your required direction, so have
hit a local optimum. But you can still re-set your rig, and go and find
another local optimum. And somewhere you will find that best
combination of AOA/rig setting & course.
You can replay all those considerations for an oarblade at the catch,
plus a few more relating to the advantages of longer stroke duration.
Then with the landyacht you have also, as the craft accelerates, to
constantly re-trim & adjust course, because the apparent wind is moving
around & coming ever closer to dead ahead the faster you go.
>
> Let's focus on the instant before the catch. Pause the frame right
> where the squared blade is about to enter the water at the catch. Say
> the angle of the shaft to the water flow is 30deg. So we have a blade
> about to enter water that is flowing past it at an angle of 30deg. But
> *this* angle of the flow to the blade is meaningless, for as soon as
> the blade enters the water, it moves along the well described path
> that's been linked to and mentioned previously. IF ANYTHING that 30deg
> angle of flow to the blade right before the catch would produce NEGATIVE
> lift and slow the boat (as is the case with "backing it in"). The flow
> toward the blade that produces lift is due ONLY to the blade traveling
> away from the boat and through the water, and the resultant "flow"
> approaches the blade from the concave (face) side of the blade. This is
> why I've been saying that it matters not how fast the flow of water
> around the hull is (boat speed) - because that is not the flow that
> causes lift forces on the blade. Indeed, the blade virtually never sees
> that flow, for as soon as it enters the water, it is moving WITH *that*
> flow (negating it) and the only resultant flow about the blade is due to
> the blade's travel away from the hull.
>
> -Kieran
>
> -Kieran
That's me lot for tonight!
Cheers -
KC
>>>> Well then your point was exactly wrong. At least if it relies on ME
>>>> being inconvincible, as I clearly am convincible when being convinced
>>>> to a correct conclusion. i.e. I was convinced that: Sliding Riggers do
>>>> not produce faster times over 2000m race courses by any inherent
>>>> benefit contained in the system.
>>>> - Paul Smith
>>> Glad to hear it. Maybe this means then that some day, some one will
>>> convince you that the net force on a blade is nearly always
>>> perpendicular to the shaft at virtually all instances during the drive,
>>> since that is a true statement.
>
> Oh, so now it's "nearly" is it?
>
> How about this then: "The propulsive force at the Pin is nearly
> always in the direction of hull travel, and that force is parallel to
> the force acting at the Blade." (Note, this is exactly in line with
> what I have been saying all along. Just a new bunch of words strung
> together.)
I don't disagree with that at all. The *propulsive* (component of)
force is by definition in the direction of propulsion. It is only one
component of the total force on the blade (and pin). The rest is
perpendicular to the propulsive force. What I *think* you have disputed
with me is that the resultant force (i.e. total, net, sum force) on the
blade is NEARLY perpendicular to the oar shaft at virtually all
instances during the drive. Is that a fair statement? I don't want to
misrepresent your claims.
I would think you could come up with something more profound than "the
propulsive force is in the direction of hull travel."
>>> -Kieran
>> I'd like to also point out the irony in your last claim. "I am ...
>> convinceable [sic] when being convinced to a correct conclusion."
>>
>> One can not be "convinced" to a conclusion they see as correct, for if
>> they see it as correct they need no convincing. So, that *you* see my
>> claims as incorrect should not affect your ability to be convinced,
>> unless you truly are "unconvinceable" (to continue to use our new non-word).
>>
>> -KC- Hide quoted text -
>>
>> - Show quoted text -
>
> Why in the heck do I try?
I hadn't noticed that you had.
> You can't even stick with the word (real word) that I was using.
> I bothered to look it up, though one can only wonder why.
> inconvincible
> Main Entry: in搾on暇inc搏ble
> Pronunciation: "in-k&n-'vin(t)-s&-b&l
> Function: adjective
> : incapable of being convinced
Congrats on being a better speller than I. I concede and bow to your
spellingbeeness.
> Though there doesn't seem to be the opposition term, "convincible".
> But it communicates the required meaning well enough.
>
> Yes, the irony is thick, but I think you miss it completely. At least
> I am convinced so.
>
> Socrates is rolling in his grave as this continues. My confidence in
> reaching critical mass is still high, hopefully a sign of my
> persistent optimism.
I don't know much about Philosophy. Feel free to flex your BA degree
and fill me in on the reference to Socrates here. Maybe it'll make you
feel even better... TWO areas where you are superior! Keep it up, I
shiver in the presence of your correctly spelled philosophical superiority.
The irony NOW, is that my mother was an English teacher and my father
has a BA in philosophy. :^P
-Kieran
KC
> KC wrote:
>> Carl Douglas wrote:
>>
>>>
>>>
>>> The blade path has to be different, depending on whether there is a
>>> standing start or what passes in rowing for steady state. The
>>> question is; "How different?"
>>>
>>> At the static start you do not normally reach so far forward because,
>>> there being no forward motion, there can not yet be any component in
>>> the intended direction of the boat to give you a resolved flow along
>>> the blade. However, this being a transient flow situation, you do
>>> not get stall either - that takes some time to become established.
>>
>>
>> The confounding part of what you said Carl, is "the intended direction
>> of the boat". The boat must move for the blade to slide through the
>> water on it's lift-producing path, but it is not the motion of the
>> boat that is imperative to producing lift, it is the motion of the
>> blade away from the hull, through the water, that gives us flow over
>> the blade. Then of course there must also be some load applied, for a
>> lifting body in flow sans load produces no lift.
>
> You won't get flow _along_ the blade at a modest angle of attack, the
Okay... but I'd say that a modest AoA occurs *almost* instantaneously
once the oars are pulled upon by the rower.
Hmmm... well, the way I recall it is that "lift" is *defined* as the sum
of the force components that are in the directions perpendicular to the
free stream flow.
>>
>> However, the total resultant force acting on the blade (resultant of
>> drag and lift) is very nearly perpendicular to the cord of the blade,
>> and therefore also to the oar shaft.
>
> Sure, provided that lift predominates over drag, but nothing is for
> nothing.
Well, I was talking about the case of a rowing oar blade in the early
stages of the drive phase of a stroke. But anyway, drag predominates
over lift in the mid-stroke, yet the net force on the blade is still
perpendicular to the cord. AIUI, as lift and drag gradually exchange
dominance throughout the stroke, the net force on the blade remains
perpendicular to the cord. Is this not correct? Most of the graphs of
blade force I've seen show the resultant vector to be nearly
perpendicular to the blade cord at all times during the stroke.
>>
>>> liken the early & late parts of the blade action to those of a
>>> section of a propeller, then that side force relates directly to the
>>> torque applied to drive the propeller. As with aircraft, rowing has
>>> ways to cancel the effect of the side force, leaving only the axial,
>>> propulsive force - in boats by having a blade on the other side,
>>> working in mirror fashion.
>>
>>
>> The side forces cancel out in that they (ideally) do no work. But
>> they do not disappear insofar as stresses on the boat parts are
>> concerned. Please correct me if I'm wrong here, Carl.
>
> But I've never noticed a boat getting hot, nor do they change
> perceptibly in shape through the stroke. In fact, it is hard to measure
> stresses in a rigger because they produce relatively small amounts of
> elastic deformation. Much easier to strain-gauge an oar.
That the boat elements do not deform, nor heat up, says more about the
design of the boat & materials than the significance (or lack thereof)
of the force, does it not? For level flight of an aircraft, the Lift
force does no work. It is perfectly in opposition to the force of
gravity on the aircraft. That Lift does no work does not mean that the
aircraft does not endure the effects of its application.
Your perspective is (respectfully) from that of a boat builder here. I
am not trying to say that a "pinching" force exists to say that boats
should be designed differently. Boats are obviously designed well
enough to withstand it.
However, if the net resultant force on the blade is perpendicular to the
shaft (as you have agreed) then...
Given a catch angle of 30deg,
with an applied handle force of 577.5N
With force applied at 30deg to the oar handle (due to reach of the
rowers arms)
perpendicular force on the oar handle from the rower is then
Fph = 577.5cos(30) = 500N perpendicular to the handle.
given a scull length of 280cm
inboard of 80cm
the force at the blade is then 200N, perpendicular to the shaft.
At 30 deg oar angle, there will be
Fsb = cos(30)*200N = 173.2N directed toward the boat, and
Fdb = sin(30)*200N = 100N directed "forward" (direction of travel)
So, If I'm wrong I really want to know, because I don't see anything
erroneous there. If I've made a mistake somewhere in some assumption,
or calculation I'd like to know exactly what it is, so that I can get it
right for the future. Paul can't seem to point out my errors, so
hopefully someone else can.
To continue that calclulation... With a perpendicular oar force of 500N
at the handle and 200N at the blade, the reaction of those forces at the
pin is 700N with a direction perpendicular to the oar shaft.
That means, at a 30deg oar angle, there is 350N applied to one pin that
is driving the boat forward, and 606.2N applied to that pin pointing in
toward the boat.
I don't consider ~135lb of force at the pin pointed perpendicular to the
long axis of the hull to be insignificant. Maybe it's not enough to
deflect the pin or stress the riggers much, but it's about a coxswain's
worth of force! :-) So, where is(are) my mistake(s)?
(There is also a reaction from the axial component of the rower's
applied force ("pressure against the pin"). But since it has no
reaction against the water, it can't drive the boat and should be
ignored for calculations of boat propulsion, but for calculations of pin
deflection it could be included. It resolves to Frpa = 144.4i + 250j N
where i is positive away from the hull, and j is positive from stern to
bow.)
>>
>>> The side force, of itself, does no work because it (normally) moves
>>> nothing. And there is no such thing as pinching the boat at the catch.
>>
>>
>> Whether you want to call it "pinching" or "two opposing forces that do
>> no work" the forces do exist, and since the boat parts are not
>> perfectly incompressible, some lateral compression must occur. If
>> not, please explain in more detail how this is so.
>
> As above, the movements are insignificant so the losses are similarly
> insignificant.
Well, I'm not really worried about the losses, or deflections. Only
that there is a significant component of the blade force pointed at the
hull, and that for oar angles greater than 45 degrees, this component is
greater than the component of force which drives the boat forward.
Paul has said that the component of force directed at the boat is "so
small as to be insignificant". If my math, and mechanics have failed
me, I'd sure like to know where. Because as far as I can tell, at the
catch, the amount of force pointing toward the hull is greater than the
amount pointing in the direction of travel.
snipping the part on sailboats.... (great though it was).
-Kieran
marco...@gmail.com
have rowed the only 2 resolute with enclosed bulkhead. Resolute are
famous here because if the stroke drop the water-bottle on the bottom
of the boat it always end up getting stock on the bow (we had several
time to do the "water-bottle dance" with the boat over hour head
trying un-stuck them from the bow and we had a 2004).
Last time I spoke with their reps at the 2006 Head of the Charles they
were planning to follow the FISA rules for the next production, but I
have not seen any of the enclosed 8+ yet.
The "testing" are actually coming from real life experience. Brown
many time race in one of the worse body of water (always really choppy
and windy) and their boat usually don't swamp completely.
I think is because the gunnels (I think it's gunwhales or something
like that in the UK right?) of the Resolute trap a big amount of air
under neat creating buoyancy. Although what really matter is the
result it's not quite the same of having a fully buoyant boat.
On the rest I agree with you they are really great boats:
my favorites when initial price is not a problem and my crew know what
it means to catch. 8^)
Ciao,
Marco
On Feb 21, 5:13 pm, "Lawrence Edwards"
<Lawrence.Edwa...@btinternet.com> wrote:
> <marco.b...@gmail.com> wrote in message
>
> news:1171905951.2...@p10g2000cwp.googlegroups.com...> Neither of the 2 boat builders at the moment are producing a FISA
> > standards compliant boat. I assume though that at least Resolute will
> > have no problem to fit their boats with sealed bulkhead.
>
> The Resolute 8 I steered at the Head of the Charles a couple of years ago
> was fully bouyant, as were all of the other Resolutes I saw. They even had a
> section in their brochure explaining that they have tested the boats with
> over-weight crews full of water and they are perfectly rowable. I also saw a
> US crew in a Resoltue 8 at Women's Henley last year, which was enclosed
> under the seats. I've just looked on their website and it is not clear
> whether the current designs have enclosed underseat areas, but I would be
> surprised if they didn't after having spoken to the owner of the company
> about this issue!
>
> They are also very well made, with an excellent finish and very high quality
> fittings.
>
> It is the quality of the fittings and finish on the Chinese boats that I
> would question. At the World Masters Row Show 2 years ago there were some
> Swift boats, the basic hulls and bulkheads looked fine but the riggers,
> seats, rails, and nuts and bolts looked as though they had been selected on
> the basis of price not quality.
>
> It would be a false economy to save a few grand on a new 8 only to find that
> after a moderate amount of use everything started to break.
>
> I apply the same logic to purchasing cars. I've tried many makes over the
> years, and settled on Mazdas because they seem to be the only cars which are
> both well made and good value for money. NB I have no connetion with either
> Mazda or Resolute!
>
> P.S. following the SARA ruling we have retrofitted bulkheads to our Empacher
> 8. It was not that difficult to do and has not affected the performance
> enough to notice. Perhaps at Olympic or World Championship level it might
> make a difference, but the quality of the crew is far more important than a
> few extra kilos in an 8. It also means you can keep your kit dry....
>
> Lawrence
marco...@gmail.com
paul_v...@hotmail.com
Be as semanitc pedantic as you want, I asked you which way the "soft
pin" would bend and you didn't answer, you know what I mean. Or if
you don't, I'm serioulsy not equipped to explain it to you.
> >>> -Kieran
> >> I'd like to also point out the irony in your last claim. "I am ...
> >> convinceable [sic] when being convinced to a correct conclusion."
>
> >> One can not be "convinced" to a conclusion they see as correct, for if
> >> they see it as correct they need no convincing. So, that *you* see my
> >> claims as incorrect should not affect your ability to be convinced,
> >> unless you truly are "unconvinceable" (to continue to use our new non-word).
>
> >> -KC- Hide quoted text -
>
> >> - Show quoted text -
>
> > Why in the heck do I try?
>
> I hadn't noticed that you had.
>
> > You can't even stick with the word (real word) that I was using.
> > I bothered to look it up, though one can only wonder why.
> > inconvincible
> > Main Entry: in·con·vinc·ible
> > Pronunciation: "in-k&n-'vin(t)-s&-b&l
> > Function: adjective
> > : incapable of being convinced
>
> Congrats on being a better speller than I. I concede and bow to your
> spellingbeeness.
>
> > Though there doesn't seem to be the opposition term, "convincible".
> > But it communicates the required meaning well enough.
>
> > Yes, the irony is thick, but I think you miss it completely. At least
> > I am convinced so.
>
> > Socrates is rolling in his grave as this continues. My confidence in
> > reaching critical mass is still high, hopefully a sign of my
> > persistent optimism.
>
> I don't know much about Philosophy. Feel free to flex your BA degree
> and fill me in on the reference to Socrates here. Maybe it'll make you
> feel even better... TWO areas where you are superior! Keep it up, I
> shiver in the presence of your correctly spelled philosophical superiority.
You'll surely appreciate that I was conferred a BS degree, rather than
a BA. Though I like the attempted matriculatory swipe. And you calim
to be the civil participant? Pshaw!
> The irony NOW, is that my mother was an English teacher and my father
> has a BA in philosophy. :^P
>
> -Kieran
Good humor there, I gotta admit.
Here is another bit of irony for you, that relates to what may
eventually dawn on you as you discuss further with Carl. When the
Sail is providing lift, the vessel is being dragged along. Now think
of the Pin as a mast. [;o)
- Paul Smith
martin+x@y.z
books off the shelf and dive into some of this very interesting
material.
The term "lift" is being used a lot in this thread. I have never
thought of an oar generating propulsive force (at least the bulk of
it) through lift. I just can't see it. I see an oar as generating
propulsive forces purely by drag. To go to one of my
simplifications: Where do you generate lift if you run into the wind
with a 4 x 8 foot sheet of plywood in front of you (and normal to the
wind)? The force you will experience is 100% drag, right?
Isn't this how an oar blade generates force to propel a boat?
Where does lift come into this equation? The kind of lift that
sustains an airplane or propels a sailboat. Most of the work I've
done in the field is in aerodynamics for model airplane airfoils.
Quite a bit of it, in fact. I even built my own wind tunnel to test
airfoils a while back. I may need to develop the right perspecitve to
see what is happening with an oar blade.
I must admit, RSR is starting to peak my interest in hydrodynamics and
boat design. I know myself. I know where this is leading. I have a
garage full of fiberglass, kevlar and carbon fiber. I see myself
building a boat by the end of the year if this trend continues. :-)
-Martin
Jonny
> OK, I'm confused here.
So are most of the rest of us. Welcome to the club
>
> The term "lift" is being used a lot in this thread. I have never
> thought of an oar generating propulsive force (at least the bulk of
> it) through lift. I just can't see it. I see an oar as generating
> propulsive forces purely by drag. To go to one of my
> simplifications: Where do you generate lift if you run into the wind
> with a 4 x 8 foot sheet of plywood in front of you (and normal to the
> wind)? The force you will experience is 100% drag, right?
You only really have 100% drag when the face of the blade is
perpendicular to the direction of travel (ie 90deg from side of boat,
mid-drive).
>
> Isn't this how an oar blade generates force to propel a boat?
>
> Where does lift come into this equation? The kind of lift that
> sustains an airplane or propels a sailboat. Most of the work I've
> done in the field is in aerodynamics for model airplane airfoils.
> Quite a bit of it, in fact. I even built my own wind tunnel to test
> airfoils a while back. I may need to develop the right perspecitve to
> see what is happening with an oar blade.
http://www.concept2.com/us/products/oars/testing/default.asp
Check out the diagrams and explanatory stuff here. They are trying to
promote a particular product here that many people 'poo-poo' but the
pictures are easy to understand. The physics makes sense for aircraft
wings, but the benefit to rowing is not proven. I have heard many
rowers/coaches refer to the product in question as the "Vortex Edge
Protector".
>
> I must admit, RSR is starting to peak my interest in hydrodynamics and
> boat design. I know myself. I know where this is leading. I have a
> garage full of fiberglass, kevlar and carbon fiber. I see myself
> building a boat by the end of the year if this trend continues. :-)
>
I've got the same dream. I have done a traditional timber boat
building short course and made a lapstrake dingy. Even Carl said it
was quite nice (but I didn't send him any close up pictures)!
martin+x@y.z
Ah, that's some of the context that may have been missing for me. The
path and motion of the blade are interesting.
Some quick math looking at these pictures tells me that the blade
might be moving at as much as 60 MPH during the first quarter of the
stroke. This is the part of the stroke where it is being proposed
that the tip of the blade is, effectively, the leading edge of a
wing. On a quick first inspection this makes a lot of sense. And 60
MPH is not trivial (in terms of generating lift) due to the density of
water.
This tells me that you should not apply much force to the oar handle
during the first 1/4 of the drive. If your blade is truly a lift-
generating device...well...blade shape...60MPH in water is a big
deal...
I need to think about this a little. Got a few ideas to play with at
the lake tomorrow.
-Martin
Carl
> OK, I'm confused here. Which I Iike because it forces me to pull
> books off the shelf and dive into some of this very interesting
> material.
>
> The term "lift" is being used a lot in this thread. I have never
> thought of an oar generating propulsive force (at least the bulk of
> it) through lift. I just can't see it. I see an oar as generating
> propulsive forces purely by drag. To go to one of my
> simplifications: Where do you generate lift if you run into the wind
> with a 4 x 8 foot sheet of plywood in front of you (and normal to the
> wind)? The force you will experience is 100% drag, right?
No. It is "lift" which takes off roofs in gales, lifts sheets of ply,
steers boats (even with those silly flat tin rudders, or the big
barn-door jobs we used to have. And in the first & last thirds of the
stroke it is lift which does most to propel the boat.
With the sheet of ply, it is lift which wrenches it out of your grasp,
but drag which blows it down-wind. Lift is greatest in the part closer
to the leading edge so, since you must hold your sheet of ply roughly in
the centre, as soon as the AOA is enough to allow lift generation the
leading edge flips up & she starts to fly. It is your efforts to hold
the ply still which maintain sufficient wind velocity across the sheet,
As the wind gets her, the sheet first tilts leading-edge up, then as
AOA increases the lift increases proportionately until she hoists up out
of your grip. Then the sheet is carried away by the wind until it
reaches something close enough to wind speed that the lift in whatever
attitude it then has as it tumbles in the air is insufficient to defy
gravity. At which point the sheet falls to the ground, which removes
its leeward motion &, if it bounces or cartwheels on impact, may cause
it to lift & fly some more.
>
> Isn't this how an oar blade generates force to propel a boat?
>
> Where does lift come into this equation? The kind of lift that
> sustains an airplane or propels a sailboat. Most of the work I've
> done in the field is in aerodynamics for model airplane airfoils.
> Quite a bit of it, in fact. I even built my own wind tunnel to test
> airfoils a while back. I may need to develop the right perspecitve to
> see what is happening with an oar blade.
Consider the triangle of velocities at the blade near the catch: a
large component parallel to the boat's axis & a small component
perpendicular to the oar's axis generate a resultant flow (as seen by
the blade) which is roughly parallel with the oar's axis. Then apply
load & you get, as ever, matching lift. That would work fine for a flat
blade, but works rather better for one which is curved along its length
as an oarblade is curved.
You don't push water sideways with the blade - the blade cuts tip-first
through the water. You don't drag through a small packet of water - the
blade slices past large volumes of water (until near mid-stroke, where
it stalls & drags (see the triangle of velocities).
>
> I must admit, RSR is starting to peak my interest in hydrodynamics and
> boat design. I know myself. I know where this is leading. I have a
> garage full of fiberglass, kevlar and carbon fiber. I see myself
> building a boat by the end of the year if this trend continues. :-)
>
That way, madness lies ;)
Carl
It was nice, Jonny.
C
Carl
Dunno where you get 60mph from, Martin. More like 4 or 5 m/sec, or 9 to
11 mph. But still oodles to provide all the lift you need.
Next, the lift force on a foil is only the same as the force applied to
that foil, with the AOA self adjusting accordingly. There is no free
lift. If you want propulsion, you must apply load. Otherwise you have
a kind of perpetual motion machine, which is a thermodynamically illegal
version of a free lunch.
The virtue of lift is that it makes a seemingly soft & compliant fluid
remarkably hard. Whereas a body supported in a fluid by drag alone will
fall through it, without options, a lifting foil slicing through a fluid
can generate many times more lift than the drag it incurs, & that lift
can be used to move the foil about in the fluid or at least to prevent
if from falling (as with a wing).
Don't expect fluid dynamics to be intuitive!
B3
> URLs: www.carldouglas.co.uk(boats) &www.aerowing.co.uk(riggers)
Well, I've learned something here. I am a fluid dynamicist, and I
never thought of the oar producing lift either, only as a drag
device. What I guess I was missing is the rather large (at the catch)
outward motion of the blade relative to the boat and the water. It is
hard to visualize that there is enough relative velocity along the
blade axis that you create much lift. This implies that if you stop
pulling mid stroke you will not get a puddle (if you stop pulling
before stall). I'll have to try it out sometime. I just figured the
blade was always stalled.
But, what does this imply about the finish? Once you have no flow
along the axis (mid-stroke) the blade stalls. Is there enough forward
velocity that you want to pull enough less at the finish to establish
attached flow going the opposite way along the oar axis? That is
certainly counter-intuitive!
Bob
martin+x@y.z
> 11 mph. But still oodles to provide all the lift you need.
>From attempting even basic math at 1:30AM!
Assuming a blade tip displacement of 24 inches during the first 0.2
seconds of the drive (phase 1 in the C2 page) I get 6.8mph...60 didn't
feel right, even at 1:30AM, but I was too tired to bother checking my
numbers.
> Next, the lift force on a foil is only the same as the force applied to
> that foil, with the AOA self adjusting accordingly. There is no free
> lift. If you want propulsion, you must apply load. Otherwise you have
> a kind of perpetual motion machine, which is a thermodynamically illegal
> version of a free lunch.
Ture enough. I was thinking of a boat already in motion. I was
thinking about the first 0.2s or so of the drive, when the drag
generate by the blade is at a minimum. I wonder if it might be better
to conserve energy and not drive very hard --let pre-existing flow
create lift for you. It's not free energy. The idea being to apply
enough force to produce a little net thrust but not waste most of it.
If, on the other hand, the first phase of the drive is the most
efficient from a lift generating standpoint, this might very well be
where you want to apply the most effort.
It's a question, really: Where is your energy best used? I you can
apply the right amount of force at the segment of the drive when it
will generate the most results this might provide a real and
measureable improvement in performance.
If I read the paper correctly it seems that the last phase, just prior
to release, is where blade designers have to thow their hands up and
simply accept sub-optimal performance. They even go as far as
suggesting that one might want to end the stroke sooner.
Do you agree with most of that paper? I can't see the logic behind
even talking about vortices. I always thought that flow separation
required a compressible fluid. Or, in the case of water, cavitation.
Does that happen with an oar?
> Don't expect fluid dynamics to be intuitive!
My friends in aerospace like to say that the only people who believe
the results of simulations are those who wrote the software.
My favorite aero-engineering mental image: Burt Rutan tossing a foam
and cardboard model of his Spaceship One to see how it will fly when
feathered. I love it.
It's an interesting realization that an oar is operating mostly on
lift.
Lots to think about today at the lake.
-Martin
martin+x@y.z
I'd prefer to call that a pressure differential.
I've heard from those who propose that lift (as in propeller or
aircraft wing lift) is really about how much air mass is being "pushed
down" by the airfoil rather than the pressure differential concept.
In other words, in a way, the goal is to make a good air pump. When
this "pump" moves air in such a way that it generates more upward
force than drag, you have lift and flight.
-Martin
paul_v...@hotmail.com
During mid stroke, there is still a remnant of circulation about the
blade that can provide lift, as long as cavitation is kept to a
minimum through either a reduction of pressure or depth of the blade
("don't tear the water"), then the blade reverses direction and
reestablishes lift forces. If the water is torn during mid stroke,
the finish will not be able to take advantage of lift forces, but the
puddle will look very impressive.
When trying to get this point across to rowers I tell them to produce
dark puddles, not light frothy ones. But of course you still need to
be putting in the required work to move the system along at the
desired speed. To get the point across in a more hands on way, simply
go out in a 2x and row individually with an objective speed measuring
device on board. Have the "frothy" rower establish the baseline
speed, obvserve the puddles, then swap rowers and see what it takes
for "Mr. Dark" to produce the same speed and observe the puddles. The
input is apparently equal, but I guarantee that "Mr. Frothy" thinks he
was doing more.
- Paul Smith
martin+x@y.z
> never thought of the oar producing lift either, only as a drag
> device. What I guess I was missing is the rather large (at the catch)
> outward motion of the blade relative to the boat and the water. It is
> hard to visualize that there is enough relative velocity along the
> blade axis that you create much lift.
Well, that makes me feel better. I was feeling pretty dumb for not
having realized this. Embarrassed really. I too was absolutely
convinced that it was thrust-by-drag.
Time to stop thinking ang go row!
-Martin
Carl
Lift can be described as the pressure differential, over the area of the
foil, acting perpendicular to the fluid flow due to a reduction in the
pressure in the fluid flowing over one surface compared to that flowing
over the other surface). You can generate substantial lift with flat
"foils" but get more lift over wider angles of attack, & less
stall-sensitivity, from cambered foils.
You will get fluid displacement perpendicular to the flow, of course,
but that can be seen as a consequence of the pressure differential which
cause lift, rather than as a cause of lift. And in generating lift the
objective is to move as large a mass of the surrounding fluid as
possible so that, by keeping its vertical velocity small you minimise
the resulting kinetic energy absorption.
So I see what you mean about the air pump analogy, but I don't think it
represents the process too well
Chris Kerr
Me too. I can't remember whether I read it here or in "Rowing Faster"
first, but either way I spent six years thinking that the most efficient
part of the stroke was when the oar was perpendicular to the boat.
paul_v...@hotmail.com
I think you can probably read "the most efficient part of the stroke
was when the oar was perpendicular to the boat." in many places, and
being proclaimed by what could be called "influential" coaches. It's
just wrong, and quite likely not the way their rowers actually move a
boat quickly. Isn't it nice that a real understanding of what is
happening is not required to still do a good job where it needs
doing. Though I do think that having an understanding will eventually
lead to performance improvements at the highest levels when the
athletes are not only superbly conditioned but are at the technical
peak also.
- Paul Smith
Kieran
Now that we have a fluid dynamicist, an(other) engineer, and an
engineer/boat builder active in this thread... I'd really appreciate it
if any of you could offer some feedback on where if anywhere, I've
screwed up such that the sideways (inward) components of the forces on
the blade (and thus the pin) in the early part of the stroke are less
than what I've figured.
Thanks!
-Kieran
Kieran
The blade is more efficient when it's producing lift, but it is very
likely (depending on the rower) that the rower is most powerful midway
through the stroke.
This is unfortunate, because that is when the extra force is very likely
wasted in "ripping" the blade through the water (slippage).
On the other hand, at least during mid stroke almost all the force on
the blade (and pin) is pointing in the direction of travel. Whereas
earlier in the stroke at steeper oar angles when the blade is producing
lift most effectively, much of the force is directed at the boat and
less in the propulsive direction.
-Kieran
paul_v...@hotmail.com
I'm still finding your "quest" less than believable, when you still
manage to post things like this (after the above posting):
"On the other hand, at least during mid stroke almost all the force on
the blade (and pin) is pointing in the direction of travel. Whereas
earlier in the stroke at steeper oar angles when the blade is
producing
lift most effectively, much of the force is directed at the boat and
less in the propulsive direction. - KC"
You seem to be continuing to ask "Where have I gone wrong?", yet you
are the one with just the education to figure that out, aren't you?
I'm certainly not! Though my suspicions are in the direction that you
are not seeing the difference between the static and dynamic
situations. Since you have told me "They are not different." in so
many words. There is some additional irony in that you do see "clear
differences" when this sort of thing is discussed regarding the Erg,
and we can't agree there either. [:o)
Maybe this is not only "thick" irony, but deep too.
- Paul Smith
Kieran
Which is why I did not ask you (that time), and do not believe you when
you claim I'm wrong. I HAVE asked you to show where I am wrong, and you
are not capable, you only are capable of claiming I'm wrong without any
reasonable explanation as to how or why.
> Though my suspicions are in the direction that you
> are not seeing the difference between the static and dynamic
The first stroke vs. any other stroke, is less important to me. I'd be
happy to ignore the case of the first stroke and just have you prove
your claims for the "dynamic" case. Can you do that?
> situations. Since you have told me "They are not different." in so
Now who's misrepresenting whom? I stated very clearly that they are
similar, as in not very different. I never implied they were literally
identical.
> many words. There is some additional irony in that you do see "clear
> differences" when this sort of thing is discussed regarding the Erg,
> and we can't agree there either. [:o)
Hmmm... ironic indeed. Although I thought you said you did agree with
me (or part of what I said) after reading what I posted in that blog a
few weeks back. Still waiting for that reply/response, too. Email
would suffice.
-Kieran
B3
I will not try to quantify the forces, as I do not think I would
hazzard trying to estimate the various forces. However, I would point
out that the forces on the blade are not the only ones on the oar or
pin. At the catch I expect that there is a rather large force applied
along the axis of the oar from the rowers hand toward the pin. A
portion of this axial component will be outboard, and perhaps cancels
some of the inward force created by the blade at a steep initial angle
to the boat. If the oar did not have a collar, I suspect that it
would shoot forward (and outboard) rather dramatically at the catch.
Perhaps this is the missing force component on the pin y'all have been
arguing about?
Bob
Charles Carroll
> This is unfortunate, because [midway through the stroke] is when
> the extra force is very likely
> wasted in "ripping" the blade through the water (slippage).
Kieran,
Is this really true? I find that midway through the stroke is the only place
I don't have to worry about slippage. Of course I have to add, 'providing
the blades are immersed at the correct depth.'
I remember Chris Dadd's, one of the coaches at OWRC, once telling me when we
were rowing a double, “Squeeze! Don’t chop at it. Don’t jerk. Squeeze!”
It isn't until somewhere around "midway through the stroke" that I can get a
sense of whether I have succeeded in squeezing. When I have it is always
accompanied with the sensation that I am really holding on to the water and
sending the boat away.
Cordially,
Charles
Kieran
Charles,
It depends on how well buried the blade is. If we assume a constant
depth throughout (not saying that's preferable necessarily, just to
simplify) then the blade slip/"rip" more when it is stalled, which is
during the mid-stroke, than when it is effectively generating lift,
which is in the first 1/3 (or so) and last 1/4 (or so) of the drive.
-Kieran
Kieran
Bob,
I did do the math, with some guestimated applied force levels at a catch
angle of 30deg. I grant that those force levels are high, but I wanted
to use 500N normal to the handle to simplify the arithmetic (I suppose
underestimating with 100N would have made the math even easier!) Anyway,
the numbers all scale, but the trig is laid out in detail, and I did
account for the axial force (outward against the collar) that the rower
supplies on the handle. The inward component of that force does not
fully negate the inward force at the pin due to the blade/handle normal
forces.
Please check my math if you're interested. It was posted on 2/23/07 at
5:50PM EST, and quoted (by me) again today at 5:12PM EST.
-Kieran
Charles Carroll
> depth throughout (not saying that's preferable necessarily, just to
> simplify) then the blade slip/"rip" more when it is stalled, which is
> during the mid-stroke, than when it is effectively generating lift,
> which is in the first 1/3 (or so) and last 1/4 (or so) of the drive.
Which explains why I am so often reminding myself midway through the stroke
not to let go of the water. Thanks, Kieran.
Charles
paul_v...@hotmail.com
>
> > I'm still finding your "quest" less than believable, when you still
> > manage to post things like this (after the above posting):
>
> > "On the other hand, at least during mid stroke almost all the force on
> > the blade (and pin) is pointing in the direction of travel. Whereas
> > earlier in the stroke at steeper oar angles when the blade is
> > producing
> > lift most effectively, much of the force is directed at the boat and
> > less in the propulsive direction. - KC"
>
> > You seem to be continuing to ask "Where have I gone wrong?", yet you
> > are the one with just the education to figure that out, aren't you?
> > I'm certainly not!
>
> Which is why I did not ask you (that time), and do not believe you when
> you claim I'm wrong. I HAVE asked you to show where I am wrong, and you
> are not capable, you only are capable of claiming I'm wrong without any
> reasonable explanation as to how or why.
I really have tried, but you apparently do not appreciate the effort.
Can you prove I'm wrong in the claim I'm making?
Since you were unaware of my lnwledge of Ken Youngs Pages, perhaps you
are unaware of this one, that discusses the "worrisome "pinch-point"".
http://www.atkinsopht.com/row/bowangle.htm
Please do not take this as any sort of wholesale endorsement of the
entire Atkinsopht site, there is a lot there to disagree with, but
this particular bit might be applicable to our current discussion.
(His section regarding lift and drag doesn't seem to make sense to me,
so you might like it a lot, with your advanced level of understanding
and all.)
> > Though my suspicions are in the direction that you
> > are not seeing the difference between the static and dynamic
>
> The first stroke vs. any other stroke, is less important to me. I'd be
> happy to ignore the case of the first stroke and just have you prove
> your claims for the "dynamic" case. Can you do that?
Apparently not to your satisfaction, no. I can/do prove it to my own,
every time I am out in a boat with the tools in question at hand.
> > situations. Since you have told me "They are not different." in so
>
> Now who's misrepresenting whom? I stated very clearly that they are
> similar, as in not very different. I never implied they were literally
> identical.
--yawn-- Have you already forgotten about your claim that the blade
paths through the water were the same whether the boat is moving or
not. Which is a lot like arguing that since 2+2=4 and 2x2=4, that
3x3=6. Perhaps a "reasonable" argument could be made, it would just
be wrong!
This is tiresome, let's try to stay consistent, or it gets difficult
for me to follow.
> > many words. There is some additional irony in that you do see "clear
> > differences" when this sort of thing is discussed regarding the Erg,
> > and we can't agree there either. [:o)
>
> Hmmm... ironic indeed. Although I thought you said you did agree with
> me (or part of what I said) after reading what I posted in that blog a
> few weeks back. Still waiting for that reply/response, too. Email
> would suffice.
>
> -Kieran
Perhaps our agreement was insignificant to me. i.e. I think that you
mentioned something about how the mass of the moving head wasn't all
that critical in creating a "rowing simulation", which is what I had
argued long ago (remember, it's why I have a problem with using the
"static" and "dynamic" labels for Ergs), so of course I agree with
things that I have said, otherwise I keep my yap shut (and think about
it some more), for the most part. [;o)
- Paul Smith
Kieran
> Big snip--
>
>>> I'm still finding your "quest" less than believable, when you still
>>> manage to post things like this (after the above posting):
>>> "On the other hand, at least during mid stroke almost all the force on
>>> the blade (and pin) is pointing in the direction of travel. Whereas
>>> earlier in the stroke at steeper oar angles when the blade is
>>> producing
>>> lift most effectively, much of the force is directed at the boat and
>>> less in the propulsive direction. - KC"
>>> You seem to be continuing to ask "Where have I gone wrong?", yet you
>>> are the one with just the education to figure that out, aren't you?
>>> I'm certainly not!
>> Which is why I did not ask you (that time), and do not believe you when
>> you claim I'm wrong. I HAVE asked you to show where I am wrong, and you
>> are not capable, you only are capable of claiming I'm wrong without any
>> reasonable explanation as to how or why.
>
> I really have tried, but you apparently do not appreciate the effort.
> Can you prove I'm wrong in the claim I'm making?
I have posted several links to sites which state that the net force on
the blade during the drive is nearly perpendicular to the blade cord.
That is all that's needed. If net force on the blade is perpendicular
to the blade, then there is a significant component of that force vector
pointing directly at the boat at steep oar angles. That's it, there's
nothing else to say really. Everything I know about the path of the
blade through the water, and how lift is generated, and every reference
I have read on the topic agrees with this. So, I have nothing else to
offer, unless someone can show me where my reasoning is wrong. If you
have done so, please re-state it, as I can think of nothing you've said
that indicates to me that net force on the blade is not perpendicular to
its cord. Your comments about back stays are pointless because back
stays are largely unnecessary, and most people could get by without
them. In fact, Vespoli delivered a brand new M2 pair (with wing
riggers) to me (my club really, but I was the coach using it) a couple
years ago, without back stays. I unwrapped the boat and riggers at US
Nationals in Indy, to find no back stays. I went to the Vespoli rep and
asked what was up, and he said they don't ship these boats with back
stays unless specifically requested, because they feel they are not
necessary. This was a mens heavyweight M2 2-, mind you... the same boat
they would have supplied to Olympic men had those men wanted a Vespoli.
So, the deflection of the pin is almost negligible.
Your thought experiment about a flexible pin is pointless unless you've
ever really rowed with such a pin. Regardless, since the force on the
pin is almost perpendicular to the oar shaft at all times ("almost"
because of the axial force from the rower pushing the button against the
gate) your "flexible pin" would deflect in a constantly changing
direction throughout the drive.
> Since you were unaware of my lnwledge of Ken Youngs Pages, perhaps you
> are unaware of this one, that discusses the "worrisome "pinch-point"".
> http://www.atkinsopht.com/row/bowangle.htm
I'm quite familiar with that page, and agree with its general point
about the pinch force. Note he does not say that it does not exist,
only that it does not waste any power. This is true because as Carl
pointed out, there's no work done by the pinch force, since there's no
deflection of the pin, and the pinch force on the other side of the boat
equally opposes it, so the boat does not move sideways, either.
Therefore since the applied force results in no motion, no work is done,
and therefore no power dissipated, and therefore there are no losses (no
waste) as a result of the pinch forces. But they do exist, and at steep
catch angles they are much larger than the forces pointing in the
propulsive direction.
> Please do not take this as any sort of wholesale endorsement of the
> entire Atkinsopht site, there is a lot there to disagree with, but
> this particular bit might be applicable to our current discussion.
> (His section regarding lift and drag doesn't seem to make sense to me,
> so you might like it a lot, with your advanced level of understanding
> and all.)
If his section on lift and drag doesn't make sense to you, where do you
get off making statements that imply you have a solid understanding of
lift and how it works? You spout off that "lift acts in the direction
we need it to" as if you're some expert on the subject. I don't claim
to be an expert but at least what I do say is based on textbook
knowledge of the phenomena of which I speak. You may think that your
observations back up your ideas of how something works, but don't forget
that two technical people here, one a fluid dynamicist just admitted to
not having realized that oars did not work primarily due to drag. You
may not see the pinch force, or its effects, or understand how or why it
is there, but that doesn't mean it's not significant and real.
>>> Though my suspicions are in the direction that you
>>> are not seeing the difference between the static and dynamic
>> The first stroke vs. any other stroke, is less important to me. I'd be
>> happy to ignore the case of the first stroke and just have you prove
>> your claims for the "dynamic" case. Can you do that?
>
> Apparently not to your satisfaction, no. I can/do prove it to my own,
> every time I am out in a boat with the tools in question at hand.
Great, then I ask politely and respectfully, please would you (again
since I've missed it sorry) state a detailed description of this proof
you see when you are rowing?
>>> situations. Since you have told me "They are not different." in so
>> Now who's misrepresenting whom? I stated very clearly that they are
>> similar, as in not very different. I never implied they were literally
>> identical.
>
> --yawn-- Have you already forgotten about your claim that the blade
> paths through the water were the same whether the boat is moving or
Please cut and paste where I said they are literally identical. I don't
think I ever said that. Much how you may have never said that the pinch
force doesn't exist, just that it's "so small as to be insignificant".
I did say that the SHAPE of the paths was VERY *similar*. But anyway,
let's let the case of the first stroke go by the way side. It doesn't
matter right now. I'm happy to focus on a moving-boat stroke.
> not. Which is a lot like arguing that since 2+2=4 and 2x2=4, that
> 3x3=6. Perhaps a "reasonable" argument could be made, it would just
> be wrong!
> This is tiresome, let's try to stay consistent, or it gets difficult
> for me to follow.
>
>>> many words. There is some additional irony in that you do see "clear
>>> differences" when this sort of thing is discussed regarding the Erg,
>>> and we can't agree there either. [:o)
>> Hmmm... ironic indeed. Although I thought you said you did agree with
>> me (or part of what I said) after reading what I posted in that blog a
>> few weeks back. Still waiting for that reply/response, too. Email
>> would suffice.
>>
>> -Kieran
>
> Perhaps our agreement was insignificant to me. i.e. I think that you
Well, it seemed significant to you at the time, and you said you were
going to post a follow up to my post.
-Kieran
Lawrence Edwards
news:1172274728....@8g2000cwh.googlegroups.com...
> Sorry to contradict you Lawrence, but you must be either mistaken or
> have rowed the only 2 resolute with enclosed bulkhead.
It was a couple of years ago, so perhaps my memory is faulty. Do Resolute
manufacter different specification 8s perhaps? The boat we used was a few
years old, so I would guess it was made around 2000.
KC
> On Feb 23, 2:07 pm, KC <kc_n...@sonic.net> wrote:
>> paul_v_sm...@hotmail.com wrote:
>>> On Feb 23, 12:10 pm, KC <kc_n...@sonic.net> wrote:
>>>>>> Well then your point was exactly wrong. At least if it relies on ME
>>>>>> being inconvincible, as I clearly am convincible when being convinced
>>>>>> to a correct conclusion. i.e. I was convinced that: Sliding Riggers do
>>>>>> not produce faster times over 2000m race courses by any inherent
>>>>>> benefit contained in the system.
>>>>>> - Paul Smith
>>>>> Glad to hear it. Maybe this means then that some day, some one will
>>>>> convince you that the net force on a blade is nearly always
>>>>> perpendicular to the shaft at virtually all instances during the drive,
>>>>> since that is a true statement.
>>> Oh, so now it's "nearly" is it?
>>> How about this then: "The propulsive force at the Pin is nearly
>>> always in the direction of hull travel, and that force is parallel to
>>> the force acting at the Blade." (Note, this is exactly in line with
>>> what I have been saying all along. Just a new bunch of words strung
>>> together.)
>> I don't disagree with that at all. The *propulsive* (component of)
>> force is by definition in the direction of propulsion. It is only one
>> component of the total force on the blade (and pin). The rest is
>> perpendicular to the propulsive force. What I *think* you have disputed
>> with me is that the resultant force (i.e. total, net, sum force) on the
>> blade is NEARLY perpendicular to the oar shaft at virtually all
>> instances during the drive. Is that a fair statement? I don't want to
>> misrepresent your claims.
>>
>> I would think you could come up with something more profound than "the
>> propulsive force is in the direction of hull travel."
>
> Be as semanitc pedantic as you want, I asked you which way the "soft
> pin" would bend and you didn't answer, you know what I mean. Or if
> you don't, I'm serioulsy not equipped to explain it to you.
>
>>>>> -Kieran
>>>> I'd like to also point out the irony in your last claim. "I am ...
>>>> convinceable [sic] when being convinced to a correct conclusion."
>>>> One can not be "convinced" to a conclusion they see as correct, for if
>>>> they see it as correct they need no convincing. So, that *you* see my
>>>> claims as incorrect should not affect your ability to be convinced,
>>>> unless you truly are "unconvinceable" (to continue to use our new non-word).
>>>> -KC- Hide quoted text -
>>>> - Show quoted text -
>>> Why in the heck do I try?
>> I hadn't noticed that you had.
>>
>>> You can't even stick with the word (real word) that I was using.
>>> I bothered to look it up, though one can only wonder why.
>>> inconvincible
>>> Main Entry: in搾on暇inc搏ble
>>> Pronunciation: "in-k&n-'vin(t)-s&-b&l
>>> Function: adjective
>>> : incapable of being convinced
>> Congrats on being a better speller than I. I concede and bow to your
>> spellingbeeness.
>>
>>> Though there doesn't seem to be the opposition term, "convincible".
>>> But it communicates the required meaning well enough.
>>> Yes, the irony is thick, but I think you miss it completely. At least
>>> I am convinced so.
>>> Socrates is rolling in his grave as this continues. My confidence in
>>> reaching critical mass is still high, hopefully a sign of my
>>> persistent optimism.
>> I don't know much about Philosophy. Feel free to flex your BA degree
>> and fill me in on the reference to Socrates here. Maybe it'll make you
>> feel even better... TWO areas where you are superior! Keep it up, I
>> shiver in the presence of your correctly spelled philosophical superiority.
>
> You'll surely appreciate that I was conferred a BS degree, rather than
> a BA. Though I like the attempted matriculatory swipe. And you calim
> to be the civil participant? Pshaw!
It was not a swipe at all. Some of my best friends have BA degrees!
;-) No seriously, I only guessed it was a BA, but it makes no
difference. I know an engineer who's ENGINEERING degree is a BA,
because the college he went to was a lib.arts college that happened to
have one or two very good engineering majors too (can't recall which
school now) but all degrees from the college were BA's because the GE
core was so lib.arts heavy. On the other hand, anyone graduating from
the Naval Academy gets a BS degree, even if they majored in English (for
example) because the core GE coursework is so science-heavy. So I
really wasn't taking a swipe. It was just a guess at the degree,
because I couldn't remember the exact major (Psych?).
>> The irony NOW, is that my mother was an English teacher and my father
>> has a BA in philosophy. :^P
>>
>> -Kieran
>
> Good humor there, I gotta admit.
Indeed. But really, what's the Socrates reference that you made? I'm
curious.
> Here is another bit of irony for you, that relates to what may
> eventually dawn on you as you discuss further with Carl. When the
> Sail is providing lift, the vessel is being dragged along. Now think
> of the Pin as a mast. [;o)
>
> - Paul Smith
>
I fully understand that analogy. But it seems that maybe you don't.
The difference is that if you took a moving rowing boat and just placed
the "sails" (blades) in the "blowing wind" (moving water) and held on,
you'd STOP the boat: no lift would be produced from the "blowing wind"
(water moving relative to the blades due to forward motion of the hull).
That was my main point in saying that it doesn't *really* matter if
the boat is moving or not. (Small differences occur, like how quickly
appreciable lift builds up from a dead stop: "quickly" according to
Carl, "almost instantly" according to me) The reason it doesn't matter
if the boat is moving or not, is because except for some (relatively
small in the grand scheme) slippage of the blade through the water, the
blade does not move relative to the water in the direction of travel.
The blade PRIMARILY moves out away from, and back in toward the hull
(transverse axis). So as I've said there ARE small differences between
moving boat strokes and first strokes, but they are small, since in both
cases, the blade doesn't move (much) relative to the water in the travel
axis.
But none of this talk directly addresses the primary misconception that
you have, or rather disagreement that we have, which is (correct me if
I'm wrong) that during the first 1/3 of the stroke
A) As I have said, the net force on the blade is virtually perpendicular
to the shaft.
B) Or as you see it, that the net force on the blade is nearly parallel
with the hull's long axis.
I'd like it if someone could explain which one of those statements is
most correct and why, as you and I just seem to go back and forth with
no progress, and if I *am* wrong, I'd really like to understand how/why
and no offense, but you don't seem to be able to explain it well enough
to convince me.
-Kieran
Walter Martindale
> paul_v...@hotmail.com wrote:
>> On Feb 23, 2:07 pm, KC <kc_n...@sonic.net> wrote:
>>> paul_v_sm...@hotmail.com wrote:
>>>> On Feb 23, 12:10 pm, KC <kc_n...@sonic.net> wrote:
Not much of a tangent from Vespoli Boats, which started this thread, is it?
Volker talked about lift forces on the blade during his presentations a
the Canuck NCC in January. In Vancouver a few years ago he spoke about
the similarity between the path of the blade in the first part of the
drive "at speed" to the path of the blade in speedskating (both push
outwards, generating motion towards the finish line). Very small force
component opposite direction of travel, large force component almost
perpendicular to direction of travel. Large movement component in
direction of travel, very small motion "outwards" - both "blades"
generate a lot of forward motion. Another analogy could be when you
squeeze a wet bar of soap between your fingers, and it squirts out
perpendicular to the direction of your squeeze. If the water is your
fingers, the blades / boat are the bar of soap and the reaction force of
the water is "inwards" but with a small component in the direction of
travel, does the boat "squirt" towards the finish line like that bar of
soap, or like the stone pinched under a tire?
\Gotta go stretch my neck - darned pinched nerve acting up..
W
paul_v...@hotmail.com
<wmart...@telusSPAMSTOPplanet.net> wrote:
I've seen the speedskating analogy before, but could never relate it
to rowing, other than perhaps a similar type of transistion from what
is done in the early stages of the start being quite different from
what is happening once under way. There is no "lift" being generated
with a skating blade tracking a course on the ice, though it does
trace a path refered to as a "tractrix" described on the Atkinsopht
web site that discusses lift. The skating and soap squeezing
exercises may well be analogous, since there is no "lift" component in
either of them, and all surfaces being engaged are more solid than
fluid (no I do not care about the thin layer of water that is being
created by heat and refrozen quickly, as that is a lubrication issue
rather than hydrodynamic effect.).
Kieran states above, with contextual qualifications that I think can
be removed while the declaration maintains his meaning, "the blade
does not move relative to the water in the direction of travel.".
Which especially during the first 1/3 of the stroke is completely
wrong. Perhaps that is the root of the error that is being made.
The blade clearly travels in the direction of travel for a meter or
more in the early stages of the drive. That is, if it has been placed
in the water at the apex of reach. It will also continue to advance
in the direction of travel until the blade stalls, regardless of how
much force is exterted at the handle. I can also argue that due to the
flexibility of the shaft, that a very quick rise in handle force would
keep the blade advancing for an even longer period of time due to the
increased angle of attack that early flex would cause. Ever notice
that an increase in handle pressure seems to increase the time in the
water even though the system is moving faster. I used to think this
was an illusion caused by the onset of stress, much like time
dialation when encountering danger, but it does in fact seem to be
quite real when focus is directed specifically at observing it. Some
accurately gathered video would be an excellent way to prove this,
however that is more difficult to arrange than I thought it might be.
I did it fairly often when I was coaching, but then I was in control
of the equipment and situation. Anyway, Ken Youngs pages contain all
the information required to show the blade advance in the direction of
travel, he calls it displacement on his site and makes a mistake in
the total distance because of an oversite in accounting for the very
end of the drive.
- Paul Smith
KC
You're right, that was a very poor choice of words on my part. Not only
does it travel a bit forward, it also travels a fair amount back when
stalled. As I pointed out recently to Martin, engineers often over
simplify to get to the crux of the situation, especially for
"estimations". My point is that the amount it travels forward is not as
important as the distance it moves along it's path, for that is where
the lift-inducing flow comes from. In other words, it's traveling on a
curved diagonal (at first) path away from the boat. It is hardly "a
meter or more" though. Again I reference Ken Young's image:
http://tinyurl.com/2cn3ma
The vertical distance (in the bow-ward direction) from the first red dot
to the highest red dot (that one that is a little further out than the
rest) is not even the length of the scull's blade. I don't know what
kind of sculls you use, but mine do not have meter-or-more-long blades.
;^)
> It will also continue to advance
> in the direction of travel until the blade stalls, regardless of how
> much force is exterted at the handle.
With zero force, there will be no lift and no stall, and the blade will
probably not translate much at all in the travel-direction.
Anyway, let's try not to keep beating each other down over semantics.
The POINT was (poorly made by me) that the majority of travel is OUTWARD
away from the hull, and that the blade doesn't move a whole lot (in the
grand scheme of things... i.e. in comparison to the distance the boat
travels in a stroke) in the direction of travel.
To reference another picture, based on Ken Young's work, take a look
again at the C2 site on the topic:
http://www.concept2.com/us/products/oars/testing/default.asp
and more specifically at this image:
http://www.concept2.com/us/images/products/op1e240.gif
That one is a great image of four consecutive 15fps frames of a blade's
position during the early phases of a stroke. If the black background
is still water, then the direction of flow of water about the blade is
clear, because ONLY the blade is moving (which is what I was trying to
get at before). It is very clear in that image that the AoA is quite
small. And, as I've said numerous times, if you take the net resultant
vector of lift and drag (the diagonal of a rectangle formed by L & D in
that image) is very nearly perpendicular to the blade. Which, I'd also
like to point out is just as it should be. The L vector is roughly
perpendicular to the path of the blade; actually it is perpendicular to
the relative stream flow, which again comes for the DEFINITION OF LIFT.
But if I understand Paul correctly, he believes that the L vector in
that image should be pointing almost directly to the right (maybe
slightly downward). If that is what you think Paul, what causes this
magical shift in the Lift vector? Where do these other components of
force come from that are not depicted in the above linked image?
Please please please please *PLEASE* enlighten me. (pretty please with
sugar on top)
-Kieran
paul_v...@hotmail.com
This is correct, I was remembering the ken Young Graphic as having a
1m scal but it is only half of that, so the forward track in his case
was 0.4m, though that would change based on the speed of travel. The
degree of "backward" travel is also subject to controllable variables
(i.e. blade depth) and though it will happen, I do not believe that it
must happen to the extent shown in Ken Youngs graphic.
Here is what I would say is going to happen with the blade:
The drive is divided in quarters.
1- blade advances
2- blade hold position
3- blade retreats
4- blade advances
> > It will also continue to advance
> > in the direction of travel until the blade stalls, regardless of how
> > much force is exterted at the handle.
>
> With zero force, there will be no lift and no stall, and the blade will
> probably not translate much at all in the travel-direction.
Well, that would depend on the catch angle, if the blade were parallel
with the boat, then it would travel quite a bit in the direction of
travel without any handle pressue, though there would also be no input
to accelerate the system either. But there must be a continuum of
advance distance based on catch angle, or maybe more accurately angle
of attack.
> Anyway, let's try not to keep beating each other down over semantics.
> The POINT was (poorly made by me) that the majority of travel is OUTWARD
> away from the hull, and that the blade doesn't move a whole lot (in the
> grand scheme of things... i.e. in comparison to the distance the boat
> travels in a stroke) in the direction of travel.
>
> To reference another picture, based on Ken Young's work, take a look
> again at the C2 site on the topic:http://www.concept2.com/us/products/oars/testing/default.asp
> and more specifically at this image:http://www.concept2.com/us/images/products/op1e240.gif
Yep, I've seen everything that C2 has published on the matter also.
Unfortunately it does not sway what I believe to be happening.
> That one is a great image of four consecutive 15fps frames of a blade's
> position during the early phases of a stroke. If the black background
> is still water, then the direction of flow of water about the blade is
> clear, because ONLY the blade is moving (which is what I was trying to
> get at before). It is very clear in that image that the AoA is quite
> small. And, as I've said numerous times, if you take the net resultant
> vector of lift and drag (the diagonal of a rectangle formed by L & D in
> that image) is very nearly perpendicular to the blade. Which, I'd also
> like to point out is just as it should be. The L vector is roughly
> perpendicular to the path of the blade; actually it is perpendicular to
> the relative stream flow, which again comes for the DEFINITION OF LIFT.
>
> But if I understand Paul correctly, he believes that the L vector in
> that image should be pointing almost directly to the right (maybe
> slightly downward). If that is what you think Paul, what causes this
> magical shift in the Lift vector? Where do these other components of
> force come from that are not depicted in the above linked image?
That looks to be shoing the reaction force off the face of the
bladeand to not include lift at all. Perhaps this is similar to
another argument point between the relative non-advantage of an Erg
+Slides over the grounded Erg. If there is a difference in having to
more the body mass back and forth Vs the Erg mass back and forth it is
such a small thing that it makes no difference in the actual practice
of the two situations. The relatively small reaction off the face of
the blade (due to drag) can be dispersed as you suggest, they are
merely insignificant. The lift, which is the major force being
capitalized on to overcome the drag of the system is directed exactly
opposite that drag (in the direction of travel).
> Please please please please *PLEASE* enlighten me. (pretty please with
> sugar on top)
>
> -Kieran
I would if I could, but you have already resigned yourself to the fact
that I am probably not the one to be able to do that. You tell me
that at 30deg catch angle, 87% of the force is being directed
perpendicular to the hull, yet that can not apparently be measured by
instrumenting a rigger (as Carl mentioned). I think it could be
measured with a instrumented pin, which is what at least one system
I've seen, appeared to do.
And just to show that several strokes must take place prior to the
forward velocity truly building up to something meaningful.
http://neutrino.phys.washington.edu/~wilkes/post/temp/phys208/shell.acceleration.html
Scroll almost to the bottom.
- Paul Smith
Walter Martindale
> On Feb 26, 12:33 pm, Walter Martindale
> <wmart...@telusSPAMSTOPplanet.net> wrote:
>> KC wrote:
>>> paul_v_sm...@hotmail.com wrote:
>>>> On Feb 23, 2:07 pm, KC <kc_n...@sonic.net> wrote:
>>>>> paul_v_sm...@hotmail.com wrote:
>>>>>> On Feb 23, 12:10 pm, KC <kc_n...@sonic.net> wrote:
>> W
>
> I've seen the speedskating analogy before, but could never relate it
> to rowing, other than perhaps a similar type of transistion from what
> is done in the early stages of the start being quite different from
> what is happening once under way. There is no "lift" being generated
> with a skating blade tracking a course on the ice, though it does
> trace a path refered to as a "tractrix" described on the Atkinsopht
> web site that discusses lift.
Perhaps it's not "just" lift going on - I'm not dismissing the lift, as
I believe it's there, too, but something that's independent of the lift
(or perhaps dependent on it being there but hard to describe) is the
**small lateral movement component, big forces, with small forces in the
direction of the finish line, and big movement** analogy. Speedskaters,
when getting tired, are said to be pushing "back" when all their
coaching is to push sideways. Scullers go faster who have long catches,
and when they re-rig to go shorter, they go slower (I'm not naming names
because one in particular who did this is still around). I can't explain
this with drawings, previous research from others, equations, and don't
care to - I have this gut feeling that it works, when I explain it to
kids that I'm coaching, and they get it, and then they get longer at the
catch, they go faster, feel the blades getting "sticky" in the early
part of the drive, and, well, go faster (i.e., I have to throttle up in
the motor boat to keep with them)
Example - was coaching a young woman who'd been out of the 1x for a
while - she was rowing quite short at the catch (esp. for someone over
6'tall). After being asked to reach wider at the catch, she looked at
her speedcoach monitor and her eyes bugged out - she'd knocked 15 s off
her /500 number without pulling harder, just by getting longer/wider
reach, tighter angles, more "gearing" or more "lift" or more of both, or
whatever.
See - we're all a little different - you coach people one way, it works,
they get faster. KC coaches people another way, it works, they get
faster. I coach people yet a different way, it works, they get faster.
Well - that's the idea, anyway. I'm not saying "you're wrong" or
anything like that because it doesn't matter enough to me to get into an
argument or to even think hard enough about it to figger out if you're
wrong, or right, or leftist, or, or... 8-) I've just put my oddball
take into the discussion (which, by the way, started as a thread on
"vespoli boats" or something like that...) Bedtime for bonzo - well,
actually dinner time - Lake Karapiro is just now emptying of schoolkid
rowing shells from the Cambridge area, and I'm going to go look for used
cars, kitchen appliances, and beds for when I get to move in next week.
Walter
Opinions expressed here are mine, and not reflective of any policy of my
employer.
Paul
I am quite happy to admit that I am being really stupid here, but
surely what ever lift is created by the spoon as it moves sidways
(relative to the boat) in the water is pushing the spoon towards its
back side, ie towards the bow. This seems to me to be the opposite of
what we want, ie lift is our enemy not our friend. Please explain in
very sinple terms what is actually going on?
Ewoud Dronkert
> surely what ever lift is created by the spoon as it moves sidways
> (relative to the boat) in the water is pushing the spoon towards its
> back side, ie towards the bow. This seems to me to be the opposite of
> what we want
No, it's exactly what we want. If you hold the oar and someone else
pushes the spoon towards the bow, the boat moves forward.
--
E. Dronkert
kda...@kidare.com
You want the spoon to act as solid a fulcrum as possible for the
effort of the rower to be converted to load on the pin. So you don't
want it to move stern-wards as this would reduce the load on the pin
and thus be a waste of effort. In the first part of the stroke, it is
partly (largely?) the low pressure on the back side of the spoon that
provides the solidity we want. This is why it is important to keep the
blade buried, to keep the low pressure up for as long as possible.
That's my simplistic and not particularly technical response. Hope
it's useful.
Kit
Paul
That makes perfect sense, the lift opposes the slip. I had assumed
that people were arguing that the lift had a direct propulsive effect
on the boat, which was what was confusing me.
Ewoud Dronkert
> I had assumed [...] that the lift had
> a direct propulsive effect on the boat
Correct assumption.
--
E. Dronkert
kda...@kidare.com
> Thank you
>
> That makes perfect sense, the lift opposes the slip. I had assumed
> that people were arguing that the lift had a direct propulsive effect
> on the boat, which was what was confusing me.
>
> > You want the spoon to act as solid a fulcrum as possible for the
> > effort of the rower to be converted to load on the pin. So you don't
> > want it to move stern-wards as this would reduce the load on the pin
> > and thus be a waste of effort. In the first part of the stroke, it is
> > partly (largely?) the low pressure on the back side of the spoon that
> > provides the solidity we want. This is why it is important to keep the
> > blade buried, to keep the low pressure up for as long as possible.
>
> > That's my simplistic and not particularly technical response. Hope
> > it's useful.
>
> > Kit
That's correct. The benefit is mainly indirect than direct. However,
there's not necessarily a clear line between the two. Considering that
at extraction the spoon is a few inches further bow-ward than when
inserted, there is a slight direct propulsive effect. Over 2000m, it
might add up to about a length of an 8+. So, with Olympic finals being
won by < 0.5 seconds, getting better hydrodynamic lift than your
opponent could be decisive, whether by altering catch angles or better
blade design.
Kit
Paul
more sense.
On 27 Feb, 13:03, Ewoud Dronkert <firstn...@lastname.net.invalid>
wrote:
Paul
small effect on the boat speed. Some of the discussion on the subject
seems to imply that it is the main effect, at least in the first part
of the drive, this is what didn't make sense to me.
> Kit- Hide quoted text -
bobmcm...@comcast.net
> On 27 Feb, 11:47, "Paul" <pgosl...@hotmail.co.uk> wrote:
>
Not to further complicate the issue, but if you drop the blade in at
the catch and apply no pressure, it ends up nearer the finish line
than it would have if you applied pressure (since there is no slip).
But since there's no pressure on the blade, and consequently no angle
of attack, there's also no lift. So it seems to me that the blade
ending up closer to the finish line than it was at the catch is not
very convincing evidence of lift. If I tied a beach ball to the end of
a shaft and dropped it in at the catch, it would also describe the
same tractrix. I'm having trouble sorting this all out.
Bob
Paul
Bob
Not sure I understand your point, but try this. While stationary in a
single pull, one blade in a couple of feet. With it square in the
water, parallel to the boat, push to outwards as fast as you can. In
theory this should creat lots of lift, and move the boat. In fact it
does nothing, which was the origin of my lack of faith tin the whole
lift thing.
Ewoud Dronkert
> How? That was kind of my question.
How what?
(A: Top posting.
Q: What's the most irritating thing on usenet?)
But to answer, see my first reply in this thread d.d. Tue, 27 Feb 2007
11:26:43 +0100.
The oar is a simple lever. Doesn't matter if the spoon is fixed and
the handle moves bowward, or the other way around, or both move
bowward; the result is that the pin moves bowward == propulsion.
--
E. Dronkert
Paul
wrote:
Ok I think I am almost there! The spoon creates a very small amount of
lift, this counteracts some of the slippage of the blade, so the net
effect is less slippage, so the boat is propelled more efficiently
than if no lift was created. A bit like an aeroplane on a glide path,
the wing is creating lift to partly counteract gravity. The lift does
not make the plane go up, but slows its descent.
KC
Amazing: we both conceded mistakes of our own and corrections by the other!
> degree of "backward" travel is also subject to controllable variables
> (i.e. blade depth) and though it will happen, I do not believe that it
> must happen to the extent shown in Ken Youngs graphic.
Well that's a debate not worth anyone's time until someone does some
testing, or CFD analysis or something. Increasing the depth of the
blade during stall will minimize aeration of the water, and thus
minimize slip. But the extent to which it is minimized is very
difficult to guess or quantify without some complicated computations or
testing. As an example, an aircraft wing could be considered to be
"fully immersed" in it's fluid. There is no analog to the air/water
interface that allows aeration which effectively decreases the density
of the water (to put it roughly). Yet, when an aircraft stalls, it
falls. When the blade stalls, it will slip regardless of depth.
Arguing the extent to which greater depth minimizes this is futile
without proper analysis and testing.
> Here is what I would say is going to happen with the blade:
> The drive is divided in quarters.
> 1- blade advances
> 2- blade hold position
> 3- blade retreats
> 4- blade advances
>
>>> It will also continue to advance
>>> in the direction of travel until the blade stalls, regardless of how
>>> much force is exterted at the handle.
>> With zero force, there will be no lift and no stall, and the blade will
>> probably not translate much at all in the travel-direction.
>
> Well, that would depend on the catch angle, if the blade were parallel
Again this is a futile argument. Lift is not produced without an
applied resistive load (Newton #3). If you take an airfoil and drop it
in a stream, without holding it steady it will just flip flop and fall
or blow away in the direction of flow. In a wind tunnel, the airfoil
x-section being tested must be fixed firmly to the transducer rig, or it
will just blow away. This fixation is replaced by gravitational force
on the aircraft in actual flight. In rowing, take a 2x for example,
have stroke row, and bow try to place the blades in the stream without
any force applied. Almost impossible. If bowman uses a little force,
to make the catch, then lets go, you might get an example of what you're
envisioning, but some force has to be applied: at first the rower's
hands do it, then if he is able to let got and move out of the way, the
blades might trace a quasi-normal path, but even then they are likely to
succumb to drag and slide up through the oarlocks. If you then try to
hold on through the stroke, but don't pull, you're likely to apply SOME
force, and even if it's only axial against the button/gate, the motion
of the boat caused by the other rower will apply some force to the pin
and thus to the blade. So no matter how you slice it, it's really not
possible to get the blade to travel it's normal path without some force
applied. This is because an unloaded foil produces no lift.
> with the boat, then it would travel quite a bit in the direction of
> travel without any handle pressue, though there would also be no input
> to accelerate the system either. But there must be a continuum of
> advance distance based on catch angle, or maybe more accurately angle
> of attack.
Don't confuse catch angle with angle of attack. The AoA is mostly
dependent on how hard the oar is being pulled upon, not by how steep the
catch angle is.
>> Anyway, let's try not to keep beating each other down over semantics.
>> The POINT was (poorly made by me) that the majority of travel is OUTWARD
>> away from the hull, and that the blade doesn't move a whole lot (in the
>> grand scheme of things... i.e. in comparison to the distance the boat
>> travels in a stroke) in the direction of travel.
>>
>> To reference another picture, based on Ken Young's work, take a look
>> again at the C2 site on the topic:http://www.concept2.com/us/products/oars/testing/default.asp
>> and more specifically at this image:http://www.concept2.com/us/images/products/op1e240.gif
>
> Yep, I've seen everything that C2 has published on the matter also.
> Unfortunately it does not sway what I believe to be happening.
Yes, indeed that is unfortunate, because that image is a VERY good
representation of what's actually going on.
>> That one is a great image of four consecutive 15fps frames of a blade's
>> position during the early phases of a stroke. If the black background
>> is still water, then the direction of flow of water about the blade is
>> clear, because ONLY the blade is moving (which is what I was trying to
>> get at before). It is very clear in that image that the AoA is quite
>> small. And, as I've said numerous times, if you take the net resultant
>> vector of lift and drag (the diagonal of a rectangle formed by L & D in
>> that image) is very nearly perpendicular to the blade. Which, I'd also
>> like to point out is just as it should be. The L vector is roughly
>> perpendicular to the path of the blade; actually it is perpendicular to
>> the relative stream flow, which again comes for the DEFINITION OF LIFT.
>>
>> But if I understand Paul correctly, he believes that the L vector in
>> that image should be pointing almost directly to the right (maybe
>> slightly downward). If that is what you think Paul, what causes this
>> magical shift in the Lift vector? Where do these other components of
>> force come from that are not depicted in the above linked image?
>
> That looks to be shoing the reaction force off the face of the
> bladeand to not include lift at all. Perhaps this is similar to
What reaction force might you be referring to? The reaction force is
equal and opposite to Lift, and is provided by the rower at the handle.
There are only two forces not shown in that image:
http://www.concept2.com/us/images/products/op1e240.gif
1)In opposition to Drag, there is thrust (provided by the rower) and
2)In opposition to Lift there is "weight" (to use the aircraft analog)
also just referred to me as the reaction force, also provided by the rower.
(BTW, the yellow vectors drawn most likely apply to the third frame in
the sequence.)
I bet if you took that image, removed the yellow stuff, and handed it to
any fluid dynamicist, and asked them to draw in the appropriate vectors
of lift, drag, thrust and "weight" (explaining to them the order of the
sequence v. time) you'd get a drawing of vectors just like (or very
similar to) the yellow ones.
Would you maybe take that image and use Paintbrush or Photoshop, and
sketch in what you think are the "correct" force vectors acting on the
blade, and why?
> another argument point between the relative non-advantage of an Erg
> +Slides over the grounded Erg. If there is a difference in having to
> more the body mass back and forth Vs the Erg mass back and forth it is
> such a small thing that it makes no difference in the actual practice
> of the two situations. The relatively small reaction off the face of
> the blade (due to drag) can be dispersed as you suggest, they are
> merely insignificant. The lift, which is the major force being
> capitalized on to overcome the drag of the system is directed exactly
> opposite that drag (in the direction of travel).
>
>> Please please please please *PLEASE* enlighten me. (pretty please with
>> sugar on top)
>>
>> -Kieran
>
> I would if I could, but you have already resigned yourself to the fact
> that I am probably not the one to be able to do that. You tell me
> that at 30deg catch angle, 87% of the force is being directed
cos(30) = 0.866025 to be precise. :^)
> perpendicular to the hull, yet that can not apparently be measured by
> instrumenting a rigger (as Carl mentioned). I think it could be
I don't think Carl said it could NOT be measured, just that it would be
difficult.
> measured with a instrumented pin, which is what at least one system
> I've seen, appeared to do.
Do you have any reports from that system? Usually, when you put strain
gauges on something, you already know the direction of force applied
(and thus the direction of deflection of the beam) and you apply the
strain gauges accordingly. With a rowing pin, the direction of the
force applied changes constantly. If they instrumented the strain
gauges to measure deflection in the direction of travel only, then
that's the only component force they would see. You could set up strain
gauges at 90deg to each other. That would record the two components of
force necessary.
>
> And just to show that several strokes must take place prior to the
> forward velocity truly building up to something meaningful.
> http://neutrino.phys.washington.edu/~wilkes/post/temp/phys208/shell.acceleration.html
> Scroll almost to the bottom.
"Something meaningful" is relative. Yes it takes several strokes to get
to the average race speed, but it doesn't take much boat velocity at all
to get an appreciable AoA on the blade at the catch.
-Kieran
KC
No, in theory pushing the blade out away from your boat like that should
do nothing, because you've supplied no reaction force to lift. An
unloaded foil creates zero lift. Now, if you did your experiment but at
the same time pulled on the oar, then you'd move the boat forward, as
lift would be created. The other thing you've not done in your
experiment is supply a proper angle of attack. You're essentially
giving the blade an AoA of zero, and thus virtually zero lift is produced.
Ever put your hand out of the car window and varied the angle of your
palm to the direction of flow? If you make your hand level with the
ground, nothing happens. Tilt it up, and away flies your hand. Balance
the angle just right, and you can "fly" your arm, where lift will
support the weight of your arm, and the only work you have to do is hold
the right AoA.
-Kieran
Carl Douglas
There's a bit of a problem with that term "lift". It has
non-engineering meanings & can be misread to suggest that, if you have
lift over a surface, then that lift sucks that surface along for free.
And hearing that lift is what moves sailboats may consolidate the notion
of "summat for nowt".
I'll try to clear up the potential confusion without (I hope!) further
confusing.
Lift is a force dependent on fluid flow. It arises when a fluid (gas or
liquid) flows along & over a (relatively) streamlined object of suitable
shape. But to make a fluid flow, or the object to move through a fluid,
always uses energy.
With an object anchored in a flow the energy to sustain flow comes from
external sources, fondly termed "renewable energy", which drive winds &
water. But if an oar is part of a system you're moving through water,
then you must provide that energy or all relative motion will cease.
Flowing fluids _can_ power unanchored systems if another solid or fluid
medium provides the necessary reaction force, as with a land-yacht or a
sail-boat. Here the wheels on the ground, or the immersed hull or
keel's resistance to side-slip, provide forces to counter the less
useful components of the lift forces. You end up with a pair of
nearly-opposed side forces, but acting slightly out of line, & the
vessel is propelled forward like the squeezed pea or soap bar described
elsewhere.
Note here that the lift forces in some situations greatly exceed the
force component which actually moves the system, just as forces
generated within a car gearbox greatly exceed those at the engine or
road surface.
But why is lift so desirable?
All fluids are soft & rather unresisting. They can't support a static
object, which must thus (unless buoyant) fall away under gravity or slip
under applied load. If you pull something face-first in water it will
start to slip & as it accelerates will dissipate large amounts of energy
in turbulence. That is called "stall", and the resistance produced is
"drag". All objects moving in a fluid experience drag, but drag is
hugely sensitive to shape, velocity, fluid density & other factors
Lift is special. It has the magical ability to make a fluid appear very
hard & resistant. The absolute prerequisite for generating lift is a
relative edge-on motion between fluid & object (the object then being
termed an aerofoil or a hydrofoil).
When fluid flows around a body, then depending on shape & alignment that
flow goes faster over some areas than others. When fluid accelerates
thus its internal pressure falls - the energy to accelerate it (kinetic
energy) being won by exchange for a loss of the fluid's internal,
pressure energy. It is that fall in pressure, exerted over an area,
which causes lift.
When a wing flies, its shape causes flows which induce a fall in
pressure over the upper surface. There's little or no increase in
pressure below the wing, so it's the difference between local
atmospheric pressure & that lower pressure above the wing which creates
the lift that keeps the plane up. The drag force on the plane (in
direction of motion) is much much less than the lift force, & the
engines have only to generate enough trust to overcome the drag force.
So, thanks to aerodynamic lift, 300 tons of plane can be kept aloft, and
moving fast, by a few tens of tons of thrust, whereas it would take a
full 300 tons of vertical thrust to make that plane hover.
The reason that this pressure drop above the wing can support the plane
so economically while it is so costly to make it hover is due to the
huge amount (weight) of air which is slightly influenced by the flying
plane, compared with the relatively small amount surrounding & affected
by the hovering plane. There must (Newton) be an equal & opposite
reaction to an applied force. The flying plane briefly tugs locally on
all of the air below which it passes & those brief tugs, being so short,
induce relatively slow vertical motion in every affected air molecule.
The important thing to understand here is that, while the reaction force
is based on the momentum given per unit time, which is proportional to
air mass x velocity change, the energy given to that air is proportional
to air mass x square of velocity. For the fast-moving plane the air
mass is huge & the velocity increment correspondingly small, whereas the
velocity increment for a hovering plane is very high because the mass of
air affected is so much less. It is that extra multiple of a much
larger velocity, despite the smaller mass, which makes hovering so much
more costly than using fluid-dynamic lift. And it is the ability to
spread load so transiently over so much fluid which makes lift such an
effective & efficient way to resist a load, & makes air then seem so
hard & supportive.
But remember: you only get lift a) when you have the right shape & flow
conditions (lift will fail otherwise) & b) when you apply a load to the
foil. Kieran, too, has pointed out that an unloaded foil generates no
lift - because it self-adjusts to that angle of attack which gives no
lift (No, I never said this was easy!). So an unloaded oarblade
generates no lift, period. And there are no free lunches from lift in
rowing, just smaller losses. Indeed, if you could abstract free energy
from lift in rowing you would actually be sailing, which isn't really
allowed (although it doesn't say so in the rules).
In conclusion, what lift does is to give you a very solid connection
with the water - one which is effectively slip-free (because the loading
on an oar in the lift phases is ridiculously small for its area). So it
is like slotting your blade into a virtual tramline in the water but
without the sparks. However, in the middle of the stroke you nolonger
have suitable conditions to sustain lift, so there you are dependent on
induced drag to resist slippage & the blade can never be large enough,
especially if it is not well buried either. Therein lies a
little-understood compromise in oar design which is the subject of any
amount of rowing & promotional bullshit!
As regards the points made about the path of the oarblade through the
water (that there tractrix!):
The unloaded oar will swing right through the stroke if the boat is kept
moving &, since the difference is so slight between angle of attack when
unloaded & when under load, there is scarcely any difference between the
paths taken in the first part of the stroke. Similarly for the last
part of the stroke. The only difference between loaded & unloaded oars'
paths is in the mid-stroke, where the loaded oar slips by a large amount
(as much as 30cm/1ft).
We owe very much to the work of, among others, the late, great Ken
Young, but I would humbly suggest that the forward displacement of the
oar tip (which is much less than the forward displacement of the blade
root) is not an indicator of the presence of lift but a consequence of
the stroke's mechanics. In short, the rowing stroke is too darned
complex for such a deduction. However, in kayaking the lift generated
in the type of stroke taken with the appropriately-named wing paddle (&
for this I'm much indebted to another very expert RSR contributor &
paddler who may be reading this) does result in its forward motion
during the stroke, & the degree of this forward displacement does
apparently correlate with the paddler's effectiveness.
Phew!
Carl
--
Carl Douglas Racing Shells -
Fine Small-Boats/AeRoWing low-drag Riggers/Advanced Accessories
Write: The Boathouse, Timsway, Chertsey Lane, Staines TW18 3JY, UK
Email: ca...@carldouglas.co.uk Tel: +44(0)1784-456344 Fax: -466550
URLs: www.carldouglas.co.uk (boats) & www.aerowing.co.uk (riggers)
Paul
>
> - Show quoted text -
Ok I understand your point about angle of attack, but I don't
understand your first point. Surely lift is produced when the foil
moves through the water. That should push it towards the bow, but it
can't because I have got hold of the blade handle, I am in effect
creating your reaction force so there should be some effect, unless
the lift is so small as to lost in all the variables, my not totally
ridgid arm etc. I guess what I am getting at is how big a component of
the boat speed is created by lift? I can't see it being very much,
though as already pointed out, even 1% is a lot in terms of race
results. Some of the discussion on this subject now and in the past
seems to imply it is the main propulsive effect, I am assuming you are
not saying that?
Carl
Paul -
I don't know if I'm missing your point, but this might help:
At the catch the oar is angled forward, it is only rotating slowly about
the pin & the boat is moving forward. So whatever else happens, when
the blade enters the water there is a large component of velocity
driving the blade tip-first into the water. So the prerequisite for
hydrodynamic lift is met - there is lengthwise flow along the blade.
Taken to the extreme - oar parallel with the boat - the water would be
flowing along the blade from tip to root at boat speed, Taken at 30
degrees to the boat & held there (I know it can't be done, but let's
assume), the water would be deflected by the blade towards the boat,
causing it to move even faster along the blade (inversely proportional
to the cosine of the angle to the boat).
In reality the oar swings sternwards, as it must, which slightly reduces
the water velocity along the blade. And when you load it (& not before)
hydrodynamic lift will develop. And just as when you pull against a
fixed object, the lift is in exact match to the load you apply, for as
long as that the blade continues to slice-tip-first through the water
(for at least the next 30 degrees).
[I know that it requires mental contortions to visualise how the water
flow appears from the blade's point of view, but that's what one has to
do to understand the hydrodynamics of blade function.]
Boat speed is not "created by lift". That was the point I had attempted
to make in my lengthy posting of last night. What lift does is to
eliminate face-first blade slip - that's it & that's all - bringing
those phases of the stroke where lift does operate very much closer to
100% propulsive efficiency than the mid-stroke can ever be.
Lift is not a propulsive effect, it is a phenomenon which prevents or
limits slip. Slip is what reduces the propulsive efficiency by throwing
away a substantial proportion (50% to 40%) of the work you do. So an
understanding of where in the stroke hydrodynamic lift is most available
(as opposed to the mid-stroke region where there is only loss-making
drag & slip) should be of real value to any well-informed coach or rower.
Cheers -
Paul
> URLs: www.carldouglas.co.uk(boats) &www.aerowing.co.uk(riggers)- Hide quoted text -
>
> - Show quoted text -
Thanks Carl (and others) I think I more or less understand it now.
Though I am not sure quite how, if at all, I am going to be able to
use this knowledge to make my boat go faster. Its interesting none the
less.
Carl
>
>
> Thanks Carl (and others) I think I more or less understand it now.
> Though I am not sure quite how, if at all, I am going to be able to
> use this knowledge to make my boat go faster. Its interesting none the
> less.
>
>
What it tells you is that the ends of the stroke are really important.
You can only expend 100% effort, but you are able to redistribute your
effort within the stroke cycle to get a better result from that same effort.
Let's do the maths crudely:
Say the stroke has 3 phases (that's already wrong, but keep it simple!).
Let the efficiency, E1 & E3, of the catch & finish phases be 100% (wrong
again, but keep simple).
Let the efficiency of the middle phase, E2, be 50% - i.e. half of what
you invest there goes to waste.
Let work done in each phase be w1, w2, w3
So stroke efficiency, Es, =(E1 x w1 + E2 x w2 + E3 x w3)/(w1+w2+w3)
If, as might be, a crew invests as follows:
w1 = 20%, w2 = 70%, w3 = 10%; then Es = 65%
But suppose they re-distribute as follows:
w1 = 25%, w2 = 60%, w3 = 15%; then Es = 70%
So those small adjustments in how you apply the same amount of
work/stroke will improve overall efficiency from 65% to 70%.
A propulsive efficiency improvement on that scale will give you an
increase in boat speed of ~2.5% which, when translated into winning
margins, is huge (50m per 2000m).
Sure, the numbers are very rough estimates, but the message is unambiguous.
HTH
Carl
--
Carl Douglas Racing Shells -
Fine Small-Boats/AeRoWing low-drag Riggers/Advanced Accessories
Write: The Boathouse, Timsway, Chertsey Lane, Staines TW18 3JY, UK
Paul
> URLs: www.carldouglas.co.uk(boats) &www.aerowing.co.uk(riggers)
Get the catch in as early as possible, stay smooth in the mid section
and as a former coach once told me tw*t the finish! I think that is
just about what I am aiming for, both when rowing and coaching, though
I arrived at this point by another route.
paul_v...@hotmail.com
> Paul wrote:
> > On 28 Feb, 11:51, Carl <c...@carldouglas.co.uk> wrote:
> <snipped>
>
> > Thanks Carl (and others) I think I more or less understand it now.
> > Though I am not sure quite how, if at all, I am going to be able to
> > use this knowledge to make my boat go faster. Its interesting none the
> > less.
>
> What it tells you is that the ends of the stroke are really important.
> You can only expend 100% effort, but you are able to redistribute your
> effort within the stroke cycle to get a better result from that same effort.
>
> Let's do the maths crudely:
> Say the stroke has 3 phases (that's already wrong, but keep it simple!).
> Let the efficiency, E1 & E3, of the catch & finish phases be 100% (wrong
> again, but keep simple).
> Let the efficiency of the middle phase, E2, be 50% - i.e. half of what
> you invest there goes to waste.
> Let work done in each phase be w1, w2, w3
>
> So stroke efficiency, Es, =(E1 x w1 + E2 x w2 + E3 x w3)/(w1+w2+w3)
>
> If, as might be, a crew invests as follows:
> w1 = 20%, w2 = 70%, w3 = 10%; then Es = 65%
> But suppose they re-distribute as follows:
> w1 = 25%, w2 = 60%, w3 = 15%; then Es = 70%
If I am to believe what Kieran is suggesting, doesn't there need to be
a trigonometric factor included for each wX?
Even though it would be said that the "pinch" does no work, we must
certainly be utilizing muscular effort to produce the force involved,
and we would like as much of that effort going toward advancing the
system as absolutley possible, which seems as if makes it "work"
afterall.
This goes back to my proposal of training a very specific habit of
producing a particular force profile, utilizing objective biofeedback
on an ergometer. Then the only thing that needs to be done in a boat
is developing the skills required to get the blade in (and out) at the
right moments, along with some balance skills. [;o)
> So those small adjustments in how you apply the same amount of
> work/stroke will improve overall efficiency from 65% to 70%.
>
> A propulsive efficiency improvement on that scale will give you an
> increase in boat speed of ~2.5% which, when translated into winning
> margins, is huge (50m per 2000m).
>
> Sure, the numbers are very rough estimates, but the message is unambiguous.
>
Yes it is. "Get your blade in at full reach, get to the peak force
required as quickly as possible, extract the blade the instant that
the force returns to zero, get back up the slide to do it again while
mucking up the even travel of the system as little as possible, repeat
as required." [:o)
- Paul Smith
KC
You could view it such that the trig is built in to Carl's example. So,
for example, using +y = stern to bow direction, +x = stbd to port
direction, F = force on pin, Fy, Fx = y,x components of that force, D =
distance traveled by pin, Dx, Dy, components of distance traveled by pin...
w1 = F1*D1 = Fy*Dy + Fx*Dx
Dx = 0 --> Fx*Dx = 0 and w1 = Fy*Dy.
The trigonometric aspect comes in when you figure out exactly what Dx,
Dy, Fx and Fy are given the instantaneous angle.
> Even though it would be said that the "pinch" does no work, we must
> certainly be utilizing muscular effort to produce the force involved,
> and we would like as much of that effort going toward advancing the
> system as absolutley possible, which seems as if makes it "work"
> afterall.
One of the things students have a hard time with when learning about
"work" in the Newtonian sense of the word, is how it differs from
metabolic energy expenditure. You can stand and hold a 10lb weight at
arms length for 10 minutes, and do now Newtonian "work" on the weight,
although you will have expended a fair amount of metabolic energy that
has done nothing but hold a weight up.
So yes, unfortunately some of our metabolic "work" goes to "waste" if
you want to view it that way.
> This goes back to my proposal of training a very specific habit of
> producing a particular force profile, utilizing objective biofeedback
> on an ergometer. Then the only thing that needs to be done in a boat
> is developing the skills required to get the blade in (and out) at the
> right moments, along with some balance skills. [;o)
That training philosophy still applies, even though some of the force we
put into the oar does no work in moving the boat.
-Kieran
Carl
Paul, my calculations refer to an energy balance, not to a force
balance. The work input must match the work output, although in my
example I have, by use of the efficiency multipliers, implicitly split
the output into useful & wasted components.
Forces are interesting & affect accelerations, but they don't affect
actual work done by the rower, nor net propulsive work output. Sure,
you can do the same sums by resolving forces & multiplying these by
distances moved, but the results must come out the same. So, being a
typical cheating engineer, I took the easy way out by using an energy
balance. And why not, say I?
I have often drawn attention to the complexities of the rowing stroke's
inbuilt, continuously-variable gearing. A gear ratio is the ratio of
force applied to force delivered (or the inverse, but I'll leave that to
the purists to decide for me), which is the same as the ratio of
velocity out to velocity in, or the ratio of the distances moved by the
boat & the hands in unit time - 3 interchangeable definitions.
Considered first without slip, the rowing stroke has a high gear ratio
at the catch (hands move slowly in relation to boat speed) changing
smoothly to a lower ratio in mid-stroke & then climbing higher as you
move towards the finish. If you then apply known information on blade
slip, you don't really change the gear ratio anywhere in the stroke but
in mid-stroke the sudden increase of slip gives the rower a false sense
that the ratio has become much easier. And the harder you pull in the
mid-stroke, the greater the slip & the easier still does the gearing
seem - especially if you don't row deepish in mid-stroke.
As you know, but others may not, slip as much use to a rower as greasing
the road is to a driver - it results in wheelspin & waste heat.
Gearing, however, does matter as it impinges on our biomechanical
effectiveness - we work best within a narrow range of muscle contraction
rates, which do of course vary according to the degree of a muscle's
extension. So we are limited in how much we can switch effort between
different parts of the stroke but we can certainly learn to row better
if we know what's going on in the water. But when I say "better" I mean
that in a very Fairbairn sense - not that it looks good but that it
enables us to move the boat faster.
This might be a good point to further complicate matters by referring
again to that other component of gearing which we habitually ignore -
i.e. rating.
Since rating can't be raised "in the water" (because boat speed varies
so little with increased work levels) & since the human body works best
if you keep it close to its optimum loadings for the course
distance/duration, you can really only raise the rate either by
shortening up (clearly a lousy idea as it canes propulsive efficiency)
or by shortening recovery duration.
By shortening recovery time you shorten stroke cycle time, so you get in
more strokes &, unless you lower pressure, you proportionately increase
work load. Since you don't want to overload, as rate rises it makes
sense to lower pressure a bit. If you wanted to vary rate without
varying boat speed you would simply adjust pressure in inverse
proportion to rate. And that brings us straight back to my 3rd
definition of gearing: you are doing more strokes per minute, keeping
their length the same as before, & the boat is going at the same speed,
so the gearing has become easier, which is demonstrated by the fact that
you are not having to pull as hard as at the lower rate.
>
> This goes back to my proposal of training a very specific habit of
> producing a particular force profile, utilizing objective biofeedback
> on an ergometer. Then the only thing that needs to be done in a boat
> is developing the skills required to get the blade in (and out) at the
> right moments, along with some balance skills. [;o)
>
>
What you need, therefore (& I suggested this many years ago to the
Rowperfect originator but it was not taken up) is the ability to
superimpose user data (shall we say "opinion"?) on how propulsive
efficiency of the oar varies through the stroke, in the form of a curve
constrained to lie between the limits of 0% & 100% efficiency. Then
have your erg monitor compute, at every part of the stroke & in sum for
the whole stroke, the net amount of propulsive (= useful) work done by
the rower. That might be a powerful tool for a coach, although I'd
prefer to see it on a dynamic erg.
>
>>So those small adjustments in how you apply the same amount of
>>work/stroke will improve overall efficiency from 65% to 70%.
>>
>>A propulsive efficiency improvement on that scale will give you an
>>increase in boat speed of ~2.5% which, when translated into winning
>>margins, is huge (50m per 2000m).
>>
>>Sure, the numbers are very rough estimates, but the message is unambiguous.
>>
>
>
> Yes it is. "Get your blade in at full reach, get to the peak force
> required as quickly as possible, extract the blade the instant that
> the force returns to zero, get back up the slide to do it again while
> mucking up the even travel of the system as little as possible, repeat
> as required." [:o)
>
Don't they call that "rowing"?
Cheers -
KC
> This might be a good point to further complicate matters by referring
> again to that other component of gearing which we habitually ignore -
> i.e. rating.
>
> Since rating can't be raised "in the water" (because boat speed varies
> so little with increased work levels) & since the human body works best
> if you keep it close to its optimum loadings for the course
> distance/duration, you can really only raise the rate either by
> shortening up (clearly a lousy idea as it canes propulsive efficiency)
> or by shortening recovery duration.
>
> By shortening recovery time you shorten stroke cycle time, so you get in
> more strokes &, unless you lower pressure, you proportionately increase
> work load. Since you don't want to overload, as rate rises it makes
> sense to lower pressure a bit. If you wanted to vary rate without
> varying boat speed you would simply adjust pressure in inverse
> proportion to rate. And that brings us straight back to my 3rd
> definition of gearing: you are doing more strokes per minute, keeping
> their length the same as before, & the boat is going at the same speed,
> so the gearing has become easier, which is demonstrated by the fact that
> you are not having to pull as hard as at the lower rate.
I must say, while I know your overall message was not different, the
break-down of it, and the example at the end were much better this time,
IMO. Last time I agreed that this was a beneficial thing, but that it
wasn't really a change in "gearing". Having read this most recent
description though, I agree that it is a form of gearing. Interesting
that an effective gear change can be achieved without changing the
physical levers (inboard/outboard) used (which is what had me stuck last
time.)
-Kieran
paul_v...@hotmail.com
>
> - Show quoted text -
The "physical levers" can be thought of as changing. This is
addressed in A Textbook of Oarsmanship in a discussion of what Bourne
refers to as the "turning point", the point along the shaft between
the pin and blade tip which neither advances nor retreats from the
finish line. How the rower goes about executing a particular force
profile has much to do with the sliding of the outboard along the
turning point.
We occassionally do "zero pressure reverse ratio" rowing, which of
course does not entail "zero" pressure, but it is relatively low, but
does entail a very quick recovery. The fun part of the exercise is to
see the speed that is generated even under the low input force,
especially if all the timing lines up, which is a bit more challenging
than when rowing at more conventional ratios. It also looks a bit
funny and an onlooking coach might feel the need to drive themselves
across the lake to explain how to do things "right"... [;o)
Carl, I truly would like to believe Kieran's notion of "Boat Pinch"
as he has explained it, and since you have been relatively objective
over the years of discussion we have had, is there a way that you can
put it simply for me. Perhaps, "Paul, Kieran has accounted for all
the forces accurately, and if the pin was made of brittle/weak
material it would shear directly perpendicular to the oar shaft at the
catch when the boat is under way." (proper catch and close to instant
handle pressure assumed)
And even though that contradicts what my playing about with an oar
along the length of our dock has tactilely demonstrated to me any
number of times I've tried, I'll stop pestering him to explain why his
theory doesn't explain the results of my experiment. BTW - what Kieran
says makes plenty of sense intuitively, it merely contradicts what
I've found in actual testing.
Good enough Kieran?
- Paul Smith
KC
That's a good point, as you adjust the pressure on the stroke, the
amount of slip will change and thus the turning point will shift.
Although I think the "turning point" is an instantaneous thing, and that
it varies along the shaft throughout the stroke. This can be seen in
this image:
http://www.concept2.com/us/images/products/op240.gif
If you look at any few consecutive frames, they have similar turning
points, but if you compare a few at the begining, versus a few from mid
stroke, it looks *to my eye* (hard to really tell with that small image)
that the turning point is not at the same place along the length of the oar.
> We occassionally do "zero pressure reverse ratio" rowing, which of
> course does not entail "zero" pressure, but it is relatively low, but
> does entail a very quick recovery. The fun part of the exercise is to
> see the speed that is generated even under the low input force,
> especially if all the timing lines up, which is a bit more challenging
> than when rowing at more conventional ratios. It also looks a bit
> funny and an onlooking coach might feel the need to drive themselves
> across the lake to explain how to do things "right"... [;o)
>
> Carl, I truly would like to believe Kieran's notion of "Boat Pinch"
> as he has explained it, and since you have been relatively objective
> over the years of discussion we have had, is there a way that you can
> put it simply for me. Perhaps, "Paul, Kieran has accounted for all
> the forces accurately, and if the pin was made of brittle/weak
> material it would shear directly perpendicular to the oar shaft at the
> catch when the boat is under way." (proper catch and close to instant
> handle pressure assumed)
>
> And even though that contradicts what my playing about with an oar
> along the length of our dock has tactilely demonstrated to me any
> number of times I've tried, I'll stop pestering him to explain why his
> theory doesn't explain the results of my experiment. BTW - what Kieran
> says makes plenty of sense intuitively, it merely contradicts what
> I've found in actual testing.
I don't remember you describing this experiment with an oar along the
length of a dock. Could you re-post that, please? This rings no bells
whatsoever to me. Maybe it was from a past thread that I didn't read?
> Good enough Kieran?
Sure. I'm sorry you need Carl's stamp of approval, though. Maybe if
you describe this experiment and the observed results to me, I can
address that. I am curious what it was.
-Kieran