form question: overcompression

[Nosmo]

The main resistance provided by the blade near the catch is not off the face of the blade, but off the back of the blade just as off the upper surface of a wing.

The standard explaination of the airplane wing involves a pressure difference between the top and bottom. I don't see how the force is on the upper surface rather then the bottom.
Furthermore the standard explaination is overly simplistic: An airplaine wing must actually push air downward. (One can good at decent approximation of lift on a "flat plate wing" just by considering momentum. Newton calculated it before Bernollui)
I've run accross a few good discusions of lift and airplaine wings on the web. If I get a chance to find them I will post a link or excerpt.
Also according to someone I spoke to at CII, the back side of the blade actually has a detached boundary layer.

Its going to take some research and thought before I am satified I understand what is going on. The more I think about it the harder the problem seems.

Since you are an experimentalist perhaps this would help...

I twice attempted to to feel the lift with a blade on the dock but not exactly the way you describe. I will try your experiment and hopefully I can get into a double before too long and try that too.

If this goes on, we should proably move this to a seperate thread. Appologies to all who wanted to remain on topic.

[RR1 Kirk]

Hello, I experienced the same banging the heels and shins past vertical.
The banging of the heels takes care of itself, it seems the body stops doing that after a while all by itself.

To reduce the overcompression I've raised the footpads to a point where my ankles are bisected by the plane of the rail. I'm working with the idea that if I raise my feet it will tilt my body rearward and change the shin to vertical, and hopefully I can maintain the same reach/ stroke length.

After more thought I don't see where the idea of overcompression of the thighs comes from. You are either flexible in your knees and thighs or you are not. The shin angle going past vertical seems more a matter of flexibility in the ankles.

Cheers,

[atheist]

so is going past vertical a problem if it doesn't cause any pain? i mean, is the stroke less efficient when i go past vertical?

-steve

[RR1 Kirk]

Good question. I'm looking forward to the opinions.

[PaulS]

so is going past vertical a problem if it doesn't cause any pain? i mean, is the stroke less efficient when i go past vertical?

-steve

You will have to define "efficient" to elicit a good answer.
Is this for long slow rowing or what one would consider "race paced" levels?
And also are you talking about when spinning a flywheel or moving a boat?

[atheist]

well, my form is pretty much the same in both long slow and short fast. i only row on the erg. i can only try to define 'efficient' in relative terms- i want to know if one in principle stands to lose anything by going past vertical. this would give me an incentive to put the effort into changing my form, or buying slides (which i understand lessen the problem of overextension/overcompression).

-steve

[RR1 Kirk]

I am only interested in making the flywheel go faster. My preferred on the water vessel is a kayak.

I am interested in PB levels of effort, both short and longer distances.

Sorry to hijack your thread.

Thanks,

[PaulS]

well, my form is pretty much the same in both long slow and short fast. i only row on the erg. i can only try to define 'efficient' in relative terms- i want to know if one in principle stands to lose anything by going past vertical. this would give me an incentive to put the effort into changing my form, or buying slides (which i understand lessen the problem of overextension/overcompression).

-steve

1) Slides do tend to shorten drive length, especially for those that pull themselves aggressively up to the catch, which uses the additional momentum of their body to aid in the compression. Since you are only on the Erg, the detrimental effects of this WRT OTW rowing can be ignored and we only address what concerns you.

The Erg is concerned with input power, as that's all it can measure. Since we are limitted by stroke length (even if over compressed) it makes sense that getting a quick rise in force (even a low peak force) is desireable. When in an overcompresed position, it will be virtually impossible to get to a high peak force quickly, so for the faster paces where that is required the efficiency is lost. On the other hand, if one is rowing at low rate, with a high ratio and a lower peak force requirement for the target pace, it may be just as well to go ahead with the over-compression since it will not be something that is rushed into but rather allows for a natural stopping point at the elastic limit of the muscles and then the drive begins slowly in concert with a much slower moving flywheel. Raising the DF may work out well in this situation also, since that gives more distance credit per revolution and you will will be getting slightly more revolutions per drive. The usual "shock load" concern is not there, because you are in a weak biomenchanical position to generate it.

Just a side not as to why this doesn't work for a boat. There is no such thing as "rowing slowly in a fast moving boat", but that very thing can be done with the virtual boat that is an Erg. Low Rate and Long Strokes at High DF's can produce paces that would never happen in a boat on the water at the same stroke rate.

Finally, efficiency on it's own is great for training (both in a boat and on an Erg), but in a boat, it is the ability of the athlete to endure the stress of making the system move the fastest over a given distance that wins races, and that generally entails absorbing the costs of some inefficiencies. i.e. Kicking in the After-burner on a jet increases the speed, but not the efficiency of the system.

[yehster]

For boat travel as a function of other parameters, I think it only depends on outboard and arclength. For a given outboard, you can change the inboard, but if you have the same arclength, the boat will still travel the same distance during the drive. The hands will just have to follow a longer or shorter arc, and the load will feel lighter or heavier depending on the change that was made. Any changes to arclength arising from a change in inboard would be more of a biomechanical thing (i.e. a function of a rower's arm length, flexibility, etc.). From the formula for a circular arc segment, I think the distance of travel will just be

outboard * sin(arclength/2). /

This looks like a formula for the distance the blade moves parallel to the direction of travel, not the distance the boat necessarily moves for a given stroke. The force applied to the water (which is related to the force you put on the oar) would have to be a factor some where, as would many other variables including the various forms of drag.

[becz]

For boat travel as a function of other parameters, I think it only depends on outboard and arclength. For a given outboard, you can change the inboard, but if you have the same arclength, the boat will still travel the same distance during the drive. The hands will just have to follow a longer or shorter arc, and the load will feel lighter or heavier depending on the change that was made. Any changes to arclength arising from a change in inboard would be more of a biomechanical thing (i.e. a function of a rower's arm length, flexibility, etc.). From the formula for a circular arc segment, I think the distance of travel will just be

outboard * sin(arclength/2).

This looks like a formula for the distance the blade moves parallel to the direction of travel, not the distance the boat necessarily moves for a given stroke. The force applied to the water (which is related to the force you put on the oar) would have to be a factor some where, as would many other variables including the various forms of drag.

I think the original question was how far the boat moves during the drive phase of the stroke, which is what the formula represents (distance the pin moves as a function of outboard and arc). If the question was in regard to how far the boat moves in total from the start of one drive to the start of the next, then yes there are many other factors to consider.

[PaulS]

For boat travel as a function of other parameters, I think it only depends on outboard and arclength. For a given outboard, you can change the inboard, but if you have the same arclength, the boat will still travel the same distance during the drive. The hands will just have to follow a longer or shorter arc, and the load will feel lighter or heavier depending on the change that was made. Any changes to arclength arising from a change in inboard would be more of a biomechanical thing (i.e. a function of a rower's arm length, flexibility, etc.). From the formula for a circular arc segment, I think the distance of travel will just be

outboard * sin(arclength/2).

This looks like a formula for the distance the blade moves parallel to the direction of travel, not the distance the boat necessarily moves for a given stroke. The force applied to the water (which is related to the force you put on the oar) would have to be a factor some where, as would many other variables including the various forms of drag.

I think the original question was how far the boat moves during the drive phase of the stroke, which is what the formula represents (distance the pin moves as a function of outboard and arc). If the question was in regard to how far the boat moves in total from the start of one drive to the start of the next, then yes there are many other factors to consider.

Yep, that was the original question, but if I take 100deg arc and 2.1m outboard, that only gives 1.6m hull travel during the drive, so there must be something more to it since that is far to short to be anwhere close to accurate. Because even if the Drive lasted for only 0.7 seconds the hul speed would be only 2.3m/sec, and frankly I don't think I can paddle that slowly. 217.6sec/500 or a 3:37.6 pace. Though I could be working the calculator button wrong, I suppose. If I add in the additional distance that my ceneter of mass was advanced (~45cm) then the system was advanced 2.05m during the drive, so a speed of 2.9m/s or 170.7s/500m or a 2:50.7 pace, which is still a lot slower than I would have thought possible.

I think what is being left out is the additional distance travelled due to lift acting on the blade, allowing it to advance through the water better than a toothpick, and that is where it starts to get complicated.

[becz]

This looks like a formula for the distance the blade moves parallel to the direction of travel, not the distance the boat necessarily moves for a given stroke. The force applied to the water (which is related to the force you put on the oar) would have to be a factor some where, as would many other variables including the various forms of drag.

I think the original question was how far the boat moves during the drive phase of the stroke, which is what the formula represents (distance the pin moves as a function of outboard and arc). If the question was in regard to how far the boat moves in total from the start of one drive to the start of the next, then yes there are many other factors to consider.

Yep, that was the original question, but if I take 100deg arc and 2.1m outboard, that only gives 1.6m hull travel during the drive, so there must be something more to it since that is far to short to be anwhere close to accurate. Because even if the Drive lasted for only 0.7 seconds the hul speed would be only 2.3m/sec, and frankly I don't think I can paddle that slowly. 217.6sec/500 or a 3:37.6 pace. Though I could be working the calculator button wrong, I suppose. If I add in the additional distance that my ceneter of mass was advanced (~45cm) then the system was advanced 2.05m during the drive, so a speed of 2.9m/s or 170.7s/500m or a 2:50.7 pace, which is still a lot slower than I would have thought possible.

I think what is being left out is the additional distance travelled due to lift acting on the blade, allowing it to advance through the water better than a toothpick, and that is where it starts to get complicated.

I actually put a slightly wrong equation. It's twice what I put previously. The distance the pin will travel is 2*outboard*sin(arc). So for a 2.1m outboard and 100 degree arc, the pin travels 4.14m. If the drive takes 0.7 seconds, that gives an average speed of 5.9m/s. This will be higher than the average speed since it is the energy input phase of the stroke cycle.

I agree there is some lift on the blade, but overhead photos of rowers show pretty clearly that there isn't much blade movement in the direction of travel during the drive.

[yehster]

I think the original question was how far the boat moves during the drive phase of the stroke, which is what the formula represents (distance the pin moves as a function of outboard and arc). If the question was in regard to how far the boat moves in total from the start of one drive to the start of the next, then yes there are many other factors to consider.

Even answering the question of how far the boat moves during just the drive phase, you still have to consider other factors, including things like the speed of the boat.

Hypothetical Scenario 1. The boat is "anchored" like at a start, and the guy holding the boat is strong enough to resist against all the oarsmen and doesn't let go. With the first stroke, the blades will move your computed distance, but the boat will go nowhere during the drive.

Hypothetical Scenario 2. The boat is moving at speed already, everyone takes a a stroke, but don't drop their blades completely into the water at the catch. Their blades will move much faster during the drive than the pins will move.

Hypothetical Scenario 3. When the oarsmen drop their blades into the water, a huge spike drops down and rigidly fixes them into the river bed. I this case, because everything is rigid, and the oars are fixed, the boat would have to move the distance completely specified by the geometry of the oars/riggers.

The more that I think about it, I believe that the parallel distance you are providing a formula for is in fact an "upper limit" to how far the boat can travel during the drive. Baring the boat being towed, driven by some other means/external force.

[Byron Drachman]

**lift on the blade is concept I am not satisfied is being applied correctly to rowing but that is another whole thread.

Hi Nosmo,

Sorry to go off topic a little, but maybe it's not off topic because if lift forces are as significant as some are saying then it might not be so bad to overreach at the catch on the water. Lots of elite rowers do seem to do that. I am also suspicious of the discussions on lift forces (Just Bernoulli's Principle or am I missing something?) applied to rowing

When I first read about lift forces in Rowing Faster, I didn't believe the Figure 10.4 on page 117, which was from an article by Affeld, Schichl, and Ziemann, especially when I look at the magnitude of the vectors that supposedly represent the lift forces compared to those representing drag forces. It looked like "dry labbing" to me, but I'm not an expert and I didn't read the quoted article, so I can't say there is anything wrong.

I did the simple experiment of walking quickly along the dock while holding the blade of an oar parallel to the direction of travel, and trying to get a feel for the "lift" forces. This is all very unscientific I confess, because I don't know how much slip there is. This just made me more suspicious.

If you do some experiments and actually measure lift forces, I would love to hear about them.

My background is mathematics and engineering. I did write one article on fluid dynamics a long time ago, but it was because I was able to compute some special functions from physics in a range that hadn't been done before, not because I had any good intuition about what was going on.

I think it would be great if a physicist got interested in this subject.

Byron

[becz]

I think the original question was how far the boat moves during the drive phase of the stroke, which is what the formula represents (distance the pin moves as a function of outboard and arc). If the question was in regard to how far the boat moves in total from the start of one drive to the start of the next, then yes there are many other factors to consider.

Even answering the question of how far the boat moves during just the drive phase, you still have to consider other factors, including things like the speed of the boat.

Hypothetical Scenario 1. The boat is "anchored" like at a start, and the guy holding the boat is strong enough to resist against all the oarsmen and doesn't let go. With the first stroke, the blades will move your computed distance, but the boat will go nowhere during the drive.

Hypothetical Scenario 2. The boat is moving at speed already, everyone takes a a stroke, but don't drop their blades completely into the water at the catch. Their blades will move much faster during the drive than the pins will move.

Hypothetical Scenario 3. When the oarsmen drop their blades into the water, a huge spike drops down and rigidly fixes them into the river bed. I this case, because everything is rigid, and the oars are fixed, the boat would have to move the distance completely specified by the geometry of the oars/riggers.

The more that I think about it, I believe that the parallel distance you are providing a formula for is in fact an "upper limit" to how far the boat can travel during the drive. Baring the boat being towed, driven by some other means/external force.

If you assume the blade is stationary after it has been placed in the water at the catch (not a bad approximation) then it doesn't matter whether it is the first stroke or the last, the pin can only move a distance governed by the outboard and the arc swept out during the stroke. The boat moves past the oar, not vice versa (unless the boat is being held, as you point out). Initial velocity will only come into play to make the stroke lighter or heavier, and to increase the speed at which you have to drive in order to accelerate the boat. For example, if the rowed boat has been tied to a motor boat and brought up to a speed beyond which the rower couldn't maintain, there is no hope of accelerating the during the drive. But when you put the oar in the water and it is moved through the stroke by the momentum of the boat, it will still only travel the distance according to the formula. It sure will feel light, however. If on the other hand you start from a dead stop, the load will feel very heavy, the time to drive will be longer, but the distance the pin travels will still be the same.