So is distance =outboard * cosine(theta_c) + outboard * cosine(180-theta_f) = 2*outboard*sin(arc)? (Not according to my calculator, 3.94 != 2.96) [2m outboard, 100deg arc, theta_c=25, theta_f=55]becz wrote: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.PaulS wrote: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.becz wrote: 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.
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 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.
Where theta_c is catch angle and theta_f is the finish angle (zero at the bow, 180 at the stern for both)
The overhead photos I've seen (Ken Young, University of Washington) have the blade tip exiting a bit ahead of where it entered the water, so certainly that difference must be added to the pin travel. i.e. Distance due to lift - Distance of slip during stalled blade. (This is a highly variable distance based on stroke profile, IMO.)
The "Blade tip displacement" from Ken Youngs work was where I derived the Stroke Phase Guide (SPG) that is used in ErgMonitor. Showing, in basically 1/4's of the drive, 'advance' - 'hold' - 'retreat' - 'advance', WRT the finish line (or the water, if you like).
It's an interesting problem that still appears to require some work in explaining.
The reason for the SPG to assist in developing an optimal stroke force profile is to help the athlete to be putting in their force while the blade is not retreating from the finish line. Which makes sense in the context that we are advancing the boat past the blade and not tossing the lake behind the boat, something I think we agree on. i.e. Why put in your greatest driving force, when your support is moving in the opposite direction that you wish to go? (late force peak) Of course none of this matters to the Erg, as the connection of chain to cog is nothing like blade to water. In fact the Erg tends to inspire just the opposite of OTW desireable, by rewarding the high peak force on the already accelerated flywheel that results in a higher power peak. Thus the divergent technique focus' of the Rower and Erg Specialist who never plans to be in a boat.
BTW - The "ridgid post where the blade is planted" idea, doesn't work with an oar, as the shaft would have to slide along that post creating a variable outboard through the drive. The "zero slip" path has been described as a 'tractrix', which I'm not versed well enough in maths to comment on.
