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In Part 2 I am looking at a series of strokes which can be maintained for an extended time. The emphasis here will be on the skater's power level and not the applied force as there is considerable data on the power levels skaters can maintain. For simplicity Part 2 will deal only with a constant power level.
| Es + Ef = Elin; or | Ps*dt | + Ef = Plin*t. |
Stroke Model Details and Power Per Stroke
The stroke used here will be oscillating sinusoidally with a fixed (maximum) stroke width (Xsmax). Its sideways velocity pattern is defined by the cosine time dependence:Vs(t)=Vsmax*cos[Pi*t/(2T)].
Here Vsmax is the maximum sideways velocity of the stroke (3.70 m/s) here, Pi =3.14..., t=time (seconds), T=time of maximum stroke width (T=0.4 seconds in this model). In the figure below the sideways velocity is plotted in blue and the sideways position is plotted in red. The aqua region fills the velocity curve between adjacent forward-facings of the skate (T to 3T). At T and 3T the sideways velocity is zero, The stroke width is maximum, and the forward velocity (Vf) is maximum. In the middle of this range (2T) the sideways velocity is maximum at Vsmax while the skate is crossing the center line at Xs=0.
The aqua region is used to compute the average sideways energy per stroke (Es).
Es =
3T
T(M/2)*Vsmax2*cos2[Pi*t/(2*T)]*dt __________________________________
3T
Tdt = (M*Vsmax2)/4
Energy and Power Balance
From the previous integral we find the average sideways stroke power Ps = Es /T and set it equal to the power level of the matching linear accelerator:
Ps = Plin = (M*Vsmax2)/(4*T)
Next, the energy balance equation from the beginning of this section is solved for the forward energy, Ef:
Ef = Plin*t - Es
or,
Ef = (M*Vsmax2)*t/(4*T) - (M/2)*(Vs(t))2*cos2[Pi*t/(2*T)]
and the forward velocity then is given by:
Vf(t) = SQRT(Ef).
Next the results will be presented graphically.
The following results are based on the case: Power, Ps = 600 watts; maximum sideways velocity, Vsmax = 3.7 m/s; all drag forces = 0; skater's mass, M = 70 kg; time from center lime to maximum stroke width , T = 0.4 seconds. The chosen power is low for the start but high for an extended period of time where elite skaters might be more likely to generate 350-500 watts.
energy. Because the energy is transferred from sideways (red) to forward at constant power the energy rises linearly on average. The situation is similar to that described by Mike Ryan (Accumulation of Speed, FaSST, Autumn 1999, Page 16) but the two-dimensional fully-efficient stroke accumulates energy only on average. Instantaneously the forward energy can decrease a little as the stroke turns sideways. The skater has some control over the peaks and valleys in the skating curve however as the energy is technically based on the position of the skate while the skater can maneuver his body to average out the bumps.
The energy equation above is plotted on the right. The blue line is the linear accelerator's energy as a function of time while the black curve shows the skater's forward
| The square root of the energy yields the velocity which is plotted here vs. time. The blue curve is for the linear accelerator while the black curve is the skater's forward velocity. The |
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| Here the forward velocity has been integrated over time to give the distance travelled as a function of time. The time to travel 100 meters is a little under 8 seconds at this power level which is barely below Chad Hedrick's time. The time to travel 500 |
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| To see if this model had anything in common with real skating I examined some video taken of Chad Hedrick. The image to the right was prepared from a video of Chad kindly supplied by Kim Hendrikse. The video was taken with a zoom lens from directly in front of Chad to record his sideways motion and here I cropped it down to show only his boots, frames, and wheels. The video comprising almost one full time cycle of one version of his double-push was exported to a vertical filmstrip and compressed vertically so you can't pick out much detail in this form. Here the serpentine curves are the time-tracks of Chad's feet with time increasing downwards. The apparent deviation from sinusoidal at the far left occurs when Chad lifts his right skate off the ground and that skate stops developing cornering force. The inside tracks from the two feet merge together to form a track close to sinusoidal and looking a lot like the fully efficient stroke. From these tracks it is clear that the sideways velocity goes to zero only at the left and right extreme points. Clearly the skates continue to move across the center line (not drawn) and from my examination of data read from a larger version of this image I was able to construct a sideways velocity pattern in reasonable agreement with the |
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