Kinetic energy dissipated by the moving rower

Kinetic energy dissipated by the moving rower.<= /p>

Revised September 2009=

homepage

Introduction=

The boat is moving through the wa= ter and the rower is moving on the seat with respect to the boat. Better: rower and boat are moving with respect to each other. Discussed is the question how m= uch energy is dissipated as a result of the repeating acceleration and decelera= tion of the rower and the boat. To introduce the problem the discussion will be = introduced by considering the situation on the ergometer. The subject is also covered = by Bill Atkinson= .

On the Concept 2 ergometer

On the Concept 2 ergometer the si= tuation is somewhat simpler as there is no relative motion of the seat with respect= to the boat but motion with respect to the earth only.

Situatio= n 1.

The rowe= r is moving to and fro without pull= ing the chain. The following explanation is valid for moving in both directions= .

Sliding = distance s =3D 0.6m

Velocity= profile: triangular, max speed reached at 0.3m

Max. speed u₁ =3D 1 m/s=

Average = speed u_a =3D 0.5 m/s

Time to = go one way T₁ =3D 0.6/0.5 =3D 1.2s

Cycle ti= me T =3D 2.T₁ =3D 2.4s

Stroke r= ate SR =3D 60/2.4 =3D 25/s

Moving m= ass m =3D 60kg

Kinetic = energy at top speed E =3D 0.5 m u₁² =3D 0.5 60 1² = =3D 30 J

Kinetic = energy loss expressed as mean power P =3D E/T₁ =3D 30/1.2 =3D 25W<= /o:p>

The same calculation yields for the movement in both direction so we can say the row= er is constantly (but averaged over the cycle) delivering 25W.

&nb= sp;

Some fur= ther considerations.

Maximum = seat speed is reached after T₁/2 s.

The acce= leration of the mass is a=3Du₁/(T₁/2) =3D 1/0.6 =3D 5/3 m/s2

The leg = force F =3D m a =3D 60 5/3 =3D 100 N.

The work= done by the leg to reach the top speed is W =3D F s/2 =3D 100 0.3 =3D 30 J

Of cours= e E =3D W.

During t= he acceleration of the mass, force and velocity have the same direction; the w= ork during this period is positive. During the deceleration of the mass, force = and velocity have opposite direction, the work done is negative. In a conservat= ive system (with springs e.g.) the energy is stored by the system and is not lo= st. The system under consideration is not conservative and the kinetic energy is simply lost. Besides the deceleration of the mass requires some effort of t= he rower, he has to do some work. This work is neglected here. Similar situati= ons: a person descending a staircase or lowering a lifted weight. I did not find= a method to calculate the work involved in these situations in the biomechani= cal literature but the problem must have been addressed somewhere. In the follo= wing positive work will be considered as delivered by the rower and used in a us= eful manner: to increase the kinetic energy in the system or to overcome the external friction. Negative work will be considered as internal dissipation, useless but not to be delivered by the rower.

As found= above: without pulling the chain the rower is producing 30J per cycle or an average power of 25W.

Situatio= n 2

This is = the common situation. When the rower moves away from the flywheel he pulls the chain a= nd when he returns (recovery) no interaction with the flywheel exists. The recovery equals Situation 1. During the recovery (same parameters as above)= the rower produces 25W and this output is not recorded on the monitor.

During t= he drive phase the rower maintains a pull on the chain. In the first part of the dri= ve the forces exerted by the rower are used to accelerate his own body and to = pull the chain. His leg force is the sum of the chain pull and the force necessa= ry to accelerate his body. In the second phase of the pull his body is deceler= ated by the chain pull. His leg force is the chain pull minus the force to decelerate his body. This means that under normal conditions during the who= le drive the rower’s power is positive and all the energy (power) he delivers is recorded on the monitor. The figure below shows the forces on t= he rower (greyed rectangle) and m.a is the inertia force mass times accelerati= on (deceleration).

        =             &nb= sp;

        =    F_legs =3D F_chain + m.a        =            =             &nb= sp;  F_legs =3D F_chain - m.a        =
        =             &nb= sp;

In this situation the rower produces 30J during the recovery or 30/2.4 =3D 12.5W more than the monitor indicates. This is a rough estimate. Usually the times for the reco= very and the drive are different.

In the boat

Situation 1.

The situation in the boat is diff= erent. First we consider the boat and rower as a system without external forces. T= his is

&nb= sp;            =             &nb= sp;           =    m₁        =             &nb= sp;            =

m₁ =3D mass of boat and part o= f the mass of rower fixed to the boat [kg]

m₂ =3D moving mass of rower [k= g]

u =3D speed of m₂ with respect= to m₁ [m/s]

U =3D speed of m₁ [m/s] with r= espect to earth

At the start of the seat motion u =3D 0 a= nd U =3D U_o, the maximum value of u is u =3D u₁, and the corresponding speed of m₁ is then U =3D U₁(minimum v= alue). After the seat has stopped again U =3D Uo, this follows from the principle = of conservation of momentum.

The kinetic energy in the system before a= nd after the seat motion (motion of m2) is the same because the velocities are= the same. When u reaches its maximum value, the kinetic energy in the system reaches its maximum values. This difference in the amount of kinetic energy= has to be delivered by the rower to the system and is then lost: dissipated in = the muscles and articulations of the rower as in the case of the ergometer but = the amount lost energy is different as will be shown.

Momentum at the start and at the end of t= he seat motion:

<= !--[if gte mso 9]>

Momentum at maximum value of u:

<= !--[if gte mso 9]>

Conservation of momentum:

<= !--[if gte mso 9]>

or

&nb= sp;

<= !--[if gte mso 9]>

this yields:

<= !--[if gte mso 9]>

Kinetic energy at the start (and at the e= nd) of the seat motion is:

<= !--[if gte mso 9]>

Kinetic energy when u =3D u₁ i= s:

<= !--[if gte mso 9]>

The difference ΔE =3DE₁ -= E_o is after some (and tedious) algebraic manipulation

<= !--[if gte mso 9]>

Remark:

·      =    In this expression m₁ an m₂ appear in the same way. Interchange of m₁and m₂does= not influence the result

·      =    U₀ and U₁ are absent<= /o:p>

In this derivation was u>0. When u<0 the result appears to be the same.

For the single scull we take as for the p= erson on the ergometer for the mass moving with the seat m₂ =3D 60 kg.= The mass moving with the boat is the remaining mass of the rower plus the mass = of the boat or m₁ =3D 30 kg. When u1 =3D 1 m/s again, the energy lo= ss is now:

<= !--[if gte mso 9]>

Expressed as average power this is 20/1.2 =3D 16.7W

Compared with 30J or 25W on the erg.

The mass distribution is that of a “normal” boat. When a roll rigger is considered and we assume m= ₁ =3D 60 kg and m₂ =3D 30 kg the result is the same. For the type = of energy loss as discussed here the roll rigger does not offer any advantage.=

Situat= ion 2

As situa= tion 1, but rower is pulling the oars. For the return it does not make any differen= ce. For the drive it may be expected that there is a situation as on the erg wh= en pulling the chain. Force and (relative) velocity have the same direction. T= he rower is during the whole drive generating power. Kinetic power is transfer= red to the oar, no power is dissipated.

The next= figure is a result of the simulation model. (Different parameters as used above). It shows the leg force during the whole cycle. Only in the second part of the return force and velocity have an opposite sign but the force is here very small. A very small amount of energy is dissipated. In the small transition zones the situation is not quite clear but that is for this discussion less important.

&nb= sp;

Conclusion

In rowin= g in racing shells the energy loss due to momentum transfer is small. This is another conclusion than Atkins= on’s. For a mechanical system, simpler = than rowing it is shown that energy loss due to momentum transfer can reach non-= negligible values. Click here.<= /o:p>

Leg force during drive and return