Summary: Positive work in cyclic motion

[Paolo De Leva - Sport Biomechanics, Rome,italy]

Dear subscribers,

the "volcanic" discussion about "POSITIVE work in cyclic motion" was
born actually as a branch of another passionate discussion about "Work in
tethered swimming" (see summary, already posted by Robin Burgess-Limerick).
I changed the subject heading because I was convinced the concept of work
in cyclic motion deserved generalization.
I am very happy for the results of the discussion, and I thank
everybody for their contributions. I also thank our moderator Ton van den
Bogert and the others who stressed the need for this discussion to stay
public, and Rick Hinrichs for his friendly advices about the discussion
tone, which had certainly become too passionate and aggressive, at least
on my part.
For the sake of completeness and clarity, my summary contains a few
new simple concepts. Since the basic points were clarified during the
discussion, I hope these new secondary concepts won't need or elicit further
discussion.

A TRICKY TEST FOR FUTURE BIOMECHANICS STUDENTS
----------------------------------------------
My hope is that this discussion will prove that particular care
must be used when teaching the concept of work to biomechanics students.
The concept of work done ON or BY the human body is not as simple as the
concept of work ON or BY a particle. We were trying to show that there
is positive INTERNAL work done BY/ON the human body in cyclic motions.
The following test turns upside down the problem (you'll see why).
However, it requires the same deep understanding of the concept of work.
Perhaps, surprisingly, it is even more tricky and apparently unacceptable
than the object of our discussion:

THE TEST | An athlete, during a vertical jump, lifts his CM by
| 0.5 m. The athlete is standing perfectly motionless both
| at the initial and final positions. Initial and final
| positions are identical. How much is the total INTERNAL
| work done by/on the athlete's body?
| Neglect air resistance and floor surface
| deformation+displacement.

TEST ANSWER
-----------
There's no change in the body's total mechanical energy
(potential+kinetic), hence total work on the body (internal+external) is
zero. There are two EXTERNAL FORCES: (1) ground reaction force, and (2)
weight.
1) Ground reaction force does no work (motionless floor surface),
and I have reasons to believe that this is probably the part of the
solution which will be most difficult to digest for the student. Force
is done, but there's no displacement. In fact, YOU HAVE TO CONSIDER
THE DISPLACEMENT OF THE POINT(S) OF CONTACT between the system and
the external objects: the surface of contact between shoes and floor.
[NOTE: Ground reaction force can be defined a contact force. It is a
mistake to use the displacement of the center of mass (CM) to
compute work by contact forces. Contact external forces
are not applied on the center of mass. Although F=ma refers to the
acceleration of the CM, W=Fs does not refer to the displacement of
the CM. Not even if the object on which the force is applied is a
rigid body!]
2) Weight is a "conservative force" and does no net work (net
displacement of all particles=0).

Hence the total EXTERNAL work is zero. Considering that also
the total work (internal+external) is zero, the INTERNAL work (total
minus external) MUST BE ZERO!
Why then the athlete jumps 0.5m high (getting temporarily some
potential energy), if no work is done by the ground reaction force?
Internal work is responsible for that: although total internal work
is zero, positive internal work is done in the first part of the jump
(ascending phase). Negative internal work is then done during the
second part (descending phase).
Please, read the rest of the summary and the compilation of
postings if you couldn't understand my terminology or my rationale.

OPERATIONAL DEFINITIONS OF EXPRESSIONS
--------------------------------------
1) The total work done ON a system (e.g. the human body, the
swimming pool, etc.) is the sum of the works done ON all the particles
that compose the system, both BY other particles of the system and BY
external systems.
2) The best definition for cyclic motions, within the boundaries of
this discussion, is: motions that start and end exactly in the same position,
without net change in internal mechanical energy (sum of potential and kinetic
energies). When there's a net change in internal mechanical energy (for
instance when the upper limb of the swimmer is rotating faster and faster),
things become slightly more complex, and I think the simpler the better.
[NOTE: By the way, changes in heat energy are not relevant at all,
because heat does not depend on work (by definition, according to the first
law of thermodinamics)].
3) I define internal work as the total work done by internal forces.
Hence internal work is work done by parts of a system on other parts of
the same system (see my first message). I don't know if this definition
is widely accepted; however, at least I can say that it is very useful
and that nobody rejected it during this discussion.
[NOTE: Notice that the sum of internal forces is always zero,
due to the first Newton's law, whilst the sum of internal works done by
those forces may be different from zero, either positive or negative!.
I know this is confusing for someone. Remember that you have to multiply
by the displacements. Also, I am sure everybody can see that internal
mechanical energy can be changed by internal forces. Well, mechanical
energy can change only if work is done. For instance, an astronaut floating
motionless in the space will be able to increase his mechanical energy,
using positive INTERNAL work, by flexing an elbow, or turning his head
to the right]
4) I define external work as work done by the system on its
external environment, OR VICEVERSA. Again, I don't know if this definition
is widely accepted.
[NOTE: External work can be done either FROM A DISTANCE by
gravitational-magnetic-electric forces, or BY CONTACT FORCES exerted
by external objects that touch the boundary of the system. Whilst
gravitational-magnetic-electric forces can be applied on all the
particles of the system, contact forces can be applied only on the
("superficial") particles that are adjacent to the system boundary.
This is why external work by contact forces must be computed using the
displacement of contact points (not the displacement of the center of mass
of the system)!]
5) External resistance: an external force that is always opposite
to the direction of motion of the body on which it is applied (e.g. water
resistance for the hands, feet, knees, shoulders, nose, etc. of a swimmer).
External resistance is conventionally classified as a "non conservative
force", because it does negative work even when net displacement is zero
(i.e. in cyclic motion). In fact, IT KEEPS DOING NEGATIVE WORK ON THE BODY,
whatever is its direction of motion.

DEFINING WHEN WORK IN CYCLIC MOTION IS POSITIVE
-----------------------------------------------
The subject heading "POSITIVE work in cyclic motion" already contained
the conclusion on which I believe we all eventually agreed. There are cyclic
motions that require a non-negligible net positive INTERNAL work BY the system
that is performing the cyclic motion (e.g., the body of a swimmer, or a
fan - see compilation of answers).
Positive internal work during cyclic motion may also imply
a net positive work BY the system (e.g. the swimmer) ON its external
environment (external work).
[NOTE: For the sake of completeness, I wish to add that the
latter statement is not always true; positive internal work
during cyclic motion can be associated with null or negative
external work done by the system on its environment.
For instance, internal work is positive and external is
zero if the system (e.g. our austronaut) performs cyclic motion
without applying external forces. The same is true if there's
an "external resistance" against the cyclic motion, but this
resistance is only due to friction between the external surface
(boundary) of the system and a >>>motionless But the swimmer moves the water, so is not the swimmer doing work
> against the water? To determine this work, you would need to know
> the flow rate (volume (mass) and velocity)
>
> But that is only considering a system that incloses the hand
> and the whole water pushed. In the whole system, water must
> replace the water pushed back (assuming water is not compressed
> and there is no cavitation (vacuum)) and thus again the net
> work is zero. The water is just a medium that effectively
> allows the swimmer to push against the pool wall.

The truth is that the swimmer does positive work on some
particles of water. It does not matter at all that other systems at
the same time do work on other particles to refill the theoretical
gap left by the water moved by the swimmer (by the way, this is
positive work too). The important thing is that THE SWIMMER (actor)
does positive work on the particles of water that he touches and
pushes, and spends energy to do it. The work done LATER on these
particles BY THE WALL, for instance, or BY OTHER PARTICLES of water,
is not performed BY THE SWIMMER. Therefore, we are not interested in
it!
Again, I stress the importance of being extremely clear about:
(1) who or what is doing the work (THE SWIMMER, in our problem)
(2) the object(s) or particle(s) on which the considered work is done
(the weight, if the weight is moved, and every single particle of
water that is moved by the swimmer).
Who did the work to fill the theoretical gap? Was it the swimmer
or something else? Of course it was not the swimmer. I cannot discuss
exactly what happens to the infinite particles of water that are in
the pool. One thing I know for sure: the total work done by the
swimmer ON THE WATER is not zero!
Not to talk of the fact that the water increases its total
KINETIC ENERGY after every cycle, which means that the TOTAL work
done ON the water, BY swimmer, gravity, and walls is positive as
well.....

Consider also that the swimmer pushes a small amount of water
forward, e.g. when the arm splashes into the water after the arm
recovery. Somebody might think that this is negative work, and can be
subtracted from the (much larger) positive work done to push water
backward. This is of course wrong. The particles that are pushed
forward are not the same as those that are pushed backward....
Here I am stressing the importance of considering SEPARATELY
each of the particles on which the work is performed.
And let's not forget any particles! For instance, some water may
flow toward the swimmer's arm or body, hit the skin and receive
negative work FROM the swimmer, but this is negligible, compared to
the positive work done by the arm on many other particles of water
that are touching the arm at that time. Some negative work is also
done by the swimmer's legs on other particles of water.

I'll give three examples that I used to clarify the concept
in my mind before writing this message:
1) Somebody throws an object vertically upward, as high as he
can, starting from floor level. The object then falls down on the
floor and stops at its initial position. The net displacement during
the whole period of time was exactly zero.
In the time needed for the object to go from its initial to its
final position (initial=final=on the floor), what kind of work was
done BY the man ON that object (positive, negative, or zero)?
Nothwithstanding the fact that the net displacement was ZERO (!)
the work done by the man was large and positive! The TOTAL work was
zero (negative work was performed on the object BY SOMETHING ELSE,
NOT BY THE MAN), but we don't care. The man spent a lot of energy to
perform his LARGE POSITIVE work! Apparently, the work was zero.
However, the energy that was used to lift the object came from the
man's muscles, while the energy to bring the object back to its
initial position was spent by the gravitational field (potential
energy)! The work BY YOUR MUSCLES ON THE OBJECT was positive
(displacement and force have same direction). The work done BY THE
GRAVITATIONAL FORCE ON THE OBJECT was zero (first negative, then
positive, constant force). The work done BY THE FLOOR ON THE OBJECT
to stop its fall was negative! Total work ON THE OBJECT=ZERO.
If what Peter wrote were true, we could say that a chairman who
is bringing a suitcase upstair at the third floor, then throws it out
through the window does the same work as another chairman who covers
the same horizontal distance on a completely horizontal path. I
disagree. The chairman problem is not so simple as it seems, though
(and this is what always happens when you deal with work and energy,
in my opinion). In fact, what if the suitcase were not thrown out of
the window, but simply brought back by the chairman himself
downstairs, to its initial position? Would the chairman do a
positive, negative or null work in the vertical direction? I leave
the answer to you.
Here are two other problems:
2) In the last three minutes you have been pushing a car . The
car has been running at constant velocity. There has been no change
in its kinetic energy. Therefore, NO WORK has been done on the car.
Did you do some work during these three minutes? YES, of course! The
TOTAL WOK was zero, but the "component" done by you was LARGE and
POSITIVE. Friction did negative work, but you don't care, you spent a
lot of your energy (some lost into heat, the rest used to perform
work), and this is what you are interested in.
3) Eracles is pushing two cars at the same time in two opposite
directions. The forces he is applying on the two cars have exactly
the same magnitude. At every given instant, the two cars move exactly
at the same speed, and their accelerations have exactly the same
magnitude. Hence during the considered period of time the two cars
covered exactly the same space (the magnitude of their displacements
was exactly the same).
How much was the total work done BY Eracles ON BOTH CARS
(positive, negative, or zero)? Consider that the center of mass of
the two cars together (car 1 + car 2) DID NOT MOVE! Its displacement
was ZERO! Even Eracles' center of mass did not move at all. Its
displacement was ZERO!.....

Peter also wrote:

> 4. The work done by the movement of the body
> segments is called internal work and over one cycle,
> is equal to zero. Swimming is just a series of movement
> cycles and thus the net internal work is zero.

I totally disagree. Internal work is not zero over one cycle,
EVEN THOUGH ANGULAR ACCELERATION IS ZERO, and final velocity+position
are the same as initial velocity+position! (Not to consider that,
when final velocity after one cycle is larger or smaller than initial
velocity, even the TOTAL work is not zero, but this is well known,
and my point is different).
By the way, internal work, in my opinion, should be defined as
work done BY parts of the body ON other body parts (muscles on bones,
bones on bones, etc.). Peter's definition is not clear, and I don't
know what exactly he meant. I just know that my own definition comes
directly from that of internal FORCES.
Peter, again, as others did before him, forgot to analyze the
different components, and isolate what we are talking about. In this
case we are dealing with what I would call the TOTAL "ANGULAR" WORK
performed ON the upper arm (actually, the work done in the tangential
direction on the rotating particles of the upper arm). In spite of
the fact that initial (linear) position = final (linear) position,
the angle is always increasing, and never goes back to zero. This
means that final ANGULAR position=initial ANGULAR position + 1 turn
(2*PI radians). (Of course I simplified the situation, by assuming
that the arm is moving on a plane instead of a 3D space; again, I
point out that the situation is extremely complex). Therefore, here
we are not going back and forth (as the chairman with the suitcase).
The arm keeps moving always in the same (angular) direction. It
follows that the work done by the muscles to keep the arm rotating at
constant speed is always POSITIVE, and becomes larger and larger, as
the arm rotates.
And why the swimmer needs to apply WITH HIS MUSCLES a torque
(tangential force=positive work) on the arm to keep the arm rotating
at constant angular velocity? Of course, because there is a QUITE
INTENSE, not negligible equal and opposite EXTERNAL torque applied on
the arm due to water resistance! And this external torque means that
there is an external NEGATIVE WORK done BY the water ON the arm.
Hence:
- TOTAL torque is zero,
- angular acceleration is zero (I am simplifying, of course),
- TOTAL work is also zero;
- however, INTERNAL WORK is large and positive.

The following examples can be used to better understand cyclic
motion. Consider for example the cyclic motion performed by the
forearm of a man who is shaking a coctail. Let's say the man performs
that action by cyclically flexing and extending the elbow, and let's
assume for the sake of simplicity that the force of gravity is zero
(the man is on a spaceship). Let's analyze one cycle (simplifying, of
course):
1) Flexion - first half: BICEPS (elbow flexor) does positive work ON
the forearm+shaker.
2) Flexion - second half: TRICEPS (elbow extensor) does negative work
ON the forearm+shaker
3) Extension - first half: TRICEPS does positive work ON the
forearm+shaker.
4) Extension - second half: BICEPS does negative work ON the
forearm+shaker.
Conclusion:
Total work BY BICEPS=zero.
Total work BY TRICEPS=zero.
Total work (BICEPS+TRICEPS)=zero.
Now let's consider a very similar motion: hammering a nail into
the spaceship wall. Can you see that in this case the work done by
BICEPS over one cycle is not zero anymore? Some of the negative work
is performed by the wall+nail on the forearm+hammer system.
Conclusion:
TRICEPS work is zero
Total work is zero
BICEPS work is positive!
TRICEPS+BICEPS work is positive!.

This is Peter's conclusion:
>In summary, the tethered swimmer is a system that isn't
>designed to produce work, so one shouldn't expect to
>be able to measure any work done. The system is 100
>percent inefficient.

Nonsense, in my opinion.

By the way, it should be also noticed that the energy spent by
the swimmer is much more than the positive work done by the swimmer
on external bodies! Efficiency is low, but not zero.

P.S. I admit I did a terrific misktake while I was writing the draft
of this text. Luckily, I asked an opinion to my friend Jesus Dapena,
who spotted my mistake immediately. Thanks a lot to Jesus. Now, I
believe that everything is correct, but who knows? I am not one of
those who can be called an expert of work and energy.

__________ _________ ___________~___ ________ _________________~___
/ ~ ~ ~ ~ \
/______________~______~__________ _______~_____~______________~_____~_____\
| Paolo de Leva ~ \ Tel.+ FAX: (39-6) 575.40.81 |
| Istituto Superiore di Educazione Fisica > other FAX: (39-6) 361.30.65 |
| Biomechanics Lab / |
| Via di Villa Pepoli, 4 < INTERNET e-mail address: |
| 00153 ROME - ITALY \ deLEVA@RISCcics.Ing.UniRoma1.IT |
|_____________________~________~__________________ __________________ _____|
challenging entropy :-)


************************************************** **************************
Date: Tue, 5 Sep 1995 11:59:07 SAST-2
From: Craig Nevin
Subject: Re: POSITIVE work in cyclic motion

Hi netters,

Much has been said about the work done by a swimmer tethered at the
waist swimming _stationary_ in a pool. In this case I would like
to echo Paolo de Leva's comments regarding the principle of work.

> ... the concepts of work ... in ... application to the study of
> human motion is so complex that deserves either:
> a) a lot of study and a humble, doubtful attitude.
> or
> b) a lot of study, tons of effective thinking, and a good mind.

For practical reasons, I would suggest that it is safer to stick to
(a) :-)

As for the _study_ here are some useful references.

1. Knuttgen HG. Force, work, power and exercise. Medicine and
Science in Sports (and Exercise), 10(3):227-228, 1978.

Here the basic distinction is drawn that work has a particular
definition which is often confused with exercise or "effort".

Work = force x displacement.

Displacement is a vector quantity therefore in cyclic movements the
work should be zero. (Note the _should be_ :-)

The mechanical definition of the term work differs from the common
usage of the term work; for example to "work-out" in the gym.
If the weight-lifter returns the weights to the rack where he found
them, he (should have done) done zero work. He may have performed
many hours of exercise, but has (probably) done zero work. It is
perhaps unwise to insist too loudly/repeatedly that this is only a
"small semantic technicality", particularly within earshot of a
steriod-filled weight-lifter. (This is where the humble attitude
comes into it :-).

> Peter Davidson wrote:
> > ...Regarding your question on the amount of work done
> > during tethered swimming, I believe the answer is zero.

> Paulo de Leva wrote:
> I totally disgaree. It is true that the work done BY THE SWIMMER
> ON THE WEIGHT is zero, because the weight does not move. However,
> there's a lot of positive work done BY THE SWIMMER ON THE WATER.

> Peter wrote:
> > But the swimmer moves the water, so is not the swimmer doing work
> > against the water? To determine this work, you would need to know
> > the flow rate (volume (mass) and velocity)

This brings in the second point of confusion about work -- it is time
independant. Power is work x time. Again from the reference
material, it is better to refer to power when time is involved. By
introducing time (flowrate, velocity, instantaneous dislacements...)
we have digressed, and are no longer discussing work, but power.

The distinction needs to be borne in mind. Taken to its logical
conclusion imagine the swimmer swimming so slowly that the water is
hardly disturbed. The water that is pushed aside returns to fill
the gap. The _water_ has not displaced but individual molecules
have. You therefore you have to consider whether the swimmer is doing
work on the homogeneous _water_ or on the water molecules! The work
done on the _water_ is zero because it has not displaced (except for
that splashed out of the pool, of course ;-). However, this IS
measureable as a displacement of the weight.

The ultimate end point of any work is motion between perfectly
frictionless movements between molecules, which has another
name - - HEAT. Heat seems to come into it quite a bit (resulting in
heated debates, hot and sweaty exercises, people getting hot under
the collar, etc etc..) therefore let's look at the thermodynaic
definition of work as contained in the article:-

2. Webb P, Saris WHM, Schoffelen PFM, Van Ingen Schenua GJ,
Ten Hoor F. The mechanical work of walking: a calorimetric
study. Medicine and Science in Sports and Exercise, 20(4):331-
337, 1988.

Work = the energy transferred from a system NOT in the form of heat.

Now things get complicated because we cannot easily determine what
portion of the energy content of any given system ultimately
downgrades to heat (the term ultimately is safely used here because
we do know that work in independant of time). We also do
not know how much energy is stored in the system as body
fat, muscle glucogen etc. For this purpose, a thorough energy
balance here is essential; if one term is neglected or erroneously
estimated, then the whole calculation may be invalidated. (This is
where the tons of thinking and a good mind become necessary!)

I will pass on that one, but refer to another reference which
contains as complete an energy balance as I have seen anywhere.

3. Ward-Smith AJ. A mathematical theory of running, based on the
first law of thermodynamics, and its application to world-class
athletes. Journal of Biomechanics, 18(5):337-349, 1985.

The measurement of these quanties can be estimated or measured in a
whole-room calorimeter over a period of days (which allows the
energies to reach equilbrium).

The interesting conclusion of study [2] is that work IS done
during walking, but it is not done during cycling (please excuse any
misinterpretations here, as I am working (sic) from memory).
These results imply that not all the energy released from the human
body reappears as heat while walking; but all the energy reappears as
heat during cycling. Work can be seen as the mechanical transfer
of work across a boundary; it probably will dissipate as heat
within the receiving body, but may not :-).

At this stage, based on all this evidence -- and a great deal more
that I have not discussed -- I would hazard a guess that the work
done during swimming is also zero.

I base this assessment (in very, very, very, very broad terms on
the fact that cycling consists of cyclic motions as does swimming,
whereas walking is an oscillating movement :-). This conclusion is
based entirely on speculation -- and the wise reader should treat it
with the utmost sceptism... however I believe it to be correct.

If you want to test this assumption on a swimmer, you would have to
dress them in a thermal wetsuit and do the whole body calorimeter
experiment. The emersion of the body in water is so well suited
(sic) to a calorimetry experiment, that it would not surprise me if
it has not been done already.


Craig Nevin
Biomedical Engineer
Department of Physiology/Sports Science
University of Cape Town, South Africa
CNEVIN@anat.uct.ac.za

************************************************** **************************
Date: Tue, 5 Sep 1995 09:21:52 -0400
From: "Heather L. Beck Abushanab"
Subject: Re: POSITIVE work in cyclic motion

Actually, work has alot more definitions than the one being described
in most of the tethered swimmer analysis. Yes, in a LINEAR mechanical
system work equals the integral of force acting through a distance, which
for a CONSTANT force is

Work = force x displacement.

In a ROTATIONAL mechanical system, the work is found by the integral of
torque acting through an angular displacement. Again for constant torque

Work = Torque x angular displacement.

There are other types of systems, so of course there are other types of
work, mainly electrical work and fluid work.

The tethered swimmer can be analyzed in a variety of ways depending on what
the real question is in regard to work. Net translational work done on the
mass? Rotational work produced by the arms? Biomechanical work produced by
the muscles? Total work done on the water? As others have said, the first
job in analyzing a problem of this sort is to identify the system of
interest.

The net translational work is easy to identify (zero displacement, zero
work), but all of the other calculations are extremely complex since the
torques, forces, local water pressure, etc are changing with time.

- Heather Abushanab
BU NeuroMuscular Research Center


************************************************** **************************
Date: Tue, 5 Sep 1995 19:15:33 +0200
From: "Paolo de Leva - Sport Biomechanics, Rome, Italy"
Subject: Re: POSITIVE work in cyclic motion

Dear subscribers,

after spending an entire Sunday for writing my latest
message, I feel it difficult not to react with dismay to some of Craig
Nevin's statements. Craig didn't understand at all my point.
He just repeated the same mistakes against which I wrote.

Are there any other subscribers who want to defend Newton?
I quit. I feel I have already done more than I should.

Craig Nevin wrote:

> The _water_ has not displaced but individual molecules
>have. You therefore you have to consider whether the swimmer is doing
>work on the homogeneous _water_ or on the water molecules! The work
>done on the _water_ is zero because it has not displaced (except for
>that splashed out of the pool, of course ;-). However, this IS
>measureable as a displacement of the weight.

The work done

>>>>>>>>>>>>>>>>>>>>>>>BY THE SWIMMER 2) In the last three minutes you have been pushing a car . The
>car has been running at constant velocity. There has been no change
>in its kinetic energy. Therefore, NO WORK has been done on the car.
>Did you do some work during these three minutes? YES, of course!
>... you spent alot of your energy ...

See point 1. Besides, one of the conditions of the question is
that no work is done.

> 3) Eracles is pushing two cars at the same time in two opposite
>directions. The forces he is applying on the two cars have exactly
>the same magnitude.....

Again no work is done. At a constant velocity any resistance
along a level ground would be to overcome friction. (See Newton's Laws)
Overcoming friction is not a form of work just a form energy
lost to disorder.

I wrote:

4. The work done by the movement of the body
segments is called internal work and over one cycle,
is equal to zero. Swimming is just a series of movement
cycles and thus the net internal work is zero.

> I totally disagree. Internal work is not zero over one cycle,

This implies that there is some form of internal energy gain
from each cycle. I think the biggest problem of exercise is energy
drain, not accumulation of energy.

>By the way, internal work, in my opinion, should be defined as
>work done BY parts of the body ON other body parts (muscles on bones,
>bones on bones, etc.). Peter's definition is not clear, and I don't
>know what exactly he meant. I just know that my own definition comes
>directly from that of internal FORCES.

It is common knowledge that the body moves its segments by forces
between body "parts".


>...direction on the rotating particles of the upper arm). In spite of
>the fact that initial (linear) position = final (linear) position,
>the angle is always increasing, and never goes back to zero. This
>means that final ANGULAR position=initial ANGULAR position + 1 turn
>(2*PI radians). (Of course I simplified the situation, by assuming...

Work done is compared between one state relative to another.
The fact that you can refer the same relative angle as 30 degrees, 390 degrees
and 3630 degrees does not create more energy between these states.

>And why the swimmer needs to apply WITH HIS MUSCLES a torque
>(tangential force=positive work) on the arm to keep the arm rotating
>at constant angular velocity?

To overcome friction and to create the resisting force.
See previous point on friction.

I have no further comments on Paolo de Leva's reply. I did not understand
the point of the spaceship examples and I cannot reply to the
"Nonsense, in my opinion" remark when no details were given.


------------Peter Davidson----------- *********
Doctorate Student, Biomechanics .:***********>*****
Health Sciences *:@*************>****
Universiy of Otago *** ******>****
PO Box 913, Dunedin ** *********
New Zealand * I I
peterd@gandalf.otago.ac.nz ~ ~

understanding entropy


************************************************** **************************
Date: Tue, 5 Sep 1995 23:54:02 -0500
From: "Leonard G. Caillouet"
Subject: Re: Positive Work in Cyclic Motion

Peter Davidson made two points that were close to being correct but are
in fact examples of his own mistaken description of the problem of
describing work. He states that "work is a form of energy..." In fact
work is the transfer of energy. When describing the work done by an
object, or the work done on an object one must define the question of
interest carefully. As Peter stated, the system must be defined. His
mistake can be illustrated most simply by the example of the object
thrown straight up which falls to rest in exactly the same location. The
problem is that the object doing the work we are interested in measuring
was not defined carefully. There was indeed no net work done on the
object thrown BY ITS ENVIRONMENT, but there certainly was work done by
the thrower on the object. The thrower displaced the object against the
force due to gravity and changed its mechanical potential by the distance
above its resting position it was lifted. The fact that gravity caused
the object to be displaced in the other direction does not make the work
done by the thrower zero.
Similarly, the fact that the kinetic energy transfered to the
particles of water is ultimately dissipated to the environment does not
make the work done on the water BY THE SWIMMER zero. This work would
seem to be the transfer of energy that is of interest in this
experimental condition.
To quickly summarize, the discussion of work must include the
definition of the bodies of interest because work is not energy but the
transfer of energy. If the bodies of interest are not described in the
statement of the problem we can expand the system infinitely to the level
of the universe and the problem becomes one of "understanding entropy."
Careful use of terms such as work and energy is essential.
Considering them to be equal would be like using heat and thermal energy
to mean the same thing. A final note to Peter on this point: You
suggested that there may be confusion on the list concerning "effort",
"work", and "energy". I suspect that few of us are confusing effort and
energy, but the latter terms may be used carelessly. The more important
lesson is to carefully define the objects of study between which energy
is being transfered. It is one of the first lessons in mechanics that
Newton would have taught, I'm sure. It's certainly one that most of us
who studied physics or mechanics had to learn.

Leon.

Leonard G. Caillouet
Louisiana State University
MS Candidate, Dept. of Kinesiology


************************************************** **************************
Date: Wed, 6 Sep 1995 16:11:07 +1000
From: "NAME \"Ross Armstrong\""
Subject: Newton and +ve work in cyclic motion

I'm not a biomechanist, per se, but I have used biomechanice
in my work and appreciate it as a discipline. I have a sneaking suspicion
that there are a few biomechanists out there who don't
understand properly the basic physics of their discipline.

eg at a conference once I witnessed the following overhead presented in an
almost reverent way:

F=ma (Winter, 1979)

Admittedly the paper was based on a masters' thesis but the day someone
presented one of Newton's laws and attributed it to Winter (1979) is still
discussed in ergo circles here.
The masters' topic by the way was on biomechanics and being supervised by
a biomechanist.

If there are biomechanists out there who don't know there mechanics and
dynamics that well what does this say about those who taught them and
those they teach?

Ross Armstrong
Lecturer in Ergonomics
Manager Centre for Ergonomics and Human Factors
La Trobe University
tel 61 3 285 5311 fax 61 3 285 5184
email hubrga@lure.latrobe.edu.au
Snail Mail: Locked Bag 12, Carlton Sth PO
Victoria, 3053, Australia


************************************************** **************************
Date: Thu, 7 Sep 1995 16:33:49 MDT
From: Ton van den Bogert
Subject: Summary vs. discussion

Dear subscribers,

It is certainly not the policy of Biomch-L to allow only
questions and summaries. Some topics require a public
discussion, and I personally feel that the work-energy discussion
is one of those topics.
[omissis]

-- Ton van den Bogert, Biomch-L co-moderator


************************************************** **************************
Date: Thu, 7 Sep 1995 16:22:26 -0700
From: "Rick Hinrichs (Ariz. State Univ. USA)"
Subject: open dialog

Dear Colleagues:

OK, I've been outvoted!

I agree that we need to preserve the option of dialog on this list.
However, the discussion was starting to appear in the form of personal attacks
rather than healthy dialog, hence I found it necessary to say something.
Don't get me wrong, I am just as interested in work and energy discussions
as anyone on this list. I didn't want to simply delete the postings
before reading them. I wanted the tone to change and thought that a summary
might help set the stage for a more intelligent discussion.

Remember, when you are contemplating a reply to a biomch-l posting, please
communicate with the person (directly or through the entire list if
appropriate) in a professional manner, as if you were making comments to or
asking questions of a presenter at a professional meeting..

(As Herman always used to say): With warm regards,

--Rick

----------------------------------------------------------------------------
| Richard N. Hinrichs, Ph.D. | email: Hinrichs@ASU.EDU |
| Associate Professor | or Hinrichs@ESPE1.LA.ASU.EDU |
| Dept. of Exercise Science & PE | or atrnh@ACVAX.INRE.ASU.EDU |
| Arizona State University | Phone: (602) 965-1624 |
| Tempe, AZ 85287-0404 USA | FAX: (602) 965-8108 |
----------------------------------------------------------------------------


************************************************** ****************
Date: Fri, 15 Sep 1995 09:51:28 -0300
From: Jianyu Cheng
Subject: Tethered swimming and work in cyclic motion

Dear Biomch-L,
There have been plenty of thoughtful discussions on the work
done by the tethered swimmer. I agree with Paolo de Leva and some
others, the work done by the tethered swimmer on water per cycle
is not equal to zero and is positive generally. The work can be
calculated, as we often did in the studies of aquatic animal
locomotion, if you know hydrodynamic forces acting on the swimmer
(basically the fluid pressure field near the swimmer) at any
instant. Multiplying the distributive hydrodynamic force by
the corresponding differential displacement, and summerizing
(integrating) it over the whole body surface and over one cycle
give a value, negative of which is the work done by the swimmer
on surrounding fluid and is supplied by the mechanical energy
produced by muscle contraction. But this method probably won't
help much in solving the problem posted by the originator of
the debate. The reason is that, I believe, little about the
fluiddynamics of human swimming has been known. I would expect
that the fluid flow around a normal adult human swimmer is
associated with large or middle Reynolds numbers (during swimming).
Thus the hydrodynamic forces acting on the body or body parts at
any instant is path dependent and even history dependent
(we need to know complete kinematics in order to obtain
the hydrodynamic forces).
I would say that many arguments about the fluid flow in
previous posting are wrong, but this can be excused as those
people probably have not much backgroud in the mechanics
of continuous media.
For those of you interested in swimming, I have listed below
some of our work on aquatic animal swimming. These studies
include hydrodynamics, swimming mechanics, and dynamics of
locomotor system. We can quantify the mechanical energy
generated by muscle contraction, and its partition into
water for propulsion, to deform internal soft tissues,
and maintaining cyclic motions of the body parts.


Cheng J-Y., Zhuang L-X., Tong B-G., Analysis of swimming three-dimensional
waving plates, J. Fluid Mech., Vol .232: 341-355, 1991
Blickhan R. & Cheng J-Y., Energy storage by elastic mechanisms in the
tail of large swimmers-- a re-evaluation, J. Theor. Biol., Vol.168:315-321,
1994
Cheng, J-Y & Blickhan R., Bending moment distribution along swimming
fish, J. Theor. Biol., Vol.168:337-348, 1994
Cheng, J-Y & Blickhan R., Note on the calculation of propeller
efficiency using elongated body theory, J. Experimental Biology, Vol.192:
p169-177, 1994
Cheng, J-Y., DeMont, M.E., Jet propelled swimming in scallops:
swimming mechanics and ontogenic scaling, Canadian Journal of Zoology
(submitted)
Cheng, J-Y., DeMont, M.E., Hydrodynamics of scallop locomotion:
unsteady fluid force acting on clapping shells, J. Fluid Mech.
(submitted)
Cheng, J-Y., Davison, I & DeMont, M.E., Dynamics and energetics
of scallop swimming, J. Experimental Biology, (submitted)

Regards!

Jianyu Cheng
Biology Dept, STFX University, Canada

************************************************** ****************
Date: Fri, 15 Sep 1995 11:43:31 PST
From: Dave Maurice
Organization: U.S. Bureau of Mines
Subject: work in cyclic motion

Hello friends:

I am sorry to be entering into this so late, and I must confess I
have no knowledge of what started this. However, I have seen several
well-intentioned postings the last few days which have reflected some
of the confusion which commonly arises from considerations of cyclic
motion. It's very easy to get confused with this, but I hope to try
to clarify some of the points that seem to confuse people most
frequently. In so doing I hope to avoid offending anybody; if I
point out a (perceived?) error please don't think I am singling
somebody out.

The most common misconception that arises is that if the final
position is the same as the initial position, no work has been done.
This stems from an (sorry, every word I think of for here sounds
unfairly critical) interpretation of the definition of work, which
most people take to be W = F delta X. However, a more illustrative
definition in my opinion is that W = the integral of FdX. This
seemingly simplistic (even semantic) distinction makes a profound
difference.

Rather than insult your intelligence with a review of this (because I
can hear the lightbulbs clicking on all over the world I know it
isn't necessary) I'll instead risk the wrath of a few courageous
folks who unfortunately overlooked this in their attempts to explain.
Friends, this isn't to draw attention to your errors, but is done
because those errors are the ones we ALL are inclined to make, and
because the examples will clarify the necessity of integration over
the path rather than mere attention to the endpoints.

If a hand presses against a spring and compresses it some distance,
then the hand performs work on the spring. If the hand relaxes and
the spring presses the hand back to the original position, then the
spring perofms work on the hand. Now that they are returned to the
original position, the temptation is to say that the net wok done on
the spring is zero. But it isn't; in fact the return to the original
position in no way reduces the work done on the spring. If you wish
to see this mathematically you can graph it; for a constant rate
spring you will get a triangle whose area is a measure of the work
done. An "easier" way is to make the situation initially more
complex. Imagine that we had a thingamajig (all inclusive term for
mechanical translation device - e.g. a gear) to transform that spring
compression (the motion thereof) into the movement of yet another
piece. When the spring is allowed to return to its initial position,
that other movement would still have taken place, right? And in
fact, this principle is, I am sure you will agree, rather closely
related to the principle behind the Carnot cycle (and the internal
combustion engine).

Back to our example. Upon return to the starting point, the net work
done on the spring is the work done in compressing it. The net work
done on the hand by the spring is the work done in moving the hand
back to the starting position. These do not "sum" to zero.

The swimmer will be the second example. If we cling to the idea that
work is defined by F delta X, then one rotation of an arm would seem
to yield zero. We can't do that. In fact, we can't directly apply
the definition I gave, as it is "phrased" for direct application to
rectilinear motion. For rotational motion it becomes W = integral of
tau d theta, where tau is the torque and theta the angular
displacement. And this will give the type of results that are
desired; if we rotate the arms twice with a constant stroke we get
twice the work, not a return to zero. Now admittedly the tethered
swimmer problem is fairly complex (how to determine tau?) but this is
the approach that will give the mechanical work done by the swimmer
on the water.

Thank you for your attention. I hope I haven't bored or offended to
many of you; if any clarification is desired I will do what I can.

Dave


************************************************** ****************
Date: Mon, 18 Sep 1995 20:35:15 +0200
From: "Paolo de Leva - Sport Biomechanics, Rome, Italy"
Subject: Re: +Work in cyclic motion - D.Lemmon's contribution

Dear subscribers,

I am forwarding a message by David Lemmon, with the subject "My
insight into the work/energy discussion". It was sent directly to my address
immediately after it had been suggested by Hinrichs to summarize this
particular discussion, and before it was eventually decided to keep it
public.
I am very sorry for the delay. I tried to forward David's message
last week, then something went wrong, and I eventually forgot to try again.
I apologize.

As well as the interesting contents of David's message, you may enjoy
his successfull effort of diplomacy. These two things together make this
message an excellent contribution. I hope it will help to reach the final
agreement that I have been seeking (not as patiently as David,
unfortunately). Notice that David used the same (classic)
example used by Dave Maurice in his recent message. However, the
conclusions are different. For those who are interested, I fully agree
with David's conclusions (although he did some not declared, yet totally
acceptable and commonly used, simplifying assumptions).

Maybe there's someone else who wants to post her/his contribution. If
you are one of them, please allow me to insist suggesting you to make always
clear "by what" and "on what", whenever they use the word "work". I feel
this is crucial. We have been discussing several types of work (internal,
external, PARTIAL, TOTAL, on the water, on the weight, by the swimmer, by
the water, etc.), and it is very easy to get mixed up.

I would like to apologize again for my too passionate replies.
Biomechanics has dozens of branches. There's one thing that makes us a
single comunity: our sound common backround in physics (the "mechanics"
component of BIO-mechanics). My intention was only to promote the
developement of this common background, and underline the absolute need of
its soundness. Although I did that too aggressively, and apparently in some
cases I was not very successfull, I still hope that the overall effect was
positive. Anyway, some excellent contributions have been posted recently,
and they showed both the truth and the great and magnificent nature
of BIOMCH-L.

I still have a high esteem of many of you, but now I have a feeling
that several people just needed to participate, directly or indirectly, to
this public discussion about simple basic physics, though I might be wrong.
Let me know if you agree. Personally, I enjoyed discussing this topic, but
at the same time I would have preferred not to participate in this discussion
at all. I just felt forced to do so (and I did it with pleasure), for the
reasons I tried to explain above.

With the kindest regards,

---------------------------------------------------------------
Paolo de Leva (DELEVA@Risccics.ing.uniroma1.it)
Sport Biomechanics
Istituto Superiore di Educazione Fisica
Rome, Italy
---------------------------------------------------------------

Here's David's contribution:
----------------------------

************************************************** **************************
Date: Thu, 07 Sep 1995 15:27:39 -0400
From: drl10@psu.edu (David Lemmon)
Subject: My insight into the work/energy discussion
************************************************** **************************

Dear Dr. Paolo de Leva:

I would like to send this message to the original poster of this topic, but
by now I have lost the address of that person, so I am sending it to you.
If you would like to post this response to BIOMCH-L, please feel free.

Like many others, I have been giving the matter considerable thought. I
believe the reason there are such heated emotions on both sides is that both
sides are correct, in a sense. Let me explain.

One side seems to be saying that as the swimmer swings his (or her) arms
around and kicks his feet, he is performing positive work on the water in
which he is immersed. The other side claims that the whole process is
cyclical and thus, if there is no net motion of the swimmer (or motion of an
attached, pulleyed weight), there is no net work being done. I can see
where each is coming from.

Consider a cyclical biomechanical problem in which a person is pressing
against a spring and then releasing back to the unloaded condition, say,
with simple harmonic motion. During compression, the person's hand is doing
positive work on the spring. However, during release, the spring is doing
positive work on the person's hand [NOTE by P. de Leva: although this is
true, I believe that here it would be better to highlight that during
release the work done BY the hand on the spring is negative; that's why at
the end total work on the spring is zero, if you neglect the trifling
effects of friction and spring inertia].
We all agree that work is force times displacement. If we
consider both the force and displacement to be positive on the "in" stroke,
the force on the return stroke is still positive, while the displacement is
negative. Thus, there is negative work being done by the hand on the return
stroke. The net work performed by the hand in this cyclical problem is
zero. Some call these forces "conservative", referring to the principle of
conservation of energy.

Now, let's replace that spring with a damping device such as an automobile
shock absorber, with the same simple harmonic motion. The person's hand
performs positive work in the "in" stroke. However, on the return stroke,
the damping device is not performing work on the hand. Because the signs of
the force and the displacement become negative, the work is still positive.
Thus the net work is positive. So these forces are called "non-
conservative" forces.

In viewing these two scenarios, I believe that the tethered swimming problem
is related to the latter, rather than the former. The forces applied by the
swimmer are non-conservative, meaning, if the swimmer relaxes, the water
will never push the hand back to where it was before. Thus, I would say
that, even if the swimmer is held stationary by a tether and performs
cyclical motion, he (or she) is still performing net positive work.

I hope this is helpful information. I believe in the existence of truth, and
hope that through this discussion, we may somehow come a little closer, at
least in the realm of biomechanics.

Regards,
=====================================
David R. Lemmon, Ph.D.
Center for Locomotion Studies (CELOS)
10 IM Building
The Pennsylvania State University
University Park, PA, 16802 USA
Phone: 814-865-1972
FAX: 814-863-4755
=====================================



************************************************** ********************
From: "Craig Nevin"
To: deleva@risccics.ing.uniroma1.it
Date: Wed, 20 Sep 1995 21:26:10 SAST-2
Subject: Re: personal message about cyclic motion

Paolo:

[omissis]
________________________

Here is a summary of the "work" debate as I see it.

The first law of thermodynamics states

dE = dW + dQ

where dE is the CHANGE in internal energy with a real or imaginary
boundary, dW is the mechanical work transferred across that boundary
and dQ is the heat transfer across the same boundary.

Positive mechanical work done by the system is transferred across
the boundary and does an equalivent amount of negative work on the
other side. By shifting the boundary to include the destination of
the energy transfer, a condition can always be found where zero work
is done. Therefore, whether the work done is positive, negative or
zero depends soley on the choice of the position of the boundary,
and the orientation of the observer in relation to that boundary.

The summation of dW and dQ implies a dimensional equivalence of those
terms. The thermal energy within the boundary comprises frictionless
momentum preserving collisions at a molecular level. The frictionless
properties of these interactions prevent the molecules from gripping
any object, and hence all "mechanical" work eventually downgrades to
thermal energy.

Since "mechanical work" is considered to be of greater practical
value than heat, engineers tend to choose to define the boundary in
such a way as to optimise the positive work component of the
equation. The fact that the swimmer may be doing positive work on
the water is really a technicality. The issue at hand is really one
of how much power he is expending.

Let us make an analogy with the tethered swimmer. Take for example
an electric fan turning in a closed room.

While the fan is rotating, it IS doing work on the surrounding air.
However, if it stops the air will eventually become still, then no
work is being done. The total amount of the work done remains
unchanged, but the work has now been transformed into thermal energy.
How much work was done on the air is little importance as it is no
longer available to do anything practical with anyhow. Work is
therefore, apparently, only of value at the time it is being
performed. To calculate it at any other time may seem pointless
because it will always be zero.

So what is the value of measuring work? Surely power is a more
useful term. The idea is that if we can measure the work, then we
can determine the power from the equation

power = work / time

The power consumed by the FAN can be calculated from the amps x
volts. The power rating of the fan is also usually printed
on a little label on the side or back.

Contrasting the fan example with the tethered swimmer, a direct
comparison can be made. The swimmer's arm flail through the water
doing an indeterminate amount of work. However, when he stops the
water will settle and its temperature rise a little bit.

The difficulty is that there is no electric input to determine how
much work the swimmer has done, and there is no label on his backside
either.

Now we have reduced the problem down to a small a teaser.

We can always ask (1) an electrical engineer to determine the rating
of the fan, or
(2) a mechanical engineer to determine the rating
of the steam or gas turbine that is driving
the electric generator,

But who do we ask to determine the rating of the swimmer?

The answer is, of course, a biomechanical engineer!

Good luck to you,

The issue reminds me of the class where we as engineers are taught the
concept of the "coefficient of friction". It all sounds very
simple. It physics we learn about how friction occurs at the end of
an applied force. Now the time comes to learn how much force is
applied to any object, provided we know the coefficient of friction!

So where do we find these magic coefficients of friction. No one
seems to have any good answers to that question, so we visit the
library and find the biggest technical manual we can find, but there
is nothing much of use there either... Fortunately, the topic in
class changes and we soon forget all about it.

Then we go ergonomics class, and repeat the whole story looking for
the magic "anthropometric data table"! (You would have thought we
would have learnt our lesson by now). But no, we still need to
determine how much work a swimmer does, afterall if we can determine
how much work the swimmer does, then we can use the equations we have
been taught to find the "biomechanical power" of the swimmer.

The problem with real life is that it is not a test, and there is no
one to give us the data to complete the equations. THAT we have to
figure out ourselves, or alternatively ask any biomechanist to explain
it to you.

Good luck,
and have a nice day

Craig Nevin
Biomedical Engineer
Department of Physiology/Sports Science
University of Cape Town, South Africa
CNEVIN@anat.uct.ac.za




*********************
END OF COMPILATION ABOUT WORK IN CYCLIC MOTION
**********************************************