On 4/6/2004, Trot of "Terry and Shirley" posted:

I am sure that some of the clever people on this forum can answer this query for me.

If two people of the same weight ride similar bikes 20 km and one rides it in say 30 minutes and the other in 60 minutes is the same amount of energy used?


David Martin answered:

Good question. Without wishing to appear clever, I'll have a go at it :-)

Let's look at a very much simplified system first, to give an insight into the real problem which is otherwise quite tricky.

Assume a level road and the absence of any friction in the bike / air / road / muscles / joints. The only energy involved in this case is in bringing the bike plus rider from rest up to the final speed. The energy expended is then the kinetic energy (mv²)/2. (Exactly how a bike is brought up to speed in the absence of any friction is an exercise (!) left to the reader :-)

The faster rider in this case will have expended more energy: four times as much if she's going twice as fast.

Back in the real world, the kinetic energy will be negligible compared to the work done against friction, which ends up as heat. The energy expended over any stage of the journey is the force times the distance moved. The way in which the friction force varies with speed is quite complicated and depends on the system involved (e.g. muscles, wind resistance etc.). A good approximation to many systems is that friction varies as the speed squared.

If this assumption is made then, again, the faster rider expends four times as much energy if she goes at twice the speed, since both riders travel the same distance.

The problem is more complicated if the speeds vary over the journey (only the average speeds are given in the problem) but I'll leave it there for now.

In summary, the faster rider would expend *at least* four times as much energy.


Zero Sum responded:

The original post was about "energy used" ie. work...

Kinetic energy is not an appropriate measure of work.  Yes there is a higher energy output for higher speeds, but more ground is traversed.

When all is said and done, a certain number of kilograms have been moved a certain number of metres to a possibly different elevation.  The same work is performed at either speed.

The only difference in the work concerned can be in the "losses".  We have friction and air resistance as those factors.  The friction losses will likely decrease (to a point) with accelerated pace but the air resistance would increase by second power.

The differences in the start and end conditions are not likely contribute a great deal to the overall effort.


Ray commented:

As an aside...

Way back in early secondary school we used physics to prove that homework should be reduced (not that the teacher paid any attention, but it was worth a try).

Since F = ma, and
work (w) = Fd (displacement or distance) then substituting for F,
work = mad.

QED  :)



Zero Sum replied:

For those interested the matter is well covered at
<http://www.physicsclassroom.com/Class/energy/U5L1a.html>

Consider the waiter and tray example....
 
and:


Feel free, you should know better than I anyway...



Yep.  From memory: Work = force * distance * cos (angle of applied force)


Yep.


Yep to the first again but...

No.  There I do not have it.  While I have no issue with your precise definition I do have difficulty with its application in this instance.


Yep.  The kinetic energy represents stored energy.  So?  With regard to overall work done in travelling the distance the only difference in kinetic energy is the difference in energy as the bike traverses the finish line (assuming they both start at zero).  If the journey is merely one metre then this will totally overshadow the work involved in the whole trip.  In a journey of a thousand kilometres it is utterly ignorable (at any speed attainable by human muscles).


Let us look at an example based on your (above) paragraph combined with the problem posed in the original post. ie. "If two people of the same weight ride similar bikes 20 km and one rides it in say 30 minutes and the other in 60 minutes is the same amount of energy used?"

We can assume the same elevation because any differences in elevation will result in the same stored energy potential.   In both cases the work performed to achieve the storage of the potential energy will be the same.

Although it isn't specified it is a reasonably fair assumption that initial velocities and final velocities are zero (ie. the rider coasts to a stop).

This is a reasonable assumption because the specified problem asks about the energy consumed in the journey not how much energy is stored as kinetic energy.  Thus overall there is no change in kinetic energy averaged over the whole journey for either case.  They go from stopped to stopped, no change in kinetic energy.

You said "in the absence of any friction".  With that being the case, there is no change in the energy state of the system from start to end and no work has been done.  End Of Story.  No work performed in either case.

You brought up the integration.  That's fair enough.  But not if you change your interval of integration to suit the argument.  The interval of integration is the whole journey from start to finish.  By integrating over less than that interval you do not answer the problem posed.


Energy and work are identical?  Interchangeable but not the same.  The dimensions are the same
( k*m²/s² ).  But that is like saying energy and mass are the same thing.  They are convertible not identical.


You are changing the problem to suit your argument.  Yes, there is more stored energy in a higher velocity and that energy is a needed input to attain the speed.  SO?  It simply is not relevant to the problem at hand.

In physics, work is defined as a force acting upon an object to cause a displacement.

Work = force * distance * cos (angle of applied force).   We can ignore the angle because it is in the same direction (and the cosine is 1) as the displacement  so we just wind up with force times distance.



Try 'increased speed".


If you visit your local professor of engineering he will tell you that bearings have an optimal speed.  If not rotating fast enough the bearings bump from ball to ball rather than perform a perfect rotation.  This is barely noticeable but does result in increased losses when a bearing runs too slow.

Since having written the above I have read Gerald's post on bearings and I think it at least party illustrates my case.  Until up to optimum operating speed there will be a decrease in friction with speed, after that optimum speed there is an increase.  The properties of the curves depend on the properties of the bearings and lubricants but there is always an optimum operating speed for any bearing.


Neither do I.

I am finding this reply rather hard to write.  I am disinclined to argue with someone in their speciality as other than a provocation (in the logical not emotional sense) but it seems to me that what is going on here is not being clearly stated.  I'll try once more, to explain it.  I am sorry that I lack the education to do it well.

The bikes and rider are going from one state to another state.  If we assume the same elevation for those states there is no difference in potential energy and even if they were, being that it would be the same measure of potential energy.  So speed makes no difference there. If we assume that both states are stationary (stationary at the start and stationary at the finish) there is no difference in the kinetic energy between the two states either, so again speed is not an issue.

In fact there is no difference in the states however long or however short the transition - and I think that that is the answer the poser of the problem was seeking.

It is like sliding a cup across a table, the only work performed is that needed to overcome frictional losses (including air friction).

Now the issue of whether moving the cup (or the bike and rider) against those losses takes more or less work is a different issue - and one we are not likely to be able to settle here.  But since they all go from zero velocity to zero velocity the kinetic energy at any point is irrelevant. Any energy getting stored as kinetic energy will be returned by the time transition from the original state to the final is complete.

In riding a bike there are two things that need to be overcome, rolling resistance and air resistance.  By and large rolling resistance decreases with speed (for as far as human muscles can push a bike anyway) and air resistance increases with the second power of velocity.  If those graphs are summed for any given bike and journey, there will be a low point where the rider has to make the least effort, do the least work, but where that point is none of us can say because the original poser did not provide enough information in the problem.  SO... I don't think that was what they were asking.

The problem as posed was "If two people of the same weight ride similar bikes 20 km and one rides it in say 30 minutes and the other in 60 minutes is the same amount of energy used?"

I can only interpret that as "In a change of one state to another, does it use more energy to transition to that state faster", to which the answer is clearly "No".  Conservation of energy provides that answer.

If you want to pose a more practical problem, fair enough, but I need to know a lot more than was provided, starting with the distance (#1),  rolling resistance, height graph of course, air resistance at all velocities, and that is before we start talking about the rider (#2).

(#1) I'd like to see anyone ride a one kilometer course smoothly in 60 (or even 30) minutes.  I don't think anyone has that sort of balance....

(#2) The calorific content of the rider before and after would be needed as a minimum.  I don't think we would get any volounteers.

Mind you if we could get the model right, it would make us a lot of money because it would apply to racehorses too....


Angus wrote:

I meant to send this earlier but it missed... I don't know if it confuses the issue or not, but more personal experience than theoretical. As an unfit recent convert to commuting by bike, there seems to be a big difference made by conditions such as road surface and head wind. If I am riding quickly, there is a stronger awareness of wind resistance, but when going slowly with a head wind it feels that much more effort is required to keep up speed. Likewise, going fast on a particular road surface seems to require little extra effort to maintain, whereby going slowly takes more. Is it possible that energy use/work done/power required is related to the ratio between friction and velocity? The slower rider uses more energy because a greater proportion of power exerted is ovecoming friction compared to providing speed?



Kevin Phyland replied to Angus:

Don't take this as gospel (considering my recent posts) but if memory serves at least air resistance is also a squared relationship (i.e. twice the speed, four times the air resistance force)...with regard to friction (as in two solid surfaces in contact) I vaguely recall it has to do with the two substances in contact (something called coefficient of friction)...

and to Paul:

I think it can be deconstructed quite easily (at least from a humanistic point of view)...

You have to climb three flights of stairs.

One person sprints up in a minute, while another leisurely strolls in five minutes.

Who is more tired?

The faster one has expended exactly the same amount of energy in raising their elevation as the other...

however...the faster person has expended their energy much faster...(i.e. much more power required)...as well as the kinetic energy required being speed-squared...

In humans expenditure of energy over a shorter period of time takes more chemical energy (muscles etc...) so...

my best guess is as follows...

the total energy expenditure of the faster person is not only 25 times (from my example - 5 squared) but much more than that due to conversion of body resources (clearly biol is not my field) to thermal energy etc...

Any thoughts?

Geez...and I thought this was an ANSWER...:))


Angus replied:

I am familiar with the coefficient of friction idea, but recall it being two values - stationary and moving - but is there a 'sliding scale' of coefficient values as velocity changes. Alternatively, does frictional resistance remain constant for a given contact, thereby proportionally less for the faster moving object? I asked some (fitter) cyclists today of their thoughts on the problem, and they also felt that riding slower consumes more energy, which is how it feels to me...sorry for the subjectivity.


Anthony Morton replied:


They are identical, but as you say, only in the absence of friction.  Most of the work done on a one-hour bike ride is against friction, and work done against friction does not affect one's kinetic energy.  So the two can't be equated.  The work-energy theorem is a beautiful theoretical finding, but unfortunately not relevant to terrestrial
conditions.

Let's return to the original scenario: two cyclists who cover the same distance in different times due to riding at different average speeds. Who has used the most energy?  For simplicity, let's assume each cyclist accelerates from stop to a given speed V, maintains that constant speed throughout the trip on a flat route, then decelerates to a stop at the end.  So the kinetic energy increases from zero to MV²/2, remains at MV²/2 for most of the trip, then drops to zero again.

During the initial acceleration, two kinds of work were done: 'inertial work' to change the rider's speed, and 'dissipative work' to overcome friction (I use the term 'friction' to include all dissipative forces, including air resistance).  Energy has to be contributed by the rider to perform both kinds of work, but only the inertial work is expressed in the kinetic energy.  The energy associated with the dissipative work is dispersed as heat in the mechanical system and the atmosphere.

Once the rider is up to speed, all work is dissipative in nature.  (Remember, I'm assuming a flat route for the time being.)  The net forward thrust is zero, the speed and hence the kinetic energy remain constant, but work must still be contributed to overcome friction in order to maintain that speed.  As frictional forces increase with speed at the kinds of speeds we're talking about, the faster rider will be doing more work and hence expending more energy.  (But at slower speeds other factors come into play, of course.)

At the end of the ride, the rider must get rid of that MV²/2 kinetic energy in order to come to a stop.  The work required of the rider in applying the brakes is negligible; the kinetic energy is dissipated as heat in the brake pads, and other dissipative forces continue to operate, helping slow the whole system down.

Net result: the faster rider has expended slightly more energy in accelerating to the higher speed, and has expended more energy maintaining that higher speed.  In practice, the energy expended in speed maintenance is much greater than that expended in acceleration. In a frictionless world, the only difference would be in the energy of acceleration and would be of much lesser magnitude.

The above analysis must be corrected in two ways: to account for speed variations along the way, and to account for gravity if the route is not flat.  Both effects generate extra energy inputs and outputs, but as gravity is a conservative force it is only the speed variations that affect the difference in energy expenditure between the two riders.


To put this another way: work is not mass times distance, but force times distance.  The only force with a direct relation to mass is gravity: thus, if two objects of the same mass are catapulted vertically into a vacuum and both reach the same maximum height, you know that the same amount of work was done in each case, and so the same amount of energy must have been released from the catapult.  But then you can also deduce from basic kinematics that the two objects must have risen at the same speed.

The only way to project two equally massive objects upwards at differing speeds (in the same gravitational field and with no friction) is to impart a greater (non-gravitational) thrust force on one than the other, doing more work and hence imparting more kinetic energy.  Say the slower object reaches a maximum height H.  When the faster object reaches height H it will still be travelling upwards at some nonzero speed V.  To relate this to the two-cyclist example, bringing the faster object to a stop at height H would require dissipating the extra kinetic energy MV²/2 at that point.  The same work has not been performed on both objects.


Steve commented:

Yep, during my stint as security guard I had to climb lots of stairs and taking those 2 at a time made me far less tired  than doing them one by one by one!


Ray noted:


Pedant mode.

Actually 4 times is an increase of 300%, I reckon.
Or 400% times as much.

One sounds better that the other if you're advertising something or trying to increase the 'spin' of data



David Martin wrote:


Actually, David has been wrong on many occasions, it's just that no-one has noticed until now :-)

My last post was a (perhaps rather picky) response to Zero's comment that "Kinetic energy is not an appropriate measure of work". I stick by what was said there, since I went to great pains to point out that the energy involved in accelerating a bike up to speed would be negligible in comparison to the work done against friction on a long journey.

Anthony said "The work-energy theorem is a beautiful theoretical finding, but unfortunately not relevant to terrestrial conditions". Actually, Anthony, it's of great relevance to many physical systems, terrestrial and otherwise, but I agree that a bike ride is not one of them. I really did try to be very careful to point this out in my emails.

Now, what about the energy lost to friction? I based my original factor of four increase on the (widely assumed) variation of wind resistance with speed squared. This was based on twice the *average* speed. In practice, the
speed would vary over the trip and this would further increase the total resistance. Hence my comment "at least four times".

Having thought more carefully, and read all the other posts on this topic,
my assumption now seems somewhat naïve.

Firstly, wind drag is one of those properties which is very difficult to calculate from first principles. There is no fundamental law which leads to the square dependence on speed, it's just an assumption; widely found in many textbooks and websites, however.

Alan has pointed out that "The drag of an aeroplane in flight decreases markedly as speed increases up to what is known as "min drag speed". Thanks for that information Alan, could you let me have a reference?

There are also other sources of energy loss (i.e. as heat): rolling resistance in the bike and the metabolic rate of the rider. These don't seem to vary as the speed squared and I have no idea how much they would contribute to overall energy loss, compared with wind resistance.

I think it's still probably safe to say that the faster rider would use more energy; how much more depends on a variety of factors and would have to be measured, as opposed to calculated.

This has been a very interesting thread for me, thanks to you all.


Chris Forbes-Ewan commented:

David Martin will be able to explain the physics better than I can, but riding slowly, walking slowly, swimming slowly and so on all involve a lower rate of energy expenditure than doing the same activity quickly.

In fact, doing these activities very quickly (e.g. anaerobically) is far more tiring than doing them aerobically for the same total distance covered in the same amount of time.

As an example, if you are fairly fit, try running 5,000 m at a leisurely pace (say 10 km/h). This will take 30 minutes.

When fully recovered (e.g. 2-3 days later) try doing 50 x 100 m sprints at maximum pace over a period of thirty minutes (i.e. starting a new sprint every 36 seconds).

Anyone silly enough to actually do this experiment might like to report back to the list from their hospital bed (or perhaps the ICU) on which was the more tiring :-)



Anthony Morton posted:


True.  My apologies for being overly terse with this one.  By 'terrestrial' I was referring to those problems where ignoring friction gives misleading answers.  This includes most problems of interest to mechanical engineers.  I could certainly have spelled this out more clearly.

and:

One last stab at the two-cyclist problem, just because I find it so interesting.....


That's the same thing that concerned me about this finding.  It implies that the net force acting on a cyclist does not vary with speed, in defiance of reality.

There are two possible explanations: either the observations were made with a stationary bike, as Alan guesses, or else the human motor system is so inefficient that the variations in drag force with speed are swamped by the muscular effort required just to move the bicycle.  As a cyclist myself I don't believe the latter.

It's worth noting a couple more caveats with the two-cyclist thought experiment.  First, we've implicitly assumed that the wind speed and direction are constant for the duration of the experiment.  Winds have a huge effect on the energy used by a cyclist.  To determine a cyclist's effective speed for assessing drag you need to subtract the component of wind velocity in the direction of travel.

Secondly, the fact that gravity is a conservative force and gravitational energy changes are 'notionally recoverable' might give people the misleading impression that a cyclist will use the same energy riding on hilly terrain as on flat terrain provided the initial and final elevations are the same.  For this to be true requires rather
special conditions: the speed profiles must be identical, *and* there must be no braking.

The reasoning here is nontrivial, so it's worth looking at some examples.  Picture two identical cyclists, A and B.  Cyclist A rides down an incline and then up another incline to finish at the same elevation as the starting point.  Cyclist B covers the same ground distance on the flat.  Seen from above, A and B both ride along
straight lines (though A's line is very slightly shorter due to the inclines), and there is no wind to complicate things.

We suppose that B at all times knows A's instantaneous speed and accelerates or decelerates as necessary to maintain the exact same speed.  Thus, A and B's speed profiles are identical and they experience the same frictional forces.  (This is not precisely true, because rolling friction decreases slightly on a slope, but the resulting difference is negligible provided the slope isn't very steep.)
 
Under these conditions, the forward thrusts required from A and B differ at all times by precisely the gravitational contribution m*g*sin(theta) where m is mass, g the coefficient of terrestrial gravity and theta the angle of incline.  On the way down, B must provide this extra thrust to match what A gets from gravity; on the way up, A's thrust must exceed B's by this amount.  *Provided* neither cyclist applies the brakes, the energy contributed by each cyclist is the path integral of the thrust.

Now let's consider some specific scenarios.  (Readers not interested in the detail can skip to the end, where the conclusions are justified by a simple energy argument.)

Scenario 1: A coasts down the incline, reaching a maximum speed v₁ at the bottom, then pedals to maintain this speed v₁ all the way up the other side.

On the downward leg, B must accelerate on the flat to match A's acceleration due to gravity.  To do so B supplies an additional thrust m*g*sin(theta) over the equivalent distance d.  The additional energy that must be supplied by B over this leg is m*g*d*sin(theta), or m*g*h where h is the vertical distance from the top to the bottom of the incline.  On the upward leg, B must supply a continuous thrust to oppose friction, but A must supply an additional thrust m*g*sin(theta) over that of B to overcome gravity.  The additional energy that must be supplied by A over this leg is again m*g*h.  The energies balance out in this case, and we recognise m*g*h as the difference in gravitational potential that A gains on the way down and loses on the way up.

Scenario 2: A and B both ride so as to maintain a constant speed v₂ throughout.  (Assume  v₂ is also their initial speed, and that  v₂ is small enough that A would accelerate upon hitting the downward incline.)

On the downward leg, A must apply the brake in order not to accelerate with gravity.  The brake must provide a continuous negative thrust equal to m*g*sin(theta) less the thrust required to overcome friction at speed v₂.  Assuming friction scales as speed squared, we can write this as m*g*sin(theta) - k*v₂^2. 
The total energy dissipated in the brake is m*g*h - k*v₂², and this energy is not recoverable, being lost as heat in the brake pads.  On the upward leg the situation is as in scenario 1: A must provide an additional thrust m*g*sin(theta) over that of B to overcome gravity.  B's thrust is just the k*v₂² required to overcome friction at speed v₂.  So at the finish point A has put in an amount of energy equal to m*g*h + k*v₂²*d, while B has put in k*v₂²*(2d).  A's energy input exceeds B's by m*g*h - k*v₂²*d, which is also the amount of energy lost to braking.

Scenario 3: A coasts down the incline and part way up the other side, until A's speed decreases to some comfortable value v₃.  A then pedals to maintain this speed v3 until reaching the top.

On the downward leg, this is identical to Scenario 1.  B must contribute an additional amount of energy m*g*h in order to match A's free acceleration.  But on the upward leg, A and B are experiencing the same opposing frictional force but A has in addition an opposing gravitational force.  The only way B can decelerate enough to match A is by applying the brake, with a braking force m*g*sin(theta) to match the gravity force on A.  B continues to apply the brake until A reaches speed v₃, say at a vertical distance h₀ from the top of the incline.
 From this point we are back at Scenario 1.  The additional energy A must supply over B to ascend the final distance h₀ is m*g*h₀.  But B's additional energy from the downward leg is m*g*h, so the conclusion in
this case is that B's energy input exceeds A's by m*g*(h-h₀).  Again, this is exactly equal to the total energy dissipated in the brakes.

Of these three scenarios the last is probably the most realistic as far as A is concerned, but B's situation is rather contrived (deliberately so, in order to illustrate a point).  It's also worth comparing A's effort with that of B if B just rides at a constant speed v_B, as one ordinarily would on a flat straight route.  By solving the equation of
motion for A it's possible to determine the height h₀ as a function of the friction coefficient k, the speed v₃ and the geometry of the problem; one can then work out the conditions on <theta, h, k, v₃, v_B> under which A's energy input is greater than or less than B's.  This is rather too mathematically involved to go into here, so I'll leave it as a homework problem for the interested reader.  :-)

Conclusion:

Obviously what is going on here is simply that human-powered bikes have what us engineers call 'non-regenerative braking'.  Any time you apply the brakes, you are dissipating energy that cannot then be recovered by the system.

The energy conservation law for each cyclist looks like this:

   Energy in
   = Energy lost to friction  + Energy lost to braking  + Net energy to gravitational field+ Change in kinetic energy

As the cyclists have the same speed profile, the net change in kinetic energy is the same for each, as is the net energy lost to friction (to a good approximation).  As the initial and final elevations are the same for each cyclist, the difference in energy input must be the same as the difference in energy lost to braking.

If this braking energy could instead be stored (such as in a flywheel) and released at an appropriate time later, then cycling up and down hills would indeed be no worse than cycling on the flat.


Zero Sum wrote:


Alan, my comment was...

"Kinetic energy is not an appropriate measure of work".

The word was "appropriate" not "good measure".  And it remains the case that that measure was inappropriate for the argument being used because the argument invoked "no friction".  In which case, the only energy consumed is that energy difference between the final and end state.  Which is zero.


Chris Forbes-Ewan replied:


Zero, This is actually one of the things that you can say is not correct!

Unless the increase in speed results from the force due gravity (i.e. the rider is going downhill and has to apply the brakes to reduce speed) any increase in speed leads to an increase in the rate of work (power output).

This is not only intuitively correct, it has been demonstrated experimentally countless times (see my other messages on this, tonight and previously).

and:


Pardon?

Other than as depicted in the typical late night SBS movie, I have no idea of the meaning of 'thrust x speed'. Is this an aeronautical term? If so, I'm not sure what relevance it has to bicycling.

Power is work per unit time. The unit of work is the joule (J). The unit of power is the watt (W), with one watt equal to one joule per second. Basal metabolic rate (BMR) for an average adult human is of the order of 100W.

So a power output of 6 METS (as applies to leisurely riding) means energy is expended at a rate of ~600 W (for the average adult). This would involve ~100 W for normal maintenance of homeostasis (i.e. BMR) and ~500 W for the physical activity.

Of the 500 W attributable to the physical activity, ~100 W would be for 'external' power output with ~400 W converted to body heat (because the body is only ~20% efficient at doing physical work).


How about one million dollars ... and I'll give you odds of a hundred to one :-)

As mentioned in my original message, the results were for bicycling at various speeds, including 'leisurely' at one extreme and 'racing, not drafting' at the other.

Results in the reference previously cited for various levels of power output on the bicycle ergometer include:

bicycling, stationary, 50 W, very light effort 3.0 METS
bicycling, stationary, 100 W, light effort 5.5 METS
bicycling, stationary, 150 W, moderate effort 7.0 METS
bicycling, stationary, 200 W, vigorous effort 10.5 METS
bicycling, stationary, 250 W, very vigorous effort 12.5 METS


Daya Papalkar commented:


Does B always have to brake in this scenario (no matter what the incline of A's slope)? This doesn't intuitively make sense to me - why can't B just stop pedalling and allow friction to cause the reduction in speed in order to match A.


Can I introduce another variable? What about the gear ratio of the bike?
I am not much of a cyclist (especially not now) but I recall being told that there is a certain number of rpm of the pedals that creates maximum efficiency - hence staying in too high a gear when cycling up a slope become
disadvantageous. On the other hand, riding in too low a gear just leads to spinning the pedals and loss of efficiency. I noticed (on a mountain bike) that a lot of downhills I could no longer actively contribute to speeding up because the gear ratios wasn't right - so you have to coast until the bike slows down. Would the above statement still apply?

and:

Alan wrote:


I blame William F Ganong for that... "Review of Medical Physiology" actually uses the term "negative work".

What I think is meant is: the muscle provides a force in one direction slowing the movement (due to gravity) of the object in the opposite direction.

We use our muscles like this all the time eg walking downstairs, in which case the quadriceps are active when the knee is flexing (bending) (the quadriceps are extensors ('straighteners') of the knee).

and:


Sounds fantastic!

I should point out that running downhill with a pack probably takes a toll on the joints if the ground is hard.

and:


I think the range Chris was referring to didn't go to these very slow levels. It is difficult to ride very slowly, but this is mainly a balance problem (and could be corrected with training wheels). It would be interesting to know how much energy is used in riding very slowly, though.

I think there is a difference between riding slowly and walking slowly (because of the way we use our muscles to oppose gravity).


Anthony Morton responded:


To expand further on this: the instantaneous power - meaning the rate at which energy is being supplied from the propulsion source - is equal to the dot product of thrust (propulsion force) and instantaneous velocity.  Because in any small time interval dt, the system moves a distance dx = v dt and the work done by the thrust force F is F • dx = (F • v) dt.  So if p denotes the instantaneous power, we have p = F • v.

If a vehicle is moving at a constant speed V, then the power input is constant and equal to thrust times speed:
P = F V.

By Newton's first law, if a body is moving at constant speed then the net force on it must be zero.  This entails that the thrust force must  be just enough to balance the external forces due to gravity and friction.  For level motion the only external forces are due to friction (of one sort or another), and so the thrust required at a given speed is a direct measure of the frictional force at that speed.

The somewhat counter-intuitive consequence of this has already been noted: if the frictional force on a body is independent of speed, then the power input required is *not* independent of speed, but goes up linearly with speed.  One approach to understanding this intuitively is that while the energy per unit *distance* is the same, the energy per unit *time* must increase because more distance is being covered in the same time.

Engineering folklore has it that air resistance on a moving body increases roughly as speed squared: F=KV².  It follows that for motion at constant speed, the power input is proportional to speed cubed: P=KV³.

Looking again at your figures I realise that they don't show a linear increase with speed after all - my earlier assumption was erroneous.  There is an approximate incremental linear increase, but that's a different thing entirely.  Let's just take the two extreme points noted:
        17.5 kph        6 METS
        28.0 kph        12 METS
We have a twofold increase in power for a 60 per cent increase in speed.  Taking logs, the power law index is approximately 1.5.  In other words, a crude approximation to the power-speed law is P = K V^(3/2).  This in turn implies that the friction force on the moving cyclist scales roughly as the square root of speed in this range.  So
for double the speed, friction goes up by about 40 per cent and power almost threefold.

It seems appropriate to draw the conclusion that the square law for friction may not apply at the speeds typical of cycling.  The truth is that friction is a highly nonlinear phenomenon and doesn't obey any neat law across the whole range of possible conditions.  For speeds in the 100kph range a square law may be the best approximation, but in the 10kph range we have evidence for a smaller rate of increase with speed.

and:


The experiment was set up in such a way that A and B experience the same frictional force.  When A hits the bottom of the incline, both A and B stop pedalling.  But B's friction alone isn't enough to decelerate B as much as A, because A has the same friction and gravity acting as well.  So B will always have to apply the brakes under the conditions of the problem.


Yes, gear ratios are important for riding with maximum efficiency.  Cycling folklore has it that 'spinning' in a lower gear (but not too low) is more efficient than 'grunting' in a higher gear.  The optimum spinning rate ('cadence') is thought to be around 1-2 revs per second.

Implicit in all these thought experiments is the assumption that the cyclist selects the appropriate gear to match their speed.  (And when comparing two cyclists, it's assumed the gearing on the two bikes is identical.)  As long as the two cyclists both change gears when appropriate, it doesn't affect my conclusions.


Alan Emmerson noted:

Tony wrote, on 10 June


There is a misaprehension here Tony.  In Chris' data, the power in units of MET is the input power to the human engine. To estimate the drag law you need the output power from the human engine.  I tabulated that in my previous post. Mind you the difference in efficiency between the speeds above is not great.

My recollection is  is that there is a near constant component in consequence of the continuous deformation of the tyres and wheels plus a near square law due to pressure drag  and skin friction. and  I think the churning effect caused by spokes and pedals can be subsumed into the square law.








FRICTION

Gerald Cairns started this thread, as an offshoot of the discussion on Energy, by posting:

In response to the energy thread I thought I might throw up a related one since not only have I been working on related matters for some time nor have I made any serious posts of late. Thus as penance for me and sufferance for you lot here are some thoughts.

Moving components require some form of lubrication to reduce wastage of energy between opposing surfaces. Friction is overcome by providing a film of lubricant that reduces this surface to surface interaction in two broad
mechanisms.

1. lubricants form a micron thick boundary layer that conforms to the surface profile of the individual surfaces as no surface is perfectly smooth. This prevents grabbing and "welding" of the high points of the respective surfaces but does not prevent surface to surface contact at these high points therefore there is a characteristic resistance to movement. While this contact results in wear and significant friction the boundary layer is critical in preventing the surfaces from seizing.

2. lubricants also provide and additional mechanism that assists in the reduction of friction and while in operation can reduce wear to almost negligible limits. This mechanism is referred to as hydrodynamic operation and is similar to aquaplaning of car tyres on a wet road where a film of water builds up under the right conditions to separate the tyre from the road. This is achieved by the creation of a thicker and separate layer to
the boundary layer.

Hydrodynamic operation is dependent on another characteristic of lubricants and this is laminar flow whereby the flowing lubricant tends to flow in layers that resist mixing. I have some video clips of the laminar flow mechanism and what happens when this mechanism breaks down, but these are confidential for the time being. Some claim that modern bearings operate hydrodynamically and therefore there is no metal to metal contact. Unfortunately this claim is false because hydrodynamic conditions require certain parameters to operate. These parameters are;

1. temperature of the system
2. viscosity and surface affinity of the lubricant
3. type of solids components in the lubricant
4. load on the bearing
5. rotational speed of the relative surfaces
   

This is a very complicated subject that I do not pretend to understand fully, by a considerable margin and even some experts with whom I have discussed this are forced to rely heavily on theory for explanation and requiring mathematics that reach places I once could go. Such skills have degenerated through lack of use over a long period and I would have to do a refresher before attempting to tackle this, unfortunately time and other matters prevent this. However Toby may have some thoughts to shed on this subject?

There is no possibility of avoiding metal to metal contact in a heavily loaded vehicle bearing at standstill, no grease can hold surfaces totally separate in static mode, if it could it would be unlikely to perform as a lubricant. As the bearing rotates it reaches a point where the excess lubricant above the boundary layer begins to form a layered high pressure buffer between the surfaces so achieving surface separation and minimum friction. The remaining friction is largely fluid friction and drag. Thus it is desirable to maintain certain minimum speeds that achieve hydrodynamic and most efficient operation.

The importance of the boundary layer cannot be over estimated. Even in seemingly well lubricated systems it is possible for bearings to seize rock solid if the boundary layer is degraded or lost as a separate entity.

Similar processes are believed to be responsible at least in part for the burning off 150 mm diameter axles on heavy equipment. I have demonstrated this seizure in bearings so well lubricated that they are so slippery it is
virtually impossible to pick them up or even rotate them by hand. I have found a number of apparently lubricious materials that disturb the boundary layer, that cause this seizure effect profoundly in the apparent presence of adequate quantities of lubricant. It behoves those formulating and maintaining bearings to keep these issues in mind. While I will not take this aspect further as it impinges on patent applications now in preparation among other matters, I recommend that no solvents of any kind be permitted anywhere near greases in storage or maintenance of such equipment. If they have been used then they must be completely removed before replacing fresh grease.

How to do this is a subject I am involved in at present and if I can get paid for our IP it may well even get published but that is an unlikely scenario in Australia and I talk from bitter experience. We are on the verge of taking this IP overseas because of the lack of government support at all levels, most of this support revolves around public service self justification and double dipping i.e. we pay for the services in taxes then get slugged a second time with unnecessarily massive registration, patent fees etc. etc. etc. Of course phenomenon this is not restricted to Australia but we have refined the techniques to a high level indeed.

How does this help the two cyclists - not much I think at their loads, temperatures and speeds but I leave that to others who may understand more and are more mathematically agile than I am.


Angus responded:

All sounds intersting and the mystery makes it intriguing...
"The importance of the boundary layer cannot be over estimated" - what do you mean by boundary layer? I understand there is some debate over the natuer of the solid-liquid interface (ie first 1-5 nm) regarding whether the undermost fluid particles are stationary or if in fact they can "slip" across the surface. Is this the scale you are referring to?


Gerald Cairns answered:

The boundary layer scale is indeed very small but it is dependent on a number of factors as I noted, however the single most important one is that of affinity for the surface, to put it crudely adhesive force of attraction that maintains it in contact with the surface aided by the hills and valleys of the surface. Coupled with this is the fact that lubricants flow in a laminar fashion so there is little tendency for the boundary layer to be removed by mixing in motion as the lubricants, greases or oils, tend to flow in non mixing layers. Therefore the boundary layer does flow but slowly and remains close to the surface.  As I noted also I have crudely video taped some of these effects and they are quite observable in macro scale. The persistence of these properties are quite remarkable even at high pressures, temperatures and externally induced turbulence. It is this among other things that makes lubricants what they are.

I do not have the equipment to visualise or measure the movement at the boundary layer level but the principles have been encountered most clearly in certain experiments that are difficult to explain in any other fashion. The macro observations are pretty compelling and supported by other aspects about which I must remain silent for now. I no longer have the benefit of my own Uni Lab at the QUT which we had to shut down in 1991, so all my work is done with no budget and hand made testing equipment that has nonetheless received favourable compliments from independent consultants. My findings for instance reduce certain procedures from about 8 hours to between 3 and 5 minutes and the wastes are compostable, no toxics or objectionable odours and excellent workplace safety etc. etc.

As the subject matter is concerned with work I have been involved with for the last two years or more, a fair amount of it wasted by the failure of others to accept the nature of the systems or the advice, I am not yet able to divulge more information. Some of this is very sensitive commercially and politically and I have no wish to become a meal for certain large organisations who may be significantly embarrassed though I would happily give the pollies a serve. I have had to verify my results personally at our own cost which has taken me full time almost seven months of long hours and I am expected to hand over the results for nix and a promise that something might happen in the future - and pigs might fly!! I have also taken the precaution of having the observations confirmed by independent consultants. Thus for the time being my hands are tied but the doors are locked and I have the keys. :-).

and:

Re reading my response I realise that I did not explain what I meant by the boundary layer. It is that thickness of lubricant to which you refer but largely stays in position or moves only very slowly while the upper layers of the lubricant flow across it with minimal disturbance of it. The same happens within macro layers of the bulk lubricant above the boundary layer. It is in effect a multi layered system generated from the same material.

Various additives are introduced to the formulation to improve high pressure performance and these appear to have higher surface attractive properties and internal cohesion. I believe these are predominant in the boundary layer. I have photographs of the macro effects that I believe confirms my views but really need a laboratory environment to bring out the fine detail and alas I no longer have those facilities.

and to Paul:

Imagine a railway wagon weighing around 120 + tonnes this translates into approximately 7 tonnes per bearing. In a spherical bearing for example (not a conical type)The load is carried at the top of the bearing over about 12 rollers the rest are free waiting to rotate into the load zone. For the sake of this discussion assume each roller has about .1 mm square in contact with the races (2) therefore there is about  2(12 x 0.1) sq mm or
2.4 sq mm carrying 7 tonnes. I haven't allowed for elastic deformation of the rollers that would increase this area by a small amount but I think it unlikely that any grease would be able to keep the metal surfaces apart
standing still under such conditions but the boundary layer is restricted to the valleys in the surface under these conditions. Until the vehicle is moving at a particular speed depending on temperatures and viscosity in particular a lubricating layer or layers cannot form between the rollers and the races for which Alan Emmerson has correctly pointed out a degree of ambiguity in my post. Thus until hydrodynamic conditions are achieved there will be metal to metal contact and wear.

The point I was making that some people claim that there is no metal to metal contact in hydrodynamic bearings which is arrant nonsense, a standing vehicle has no dynamism about it at least in respect of the bearings. With loads as described above all the grease will be squeezed away from the contact surfaces. I know of no grease that has such physical strength to resist such loads. If it were possible to have a bearing with no metal to metal contact then there would be no wear and I can tell you analyses show the metals in used greases. If greases are contaminated with solvents the lubricant films can break down, i.e. lose adhesion with the surface and viscosity changes can also take place at the transient load points resulting in metal to metal contact at high speed. The heat generated is already high but would be likely to go critical, very rapidly proceeding to catastrophic failure.

Bearings do run hot, very hot and I have images to prove it, however, there is scurrilous MS Power Point Show that demonstrates what can happen when a bearing runs hot.  I have not been able to confirm the authenticity although it looks authentic and it appears to have been at least re-edited on a Department of Primary Industries computer here in Brisbane. I have spoken to owner and he is unaware of it but the file history shows his name etc. He is geographer with no interest in this sort of thing, I believe him, others have access
to his machine. The results are not inconsistent with other reports from the US and Canadian Departments of transportation however. I will leave the first frame in place because it may have relevance to someone who can
confirm this accident but may be indulging in unfair criticism. The Rail Road Operators named refuse to respond to my requests for confirmation.

Rail roads operate track side infra red detectors called Hot Box Detectors. The US and Canadians say that they place them at 25 mile intervals but that most over heating bearings fail between 9 and 16 miles after the last detector! The problem seems to me to be the temperature gradient within the bearing housing and the measurement of the temperature at the surface. There are better temperature sensing systems that radio the state of individual bearings constantly from the interior but these are expensive and generally not applied. When a hot bearing, 80 degree threshold, is detected it is flashed up on the driver's console and the train must be stopped for a replacement bogie. I have much more to say on this but it must remain confidential for the time being. The position of all bearings at any time are known and monitored constantly. Well lubricated bearings
such as these under good conditions should last up to 20 years.

What I have not referred to is the fact that bearings have not only rolling elements but also sliding elements such as cages and dividers and these add to the complexity. Again returning to our cyclists these issues would pale
into insignificance against the example above.

With respect to maglev technology that is different again but at rest there would be no point in wasting energy supporting the vehicle. As far as the pressures involved some US experts quote the hydrodynamic lubricating films as being exposed to transient loads of up to 200,000 Lbs/sq in. I suspect if the bearing were to be pressurised the labyrinth seals would either collapse or blow out completely. That does not eliminate the possibility that it could be done but I doubt that it would be practical or economic. As to buckyballs I have no information to offer but doubt that they would have much to offer. You have to choose between a wet, i.e. grease, system or a dry one and I suspect buckyballs would ignite under these conditions not to mention static problems.


Alan Emmerson wrote:

Those who are following this thread might be interested to know that  one of the functions of a lubricant can be  to remove heat from the bearing.  For heavily loaded plain bearings, railway waggon journals for example,
temperature rise used to be a design criterion.

and:


The drag of an aeroplane  in flight decreases markedly as speed increases up to what is known as "min drag speed".


Alan Emmerson replied:

The drag of a complete aeroplane at low speeds is dominated by what is called lift induced drag or just "induced drag:"

Induced drag depends on the square of the wing loading  ( (lift or weight)/wing area)  and the reciprocal of the velocity squared, and the reciprocal of the air density..


More generally, the relationship between drag, surface area and speed changes with the Reynolds number of the flow ( Re representing the ratio of inertia forces to viscous forces in the fluid)

Very generally, drag tends to be proportional to surface area, proportional to speed at low Re and to the square of speed at higher Re.  Very thick books have been written on this subject.

In the case of a bicycle and rider, or motor cycle for that matter, it is hard to know what to choose as the characteristic area.  But, as the "area" of a bike and rider doesn't change too much, it is usual to drop the area
dependency and subsume it in a constant.  Standard texts say that the drag of a bike and rider is proportional to the speed squared.