Drag force

It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from clear.
--
Dieter Britz (oldnobatyahoo.dk)

On Wed, 29 Apr 2009 11:28:19 +0200, Dieter Britz wrote:

It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from
clear.

Force is the time rate of change of momentum. The only mass being moved
is a volume of air, on a level course. So, it's simply calculating the
mass of that air, multiplying it by velocity and dividing it by time to
calculate the force.

Assume the air has a constant density per unit volume: r.
Assume there is no air velocity.
Assume the rider is traveling at a constant velocity: v.
Assume the rider profile presents a cross-sectional area perpendicular to
the direction of motion: A.

Then in time T, the rider will travel a distance vT. This is also the
depth of the volume of air that must be displaced.

So, the volume of the air is A vT. Its mass is r AvT. Is momentum is
rAvTv or rATv^2.

Finally, this is the momentum per unit time:T so the force exerted is:
rAv^2.

Dieter Britz wrote:
It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from clear.

Here you go:
http://www.newton.dep.anl.gov/askasc...0/phy00200.htm

--
Andrew Muzi
www.yellowjersey.org/
Open every day since 1 April, 1971

On Apr 29, 7:46*am, Stephen Bauman wrote:
On Wed, 29 Apr 2009 11:28:19 +0200, Dieter Britz wrote:
It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from
clear.

Force is the time rate of change of momentum. The only mass being moved
is a volume of air, on a level course. So, it's simply calculating the
mass of that air, multiplying it by velocity and dividing it by time to
calculate the force.

Assume the air has a constant density per unit volume: r.
Assume there is no air velocity.
Assume the rider is traveling at a constant velocity: v.
Assume the rider profile presents a cross-sectional area perpendicular to
the direction of motion: A.

Then in time T, the rider will travel a distance vT. This is also the
depth of the volume of air that must be displaced.

So, the volume of the air is A vT. Its mass is r AvT. Is momentum is
rAvTv or rATv^2.

Finally, this is the momentum per unit time:T so the force exerted is:
rAv^2.

One result of this is that the power required to overcome air drag, in
watts of course, is proportional to the cube of the airspeed. This is
the basis of the fact that the resistance to rider propulsion, the
limit of speed, is aerodynamic, not rolling, resistance.

Ken Freeman

On Apr 29, 7:46*am, Stephen Bauman wrote:
On Wed, 29 Apr 2009 11:28:19 +0200, Dieter Britz wrote:
It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from
clear.

Force is the time rate of change of momentum. The only mass being moved
is a volume of air, on a level course. So, it's simply calculating the
mass of that air, multiplying it by velocity and dividing it by time to
calculate the force.

Assume the air has a constant density per unit volume: r.
Assume there is no air velocity.
Assume the rider is traveling at a constant velocity: v.
Assume the rider profile presents a cross-sectional area perpendicular to
the direction of motion: A.

Then in time T, the rider will travel a distance vT. This is also the
depth of the volume of air that must be displaced.

So, the volume of the air is A vT. Its mass is r AvT. Is momentum is
rAvTv or rATv^2.

Finally, this is the momentum per unit time:T so the force exerted is:
rAv^2.

One result of this is that the rider power in watts required to
overcome aero drag is the cube of the airspeed.

Ken Freeman

On 29 Apr, 11:28, Dieter Britz wrote:
It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from clear.
Dieter Britz (oldnobatyahoo.dk)

One should admit that this one, like many other 'empirical laws' in
physics, is largely phenomenological. Whether 'true' or not it rests
on its accuracy to describe a phenomenon, not in our ingenuity to
explain and to justify it within a model from first principles.

So, better take it as a fair description, with inevitable inaccuracy,
of what actually happens.
That is, after all, where it does come from.

Sergio
Pisa

On Apr 29, 10:47*pm, Sergio wrote:
On 29 Apr, 11:28, Dieter Britz wrote:

It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from clear.
Dieter Britz (oldnobatyahoo.dk)

One should admit that this one, like many other 'empirical laws' in
physics, is largely phenomenological. Whether 'true' or not it rests
on its accuracy to describe a phenomenon, not in our ingenuity to
explain and to justify it within a model from first principles.

So, better take it as a fair description, with inevitable inaccuracy,
of what actually happens.
That is, after all, where it does come from.

That's fine and good as long as everyone understands we're dealing
with an approximation, albeit a rather good one. For instance, the
science from which all these formulae are derived, not just the ones
presented by Stephen and Jobst, assumes a hard body, and the
calculation is made as if it is a hard-edged flat plate of so much
area. In real life, if a multimillion dollar wind tunnel can be called
real life, the first fact of real life that one discovers is that even
hard bodies can shift the results up and down appreciably if their
edges are radiused *as little as 3 to 6mm*. A human body, even that of
a roadie, is of course both radiused and soft, which adds another
layer of complication.

But that's merely for the record. I don't see how we would control the
circumstances of a cyclist out in nature sufficiently closely to prove
that the theoretical model is only an approximation to a real cyclist.
That by itself, even if a lot of work in other fields didn't already
validate the formula for practical purposes, tells us it is a workable
formula.

Andre Jute
Relentless rigour -- Gaius Germanicus Caesar

"Andre Jute" That's fine and good as long as everyone understands we're
dealing
with an approximation, albeit a rather good one. For instance, the
science from which all these formulae are derived, not just the ones
presented by Stephen and Jobst, assumes a hard body, and the
calculation is made as if it is a hard-edged flat plate of so much
area.(clip)
^^^^^^^^^^^^^^^^^^^^
That may be true in a high school physics class, but I went on and took
engineering in college. Just as Einstein introduced the "cosmological
constant" into his equation, engineers introduce the "drag coefficient."
The drag coefficient handles the differences between round-edged plates and
sharp edged plates. It handles the difference between an old car with a
vertical radiator and windshield and a streamlined racer. It handles the
difference between a cyclist sitting bolt upright and a rider crouched
behind a fairing. How do we know what drag coefficient to use? We guess.

Even though we can't calculate accurately what the drag will be in a
particular case, it is very useful to derive the equations from first
principles. It gives us an understanding of why the data come out the way
they do. It helps when we're curve-fitting scattered data. It helps us
understand what is possible and what is not worth trying. An excellent
example was given in an earlier post. If you are trying to ride fast, it is
much more useful to get rid of your billowing windbreaker than to look for a
lighter wheel bearing grease. (A tight-fitting riding suit lowers the drag
coefficient, which is multiplied by the cube of the speed.)

Andre Jute wrote:
On Apr 29, 10:47 pm, Sergio wrote:
On 29 Apr, 11:28, Dieter Britz wrote:

It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from clear.
Dieter Britz (oldnobatyahoo.dk)
One should admit that this one, like many other 'empirical laws' in
physics, is largely phenomenological. Whether 'true' or not it rests
on its accuracy to describe a phenomenon, not in our ingenuity to
explain and to justify it within a model from first principles.

So, better take it as a fair description, with inevitable inaccuracy,
of what actually happens.
That is, after all, where it does come from.

That's fine and good as long as everyone understands we're dealing
with an approximation, albeit a rather good one. For instance, the
science from which all these formulae are derived, not just the ones
presented by Stephen and Jobst, assumes a hard body, and the
calculation is made as if it is a hard-edged flat plate of so much
area. In real life, if a multimillion dollar wind tunnel can be called
real life, the first fact of real life that one discovers is that even
hard bodies can shift the results up and down appreciably if their
edges are radiused *as little as 3 to 6mm*. A human body, even that of
a roadie, is of course both radiused and soft, which adds another
layer of complication.

But that's merely for the record. I don't see how we would control the
circumstances of a cyclist out in nature sufficiently closely to prove
that the theoretical model is only an approximation to a real cyclist.
That by itself, even if a lot of work in other fields didn't already
validate the formula for practical purposes, tells us it is a workable
formula.

Andre Jute
Relentless rigour -- Gaius Germanicus Caesar

Just to add a nice twist to this thread, word was going around about a
month ago that some riders at the track world championships were wearing
a shoulder girdle of sorts to help put their shoulders into a more
aerodynamic position and help hold it there as they tired from their
efforts and their form goes away.

Stephen Bauman wrote:

On Wed, 29 Apr 2009 11:28:19 +0200, Dieter Britz wrote:

It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from
clear.

Force is the time rate of change of momentum. The only mass being moved
is a volume of air, on a level course. So, it's simply calculating the
mass of that air, multiplying it by velocity and dividing it by time to
calculate the force.

Assume the air has a constant density per unit volume: r.
Assume there is no air velocity.
Assume the rider is traveling at a constant velocity: v.
Assume the rider profile presents a cross-sectional area perpendicular to
the direction of motion: A.

Then in time T, the rider will travel a distance vT. This is also the
depth of the volume of air that must be displaced.

So, the volume of the air is A vT. Its mass is r AvT. Is momentum is
rAvTv or rATv^2.

Finally, this is the momentum per unit time:T so the force exerted is:
rAv^2.

This seems reasonable, but it ignores sideways flow. Also, the
argument must hold for all Reynolds numbers, and out of Bird, Stewart
& Lightfoot I get that the drag force for creeping flow around a
sphere is proportional to v up to Re = 0.1 or so. Then there is a
complicated expression up to Re = 1, and after that, it's empirical
friction factors. Re for a cyclist going at, say, 10 m/s is about 50,
so we are in the range where things are decidedly complicated, and
BSL do not show a result. Which is why I asked where this dependence
on v^2 comes from. I guess it has been found to be so by experiment.
--
Dieter Britz (oldnobatyahoo.dk)