More on fat people going downhill
Clive George
which caused a lot of confusion, so I shall see what people here (and on
cyclingforums) make of it:
My mate Arthur Pedaller has a really great bike. It has absolutely no
rolling resistance whatsoever - the sole retarding force is air resistance.
On a windless day he can happily pedal at 25mph, and he covers the 25 miles
to work on the marvellous straight road the council installed for him at
that speed.
But today's a bit windy - 10mph headwind to be precise. He decides to ride
at 15mph in order to make sure his hair doesn't look abnormal when he
arrives, and off he goes. Obviously he takes 1 hour 40 minutes to cover the
distance, but the question is, has he put in more or less work than on a
windless day?
cheers,
clive
Pete Biggs
Gordon Bennett, I dunno, but I bet the answer can be worked out via
http://www.analyticcycling.com
~PB
David Martin
415d5e87$0$69729$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
Work done is Force applied times distance moved, W=Fd.
And one would presume that the distance is over the ground.
So no on a first glance.
But this is not the whole story. He has only done work in moving the wind so
we have to consider teh distance travelled through the wind.
In the first instance this is 25x1hours worth of air movement or 25 air
miles.
In the second case this is (15+10)x1hr40 worth of air miles, or 41.6 air
miles.
Going home he rides at 35mph all the way.. which takes him 43 minutes. His
work done is (35-10)x43mins worth of air miles.
So on the still day his return journey is 50 air miles. On the windy day his
return journey is 60 air miles (or near enough).
..d
MSeries
Clive, if you weren't a regular I'd suspect someone was trying to get
some help with their homework. Have you started a GCSE physics course ?
Clive George
news:BD8322A2.221FB%d.m.a....@dundee.ac.uk...
> But this is not the whole story. He has only done work in moving the wind
so
> we have to consider teh distance travelled through the wind.
Are you sure? :-)
cheers,
clive
David Martin
415d673c$0$80627$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
Yes. It is the same as the problem where the bloke in the boat rows upstream
and then downstream. It is Force x Distance moved. The Force is through the
air and the plane of reference for distance is to the moving air, not the
ground.
..d
Clive George
Are you really really sure? :-)
cheers,
clive
David Martin
415d6c19$0$69728$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
> "David Martin" <d.m.a....@dundee.ac.uk> wrote in message
> news:BD8327EA.22210%d.m.a....@dundee.ac.uk...
>> On 1/10/04 3:22 pm, in article
>> 415d673c$0$80627$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
>> <cl...@xxxx-x.fsnet.co.uk> wrote:
>>
>>> "David Martin" <d.m.a....@dundee.ac.uk> wrote in message
>>> news:BD8322A2.221FB%d.m.a....@dundee.ac.uk...
>>>
>>>> But this is not the whole story. He has only done work in moving the
> wind
>>> so
>>>> we have to consider teh distance travelled through the wind.
>>>
>>> Are you sure? :-)
>>
>> Yes. It is the same as the problem where the bloke in the boat rows
> upstream
>> and then downstream. It is Force x Distance moved. The Force is through
> the
>> air and the plane of reference for distance is to the moving air, not the
>> ground.
>
> Are you really really sure? :-)
Are you bored?
..d
Clive George
news:1096640121.3...@k17g2000odb.googlegroups.com...
> Clive, if you weren't a regular I'd suspect someone was trying to get
> some help with their homework. Have you started a GCSE physics course ?
Hey - I've got an O level in physics - none of that GCSE nonsense!
cheers,
clive
David Martin
415d6d47$0$69728$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
Likewise. and an A level. Though this is more applied maths (for which I
have an old style AO).
..d
Dave Larrington
> But this is not the whole story. He has only done work in moving the
> wind so we have to consider teh distance travelled through the wind.
You and Ian SMith are one and teh same AICMFP ;-)
--
Dave Larrington - http://www.legslarry.beerdrinkers.co.uk/
===========================================================
Editor - British Human Power Club Newsletter
http://www.bhpc.org.uk/
===========================================================
Clive George
I'm just waiting to see if anybody else is going to offer an opinion, while
at the same time dropping hints that your analysis may not be entirely
correct.
cheers,
clive
David Martin
Larrington" <m...@privacy.net> wrote:
> David Martin wrote:
>
>> But this is not the whole story. He has only done work in moving the
>> wind so we have to consider teh distance travelled through the wind.
>
> You and Ian SMith are one and teh same AICMFP ;-)
Not true.. honest.
I am just used to being a little portly and rolling rapidly in a
gravitationally assisted direction.
and mistyping het frequently.
..d
David Martin
415d6ef3$0$69730$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
It is an approximation that does for me. The answer is that it is more work
when it is windier.
So what would be your answer?
..d
soup
and said
I have an "O" grade in physics, a higher and a CSYS
(certificate of sixth year studies). :-P
That was years ago tho' I wouldn't like to analyse
that situation now.
--
yours S
Nihil curo de ista tua stulta superstitione
MSeries
Just trying to be modern and trendy !!!
FWIW
I have O Level Physics, A Level Physics and A Level Applied Maths
(amongst others) Can't be arsed to think about your problem though !
Jack Ouzzi
Crikey ............ is there any formula for distance travelled at
18mph (with a very slight 23 degree sidewind) giving the amount of
fruit pies I can eat to replace the calories consumed during said
distance covered ??
(Thats 45 gram Bramley Apple pies)
All this 4 people build 6 houses in 7 days how many in 28 days stuff.
I NEVER understood it at school .......... and why does the sound of a
train whistle change as it passes you ????
Clive George
news:5gsql0p4cfve38uvj...@4ax.com...
> All this 4 people build 6 houses in 7 days how many in 28 days stuff.
> I NEVER understood it at school .......... and why does the sound of a
> train whistle change as it passes you ????
The second one's easy - they have two whistles and are watching you.
cheers,
clive
Peter Clinch
> I'm just waiting to see if anybody else is going to offer an opinion
Oh, let's make a fool of myself then...
ISTM we're looking at total energy expended for the journey. He's
working at the same basic output in joules/second in each case because
he'll need the same amount of energy to make headway in what is
effectively the same headwind, so he's burning energy at the same rate,
but he's burning it for longer on the slower journey.
If that's cobblers I blame the fact that it's Friday afternoon and
having long since moved into computing my physics is horrible these days...
Pete.
--
Peter Clinch Medical Physics IT Officer
Tel 44 1382 660111 ext. 33637 Univ. of Dundee, Ninewells Hospital
Fax 44 1382 640177 Dundee DD1 9SY Scotland UK
net p.j.c...@dundee.ac.uk http://www.dundee.ac.uk/~pjclinch/
Juliette
reader01.plus.net>, cl...@xxxx-x.fsnet.co.uk says...
> Obviously he takes 1 hour 40 minutes to cover the
> distance, but the question is, has he put in more or less work than on a
> windless day?
Less because he got into work 40 minutes later than usual.
Juliette
--
David Martin
"Jack Ouzzi" <nos...@nowayhose.co.au> wrote:
> Crikey ............ is there any formula for distance travelled at
> 18mph (with a very slight 23 degree sidewind) giving the amount of
> fruit pies I can eat to replace the calories consumed during said
> distance covered ??
> (Thats 45 gram Bramley Apple pies)
one Ulrich per Alp is the unit in which to measure pies...
..d
PS. Was that with or without custard?
soup
said
> why does the sound of a
> train whistle change as it passes you ????
Sound seems to change because of the "Doppler shift"
Very simplified description of the "Doppler effect" follows:-
"all the waves that make up the whistles noise come at
you quicker when the train is coming towards you so
the note seems higher, when the train has passed you
and is going away the waves are coming at you slower
so the note seems lower."
NOTE that was simplified so don't take it to bits.
Dave Larrington
> All this 4 people build 6 houses in 7 days how many in 28 days stuff.
> I NEVER understood it at school ..........
Likewise. "If it takes a man and a half a day and a half to dig a hole and
a half, how long does it take a fly with clogs on to walk through a barrel
of treacle?"
wot use is /that/ in the new elizabethan age, eh?
Clive George
news:2s5atvF...@uni-berlin.de...
> Clive George wrote:
>
> > I'm just waiting to see if anybody else is going to offer an opinion
>
> Oh, let's make a fool of myself then...
>
> ISTM we're looking at total energy expended for the journey. He's
> working at the same basic output in joules/second in each case because
> he'll need the same amount of energy to make headway in what is
> effectively the same headwind, so he's burning energy at the same rate,
> but he's burning it for longer on the slower journey.
>
> If that's cobblers I blame the fact that it's Friday afternoon and
> having long since moved into computing my physics is horrible these
days...
If it's any consolation I thought the same at first, and it was only when I
worked it all out from very simple principles for myself that I realised the
error of my ways.
cheers,
clive
soup
on and said
> Jack Ouzzi wrote:
>
> > All this 4 people build 6 houses in 7 days how many in 28 days
> > stuff. I NEVER understood it at school ..........
>
> Likewise. "If it takes a man and a half a day and a half to dig a
> hole and a half, how long does it take a fly with clogs on to walk
> through a barrel of treacle?"
>
> wot use is /that/ in the new elizabethan age, eh?
Can't remember the chaps name but he asked
questions along the lines of
"if it takes 100 men 10 days to build a wall how
long will it take a million men"
of course it can be worked out mathematicaly (28.8 secs)
but this chap was of the opinion that a million men
would spend all their time getting in each others road
so a million men would never actually get a wall built.
Gawnsoft
less):
...
>Crikey ............ is there any formula for distance travelled at
>18mph (with a very slight 23 degree sidewind) giving the amount of
>fruit pies I can eat to replace the calories consumed during said
>distance covered ??
>(Thats 45 gram Bramley Apple pies)
Yes - and you can probably find it at www.analyticcycling.com
But at a rough estimate, I'd use
~150kcal per 45g bramley apple pie (based on UK Safeway's own brand)
And ~58kcal/mile at 18mph+slight sidewind for a 90kg man,
down to ~45kcal at 18mph+slight sidewind for a 70kg man.
So for me, I'd need to cycle nearly 3 miles / Bramley apple pie. YMMV
(quite literally).
--
Cheers,
Euan
Gawnsoft: http://www.gawnsoft.co.sr
Symbian/Epoc wiki: http://html.dnsalias.net:1122
Smalltalk links (harvested from comp.lang.smalltalk) http://html.dnsalias.net/gawnsoft/smalltalk
bugbear
> I remembered a thread from the tandem mailing list a couple of years back
> which caused a lot of confusion, so I shall see what people here (and on
> cyclingforums) make of it:
>
> My mate Arthur Pedaller has a really great bike. It has absolutely no
> rolling resistance whatsoever - the sole retarding force is air resistance.
> On a windless day he can happily pedal at 25mph, and he covers the 25 miles
> to work on the marvellous straight road the council installed for him at
> that speed.
> But today's a bit windy - 10mph headwind to be precise. He decides to ride
> at 15mph in order to make sure his hair doesn't look abnormal when he
> distance, but the question is, has he put in more or less work than on a
> windless day?
Since the only work (as stated) is done overcoming air resistance,
his work rate (power expenditure) is the same in both cases.
Total Energy = power * time
BugBear
Clive George
news:415d8056$0$82273$ed26...@ptn-nntp-reader03.plus.net...
Simple, succinct, and unfortunately wrong!
cheers,
clive
David Martin
415d7e09$0$69730$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
Another way to look at it:
The work done is force x distance moved.
At a constant airspeed, the work done is that to move the air molecule
sideways enough to get round the rider.
At a constant airspeed (and air density) the number of air particles
requiring to be moved is constant.
so F is constant but d is proportional to the number of air molecules.
The number of air molecules is proportional to the time taken.
So from this the distance is proportional to the time taken (as the only
effective force is the force required for movement of the air, Arthur is
otherwise in steady state).
So the work done on moving Arthur is zero, but the work done on moving air
our of Arthur's way is proportional to the time taken for a given air speed
and density.
It does take more work to go through moving air than still air.
..d
..d
Ian Smith
> My mate Arthur Pedaller has a really great bike. It has absolutely no
> rolling resistance whatsoever - the sole retarding force is air resistance.
> On a windless day he can happily pedal at 25mph, and he covers the 25 miles
> to work on the marvellous straight road the council installed for him at
> that speed.
> But today's a bit windy - 10mph headwind to be precise. He decides to ride
> at 15mph in order to make sure his hair doesn't look abnormal when he
> arrives, and off he goes. Obviously he takes 1 hour 40 minutes to cover the
> distance, but the question is, has he put in more or less work than on a
> windless day?
I'll assume it's not a pun (in which case the answer is 40 minutes
less). I did once manage to use the old faithful - "how many people
work here?" said visitor "oh, about half of them" says I, having been
saving it up for years.
I'll assume we're assuming teh force required to ride at 25mph on a
still day is teh same as that to ride at 15mph into a 10mph headwind.
(There are mechanisms by which it would not be, even given this
marvelous bike - the spoke speeds through teh air differ, for example,
in one case peaking at 50mph, and in teh other at 40mph).
Exerted force travels same distance, so I'd struggle to conclude
anything other than that they do teh same work (in the strict
forcexdistamce sense - the windy day will be a harder ride).
regards, Ian SMith
--
|\ /| no .sig
|o o|
|/ \|
Clive George
news:slrnclr2n...@phlegethon.smithnet...
> Exerted force travels same distance, so I'd struggle to conclude
> anything other than that they do teh same work (in the strict
> forcexdistamce sense - the windy day will be a harder ride).
Almost a gold star to that man (and bonus points for getting the different
spoke speeds - yes, I am ignoring those).
Actually on the windy day Arthur won't be pedalling as hard - he'll probably
be in a lower gear for the same wheel force and the same cadence, which
gives the same work against the wind. He'll pedal at 60% effort for 1 2/3 as
long.
For a followup question, assuming wind resistance is proportional to wind
speed squared, how fast does Arthur have to go in a 10mph headwind to be
working at the same rate as he does going 25mph in still wind?
(the latter case is of course more like how real people ride)
cheers,
clive
Clive George
What happens if Arthur hits a 25mph headwind, and just decides to sit there
with the brakes on? All the work moving the air out of Arthur's way is still
done, yet he's obviously doing no work.
If Arthur's bike was propellor driven rather than wheel driven (ie against
the air rather than the ground) then you'd be right. But it isn't.
cheers,
clive
daren_RAIN...@hotmail.com
In the cyclists frame of reference (sat on the saddle), riding at 25mph
requires more force on the pedals than 15mph into the headwind because
the wind is stronger. Eventually if the wind provides a force of
greater than ~10*BWT he won't move forward (unless on a bent and
benchpressing into the seat).
He is working 15/25 as hard and his journey takes 100/60th's so he
expends 100% of the work at 25mph.
Last time I mentioned going downhill I started a flame war...
kind regards,
Daren
--
remove outer garment for reply
Clive George
news:1096652052....@k17g2000odb.googlegroups.com...
> Force is applied to pedal to overcome retarding force of wind.
>
> In the cyclists frame of reference (sat on the saddle), riding at 25mph
> requires more force on the pedals than 15mph into the headwind because
> the wind is stronger.
Not convinced about this bit of the argument. The wind the cyclist feels is
identical in both cases (25mph into no wind, 15mph into a 10mph headwind).
> Eventually if the wind provides a force of
> greater than ~10*BWT he won't move forward (unless on a bent and
> benchpressing into the seat).
(Is that 10 * bodyweight?)
> He is working 15/25 as hard and his journey takes 100/60th's so he
> expends 100% of the work at 25mph.
Tick vg.
> Last time I mentioned going downhill I started a flame war...
I think the gentleman in question has gone now..
cheers,
clive
Ian Smith
> news:slrnclr2n...@phlegethon.smithnet...
>
> > Exerted force travels same distance, so I'd struggle to conclude
> > anything other than that they do teh same work (in the strict
> > forcexdistamce sense - the windy day will be a harder ride).
>
> Almost a gold star to that man (and bonus points for getting the different
> spoke speeds - yes, I am ignoring those).
>
> Actually on the windy day Arthur won't be pedalling as hard - he'll probably
> be in a lower gear for the same wheel force and the same cadence, which
> gives the same work against the wind. He'll pedal at 60% effort for 1 2/3 as
> long.
I still think the windy day will be a 'harder' ride, in some sort of
entirely vague and qualitative sense, however.
Jon Senior
> Likewise. and an A level. Though this is more applied maths (for which I
> have an old style AO).
My mum once described applied maths in terms of an example exam
question:
"A train of 10 carriages, where each carriage has a mass of 5 tonnes
starts at the top of a 1 mile long, 14% incline. It accelerates down the
incline from a stationary start.
Given a large solid object at the bottom of the hill, how many
passengers will be killed?"
Jon
David Martin
415d8fde$0$80628$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
And he isn't moving. Exactly the same as a stone on the ground is still
subject to gravity but is not moving. The work done by the air will slow it
down.
If he takes the brakes off will he have to apply force to keep himself from
going backwards?
> If Arthur's bike was propellor driven rather than wheel driven (ie against
> the air rather than the ground) then you'd be right. But it isn't.
Propellor driven makes no difference. It is still a question of the force
required to get acceleration of the air particles out of the way.
Arthur is in steady state. His speed is constant.
There is no acceleration, hence no force pertaining to his body.
There is a force he puts in which is effective on the air molecules and
moving them. This force displaces each molecule around his body. There is a
change in velocity and hence an acceleration, so there is a force on each
air molecule.
We assume for simplicity that each molecule requires the same amount of
force and that the number of molecules he will need to displace in unit time
is constant (at zero effective air speed that would be zero).
The energy expended per air molecule is a very complex summation of the
forces and distances involved, we can call it x.
The total energy expended will be x times the number of molecules.
The number of molecules is proportional to the density of the air (a
constant) the relative speed of the air mass (again constant) and the
exposure time.
So we end up with the energy expended being proportional to the time taken.
If indeed he was doing only 60/100 of the work but taking 100/60 as long,
then headwinds would have no effect on the speed a cyclist can travel in
your model. Work rate would be constant at a constant speed irrespective of
the headwind.
Which, if you have ridden a bike, you would realise is rubbish.
I'm still perplexed as to where the non-zero work rate comes from when
Arthur is at steady state if it is not from interaction with the air. Would
you care to explain how F=ma can be positive when a is zero? (oh, and the
equivalent for the angular velocity of the wheels)
..d
Clive George
> >> It does take more work to go through moving air than still air.
> >
> > What happens if Arthur hits a 25mph headwind, and just decides to sit
there
> > with the brakes on? All the work moving the air out of Arthur's way is
still
> > done, yet he's obviously doing no work.
>
> And he isn't moving. Exactly the same as a stone on the ground is still
> subject to gravity but is not moving. The work done by the air will slow
it
> down.
>
> If he takes the brakes off will he have to apply force to keep himself
from
> going backwards?
Yes. Force, but not work, because he isn't moving.
> > If Arthur's bike was propellor driven rather than wheel driven (ie
against
> > the air rather than the ground) then you'd be right. But it isn't.
>
> Propellor driven makes no difference. It is still a question of the force
> required to get acceleration of the air particles out of the way.
No, propellor driven makes _all_ the difference.
Consider a boat with a little engine which can pootle along at 5 knots. Put
it in a 5 knot current going the wrong way, and the boat will just sit
still.
Now put a pulley and rope system attached to the bridge a few miles upstream
in instead of the prop, and some suitably low gears. The boat will be able
to move forwards, despite there being no extra power generated by the engine
and no less force exerted by the water.
<discussion of air molecules>
> So we end up with the energy expended being proportional to the time
taken.
But where does the energy come from? The answer is that Arthur doesn't
provide it. Just as if he is standing still in a 25mph wind for an hour, the
same energy is dissipated. Where does a windmill get its energy from?
> If indeed he was doing only 60/100 of the work but taking 100/60 as long,
> then headwinds would have no effect on the speed a cyclist can travel in
> your model.
Er - no! Yes, he is doing 60% of the work by travelling at 60% of the speed
with the same net windspeed. My model does not consider anything apart from
the one windspeed, which means the following:
> Work rate would be constant at a constant speed irrespective of the
headwind.
cannot be deduced from it.
> Which, if you have ridden a bike, you would realise is rubbish.
The thing is that in real life Arthur won't cycle at a constant windspeed -
he will in fact cycle at something closer to constant effort. Compared to
riding in still air, with a headwind his roadspeed will be lower but his
windspeed will be higher. That's one of the tricksy bits of the original
problem!
> I'm still perplexed as to where the non-zero work rate comes from when
> Arthur is at steady state if it is not from interaction with the air.
No, it is from interaction with the air. But you've also got to consider
what you're pushing against, viz the ground. Eg clearly when Arthur is
standing still in a 25mph wind, he's doing the same to the air that he would
be riding at 25mph in still air. Yet in one case he's working quite hard, in
the other he just needs to brace himself appropriately. (ignoring the fact
that muscles do in fact take energy just to stand still). Standing still
isn't a special case here - it's just one point on a continuum.
> Would you care to explain how F=ma can be positive when a is zero?
I don't believe I'm saying any such thing.
> (oh, and the equivalent for the angular velocity of the wheels)
I'm ignoring wheels - it's complicated enough as it is!
cheers,
clive
David Martin
415da915$0$69739$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
> No, propellor driven makes _all_ the difference.
> Consider a boat with a little engine which can pootle along at 5 knots. Put
> it in a 5 knot current going the wrong way, and the boat will just sit
> still.
> Now put a pulley and rope system attached to the bridge a few miles upstream
> in instead of the prop, and some suitably low gears. The boat will be able
> to move forwards, despite there being no extra power generated by the engine
> and no less force exerted by the water.
But according to your argument, said boat (assuming appropriate efficiency)
should consume just as much fuel doing 3knots into a 2 knot current over
1hr40 as doing 5 knots on slack water, and at almost 5 knots current it will
burn exactly the same amount of fuel in almost infinite time.
What is the work you are doing? It isn't maintaining a body at constant
speed because that takes no work. So it must be moving the air.
For the boat, it is proportional to the amount of water moved around the
hull.
The fluid through which the object travels is slowed down by it's
interaction with the object.
..d
Note that even in a vacuum it will take more energy for Arthur to travel at
a fixed speed as it takes longer distance or a larger force to accelerate to
the faster speed from rest.
Clive George
> On 1/10/04 8:03 pm, in article
> 415da915$0$69739$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
> <cl...@xxxx-x.fsnet.co.uk> wrote:
>
> > No, propellor driven makes _all_ the difference.
>
>
>
> > Consider a boat with a little engine which can pootle along at 5 knots.
Put
> > it in a 5 knot current going the wrong way, and the boat will just sit
> > still.
> > Now put a pulley and rope system attached to the bridge a few miles
upstream
> > in instead of the prop, and some suitably low gears. The boat will be
able
> > to move forwards, despite there being no extra power generated by the
engine
> > and no less force exerted by the water.
>
> But according to your argument, said boat (assuming appropriate
efficiency)
Which boat? The one driven by the propellor or the one driven by the pulley
onto the bridge?
cheers,
clive
Clive George
> Note that even in a vacuum it will take more energy for Arthur to travel
at
> a fixed speed as it takes longer distance or a larger force to accelerate
to
> the faster speed from rest.
This can be safely ignored for the purposes of this discussion - the KE
required is sufficiently small compared to the total energy required for the
original 25 mile trip.
cheers,
clive
David Martin
415db06a$0$69735$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
But the body is at constant speed so where is all the rest of this energy
going?
If we can ignore the speed over the ground then it is just the speed through
the air that matters. In which case it is work rate that matters. It is the
acceleration of the air that is the difference, not the acceleration of the
body (as it is not accelerating.) The distance the air is moved is greater
so the work is greater.
..d
David Martin
415db06a$0$69735$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
lets try an analogy.
You are standing on movable platform by a conveyor belt. This belt is 10m
long. It brings towards you a series of even weights. You have to move the
weights to one side.
You can set the platform to move at a constant speed up the conveyor belt.
The job finishes when you get to the other end of the conveyor belt.
now, assuming you can move weights at a constant rate, will you do more work
when the conveyor belt is moving against your direction of movement, or when
it is stationary?
What about travelling back again with the conveyor moving in the same
direction as you?
..d
Clive George
Obviously when it's moving against you.
BUT - This isn't analogous to Arthur on his bike. It's analogous to Captain
Haddock in his little boat (the one with the propellor, not the one with the
rope - btw if you tell me which one you meant I'll finish my answer to your
other post).
> What about travelling back again with the conveyor moving in the same
> direction as you?
ditto.
I'll bring the weights and string analogy in now.
(For the benefit of the public:
Arthur's feet hurt, so he can't pedal. Bruce, his boss, still wants him to
get to work. Fortunately work is on the edge of a 25 mile high cliff and
Bruce happens to be a maker of magic weightless rope. So Bruce works out how
much force is required to move Arthur through the air at 25mph, and gets a
lump of metal which weighs that much. He ties one end to a 25 mile long
rope, and takes the other end to Arthur's house. In the morning, Arthur
clips onto the rope, phones Bruce up who then drops the weight over the edge
(the rope runs over a pulley).)
Pretty soon the system reaches a steady state, with the Arthur moving on his
bike with a windspeed of 25mph, since that's has the same force as that
exerted by the rope.
This happens twice - once on a windless day, and once when there is a 10mph
headwind.
Obviously on the first day, Arthur goes at a windspeed and roadspeed of
25mph, and reaches work in about an hour.
On the second day, Arthur still goes at a windspeed of 25mph, but his
roadspeed is 15mph, so it takes him 1h40m to get to work.
But the sole energy input is the weight falling, and it has fallen the same
distance in both cases, so the total energy required for both trips must be
the same.
Therefore if Arthur repeats these journeys, except this time pedalling
himself, he must put the same amount of energy in in both cases.
The thing is, when he has the headwind, he isn't pedalling as hard (!), so
he ends up doing the same amount of work, except over a longer time. If he
was to pedal as hard he would go faster than 15mph.
(I am ignoring end effects and wheel rotation!)
cheers,
clive
the.Mark
> Clive George popped their head over the parapet saw what was
> going on and said
>> "MSeries" <skank...@hotmail.com> wrote in message
>> news:1096640121.3...@k17g2000odb.googlegroups.com...
>>
>>> Clive, if you weren't a regular I'd suspect someone was
>>> trying to get some help with their homework. Have you
>>> started a GCSE physics course ?
>>
>> Hey - I've got an O level in physics - none of that GCSE
>> nonsense!
>>
>> cheers,
>> clive
>
> I have an "O" grade in physics, a higher and a CSYS
> (certificate of sixth year studies). :-P
> That was years ago tho' I wouldn't like to analyse
> that situation now.
I'm a chemist am I in the wrong newsgroup.
--
Mark
1x1 wheel, 3x2 wheels & 1x3 wheels.
daren_RAIN...@hotmail.com
frame of reference.
A simpler way of looking at the problem is as follows:
It's the annual Open Alpe d'Huez mountain climb (not time trial). Today
they have opened a new route for the 1000m climb (vertical rise). The
road, at an incline of 1 in 8 and the new wimp-out route at 1 in 16.
Which to take to assure success?
The force of resistance climbing the hill is approximately 10 x weight
x incline (mg Sin(x)). So the wimp-out route will require half the
force on the pedals to make progress.
However in terms of energy needed to climb the mountain, either route
is fine because the total amount of energy to get to the top is the
same.
What differs is POWER.
Here comes Lance, he's lean mean and 80kg including that nice bike.
Energy required to get to top = 80 (mass) x 10 (g) x 1000 (h) = 800,000
Joules
Lance is good for a steady 300W (Joules/sec) So the time to get to the
top will be roughly
(Work/Power) 800,000/300 = 2667 secs = 44min 27sec (he made it in 39
this yr, but 1000m is a guess at vertical climb).
Now here come Daren. He's a lean, not very mean 90kg (bike is twice as
heavy with panniers and flask of tea). Average power output 100W
(including tea stop).
Being heavier will require more energy (but not much, unless I'm on the
kiddyback with stokid), but the lower power means that he is going to
take a lot longer to expend those 900,000 Joules, in fact a whopping
9000 secs = 150min or 2hr 30min :-(
Marco Pantani's record is ~37min, and he weighed approx 65kg (with
bike).
Power output approx 650,000/(37*60) = 292Watts or 4.5Watts/kg
This time will be hard to beat as Lance has to do 20% more work for the
same time, but his power output is broadly similar. Of course Marco's
extra 20% may have come from a pharmacopia to which Lance has no
access. EPO provides an approx increase of 6-10% in VO2max (say power)
Anyway back to the problem; the slope of the hill is equivalent to the
strength of headwind, and the fact that the vertical height is the same
means you need to do the same amount of work to get there, it just
takes longer.
kind regards,
Daren (IUTBAP) - I used to be a Physicist :-)
--
remove outer garment for relply
Jon Senior
> I'm a chemist am I in the wrong newsgroup.
I'm a (lapsed) biochemist. But I'm as stubborn as anyone else here so I
guess that must count for something! ;-)
Jon
David Martin
415dc40b$0$69729$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
No, it is analagous to the one with the rope.
We could make it more obvious, joe on the conveyor is moving forward at a
constant speed with respect to the floor, and as he is at constant speed he
is doing no work to move. the only work he is doing is in moving things on
the conveyor.
The propellor analogy is flawed because to propel oneself with a propellor
means dealing with the absolute square of the speeds rather than the
difference in speed.
The only input from the weight is the acceleration to terminal velocity.
After this time there is no further acceleration. So the weight does no work
as it is not accelerating. The force of gravity is exactly balanced by the
air resistance. The weight and Arthur move at constant speed.
At the end of the drop the weight has accrued a certain amount of kinetic
energy. In fact, it is the same amount of kinetic energy as it had when it
got to the terminal velocity so there only difference between the faster and
slower situations is the distance taken to get up to the terminal velocity.
Assuming that the weight is constant, then the initial acceleration will be
constant. The distance moved under the influence of the force (until the
forces balance) is a function of the change in acceleration but without
needing to go into details, this can be expressed as a larger amount of work
giving rise to a higher speed. Either more force or more distance. Either of
which is more work. Once he is at steady state he is essentially at rest as
all the forces on him are balanced.
So the difference is purely the acceleration to his cruising speed.
Irrespective of how far he goes.
Or maybe there is something in this moving air around absorbs energy deal?
And the longer you move air around, the more energy it absorbs.
Arthur carries out work on the air and the longer he is travelling through
the air the more work is done. This is equivalent to the difference in
kinetic energy between the heavy weight in the still and windy conditions.
It may not be much but it is present.
> Therefore if Arthur repeats these journeys, except this time pedalling
> himself, he must put the same amount of energy in in both cases.
The kinetic energies are different at the end. Therefore less work has been
done in the still conditions.
> The thing is, when he has the headwind, he isn't pedalling as hard (!), so
> he ends up doing the same amount of work, except over a longer time. If he
> was to pedal as hard he would go faster than 15mph.
he ends up doing more work over the longer time. Not by much but it is more.
Once at steady state there is no difference (if you assume the work done on
the wind to be negligible by comparison to the acceleration, which it
shouldn't be.)
..d
David Martin
1096666329.5...@k26g2000oda.googlegroups.com,
"daren_RAIN...@hotmail.com" <daren_RAIN...@hotmail.com> wrote:
> Not all of the wind force is providing resistance in the cyclist's
> frame of reference.
>
> A simpler way of looking at the problem is as follows:
> It's the annual Open Alpe d'Huez mountain climb (not time trial). Today
> they have opened a new route for the 1000m climb (vertical rise). The
> road, at an incline of 1 in 8 and the new wimp-out route at 1 in 16.
>
> Which to take to assure success?
>
> The force of resistance climbing the hill is approximately 10 x weight
> x incline (mg Sin(x)). So the wimp-out route will require half the
> force on the pedals to make progress.
>
> However in terms of energy needed to climb the mountain, either route
> is fine because the total amount of energy to get to the top is the
> same.
but the distance is different.. Vertical energy gain is identical, over the
ground distance is different.
Energy expediture is the same.
> What differs is POWER.
> Here comes Lance, he's lean mean and 80kg including that nice bike.
> Energy required to get to top = 80 (mass) x 10 (g) x 1000 (h) = 800,000
> Joules
>
> Lance is good for a steady 300W (Joules/sec) So the time to get to the
> top will be roughly
> (Work/Power) 800,000/300 = 2667 secs = 44min 27sec (he made it in 39
> this yr, but 1000m is a guess at vertical climb).
> Now here come Daren. He's a lean, not very mean 90kg (bike is twice as
> heavy with panniers and flask of tea). Average power output 100W
> (including tea stop).
>
> Being heavier will require more energy (but not much, unless I'm on the
> kiddyback with stokid), but the lower power means that he is going to
> take a lot longer to expend those 900,000 Joules, in fact a whopping
> 9000 secs = 150min or 2hr 30min :-(
>
> Marco Pantani's record is ~37min, and he weighed approx 65kg (with
> bike).
> Power output approx 650,000/(37*60) = 292Watts or 4.5Watts/kg
>
> This time will be hard to beat as Lance has to do 20% more work for the
> same time, but his power output is broadly similar. Of course Marco's
> extra 20% may have come from a pharmacopia to which Lance has no
> access. EPO provides an approx increase of 6-10% in VO2max (say power)
Very good..
> Anyway back to the problem; the slope of the hill is equivalent to the
> strength of headwind, and the fact that the vertical height is the same
> means you need to do the same amount of work to get there, it just
> takes longer.
No it isn't. It is equivalent to having a hill of a steeper slope but the
same length. So if you climb it at the same metres per minute, you will take
longer to climb overall.
The analogy is better described with the wind being added weight. If you
climb at the same power it will take longer with more weight for the same
distance. The slope of the hill is best described in terms of the riders
desired power output.
However, the descent will be faster;-)
In all of this you only assess the rider and not hte work done on the fluid
through which the rider is passing.
..d
DavidR
"Clive George" <cl...@xxxx-x.fsnet.co.uk> wrote
> (For the benefit of the public:
> Arthur's feet hurt, so he can't pedal. Bruce, his boss, still wants him
> to
> get to work. Fortunately work is on the edge of a 25 mile high cliff and
> Bruce happens to be a maker of magic weightless rope. So Bruce works out
At the top of the cliff the weight has a quantity of potential energy. Let
it drop in free fall, it gains kinetic energy which is released as heat &
smoke when it hits the bottom.
We now add a mass (our cyclist) on a horizontal track connected to
the falling mass with our magic rope. At first we let it run frictionless
and bring both masses to a stop at the same time (we don't want Arthur
going over the cliff). The same KE is now shared between the
two masses - and since there was no friction, no energy could be lost to
it.
Now repeat with friction. Same total energy but it can not run as fast
so that at the end of the run there is less heat & smoke. The difference
went into friction. Add more friction, less KE, more energy lost
to friction.
The laws of physics are unbroken. Phew.
James Annan
> I remembered a thread from the tandem mailing list a couple of years back
> which caused a lot of confusion, so I shall see what people here (and on
> cyclingforums) make of it:
>
> My mate Arthur Pedaller has a really great bike. It has absolutely no
> rolling resistance whatsoever - the sole retarding force is air resistance.
> On a windless day he can happily pedal at 25mph, and he covers the 25 miles
> to work on the marvellous straight road the council installed for him at
> that speed.
> But today's a bit windy - 10mph headwind to be precise. He decides to ride
> at 15mph in order to make sure his hair doesn't look abnormal when he
> arrives, and off he goes. Obviously he takes 1 hour 40 minutes to cover the
> distance, but the question is, has he put in more or less work than on a
> windless day?
Much less work, but perhaps I should not have posted until a few more
people had got it wrong...
James
--
If I have seen further than others, it is
by treading on the toes of giants.
http://www.ne.jp/asahi/julesandjames/home/
James Annan
James Annan wrote:
>> the question is, has he put in more or less work than on a
>> windless day?
>
>
> Much less work, but perhaps I should not have posted until a few more
> people had got it wrong...
Sorry, misread the question - thought it was the standard one about the
effort required for 25mph in still air vs 15 in a 10 headwind...
Russell
Sorry but I'm sure you are both wrong:-) In this (unreal) example
Arthur will have to generate the same amount of force to travel
through the still air at 25 mph as he does to travel through the 10mph
headwind at 15mph. Remember "the sole retarding force is air
resistance". So either way the air is hitting him at 25mph. Thus due
to the extra time taken on the windy day, the work done will be
66.667% greater. I can't see what all complex maths & confusion is
about its simple! Oh and as for sitting in a 25mph wind with his
brakes on. The force of the wind on Arthur is merely being transferred
into turning the Earth slightly, apart from a small amount of friction
warming his clothes which would probably be negated by the windchill
anyway.
Now looking at the above, I can see that Arthur would have to use far
greater (driving wheel) force to ride against a 30mph wind and
therefore a much lower gear and therefore his journey would take much
longer again for the same level of effort. Now what troubles me is
that now he is now travelling though the air at 30MPH for the same
level of effort! So with a low enough gear, could Arthur ride against
the 30MPH wind?
Hmmm ... I think we're all wrong.
Russell (O level grade C).
Simon Brooke
('xlu...@users.sourceforge.remove.this.antispam.net') wrote:
> On Fri, 01 Oct 2004 16:15:00 +0100, Jack Ouzzi ... wrote (more or
> less):
>
> ...
>>Crikey ............ is there any formula for distance travelled at
>>18mph (with a very slight 23 degree sidewind) giving the amount of
>>fruit pies I can eat to replace the calories consumed during said
>>distance covered ??
>>(Thats 45 gram Bramley Apple pies)
>
> Yes - and you can probably find it at www.analyticcycling.com
>
> But at a rough estimate, I'd use
> ~150kcal per 45g bramley apple pie (based on UK Safeway's own brand)
>
> And ~58kcal/mile at 18mph+slight sidewind for a 90kg man,
> down to ~45kcal at 18mph+slight sidewind for a 70kg man.
>
> So for me, I'd need to cycle nearly 3 miles / Bramley apple pie. YMMV
> (quite literally).
That's it! We have a new measure of fuel efficiency: MAP - Miles per
Apple Pie.
--
si...@jasmine.org.uk (Simon Brooke) http://www.jasmine.org.uk/~simon/
my other car is #<Subr-Car: #5d480>
;; This joke is not funny in emacs.
Clive George
> So the difference is purely the acceleration to his cruising speed.
> Irrespective of how far he goes.
Wahey!
> Or maybe there is something in this moving air around absorbs energy deal?
> And the longer you move air around, the more energy it absorbs.
> Arthur carries out work on the air and the longer he is travelling through
> the air the more work is done. This is equivalent to the difference in
> kinetic energy between the heavy weight in the still and windy conditions.
> It may not be much but it is present.
>
>
> > Therefore if Arthur repeats these journeys, except this time pedalling
> > himself, he must put the same amount of energy in in both cases.
>
> The kinetic energies are different at the end. Therefore less work has
been
> done in the still conditions.
I'll repeat what I wrote originally - you've snipped it, whuch is remarkably
unhelpful in this case:
"(I am ignoring end effects and wheel rotation!)"
The acceleration to the steady state at the beginning, and the KE present at
the end are the end effects - over the 25 mile trip, they are a sufficiently
small proportion of the energies involved as to be ignorable.
> > The thing is, when he has the headwind, he isn't pedalling as hard (!),
so
> > he ends up doing the same amount of work, except over a longer time. If
he
> > was to pedal as hard he would go faster than 15mph.
> he ends up doing more work over the longer time. Not by much but it is
more.
By about 0.1%. That's not much at all. If we ignore the first and last miles
what is the answer?
> Once at steady state there is no difference (if you assume the work done
on
> the wind to be negligible by comparison to the acceleration, which it
> shouldn't be.)
Phew. I think we're agreeing.
Once at a steady speed, it takes Arthur the same amount of work to travel 20
miles whether he does it at 25mph in still air or 15mph in a 10mph headwind.
Or 35mph in a 10mph tailwind, etc etc. And Arthur's hair need not get out of
place.
cheers,
clive
Clive George
news:1656ad32.04100...@posting.google.com...
> Sorry but I'm sure you are both wrong:-) In this (unreal) example
> Arthur will have to generate the same amount of force to travel
> through the still air at 25 mph as he does to travel through the 10mph
> headwind at 15mph. Remember "the sole retarding force is air
> resistance". So either way the air is hitting him at 25mph. Thus due
> to the extra time taken on the windy day, the work done will be
> 66.667% greater.
Russell - go and have a look at the message where I give an example with a
weight pulling Arthur, to see why you're wrong. Alternatively there's
another explanation below.
> I can't see what all complex maths & confusion is about its simple!
Why do you think I brought the question up? :-)
> Oh and as for sitting in a 25mph wind with his
> brakes on. The force of the wind on Arthur is merely being transferred
> into turning the Earth slightly, apart from a small amount of friction
> warming his clothes which would probably be negated by the windchill
> anyway.
So plot work done against groundspeed, with windspeed being kept at a
constant 25mph. Is 0mph a discontinuity? (hint - no)
> Now looking at the above, I can see that Arthur would have to use far
> greater (driving wheel) force to ride against a 30mph wind and
> therefore a much lower gear and therefore his journey would take much
> longer again for the same level of effort. Now what troubles me is
> that now he is now travelling though the air at 30MPH for the same
> level of effort! So with a low enough gear, could Arthur ride against
> the 30MPH wind?
Indeed he could. You're getting closer to the way I originally proved the
answer to myself.
Back to the 25mph and 15 + 10mph cases:
Arthur is a bit picky - he has his preferred cadence and hates deviating
from it. Fortunately he has a 100" gear he uses at 25mph and a 60" gear he
uses at 15mph.
Now we're looking for the same force against the wind in both cases. Force
against wind is proportional to pedal force divided by gear inches (*). So
at 15mph with a 10mph headwind he's using 60% of the pedal force that he
would be using at 25mph in still air. Which conveniently cancels out with
the extra time spent pedalling to leave the same work done.
(*) This is the statement you're going to have to pick holes in - but it is
pretty much true for Arthur.
cheers,
clive
Ian Smith
> Sorry but I'm sure you are both wrong:-) In this (unreal) example
> Arthur will have to generate the same amount of force to travel
> through the still air at 25 mph as he does to travel through the 10mph
> headwind at 15mph. Remember "the sole retarding force is air
> resistance". So either way the air is hitting him at 25mph.
Yes.
> Thus due
> to the extra time taken on the windy day, the work done will be
> 66.667% greater.
No. Work is force x distance, not force x time.
If you want to use time, you need power x time, but the power output
in the headwind case is lower, exactly as much lower as teh time is
longer.
You can exert a million tonnes force, for a thousand years, and if it
does not move you will have done no work.
Whingin' Pom
I'm a physiology researcher working in membrane transport.
The answer is that he lives in a salt mine and travels to work down
the electrochemical gradient.
What do I win?
--
Matt K.
"And on the seventh day, He exited from append mode."
James Annan
>
> For a followup question, assuming wind resistance is proportional to wind
> speed squared, how fast does Arthur have to go in a 10mph headwind to be
> working at the same rate as he does going 25mph in still wind?
about 19mph, being the solution to x(x+10)^2=25^3
Dave Kahn
<whing...@donotreply.com> wrote:
>I'm a physiology researcher working in membrane transport.
Every so often a huge container of membranes arrives and you have to
unload it?
--
Dave...
Get a bicycle. You will not regret it. If you live. - Mark Twain
Whingin' Pom
wrote:
>On Sat, 02 Oct 2004 19:24:45 +1200, Whingin' Pom
><whing...@donotreply.com> wrote:
>
>>I'm a physiology researcher working in membrane transport.
>
>Every so often a huge container of membranes arrives and you have to
>unload it?
Bingo. :-)
It's mainly testing drugs via the patch-clamping technique at the
moment. I can go into more technical detail if you really want...
Big picture is that it's working towards a potential pharmacological
treatment for cystic fibrosis. I do the experiments and analyse the
data, the boss writes the papers.
Jon Senior
> I'm a physiology researcher working in membrane transport.
> The answer is that he lives in a salt mine and travels to work down
> the electrochemical gradient.
But how much energy is used by ion pumps, to maintain the gradient?
Jon
Whingin' Pom
Given that each cell shifts about, ooh, 4 billion-odd Na+ ions per
minute in the small intestine, that works out as....
<scribbles on back of envelope>
1 apple pie for every 3 miles. Or a cafe stop for cake and a nice hot
cup of a phosphodiesterase inhibitor.
:-)
I take it drug use in order to get home from work would lead to him
being suspended?
soup
and said
> constant 25mph. Is 0mph a discontinuity? (hint - no)
Think you are on a sticky wicket here at 0 the work done is zero
Work = force times distance moved( in the direction of force)
A man could push sideways on a railway wagon for two days,
while exhausted he has done no work.
http://www.sidtech.co.uk/iu/soup3203636634.JPG
--
yours S
Nihil curo de ista tua stulta superstitione
Clive George
news:k6v7d.179510$hZ3.1...@fe2.news.blueyonder.co.uk...
> Clive George popped their head over the parapet saw what was going on
> and said
> >> So plot work done against groundspeed, with windspeed being kept at a
> > constant 25mph. Is 0mph a discontinuity? (hint - no)
>
> Think you are on a sticky wicket here at 0 the work done is zero
> Work = force times distance moved( in the direction of force)
> A man could push sideways on a railway wagon for two days,
> while exhausted he has done no work.
That's at least part of my point.
Does the plot (of work done per unit time) look like this:
*********************
*
or this:
***
***
***
***
***
(excuse ascii art resolution errors!)
cheers,
clive
David Martin
415df324$0$80095$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
> I'll repeat what I wrote originally - you've snipped it, whuch is remarkably
> unhelpful in this case:
>
> "(I am ignoring end effects and wheel rotation!)"
You brought in the bit about end effects later, not in the original
question.
So the answer is he does a miniscule amount more work due to the end
effects..
It's been fun, but no one else wanted to post imaginatively wrong answers..
..d
DavidR
> "Russell" <rcat...@yahoo.co.uk> wrote
>
>> Sorry but I'm sure you are both wrong:-) In this (unreal) example
>> Arthur will have to generate the same amount of force to travel
>> through the still air at 25 mph as he does to travel through the 10mph
>> headwind at 15mph. Remember "the sole retarding force is air
>> resistance". So either way the air is hitting him at 25mph. Thus due
>> to the extra time taken on the windy day, the work done will be
>> 66.667% greater.
>
> Russell - go and have a look at the message where I give an example with
> a
> weight pulling Arthur, to see why you're wrong. Alternatively there's
> another explanation below.
But *you* got that wrong.
> Arthur is a bit picky - he has his preferred cadence and hates deviating
> from it. Fortunately he has a 100" gear he uses at 25mph and a 60" gear
> he
> uses at 15mph.
>
> Now we're looking for the same force against the wind in both cases.
> Force
> against wind is proportional to pedal force divided by gear inches (*).
> So
> at 15mph with a 10mph headwind he's using 60% of the pedal force that he
> would be using at 25mph in still air. Which conveniently cancels out with
> the extra time spent pedalling to leave the same work done.
You have just said "same force against the wind in both cases". Excellent.
Energy is force times time. Time increases in a head wind.
> (*) This is the statement you're going to have to pick holes in - but it
> is
> pretty much true for Arthur.
Tricky.
Clive George
news:2s7pm3F...@uni-berlin.de...
> "Clive George" <cl...@xxxx-x.fsnet.co.uk> wrote
> > "Russell" <rcat...@yahoo.co.uk> wrote
> >
> >> Sorry but I'm sure you are both wrong:-) In this (unreal) example
> >> Arthur will have to generate the same amount of force to travel
> >> through the still air at 25 mph as he does to travel through the 10mph
> >> headwind at 15mph. Remember "the sole retarding force is air
> >> resistance". So either way the air is hitting him at 25mph. Thus due
> >> to the extra time taken on the windy day, the work done will be
> >> 66.667% greater.
> >
> > Russell - go and have a look at the message where I give an example with
> > a
> > weight pulling Arthur, to see why you're wrong. Alternatively there's
> > another explanation below.
>
> But *you* got that wrong.
<panto> Oh no I didn't! </panto>
> > Arthur is a bit picky - he has his preferred cadence and hates deviating
> > from it. Fortunately he has a 100" gear he uses at 25mph and a 60" gear
> > he
> > uses at 15mph.
> >
> > Now we're looking for the same force against the wind in both cases.
> > Force
> > against wind is proportional to pedal force divided by gear inches (*).
> > So
> > at 15mph with a 10mph headwind he's using 60% of the pedal force that he
> > would be using at 25mph in still air. Which conveniently cancels out
with
> > the extra time spent pedalling to leave the same work done.
>
> You have just said "same force against the wind in both cases". Excellent.
> Energy is force times time. Time increases in a head wind.
But you're thinking of the energy required to shift the wind. As the
standing still case illustrates, Arthur doesn't need to provide that.
> > (*) This is the statement you're going to have to pick holes in - but it
> > is
> > pretty much true for Arthur.
>
> Tricky.
Yup. But can you do it? I can't - but then I'm proposing it.
cheers,
clive
DavidR
> "DavidR" <david.nospam.r_a@t_cwcom.net> wrote
>> > Russell - go and have a look at the message where I give an example
>> > with a weight pulling Arthur, to see why you're wrong. Alternatively
>> > there's another explanation below.
>>
>> But *you* got that wrong.
>
> <panto> Oh no I didn't! </panto>
OK then, where did *I* go wrong? Or has my original message not passed
through your ISP?
>> > Arthur is a bit picky - he has his preferred cadence and hates
>> > deviating from it. Fortunately he has a 100" gear he uses at 25mph and
>> > a 60" gear he uses at 15mph.
>> >
>> > Now we're looking for the same force against the wind in both cases.
>> > Force against wind is proportional to pedal force divided by gear
>> > inches (*). So at 15mph with a 10mph headwind he's using 60% of the
>> > pedal force that he would be using at 25mph in still air. Which
>> > conveniently cancels out with the extra time spent pedalling to leave
>> > the same work done.
>>
>> You have just said "same force against the wind in both cases".
>> Excellent. Energy is force times time. Time increases in a head wind.
>
> But you're thinking of the energy required to shift the wind. As the
> standing still case illustrates, Arthur doesn't need to provide that.
But we stipulated that Arther was trying to do a journey of 25 miles.
If Arthur is putting an exact force on the pedal to overcome the wind
without moving his journey is time is infinite. Energy requred = force
times infinity. Power is zero. Sorry.
>> > (*) This is the statement you're going to have to pick holes in - but
>> > it is pretty much true for Arthur.
>>
>> Tricky.
>
> Yup. But can you do it? I can't - but then I'm proposing it.
No. Otherwise I would have written it. You do realise I won't get any sleep
now :)
DavidR
> But we stipulated that Arther was trying to do a journey of 25 miles.
> If Arthur is putting an exact force on the pedal to overcome the wind
> without moving his journey is time is infinite. Energy requred = force
> times infinity.
That's my credibility blown out of the water
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
energy = force times distance
Clive George
> "Clive George" <cl...@xxxx-x.fsnet.co.uk> wrote
> > "DavidR" <david.nospam.r_a@t_cwcom.net> wrote
>
> >> > Russell - go and have a look at the message where I give an example
> >> > with a weight pulling Arthur, to see why you're wrong. Alternatively
> >> > there's another explanation below.
> >>
> >> But *you* got that wrong.
> >
> > <panto> Oh no I didn't! </panto>
>
> OK then, where did *I* go wrong? Or has my original message not passed
> through your ISP?
I've just read your message about the weights dropping - you're just
considering the end effects, which are essentially irrelevant.
Carol, Arthur's wife, has given him a shove at the beginning, and Bruce
catches him at the end (after all, losing one's prize employee over a 25
mile high cliff could be a mite embarrassing). So now we're only interested
in his state while moving.
We have carefully chosen the weight to provide the same amount of force as a
25mph wind provides. (you can work this out while stationary with a spring
balance, a big fan and an anaemometer). This force will be the same whether
it's 15mph into a 10mph headwind, 25mph into still wind or stationary in a
25mph wind.
Obviously the steady state speed for these three cases is 15mph, 25mph and
0mph.
Equally obviously, to move Arthur a distance X at the steady state speed
will take the same amount of energy, as the weight will have fallen the same
distance. With a headwind, it will take a longer time, so the power required
must be lower.
The thing is, the energy lost to friction depends on ground distance and the
windspeed - it has nothing to do with how long it takes. More headwind
possibly = more friction (well, Arthur is going slower), but not more energy
lost to it.
> >> > Arthur is a bit picky - he has his preferred cadence and hates
> >> > deviating from it. Fortunately he has a 100" gear he uses at 25mph
and
> >> > a 60" gear he uses at 15mph.
> >> >
> >> > Now we're looking for the same force against the wind in both cases.
> >> > Force against wind is proportional to pedal force divided by gear
> >> > inches (*). So at 15mph with a 10mph headwind he's using 60% of the
> >> > pedal force that he would be using at 25mph in still air. Which
> >> > conveniently cancels out with the extra time spent pedalling to
leave
> >> > the same work done.
> >>
> >> You have just said "same force against the wind in both cases".
> >> Excellent. Energy is force times time. Time increases in a head wind.
> >
> > But you're thinking of the energy required to shift the wind. As the
> > standing still case illustrates, Arthur doesn't need to provide that.
>
> But we stipulated that Arther was trying to do a journey of 25 miles.
> If Arthur is putting an exact force on the pedal to overcome the wind
> without moving his journey is time is infinite. Energy requred = force
> times infinity. Power is zero. Sorry.
Nothing to apologise about. I know the power Arthur puts in when he's
standing still is zero. (did you seem my little graph?). However I also know
the power Arthur puts in when he's riding at 12.5mph into a 12.5mph headwind
is half that he puts in when he's riding at 25mph in still air - it takes
him twice as long with half the power, resulting in the same energy.
It can be a bit counterintuitive that Arthur puts in half the power when he
is in the headwind - this is the core of the problem. If Arthur was to pedal
as hard as he does normally he would actually go faster, and the net wind he
experiences would be more than 25mph.
cheers,
clive
Jack Ouzzi
<xlu...@users.sourceforge.remove.this.antispam.net> wrote:
>On Fri, 01 Oct 2004 16:15:00 +0100, Jack Ouzzi ... wrote (more or
>less):
>
>...
>>Crikey ............ is there any formula for distance travelled at
>>18mph (with a very slight 23 degree sidewind) giving the amount of
>>fruit pies I can eat to replace the calories consumed during said
>>distance covered ??
>>(Thats 45 gram Bramley Apple pies)
>
>Yes - and you can probably find it at www.analyticcycling.com
>
>But at a rough estimate, I'd use
>~150kcal per 45g bramley apple pie (based on UK Safeway's own brand)
>
>And ~58kcal/mile at 18mph+slight sidewind for a 90kg man,
>down to ~45kcal at 18mph+slight sidewind for a 70kg man.
>
>So for me, I'd need to cycle nearly 3 miles / Bramley apple pie. YMMV
>(quite literally).
Wow, I did 40 miles this morning ............ now looking forward to
eating 13 pies ( maybe a mixed berry fruit pie in between, so's not to
get bored)
Yummy :-)
DavidR
> "DavidR" <david.nospam.r_a@t_cwcom.net> wrote in message
> I've just read your message about the weights dropping - you're just
> considering the end effects, which are essentially irrelevant.
> Carol, Arthur's wife, has given him a shove at the beginning, and Bruce
> catches him at the end
But Bruce only absorbs the energy that Alice put in.
OK, the problem comes down the quantity of energy needed to cover a middle
fraction of that 25 miles.
Brings to mind a similar sort issue about being in a rocket travelling
through space at Newtonian speed. The rocket engine is producing thrust.
How much power?
Power = force x speed. What is its speed measured against?
Russell
Ah Clive, you've just hit the nail on the head but explained dit
rather badly (hope you don't mind me saying so). Either that or you
are still wrong:-)
As you say, the reason I was getting to confused is we are all
worrying about the distance travelled by Arthur and his bike. What we
should be thinking about is the distance his feet & pedals have
travelled. He is not using 60% of the pedal force in a lower gear
riding agaisnt the wind as you say, he is using the same force and the
same cadence thus producing the same energy. The issue is his FEET
have travelled much further by the end of his journey, not him. So the
work done (force * pedal distance) is much greater as is the time. Due
to the lower gearing, the force on the road wheels is greater and the
distance travelled is less so the work done is still the same. This
also explains how he can ride against a 30mph wind whilste still
producing the same amount of energy, because the FORCE transferred to
the wheels in a much lower gear is much greater, so he can overcome
the extra wind resistance although the distance travelled by the
wheels is much smaller.
The issue of work done whilste sitting stationary in the wind with his
brakes on is a red herring as his infinite friction with the road
overcomes the wind without him having to do any work.
Phew!
Russell
Clive George
news:1656ad32.04100...@posting.google.com...
> > Back to the 25mph and 15 + 10mph cases:
> >
> > Arthur is a bit picky - he has his preferred cadence and hates deviating
> > from it. Fortunately he has a 100" gear he uses at 25mph and a 60" gear
he
> > uses at 15mph.
> >
> > Now we're looking for the same force against the wind in both cases.
Force
> > against wind is proportional to pedal force divided by gear inches (*).
So
> > at 15mph with a 10mph headwind he's using 60% of the pedal force that he
> > would be using at 25mph in still air. Which conveniently cancels out
with
> > the extra time spent pedalling to leave the same work done.
>
> Ah Clive, you've just hit the nail on the head but explained dit
> rather badly (hope you don't mind me saying so). Either that or you
> are still wrong:-)
I'm quite willing to accept I haven't explained it sufficiently well - the
fast that you still haven't understood my explanation demonstrates this.
> As you say, the reason I was getting to confused is we are all
> worrying about the distance travelled by Arthur and his bike. What we
> should be thinking about is the distance his feet & pedals have
> travelled.
Be very careful here - you're in danger of being confused.
> He is not using 60% of the pedal force in a lower gear
> riding agaisnt the wind as you say, he is using the same force and the
> same cadence thus producing the same energy.
Nope. Let's take it step by step, considering the two cases of 25mph in no
wind, and 15mph in a 10mph headwind, with equal cadence being used by Arthur
in both cases.
Tell me which of these steps you disagree with.
1) The force due to the wind is the same in both cases.
2) That force must be equalled by a drive force on the wheel.
3) So wheel force is the same in both cases.
4) Equal cadence, different speeds means he must be in a gear proportional
to the speed - eg 100" and 60" for 25mph and 15mph.
5) We know that wheel force = pedal force / gear ratio, or pedal force =
wheel force * gear ratio
(eg higher gear for a given pedal force gives us less acceleration)
6) The two wheel forces are the same, so we have
pedal force 1 / gear ratio 1 = pedal force 2 / gear ratio 2
Gear ratio 1 is 100", gear ratio 2 is 60", so pedal force 1 is 100/60 that
of pedal force 2.
Ie at 15mph he's pedalling at 60% of the effort that he puts in at 25mph.
Since he isn't using the same pedal force, the rest of your argument falls
down. Sorry!
> The issue of work done whilste sitting stationary in the wind with his
> brakes on is a red herring as his infinite friction with the road
> overcomes the wind without him having to do any work.
No, it's merely the continuation of the line.
> Phew!
I understand what you're trying to say (as I said earlier, I made the same
mistake myself), but I'm afraid you're wrong. Can you work through my 6
steps above and see if they make it any clearer?
If they don't, here's another question:
Consider the first hour of Arthur's ride. On the no wind day, he covers 25
miles. On the windy day he covers 15 miles. Both days have the same wind
force.
Has he done more work covering the 25 miles or the 15 miles, or are they the
same?
(Remember, I say he's done 60% of the work in the 15 miles, as when he gets
to the end of the journey he's done the same amount of work)
cheers,
clive
AndyMorris
>
> You have just said "same force against the wind in both cases".
> Excellent. Energy is force times time. Time increases in a head wind.
>
No, no, no.
Energy is Force times distance.
From Google -> define Joule
The SI unit of energy, work, or quantity of heat. One Joule is the energy
expended when a force of one newton is applied over a displacement of one
meter in the direction of the force.
So, If Arthur is riding against the same relative headwind, he will be
experiencing the same force, the work he does is the force time the distance
and it doesn't matter how long it takes him.
--
Andy Morris
AndyAtJinkasDotFreeserve.Co.UK
Love this:
Put an end to Outlook Express's messy quotes
http://home.in.tum.de/~jain/software/oe-quotefix/
Ambrose Nankivell
Clive George <cl...@xxxx-x.fsnet.co.uk> typed:
>> He is not using 60% of the pedal force in a lower gear
>> riding agaisnt the wind as you say, he is using the same force and
>> the same cadence thus producing the same energy.
>
> Nope. Let's take it step by step, considering the two cases of 25mph
> in no wind, and 15mph in a 10mph headwind, with equal cadence being
> used by Arthur in both cases.
>
> Tell me which of these steps you disagree with.
>
> 1) The force due to the wind is the same in both cases.
> 2) That force must be equalled by a drive force on the wheel.
> 3) So wheel force is the same in both cases.
> 4) Equal cadence, different speeds means he must be in a gear
> proportional to the speed - eg 100" and 60" for 25mph and 15mph.
> 5) We know that wheel force = pedal force / gear ratio, or pedal
> force = wheel force * gear ratio
> (eg higher gear for a given pedal force gives us less acceleration)
> 6) The two wheel forces are the same, so we have
> pedal force 1 / gear ratio 1 = pedal force 2 / gear ratio 2
> Gear ratio 1 is 100", gear ratio 2 is 60", so pedal force 1 is 100/60
> that of pedal force 2.
> Ie at 15mph he's pedalling at 60% of the effort that he puts in at
> 25mph.
You do realise you're wrong here, don't you?
A
Clive George
news:2s90o7F...@uni-berlin.de...
Au contraire, I realise I'm quite right! Since you think otherwise, which of
my steps above do you disagree with?
cheers,
clive
Ambrose Nankivell
Clive George <cl...@xxxx-x.fsnet.co.uk> typed:
> "Ambrose Nankivell" <$firstname+n$@gmail.com> wrote in message
> news:2s90o7F...@uni-berlin.de...
>> In news:415f46f9$0$17960$ed26...@ptn-nntp-reader02.plus.net,
>> Clive George <cl...@xxxx-x.fsnet.co.uk> typed:
>>>> He is not using 60% of the pedal force in a lower gear
>>>> riding agaisnt the wind as you say, he is using the same force and
>>>> the same cadence thus producing the same energy.
>>>
>>> Nope. Let's take it step by step, considering the two cases of 25mph
>>> in no wind, and 15mph in a 10mph headwind, with equal cadence being
>>> used by Arthur in both cases.
>>>
>>> Tell me which of these steps you disagree with.
>>>
>>> 1) The force due to the wind is the same in both cases.
>>> 2) That force must be equalled by a drive force on the wheel.
>>> 3) So wheel force is the same in both cases.
I disagree with this one. Wheel force is the same but wheel speed is
different.
>>> 4) Equal cadence, different speeds means he must be in a gear
>>> proportional to the speed - eg 100" and 60" for 25mph and 15mph.
>>> 5) We know that wheel force = pedal force / gear ratio, or pedal
>>> force = wheel force * gear ratio
>>> (eg higher gear for a given pedal force gives us less acceleration)
>>> 6) The two wheel forces are the same, so we have
>>> pedal force 1 / gear ratio 1 = pedal force 2 / gear ratio 2
>>> Gear ratio 1 is 100", gear ratio 2 is 60", so pedal force 1 is
>>> 100/60 that of pedal force 2.
>>> Ie at 15mph he's pedalling at 60% of the effort that he puts in at
>>> 25mph.
>>
>> You do realise you're wrong here, don't you?
>
> Au contraire, I realise I'm quite right! Since you think otherwise,
> which of my steps above do you disagree with?
The question is, why is he cycling at only 60% of the effort that he could
be?
A
Martin Wilson
>eating 13 pies ( maybe a mixed berry fruit pie in between, so's not to
>get bored)
>
>Yummy :-)
Just feel sorry for me I've had no pies in about 4 months :-(
I did have half a cornish pasty and some apple crumble but pie wise
I'm down to 0 from probably a 5-6 average a month!
Jon Senior
> The question is, why is he cycling at only 60% of the effort that he could
> be?
Because he's not trying hard enough!
Jon
Ian Smith
>
> As you say, the reason I was getting to confused is we are all
> worrying about the distance travelled by Arthur and his bike. What we
> should be thinking about is the distance his feet & pedals have
> travelled. He is not using 60% of the pedal force in a lower gear
> riding agaisnt the wind as you say, he is using the same force and the
> same cadence thus producing the same energy.
So if he's using teh same force and teh same cadence, but a lower gear
(as he must be to get there slower), he's exerting more force at the
road, yes?
So he's exerting more force, against the same resistance? (Must be
the same resistance - he's got the same 25mph wind speed)
So why does he not accelerate?
Clive George
> In news:415f4e4e$0$17922$ed26...@ptn-nntp-reader02.plus.net,
> Clive George <cl...@xxxx-x.fsnet.co.uk> typed:
> > "Ambrose Nankivell" <$firstname+n$@gmail.com> wrote in message
> > news:2s90o7F...@uni-berlin.de...
> >> In news:415f46f9$0$17960$ed26...@ptn-nntp-reader02.plus.net,
> >> Clive George <cl...@xxxx-x.fsnet.co.uk> typed:
> >>>> He is not using 60% of the pedal force in a lower gear
> >>>> riding agaisnt the wind as you say, he is using the same force and
> >>>> the same cadence thus producing the same energy.
> >>>
> >>> Nope. Let's take it step by step, considering the two cases of 25mph
> >>> in no wind, and 15mph in a 10mph headwind, with equal cadence being
> >>> used by Arthur in both cases.
> >>>
> >>> Tell me which of these steps you disagree with.
> >>>
> >>> 1) The force due to the wind is the same in both cases.
> >>> 2) That force must be equalled by a drive force on the wheel.
> >>> 3) So wheel force is the same in both cases.
>
> I disagree with this one. Wheel force is the same but wheel speed is
> different.
What are you disagreeing with? I think we agree that wheel force is the same
and we agree that wheel speed is different, don't we?
> >>> 4) Equal cadence, different speeds means he must be in a gear
> >>> proportional to the speed - eg 100" and 60" for 25mph and 15mph.
> >>> 5) We know that wheel force = pedal force / gear ratio, or pedal
> >>> force = wheel force * gear ratio
> >>> (eg higher gear for a given pedal force gives us less acceleration)
> >>> 6) The two wheel forces are the same, so we have
> >>> pedal force 1 / gear ratio 1 = pedal force 2 / gear ratio 2
> >>> Gear ratio 1 is 100", gear ratio 2 is 60", so pedal force 1 is
> >>> 100/60 that of pedal force 2.
> >>> Ie at 15mph he's pedalling at 60% of the effort that he puts in at
> >>> 25mph.
> >>
> >> You do realise you're wrong here, don't you?
> >
> > Au contraire, I realise I'm quite right! Since you think otherwise,
> > which of my steps above do you disagree with?
>
> The question is, why is he cycling at only 60% of the effort that he could
> be?
As I said originally,
"He decides to ride at 15mph in order to make sure his hair doesn't look
abnormal when he
arrives, and off he goes."
James has answered my other question elsewhere in this thread, which was
'How fast will Arthur go if he does decide to pedal as hard'.
You could then ask, 'Why did I pose the question in such a way?'. The answer
is because it's slightly counterintuitive, and hence an interesting (well,
to me :-) ) problem.
cheers,
clive
Gawnsoft
<cl...@xxxx-x.fsnet.co.uk> wrote (more or less):
>"David Martin" <d.m.a....@dundee.ac.uk> wrote in message
>news:BD8343EB.2227C%d.m.a....@dundee.ac.uk...
>> On 1/10/04 4:57 pm, in article
>> 415d7e09$0$69730$ed26...@ptn-nntp-reader01.plus.net, "Clive George"
>> <cl...@xxxx-x.fsnet.co.uk> wrote:
>>
>> > "Peter Clinch" <p.j.c...@dundee.ac.uk> wrote in message
>> > news:2s5atvF...@uni-berlin.de...
>> >> Clive George wrote:
>> >>
>> >>> I'm just waiting to see if anybody else is going to offer an opinion
>> >>
>> >> Oh, let's make a fool of myself then...
>> >>
>> >> ISTM we're looking at total energy expended for the journey. He's
>> >> working at the same basic output in joules/second in each case because
>> >> he'll need the same amount of energy to make headway in what is
>> >> effectively the same headwind, so he's burning energy at the same rate,
>> >> but he's burning it for longer on the slower journey.
>> >>
>> >> If that's cobblers I blame the fact that it's Friday afternoon and
>> >> having long since moved into computing my physics is horrible these
>> > days...
>> >
>> > If it's any consolation I thought the same at first, and it was only
>when I
>> > worked it all out from very simple principles for myself that I realised
>the
>> > error of my ways.
>>
>> Another way to look at it:
>>
>> The work done is force x distance moved.
>>
>> At a constant airspeed, the work done is that to move the air molecule
>> sideways enough to get round the rider.
>>
>> At a constant airspeed (and air density) the number of air particles
>> requiring to be moved is constant.
>>
>> so F is constant but d is proportional to the number of air molecules.
>>
>> The number of air molecules is proportional to the time taken.
>>
>> So from this the distance is proportional to the time taken (as the only
>> effective force is the force required for movement of the air, Arthur is
>> otherwise in steady state).
>>
>> So the work done on moving Arthur is zero, but the work done on moving air
>> our of Arthur's way is proportional to the time taken for a given air
>speed
>> and density.
>>
>> It does take more work to go through moving air than still air.
>
>What happens if Arthur hits a 25mph headwind, and just decides to sit there
>with the brakes on?
The rolling resistance is no longer zero, is what happens.
>All the work moving the air out of Arthur's way is still
>done, yet he's obviously doing no work.
>
>If Arthur's bike was propellor driven rather than wheel driven (ie against
>the air rather than the ground) then you'd be right. But it isn't.
Except that you started this all off by saying the rolling resistance
was zero.
--
Cheers,
Euan
Gawnsoft: http://www.gawnsoft.co.sr
Symbian/Epoc wiki: http://html.dnsalias.net:1122
Smalltalk links (harvested from comp.lang.smalltalk) http://html.dnsalias.net/gawnsoft/smalltalk
Clive George
message news:vfqvl0phnudub04hh...@4ax.com...
> >> It does take more work to go through moving air than still air.
> >
> >What happens if Arthur hits a 25mph headwind, and just decides to sit
there
> >with the brakes on?
>
> The rolling resistance is no longer zero, is what happens.
There's no rolling, so how can there be rolling resistance? If you prefer,
think of it as Arthur bracing against the pedals rather than using the
brakes.
> >All the work moving the air out of Arthur's way is still
> >done, yet he's obviously doing no work.
> >
> >If Arthur's bike was propellor driven rather than wheel driven (ie
against
> >the air rather than the ground) then you'd be right. But it isn't.
>
> Except that you started this all off by saying the rolling resistance
> was zero.
See above.
cheers,
clive
Gawnsoft
<cl...@xxxx-x.fsnet.co.uk> wrote (more or less):
>"Gawnsoft" <xlu...@users.sourceforge.remove.this.antispam.net> wrote in
>message news:vfqvl0phnudub04hh...@4ax.com...
>> >> It does take more work to go through moving air than still air.
>> >
>> >What happens if Arthur hits a 25mph headwind, and just decides to sit
>there
>> >with the brakes on?
>>
>> The rolling resistance is no longer zero, is what happens.
>
>There's no rolling, so how can there be rolling resistance?
If there's no rolling, that's because the resistance to rolling (i.e.
rolling restitance) is so high
>If you prefer,
>think of it as Arthur bracing against the pedals rather than using the
>brakes.
He's still providing a rolling resistance. (i.e. a force)
>
>> >All the work moving the air out of Arthur's way is still
>> >done, yet he's obviously doing no work.
>> >
>> >If Arthur's bike was propellor driven rather than wheel driven (ie
>against
>> >the air rather than the ground) then you'd be right. But it isn't.
>>
>> Except that you started this all off by saying the rolling resistance
>> was zero.
>
>See above.
>
>cheers,
>clive
>
--
Gawnsoft
wrote (more or less):
>in message <050rl01ldi711htu7...@4ax.com>, Gawnsoft
>('xlu...@users.sourceforge.remove.this.antispam.net') wrote:
>
>> On Fri, 01 Oct 2004 16:15:00 +0100, Jack Ouzzi ... wrote (more or
>> less):
>>
>> ...
>>>Crikey ............ is there any formula for distance travelled at
>>>18mph (with a very slight 23 degree sidewind) giving the amount of
>>>fruit pies I can eat to replace the calories consumed during said
>>>distance covered ??
>>>(Thats 45 gram Bramley Apple pies)
>>
>> Yes - and you can probably find it at www.analyticcycling.com
>>
>> But at a rough estimate, I'd use
>> ~150kcal per 45g bramley apple pie (based on UK Safeway's own brand)
>>
>> And ~58kcal/mile at 18mph+slight sidewind for a 90kg man,
>> down to ~45kcal at 18mph+slight sidewind for a 70kg man.
>>
>> So for me, I'd need to cycle nearly 3 miles / Bramley apple pie. YMMV
>> (quite literally).
>
>That's it! We have a new measure of fuel efficiency: MAP - Miles per
>Apple Pie.
I'm already using the km/Jaffa cake. :-) With my current level of
obesity, when walking it's almost exactly 1 km /Jaffa Cake :-(
I look forward to the day when I've lost enough weight that there's
only 3/4 km/ Jaffa Cake !
Jack Ouzzi
I wish to complain Gawnsoft ............. your bramley pie calculation
is incorrect, and I have subsequently put on 3kg in weight over the
weekend following your calculations. A letter from my lawyer is in the
post !!
Burp ..............
Clive George
message news:2o00m0hk7imonp7ef...@4ax.com...
> On Sun, 3 Oct 2004 13:05:43 +0100, "Clive George"
> <cl...@xxxx-x.fsnet.co.uk> wrote (more or less):
>
> >"Gawnsoft" <xlu...@users.sourceforge.remove.this.antispam.net> wrote in
> >message news:vfqvl0phnudub04hh...@4ax.com...
> >> >> It does take more work to go through moving air than still air.
> >> >
> >> >What happens if Arthur hits a 25mph headwind, and just decides to sit
> >there
> >> >with the brakes on?
> >>
> >> The rolling resistance is no longer zero, is what happens.
> >
> >There's no rolling, so how can there be rolling resistance?
>
> If there's no rolling, that's because the resistance to rolling (i.e.
> rolling restitance) is so high
>
> >If you prefer,
> >think of it as Arthur bracing against the pedals rather than using the
> >brakes.
>
> He's still providing a rolling resistance. (i.e. a force)
What's the difference between him providing a force holding him still and
him providing a force propelling him forwards?
Getting back to the original question, do you think Arthur is using more
power at 25mph in no wind or 15mph in a 10mph headwind? Or rather, how do
these two power outputs compare?
cheers,
clive
David Martin
415f46f9$0$17960$ed26...@ptn-nntp-reader02.plus.net, "Clive George"
<cl...@xxxx-x.fsnet.co.uk> wrote:
> Consider the first hour of Arthur's ride. On the no wind day, he covers 25
> miles. On the windy day he covers 15 miles. Both days have the same wind
> force.
> Has he done more work covering the 25 miles or the 15 miles, or are they the
> same?
<guy>
They are the same. He has done no work. He will continue to do no work until
he gets off the bike and starts doing the work he is travelling to.
</guy>
And I still think the best answer for the original question was '40 minutes
less'.
..d
Gawnsoft
<cl...@xxxx-x.fsnet.co.uk> wrote (more or less):
>"Gawnsoft" <xlu...@users.sourceforge.remove.this.antispam.net> wrote in
>message news:2o00m0hk7imonp7ef...@4ax.com...
>> On Sun, 3 Oct 2004 13:05:43 +0100, "Clive George"
>> <cl...@xxxx-x.fsnet.co.uk> wrote (more or less):
>>
>> >"Gawnsoft" <xlu...@users.sourceforge.remove.this.antispam.net> wrote in
>> >message news:vfqvl0phnudub04hh...@4ax.com...
>> >> >> It does take more work to go through moving air than still air.
>> >> >
>> >> >What happens if Arthur hits a 25mph headwind, and just decides to sit
>> >there
>> >> >with the brakes on?
>> >>
>> >> The rolling resistance is no longer zero, is what happens.
>> >
>> >There's no rolling, so how can there be rolling resistance?
>>
>> If there's no rolling, that's because the resistance to rolling (i.e.
>> rolling restitance) is so high
>>
>> >If you prefer,
>> >think of it as Arthur bracing against the pedals rather than using the
>> >brakes.
>>
>> He's still providing a rolling resistance. (i.e. a force)
>
>What's the difference between him providing a force holding him still and
>him providing a force propelling him forwards?
The distance which he exerts the force over - i.e. the work done /
effort involved / energy expended.
>Getting back to the original question, do you think Arthur is using more
>power at 25mph in no wind or 15mph in a 10mph headwind? Or rather, how do
>these two power outputs compare?
The two power outputs are the same (assuming as you did that there is
no rolling resistance involved, and the sole retarding force is due to
air resistance).
Just zis Guy, you know?
<d.m.a....@dundee.ac.uk> wrote in message
<BD85F270.225DC%d.m.a....@dundee.ac.uk>:
><guy>
>They are the same. He has done no work. He will continue to do no work until
>he gets off the bike and starts doing the work he is travelling to.
></guy>
That would be <james></james> acksherly ;-)
Guy
--
May contain traces of irony. Contents liable to settle after posting.
http://www.chapmancentral.co.uk
88% of helmet statistics are made up, 65% of them at Washington University
Clive George
message news:c7g0m0h0nkr7lskim...@4ax.com...
> >Getting back to the original question, do you think Arthur is using more
> >power at 25mph in no wind or 15mph in a 10mph headwind? Or rather, how do
> >these two power outputs compare?
>
> The two power outputs are the same (assuming as you did that there is
> no rolling resistance involved, and the sole retarding force is due to
> air resistance).
Oh dear! So, you don't believe my example with the weights (same force,
different time -> different power), nor do you believe my explanation with
gears.
So, riding along at 15mph into a 10mph headwind takes the same power as
riding at 25mph with no wind does it? Obviously I think you really should
believe my earlier reasons - there is nothing wrong with them, after all,
and you haven't managed to point out anything wrong with them either, but
how about considering this:
Who is working harder (ie what are the relative power outputs), somebody
riding at 25mph into still air, or somebody who has a 10mph conveyor belt
underneath him helping him to a speed of 25mph into still air?
When you've answered that, explain how the guy on the conveyor belt can tell
the difference between being on the conveyor belt and riding at 15mph into a
10mph headwind purely based on the forces on his bike. (ie his eyes are shut
and there's no strange humming sounds!)
cheers,
clive
DavidR
"AndyMorris" <AndyM...@DeadSpam.com> wrote
> DavidR wrote:
>>
>> You have just said "same force against the wind in both cases".
>> Excellent. Energy is force times time. Time increases in a head wind.
>>
>
> No, no, no.
>
> Energy is Force times distance.
Hangs head in shame. Thankfully I managed to post a correction before
anybody noticed.
bugbear
> "bugbear" <bugbear@trim_papermule_trim.co.uk> wrote in message
> news:415d8056$0$82273$ed26...@ptn-nntp-reader03.plus.net...
>
>>Clive George wrote:
>>
>>>I remembered a thread from the tandem mailing list a couple of years
>
> back
>
>>>which caused a lot of confusion, so I shall see what people here (and on
>>>cyclingforums) make of it:
>>>
>>>My mate Arthur Pedaller has a really great bike. It has absolutely no
>>>rolling resistance whatsoever - the sole retarding force is air
>
> resistance.
>
>>>On a windless day he can happily pedal at 25mph, and he covers the 25
>
> miles
>
>>>to work on the marvellous straight road the council installed for him at
>>>that speed.
>>>But today's a bit windy - 10mph headwind to be precise. He decides to
>
> ride
>
>>>at 15mph in order to make sure his hair doesn't look abnormal when he
>
> the
>
>>>distance, but the question is, has he put in more or less work than on a
>>>windless day?
>>
>>Since the only work (as stated) is done overcoming air resistance,
>>his work rate (power expenditure) is the same in both cases.
>>
>>Total Energy = power * time
>
>
> Simple, succinct, and unfortunately wrong!
Well, I'm NOT open to negotiation (and neither is my electricity
provider ;-) on the equation, so where
(in you opinion) is the error?
BugBear
Clive George
Obviously I'm not going to disagree with the energy = power * time bit.
Which leaves only one other place for the error to be.
His work rate (power expenditure) _isn't_ the same in both cases. At 15mph
he's putting in 60% of the power. See my other posts for more detail.
cheers,
clive
David Martin
41610af2$0$80306$ed26...@ptn-nntp-reader01.plus.net, "bugbear"
<bugbear@trim_papermule_trim.co.uk> wrote:
>>>
>>> Total Energy = power * time
>>
>>
>> Simple, succinct, and unfortunately wrong!
>
> Well, I'm NOT open to negotiation (and neither is my electricity
> provider ;-) on the equation, so where
> (in you opinion) is the error?
There isn't one.
Work is energy/time.
Energy is force times distance moved.
Speed is distance over time
SO if you use 60% of the power for 100/60 of the time. you use the same
energy.
..d
Michael MacClancy
> On 4/10/04 9:33 am, in article
> 41610af2$0$80306$ed26...@ptn-nntp-reader01.plus.net, "bugbear"
> <bugbear@trim_papermule_trim.co.uk> wrote:
>
>>>>
>>>> Total Energy = power * time
>>>
>>>
>>> Simple, succinct, and unfortunately wrong!
>>
>> Well, I'm NOT open to negotiation (and neither is my electricity
>> provider ;-) on the equation, so where
>> (in you opinion) is the error?
>
> There isn't one.
Someone hasn't had his coffee this morning.
> Work is energy/time.
No, it's not. Energy/time is Power (W).
> Energy is force times distance moved.
This is Work (J) and is energy expended.
>
> Speed is distance over time
This is correct.
>
> SO if you use 60% of the power for 100/60 of the time. you use the same
> energy.
>
This is correct and suggests that your definition of Work was an
aberration.
--
Michael MacClancy
Random putdown -
www.macclancy.demon.co.uk
www.macclancy.co.uk
David Martin
"Michael MacClancy" <her...@nospamo2.co.uk> wrote:
> On Mon, 04 Oct 2004 10:02:41 +0100, David Martin wrote:
>
>> On 4/10/04 9:33 am, in article
>> 41610af2$0$80306$ed26...@ptn-nntp-reader01.plus.net, "bugbear"
>> <bugbear@trim_papermule_trim.co.uk> wrote:
>>
>>>>>
>>>>> Total Energy = power * time
>>>>
>>>>
>>>> Simple, succinct, and unfortunately wrong!
>>>
>>> Well, I'm NOT open to negotiation (and neither is my electricity
>>> provider ;-) on the equation, so where
>>> (in you opinion) is the error?
>>
>> There isn't one.
>
> Someone hasn't had his coffee this morning.
Correct. My brain knew what it was talking about, even if my fingers
didn't..
>> Work is energy/time.
>
> No, it's not. Energy/time is Power (W).
>
>> Energy is force times distance moved.
>
> This is Work (J) and is energy expended.
>>
>> Speed is distance over time
>
> This is correct.
>
>>
>> SO if you use 60% of the power for 100/60 of the time. you use the same
>> energy.
>>
>
> This is correct and suggests that your definition of Work was an
> aberration.
Where's the coffee.....?
..d