Hills?
Thomas Kowalski
I live in south Jersey with relatively few hills to train on.
Unfortunately some of the races I did last year took me to places where
they had some. How can I effectively improve my hill climbing ability. I
have a stair stepper at home, is this a help or a hindrance?
Thanks in advance
Tom James
On 3 Apr 1997, Aarron Canino wrote:
> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> > sounds very dangerous)
> >
>
> Inclining rollers does absolutely nothing because no part of the bike or
> your body gains any altitude. You are only suggesting something very
> dangerous.....
well, I've often wondered about this too, but if it is good enough for a
certain Christopher Miles Boardman....... I think one think it might do
is get you used to riding with your body angled at 10% or so to what it
is used to, even if it is simply a case of retraining your neck muscles :-)
Tom
Al...@earth.com
Stair stepper might provide some increase in power but it is not
cycling specific. Unless, of course, you find some way to *spin* with
it and use all of the spinning muscles.
I suggest: (in order of importance)
1. Move to a hilly place
2. Buy a Computrainer to simulate the *long* climbs
3. Drive your bike to upstate NJ/NY or PA and train there several
times a week.
4. Buy a set of rollers, and incline them at 5%-10% grade (this
sounds very dangerous)
5. Race only on flat terrain.
Unfortunately, there is not much of a substitute to be good on long
climbs, except to do a lot of long climbs.
AlienOnWheels
Aarron Canino
> 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> sounds very dangerous)
>
Inclining rollers does absolutely nothing because no part of the bike or
JBMosher88
A previous note to this posting suggested putting rollers on an incline.
Since the resistance on rollers is purely due to roller resistance and any
brake (magnetic, hydraulic, or otherwise) connected to the rollers,
placing the rollers on an incline will not increase the power input
required for a given speed at all.
Physics refresher--work is force over distance, power is work over time.
Hill climbing work (at lower speeds) is primarily due to the effort
required to move mass a given elevation (vs. work on flats, which at
higher speeds, is dominated by the force of drag). Raising rollers on an
incline doesn't result in any requirement to increase the work input by
inducing a requirement to lift mass: no additional work (or power) is
required.
Roger Thomas
On 2 Apr 1997 19:23:40 -0800, Tom James <tomj...@chem1.usc.edu>
wrote:
>On 3 Apr 1997, Aarron Canino wrote:
>
>> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
>> > sounds very dangerous)
>> >=20
>>=20
>> Inclining rollers does absolutely nothing because no part of the bike =
or
>> your body gains any altitude. You are only suggesting something very
>> dangerous.....
>
>well, I've often wondered about this too, but if it is good enough for a=
=20
>certain Christopher Miles Boardman....... I think one think it might do=20
>is get you used to riding with your body angled at 10% or so to what it=20
>is used to, even if it is simply a case of retraining your neck muscles =
:-)
>
> Tom
Boardman does genuine climbing on his because it's not rollers but a
bike ridden on a running machine inclined at a gradient. Somehow he's
worked out how to balance doing this. So there's a rolling road to
climb.
John Forrest Tomlinson
In <Pine.A32.3.91.970402...@chem1.usc.edu> Tom James
<tomj...@chem1.usc.edu> writes:
>
>On 3 Apr 1997, Aarron Canino wrote:
>
>> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
>> > sounds very dangerous)
>> >
>>
>> your body gains any altitude. You are only suggesting something
very
>> dangerous.....
>
>well, I've often wondered about this too, but if it is good enough for
a
do
it
>
Boardman doesn't ride rollers like that, he rides a big treadmill.
It's quite a bit different.
JT
Jim McMillan
Put it in a 53*11 on the flats to simulate the power and rpm's.
Steven L. Sheffield
In article <Pine.A32.3.91.970402...@chem1.usc.edu>, Tom
James <tomj...@chem1.usc.edu> wrote:
> On 3 Apr 1997, Aarron Canino wrote:
>
> > > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> > > sounds very dangerous)
> > >
> >
> > Inclining rollers does absolutely nothing because no part of the bike or
> > your body gains any altitude. You are only suggesting something very
> > dangerous.....
>
> well, I've often wondered about this too, but if it is good enough for a
> certain Christopher Miles Boardman....... I think one think it might do
> is get you used to riding with your body angled at 10% or so to what it
> is used to, even if it is simply a case of retraining your neck muscles :-)
If I recall correctly, Boardman was using an inclining treadmill which
offers resistance, instead of just using rollers (which typically do
not offer resistance).
--
+-------------------------------------------------------------------+
| Steven L. Sheffield (BOB #1765/IBOB #3) Disclaimer? What's that? |
+-------------------------------------------------------------------+
| E-mail: ste...@veloworks.com / ste...@winterland.com |
| WWW: http://www.veloworks.com/rivendell/ |
| Voice: +1 415 296-9893 / Fax: +1 415 597-9849 / Ride yer bike! |
+-------------------------------------------------------------------+
Jim McMillan
>
>> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
>> > sounds very dangerous)
>> >
>>
>> Inclining rollers does absolutely nothing because no part of the
bike or
>> your body gains any altitude. You are only suggesting something
very
>> dangerous.....
>
>well, I've often wondered about this too, but if it is good enough for
a
>certain Christopher Miles Boardman....... I think one think it might
do
>is get you used to riding with your body angled at 10% or so to what
it
>is used to, even if it is simply a case of retraining your neck
muscles :-)
>
Boardman does it with a treadmill.
Jim Martin
Jim McMillan wrote:
>
> Put it in a 53*11 on the flats to simulate the power and rpm's.
Exactly! Back in the 70s and 80s Stan Blanton rode for Turin Sedis and
lived in dead flat Lubbock Texas. He was an awesome climber. When asked
how he got climbing practice he just said 'Big gears and head winds'.
Make the most of what ya got!
Jim Martin
Sundar Dorai-Raj
Steven L. Sheffield wrote:
>
> In article <Pine.A32.3.91.970402...@chem1.usc.edu>, Tom
> James <tomj...@chem1.usc.edu> wrote:
>
> >
> > > > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> > > > sounds very dangerous)
> > > >
> > >
> > > Inclining rollers does absolutely nothing because no part of the bike or
> > > your body gains any altitude. You are only suggesting something very
> > > dangerous.....
> >
> > well, I've often wondered about this too, but if it is good enough for a
> > certain Christopher Miles Boardman....... I think one think it might do
> > is get you used to riding with your body angled at 10% or so to what it
> > is used to, even if it is simply a case of retraining your neck muscles :-)
>
> offers resistance, instead of just using rollers (which typically do
> not offer resistance).
>
> --
> +-------------------------------------------------------------------+
> | Steven L. Sheffield (BOB #1765/IBOB #3) Disclaimer? What's that? |
> +-------------------------------------------------------------------+
> | E-mail: ste...@veloworks.com / ste...@winterland.com |
> | WWW: http://www.veloworks.com/rivendell/ |
> | Voice: +1 415 296-9893 / Fax: +1 415 597-9849 / Ride yer bike! |
> +-------------------------------------------------------------------+
increases power.
Gene&Pat Cottrell
Steven L. Sheffield wrote:
>
> In article <Pine.A32.3.91.970402...@chem1.usc.edu>, Tom
> James <tomj...@chem1.usc.edu> wrote:
>
> > On 3 Apr 1997, Aarron Canino wrote:
> >
> > > > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> > > > sounds very dangerous)
> > > >
> > >
> > > Inclining rollers does absolutely nothing because no part of the bike or
> > > your body gains any altitude. You are only suggesting something very
> > > dangerous.....
> >
> > well, I've often wondered about this too, but if it is good enough for a
> > certain Christopher Miles Boardman....... I think one think it might do
Does anybody know if boardman uses a treadmill???? ;-)
Gene
Andrew R. Coggan
Thomas Kowalski wrote:
>
> I live in south Jersey with relatively few hills to train on.
> Unfortunately some of the races I did last year took me to places where
> they had some. How can I effectively improve my hill climbing ability. I
> have a stair stepper at home, is this a help or a hindrance?
> Thanks in advance
A workout I've seen recommended is to ride into the wind at a good pace,
in the big ring and out of the saddle, for 10-15 min at a time. This is
followed by small-ring, downwind jaunts so that you don't forget how to
spin.
Whether this does any good or not, I'm not sure. I don't think it is a
very good simulation of actually climbing shorter, steeper hills; the
power/force requirements simply aren't high enough. It may be more
helpful in simulating extended climbs, but I think that this is a less
specialized task and has more to do with simply your power/weight ratio.
If it is any consolation to you, I live an a barrier island with a
maximum elevation of about 12 feet above sea level. The nearest real
hills are in central Texas, about a 4 hour drive away...
Andrew R. Coggan
Tom James wrote:
> On 3 Apr 1997, Aarron Canino wrote:
> > > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> > > sounds very dangerous)
> > Inclining rollers does absolutely nothing because no part of the bike or
> > your body gains any altitude. You are only suggesting something very
> > dangerous.....
> well, I've often wondered about this too, but if it is good enough for a
> certain Christopher Miles Boardman.......
The difference is that Boardman rode his bike on a *treadmill*, not
rollers. Increasing the grade of a treadmill adds additional resistance
and simulates climbing, whereas elevating the front of rollers would do
nothing (except maybe give you a numb crotch).
FWIW, I can tell you from personal experience that riding a bike on a
*level* treadill is quite difficult. You have little or no inertia, and
(on most treadmills) very little room to move either front-to-back or
side-to-side. Consequently, you have to be really smooth, even smoother
than on rollers (and I've ridden rollers for up to 8 hours straight).
So, although I've done it for 15 or so minutes at a time, I can't
imagine using this as a routine training method, as the level of
concentration required is very high. Perhaps, like rollers, it
eventually becomes second nature, or perhaps it is much easier on an
incline. Either way, I'm impressed by Boardman's reported use of this
technique to improve his climbing...
Tom James
On Fri, 4 Apr 1997, David Albelo wrote:
> Andrew R. Coggan wrote:
> > Snip..
> > Either way, I'm impressed by Boardman's reported use of this
> > technique to improve his climbing...
>
> Yeah, but has it helped? I don't recall Boardman having a
> very good time in the mountains of the Tour last year.
>
> Dave
There are other hilly races than the Tour, though, and in last year's
Tour Boardman was ill. I seem to remember 5th in last year's Dauphine
Libere, a race based mainly in the Alps, and 2nd in 1995, including
finishing 2nd behind Virenque (but ahead of Indurain) on a stage that
took in I think the Galibier, Glandon and summit finish at Les Deux
Alpes. Maybe not a climber in the Virenque/Pantani class, but pretty
useful nonetheless, and rather better than his 1994 form as well.
Tom
Greg Lewis
> >On 3 Apr 1997, Aarron Canino wrote:
> >
> >> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
> >> > sounds very dangerous)
> >> >
> >>
Am I the only one who thought this was an April Fools joke?
Greg
David Albelo
Gene&Pat Cottrell
Andrew R. Coggan wrote:
>
> Thomas Kowalski wrote:
> >
> > I live in south Jersey with relatively few hills to train on.
> > Unfortunately some of the races I did last year took me to places where
> > they had some. How can I effectively improve my hill climbing ability. I
> > have a stair stepper at home, is this a help or a hindrance?
> > Thanks in advance
>
> A workout I've seen recommended is to ride into the wind at a good pace,
> in the big ring and out of the saddle, for 10-15 min at a time. This is
> followed by small-ring, downwind jaunts so that you don't forget how to
> spin.
Although strength is certainly needed to climb well, I think you still
need to climb lots of hills to get good at it. There is much to be said
for practicing on the real thing. You must learn to judge a hill
quickly, select proper gearing, determine pace etc.... Plus, there's
probably some psycological stuff to be overcome with practice on hills,
like gaining the confidence required.
--
,_~o ,__o
_-\_<, _-\_<, Gene & Pat
(*)/'(*) (*)/'(*)
Tim Lucia
Intervals. Perhaps you've heard of them? Do you have access to an
ergometer?
Inclining your rollers is pointless, except possibly to hurt yourself.
I have seen people put small blocks under he front feet to shift their
weight back a bit for better balance, but not anything like a noticeable
grade.
Also, do hill repeats on whatever hills you have.
--
Timothy J. Lucia Digital Ladebug Debugger Developer
Digital Equipment Corporation Cat IV Charter Systems/Wheelworks/NEBC
110 Spitbrook Road ZK2-3Q08 Email: lu...@zko.dec.com
Nashua, NH USA 03062-2698 603.881.2615(Voice)/603.881.0120(FAX)
L. Burton Hawley
Tom James wrote:
>
> On Fri, 4 Apr 1997, David Albelo wrote:
>
> Tour Boardman was ill. I seem to remember 5th in last year's Dauphine
> Libere, a race based mainly in the Alps, and 2nd in 1995, including
> finishing 2nd behind Virenque (but ahead of Indurain) on a stage that
Yes, and in fact I remember there were some comments about how well Boardman
was climbing during the Dauphine. He will get better, good climbing doesn't
come easy. It requires a lot of climbing.
-Burton
--
L. Burton Hawley
2330 NW Hummingbird
Corvallis, OR
Mark F. Flynn
ro...@alcala.demon.co.uk (Roger Thomas) writes:
>On 2 Apr 1997 19:23:40 -0800, Tom James <tomj...@chem1.usc.edu>
>wrote:
>>On 3 Apr 1997, Aarron Canino wrote:
>>
>>> > 4. Buy a set of rollers, and incline them at 5%-10% grade (this
>>> > sounds very dangerous)
>>> >=20
>>>=20
>>> Inclining rollers does absolutely nothing because no part of the bike =
>or
>>> your body gains any altitude. You are only suggesting something very
>>> dangerous.....
>>
>>well, I've often wondered about this too, but if it is good enough for a=
>=20
>>certain Christopher Miles Boardman....... I think one think it might do=20
>>is get you used to riding with your body angled at 10% or so to what it=20
>>is used to, even if it is simply a case of retraining your neck muscles =
>:-)
>>
>> Tom
>Boardman does genuine climbing on his because it's not rollers but a
>bike ridden on a running machine inclined at a gradient. Somehow he's
>worked out how to balance doing this. So there's a rolling road to
>climb.
Inclined or not, he's not doing any work against gravity, and therefore
he's not climbing.
Mark Flynn
John Forrest Tomlinson
In <5i5sm7$ql$1...@phoenix.kfu.com> fly...@quack.kfu.com (Mark F. Flynn)
writes:
>>Boardman does genuine climbing on his because it's not rollers but a
>>bike ridden on a running machine inclined at a gradient. Somehow he's
>>worked out how to balance doing this. So there's a rolling road to
>>climb.
>
>Inclined or not, he's not doing any work against gravity, and
therefore
>he's not climbing.
>
Someone made the same argument in the running newsgroup. To say
someone on an elevated treadmill is not working against gravity is
wrong. Think about it, if he/she stopped pedalling gravity would pull
him/her off the back of the treadmill. That's why running on an inclned
treadmill is harder than running on a level one. While I haven't
ridden my bike on a treadmill, it's the same story.
JT
Daniel Connelly
In article <5i6bv8$c...@sjx-ixn3.ix.netcom.com>,
John Forrest Tomlinson <jt...@ix.netcom.com> wrote:
>Someone made the same argument in the running newsgroup. To say
>someone on an elevated treadmill is not working against gravity is
>wrong. Think about it, if he/she stopped pedalling gravity would pull
>him/her off the back of the treadmill. That's why running on an inclned
>treadmill is harder than running on a level one. While I haven't
>ridden my bike on a treadmill, it's the same story.
The point about treadmills is a good one, and I admit my thinking on
the issue was wrong until you so concisely stated the reason or the
increased effort with an incline (I was tied up thinking about M*g*[d
h/d t]).
With rollers, though, there will be no acceleration in the absence of
effort if they are inclined.... unlike a runner, the bike COM is
constrained. Thus, inclining them adds nothing the perceived effort.
All it can do is change the orientation of the bike to better simulate
that encountered when climbing.
Dan
Aarron Canino
>
> >Boardman does genuine climbing on his because it's not rollers but a
> >bike ridden on a running machine inclined at a gradient. Somehow he's
> >worked out how to balance doing this. So there's a rolling road to
> >climb.
>
> Inclined or not, he's not doing any work against gravity, and therefore
> he's not climbing.
>
>
The difference is that the bike is not being held by the rollers when on
the treadmill. If he weren't doing any work against gravity, then he
wouldn't roll off the back should he stop pedaling. The difference is the
moving "road" below him which does not exist on rollers.
Aarron Canino
> Also, try one-legged intervals on a trainer. Unbelievable how well this
> increases power.
Try one legged riding on the rollers, and see what it does for balance!!!
>
Mark F. Flynn
"Aarron Canino" <bike...@iag.net> writes:
>>
>> >Boardman does genuine climbing on his because it's not rollers but a
>> >bike ridden on a running machine inclined at a gradient. Somehow he's
>> >worked out how to balance doing this. So there's a rolling road to
>> >climb.
>>
>> Inclined or not, he's not doing any work against gravity, and therefore
>> he's not climbing.
>>
>The difference is that the bike is not being held by the rollers when on
>the treadmill. If he weren't doing any work against gravity, then he
>wouldn't roll off the back should he stop pedaling. The difference is the
>moving "road" below him which does not exist on rollers.
>
I hate admitting that I'm wrong. You're right. In the cyclist's frame of
reference (if he doesn't pedal, he moves down the ramp at a constant
velocity), he still sees a gravitational field, and he is going uphill,
working against the field. Nice freshman physics problem!
Mark Flynn
Dave Bailey
Alright, that's enough. This isn't the first time I've had to
step in and set the record straight, and I'm sure it won't be
the last.
In article <5ib1bu$7a1$1...@phoenix.kfu.com>,
Mark F. Flynn <fly...@quack.kfu.com> wrote:
>"Aarron Canino" <bike...@iag.net> writes: [other people wrote...]
>>> >Boardman does genuine climbing on his because it's not rollers but a
>>> >bike ridden on a running machine inclined at a gradient.[...]
>>>
>>> Inclined or not, he's not doing any work against gravity, and therefore
>>> he's not climbing.
>
>>The difference is that the bike is not being held by the rollers when on
>>the treadmill. If he weren't doing any work against gravity, then he
>>wouldn't roll off the back should he stop pedaling. The difference is the
>>moving "road" below him which does not exist on rollers.
>
>I hate admitting that I'm wrong. You're right. In the cyclist's frame of
>reference (if he doesn't pedal, he moves down the ramp at a constant
>velocity), he still sees a gravitational field, and he is going uphill,
>working against the field. Nice freshman physics problem!
The original point of view held by Mark Flynn, Dan Connelly, and others
is correct. Every child knows that one must move a mass away from the
center of the earth in order to do positive work against its gravitational
field. A cyclist riding a treadmill or rollers tilted up at some angle
does no work against gravity, except to compensate for small variations
in center of mass height which are nowhere near those experienced when
climbing a real hill. Suppose I go out and climb a 400 meter (altitude
gain) hill in 16:40 (1000 seconds). My vertical velocity is 0.4 m/s,
my total (me + bike) mass is 90 kg, and the power output required to
do work against gravity is
P = F v = mg v = (90 kg)(9.8 m/s^2)(0.4 m/s)
= about 360 watts.
Now suppose I ride the treadmill. Typically I can expect to move
back and forth with a peak-to-peak amplitude of, say, 5 cm every
half pedal stroke. That is to say, each time one of my legs pushes
down on a pedal, I move forward 5 cm relative to stationary (and
then drift back during the low part of the power cycle). If I
pedal at 100 rpm, this occurs 200 times per minute. If the
treadmill is tilted at a 10% gradient, then 10% of the amplitude
represents vertical center of mass motion. Thus, each minute my
center of mass moves vertically a total of
N = (200)(0.05 m)(0.1) = 1 meter,
a very small quantity in comparison to the 24 vertical meters per
minute up the actual hill. We're talking about a 5% effect here.
The argument that work is done against gravity because the cyclist
needs to keep from rolling backwards is wrong because the cyclist's
center of mass position, to a good approximation, does not change
while the cyclist is still pedalling. Therefore, no work can be
done against gravity. Consider, for example, me riding up an actual
hill. If I stop pedalling, I might eventually roll backwards, but I
don't need 360 watts to keep from rolling backwards. I need 360
watts to move my center of mass vertically at 24 meters per minute.
In conclusion, the benefit to riding an inclined treadmill or
inclined rollers is simply that the rider practices generating
power in the climbing position. However, the cyclist does work
against friction, not gravity.
I can only conclude from reading this thread that none of you
understand the concepts of work and potential energy. Shame on
you all.
--
Dave Bailey
dba...@leland.stanford.edu
rodl...@aol.com
(Large snip)
>In conclusion, the benefit to riding an inclined treadmill or
>inclined rollers is simply that the rider practices generating
>power in the climbing position. However, the cyclist does work
>against friction, not gravity.
>
>I can only conclude from reading this thread that none of you
>understand the concepts of work and potential energy. Shame on
>you all.
And after Newton there came Einstein. Calculations don't all have to take place
using the earth as a fixed reference point. The treadmill is moving downwards.
If you cycle to stay in place, you are moving upwards relative to this, and this
is where work against gravity comes in.
To look at it another way, if you incline the treadmill at an angle of s degrees,
and the force of gravity is g, then there will be a force of g/sin s acting parallel
to the surface of the treadmill. This will need to be overcome in order to remain
stationary with respect to the earth, in addition to the usual frictional forces to be
overcome; and g/sin s will be far from insignificant.
So if frictional forces to be overcome are f, and you cycle distance d with respect
to the treadmill surface, work done (W) is;
W = (f x d) + ((g sin s) x d)
In short, I heartily and profoundly disagree, and suggest you check your physics
before being unnecessarily rude to other posters.
Rod.
Disclaimer; the opinions expressed above are not necessarily yours.
Dave Blake
rodl...@aol.com wrote:
>someone else wrote:
>>In conclusion, the benefit to riding an inclined treadmill or
>>inclined rollers is simply that the rider practices generating
>>power in the climbing position. However, the cyclist does work
>>against friction, not gravity.
>>
>>I can only conclude from reading this thread that none
>>of you understand the concepts of work and potential
>>energy. Shame on you all.
>And after Newton there came Einstein. Calculations don't all
>have to take place using the earth as a fixed reference point.
>The treadmill is moving downwards.If you cycle to stay in
>place, you are moving upwards relative to this, and this
>is where work against gravity comes in. To look at it
>another way, if you incline the treadmill at an angle
>of s degrees,and the force of gravity is g, then
>there will be a force of g/sin s acting parallel
>to the surface of the treadmill. This will need to be
>overcome in order to remain stationary with respect
>to the earth, in addition to the usual frictional forces
>to be overcome; and g/sin s will be far from
>insignificant. So if frictional forces to be overcome
>are f, and you cycle distance d with respect
>to the treadmill surface, work done (W) is;
>W = (f x d) + ((g sin s) x d)
>
>In short, I heartily and profoundly disagree, and suggest
>you check your physics before being unnecessarily rude
>to other posters.
Rod, in your equations above there is no work because
the rider does not move. If you suggest that the rider
is moving relative to the treadmill (but not the earth)
you actually need to redefine your gravitational
constant in the above equation to reflect the gravity
that exists between the treadmill and bike/rider
combo - which can be set to zero for purposes of
relevant significant digits. Gravitational work
must be done against an accelerational field. In
the case of the leaned treadmill there is none.
One of Einstein's large contributions was the
understanding of the relationship between accelerational
fields, gravity, and time.
Of course there are benefits to riding uphill on a
treadmill. If your pedal stroke is not smooth you
will bob up and down. Learning the proper pedal
stroke for climbing may be done very well on
an inclined treadmill.But it is much much more
difficult to ride up and down a hill of the same
gradient as those rollers.
--
Dave Blake
dbl...@phy.ucsf.edu
http://www.keck.ucsf.edu/~dblake/
David Casseres
In article <5ij07s$18...@itssrv1.ucsf.edu>,
dbl...@phy.ucsf.eduDELETETHISPART (Dave Blake) wrote:
> Of course there are benefits to riding uphill on a
> treadmill. If your pedal stroke is not smooth you
> will bob up and down. Learning the proper pedal
> stroke for climbing may be done very well on
> an inclined treadmill.But it is much much more
> difficult to ride up and down a hill of the same
> gradient as those rollers.
I CAN'T BELIEVE THIS DISCUSSION!!!!
Boardman is said to ride on a treadmill tilted up, and the argument is
whether this is physically similar to riding up an actual hill.
Do you seriously think it's just as easy to do this as it would be if the
treadmill was not tilted up? If you think so, get out of that armchair
and TRY IT.
Or you can try walking or running on a treadmill tilted up -- I did that
the last time I got a heart test. I guarantee you I was working against
gravity.
If you still think the tilt doesn't make you work against gravity, ask
yourself what would happen if you were on a bike on a tilted treadmill and
you didn't pedal. Do you think perhaps you might roll backward?
--
Cheers,
David
Dave Bailey
In article <19970410134...@ladder01.news.aol.com>,
<rodl...@aol.com> wrote:
[regarding tilted treadmills (tilting at windmills?)]
>And after Newton there came Einstein. Calculations don't all have
>to take place using the earth as a fixed reference point. The
>against gravity comes in. To look at it another way, if you
>to the surface of the treadmill. This will need to be overcome in
>order to remain stationary with respect to the earth, in addition
>to the usual frictional forces to be overcome; and g/sin s will
>be far from insignificant. So if frictional forces to be overcome
>are f, and you cycle distance d with respect to the treadmill
>surface, work done (W) is; W = (f x d) + ((g sin s) x d)
>
>In short, I heartily and profoundly disagree, and suggest you
>check your physics before being unnecessarily rude to other
>posters.
Rude? I was playful, tongue in cheek, not rude. What amuses
me the most here is the fact that you can feel wholly entitled
to foist your ignorance on those who don't know any better,
attempt to pass it off as knowledge of physics, and then have
the temerity to scold me for pointing out the truth. For
example, you made some remark about "Einstein" and "relative".
However, relativity, relative motion, and the principle of
equivalence (none of which you understand) have nothing to
do with this problem whatsoever:
(1) The effects of Einsteinian relativity are not detectable
when riding a treadmill because nothing on the treadmill
moves at a velocity near that of light.
(2) Relative motion, or Galilean relativity, would only be
instructive if you were a little man stuck to the surface
of the treadmill while the cyclist rode it.
(3) The principle of equivalence applies only to coordinate
systems accelerated with respect to one another. The
cyclist's coordinate system is not accelerating, so
equivalence does not apply. See, for example, A. Einstein,
"The Meaning of Relativity," Princeton University Press,
5th ed., p. 57 (1974); also see Dave Blake's followup to
your post.
Furthermore, the second term in your "W = ..." equation is
wrong because the "d" does not exist (the first term, as I
indicated in my previous post, is correct). Simply put,
cycling a distance "d" with respect to the treadmill does
not, last time I checked, move you a distance "d sin s"
vertically.
Unfortunately for you, everything I said in this and my
previous post is right. Most of what you said is wrong
(you got the friction part right). Now go and do your
homework.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Blake
cass...@apple.com says...
>Or you can try walking or running on a treadmill tilted up -- I did that
>the last time I got a heart test. I guarantee you I was working against
>gravity.
And ask yourself exactly how much your guarantee
is worth then. Gravitational work requires movement
against an accelerational field. Staying at the
same height on a tilted treadmill requires the same
amount of gravitational work as standing stationary, because
your accelerational field is the same.
You are not working against gravity in the steady
state - only dynamically as the lack of perfect
smoothness in the pedal stroke allows the bike to
float up and down. But there is no net gravitational
work. None. You can claim that you merely integrate
the work per pedal stroke to get the same effect,
but in climbing the rate of work being done is not
purely AC, but has a significant DC offset you do
not get on a treadmill.
Similarly, a stairclimber machine is a lot easier
than climbing stairs. Been there, done that. I
prefer the stairs, and I prefer the mountains.
You can keep your ignorance of physics and your
treadmills.
David Casseres
In article <5ijfcm$j...@itssrv1.ucsf.edu>,
dbl...@phy.ucsf.eduDELETETHISPART (Dave Blake) wrote:
How incredibly civil of you.
So, if I'm on a bike on an inclined treadmill, and I don't pedal at all,
will I roll down?
If the treadmill isn't moving, and I pedal hard enough to climb the
treadmill, then I think even a physics genius like you will agree that I'm
working against gravity. I'm moving up, and my gravitational potential
energy is increasing. Yes or no?
Now suppose the treadmill starts moving downward. It takes a moment for
the treadmill to accelerate, but let's skip that moment and consider
what's happening with the treadmill moving. Let's say it moves down at
the same speed that I was climbing, and I keep pedaling at exactly the
same rate. So now I am no longer going up, I am staying at exactly the
same height. Yes or no?
Now comes the crucial question. Even with your giant brain, no doubt
enriched by several Ph.D.'s in physics and related areas, you are going to
have to think hard about this. So here's the question: am I now suddenly
doing a whole lot less work per minute (i.e. developing a lot less power)?
If your answer is yes, then here's another question: what if, instead of
keeping my pedaling *rate* constant, I keep my *power output* constant,
doing the same amount of work per minute against a suddenly reduced
resistance? Do I suddenly shoot forward, up the moving treadmill, just
because it started to move in the *opposite* direction? Yes or no?
As you can see, your assumption that the amount of work done on the
inclined treadmill is less than what would be done by actually traveling
upward is wrong. On the moving treadmill, the work goes into pushing the
belt of the treadmill backward instead of into increasing gravitational
potential energy, so you are right in saying that there's no work "against
gravity," and I apologize for sloppy phrasing. However, THE AMOUNT OF
WORK IS EXACTLY THE SAME.
In order to keep me in the same place while I pedal, the treadmill has to
give a resistance equal to my weight (with bike) times the sine of the
angle it's tilted at. My rear wheel has to push against the belt with
that much force. It's exactly the same force required to move me uphill
at the same angle if the treadmill belt doesn't move -- or if I'm on a
real hill. My muscles can't tell any difference at all, because force is
force and energy is energy, whether it goes into increased gravitational
potential energy or into heating up the brake on the treadmill. And
that's the answer to the original question about whether Chris Boardman
gets the same training out of riding on a tilted treadmill that he would
get if he rode up a hill. The answer is yes.
Think about it.
--
Cheers,
David
Tom James
On 10 Apr 1997, Dave Bailey wrote:
> In article <casseres-100...@cassda.apple.com>,
> David Casseres <cass...@apple.com> wrote:
> [snip]
>
> >If you still think the tilt doesn't make you work against gravity, ask
> >yourself what would happen if you were on a bike on a tilted treadmill and
> >you didn't pedal. Do you think perhaps you might roll backward?
>
> Pop quiz, hotshot: You park your car on a steep hill and put on
> the emergency brake. You get out of your car and see that it is
> not sliding down the hill. What do you do? What do you do?
>
> The answer: You suddenly realize that the car is not doing any
> work against gravity, that you were wrong, and that you shouldn't
> post any more to this thread.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
What on earth has this got to do with anything? Are you saying we should
ride on treadmills with our brakes on? Or what?
With all the varying answers flying around, I decided to revisit my
A-level applied maths neurons, still believed to be just concious, and
try and work it out for myself. So here goes:-
To simplify things, let's ignore wind resistance and frictional losses in
the bicycle.
For a rider riding up a slope inclined at an angle alpha to the
horizontal at a velocity v, the gravitational force to be overcome is
F = m.g
and since the vertical velocity is
v(y) = v.sin(alpha)
then the power output required is
P = m.g.v.sin(alpha)
Now let us suppose the rider is on a treadmill, also angled at an angle
alpha to the horizontal. The gravitational force can be split into two
components parallel and perpendicular to the treadmill, the parallel one
being:-
F = m.g.sin(alpha)
So in the absence of input from the rider, he will slide backwards down
the treadmill. Let us assume in an infintesimally small period of time
dt, he slides back by an amount dx, such that
-dx/dt = v (ie, his forward speed)
Thus to remain in the same position on the treadmill, he has to move
himself forward by an amount dx, working against the force F. This
requires an input of energy equal to
E = F.dx
= m.g.sin(alpha).dx
But this energy is expended over a time dt, so the power input is
P = E/dt
= m.g.sin(alpha).dx/dt
and since dx/dt = v
P = m.g.v.sin(alpha)
ie, the power output is the same at the same speed and angle, whether the
rider is riding up a hill or on a treadmill.
Well, I wouldn't like to call myself a hotshot, and maths isn't my field
anyway, so I could be wrong.....
Tom
Dave Bailey
In article <01bc45eb$66441420$fe4f1ecf@spot>,
Aarron Canino <bike...@iag.net> wrote:
>You, sir, should do your homework. With the given situation, ifthe cyclist
>was not pedalling, the bike would roll off the back of the treadmill, so
>there is obviously a force required to hold the bike in the same place.
[snip]
>Given a cadence of 90rpm, this gives a power expenditure of 792.6W!!! A
>profound workout indeed!!! And this is before figuring the friction of the
>treadmill surface!!! So, before you take a smart alec tone with people
>who, although ignorant of the reasons, intuitively know the correct answer,
>I suggest you do your homework and learn what energy conversion is
>occurring in your machine.
Your complete lack of understanding of statics startles me. Given
the glib ease with which you bandy about your equations, one might
actually expect that they make some sense. Alas, they do not. I
will, as a favor to you and everyone else who is still hopelessly
confused, show you how you are wrong.
Suppose I run the treadmill in reverse and ride it facing down
the incline. If I stop pedalling, I'll be whisked off the top
immediately. But, if I pedal at just the right speed, I can
remain stationary with respect to the surroundings. The static
friction between the bicycle tires and the treadmill surface
makes this possible, just as it does in the forward case.
To reiterate, it is static friction which keeps me from sliding
off the treadmill due to its tilt. It is the act of pedalling
which keeps me from being whisked off the treadmill due to the
motion of the surface. You, and others who share your egregiously
wrong viewpoint, have confused the two.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
>the last time I got a heart test. I guarantee you I was working against
>gravity.
You were working against gravity because you had to lift your leg
each time you took a step. But once you plant your foot, the
treadmill takes it down to where you need to lift it again.
Running up an inclined surface, you need to take a real stride
(meaning, you need to lift your center of mass up the hill a
little bit) before you can lift your foot. This is the difference.
It is a consequence of the biomechanical inefficiency of running.
Dave Bailey
In article <mpd4-11049...@cu-dialup-0053.cit.cornell.edu>,
M Dolenga <mp...@cornell.edu> wrote:
>No matter how fast you run the treadmill, no matter
>how gradual the incline, you are accelerating forward because of gravity.
>You will accelerate beyond the speed of the treadmill surface and begin
>heading downhill. You cannot hold your position unless you use your
>brake.
The point is that it takes no power to keep from moving downhill
unless you are already moving downhill and need to stop.
--
Dave Bailey
dba...@leland.stanford.edu
John Forrest Tomlinson
In <5ijfcm$j...@itssrv1.ucsf.edu> dbl...@phy.ucsf.eduDELETETHISPART
(Dave Blake) writes:
>
>cass...@apple.com says...
>>Or you can try walking or running on a treadmill tilted up -- I did
that
>>the last time I got a heart test. I guarantee you I was working
against
>>gravity.
>
>is worth then. Gravitational work requires movement
>against an accelerational field. Staying at the
>same height on a tilted treadmill requires the same
>amount of gravitational work as standing stationary, because
>your accelerational field is the same.
>
>You are not working against gravity in the steady
>state - only dynamically as the lack of perfect
>smoothness in the pedal stroke allows the bike to
>float up and down. But there is no net gravitational
>work. None. You can claim that you merely integrate
>the work per pedal stroke to get the same effect,
>but in climbing the rate of work being done is not
>purely AC, but has a significant DC offset you do
>not get on a treadmill.
>
What does this mean? I never took physics. I don't know what an
accelerational field is, not AC nor DC.
It seems obvious to me that you are working against gravity on a tilted
treadmill -- otherwise the gravity from the earth would pull you down.
Even standing up on your feet you're fighting gravity, aren't you?
JT
Dave Bailey
In article <samiE8G...@netcom.com>,
Stephen A. Mills <sa...@netcom.com> wrote:
>Ok, so what happens when I want to work on my descending and tilt the
>treadmill the other way, say by 30 degrees? Will it be easier, harder,
[...]
The point is that whichever way you tilt the mill, it has no effect
on how hard it is. For example, you could just reach out with an
arm and keep gravity from dragging you down the mill - sort of like
leaning against a wall. This takes essentially no energy to do.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <01bc460b$aee10ba0$674f1ecf@spot>,
Aarron Canino <bike...@iag.net> wrote:
>So, you are saying that if the treadmill was not moving, but still
>inclined, that you would not have to exert any force on the pedals to
>remain on the belt???
No, that is not what I am saying. I am saying that you do no work
against gravity while riding a tilted treadmill whose track runs
at some constant velocity v. Acceleration, of course, requires
extra force, but I believe we have been discussing the case where
velocity is constant.
Your argument is wrong because you think that keeping from going
down the hill is the same as actually going up the hill. All of
your equations depend on that falsehood, and as such they are all
wrong. What made you presume yourself an authority on this
subject, bikerboy? I'm beginning to wonder if you've actually
ever taken a physics course. Of course, I don't expect you to
admit that you're wrong, since you've begun to take it personally.
However, think about how difficult this has been for me. I'm a
trained, practicing theoretical physicist, and I've taught years
of undergraduate physics courses in close collaboration with
some of the best in the field. I explain a simple problem, and
I'm faced with a bunch of arrogant hacks too wrapped up in their
egos to see the light. It's as if I went outside one morning to
get the paper, spotted a gaggle of drunken morons urinating on
my car, told them to get the hell off my property, and then got
assaulted as a result!
--
Dave Bailey
dba...@leland.stanford.edu
Aarron Canino
You, sir, should do your homework. With the given situation, ifthe cyclist
was not pedalling, the bike would roll off the back of the treadmill, so
there is obviously a force required to hold the bike in the same place.
You will find with basic statics, that this force is the total mass of the
bike and rider times g times the sin of the angle of incline (a) times the
gear ratio. For a 70kg rider on a 10kg bike pushing a 53-13 gear on a
10degree incline, this comes to a force of 480.6N.
F = mg sin (a) R (gear)
With this fact established, you make one critical mistake. You say that
there is no motion, when in fact this force which is fighting against
gravity is being applied around the circumference of the pedalling motion.
Remember, the bicycle's function is to convert this rotary motion to linear
motion, and just because the treadmill is absorbing the energy before it
becomes linear does not preclude the fact that the force applied is
battling gravity. Therefore, for 175mm crankarms, this gives a workload of
528.4J per rotation.
W = F X d = F X 2 X pi X l (cranks)
Given a cadence of 90rpm, this gives a power expenditure of 792.6W!!! A
profound workout indeed!!! And this is before figuring the friction of the
treadmill surface!!! So, before you take a smart alec tone with people
who, although ignorant of the reasons, intuitively know the correct answer,
I suggest you do your homework and learn what energy conversion is
occurring in your machine.
Aarron
PS I have included a picture as an attachment to clarify my model.
begin 600 Image1.jpg
<uuencoded_portion_removed>
)X645<JO^@__9
`
end
Dave Blake
In article <Pine.A32.3.91.970410...@chem1.usc.edu>,
>
>ie, the power output is the same at the same speed and angle, whether the
>rider is riding up a hill or on a treadmill.
>
>Well, I wouldn't like to call myself a hotshot, and maths isn't my field
>anyway, so I could be wrong.....
Well, you have to consider the V of the rider's
center of mass relative to the accelerational field -
not the V on the circumference of his tires.
Aarron Canino
> Suppose I run the treadmill in reverse and ride it facing down
> the incline. If I stop pedalling, I'll be whisked off the top
> immediately.
Only if you are holding your brakes and the velocity of the belt is great
enough to throw you off the top before the acceleration due to gravity can
put you off the bottom. With your hands off the brakes, the situation is
pretty close to frictionless -- just a small amount of rolling friction,
made less significant by the motion of the belt, and the friction of the
hub bearings, etc.
>But, if I pedal at just the right speed, I can
> remain stationary with respect to the surroundings. The static
> friction between the bicycle tires and the treadmill surface
> makes this possible, just as it does in the forward case.
Ah, but, if you have the treadmill running slowly enough, the force of
gravity will push you off on your smug little face not off the top even
though the belt is moving upward in this case. You are obviously acted on
by the force of gravity.
> To reiterate, it is static friction which keeps me from sliding
> off the treadmill due to its tilt. It is the act of pedalling
> which keeps me from being whisked off the treadmill due to the
> motion of the surface. You, and others who share your egregiously
> wrong viewpoint, have confused the two.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
>
So, you are saying that if the treadmill was not moving, but still
inclined, that you would not have to exert any force on the pedals to
remain on the belt??? If your viewpoint is correct, then this is what you
are stating. Draw a damn diagram and look at the forces. If the belt is
inclined, there is a force required to maintain the position on the belt
and this force is working against the gravitational component that runs
parallel to the belt. This force is acting over a distance because of the
rotation of the pedals. If this were not the case, then it would not be
easier to pedal a lower gear on flat land since the only thing that would
matter would be linear distance of the bike's travel. To ignore this fact
would be analagous to stating that it takes no work to accelerate a
pendulum from rest since there is no net vertical motion, or so it would
seem your claim would illustrate. The only factor that the motion ofthe
belt plays is that it limits what choice of gearing you can use at a
particular cadence since the wheel must rotate at the same tangential
velocity as the belt or else frictional forces would push the bike one way
or the other, and to make the motion of the wheel itself more closely
approximate a frictionless situation.
Try getting on your bike sometime. If you are on flat ground in a high
gear that is big enough that you can get a sense of the resistance in the
instant before the bike begins to move. Now do the same thing on a hill.
When the bike is completely at rest, it is clear that you can press harder
on the pedals when on the hill than you can on the flats. That is because
you are working against gravity, and must exert a force to maintain a
static position. That force acts along the circumference of the crank's
motion asyou pedal and begin to move. If we are to take your explanation
to be truth, then the only 'logical' conclusion would be that the hill
somehow has more friction holding the bike back even when it is at rest.
Clearly this is not the case.
What you are arguing is so clearly against common sense that it astounds
me. You argue a viewpoint which I have shown you completely ignores the
rotation of the pedals and yet you still cling to your view. And on top of
that, you are a sarcastic man who insists on treating others on this group
as ignorant fools rather than engaging in an adult and rational
conversation with proper respect. You are supposedly participating in a
group which is meant to serve as a forum for exchanging ideas and debating
honest and open questions, and yet you take a demeaning and condescending
tone which is not only disrespectful, but I feel is also inappropriate. I
only hope that your conversations with people in real life are
significantly more refined, because I wish no man to be lonely.
Sincerely,
Aarron
Dave Blake
cass...@apple.com says...
>As you can see, your assumption that the amount of work done on the
>inclined treadmill is less than what would be done by actually traveling
>upward is wrong. On the moving treadmill, the work goes into pushing the
>belt of the treadmill backward instead of into increasing gravitational
>potential energy, so you are right in saying that there's no work "against
>gravity," and I apologize for sloppy phrasing. However, THE AMOUNT OF
>WORK IS EXACTLY THE SAME.
The amount of work CAN be exactly the same. That still does
not make it like climbing the Dolomites. The work can
be similarly exactly the same on a Computrainer, or
on rollers, or in a pedal boat. A crucial point about hill
climbing is that you exert force through a larger portion
of the pedal stroke, probably because you decelerate faster
while climbing than while riding on flat ground. That is
why good relatively small time trialists can be bad
climbers (are you listening Eric Breukink?). The pedal
stroke itself is different.
>In order to keep me in the same place while I pedal, the treadmill has to
>give a resistance equal to my weight (with bike) times the sine of the
>angle it's tilted at. My rear wheel has to push against the belt with
>that much force. It's exactly the same force required to move me uphill
>at the same angle if the treadmill belt doesn't move -- or if I'm on a
>real hill. My muscles can't tell any difference at all, because force is
>force and energy is energy, whether it goes into increased gravitational
>potential energy or into heating up the brake on the treadmill. And
>that's the answer to the original question about whether Chris Boardman
>gets the same training out of riding on a tilted treadmill that he would
>get if he rode up a hill. The answer is yes.
You must have a mightly special treadmill to do all that. It
has to have a perfectly linear frictional loss (wrt velocity)
to equal the hill - yet it needs an exponential increase
in resistance at higher speeds to mimic the wind you
see over 10 MPH. I know of no such treadmill. All of your
arguments could similarly be applied to riding a variable
resistance trainer on flat ground - yet no one would argue
that is comparable to climbing. A key point in all of this
would be the distribution of force throughout the pedal
stroke while on the treadmill. Can you feel the treadmill
pushing back on your pedals at the flat spots in your
pedal strokes or do you simply mash the pedals ? Or
will it feel like every other set of magnetic resistance -
paradoxically seeming easier at higher speeds ? Therein
lies the difference - the hill seems to push back
on the pedals. A trainer does not - even a treadmill
elevated.
Stephen A. Mills
In article <5ii2rp$e...@epic9.Stanford.EDU>,
Dave Bailey <dba...@leland.Stanford.EDU> wrote:
>In conclusion, the benefit to riding an inclined treadmill or
>inclined rollers is simply that the rider practices generating
>power in the climbing position. However, the cyclist does work
>against friction, not gravity.
Ok, so what happens when I want to work on my descending and tilt the
treadmill the other way, say by 30 degrees? Will it be easier, harder, or the
same as the level case? Will you be impressed when I break the hour
record in this fashion? Care to redo your homework? :)
(Remember, we're talking treadmills here not rollers)
Steve
M Dolenga
In article <5ik63d$e...@epic9.Stanford.EDU>, dba...@leland.Stanford.EDU
(Dave Bailey) wrote:
> Suppose I run the treadmill in reverse and ride it facing down
> the incline. If I stop pedalling, I'll be whisked off the top
> immediately. But, if I pedal at just the right speed, I can
> remain stationary with respect to the surroundings. The static
> friction between the bicycle tires and the treadmill surface
> makes this possible, just as it does in the forward case.
Suppose you have an infinitely long treadmill, so coming off either end
isn't an issue. Suppose, also, that this is all happening in a vacuum.
If your bike is pointed down the treadmill, you will eventually be heading
downhill very fast. No matter how fast you run the treadmill, no matter
how gradual the incline, you are accelerating forward because of gravity.
You will accelerate beyond the speed of the treadmill surface and begin
heading downhill. You cannot hold your position unless you use your
brake.
Mike
Dave Blake
In article <01bc460b$aee10ba0$674f1ecf@spot>, bike...@iag.net says...
>So, you are saying that if the treadmill was not moving, but still
>inclined, that you would not have to exert any force on the pedals to
>remain on the belt??? If your viewpoint is correct, then this is what you
>are stating. Draw a damn diagram and look at the forces. If the belt is
>inclined, there is a force required to maintain the position on the belt
>and this force is working against the gravitational component that runs
>parallel to the belt. This force is acting over a distance because of the
>rotation of the pedals.
I agree with you about the force. However, there is no movement
relative to the force gradient and thus NO WORK DONE ON
YOUR CENTER OF MASS. The only source of resistance is the
treadmill itself and not the gradient. You are not working
against an incline - you are working against a treadmill's
resistance. With enough ingenuity you could rig a system that would
come pretty close to a hill - but it would require monitoring
both the position in the pedal stroke and the
velocity.
>What you are arguing is so clearly against common sense that it astounds
>me. You argue a viewpoint which I have shown you completely ignores the
>rotation of the pedals and yet you still cling to your view. And on top of
>that, you are a sarcastic man who insists on treating others on this group
>as ignorant fools rather than engaging in an adult and rational
>conversation with proper respect. ...
Please remember a few points. Everyone that I know is much nicer
in person than they are on Usenet. Many people take
offense to material that they feel is wrong when it
concerns a topic they care about - even more so when
the purported wrong poster baits them a little.
rodl...@aol.com
Another thought on the Dave Bailey idea.
Imagine you are on a treadmill at a 20 degree incline, pedalling steadily on the
limit.
Suddenly there is a power cut and the belt stops moving.
As soon as you start to move forward, you suddenly start moving upwards
and doing work against gravity; according to Dave this will need a sudden large
increase in power, which you can't sustain as you were already on the limit,
so you recoil backwards off the treadmill.
What? Something wrong?
Dave Bailey
In article <19970411071...@ladder01.news.aol.com>,
<rodl...@aol.com> wrote:
>Imagine you are on a treadmill at a 20 degree incline, pedalling
>steadily on the limit.
>Suddenly there is a power cut and the belt stops moving.
>As soon as you start to move forward, you suddenly start moving upwards
>and doing work against gravity; according to Dave this will need a
>sudden large
>increase in power, which you can't sustain as you were already on the
>limit,
>so you recoil backwards off the treadmill.
>What? Something wrong?
You wouldn't recoil backwards if you could downshift fast enough
to be in the right gear to be accelerating up a 20 degree incline.
Not only would you be doing work against gravity, you'd also be
doing work to increase your linear kinetic energy at a rate
sufficiently fast to get you going before you might lose your
balance and fall over. If you couldn't downshift, then it
would be pretty much like doing a standing start up a 20
degree slope in a big gear. The human body is not very well
suited to generate huge power output at super-low cadences,
so you'd probably be unable to do it in that gear and you'd
fall over after lurching forward a little bit.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <Pine.OSF.3.95.970411...@amy20.Stanford.EDU>,
Omar Sze Leung <ra...@leland.Stanford.EDU> wrote:
>If the treadmill were stopped, what force would you need on the crank to
>remain stationary?
>
>Power = Force * Velocity = 0
>
>Now if the treadmill starts moving, the rider still has to resist gravity
>with a force F. But he must pedal to stay in the same place.
>
>Power = Force * Velocity = nonzero
This is wrong because you have the wrong velocity. It is the upward
velocity of the rider's center of mass which must be considered in
both cases. This velocity is always zero.
--
Dave Bailey
dba...@leland.stanford.edu
Omar Sze Leung
If I were a Physics Instructor, I'd say,
Imagine two trained monkeys climbing ropes at constant velocity V. One
rope is held stationary, while the other is then lowered at CONSTANT
velocity -V.
I say that the two monkeys are expending the same amount of energy/time.
Correct? (The winch lowering rope is doing negative work to keep the
monkey at constant height.)
Now put these monkeys on a long stationary treadmill and a moving one
with their bikes and the situation is analagous.
Omar (sorry about the return address on the header)
Dave Bailey
In article <Pine.OSF.3.95.97041...@amy16.Stanford.EDU>,
>
>Imagine two trained monkeys climbing ropes at constant velocity V. One
>rope is held stationary, while the other is then lowered at CONSTANT
>velocity -V.
>
>I say that the two monkeys are expending the same amount of energy/time.
>Correct? (The winch lowering rope is doing negative work to keep the
>monkey at constant height.)
Well, Omar, I *am* a physics instructor, and I say:
Absolutely correct: They both expend no energy per unit time.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <01bc45eb$66441420$fe4f1ecf@spot>,
Aarron Canino <bike...@iag.net> wrote:
[..]
>there is obviously a force required to hold the bike in the same place.
>You will find with basic statics, that this force is the total mass of the
>bike and rider times g times the sin of the angle of incline (a) times the
>
> F = mg sin (a) R (gear)
>
Let's try applying this logic to actually going up a real hill. First,
I need you to agree that, by energy conservation, the work required to
ride up a hill is mgh, and the average power output is therefore mgh/t
where t is the time needed to climb the hill. This is an absolute
lower bound on the actual power output required, since there are
always some dissipative losses due to chain friction, biomechanical
inefficiency, etc. For our purposes, they may be neglected.
Now, if you stop pedalling up a real hill, you'll roll backwards
because of gravity, so according to you, you'll need to do W work
per pedal revolution to counteract this tendency. In addition,
since you raise your gravitational potential energy by mgh, you
must do that amount of work *on top of* the W per pedal revolution
in your wrong formula. Suppose you need N pedal revolutions to
go up the hill. Then your total work done is
W_tot = WN + mgh
(according to you). But the gain in gravitational potential
energy is just mgh, so there is a contradiction.
Let me put in some numbers. Let m = 90 kg, h = 400 m. Suppose
the climb is 5 km in length (gradient of 8%) and I climb it in
1000 seconds (16:40). Suppose I do this in a 42x21 gear (ratio
of 2.0). I ride 175 mm cranks. My speed is 5 m/s = 11.2 mph
(entirely reasonable); my cadence is around 72 rpm.
I get 158.3 Joules per pedal rotation. It takes about 1190
pedal rotations to get to the top assuming a 2.1 meter wheel
circumference. I did this in 1000 seconds, so the power
required was
P = (158.3)(1190)/(1000) watts = about 190 watts.
And, mgh/t is about 360 watts. I estimate 30 watts for air
resistance and 20 for internal friction. Total power required
for this effort, which, by the way, is at my anaerobic
threshold, is 190+360+30+20 = 600 watts - compared with
Boardman's 429 for the most recent hour record.
Your only way out is to somehow claim that the 190 watt
"keeping from going downhill" contribution is not present
when climbing a real hill, but *is* present on the treadmill.
The correct answer, of course, is that it is not present
in either case. However, I am looking forward to your
next attempt to bend the rules of physics (these last
couple of days have been quite titillating for me), so
I shall bid you a fond adieu and await your rejoinder.
--
Dave Bailey
dba...@leland.stanford.edu
Omar Sze Leung
If the treadmill were stopped, what force would you need on the crank to
remain stationary?
Power = Force * Velocity = 0
Now if the treadmill starts moving, the rider still has to resist gravity
with a force F. But he must pedal to stay in the same place.
Power = Force * Velocity = nonzero
Omar
M Dolenga
In article <5ikj2a$k...@epic9.Stanford.EDU>, dba...@leland.Stanford.EDU
(Dave Bailey) wrote:
> I'm a
> trained, practicing theoretical physicist, and I've taught years
> of undergraduate physics courses in close collaboration with
> some of the best in the field.
So what, I've seen enough professors and scientists who are complete
idiots; and even the smart ones make mistakes. Not to suggest you are an
idiot, but your credentials will not speak for you. I'm a trained,
practicing biochemist, but I've made my share of blunders.
Perhaps no extra work is being performed, and I'm not sufficiently
familiar with the precise terminology, but it is obvious to me that
someone riding a bike on an inclined treadmill "works" (in the colloquial
sense) harder to maintain a certain speed as opposed to a non inclined
treadmill. If you stop pedalling on an inclined treadmill (if it's long
enough), you'll roll back off the treadmill and you'll feel a tug as your
pedals try to turn in reverse; stop pedalling on a flat one, and you just
stop. That would suggest that the "forces" working against you are
greater on the inclined treadmill.
Mike
Daniel Connelly
In article <5il0cm$o...@epic9.Stanford.EDU>,
Dave Bailey <dba...@leland.Stanford.EDU> wrote:
>Your only way out is to somehow claim that the 190 watt
>"keeping from going downhill" contribution is not present
>when climbing a real hill, but *is* present on the treadmill.
>The correct answer, of course, is that it is not present
>in either case. However, I am looking forward to your
>next attempt to bend the rules of physics (these last
>couple of days have been quite titillating for me), so
>I shall bid you a fond adieu and await your rejoinder.
Dave, Dave, Dave......
You can dump work into a treadmill, but not (in the infinite Earth-mass-limit,
neglecting deformation of the road) on the road.
It's like a ThighMaster (tm).
This is the difference.
Dan
Mike Miller
Dave Blake wrote:
>
> cass...@apple.com says...
> >Or you can try walking or running on a treadmill tilted up -- I did that
> >the last time I got a heart test. I guarantee you I was working against
> >gravity.
>
> And ask yourself exactly how much your guarantee
> is worth then. Gravitational work requires movement
> against an accelerational field. Staying at the
> same height on a tilted treadmill requires the same
> amount of gravitational work as standing stationary, because
> your accelerational field is the same.
>
Hmmmmm.... when I walk on a treadmill, and the treadmill is set to 5%
incline, my heartrate is about 120. When I set the treadmill to 15%
incline, my heartrate goes up to 150.
Your explanation does not explain what I observe.
/m
Rob Hult
Dave Bailey wrote:
>
> In article <samiE8G...@netcom.com>,
> Stephen A. Mills <sa...@netcom.com> wrote:
> >treadmill the other way, say by 30 degrees? Will it be easier, harder,
>
> The point is that whichever way you tilt the mill, it has no effect
> on how hard it is. For example, you could just reach out with an
> arm and keep gravity from dragging you down the mill - sort of like
> leaning against a wall. This takes essentially no energy to do.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
All of the posts thus far have been most informative and entertaining.
As I don't *dare* try to support and prove any points, I have decided to
merely add some fuel to the fire.
First, is there any difference between running on a treadmill or cycling
on one? Why do I ask? On a treadmill, why is it easier to run at an
incline of 0 degrees while it is much harder (requires more work, watts,
calories, etc.) to run at a positive incline? Is this simply 'body
mechanics'? In other words, while running at an incline is my body
simply in a more inefficient running position? My heart rate certainly
goes up substantially and I 'feel' like I am running uphill, so what is
going on? Is the treadmill machine really 'tricking' me that well?
Second, consider the design of the treadmill. Someone earlier mentioned
'pushing against the belt'. Is this really what is happening? How does
the belt resistance matter? I mean, some cheap treadmills have no
motor, essentially you are the motor to move the tread. Others have a
motor and a brake along with a belt speed control. What's going on
here?
If you are pedaling a bicycle on a stationary treadmill, you are not
resisting gravity. Is this a true statement? If you are running on a
treadmill, you are not resisting gravity. Is this a true statement?
Are these conditions (running and cycling) the same?
-ROb
Dave Blake
dba...@leland.Stanford.EDU says...
>Omar Sze Leung <ra...@leland.Stanford.EDU> wrote:
>>
>>If I were a Physics Instructor, I'd say,
>>
>>Imagine two trained monkeys climbing ropes at constant velocity V. One
>>rope is held stationary, while the other is then lowered at CONSTANT
>>velocity -V.
>>
>>I say that the two monkeys are expending the same amount of energy/time.
>>Correct? (The winch lowering rope is doing negative work to keep the
>>monkey at constant height.)
>
>Well, Omar, I *am* a physics instructor, and I say:
>
>Absolutely correct: They both expend no energy per unit time.
They both do no gravitational work per unit time.
This statement is not the same as saying they expend
no energy. Clearly they are alive and there is
a basal metabolism associated with that, which
means they always expend energy (except in that
perfect frictionless massless world of physics
problems for first year students).
The monkey on the moving rope will expend more
energy to stay in place.
Mark F. Flynn
dba...@leland.Stanford.EDU (Dave Bailey) writes:
>In article <19970410134...@ladder01.news.aol.com>,
> <rodl...@aol.com> wrote:
>[regarding tilted treadmills (tilting at windmills?)]
>>And after Newton there came Einstein. Calculations don't all have
>>to take place using the earth as a fixed reference point. The
>>treadmill is moving downwards. If you cycle to stay in place,
>>you are moving upwards relative to this, and this is where work
>>against gravity comes in. To look at it another way, if you
>>incline the treadmill at an angle of s degrees, and the force of
>>gravity is g, then there will be a force of g/sin s acting parallel
>>to the surface of the treadmill. This will need to be overcome in
>>order to remain stationary with respect to the earth, in addition
>>to the usual frictional forces to be overcome; and g/sin s will
>>be far from insignificant. So if frictional forces to be overcome
>>are f, and you cycle distance d with respect to the treadmill
>>surface, work done (W) is; W = (f x d) + ((g sin s) x d)
>>
>>In short, I heartily and profoundly disagree, and suggest you
>>check your physics before being unnecessarily rude to other
>>posters.
>(2) Relative motion, or Galilean relativity, would only be
> instructive if you were a little man stuck to the surface
> of the treadmill while the cyclist rode it.
>Unfortunately for you, everything I said in this and my
>previous post is right. Most of what you said is wrong
>(you got the friction part right). Now go and do your
>homework.
I disagree. Galilean relativity is precisely what applies here, because
you are comparing frames which are moving at a relative constant veclocity
<< c. The acceleration due to gravity as seen in the two frames of
reference (one stationary to the surface of the earth, and one stationary
to the treadmill surface) is identical. Therefore, for someone on the
treadmill to move up it (in his frame of reference), he must do work
against gravity, and is therefore "climbing". I think someone else needs
to do their homework.
Mark Flynn
(and lest I be accussed of ignorance:)
Ph.D. in physics, Washington University, 1986.
Rob Hult
Mark F. Flynn wrote:
>
>
> I disagree. Galilean relativity is precisely what applies here, because
> you are comparing frames which are moving at a relative constant veclocity
> << c. The acceleration due to gravity as seen in the two frames of
> reference (one stationary to the surface of the earth, and one stationary
> to the treadmill surface) is identical. Therefore, for someone on the
> treadmill to move up it (in his frame of reference), he must do work
> against gravity, and is therefore "climbing". I think someone else needs
> to do their homework.
>
This is starting to make sense to me at least. If I consider myself or
my bike in motion relative to the surface of the treadmill tread,
everything makes sense. I guess the only thing missing is the wind
resistance, which is not a big deal when running or climbing a steep
hill. Okay, go ahead, ignite....
-ROb
Douglas Vaughan
In article <5ijnvg$8...@epic9.Stanford.EDU>, dba...@leland.Stanford.EDU
(Dave Bailey) wrote:
> In article <casseres-100...@cassda.apple.com>,
> David Casseres <cass...@apple.com> wrote:
>
> >Or you can try walking or running on a treadmill tilted up -- I did that
> >the last time I got a heart test. I guarantee you I was working against
> >gravity.
>
> You were working against gravity because you had to lift your leg
> each time you took a step. But once you plant your foot, the
> treadmill takes it down to where you need to lift it again.
> Running up an inclined surface, you need to take a real stride
> (meaning, you need to lift your center of mass up the hill a
> little bit) before you can lift your foot. This is the difference.
> It is a consequence of the biomechanical inefficiency of running.
I hope this issue has been settled, but if it has, I missed it. An
inclined treadmill establishes an inertial reference frame in which gravity
most assuredly does not vanish. Working against gravity in that reference
frame (i.e, remaining stationary in the rest frame) involves work.
But to consider the specific argument used above: What if you halve the
size of your strides and double the rate? Do you do less work? Halve them
again. And again. Until you are taking infinitesimally small steps at a
rate sufficient to maintain your position in the rest frame--as in, say,
riding a bike. When does the need to do work again gravity disappear?
Or consider a tilted moving sidewalk (a la most airports) that can be
turned on and off. Put your bike on at the top, turn on the sidewalk.
When you get to the bottom, turn off the sidewalk, ride to the top. Work.
Next time, simply ride your bike to maintain your position at the top of
the moving sidewalk. Same work.
-Doug
Berkeley, California
gdva...@lbl.gov
David Casseres
In article <5ikb7g$m...@itssrv1.ucsf.edu>,
dbl...@phy.ucsf.eduDELETETHISPART (Dave Blake) wrote:
> cass...@apple.com says...
>
> >As you can see, your assumption that the amount of work done on the
> >inclined treadmill is less than what would be done by actually traveling
> >upward is wrong. On the moving treadmill, the work goes into pushing the
> >belt of the treadmill backward instead of into increasing gravitational
> >potential energy, so you are right in saying that there's no work "against
> >gravity," and I apologize for sloppy phrasing. However, THE AMOUNT OF
> >WORK IS EXACTLY THE SAME.
>
> The amount of work CAN be exactly the same. That still does
> not make it like climbing the Dolomites.
Up to this point we have been talking exclusively about the amounts of
energy (work) involved. Well, if the amount of work IS exactly the same
(and it is, always, every time the laws of physics are applied) then it's
EXACTLY THE SAME. Which of those words is too tricky for you?
>The work can
> be similarly exactly the same on a Computrainer, or
> on rollers, or in a pedal boat.
Yes! Work is work. I thought that was what we were talking about.
>A crucial point about hill
> climbing is that you exert force through a larger portion
> of the pedal stroke, probably because you decelerate faster
> while climbing than while riding on flat ground. That is
> why good relatively small time trialists can be bad
> climbers (are you listening Eric Breukink?). The pedal
> stroke itself is different.
Ah, now you want to claim the amount of work is different because of
efficiency considerations. Well, the pedaling action is indeed different
on a level trainer, rollers, or a pedal boat. But on an inclined
treadmill, the pedaling action is also EXACTLY THE SAME as in hill
climbing. Just as the velocity of the bike may vary slightly during the
stroke on a hill, it will vary the same way on the treadmill and the bike
will actually move back and forth slightly if the treadmill's speed is
perfectly constant. Why would it not be the same? The power being
developed is exactly the same and so is the geometry of bike, surface, and
rider.
> >In order to keep me in the same place while I pedal, the treadmill has to
> >give a resistance equal to my weight (with bike) times the sine of the
> >angle it's tilted at. My rear wheel has to push against the belt with
> >that much force. It's exactly the same force required to move me uphill
> >at the same angle if the treadmill belt doesn't move -- or if I'm on a
> >real hill. My muscles can't tell any difference at all, because force is
> >force and energy is energy, whether it goes into increased gravitational
> >potential energy or into heating up the brake on the treadmill. And
> >that's the answer to the original question about whether Chris Boardman
> >gets the same training out of riding on a tilted treadmill that he would
> >get if he rode up a hill. The answer is yes.
>
>
> You must have a mightly special treadmill to do all that. It
> has to have a perfectly linear frictional loss (wrt velocity)
> to equal the hill - yet it needs an exponential increase
> in resistance at higher speeds to mimic the wind you
> see over 10 MPH. I know of no such treadmill.
Well apparently, Doctor Professor, sir, *Chris Boardman has one.* If he
didn't, then it would not be possible for him to stay in one place by
pedaling on it. Are you now shifting from "It isn't the same as climbing
a hill" to "He doesn't really do it at all because there's no such
machine"?
By the way, this is the first time wind has been mentioned. Just to keep
things clear, no, I do not claim that Chris Boardman's treadmill does air
resistance. It probably doesn't do birds and trees, either.
> All of your
> arguments could similarly be applied to riding a variable
> resistance trainer on flat ground - yet no one would argue
> that is comparable to climbing.
It differs in the rider's posture, but not in the energy required.
> A key point in all of this
> would be the distribution of force throughout the pedal
> stroke while on the treadmill.
B.S. This is a minor efficiency difference which would not apply to the
tilted treadmill.
> Can you feel the treadmill
> pushing back on your pedals at the flat spots in your
> pedal strokes or do you simply mash the pedals ?
Probably.
> Or
> will it feel like every other set of magnetic resistance -
> paradoxically seeming easier at higher speeds ? Therein
> lies the difference - the hill seems to push back
> on the pedals. A trainer does not - even a treadmill
> elevated.
I assure you with the utmost sincerity, the treadmill pushes back. It has
to, or else the bike either rolls off the bottom or collides with the
control pedestal at the front.
> http://www.keck.ucsf.edu/~dblake/
Checking out this web page, I see that you're in neurology, not physics.
It figures. If I were you I would not try to lecture about physics to
people who actually know some.
--
Cheers,
David
David Casseres
[I wrote]
> >If you still think the tilt doesn't make you work against gravity, ask
> >yourself what would happen if you were on a bike on a tilted treadmill and
> >you didn't pedal. Do you think perhaps you might roll backward?
>
> Pop quiz, hotshot: You park your car on a steep hill and put on
> the emergency brake. You get out of your car and see that it is
> not sliding down the hill. What do you do? What do you do?
No, I asked you first. You tell me if the bike will roll backward on the
tilted treadmill, with no pedaling, and THEN if you do it politely I'll
explain to you the difference between pedaling a bike on a moving
treadmill and setting a parking brake on a non-moving hill.
--
Cheers,
David
Kevin Metcalfe
This has been amusing. Perhaps it would help if we did a credential
check. I know that Dave Bailey has a PhD thesis in Physics waiting to be
signed at Stanford University. I only took the three standard engineering
physics classes and my measly brain can understand Dave's arguments.
Personally, I'd really want to think long and hard before I got in a
physics argument with a PhD physicist. But that's just me. We all know
that Dave's got a twisted desire for Dove Bars, but that doesn't mean he
doesn't know his physics.
--
Kevin Metcalfe
metc...@wheel.dcn.davis.ca.us
Davis, CA
Bruce Frech
There are two questions here:
-- is the power to ride a treadmill at a given slope the same as that
required to ride up a hill at that slope (ignoring wind resistance).
-- is that power expended on the treadmill "overcoming gravity".
Power = Force * velocity.
Riding on an inclined treadmill requires a force of m*g*slope
(=sin(angle of incline)) to keep from rolling down.
That leaves the question of what is the velocity. The velocity is the
speed of the treadmill. If the treadmill is stoped the velocity is zero
and no power is expended. If the treadmill is moving then you must
apply that force at the speed of the treadmill to maintain position.
Thus the power needed to ride a treadmill is the same as that needed to
overcome gravity when climbing a hill of the same slope.
That was the first part of the question.
The other part of the argument is saying that power is being expended to
overcome gravity. Here we are running into semantics and some would say
since the force is due to gravity (remember it is m*g*slope) then the
power is overcoming gravity. Call it what you want, the power required
is the same.
Dave Blake likes to say Force = m*g and that the velocity is the
velocity of the center of mass. That would give us the power needed to
gain potential energy. But no one is arguing that there is any gain in
potential energy, unlike climbiing a hill where one can get that energy
back when going down the hill.
Dave Blake wrote:
.> Well, you have to consider the V of the rider's
.> center of mass relative to the accelerational field -
.> not the V on the circumference of his tires.
There is no mention of acceleration field in P = F*v.
Dave Blake also wrote:
>> That still does
not make it like climbing the Dolomites. ... A crucial point about hill
climbing is that you exert force through a larger portion
of the pedal stroke, probably because you decelerate faster
while climbing than while riding on flat ground. ... The pedal
stroke itself is different. <<<
The momentum and statics are exactly the same on the incline treadmill
as on a hill. So the effect on the CoG/CoM will be the same. Thus
inclined treadmills are a good way to train on hills to develop that
magical pedal stroke.
BTW, is you run the treadmill in reverse (or tilt it down instead of up)
you will need to exert the same force to keep from rolling down.
Brad Harris
Dave Bailey wrote:
>
> Your complete lack of understanding of statics startles me. Given
> the glib ease with which you bandy about your equations, one might
> actually expect that they make some sense. Alas, they do not. I
> will, as a favor to you and everyone else who is still hopelessly
> confused, show you how you are wrong.
>
> Suppose I run the treadmill in reverse and ride it facing down
> the incline. If I stop pedalling, I'll be whisked off the top
> immediately. But, if I pedal at just the right speed, I can
> remain stationary with respect to the surroundings. The static
> friction between the bicycle tires and the treadmill surface
> makes this possible, just as it does in the forward case.
>
> off the treadmill due to its tilt. It is the act of pedalling
> which keeps me from being whisked off the treadmill due to the
> motion of the surface. You, and others who share your egregiously
> wrong viewpoint, have confused the two.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
Well I'd say that we had a neat little discussion going that turned
ugly... I am still a physics student (been through the first year so I
thought I knew what I was doing), I intuitively disagreed with Mr.
Bailey. However, I must say that the argument has become quite muddled
and I became confused as hell.... but when Mr. Bailey suggested the
above example, I decided to take matters into my own hands... owning a
fine example of a bicycle and having acess to a treadmill (corperate
exersize room that no one else uses... wouldn't want anyone to see me
doing something this dumb) a put the treadmill at as much of an angle as
it would go and placed me and my bike on it and immediately started
rolling down the short incline. Unfortunately the treadmill doesn't run
backwards (duh... guess I should have checked this before going through
the effort) but the experience convinced me that Mr. Bailey is mistaken
in his assumptions. To further confirm my beliefs I went and tried
walking up a down escelator (I figured since it was also work against
gravity that it was similar enough for an analagy) if I judged my pace
correctly I could stay in exactly the same place (relative to the rest
of the mall) but to do so sure FELT like I was doing work, and to make
sure I then tried the stairs that run paralel to the escelator and at
the same pace it FELT the same. I realize, of course, that my petty
experiments here are far from scientific in thier accuracy. I, of
course realize that Einstien's theories of relativity have absolutely no
relivance here, however, Galileo's work with relitive motion seems to
hold a great deal of relivance.
Just as a last note please don't add snide remarks, I really want to
understand what the hell is going on here.....
Thanks,
BJH
Dave Bailey
In article <casseres-110...@cassda.apple.com>,
David Casseres <cass...@apple.com> wrote:
>No, I asked you first. You tell me if the bike will roll backward on the
>tilted treadmill, with no pedaling, and THEN if you do it politely I'll
>explain to you the difference between pedaling a bike on a moving
>treadmill and setting a parking brake on a non-moving hill.
There can be no difference because if there were, it would violate
conservation of energy.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Blake
gdva...@lbl.gov wrote:
>Or consider a tilted moving sidewalk (a la most airports) that can be
>turned on and off. Put your bike on at the top, turn on the sidewalk.
>When you get to the bottom, turn off the sidewalk, ride to the top. Work.
>Next time, simply ride your bike to maintain your position at the top of
>the moving sidewalk. Same work.
Not same work. In one case you must apply a force which
is exerted by the tires to the real hill equal
to m g sin(theta). Any velocity up the slope requires
additional force at the tires, and additionaly pedaling
force. This is hill climbing.
In the case of a treadmill that has some coefficient of
friction, you must apply a force exerted by the tires
on the treadmill equal to m g sin(theta) to stay in the
same position. You cannot apply more force, or less
force, or you will move up or down the slope. Thus
the tires exert less or equal force on the slope, ALWAYS,
on a treadmill than on a fixed slope of the same grade.
To reiterate, if you are spinning on a treadmill, you
will always apply less work to the slope than if
you are on a real hill of the same incline. You will
apply exactly enough force to the tires that you
would apply if you did a trackstand on the same
incline of a real hill.
Heather Williams
I have been enjoying the physics debate in this thread immensely.
However,
as a biologist, I am afraid that I must step in and point out that all
discussion that came after the following exchange is invalid:
>> Suppose I run the treadmill in reverse and ride it facing down
>> the incline. If I stop pedalling, I'll be whisked off the top
>> immediately. But, if I pedal at just the right speed, I can
>> remain stationary with respect to the surroundings. The static
>> friction between the bicycle tires and the treadmill surface
>> makes this possible, just as it does in the forward case.
>
>isn't an issue. Suppose, also, that this is all happening in a vacuum.
>You cannot hold your position unless you use your brake.
None of you can be observing anything, arguing anything, or using your
brake.
All of you who are participating in this example are dead. I, however,
am
doing my regular rides up a 6km stretch of 7% climb, but am certainly not
getting as much in the way of cardiovascular benefit as Chris Boardman is
on his treadmill - so long as continues to avoid training in a vacuum.
Dave Bailey
In article <5ilg0v$b...@elaine42.Stanford.EDU>,
Daniel Connelly <djco...@leland.Stanford.EDU> wrote:
>Dave, Dave, Dave......
>
>You can dump work into a treadmill, but not (in the infinite
>Earth-mass-limit, neglecting deformation of the road) on the road.
I am assuming the treadmill is motorized - a non-motorized
treadmill would not work because there would be only one
velocity at which the rider could ride for a given incline.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <gdvaughan-1...@depdir8.lbl.gov>,
Douglas Vaughan <gdva...@lbl.gov> wrote:
>
>I hope this issue has been settled, but if it has, I missed it. An
>inclined treadmill establishes an inertial reference frame in which gravity
>most assuredly does not vanish. Working against gravity in that reference
>frame (i.e, remaining stationary in the rest frame) involves work.
This violates energy conservation. How much work does it take
to remain stationary walking down an up escalator? The same as
it takes to remain stationary walking up a down escalator - none.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <5ilhd0$kl2$1...@phoenix.kfu.com>,
Mark F. Flynn <fly...@quack.kfu.com> wrote:
>I disagree. Galilean relativity is precisely what applies here, because
>you are comparing frames which are moving at a relative constant veclocity
><< c. The acceleration due to gravity as seen in the two frames of
>reference (one stationary to the surface of the earth, and one stationary
>to the treadmill surface) is identical. Therefore, for someone on the
>treadmill to move up it (in his frame of reference), he must do work
>to do their homework.
The only thing which lives in the moving frame is the treadmill
tread. The cyclist lives in the stationary frame. If the
cyclist is "climbing" in his frame, he rides right off the
top of the treadmill. Since the cyclist is not climbing in
his frame, he can't be doing any work against gravity.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <mpd4-11049...@cu-dialup-0071.cit.cornell.edu>,
M Dolenga <mp...@cornell.edu> wrote:
[Dave B bragging about pedigree]
[Dolenga says so what]
>Perhaps no extra work is being performed, and I'm not sufficiently
>familiar with the precise terminology, but it is obvious to me that
>someone riding a bike on an inclined treadmill "works" (in the colloquial
>sense) harder to maintain a certain speed as opposed to a non inclined
>treadmill. If you stop pedalling on an inclined treadmill (if it's long
>enough), you'll roll back off the treadmill and you'll feel a tug as your
>pedals try to turn in reverse; stop pedalling on a flat one, and you just
>stop. That would suggest that the "forces" working against you are
>greater on the inclined treadmill.
Conservation of energy dictates that it requires no work to keep
one's height constant in one's frame of reference. For example,
if you hang from a pull-up bar, you aren't doing any work to
stay suspended in the air. The instant you let go, you fall to
the ground, but this doesn't mean you were doing work to keep
from doing that.
I am not saying it is not perceptibly harder to ride the inclined
treadmill. I think it would be, because my pedal stroke is not
perfectly smooth and I'd bob up and down a bit.
--
Dave Bailey
dba...@leland.stanford.edu
Dave Bailey
In article <334E64...@m.cc.utah.edu>,
Brad Harris <brad....@m.cc.utah.edu> wrote:
>To further confirm my beliefs I went and tried
>walking up a down escelator (I figured since it was also work against
>gravity that it was similar enough for an analagy) if I judged my pace
>correctly I could stay in exactly the same place (relative to the rest
>of the mall) but to do so sure FELT like I was doing work, and to make
>sure I then tried the stairs that run paralel to the escelator and at
>the same pace it FELT the same.
Try walking down the up escalator. It is just as hard to stay
in place going down the up escalator as it is going up the down
escalator. So, the effort expended is just the inefficiency of
walking up/down an incline - but without either working against
gravity or having work done on you by gravity. I am not saying
it takes no effort - only that no work is done against gravity.
--
Dave Bailey
dba...@leland.stanford.edu
'Fumi' Fumitaka Hayashi
You know what I'm thinking?
That all this is a matter of confusion in the terms being used.
By treadmill, I'm thinking of the device that moves a belt converyor
backwards at a constant velocity. This would allow runners and cyclist to
ride without moving.
I think we all agree tilting a set of rollers (or trainers, for that
matter) would result in no more work. This is because the rollers (or
trainers) physically hold the bike in place.
Treadmills, on the other hand, are different.
Besides the empirical evidence that it's harder to run up a tilted
treadmill, the way I think of it is thus:
Once you get going on the treadmill, there is no acceleration. Just
constant velocity, and the amount of work you need to fight against
frictional losses in the bike's drivetrain and tire deformation.
If one were to start tilting the treadmill at this point, all of the
sudden, there is an added force pulling the cyclist backwards. This
results in a constant acceleration backwards. The cyclist now has to
apply his own constant acceleration to fight this one.
Doesn't this sound right?
+-------------------------------------------------------------------------+
I Fumitaka Hayashi "A penny saved is only one cent." I
I <hay...@u.washington.edu> http://weber.u.washington.edu/~hayashi I
I Aderem Lab - Dept. of Immunology - University of Washington I
+-------------------------------------------------------------------------+
Dave Blake
metc...@wheel.dcn.davis.ca.us says...
>The treadmill example is similar. If you don't go anywhere you're doing
>no work. Get a bike with an SRM and ride on a treadmill at x mph at y
>degrees inclination. Record your wattage. Do the same on a hill at the
>same grade and same speed. Correct for wind resistance and compare
>wattage. The wattage on the treadmill will not be zero, but it should be
>lower than the wattage while climbing an actual hill unless the treadmill
>has a resistance setting and you adjust it to a high enough level simulate
>real climbing. I'm sure that is exactly what Boardman did.
To stay in the same place on the treadmill requires that
the force tangential to the treadmill surface be equal
to m g sin(theta) - this is usually the force of
propulsion on the road. It cannot be higher or lower
or you will either roll up or down the treadmill. This
force can be exerted at any speed depending on the
resistance of the treadmill.
When you are climbing a REAL hill, you maintain that
force to stay in the same place. You exert more
force to propel yourself up the hill. You always
exert more force climbing that you exert on a
treadmill of equal grade. Always. Always. Always.
'Fumi' Fumitaka Hayashi
On 11 Apr 1997, Dave Bailey wrote:
(snipped)
>
> Conservation of energy dictates that it requires no work to keep
> one's height constant in one's frame of reference. For example,
> if you hang from a pull-up bar, you aren't doing any work to
> stay suspended in the air. The instant you let go, you fall to
> the ground, but this doesn't mean you were doing work to keep
> from doing that.
>
But by the same argument, it takes no energy at all to stay still on a
treadmill going 30mph.
Something wrong, right?
> I am not saying it is not perceptibly harder to ride the inclined
> treadmill. I think it would be, because my pedal stroke is not
> perfectly smooth and I'd bob up and down a bit.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
>
>
+-------------------------------------------------------------------------+
Dave Bailey
In article <334E3D...@ziplink.net>, Rob Hult <rob...@ziplink.net> wrote:
>First, is there any difference between running on a treadmill or cycling
>on one? Why do I ask? On a treadmill, why is it easier to run at an
>incline of 0 degrees while it is much harder (requires more work, watts,
>calories, etc.) to run at a positive incline? Is this simply 'body
>mechanics'? In other words, while running at an incline is my body
>simply in a more inefficient running position? My heart rate certainly
>goes up substantially and I 'feel' like I am running uphill, so what is
>going on? Is the treadmill machine really 'tricking' me that well?
In a physical sense, the situations are equivalent. I think that
it is harder to run at an incline on the treadmill because it is
less efficient, because each step you take requires you to lift
your leg from a lower point to a higher point before planting your
foot. Remember that, neglecting air resistance, it takes no work
to move at a constant velocity - so, for example, it should take
no work to run a 3 minute mile as long as you have the correct
initial velocity. But everybody knows this is not true because
the runner is a dissipative system, so the runner must do work
to balance the inefficiency of the running stride. On a bike,
this isn't true because you can just coast.
>Second, consider the design of the treadmill. Someone earlier mentioned
>'pushing against the belt'. Is this really what is happening? How does
>the belt resistance matter? I mean, some cheap treadmills have no
>motor, essentially you are the motor to move the tread. Others have a
>motor and a brake along with a belt speed control. What's going on
>here?
I think that the treadmill must be motorized such that the track
moves at whatever velocity you choose. Otherwise, there would
be only one velocity at which the rider would remain stationary
(assuming the rider's pedalling force was all that moved the
tread). So I have been assuming a motorized treadmill.
>If you are pedaling a bicycle on a stationary treadmill, you are not
>resisting gravity. Is this a true statement? If you are running on a
>treadmill, you are not resisting gravity. Is this a true statement?
>Are these conditions (running and cycling) the same?
The conditions are the same. You are resisting gravity in both
cases, but in a sense which requires no mechanical work - just as
hanging from a pullup bar involves no work against gravity.
--
Dave Bailey
dba...@leland.stanford.edu
Kevin Metcalfe
Dave Bailey (dba...@leland.Stanford.EDU) wrote:
:
: Conservation of energy dictates that it requires no work to keep
: one's height constant in one's frame of reference. For example,
: if you hang from a pull-up bar, you aren't doing any work to
: stay suspended in the air. The instant you let go, you fall to
: the ground, but this doesn't mean you were doing work to keep
: from doing that.
: I am not saying it is not perceptibly harder to ride the inclined
: treadmill. I think it would be, because my pedal stroke is not
: perfectly smooth and I'd bob up and down a bit.
I think that this nails it on the head. Most of the people arguing
against Dave are confusing "Work" in physics terms with "Work" in more
effort or higher heart rates. The hanging from a pull up bar is a perfect
example. It would be hard as hell to hang from a pull up bar for even ten
minutes, but from a physics point of view you are doing absolutely no
work.
The treadmill example is similar. If you don't go anywhere you're doing
no work. Get a bike with an SRM and ride on a treadmill at x mph at y
degrees inclination. Record your wattage. Do the same on a hill at the
same grade and same speed. Correct for wind resistance and compare
wattage. The wattage on the treadmill will not be zero, but it should be
lower than the wattage while climbing an actual hill unless the treadmill
has a resistance setting and you adjust it to a high enough level simulate
real climbing. I'm sure that is exactly what Boardman did.
--
Dave Bailey
In article
<Pine.A41.3.95b.97041...@dante30.u.washington.edu>,
>
>But by the same argument, it takes no energy at all to stay still on a
>treadmill going 30mph.
>
>Something wrong, right?
No - the above statement is correct if there is no friction in the
bearings of the wheels, no air drag on the spokes, and no rolling
resistance. And, of course, you have to do a small amount of work
to get the wheels rotating at the proper speed, but once you reach
that speed, then it takes no energy to stay there.
Of course, in the real world you need to overcome all the dissipative
forces I mentioned above.
--
Dave Bailey
dba...@leland.stanford.edu
David Casseres
In article <5ilrdi$rts$1...@mark.ucdavis.edu>, metc...@wheel.dcn.davis.ca.us
(Kevin Metcalfe) wrote:
> This has been amusing. Perhaps it would help if we did a credential
> check. I know that Dave Bailey has a PhD thesis in Physics waiting to be
> signed at Stanford University. I only took the three standard engineering
> physics classes and my measly brain can understand Dave's arguments.
> Personally, I'd really want to think long and hard before I got in a
> physics argument with a PhD physicist. But that's just me. We all know
> that Dave's got a twisted desire for Dove Bars, but that doesn't mean he
> doesn't know his physics.
If this is true, then he is a pathetic troller because his physics is just
plain wrong in the most obvious way, his arguments don't even stay on the
same point from one to the next, he won't answer questions based on his
theory, and he's deliberately rude.
Hey David! Yo, you out there listening? Give us your thesis advisor's
email address, eh?
--
Cheers,
David
David Casseres
In article <5im1p1$c...@itssrv1.ucsf.edu>,
dbl...@phy.ucsf.eduDELETETHISPART (Dave Blake) wrote:
> To stay in the same place on the treadmill requires that
> the force tangential to the treadmill surface be equal
> to m g sin(theta) - this is usually the force of
> propulsion on the road. It cannot be higher or lower
> or you will either roll up or down the treadmill. This
> force can be exerted at any speed depending on the
> resistance of the treadmill.
Yes! Yes! Yes! You're not totally ignorant after all! You got it
exactly right!
> When you are climbing a REAL hill, you maintain that
> force to stay in the same place. You exert more
> force to propel yourself up the hill. You always
> exert more force climbing that you exert on a
> treadmill of equal grade. Always. Always. Always.
No! No! No! You are totally ignorant after all! You got it exactly wrong!
(hint: Mr. Newton sez, f = ma.)
--
Cheers,
David
N. Peter Armitage
But seriously folks, Dave is right for treating most of you like
ignorant clods with the cerebral power of a tree frog. Anyone, who says
that a poor english bloke with TdF ambitions riding a treadmill at an
angle is working against gravity is ABSOLUTLEY FREAKIN" WRONG!!!!!!
Imagine the limit where you have infinite frictional resistance between
your tires and the riding surface, while the treadmill is in a position
approaching vertical. The only work that you will be doing in this
contorted posistion is the work to move your joints around in circles,
overcome rolling resistance, and moving your spokes through the air.
The thing holding you on the treadmill is NOT your forward motion. It
is the frictional force between your tires and the rolling surface.
That's the facts kids.
-Peter Armitage
Another Physics Phd.
Dave Blake
In article <casseres-110...@cassda.apple.com>, cass...@apple.com
says...
>
>> http://www.keck.ucsf.edu/~dblake/
>
>Checking out this web page, I see that you're in neurology, not physics.
>It figures. If I were you I would not try to lecture about physics to
>people who actually know some.
Don't assume anything, and do not listen to me or anyone
else on Usenet because of credentials - they are often
quite wrong. Listen to the arguments. I do not post
my credentials, nor do I expect people to believe
me because of them.
Pejorative comments aside, I have a new slant.
To stay in the same place on the treadmill requires that
the force tangential to the treadmill surface be equal
to m g sin(theta). The treadmill can roll, and you
must have the treadmill act on you - through its
resistance mechanism - to keep you stationary.
The size of the force is determined by the angle
of the slope and your mass.
When you are climbing a REAL hill, you maintain that
force to stay in the same place as well - but now
you can maintain that force in a trackstand. While
pedaling,you exert more additional force through the
tires to the road to propel yourself up the hill.
You always exert more force (through the tires to the road)
climbing that you exert on a treadmill of equal
grade. Always. Always. Always.
I hope this is clear enough. Please feel free
to point out all the errors in the reasoning. And
you can even give me a go back to Freshman physics
card - do not pass go or collect $200 - if
you feel it is warranted.
Rob Hult
Kevin Metcalfe wrote:
>
> I think that this nails it on the head. Most of the people arguing
> against Dave are confusing "Work" in physics terms with "Work" in more
> effort or higher heart rates. The hanging from a pull up bar is a perfect
> example. It would be hard as hell to hang from a pull up bar for even ten
> minutes, but from a physics point of view you are doing absolutely no
> work.
>
CASE #1. On a real hill grade, if you stop your bike, how much energy
does it require to remain stationary? Okay, it is difficult to balance,
and you must also apply a force to the pedals to keep the bike from
rolling back. Now, let's say you are flying up a steep grade at race
pace, and let's say the hill is steep and you are going fast. I don't
care what the numbers are, but let's say you are putting out 500 watts.
What percent of that 500 watts is used to keep the bike upright and
moving forward? Probably a small percent of the total 500 watts. Does
this make sense? I guess what I am trying to do is break down the
rider's output into the "riding forces" and the "hill climbing gravity
hoopla forces".
CASE#2. So now let's try and do the same thing on a treadmill. First
we'll set the treadmill at the same gradient as the real hill. Then
we'll turn it on. I'll pedal at the same cadence. Between a combo of
gear selection and treadmill speed I will adjust accordingly until I am
putting out 500 watts. Heck, if I can't get enough resistance I'll use
a brake on my rear wheel. But I will NOT change my cadence or the
incline. Okay, I did it.
What is different between 1 and 2? As far as I can tell, this is
exactly what Chris B. did. He doesn't care what the formulae on paper
say, but he knows he is at an incline and his cadence is the same, and
his power output is the same. For all we know he actually turned on a
fan and now the wind is blowing in his face, like on a real climb.
Is Chris Boardman a Physics witch doctor?
-ROb
Dave Bailey
In article <334EC9...@fortlewis.edu>,
Dan Arnold <arno...@fortlewis.edu> wrote:
>when a person RUNS on an inclined treadmill, you are actually working
>against gravity. you take a step forward. this puts you at a slightly
>higher elevation then when you started because of the incline in the
>treadmill. the treadmill then moves your foot back down to where you
>started because it is motorized and moving against you. you take the
>next step and regain that elevation, etc... this means that RUNNING on
>an incline does indeed do work against gravity and is the same as
>running up a hill. (although MUCH more boring!)
It depends on the integrated vertical motion of the center of mass.
Because it integrates to zero, the only allowed motion is periodic.
The amplitude determines the amount of work done against gravity,
which is what you are saying. However, I can imagine a situation
where I run very carefully and reduce the amplitude to a very small
quantity. Then, I don't actually lift my center of mass very
much per stride. In the limit that I don't lift it at all, which
is certainly possible, I do no work against gravity running up
an inclined treadmill. However, in running up a real hill I
cannot devise any technique which will allow me to move forward
without also raising the position of my center of mass. Therefore,
I do must do work against gravity in this case.
--
Dave Bailey
dba...@leland.stanford.edu
David Ryan
N. Peter Armitage wrote:
> Anyone, who says
> that a poor english bloke with TdF ambitions riding a treadmill at an
> angle is working against gravity is ABSOLUTLEY FREAKIN" WRONG!!!!!!
>
> The thing holding you on the treadmill is NOT your forward motion. It
> is the frictional force between your tires and the rolling surface.
Friction holds the bike and rider on the treadmill until the treadmill
rolls back far enough to drop them off. Assuming we are talking about
a non-motorized resistance unit, the following is occuring:
Gravity is using the mass of the bike and rider to work against
the friction within the treadmill system. The rider, by pedaling
increases the speed of the treadmill and increases friction until
it just balances gravity at which point the rider will stay in
place. For a given incline, determining the vector for the operation
of gravity, and a given resistance in the treadmill (coefficient of
friction) there is exactly one speed that will work, unless resistance
can be varied. Unvaried resistance would not be exactly like a
hill because the rider would be forced to maintain a constant speed
or dismount.
Work is a technical term that has no application here. It doesn't
matter whether the edge of the bike wheel drags you to the finish
or drags the finish to you (excluding the wind factor, of course).
The rider is working against gravity, but only indirectly by increasing
the friction in the system. But who cares? Either way, he's just
pedaling, turning cranks, chain & wheels.
What do the PhD's think of that?
***********************************************************************
* David e /NO NO PACK TOO TIGHT NO CORNER Fast *
* Ryan /<> /HILL oO@O)o_@ __e IS TOO e Forward *
* aka '>(x)/THAT'S <>\<>>\<> ,\<> SHARP ,)> o Racing *
* PACKMAN! (x)/TOO HIGH (*)((*))(*)x)/(x) ===== /' ->~\\ Team *
********************************c1997*****************************LouKY
Dave Bailey
In article <334EEC...@adware.com>,
David Ryan <David...@adware.com> wrote:
[..]
>What do the PhD's think of that?
It'll have to wait until Sunday night; we're going bike
racing for the weekend. Until then, I remain...
--
Dave Bailey
dba...@leland.stanford.edu
Tom James
On 11 Apr 1997, Dave Bailey wrote:
OK, I'm bailing out at this point - when I made my calculation I was
working on the assumption of a non-motorised treadmill. the velocity the
rider can ride is the same as he could ride up a normal hill. Having a
driven treadmill changes things completely. As a useful bit of
clarification, 1) would everyone who has weighed in so far like to state
whether they were making the calculations for a driven or undriven
treadmill, as this might well be the source of everyone's confusion, and
2) (more importantly) does anyone know whether the Boardman treadmill is
driven or undriven?
Tom
David Ryan
David Ryan wrote:
Friction holds the bike and rider on the treadmill until the treadmill
rolls back far enough to drop them off. Assuming we are talking about
a motorized resistance unit, the following is occuring:
I wanted to completely separate the motorized concept from a
non-motorized one which has been part of the confusion.
If the treadmill is motorized at a constant pace and there is no
slippage of the tread due to the rider's weight, then the only
energy needed is the same energy it would take to trackstand
on an incline, overcoming gravity only to the extent of balancing
it, since the treadmill is supplying all the "forward" and "upward"
movement by moving "down" beneath the rider. The difference
in the rider motion is that on a non-moving hill, the energy is
expended in a static manner and on the treadmill, it is necesary
to move the tire at the same speed as the treadmill belt. Except
for the friction in the bike and rider, which at speed could be
considerable but unrelated to hill-climbing or overcoming gravity,
the energy required is the same.
Do I get a PhD?
Brad Harris
Dang it, I knew that I was doing something wrong, I didn't have either
an infinitely long treadmill, or a vacuum (good thing I guess for I yet
live... but really need to do something about the dirt in my carpet)
-bjh
Dave Bailey
In article <334E99...@ziplink.net>, Rob Hult <rob...@ziplink.net>
wrote:
>upright and moving forward? Probably a small percent of the total 500
>watts.
Right.
>CASE#2. So now let's try and do the same thing on a treadmill. [..]
>Between a combo of gear selection and treadmill speed I will adjust
>accordingly until I am putting out 500 watts. [..]
>
>What is difference between 1 and 2? [..]
As far as the rider is concerned, there is no difference - and it
is correct that the 500 watts must be used fighting something
besides gravity (e.g. a rear brake).
--
Dave Bailey
dba...@leland.stanford.edu
Dana E. Lehr
This has gone on long enough.
Unfortunately our colleague Dave Bailey has been using bad physics to bash
people with better intuition who just can't keep up with the jargon.
On behalf of physicists, I apologize.
When you stand on an inclined treadmill (stationary for now), a component
of the force which gravity exerts on you is directed down the slope of the
treadmill. As long as the belt doesn't move, no work is done, as in the
case of the car parked on a steep slope, hotshot.
But once the belt is permitted to move, in the direction you are pushing
it, you are doing work on the belt. You might suppose that you could hook
up a treadmill to a magic motor that keeps a constant speed. Even if this
is the case, you are doing work on that motor, and it must dissipate that
energy in order to keep the speed constant.
The misconception here concerns the way in which you hold yourself up against
gravity, and the physics definition of 'work'. If you hang from a stationary
object (like a pull-up bar) or stand on a stationary object (the floor), no
work is done. But if you try to climb up a rope which is paid out at the
speed 'v' at which you climb it, you do work against whatever machine pays
out the rope, and your output power is mg*v. If you 'climb' a stairmaster,
which constantly gives way under your feet as you press down, the same is
the case.
The upshot is that though you do no work against gravity in the traditional
sense (i.e. you are not getting higher off the ground), gravity makes you
do work against the treadmill belt at a rate which is EXACTLY THE SAME
as if you were biking up the hill of that slope. Of course, this is for
the bike on the free treadmill only, no rollers.
If Dave Bailey continues to abuse good intuition with bad physics in this
space, I know his thesis advisor's e-mail :-)
Mohan Rajagopal (mo...@astro.stanford.edu)
Stanford Astrophysics.
PS. For the record, Dave Bailey rides his bike much faster than I do.
PPS. Dana Lehr never really wanted me to post this.
'Fumi' Fumitaka Hayashi
So, I think we actually all agree.
1. tilting rollers does nothing
2. tilting a treadmill makes it harder
(but not as hard as real hill climbing)
Are we all in agreement?
Dave Bailey
In article <5imsd0$8...@spacetime.Stanford.EDU>,
Dana E. Lehr [actually Mohan] <dana...@spacetime.Stanford.EDU> wrote:
>The misconception here concerns the way in which you hold yourself up against
>gravity, and the physics definition of 'work'. If you hang from a stationary
>object (like a pull-up bar) or stand on a stationary object (the floor), no
>work is done. But if you try to climb up a rope which is paid out at the
>speed 'v' at which you climb it, you do work against whatever machine pays
>out the rope, and your output power is mg*v.
And if you don't try to climb the rope, the same exact amount of work
is done on the machine. However, if you aren't trying to climb the
rope, you aren't doing any work - so your output power is zero. The
work is done _by_ gravity _on_ the machine - not by you. Thusly, I
do refute thee, O Mohan of Across the Hall.
--
Dave Bailey
dba...@leland.stanford.edu
D. Lee
I was going to stay out of this, but ......
Dave Bailey wrote:
> snip
> >Or you can try walking or running on a treadmill tilted up -- I did that
> >the last time I got a heart test. I guarantee you I was working against
> >gravity.
>
> You were working against gravity because you had to lift your leg
> each time you took a step. But once you plant your foot, the
> treadmill takes it down to where you need to lift it again.
> Running up an inclined surface, you need to take a real stride
> (meaning, you need to lift your center of mass up the hill a
> little bit) before you can lift your foot. This is the difference.
>snip
It has been said before, but I somehow feel compelled to say it again.
You have to take a real stride or down the treadmill you go. Your frame
of reference is the surface of of the treadmill. The fact that
there is relative motion between the treadmill surface and the surface
of the earth is irrelevant. In fact the view that the surface of
treamill isn't moving and the surface of the earth is rushing by you
is perfectly valid. All you know is your frame of reference and you
are gaining potential energy relative to that frame of reference. I'm
really baffled by your insistence, after all the explanations, that
this isn't true - particularly after your professed understanding
of physics.
> It is a consequence of the biomechanical inefficiency of running.
>
> >If you still think the tilt doesn't make you work against gravity, ask
> >yourself what would happen if you were on a bike on a tilted treadmill and
> >you didn't pedal. Do you think perhaps you might roll backward?
>
> Pop quiz, hotshot: You park your car on a steep hill and put on
> the emergency brake. You get out of your car and see that it is
> not sliding down the hill. What do you do? What do you do?
>
> The answer: You suddenly realize that the car is not doing any
> work against gravity, that you were wrong, and that you shouldn't
> post any more to this thread.
>
> --
> Dave Bailey
> dba...@leland.stanford.edu
OK, hotshot. You park your car on a treamill and put on the emergency
brake. You get out of your car, stand on the ground outside the
the treadmill, and watch your car (not doing any work) go flying down
the treadmill. I wonder if it might take some work to keep that from
happening? The answer is yes, but it would seem that it is less energy
than it is going to take to get you to understand this.
dave
David Casseres
In article <5ilr67$r...@epic9.Stanford.EDU>, dba...@leland.Stanford.EDU
(Dave Bailey) wrote:
> In article <casseres-110...@cassda.apple.com>,
> David Casseres <cass...@apple.com> wrote:
> >No, I asked you first. You tell me if the bike will roll backward on the
> >tilted treadmill, with no pedaling, and THEN if you do it politely I'll
> >explain to you the difference between pedaling a bike on a moving
> >treadmill and setting a parking brake on a non-moving hill.
>
> There can be no difference because if there were, it would violate
> conservation of energy.
No, no, David. No biscuit. You didn't answer the question, you just
repeated your erroneous assertion. Read the question again, and if you
answer it politely, we'll continue.
--
Cheers,
David
Jamie Mikami
Now you get it mike, it is the movement up or down that equates out to
the forces everyone is talking about. With out moving up or down you
have the same potential energy. Imagine standing on a hill, are you
fighting gravity, no not until you roll down the hill or walk up it. I
really can't beleive how many people are missing Dave's point. Even if
he is a Stanford student, he can still make some sense out of all them
years in school. Just check out how many years he spent at the
Collegiate level.
Jamie (Not a Physics PhD, but I took a few credits)
M Dolenga wrote:
>
> In article <5ik63d$e...@epic9.Stanford.EDU>, dba...@leland.Stanford.EDU
> (Dave Bailey) wrote:
>
> > Suppose I run the treadmill in reverse and ride it facing down
> > the incline. If I stop pedalling, I'll be whisked off the top
> > immediately. But, if I pedal at just the right speed, I can
> > remain stationary with respect to the surroundings. The static
> > friction between the bicycle tires and the treadmill surface
> > makes this possible, just as it does in the forward case.
>
> Suppose you have an infinitely long treadmill, so coming off either end
> isn't an issue. Suppose, also, that this is all happening in a vacuum.
>
> If your bike is pointed down the treadmill, you will eventually be heading
> downhill very fast. No matter how fast you run the treadmill, no matter
> how gradual the incline, you are accelerating forward because of gravity.
> You will accelerate beyond the speed of the treadmill surface and begin
> heading downhill. You cannot hold your position unless you use your
> brake.
>
> Mike