Running outside or treadmill

"mr.b" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

On Tue, 24 Apr 2007 19:11:26 +0000, Elf wrote:

<snip>

morph away you arrogant worm...there's lots of room in my bozo-bin

You're being a dumbass. I know because I didn't understand it either the
first 2 or 3 times. Unfortunately for you Mr. Elf is correct this time. Read
the thread and learn.

-Tony

Elflord said:

Look, if you stick your nose up a duck's butt, you can't see its
beak anymore, but it's still a duck. But it's much harder to explain why
it's a duck.

I like this one a lot :-)

Quoted message said:

The earth is not imobile. EVERYTHING is moving. The earth is not a gold standard
of what is or isn't "really moving". Once you grasp this, it becomes more clear.

I did grasp it, but I figured I'd handled that by having the Earth
involved in both systems. Unfortunately I kind of forgot the fact that the
treadmill isn't trying very hard to move across the Earth, except by
transferring the force of my overstride :-P

The penny just dropped.

Quoted message said:

You haven't shown that the power source acts in the runners favour. The only
thing the power source does is (1) maintain speed of the belt in absence of
a runner, (2) maintain speed of the belt even when the silly runner brakes.
But the earth does the same thing -- it maintains its velocity regardless of how
hard the runner tries to stop it.

I forgot that the Earth is putting a damned sight more energy into the
deal than a treadmill. OK I got it. I failed the Cambridge entrance
interviews cos of my [censored] physics (they asked me about gyroscopes, God
bless 'em) :->

Quoted message said:

Quoted message said:

Explain to me how silly I am (do not use stats :-P)

I could have made a much simpler argument using Bayes law.

:-> your wit was particularly honed yesterday; you got laid or what? :->

steve common said:

I did grasp it, but I figured I'd handled that by having the Earth
involved in both systems. Unfortunately I kind of forgot the fact that the
treadmill isn't trying very hard to move across the Earth, except by
transferring the force of my overstride :-P

It gets confusing pretty quickly once you bring extra things into the picture.

Tony confused himself pretty well by looking at how potential energy factors into
the picture when using the incline.

Quoted message said:

Quoted message said:

Quoted message said:

Explain to me how silly I am (do not use stats :-P)

I could have made a much simpler argument using Bayes law.

:-> your wit was particularly honed yesterday; you got laid or what? :->

um, I had a really good weekend. Let's just leave it at that.

Cheers,
--
Elflord

In article <[email hidden]>,

Elflord said:

Robert Grumbine said:

The last two would lead towards your results. For the pace control,
find someone who can predictably jog on the road the same pace
that you do on the treadmill, and run with him. For the proper running
on the treadmill, do not let the belt carry your foot back; be
running such that your foot is never being pushed or pulled by
the belt.

This is just the equivalent of "braking" when running on the road (the road
abruptly "grabs" the runners foot and "carries" it back). But why would anyone
brake any more on the treadmill ? And why would doing that make a treadmill run
any faster ?

No, it isn't. But to see what I'm talking about, you need either the
experience or a lengthy physics-oriented description. Since you don't
have the former (and I do, plus the physics) ... here goes ...

From your followups, you're considering a person who is either running
perfectly, or is doing overstride braking. Neither is involved here.
'Perfectly' in this case means that the runner's leg is moving backwards
(relative to his center of mass) at exactly his running speed. If he
does this on a treadmill -- and continues the follow-through correctly
as well -- then the situation is identical to if he's doing the same
thing on the road. The road may as well be a treadmill, and vice versa,
and either could be made a rolling log floating in water, with no effect
on the energetics.

But that's not what is involved in the distinction I'm talking about,
or that others have described.

Consider first the runner landing his foot on ground (v. treadmill)
with the foot moving back (relative to center of mass) more slowly than
his center of mass is moving forward, and doesn't apply extra force
to accelerate its motion relative to his center of mass back to running
speed. He slows down. The earth isn't supplying any forces to accelerate
him horizontally.

Do this on a treadmill, however, and the belt certainly does supply
some force to accelerate your planted foot back to running speed (relative,
again, to your center of mass). It won't succeed if your foot is heading
back much too slowly, but it can do so if it's not a great deal off.

That input saves you some work. Not the entire workload. But there's
a fair amount work in a steady jog that is just getting the leg to
swing back and forth at the right speed, and the 'mill can supply a portion
of the energy for the backswing.

--
Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links.
Sagredo (Galileo Galilei) "You present these recondite matters with too much
evidence and ease; this great facility makes them less appreciated than they
would be had they been presented in a more abstruse manner." Two New Sciences

"Robert Grumbine" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <[email hidden]>,

Elflord said:

Robert Grumbine said:

The last two would lead towards your results. For the pace control,
find someone who can predictably jog on the road the same pace
that you do on the treadmill, and run with him. For the proper running
on the treadmill, do not let the belt carry your foot back; be
running such that your foot is never being pushed or pulled by
the belt.

This is just the equivalent of "braking" when running on the road (the
road
abruptly "grabs" the runners foot and "carries" it back). But why would
anyone
brake any more on the treadmill ? And why would doing that make a
treadmill run
any faster ?

No, it isn't. But to see what I'm talking about, you need either the
experience or a lengthy physics-oriented description. Since you don't
have the former (and I do, plus the physics) ... here goes ...

From your followups, you're considering a person who is either running
perfectly, or is doing overstride braking. Neither is involved here.
'Perfectly' in this case means that the runner's leg is moving backwards
(relative to his center of mass) at exactly his running speed. If he
does this on a treadmill -- and continues the follow-through correctly
as well -- then the situation is identical to if he's doing the same
thing on the road. The road may as well be a treadmill, and vice versa,
and either could be made a rolling log floating in water, with no effect
on the energetics.

But that's not what is involved in the distinction I'm talking about,
or that others have described.

Consider first the runner landing his foot on ground (v. treadmill)
with the foot moving back (relative to center of mass) more slowly than
his center of mass is moving forward, and doesn't apply extra force
to accelerate its motion relative to his center of mass back to running
speed. He slows down. The earth isn't supplying any forces to accelerate
him horizontally.

Do this on a treadmill, however, and the belt certainly does supply
some force to accelerate your planted foot back to running speed
(relative,
again, to your center of mass). It won't succeed if your foot is heading
back much too slowly, but it can do so if it's not a great deal off.

That input saves you some work. Not the entire workload. But there's
a fair amount work in a steady jog that is just getting the leg to
swing back and forth at the right speed, and the 'mill can supply a
portion
of the energy for the backswing.

Hmm. I want to believe you but I was already convinced that the physics of
running on a treadmill was the same as running on the road, but for the wind
resistance. Wait a minute, once your body is in motion over the earth, it
stays in motion, according to Newton's first law, and is slowed only by 2
kinds of friction: air resistance and contacting the earth. Why isn't the
weight-bearing friction of the foot on the treadmill the same as the
weight-bearing friction of the foot on the road? The foot is also 'pulled
backwards' by the friction of contacting the earth while in motion running
on the road.

-Tony

Quoted message said:

--
Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur
activities notes and links.
Sagredo (Galileo Galilei) "You present these recondite matters with too
much
evidence and ease; this great facility makes them less appreciated than
they
would be had they been presented in a more abstruse manner." Two New
Sciences

[email hidden] (Robert Grumbine) wrote

Quoted message said:

Consider first the runner landing his foot on ground (v.
treadmill) with the foot moving back (relative to center of
mass) more slowly than his center of mass is moving forward,

Good God, how is that possible? How can your planted foot be
moving backward with respect to your center of mass any slower than
your center of mass? Your planted foot is stationary with respect
to the ground and the only motion is supplied by your center of
mass (torso) moving forward past the planted foot, so the "speed"
of your foot relative to the center of your mass will always be
identical to the "speed" of your center of mass.

To all those trying to demonstrate how the treadmill really "helps
you run" because it has a motorized belt, please give it up.
You'll have as much luck as trying to demonstrate how cold fusion
really helps you generate free electricity.

Robert Grumbine said:

Consider first the runner landing his foot on ground (v. treadmill)
with the foot moving back (relative to center of mass) more slowly than
his center of mass is moving forward,

This is just an instance of braking though, isn't it ? It might not be abrupt
braking, but it's still braking.

Quoted message said:

and doesn't apply extra force
to accelerate its motion relative to his center of mass back to running
speed. He slows down. The earth isn't supplying any forces to accelerate
him horizontally.

Do this on a treadmill, however, and the belt certainly does supply
some force to accelerate your planted foot back to running speed (relative,
again, to your center of mass). It won't succeed if your foot is heading
back much too slowly, but it can do so if it's not a great deal off.

OK, so if I understand you, you're saying that

(1) your foot hits the belt,
(2) the belt slows down a bit, but
(3) the motor accelerates the belt back up to speed (with your foot still on
the belt)

But in the case of the earth, unlike the belt, it's not going to slow down at
all (not a whole lot anyway). So it doesn't need to speed up. When your foot
hits the earth's surface, it's going to accelerate, just like the way stuff
accelerates when it hits a fan.

Cheers,
--
Elflord

Tony S. said:

You're being a dumbass. I know because I didn't understand it either the
first 2 or 3 times. Unfortunately for you Mr. Elf is correct this time.
Read the thread and learn.

Mr. Elf demonstrates the netiquette of a 14 year old with an
attitude problem. Anyone who starts in with "shut up idiot" has nothing
to teach me about anything.

"mr.b" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Tony S. said:

You're being a dumbass. I know because I didn't understand it either the
first 2 or 3 times. Unfortunately for you Mr. Elf is correct this time.
Read the thread and learn.

Mr. Elf demonstrates the netiquette of a 14 year old with an
attitude problem. Anyone who starts in with "shut up idiot" has nothing
to teach me about anything.

He's run a lot but he is a lousy teacher. If you get past the attitude of a
13 year-old pms girl after drinking a quart of red bull, you can learn
something.

In article <qCIXh.8477$oo5.2446@trndny06>,

Tony S. said:

"Robert Grumbine" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <[email hidden]>,

Elflord said:

On 2007-04-24, Robert Grumbine <[email hidden]> wrote:

> The last two would lead towards your results. For the pace control,
> find someone who can predictably jog on the road the same pace
> that you do on the treadmill, and run with him. For the proper running
> on the treadmill, do not let the belt carry your foot back; be
> running such that your foot is never being pushed or pulled by
> the belt.

This is just the equivalent of "braking" when running on the road (the
road
abruptly "grabs" the runners foot and "carries" it back). But why would
anyone
brake any more on the treadmill ? And why would doing that make a
treadmill run
any faster ?

No, it isn't. But to see what I'm talking about, you need either the
experience or a lengthy physics-oriented description. Since you don't
have the former (and I do, plus the physics) ... here goes ...

From your followups, you're considering a person who is either running
perfectly, or is doing overstride braking. Neither is involved here.
'Perfectly' in this case means that the runner's leg is moving backwards
(relative to his center of mass) at exactly his running speed. If he
does this on a treadmill -- and continues the follow-through correctly
as well -- then the situation is identical to if he's doing the same
thing on the road. The road may as well be a treadmill, and vice versa,
and either could be made a rolling log floating in water, with no effect
on the energetics.

But that's not what is involved in the distinction I'm talking about,
or that others have described.

Consider first the runner landing his foot on ground (v. treadmill)
with the foot moving back (relative to center of mass) more slowly than
his center of mass is moving forward, and doesn't apply extra force
to accelerate its motion relative to his center of mass back to running
speed. He slows down. The earth isn't supplying any forces to accelerate
him horizontally.

Do this on a treadmill, however, and the belt certainly does supply
some force to accelerate your planted foot back to running speed
(relative,
again, to your center of mass). It won't succeed if your foot is heading
back much too slowly, but it can do so if it's not a great deal off.

That input saves you some work. Not the entire workload. But there's
a fair amount work in a steady jog that is just getting the leg to
swing back and forth at the right speed, and the 'mill can supply a
portion
of the energy for the backswing.

Hmm. I want to believe you but I was already convinced that the physics of
running on a treadmill was the same as running on the road, but for the wind
resistance. Wait a minute, once your body is in motion over the earth, it
stays in motion, according to Newton's first law, and is slowed only by 2
kinds of friction: air resistance and contacting the earth. Why isn't the
weight-bearing friction of the foot on the treadmill the same as the
weight-bearing friction of the foot on the road? The foot is also 'pulled
backwards' by the friction of contacting the earth while in motion running
on the road.

Let's look again at your phrase 'pulled backwards'. _Is_ the foot
'pulled backwards' in either situation?

Run along the earth. When you put your foot down, it stays exactly
where it was, in the earth-coordinate frame. Any 'pulled backwards'
is an illusion from thinking of your center of mass as a stationary
reference point (thence giving the notion of the foot being 'pulled back'😉.
The work done by the earth against your foot is F*d -- but d is exactly
zero, so the work is also exactly zero. This is true regardless of
the speed your foot is moving at touchdown relative to your center of
mass -- once down, it stays put.

Now run on a treadmill. (Better, watch someone else run on a
treadmill.) Your foot lands squarely under your hip (center of mass).
When you lift it off the treadmill surface, however, it is somewhere
behind your hip (same relative position as on land). So d is distinctly
_not_ zero.

_If_ you're running with the foot speed equal to the treadmill speed,
then the treadmill applies no force to your foot (no need to accelerate
it to running speed, F = ma, but a is zero). So, again, the work is
zero.

_But_ if your foot is moving too slowly, the treadmill accelerates
it to running speed. F is not zero, now. d is never zero on treadmill
running. So, in this case, the treadmill does F*d work on your leg.
This is work that you don't have to do yourself, so the treadmill is
saving you a bit.

If you're running properly, the treadmill belt never accelerates
your foot, so does no work and we're back to the ideal situation of
having minimal difference between treadmill and road. (Wind resistence,
surface flexibility, etc. still differ, but can be compensated for
and/or are small effects.) But if you're running improperly, it is
indeed the case that the treadmill does work for you that you otherwise
do on the road, making the road slower running.

--
Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links.
Sagredo (Galileo Galilei) "You present these recondite matters with too much
evidence and ease; this great facility makes them less appreciated than they
would be had they been presented in a more abstruse manner." Two New Sciences

"Salim" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

"mr.b" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Tony S. said:

You're being a dumbass. I know because I didn't understand it either the
first 2 or 3 times. Unfortunately for you Mr. Elf is correct this time.
Read the thread and learn.

Mr. Elf demonstrates the netiquette of a 14 year old with an
attitude problem. Anyone who starts in with "shut up idiot" has nothing
to teach me about anything.

Yea, at times I've been called names, but it's been worth it for what I
learned here about training science that I was too lazy to dig out of a book
or research on my own initially. For the most part there are several people
with pretty solid training knowledge and experience in here, and Mr. Elf is
one of them, though he is rude sometimes, as most people can be.

Aside from the treadmill fallacy, I've learned that the conconi test was
[censored], and that the whole body of work based on lactate threshold is iffy
at best. Lactic acid isn't what causes the threshold. I also had to overcome
my own ideas about training and what I thought was best for my body as a
master. Everything I was told in the beginning about training more
frequently with a higher overall volume was indeed better for me as a runner
than my then less frequent longer runs.

Attribute the 'dumbass' remark to late night, and to a habit of reading
rec.bicycling.racing too much!

Quoted message said:

He's run a lot but he is a lousy teacher. If you get past the attitude of
a 13 year-old pms girl after drinking a quart of red bull, you can learn
something.

Good line 😉

-Tony

Robert Grumbine said:

Run along the earth. When you put your foot down, it stays exactly
where it was, in the earth-coordinate frame. Any 'pulled backwards'
is an illusion from thinking of your center of mass as a stationary
reference point (thence giving the notion of the foot being 'pulled back'😉.
The work done by the earth against your foot is F*d -- but d is exactly
zero, so the work is also exactly zero. This is true regardless of
the speed your foot is moving at touchdown relative to your center of
mass -- once down, it stays put.

Now run on a treadmill. (Better, watch someone else run on a
treadmill.) Your foot lands squarely under your hip (center of mass).
When you lift it off the treadmill surface, however, it is somewhere
behind your hip (same relative position as on land). So d is distinctly
_not_ zero.

_If_ you're running with the foot speed equal to the treadmill speed,
then the treadmill applies no force to your foot (no need to accelerate
it to running speed, F = ma, but a is zero). So, again, the work is
zero.

_But_ if your foot is moving too slowly, the treadmill accelerates
it to running speed.

what happens if your foot is moving too slowly when it touches down on
the surface of the earth ? It gets accelerated back to running speed too,
right ?

Imagine what would happen if you carelessly stepped out of a moving train
unless you're fortunate enough to have tried it, as I have, in which case
you will remember the experience well enough not to rely on imagination!

When you step out of that train, your feet are "moving too slowly" -- when
you step out, do they or do they not get "accelerated" to the speed at which
the earth is moving ?

Cheers,
--
Elflord

[email hidden] (Robert Grumbine) wrote

Quoted message said:

_But_ if your foot is moving too slowly, the treadmill
accelerates
it to running speed. F is not zero, now. d is never zero on
treadmill running. So, in this case, the treadmill does F*d
work on your leg. This is work that you don't have to do
yourself,

That is absolutely true -- but only for the fraction of a second it
takes for the treadmill to accelerate your foot, your leg, your
torso and your head to "running speed" and shoot you out the back
like a cannon ball. Try it sometime.

There is only one way to run on a treadmill -- properly. If your
foot speed does not exactly match the speed of the belt, you will
either crash through the control panel in front of you, or be shot
out the back.

Assuming the OP is not a troll, there is only one way he could
perceive such a difference in effort between running on a treadmill
and on the road. A common mistake many beginners make is to hang
on to the rails on the treadmill. If you brace yourself using the
leverage of your upper body skeleton, you are essentially
unweighing your body. But if you do that then it's not the same as
running, you're just moonwalking your way over the moving belt.
And that effort will be significantly lower than in true,
unsupported running -- on or off the treadmill.

In article <[email hidden]>,

Elflord said:

Robert Grumbine said:

Run along the earth. When you put your foot down, it stays exactly
where it was, in the earth-coordinate frame. Any 'pulled backwards'
is an illusion from thinking of your center of mass as a stationary
reference point (thence giving the notion of the foot being 'pulled back'😉.
The work done by the earth against your foot is F*d -- but d is exactly
zero, so the work is also exactly zero. This is true regardless of
the speed your foot is moving at touchdown relative to your center of
mass -- once down, it stays put.

Now run on a treadmill. (Better, watch someone else run on a
treadmill.) Your foot lands squarely under your hip (center of mass).
When you lift it off the treadmill surface, however, it is somewhere
behind your hip (same relative position as on land). So d is distinctly
_not_ zero.

_If_ you're running with the foot speed equal to the treadmill speed,
then the treadmill applies no force to your foot (no need to accelerate
it to running speed, F = ma, but a is zero). So, again, the work is
zero.

_But_ if your foot is moving too slowly, the treadmill accelerates
it to running speed.

what happens if your foot is moving too slowly when it touches down on
the surface of the earth ? It gets accelerated back to running speed too,
right ?

Not by the earth. Either you supply the energy, or you slow down.
Seriously -- one of the ways I slow down at the end of the run is
exactly this method. I let my foot come through more slowly than
running speed, _not_ overstride breaking, and simply don't supply
the push to re-supply the loss. The earth is certainly not doing
work to keep me going at that initial speed.

Quoted message said:

Imagine what would happen if you carelessly stepped out of a moving train
unless you're fortunate enough to have tried it, as I have, in which case
you will remember the experience well enough not to rely on imagination!

When you step out of that train, your feet are "moving too slowly" -- when
you step out, do they or do they not get "accelerated" to the speed at which
the earth is moving ?

Not at all comparable -- you don't plant on the train jump, you roll.

F = ma
W = Fd

Show me where the nonzero d occurs while your foot is in contact
with the ground, at the same time as the ground provides a force in
the horizontal, and then I'll believe that W is not zero. I've
shown d not equal to zero for treadmill case, and shown how it is
that F can (but may not be) also be nonzero on the treadmill.

--
Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links.
Sagredo (Galileo Galilei) "You present these recondite matters with too much
evidence and ease; this great facility makes them less appreciated than they
would be had they been presented in a more abstruse manner." Two New Sciences

Robert Grumbine said:

Not by the earth. Either you supply the energy, or you slow down.

I don't believe this is correct. If you are moving at X miles and hour,
and your foot hits the ground (at Xmph relative to the ground), it will
pretty quickly accelerate to ground speed, *and* you don't need to supply
energy for this to happen.

Quoted message said:

Seriously -- one of the ways I slow down at the end of the run is
exactly this method. I let my foot come through more slowly than
running speed, _not_ overstride breaking, and simply don't supply
the push to re-supply the loss. The earth is certainly not doing
work to keep me going at that initial speed.

You can do this on a treadmill too. Move your feet at a slower pace, and
the treadmill will quite cheerfully shoot you out the back rather than
do the work for you.

Quoted message said:

Not at all comparable -- you don't plant on the train jump, you roll.

I don't understand the distinction. I do understand that regardless of what
you do to your feet on the train jump, your foot will very quickly get
accelerated to ground speed.

Quoted message said:

F = ma
W = Fd

I'm reluctant to accept the second equation, because it is only true for
very specific assumptions about the force and distance involved (work done
against a conservative force), I'm not convinced this is a proper application.
But, I can and will still argue that the treadmill and earth are the same.

Quoted message said:

Show me where the nonzero d occurs while your foot is in contact
with the ground, at the same time as the ground provides a force in
the horizontal, and then I'll believe that W is not zero. I've
shown d not equal to zero for treadmill case, and shown how it is
that F can (but may not be) also be nonzero on the treadmill.

I think you're allowing yourself to be bamboozled by the differing frame
of reference. In both cases, you're using the earths frame of reference.

In the *runners* reference frame, d is non-zero in the earth picture.

Cheers,
--
Elflord

In article <[email hidden]>,

Elflord said:

Robert Grumbine said:

Not by the earth. Either you supply the energy, or you slow down.

I don't believe this is correct. If you are moving at X miles and hour,
and your foot hits the ground (at Xmph relative to the ground), it will
pretty quickly accelerate to ground speed, *and* you don't need to supply
energy for this to happen.

My knees say otherwise, as does my experience running on sheer ice
near the freezing point.

Quoted message said:

Quoted message said:

F = ma
W = Fd

I'm reluctant to accept the second equation, because it is only true for
very specific assumptions about the force and distance involved (work done
against a conservative force), I'm not convinced this is a proper application.
But, I can and will still argue that the treadmill and earth are the same.

If you're going to ignore the math (feel free to bring in math
or physics you think are more appropriate) then we're done.

But W = Fd works regardless of whether you're working with conservative
forces or not. Shove a box across a real (but level) floor, you
will do work that this recognizes. F = the frictional force,
d = distance you shove the box. It's incomplete, in that there are
other kinds of work possible. But you're arguing that there
is indeed an F and a d, you just don't work it out.

Quoted message said:

Quoted message said:

Show me where the nonzero d occurs while your foot is in contact
with the ground, at the same time as the ground provides a force in
the horizontal, and then I'll believe that W is not zero. I've
shown d not equal to zero for treadmill case, and shown how it is
that F can (but may not be) also be nonzero on the treadmill.

I think you're allowing yourself to be bamboozled by the differing frame
of reference. In both cases, you're using the earths frame of reference.

In the *runners* reference frame, d is non-zero in the earth picture.

Since you're not going to do the math, you're not going to
un-bamboozle anything.

--
Robert Grumbine http://www.radix.net/~bobg/ Science faqs and amateur activities notes and links.
Sagredo (Galileo Galilei) "You present these recondite matters with too much
evidence and ease; this great facility makes them less appreciated than they
would be had they been presented in a more abstruse manner." Two New Sciences

Robert Grumbine said:

But W = Fd works regardless of whether you're working with conservative
forces or not.

OK, I played around with it, it checks out OK. Been a while since I've
done this.

As I pointed out, I believe there really is a non-zero 'd' in both cases.

Quoted message said:

Since you're not going to do the math, you're not going to
un-bamboozle anything.

I found a non-zero 'd' for you, you can take it from there. You're welcome (-;

Cheers,
--
Elflord

MarkH said:

You'll have as much luck as trying to demonstrate how cold fusion
really helps you generate free electricity.

Oh dang! Does this mean the 200$ I sent to that Nigerian scientist who
emailed me about his invention won't help make a better world?

rick++ said:

The treadmill is not running, thats why.

I'll have to remember to turn it on next time, maybe it will be easier.

Elflord said:

The earth is not imobile. EVERYTHING is moving. The earth is not a gold standard
of what is or isn't "really moving". Once you grasp this, it becomes more clear.

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Quoted message said:

The earth is just
like a super-duper good treadmill belt that is completely resistant to your
attempts to perturb its perfectly uniform rotation.

maybe i'm being a jerk, but i'm posting this anyway: for those having
problems understanding this, and to make sure i've got it right (it
takes me about an hour of hard thinking to force my old brain to
understand this kind of stuff), it might help to imagine running on a
big extra wide 100 mile long treadmill that has lines scored across it
every mile. the treadmill is located in a 100 mile long exercise room
with completely featureless walls with unseen fans blowing the air
random directions. this is a slightly fanciful setup, but i think most
would agree it still counts in every way as a treadmill.

ok, now suppose you are standing on the middle of it somewhere, so that
you can't see either end. how will you be able to tell that it's moving?
or what direction it's going? to my mind, there will be no way to
tell. if you stand still, you will cover no distance on the treadmill.
and if you run for an hour - in either direction - on the 100 mile long
treadmill at your trusty cruising pace, say 6 miles/hour, you will cover
6 miles on the treadmill (using the mile lines as a way to keep track).
in the middle of a 100 mile long treadmill there is no way to even
know how fast the machine is set to run at, you can only tell how fast
you are moving relative to it's surface.

if this isn't the case, then donovan is right and people should always
try to run in the most favorable direction relative to the earth's spin
when trying to set a personal best 🙂

Quoted message said:

Quoted message said:

Explain to me how silly I am (do not use stats :-P)

I could have made a much simpler argument using Bayes law.

don't you call steve mean names!
😉
h