Re: ISIS getting dropped?

In article <[email hidden]>,
[email hidden] wrote:

[...]

Quoted message said:

Dear BB,

I think that we disagree.

To raise or lower himself and the bike, the rider adjusts
his balance, speed, or turn so that there's a mismatch
between the centripetal force of the turn and the downward
force of his lean.

Scarcely any effort is required from the rider in terms of
balance--his tiny changes only create an initial mismatch
between the two opposing forces, much like opening a valve.

But that effort does, indeed, come from the rider. The fact that it
is small does not make it vanish.

Quoted message said:

Tip the bike and rider only slightly past the point that the
downward force of their lean is balanced by the centripetal
force against the contact patch, and the bike and rider will
start to fall faster and faster to the ground unless other
changes balance the two forces again.

|
| gravity, trying to rotate bike and rider
V downward because center of mass (x) is past
vertical from contact patch (c)

\
\
x
\
c <--centripetal force of tire in turn
shoving at contact patch and
trying to rotate bike and rider's
center of mass (x), back up around
contact patch (c)

If the centripetal force is greater (tighter or faster
turn), the bike and rider start to rise.

If the downward force is greater (more lean, less turn,
lower speed), the bike and rider start to fall.

To simplify things, let's omit speed, since it's pretty much a
component of the momentum. It's a bit of redundancy that could cause
confusion later.
Also, downward force does not change--the rider's weight and the
force of gravity remain the same throughout. You only have a variation
in the horizontal components.
Another way to consider this is that a lean on a flat surface has the
effect, in a way, of "faking" two hills from the rider's frame of
reference. You're effectively falling down a hill along your
originating heading due to leaning away from it, and using your tires'
traction to climb another hill along your new heading. Naturally, this
isn't really what's happening, but the energy transfer is similar
because you dump energy from one vector to another continuously and
simultaneously.

Quoted message said:

At a steady speed, a steady lean, and a constant radius
turn, the rider can go forever in a circle, constantly
accelerating toward the center of the circle. The forces are
balanced, so there's no spiral problem.

That is true, but keep it mind that I was quibbling your comment that
"momentum" was somehow transmorgified into a lateral "force" by some
action of the bicycle or rider.

Quoted message said:

He is using the sideways traction of the tires to convert
his current "forward speed" (the tangent at that point of
the circle) to an acceleration toward the center.

If he were converting his forward momentum into something else (even
acceleration around a circle) he would lose forward speed. You cannot
subtract anything greater than zero from a real number and wind up with
the same number. Your explanation seems to have a slight conflict.
Could you explain where that momentum is added back in?

Quoted message said:

At a steady speed (20 mph), the velocity ("forward speed"😉
is still constantly changing (an acceleration) because it's
a vector, with both magnitude (20 mph) and direction (90
degrees east, 91, 92, 93, and so on around the 360-degree
compass).

Half-way around the circle, the bicycle can still be doing
20 mph (speed), but it's no longer "forward"--it's heading
in the completely opposite direction, a considerable
acceleration (change of velocity).

If the rider has conserved his speed of 20mph, and, hopefully,
conserved his mass, then it would seem he did not convert any of his
forward momentum into anything else during the turn. So, the work used
to initiate the turn (however small) must have been added to the system
at some point. It would seem reasonable to assume that work came from
the rider.

Quoted message said:

Now heading back the way that he came, the rider can
simultaneously lean slightly upright, straighten out of the
turn, and speed up or slow down. All that he does is faintly
alter the two opposing forces--downward rotation around the
contact patch from gravity, upward rotation around the
contact patch due to centripetal force from the tire.

Righting back up again would require even more energy, which would
further slow the rider if it were being "converted" from forward
momentum.

Quoted message said:

This is along the lines of a satellite orbiting a planet and
completely incomprehensible until the idea of centripetal
force is understood. In the case of the bicycle or
motorcycle, only a tiny balancing effort is needed to tip
the bike into a turn that will cause it to fall faster and
faster toward the ground--except that the rider increases
the upward force by turning (and can then vary things by
changing speed, re-balancing his center of mass slightly,
accelerating, and so forth).

Satellites are more complex because they don't have any pavement to
push against and any changes to their vectors require ejecting mass,
making computations more complex.
There really is no easy way to change a satellite's orbit like
steering a bike because there's almost nothing to push against.
I'd go so far as to say "centripetal force" could be dropped from the
discussion since it arises only as a reaction to other forces. By the
time all is said and done, centripetal force and whatever it was in
opposition of should cancel each other out completely unless you want to
get into a discussion about slipping and skidding through turns. Same
way we can omit "normal force" unless you want to talk about sinking the
tires into soft surfaces. I don't--that would get messy in a hurry.

Quoted message said:

The forces raising and lowering the bicycle and rider come
from gravity's downward pull and from converting "forward
speed" (velocity with both magnitude and direction) into
upward force through the centripetal force at the contact
patch. Our small balance movements are merely the valve
controlling the balance between these two forces.

If you're not sure about where the force comes from, lean
through a curve or two while coasting and see how
effortlessly you rise and fall. You're just converting
"forward speed" to upward force, which happens even when the
speedometer reading remains the same--the "speed" doesn't
change, but the "forward" does, and that's an acceleration.
Only tiny changes in balance are needed to create a mismatch
between the effects of gravity and centripetal force, which
then raise and lower us easily.

Now stop and lean against a wall or tree and see how much
effort it takes to heave yourself upright again with your
hands on the bars. With no "forward speed," you have to do
all the work and it's immensely harder, if not impossible.

You're omitting displacement. Work is going to be the product of
force and displacement and should remain the same in both cases you
described. While in motion you apply far less force, but get far more
displacement, whereas the little displacement of leaning side to side
without coasting/pedaling would require a larger force to get the same
work output. Like low gear vs. high gear when climbing.
Furthermore, leaning uses all the muscles in your body to shift your
weight around, whereas pushing up with one arm uses only the relatively
puny arm muscles. So you wind up with an artificial perception of
difficulty.

Quoted message said:

That's why balancing a motionless bike is so hard. If your
center of mass tips outside the tiny base defined by the
rear tire and usually sharply angled front tire, you're
likely to fall over. There is no convenient centripetal
force generated by an automatic turn to shove against your
fall.

It's harder because you have such a small work envelope to occupy.
(ie, directly above your bicycle) It's equally hard to ride along an
arrow-straight line even though you've got loads of forward momentum.
Try it--stationary, your wheels won't track horizontally at all. Can
you ride with absolutely zero horizontal tracking of your tires?

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

In article <[email hidden]>,

Frank said:

Good Lord!!!!!!!!!

What? I'm bored. (: I need to kill time while waiting for NASA to
annihilate a small fraction of history.

http://www.nasa.gov/mission_pages/deepimpact/images/index.html

[...huge snip...]

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

Good Lord!!!!!!!!!

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

Quoted message said:

In article <[email hidden]>,
[email hidden] wrote:

[...]

Quoted message said:

Dear BB,

I think that we disagree.

To raise or lower himself and the bike, the rider adjusts
his balance, speed, or turn so that there's a mismatch
between the centripetal force of the turn and the downward
force of his lean.

Scarcely any effort is required from the rider in terms of
balance--his tiny changes only create an initial mismatch
between the two opposing forces, much like opening a valve.

But that effort does, indeed, come from the rider. The fact that it
is small does not make it vanish.

Quoted message said:

Tip the bike and rider only slightly past the point that the
downward force of their lean is balanced by the centripetal
force against the contact patch, and the bike and rider will
start to fall faster and faster to the ground unless other
changes balance the two forces again.

|
| gravity, trying to rotate bike and rider
V downward because center of mass (x) is past
vertical from contact patch (c)

\
\
x
\
c <--centripetal force of tire in turn
shoving at contact patch and
trying to rotate bike and rider's
center of mass (x), back up around
contact patch (c)

If the centripetal force is greater (tighter or faster
turn), the bike and rider start to rise.

If the downward force is greater (more lean, less turn,
lower speed), the bike and rider start to fall.

To simplify things, let's omit speed, since it's pretty much a
component of the momentum. It's a bit of redundancy that could cause
confusion later.
Also, downward force does not change--the rider's weight and the
force of gravity remain the same throughout. You only have a variation
in the horizontal components.
Another way to consider this is that a lean on a flat surface has the
effect, in a way, of "faking" two hills from the rider's frame of
reference. You're effectively falling down a hill along your
originating heading due to leaning away from it, and using your tires'
traction to climb another hill along your new heading. Naturally, this
isn't really what's happening, but the energy transfer is similar
because you dump energy from one vector to another continuously and
simultaneously.

Quoted message said:

At a steady speed, a steady lean, and a constant radius
turn, the rider can go forever in a circle, constantly
accelerating toward the center of the circle. The forces are
balanced, so there's no spiral problem.

That is true, but keep it mind that I was quibbling your comment that
"momentum" was somehow transmorgified into a lateral "force" by some
action of the bicycle or rider.

Quoted message said:

He is using the sideways traction of the tires to convert
his current "forward speed" (the tangent at that point of
the circle) to an acceleration toward the center.

If he were converting his forward momentum into something else (even
acceleration around a circle) he would lose forward speed. You cannot
subtract anything greater than zero from a real number and wind up with
the same number. Your explanation seems to have a slight conflict.
Could you explain where that momentum is added back in?

Quoted message said:

At a steady speed (20 mph), the velocity ("forward speed"😉
is still constantly changing (an acceleration) because it's
a vector, with both magnitude (20 mph) and direction (90
degrees east, 91, 92, 93, and so on around the 360-degree
compass).

Half-way around the circle, the bicycle can still be doing
20 mph (speed), but it's no longer "forward"--it's heading
in the completely opposite direction, a considerable
acceleration (change of velocity).

If the rider has conserved his speed of 20mph, and, hopefully,
conserved his mass, then it would seem he did not convert any of his
forward momentum into anything else during the turn. So, the work used
to initiate the turn (however small) must have been added to the system
at some point. It would seem reasonable to assume that work came from
the rider.

Quoted message said:

Now heading back the way that he came, the rider can
simultaneously lean slightly upright, straighten out of the
turn, and speed up or slow down. All that he does is faintly
alter the two opposing forces--downward rotation around the
contact patch from gravity, upward rotation around the
contact patch due to centripetal force from the tire.

Righting back up again would require even more energy, which would
further slow the rider if it were being "converted" from forward
momentum.

Quoted message said:

This is along the lines of a satellite orbiting a planet and
completely incomprehensible until the idea of centripetal
force is understood. In the case of the bicycle or
motorcycle, only a tiny balancing effort is needed to tip
the bike into a turn that will cause it to fall faster and
faster toward the ground--except that the rider increases
the upward force by turning (and can then vary things by
changing speed, re-balancing his center of mass slightly,
accelerating, and so forth).

Satellites are more complex because they don't have any pavement to
push against and any changes to their vectors require ejecting mass,
making computations more complex.
There really is no easy way to change a satellite's orbit like
steering a bike because there's almost nothing to push against.
I'd go so far as to say "centripetal force" could be dropped from the
discussion since it arises only as a reaction to other forces. By the
time all is said and done, centripetal force and whatever it was in
opposition of should cancel each other out completely unless you want to
get into a discussion about slipping and skidding through turns. Same
way we can omit "normal force" unless you want to talk about sinking the
tires into soft surfaces. I don't--that would get messy in a hurry.

Quoted message said:

The forces raising and lowering the bicycle and rider come
from gravity's downward pull and from converting "forward
speed" (velocity with both magnitude and direction) into
upward force through the centripetal force at the contact
patch. Our small balance movements are merely the valve
controlling the balance between these two forces.

If you're not sure about where the force comes from, lean
through a curve or two while coasting and see how
effortlessly you rise and fall. You're just converting
"forward speed" to upward force, which happens even when the
speedometer reading remains the same--the "speed" doesn't
change, but the "forward" does, and that's an acceleration.
Only tiny changes in balance are needed to create a mismatch
between the effects of gravity and centripetal force, which
then raise and lower us easily.

Now stop and lean against a wall or tree and see how much
effort it takes to heave yourself upright again with your
hands on the bars. With no "forward speed," you have to do
all the work and it's immensely harder, if not impossible.

You're omitting displacement. Work is going to be the product of
force and displacement and should remain the same in both cases you
described. While in motion you apply far less force, but get far more
displacement, whereas the little displacement of leaning side to side
without coasting/pedaling would require a larger force to get the same
work output. Like low gear vs. high gear when climbing.
Furthermore, leaning uses all the muscles in your body to shift your
weight around, whereas pushing up with one arm uses only the relatively
puny arm muscles. So you wind up with an artificial perception of
difficulty.

Quoted message said:

That's why balancing a motionless bike is so hard. If your
center of mass tips outside the tiny base defined by the
rear tire and usually sharply angled front tire, you're
likely to fall over. There is no convenient centripetal
force generated by an automatic turn to shove against your
fall.

It's harder because you have such a small work envelope to occupy.
(ie, directly above your bicycle) It's equally hard to ride along an
arrow-straight line even though you've got loads of forward momentum.
Try it--stationary, your wheels won't track horizontally at all. Can
you ride with absolutely zero horizontal tracking of your tires?

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

On Sun, 03 Jul 2005 23:16:16 -0500, "B.B."
<[email hidden]> wrote:

[snip]

Dear BB,

I think that you need to look into centripetal force rather
than dropping it. It's fundamental to circular motion, which
is what turning is.

I also think that you need to look into the difference
between speed and velocity.

Consider a bicycle being pedalled at 20 mph westward on a
very large parking lot and turning left at 16.5 degrees per
second, an easily attainable speed and turn rate.

After 5 seconds, the speedometer still reads 20 mph, but the
bicycle is now going 0 mph west--you've turned 90 degrees
and are heading north now.

The westward deceleration force is exactly the same as if
you had braked to a halt from 20 mph in a straight line in 5
seconds--that's the kind of force pushing you upright in a
turn while gravity tries to pull you and your tilted bike
down.

After another 5 seconds, the speedometer still reads 20 mph,
but the bicycle is now going -20 mph west--you've turned 180
degrees and are heading east now.

The eastward acceleration is exactly the same as if you had
pedalled up from a standing start to 20 mph eastward in a
straight line in only 5 seconds, the reverse of the braking
scenario.

Another 5 seconds pass, the speedometer stubbornly reads 20
mph, but the bicycle is again going 0 mph west--you've
turned 270 degrees and are now heading south.

Your eastward deceleration is again exactly the same as if
you had braked to a halt from 20 mph in a straight line in 5
seconds.

After 20 seconds, the bicycle is momentarily going 20 mph
west again, having sped up from 0 mph westward to 20 mph
westward--you've turned 360 degrees and could do so
indefinitely at the same speed and lean angle.

Whenever you curve through a right-angle corner at 20 mph on
a bicycle, in a car, aboard a boat, or upon an extremely
small satellite orbiting 90 degrees around a very small
planet, the acceleration is the equivalent of coming to a
dead halt in a straight line.

The force that you feel when you grab a handful of brake to
stop in 5 seconds in a straight line is the same force that
lets you lean at an otherwise impossible angle while
turning.

Slight adjustments in balance let us adjust this centripetal
force to oppose gravity and control our lean during
cornering, but our balance movements contribute little force
in themselves.

This is why even very small riders can easily tilt and
corner on 600-lb motorcycles at speed, but fear being
crushed if a parked machine tip even a little bit--even a
slight lean when motionless poses a serious danger because
no centripetal force is available to oppose gravity.

Cheers,

Carl Fogel

On Mon, 04 Jul 2005 00:28:54 -0600, [email hidden]
wrote:

[snip]

D'oh!

A less absent-minded turn-rate figure would be 18 degrees
per second for 5 seconds (not 16.5 degrees) to turn 90
degrees every 5 seconds.

I simply must start taking my socks off when I make these
intricate calculations. Eight-finger precision just isn't
good enough.

H. Simpson

Quoted message said:

On Sun, 03 Jul 2005 23:16:16 -0500, "B.B."
<[email hidden]> wrote:

[snip]

[snip]

Quoted message said:

Whenever you curve through a right-angle corner at 20 mph on
a bicycle, in a car, aboard a boat, or upon an extremely
small satellite orbiting 90 degrees around a very small
planet, the acceleration is the equivalent of coming to a
dead halt in a straight line.

Dude, acceleration is a vector. Pick up a physics book, ok? Your
posts are making me cringe.

In article <[email hidden]>,

Quoted message said:

On Sun, 03 Jul 2005 23:16:16 -0500, "B.B."
<[email hidden]> wrote:

[snip]

Dear BB,

I think that you need to look into centripetal force rather
than dropping it. It's fundamental to circular motion, which
is what turning is.

Yes and no. Centripetal force in a turn on pavement is a reaction
force. For stuff like satellites and hovercraft it's a generated force;
gravity in the case of satellites, and fan thrust in the case of
hovercraft. But a bicycle on pavement is pushing against a practically
non-deformable surface and the centripetal force arises only because the
pavement is too big and heavy to move. You'll note that leaning into a
curve on ice is doom since there isn't enough friction for the
centripetal force to develop, so you fall over and continue in the
original direction. Same reason that cars move in dead-straight lines
(the center of mass does--individual wheels might not if the car spins)
when the lock their wheels up on dry pavement.
Our scenario is more comparable to a roller coaster fixed to a track,
which does, indeed, coast through quite a few turns, conserving momentum
all the while. The centripetal force is merely the force of the track
trying to not bend and flex as to coaster falls down the hills and
coasts up and around bends. Though it's still not that good a
comparison because the riders cannot steer a roller coaster. Perhaps
the best comparison would be to a bicycle. (:
Also consider the circular motion of a spinning flywheel--the
centripetal force is the strength of the material trying to not let
itself be torn apart. It does not drain the flywheel's rotational
momentum to generate that force and hold itself together. You simply
develop a force proportional to the speed and mass of the wheel's rim
while you spin the wheel up and that force remains constant as long as
the wheel's RPM remains constant. That force will go away as the wheel
slows down. Or, in another way, the force is a side-effect of the
rotation of the wheel.
I suppose you could represent the energy stored in a flywheel by
calculating the centripetal force to hold it together, but that's
unusual outside of asteroid belts or Dyson spheres; most speak in terms
of rotational momentum.
In the end, since centripetal force is a reaction force it should
have a net cancellation with whatever generates it so you can throw it
out to simplify the calculations as long as you're sure the materials
and friction involved can take the stress. It would simplify the
discussion.

Quoted message said:

I also think that you need to look into the difference
between speed and velocity.

Consider a bicycle being pedalled at 20 mph westward on a
very large parking lot and turning left at 16.5 degrees per
second, an easily attainable speed and turn rate.

Let's just coast the bike since pedaling adds energy to the system
and screws the experiment up. After all, we're trying to determine if
the rider has to add energy to turn or if he gets it by stealing away
from his forward momentum.

Quoted message said:

After 5 seconds, the speedometer still reads 20 mph, but the
bicycle is now going 0 mph west--you've turned 90 degrees
and are heading north now.

The westward deceleration force is exactly the same as if
you had braked to a halt from 20 mph in a straight line in 5
seconds--that's the kind of force pushing you upright in a
turn while gravity tries to pull you and your tilted bike
down.

After another 5 seconds, the speedometer still reads 20 mph,
but the bicycle is now going -20 mph west--you've turned 180
degrees and are heading east now.

The eastward acceleration is exactly the same as if you had
pedalled up from a standing start to 20 mph eastward in a
straight line in only 5 seconds, the reverse of the braking
scenario.

Another 5 seconds pass, the speedometer stubbornly reads 20
mph, but the bicycle is again going 0 mph west--you've
turned 270 degrees and are now heading south.

Your eastward deceleration is again exactly the same as if
you had braked to a halt from 20 mph in a straight line in 5
seconds.

After 20 seconds, the bicycle is momentarily going 20 mph
west again, having sped up from 0 mph westward to 20 mph
westward--you've turned 360 degrees and could do so
indefinitely at the same speed and lean angle.

Whenever you curve through a right-angle corner at 20 mph on
a bicycle, in a car, aboard a boat, or upon an extremely
small satellite orbiting 90 degrees around a very small
planet, the acceleration is the equivalent of coming to a
dead halt in a straight line.

Coming to a halt in one direction and reaccelerating to the same
speed in another direction (or even the same direction) would require
the mass of the rider times the top speed squared times two worth of
energy. That's a lot. OTOH, riding around a corner obviously takes
very little effort.
The acceleration you speak of in a turn is the force the tire exerts
laterally on the road surface countered exactly by the centripetal force
you mentioned. The two cancel each other out and the result is that you
can ride around a bend for "free."
OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds the
force exerted by your momentum decelerating. When taking off again the
force of your pedaling exceeds the force created by your inertia
accelerating. Both actions, because of the unequal forces involved,
require work.
If the two scenarios (coasting around a corner and stopping then
starting again) were equivalent both would require the same amount of
work.

Quoted message said:

The force that you feel when you grab a handful of brake to
stop in 5 seconds in a straight line is the same force that
lets you lean at an otherwise impossible angle while
turning.

It's the same amount, but not the same force. The distinction is
actually important. You were getting at the difference between speed
and velocity; to me, velocity is speed with a vector. All forces
likewise have vectors. You cannot interchange forces of equal magnitude
[censored]-nilly if their vectors don't match.

Quoted message said:

Slight adjustments in balance let us adjust this centripetal
force to oppose gravity and control our lean during
cornering, but our balance movements contribute little force
in themselves.

I'm trying to make two points: one, that those "small adjustments"
you mentioned here take work on the part of the rider, and, two, that
the work for those "small adjustments" cannot be taken from the forward
momentum of the bicycle. That's all that I've been trying to say.
I've also noticed that you seem to consider things like work,
momentum, force, and acceleration to be interchangeable to some degree.
They aren't; and I'm trying to eliminate as many superfluous elements as
possible from the conversation so you are less likely to mix them.
Centripetal force causes a turn no more than brake heat causes you to
stop. Yeah, it happens, and it plays a role, but it's just a side-note
in the actual process. And, unless it gets out of parameters, can be
ignored.

Quoted message said:

This is why even very small riders can easily tilt and
corner on 600-lb motorcycles at speed, but fear being
crushed if a parked machine tip even a little bit--even a
slight lean when motionless poses a serious danger because
no centripetal force is available to oppose gravity.

You'll need to explain this paradox in your description: you talk
about a rider conserving his speed through a turn, implying that his
momentum is conserved. In other words, no energy is added to the system
or removed for the duration of the turn. (allowing, of course, for
frictional losses) Yet you also talk about a bicycle or motorcycle
taking that forward momentum away and redirecting it to perform the work
of leaning or righting the rider and vehicle.
Your example of leaning a motorcycle is no different than the prior
example of leaning a stationary bicycle--you need to account for
displacement or the comparison is totally meaningless.
Also, understand that the longer and more repetitious your posts are
the less likely I am to read all the way through to glean any new
information. If, when writing your post, you find yourself repeating
yourself, think about this: do you consider me such an idiot that I
didn't get it the first time through? Or are you glazing over what I
wrote in an effort to push your own idea? You can briefly refer to what
you've written previously without writing it all over again and I think
I'll be able to sort it all out. If you really think you're being clear
and that I'm just not "getting it" ask some appropriate questions and I
will answer them to my best ability.

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

B.B. said:

OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds the
force exerted by your momentum decelerating. When taking off again the
force of your pedaling exceeds the force created by your inertia
accelerating. Both actions, because of the unequal forces involved,
require work.

Guys, please stop. You are hurting America. Going around the corner
requires no energy because the force is always at right angles to the
direction of motion so F dot ds is always zero, ok? This talk of
force "mismatch"... I can't go on, it is just too horrible. Please,
for America's sake, just stop.

In article <[email hidden]>,

Jim Smith said:

B.B. said:

OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds the
force exerted by your momentum decelerating. When taking off again the
force of your pedaling exceeds the force created by your inertia
accelerating. Both actions, because of the unequal forces involved,
require work.

Guys, please stop. You are hurting America. Going around the corner
requires no energy because the force is always at right angles to the
direction of motion so F dot ds is always zero, ok? This talk of
force "mismatch"... I can't go on, it is just too horrible. Please,
for America's sake, just stop.

I have a question. How was my explanation wrong? Yours is more
concise, but I didn't figure Carl was familiar with the algebra
involved, so I explained it the long way.

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

Carl, have fun with these:
http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

Andrew Lee said:

Carl, have fun with these:
http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

Hey, folks, some of us would like to watch the evening (US) coverage of
the Tour _without_ knowing in advance who was going to win.

If you're going to post something about the current day's Tour stage,
please put a "spoiler alert" in the Subject: heading as a courtesy.

Sheldon "Didn't Want To Know That Yet" Brown
Newtonville, Massachusetts
+----------------------------------------------------------+
| If only God would give me some clear sign! |
| Like making a large deposit in my name at a Swiss bank. |
| --Woody Allen |
+----------------------------------------------------------+
Harris Cyclery, West Newton, Massachusetts
Phone 617-244-9772 FAX 617-244-1041
http://harriscyclery.com
Hard-to-find parts shipped Worldwide
http://captainbike.com http://sheldonbrown.com

Sheldon Brown said:

Andrew Lee said:

Carl, have fun with these:
http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

Hey, folks, some of us would like to watch the evening (US) coverage of
the Tour _without_ knowing in advance who was going to win.

If you're going to post something about the current day's Tour stage,
please put a "spoiler alert" in the Subject: heading as a courtesy.

Sheldon "Didn't Want To Know That Yet" Brown
Newtonville, Massachusetts

Sorry! It didn't cross my mind as something that happened today. I just
thinking "sprint" and "front view" photo. I didn't know about evening TV
coverage either since I don't have cable...

Andrew Lee' whatsupandrewathotmaildotcom said:

Quoted message said:

Quoted message said:

Carl, have fun with these:
http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

Hey, folks, some of us would like to watch the evening (US) coverage of
the Tour _without_ knowing in advance who was going to win.

Having watched, with some interest, the past two sprint finishes,
it seems to me that for sprinters going in a (reasonably) straight
line, the side to side motion of the bike is about what I had
measured. The bike frame doesn't sway back and forth very much.
The few that show substantial lean are swerving.

Joe

B.B. said:

In article <[email hidden]>,

Jim Smith said:

B.B. said:

OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds the
force exerted by your momentum decelerating. When taking off again the
force of your pedaling exceeds the force created by your inertia
accelerating. Both actions, because of the unequal forces involved,
require work.

Guys, please stop. You are hurting America. Going around the corner
requires no energy because the force is always at right angles to the
direction of motion so F dot ds is always zero, ok? This talk of
force "mismatch"... I can't go on, it is just too horrible. Please,
for America's sake, just stop.

I have a question. How was my explanation wrong? Yours is more
concise, but I didn't figure Carl was familiar with the algebra
involved, so I explained it the long way.

I kid. I kid.

But, the language was getting more than a little tortured. Phrases
like "inertia accelerating" don't really have any meaning. Then there
was stuff like:

"OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds
the force exerted by your momentum decelerating. When taking off
again the force of your pedaling exceeds the force created by your
inertia accelerating. Both actions, because of the unequal forces
involved, require work."

You may indeed have a grasp of the relevant physics, but it isn't
apparant from reading the previous posts. I'm not sure what you mean
by "momentum decelerating," but if we exclude secondary effects, the
force of the brakes drag is exactly equal to the time rate of change
of the momentum, that is just another way of staing Newtons 2nd law:
F=mA, same goes for the case of accelerating from a stop. There are
at least a couple ways to look at why work has to be done to slow the
bike to a stop or to accelerate from a stop. One is that the force is
acting through a distance, another is that that the kinetic energey of
the bicycle is changing. I really can't tell what you were trying to
get at with the mismatched forces thing.

On Mon, 04 Jul 2005 15:10:45 -0500, "B.B."
<[email hidden]> wrote:

[snip]

Quoted message said:

If you really think you're being clear
and that I'm just not "getting it" ask some appropriate questions and I
will answer them to my best ability.

Dear BB,

I doubt that I've been terribly clear, but here are few
related questions. Perhaps we've must been missing each
other and actually agreeing?

If a rider and his bicycle . . .

.. . . are traveling at 11.18 m/s (25 mph),

.. . . and are turning around a circle whose radius is 26
meters (85 feet)

.. . . then what is the acceleration toward the center of the
turn?

.. . . and what is the lean angle from the contact patch
through the center of mass?

***

If the same rider and his bicycle are 80 kg (160 lb rider
and 17 lb bike) . . .

.. . . what is the gravitational force in Newtons tilting the
bike and rider downward

.. . . and what is the centripetal force in Newtons at the
contact patch that opposes it?

***

If it's any help, here's a page from the same site that I
mentioned earlier:

http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l2c.html

To go back to the original picture (a rider and bike of
roughly that mass and speed tilted at roughly that angle):

http://www.cyclingnews.com/photos.php?id=photos/2005/jun05/belgium05/finish2

I think that the rider on our right is turning, not heading
straight as some people think, because he's leaned over at
an otherwise impossible angle. As far as I know, you can't
turn a bike without tilting the center of mass to one side,
and vice-versa.

The questions above work address the physics of lean angle
and opposing forces.

(For anyone curious, the page above offers equations that
suggest that an 80 kg bicycle and rider turning at 11.18 m/s
around a circle whose radius is 26 meters will be tilted at
about 26 degrees, will be accelerating toward the center of
the circle at about 0.5 G, with about 385 Newtons of
centripetal force pushing sideways at the contact patch
balancing about 785 Newtons of gravitational force, given
the 26 degree angle that stays the same for lighter and
heavier riders at that speed around that circle.)

(A higher speed requires a bigger circle to produce the same
lean angle, which can also be produced by a lower speed and
a smaller circle.)

(At 30 mph, or 13.4 m/s, the circle must be enlarged to a
radius of 37 meters for a 26-degree tilt.)

(At 25 mph, or 11.18 m/s, the circle must be a radius of 26
meters for a 26-degree tilt.)

(At 20 mph, or 8.94 m/s, the circle must be reduced to a
radius of 17 meters for a 26-degree tilt.)

(Again, the tilt angle depends only on the speed and size of
the circle.)

Carl Fogel

On Mon, 4 Jul 2005 14:14:34 -0800, "Andrew Lee"

whatsupandrewathotmaildotcom said:

Carl, have fun with these:
http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

Dear Andrew,

I like 'em--thanks.

In the first picture, they're mostly upright, with the
largest tilt of a visible rider being maybe 7-8 degrees for
the rider on our far right (the picture is tilted a little
or the camera is off-center, judging by the road stripes):

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN22

McEwen is using Grady for a pillow, and I have no idea what
the physics involved there would be.

There's a tilted wheel visible between them, but I think
that the rider's helmet is just visible above them--and he's
turning, judging by how he and his bike appear to be tilted
well past the contact patch to our left.

It's interesting how so many riders apparently sprinting for
the finish are basically upright, but maybe it was just luck
and a moment later some of them tilted to fascinating
angles.

In the second picture, they're still mostly upright:

http://www.cyclingnews.com/photos/2005/tour05/?id=tour053/BOONEN59

On our far left, the rider in blue is tilted, but obviously
turning, with his body and bike well to our left, past his
contact patch. The yellow rider on the left is tilted about
10 degrees, sprinting fashion rather than turning (again,
the angle makes it look a little more like 12 degrees or
so).

Here, to be fair, the riders are probably relaxing and not
throwing their bikes as much from side to side, but it's
still interesting how upright they are.

Thanks again.

Carl Fogel

In article <[email hidden]>,
Jim Smith <[email hidden]> wrote:

[...]

Quoted message said:

I kid. I kid.

Heh, I kind of figured with the "hurting america" bit. (:

Quoted message said:

But, the language was getting more than a little tortured. Phrases
like "inertia accelerating" don't really have any meaning. Then there
was stuff like:

"OTOH, stopping and reaccelerating are two cases where there's a
mismatch. When stopping the force of your brakes' drag exceeds
the force exerted by your momentum decelerating. When taking off
again the force of your pedaling exceeds the force created by your
inertia accelerating. Both actions, because of the unequal forces
involved, require work."

You may indeed have a grasp of the relevant physics, but it isn't
apparant from reading the previous posts. I'm not sure what you mean
by "momentum decelerating," but if we exclude secondary effects, the
force of the brakes drag is exactly equal to the time rate of change
of the momentum, that is just another way of staing Newtons 2nd law:
F=mA, same goes for the case of accelerating from a stop. There are
at least a couple ways to look at why work has to be done to slow the
bike to a stop or to accelerate from a stop. One is that the force is
acting through a distance, another is that that the kinetic energey of
the bicycle is changing. I really can't tell what you were trying to
get at with the mismatched forces thing.

Yeah, I have a pretty bad habit of mangling the hell out of language.
I blame a former existence as a (bad) poet for that one. By stuff like
"momentum decelerating" I meant more along the lines of "the force
needed to change the momentum." Or, with mathematical terms, F=delta p.
I use a physics model that all forces appear in opposed pairs that
may or may not be equal in magnitude, so that's where my comments about
"unbalanced forces" come in. For example, normal force vs. gravity,
both are equal if you're standing still, but if normal force isn't high
enough you sink into the ground. Likewise, riding forward there's a
drag force against you, and your momentum (or rather, F=dp) and pedaling
pushing with you. I like to consider both so that you've got some sort
of force vector available with a name on it to explain how someone can
go over the handlebars.
It's also handy to talk about force balance when working with things
like hydraulics since that's what will determine working pressures. And
I've had a bunch of experience with hydraulics. But I haven't touched
the math at all in about seven years now, so that could be part of why
my brain was mutilating the terminology.

--
B.B. --I am not a goat! thegoat4 at airmail dot net
http://web2.airmail.net/thegoat4/

Quoted message said:

McEwen is using Grady for a pillow, and I have no idea what
the physics involved there would be.

From Wednesday's L'Equipe:
http://anonymous.coward.free.fr/rbr/matelas.jpg

Robert Chung said:

Quoted message said:

McEwen is using Grady for a pillow, and I have no idea what
the physics involved there would be.

From Wednesday's L'Equipe:
http://anonymous.coward.free.fr/rbr/matelas.jpg

Dear Robert,

I like it--presumably the winner slept on a stiff yet
compliant mattress, while the rider nodding off spent a
stressful night on a mattress whose springs had not been
properly stress-relieved.

Carl Fogel