Re: ISIS getting dropped?

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

[...]

Quoted message said:

Quoted message said:

It must be difficult to restrain yourself to the topic - I thought you could
have spent more time with the 200N figure, which seems relevant, although,
once again, it was for comparative testing, not reproduction of a real-world
event.

Dear Sandy,

Please let us know if you find a picture of a 26-degree tilt
to one side, or test that angle while stationary with the
help of a wall.

Carl Fogel

It doesn't seem the test was meant to be constrained to real-world
circumstance. After all, if you're testing the material strength you
typically do that at a level well-beyond what you'd find in the field,
that way you're assured a "safety margin" in a sense. So this test used
an excessive angle (26deg tilt) and an excessive load (200N).
That kind of test is useful for comparing the properties of different
parts, but not for determining real-life behavior. Kind of like the
Rockwell hardness test is useful for comparing materials bikes are made
of, but not really for comparing bikes unless they're going to get shot
with supersonic BB guns at the park.

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

Robert Chung said:

Joe Riel said:

Robert Chung said:

[email hidden] wrote:

Quoted message said:

Quoted message said:

> If anyone knows of pictures that show a rider heading
> forward (rather than cornering) and tilting his bike 26
> degrees to either side, please share the links with us.

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

Quoted message said:

Quoted message said:

I don't think this is 26 degrees. I think it's closer to 24.

That, as is probably the case with any single picture, is not entirely
convincing.

Um, I'm confused. Convincing of what? Carl asked for a photo, I posted
one.

Dear Robert,

I like the photo--thanks.

It does show a bike leaning about 26 degrees in a sprint.

But what Joe and I are wondering is whether the bike is
going forward or turning to achieve that angle.

Comparing the two riders should make the question clearer.

Their bodies (b) are on the same side of the contact patch
(c), to the right side of the picture:

|b |b
| |
| |
c c

But their bike frames (f) are leaning in opposite
directions:

|b |b
| |
f| | f
c c

And their shoes (s) at the top of the pedal cycle are on
opposite sides:

|b |b
s | | s
f| | f
c c

sprint turn

The rider on our right with the big lean angle appears
either to be about to fall down or else to be turning
markedly to our right, not tipping his bike back and forth
in a straight-line sprint.

Here's a simple onscreen line and angle program for anyone
interested in slapping some quick lines on the picture:

http://www.markus-bader.de/MB-Ruler/

But a 3x5 card moved across the picture so that one edge
points down to where the contact patch would be should show
how far the rider on the right is leaning the wrong way to
be riding in a straight line.

It's not the mere size of the lean angle that makes it look
impossible for a straight line. It's that about 90% of both
bike and rider is to one side of a vertical line drawn up
from where the contact patch should be--that indicates a
turn in that direction.

Compare that to the rider on our left. His bike is leaning
to our left, while his body is leaning to our right--and the
angle is much less for what looks like straight-line tilting
of the bike back and forth.

But I still like the picture and appreciate your posting the
link. It's the kind of thing that lets everyone see what's
being discussed--and may let someone explain that no, the
rider on the right is somehow going in a straight line with
90% of the total mass to one side past the vertical.

I should add that even if I'm right about the big lean being
a turn, that doesn't mean that a bike can't be tilted back
and forth through a 52-degree arc in a straight-line sprint.
It would only mean that this particular picture fooled me at
first.

Carl Fogel

Quoted message said:

Quoted message said:

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

Quoted message said:

But what Joe and I are wondering is whether the bike is
going forward or turning to achieve that angle.

That photo was taken on Sunday at the finish line of the Belgian RR
championships. They were both going straight; Baguet (the guy leaning
24ish degrees on the right) nipped Van Impe by a few centimeters -- you
can see Baguet's shoulder just barely in front, and if Baguet had been
turning he would've lost. Sprinters need sharp elbows.

Baguet and Van Impe are friends and training partners so they ride
together all the time. I used to get a look on my face just like Van
Impe's when I did city limit sprints with my riding buddy.

Robert Chung said:

Quoted message said:

Quoted message said:

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

Quoted message said:

But what Joe and I are wondering is whether the bike is
going forward or turning to achieve that angle.

That photo was taken on Sunday at the finish line of the Belgian RR
championships. They were both going straight.

[snip social details]

Dear Robert,

Doesn't anything strike you as implausible about the idea
that the rider on the right is going straight at that
moment?

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

It sure looks like a turn--outside leg straight, inside leg
up, bike and body both leaned well past vertical the same
way into a turn.

Cover up the rider on the left with your hand and look at
the rider on the right. If he's not turning, how will he
avoid falling?

Carl Fogel

B.B. said:

That kind of test is useful for comparing the properties of different
parts, but not for determining real-life behavior. Kind of like the
Rockwell hardness test is useful for comparing materials bikes are made
of, but not really for comparing bikes unless they're going to get shot
with supersonic BB guns at the park.

What - that's never happened to you???

Mark Hickey
Habanero Cycles
http://www.habcycles.com
Home of the $695 ti frame

In article <[email hidden]>,

Mark Hickey said:

B.B. said:

That kind of test is useful for comparing the properties of different
parts, but not for determining real-life behavior. Kind of like the
Rockwell hardness test is useful for comparing materials bikes are made
of, but not really for comparing bikes unless they're going to get shot
with supersonic BB guns at the park.

What - that's never happened to you???

Only in Dallas.

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

In article <[email hidden]>,

Quoted message said:

Robert Chung said:

Quoted message said:

>

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

Quoted message said:

But what Joe and I are wondering is whether the bike is
going forward or turning to achieve that angle.

That photo was taken on Sunday at the finish line of the Belgian RR
championships. They were both going straight.

[snip social details]

Dear Robert,

Doesn't anything strike you as implausible about the idea
that the rider on the right is going straight at that
moment?

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

Holy [censored] what an annoying page. Too many animated gifs.

Quoted message said:

It sure looks like a turn--outside leg straight, inside leg
up, bike and body both leaned well past vertical the same
way into a turn.

Cover up the rider on the left with your hand and look at
the rider on the right. If he's not turning, how will he
avoid falling?

Same way people don't fall over while walking even though between
steps more than 50% of the body's weight is off to one side of the
supporting foot--momentum keeps you going forward toward your next step,
and inertia mandates that it takes a small amount of time for your body
to tip over. Get to your next step before you tip over and you'll do
fine, whereas getting there too late leads to a fall or stumble. Try
walking very, very slowly without modifying your movements to
compensate--you'll fall. With the rider--he'll swing back past center
before his body develops enough sideways speed to fall down. (he hopes,
anyway) Then his mass will have to decelerate to lose all of the
sideways velocity it picked up last lean and then reaccelerate to
develop a sideways velocity to get into the next lean. Then repeat.
The net result is that the bike, legs, and butt wiggle side-to-side
over time, and at any one instant in time the rider may be leaning to
the side. Integrated over a period of time, however, the rider's center
of mass should lie in a more-or-less straight line (humans drift) and
the sum side-to-side velocity should be zero.
This is not retained in a single still photograph of a dynamic event.
You wouldn't look at a photo taken of a sprinting runner while both feet
are airborne and assume he's capable of levitation, would you? Of
course not--you know that later, though not shown, the runner touched
down again. The cyclist will likewise make another pedal stroke, though
not shown.

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

Quoted message said:

Doesn't anything strike you as implausible about the idea
that the rider on the right is going straight at that
moment?

Nope.

Quoted message said:

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

Quoted message said:

It sure looks like a turn--outside leg straight, inside leg
up, bike and body both leaned well past vertical the same
way into a turn.

It would look like a turn if I didn't know that they were both sprinting
in a straight line for the finish. A fraction of a second later (or
earlier) and you'd've seen that Van Impe and Baguet were leaning the other
way.

Robert Chung said:

Quoted message said:

Doesn't anything strike you as implausible about the idea
that the rider on the right is going straight at that
moment?

Nope.

You might try duplicating that position on your bike (dynamically, of
course) and then rethinking the answer (I have). I suspect he is
turning, though probably not for very long.

My interest (from another thread) is in the spoke loading caused by
leaning the bike (moving body outside the plane of the frame). In
both riders the angle between the plane of the frame and the line from
the ground contact patch to the rider's center of mass is quite
small---so regardless whether we think Baguet is turning, the
effective angle appears to be small.

Joe

Joe Riel said:

Quoted message said:

Nope.

You might try duplicating that position on your bike (dynamically, of
course) and then rethinking the answer (I have). I suspect he is
turning, though probably not for very long.

Perhaps I might rethink my answer if I cared about it but I'm trying not
to get into any spoke tension argument. The last turn before the 300m long
finishing straight was a right hander.

Quoted message said:

Still, I may be wrong

.... says it all.
--
Bonne route !

Sandy
Vreneuil-sur-Seine FR

Quoted message said:

The camera angle also exaggerates his lean, though you
wouldn't think so at first--notice how much of his rear
stays can be seen.

Actually, see that yellow pole in the background? If that thing is
supposed to be vertical, you have to rotate the photo a tad clockwise.

Robert Chung said:

Quoted message said:

The camera angle also exaggerates his lean, though you
wouldn't think so at first--notice how much of his rear
stays can be seen.

Actually, see that yellow pole in the background? If that thing is
supposed to be vertical, you have to rotate the photo a tad clockwise.

Dear Robert,

Yes, I noticed the fuzzy yellow pole, but when I measured it
a few days ago, its presumable vertical angle appeared to be
about only about 1~2 degrees off the picture vertical.

(I just put this onscreen protractor at the top of the pole
on one edge and tried to drag the line down the side of the
pole:

http://www.markus-bader.de/MB-Ruler/

If anything, the pole in the picture suggests that the bike
of the rider on our left is not tilted quite as far as it
looks, while the rider on our left is leaning slightly
farther into the turn.

What I meant by the camera angle is that the rider on our
left doesn't seem to be dead head-on to the camera. We can
see one whole chainstay, the nose of his seat is clearly to
one side of the center of his handlebar stem, and the back
of his front tire is visible off to the same side.

The other rider may be off to the other side of the camera,
but his details aren't as clear. The nose of his saddle
looks almost centered on the line from his front tire
through his handlebar stem, but it might be a little bit
off. I think that his front tire is pretty much centered in
the fork, with no rear portion of the wheel visible. One
chainstay is visible, but I think that it's a curved or
bowed chainstay, with much of the upper part hidden, unlike
the other rider's chainstay, which looks like an
old-fashioned straight chainstay and is sticking out
clearly.

But I still think that you found a hell of picture--I can't
imagine getting a real camera shot of two riders any closer
to dead head-on than this--obviously, at least one rider
must be a little off to the side, if not both.

Carl Fogel

On Sat, 02 Jul 2005 11:06:41 -0500, "B.B."
<[email hidden]> wrote:

[snip]

Quoted message said:

Question: by turning are you meaning a whole round-a-corner turn or
just as little as slight side-to-side "lane shifts"? I'm explaining
that he can adopt that pose over the duration of a photograph without
necessarily going around a bend. But I'm not trying to say he could
keep his tires exactly straight doing that--I know of no-one who can
sprint and keep the bicycle moving along an arrow-straight line down the
road.

Dear BB,

This may be a clearer explanation of what I think that the
two riders are doing in this picture:

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

The rider on our left looks as if he's throwing his bike
past the vertical one way and lunging forward and throwing
his body past the vertical the other way, keeping the
contact patch under his body. Sorry about the crude
proportions of the ASCII diagrams.

front overhead
view contact
path
track--straight

/ /
/ / rider
/ /
\ \ bike
. . . contact patch

|
|
|
|
. .

\
\
\
/
. .

Note that the contact patch stays pretty much centered,
along with the center of mass, even though there must be
some small turning motions, too. (As you say, not arrow
straight, but the primary motion relative to the road is the
bike tipping back and forth, not the contact patch.)

The rider on our right looks as if he's leaning into a turn,
with the contact patch swerving back and forth as he carves
arcs--his bike and body are titled well past the same side
of the vertical:

front overhead
view contact
patch
track--turning

| |
| | rider
| |
/ / bike
. . . contact patch

|
|
|
|
. .

|
|
|
\
. .

He's not throwing the bike one way and his body the other
way--they both go the same way. Instead, he's throwing the
contact patch back and forth, like a sprinting ice skater.
This requires marked turning back and forth and is a
different gait, so to speak.

Since ASCII diagrams often show up more poorly as they grow
larger, here's a condensed version of the riders on our left
and right:

/ |
\ /
. .

| |
| |
. .

\ |
/ \
. .

Carl Fogel

Carl Fogel said:

But I still think that you found a hell of picture--I can't
imagine getting a real camera shot of two riders any closer
to dead head-on than this--obviously, at least one rider
must be a little off to the side, if not both.

Carl Fogel

I think it's pretty obvious that that picture is not representative of the
typical angles reached when a rider sprints. They are bumping each other a
bit, reacting to the bumping, or at least reacting to the proximity of the
other rider. The guy on the left's upper body is way off the center, right
shoulder over the left side of the bars (not in the best position for
power)... you don't see that in a sprint when riders are not bumping. The
guy on the right's left knee is kicked out, also not in the best position
for power. His knee is kicked out for balance because of their bumping or
near bumping. The guy on the right might be generally going in a straight
line, but there is a big swerve (rather than the turn you propose) going on.

On Sat, 2 Jul 2005 15:09:10 -0800, "Andrew Lee"

whatsupandrewathotmaildotcom said:

Carl Fogel said:

But I still think that you found a hell of picture--I can't
imagine getting a real camera shot of two riders any closer
to dead head-on than this--obviously, at least one rider
must be a little off to the side, if not both.

Carl Fogel

I think it's pretty obvious that that picture is not representative of the
typical angles reached when a rider sprints. They are bumping each other a
bit, reacting to the bumping, or at least reacting to the proximity of the
other rider. The guy on the left's upper body is way off the center, right
shoulder over the left side of the bars (not in the best position for
power)... you don't see that in a sprint when riders are not bumping. The
guy on the right's left knee is kicked out, also not in the best position
for power. His knee is kicked out for balance because of their bumping or
near bumping. The guy on the right might be generally going in a straight
line, but there is a big swerve (rather than the turn you propose) going on.

Dear Andrew,

Er, a swerve is a turn.

In case I haven't been clear enough, the rider on the right
looks as if he's turning, but he doesn't have to keep
turning--he only has to turn enough to convert his forward
momentum into enough force to get his body back up to a
reasonable position for balance, which is going to be the
blink of an eye:

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

As soon as he turns enough to get upright again, he could
switch back to the more normal gait (for lack of a better
word) that the rider next to him is using and stop swerving.

My point is that the rider on our right is acting like a
pendulum with both his body and his bike:

/|\
/ | \
c c c

This requires constant turning (swerving) back and forth,
leaning his bike and body the same way and moving the
contact patch from side to side. (He can quit very quickly
and probably did, but a turn is likely the only way to get a
bike tilted over as far as 25-27 degrees like that.)

The rider on our left is acting more like a hinged spring or
a double-jointed knee viewed from the side:

/|\
\|/
c

The contact patch stays pretty much underneath the rider on
the left with very little swerving/turning, while he throws
his body one way and the bike the other, not nearly as much
as we like to think--Joe Riel found that his bike tilted
back and forth only a total of 26 degrees, not 52:

http://groups.google.co.uk/group/rec.bicycles.tech/msg/708e9e998b35585c?dmode=source&hl=en

Joe was, I think, tilting his bike back and forth, keeping
the contact patch underneath him, not swerving/turning
enough to get the contact patch well to one side.

Most of this kind of motion is lunging forward, with the
rider's butt swinging out, but the rest of him swinging in.

Notice in the picture that the face and shoulders of the
rider on the left are on one side of a vertical line from
the contact patch, but much of his hip is swung out on the
other side of the same vertical line.

It looks like two entirely different gaits to me.

Carl Fogel

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

[...]

Quoted message said:

I'm saying that it looks as if the rider on our right in the
photograph must turn substantially to our right in order to
convert forward momentum to force himself and his bike back
up toward vertical:

Carl, you're not listening. Try this experiment: lay your bike over
at about a 30 or so degree angle while you stand on the side it is
leaning toward. Now, give the seat a shove with enough force to swing
over center and fall the other direction.
Imagine if someone took a photograph of the front of the bike between
the moment the bike lost contact with your fingers and the moment it
became vertical. That photo would show a bicycle leaning over center
entirely in your direction, which would make matters appear as if the
bicycle is falling toward you. However, only a moment later the bicycle
would be laying on the ground away from you.
How is such a feat possible? Keep in mind that there is no turning,
and no forward momentum involved.
ASCII:

Photo:
O
\ ^
\ |_/ | \ <--Carl
\/ \ | |
x ^
/ \ / \
/ _/ \_

^
|
bike

Result:

O
^
|_/ | \
| | | | What magic is this?!
|---+---- ^
| | / \
_/ \_

[...brevity snip...]

Point is: you're trying to extrapolate too much from too little data.
It's possible to argue both ways based on the photo of the cyclists:
either that the guy on the right must be turning, or that he's wiggling
substantially, but not really turning. But in the end, there is simply
not enough information available to state clearly one way or another
based only on the single photo.
I wasn't trying to say that your theory--that he is turning--is
untenable; only that you cannot exclude other possible explanations.

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

B.B. said:

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

[...]

Quoted message said:

I'm saying that it looks as if the rider on our right in the
photograph must turn substantially to our right in order to
convert forward momentum to force himself and his bike back
up toward vertical:

Carl, you're not listening. Try this experiment: lay your bike over
at about a 30 or so degree angle while you stand on the side it is
leaning toward. Now, give the seat a shove with enough force to swing
over center and fall the other direction.
Imagine if someone took a photograph of the front of the bike between
the moment the bike lost contact with your fingers and the moment it
became vertical. That photo would show a bicycle leaning over center
entirely in your direction, which would make matters appear as if the
bicycle is falling toward you. However, only a moment later the bicycle
would be laying on the ground away from you.
How is such a feat possible? Keep in mind that there is no turning,
and no forward momentum involved.
ASCII:

Photo:
O
\ ^
\ |_/ | \ <--Carl
\/ \ | |
x ^
/ \ / \
/ _/ \_

^
|
bike

Result:

O
^
|_/ | \
| | | | What magic is this?!
|---+---- ^
| | / \
_/ \_

[...brevity snip...]

Point is: you're trying to extrapolate too much from too little data.
It's possible to argue both ways based on the photo of the cyclists:
either that the guy on the right must be turning, or that he's wiggling
substantially, but not really turning. But in the end, there is simply
not enough information available to state clearly one way or another
based only on the single photo.
I wasn't trying to say that your theory--that he is turning--is
untenable; only that you cannot exclude other possible explanations.

Dear BB,

I confess that I don't follow your theory.

If the rider is on the bicycle tipped over at that angle,
with both the bicycle and his body well past the vertical,
is riding straight forward, what can he push against to go
the other way? Where is his leverage?

There's no leverage, so you seem to suggesting that the
rider on our right was finishing a turn at that instant,
straightening out, and his momentum from the turn was about
to raise his body and his bike upright--no turn, he falls
down.

In your situation, he's emphatically not tilting his bike
back and forth while his body sways the other way.

If he's "finishing" a turn, that's what I've been
saying---he's swerved as far to our left as he can and is
now at the x's leaning back toward our right:

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

[...]

Quoted message said:

Dear BB,

I confess that I don't follow your theory.

If the rider is on the bicycle tipped over at that angle,
with both the bicycle and his body well past the vertical,
is riding straight forward, what can he push against to go
the other way? Where is his leverage?

You're looking at it in that photo. He's in the process of ending
one lean and starting another. See below.

Quoted message said:

There's no leverage, so you seem to suggesting that the
rider on our right was finishing a turn at that instant,
straightening out, and his momentum from the turn was about
to raise his body and his bike upright--no turn, he falls
down.

In your situation, he's emphatically not tilting his bike
back and forth while his body sways the other way.

If he's "finishing" a turn, that's what I've been
saying---he's swerved as far to our left as he can and is
now at the x's leaning back toward our right:

In that case it seems we've been arguing (or discussing) "around"
each other. I consider a turn to be around a corner, to change the
total direction of motion. Side-to-side gyration I don't necessarily
consider to be a turn as long as the general direction of travel remains
the same, except maybe in Trevor's 10' side-to-side style.
I understand that for the rider to not fall in that photo the bike
would need to move laterally at some time, but that isn't a motion I
would consider to be a turn.
However, if we settle on the usage of the word such that "turn" also
includes any side-to-side shifting of the bike's tires on the ground,
then yes, I'll agree that he's turning.

Quoted message said:

.
.
.
x--> .

But he needs the rest of the turn to convert his "forward"
momentum to a force raising himself and his bike 26 degrees
or so.

I'm gonna quibble the concept of converting forward momentum to
somehow straighten him up. Forward is forward, not sideways. He can
use his forward travel and the grip his tires have on the ground to
steer his bike to a location under him, but that has to come from
all-new force out of him. There is no mechanism by which the bike would
be able to collect energy form his forward travel, store it briefly, and
then direct it into a lateral force.
If a bike could do such a thing you'd be able to turn any angle at
any speed, regardless of traction.

[...]

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

B.B. said:

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

[...]

Quoted message said:

Dear BB,

I confess that I don't follow your theory.

If the rider is on the bicycle tipped over at that angle,
with both the bicycle and his body well past the vertical,
is riding straight forward, what can he push against to go
the other way? Where is his leverage?

You're looking at it in that photo. He's in the process of ending
one lean and starting another. See below.

Quoted message said:

There's no leverage, so you seem to suggesting that the
rider on our right was finishing a turn at that instant,
straightening out, and his momentum from the turn was about
to raise his body and his bike upright--no turn, he falls
down.

In your situation, he's emphatically not tilting his bike
back and forth while his body sways the other way.

If he's "finishing" a turn, that's what I've been
saying---he's swerved as far to our left as he can and is
now at the x's leaning back toward our right:

In that case it seems we've been arguing (or discussing) "around"
each other. I consider a turn to be around a corner, to change the
total direction of motion. Side-to-side gyration I don't necessarily
consider to be a turn as long as the general direction of travel remains
the same, except maybe in Trevor's 10' side-to-side style.
I understand that for the rider to not fall in that photo the bike
would need to move laterally at some time, but that isn't a motion I
would consider to be a turn.
However, if we settle on the usage of the word such that "turn" also
includes any side-to-side shifting of the bike's tires on the ground,
then yes, I'll agree that he's turning.

Quoted message said:

.
.
.
x--> .

But he needs the rest of the turn to convert his "forward"
momentum to a force raising himself and his bike 26 degrees
or so.

I'm gonna quibble the concept of converting forward momentum to
somehow straighten him up. Forward is forward, not sideways. He can
use his forward travel and the grip his tires have on the ground to
steer his bike to a location under him, but that has to come from
all-new force out of him. There is no mechanism by which the bike would
be able to collect energy form his forward travel, store it briefly, and
then direct it into a lateral force.
If a bike could do such a thing you'd be able to turn any angle at
any speed, regardless of traction.

[...]

Dear BB,

Throughout a turn (however brief), the tires must provide a
force pushing the rider and bicycle into the turn and away
from the straight "forward" line that they would otherwise
take.

This "forward" line is constantly changing--it's the tangent
of the circle described by the turn.

(The circle itself can either be constant, as in a
constant-radius turn, or changing, as in an increasing or
decreasing radius turn.)

The rider and bicycle experience centripetal force as the
tires push them into the turn and force them away from the
straight line that they would take if the surface turned to
ice and there was no traction to allow the tires to exert
force.

The force is applied at the contact patch, as if some
invisible 2x4 lying on the road was being shoved against the
outside of the tire to push the bike and rider toward the
inside of the turn:

inside of turn \
x center of mass
\
c<------centripetal force
applied at contact patch

The rider and bicycle must lean toward the inside of the
turn to balance the centripetal force.

In the constant acceleration of a constant radius and speed
turn, the center of mass (well above the contact patch) is
constantly falling down toward the inside of the turn, but
is balanced by the centripetal force constantly shoving the
contact patch inward, too.

As a turn begins, the bike and rider going straight forward
start to lean over. They'd keep leaning over and fall to the
ground, but the front tire automatically starts to force a
turn. The rider unconsciously adjusts his fall to the inside
of the turn to match the force of the tire, and remains
happily balanced at an otherwise impossible angle.

He is converting his original "forward" motion to a constant
turning force applied at his contact patch in order stay
leaned over at an angle that otherwise would produce an
immediate fall to the ground.

He can keep converting his current "forward" motion into
this force forever if he continues turning at a constant
speed in a constant radius circle.

When a turn ends, the path of the bike and rider straightens
out, the centripetal force is reduced, less lean angle is
needed to balance it, and the bike and rider rise to the
vertical.

The force for raising the bike and rider is still applied at
the contact patch and is still centripetal force. The rider
alone cannot supply enough force to raise his body and bike
back to vertical.

He simply changes his balance ever so slightly so that the
angle of his lean and his fall toward the inside of the turn
are no longer a perfect match for the centripetal force
pushing at the contact patch.

He now "falls" back upward toward the vertical--the force at
the contact patch is shoving the tire back under him faster
than he is falling toward the inside of the turn.

As this happens, he automatically steers out of the turn,
until he reaches an upright position just as the last of the
centripetal force is eliminated and he's going straight
again.

Viewed from a trailer behind the bicycle, the upright rider
mysteriously leans over to an impossible angle and then just
as mysteriously rises back again.

What he's doing is adjusting the sideways centripetal force
and acceleration at his contact patch to the downward force
and acceleration provided by his tilted center-of-mass.

Because we often get things completely backwards, it can be
useful to look at a page like this:

http://www.glenbrook.k12.il.us/gbssci/phys/Class/circles/u6l1d.html

As it explains, the force involved in a turn is always
centripetal, pushing or pulling the rider (or ball on a
string or moon in orbit) toward the center.

When our senses confuse us, we end up believing in the
opposite of what's happening, but there is no centrifugal
force pushing us to the outside of the turn. It just feels
that way as a centripetal force pushes us to the inside of
the turn.

Carl Fogel