The bit I'm interested in is the gyroscope effect of the wheels.

If I get a puncture and take the wheel off, hold the axle in my two
hands and spin the wheel, it is very hard to alter the direction of
the axle if the spin is fast.

Can anyone explain this (gyroscope) phenomenon in simple English with
little or no maths ?

Hi gruffgareth and thanks for the question,

When you hold your bike wheel and spin it,  the resistance you can
feel when you try to alter the angle of the spindles is called
“gyroscopic inertia”
as Sir Isaac Newton explained  “ a body tends to continue in its state
of rest or uniform motion unless subject to outside forces.” In other
words something will stay still or continue moving unless something
else affects it, in this case you, trying to alter the angle of the
spinning wheel.

For a simple explanation using your example of the spinning bicycle
have a look at this website:
http://www.exploratorium.edu/snacks/bicycle_wheel_gyro.html

For an explanation of "gyroscopic inertia" have a look at this website
 (near the bottom)
http://www.probertencyclopaedia.com/SG.HTM#GYROSCOPE

"How Gyroscopes Work" is a detailed description without too much
maths.
http://www.howstuffworks.com/gyroscope.htm

and for the full maths version have a look at this website:
http://www.geocities.com/CapeCanaveral/2188/main.html#index

Hope this doesn´t set your head spinning too much.

THX1138

---

Clarification of Answer by thx1138-ga on 14 Jun 2002 12:50 PDT

Sorry I forgot to mention the type of searches I used to try and
explain your question:
Search Strategy:
gyroscope science explanation
://www.google.com/search?hl=pt&as_qdr=all&q=gyroscope+science+explanation&lr=

nice googling !

Although you limited your question to gyroscopes, I'd like to add
something about gyroscopes and bicycles.  The way your wheel is
connected to your frame is more important for balance than gyroscopic
action.

If you look at the frame of your bike, you'll notice that the axis of
rotation for the front wheel is not vertical.  This is called caster.
As you start to tilt right (don't fall), the wheel will turn right. 
This is because the wheel is touching the ground behind the axis of
rotation.  When the bike turns right, your body wants to keep going
straight, pulling you upright again.
This web page describes this process very well.  It also has a good
picture showing caster and trail.  It even has references.
http://www.dclxvi.org/chunk/tech/trail/

I'd also like to add that the resistance to 
the change in rotational axis due to gyroscopic effects
is *not* directly holding you up.   The tire only weighs a 
couple pounds and you weigh 50-100 times more.  People always
cite the spinning tire in the hands physics experiment, but 
the fact that you *can* still move it with your hands is proof
that it's not a significant direct effect.  We've spun bicycle
tires up to over 100mph and still were able to force them to 
change their rotational axis.

What actually happens is that gyroscopic precession (ambly described elsewhere)
causes the handlebars to turn (easier than trying to describe the right
angle nature of the change in the tire's rotational access) the 
same direction that you lean.
As the bike starts to fall to the right, the gyroscopic
precision turns the front tire to the right and that
drives the tire back underneath the center of gravity, keeping
the bike from falling over.   At all times the bicycle is attempting to balance
by means of the gyroscopic precession, plus the above mentioned castor
and rake, trail (and tire profile) which all cause it to try to keep the 
front tire underneath the Cg causing the bike to remain either upright
or at least at the same lean angle it started at.

What's more interesting is how you actually initiate
a turn on a bicycle or any balancing vehicle.  It's called
countersteering and it means that you actually have to counter
the natural balancing effects above to cause it to turn.   Practically 
speaking it means that if you want to turn right, you actually apply
pressure to the handlebars such that they briefly turn left.
Simpler said, push on the right handlebar to initiate
a turn to the right, and left bar to initiate a turn to the left.
What actually happens is the tire turns very slightly to
one way, the bike begins to fall the other way, and
then all the above stabilization mechanisms kick into place
to hold the bike at a particular lean angle which will
cause the bike to go around in a circle.   The harder you
countersteer, the quicker or further it turns.   You also
countersteer to stop a turn by turning the bars back
toward the direction you're leaned which causes the bike
to stand back upright. 

Because of the greater mass of motorcycles most motorcycle riders
are exposed to the concept of countersteering eventually,
many right from the beginning.  It's critical to making 
an effective turn.  However most people are never taught that 
countersteering works on bicycles but their bodies simple learn it.
The process of learning to countersteer is basically the
"steep learning curve" part of learning how to ride a bicycle and most
people never know they're doing it.   Watch a kid who's just
learning when they're presented with an obstacle to avoid.
You'll often see them develop wild wobble where they 
manage to traverse 30-40 feet and smack right into the
thing they're trying to avoid.  The problem is their brain 
(and possibly their parent) is telling them to turn the
handlebars one way, while their body, having just learned
countersteering knows it need to go the other way.  The conflict
is displayed as wobble and in the end they average a straight line 
toward the obstacle.

Try countersteering yourself.  Ride your bike, apply a gentle
pressure to the right bar (as though you want to turn it
to the left), and you will end up turning right.   Try 
twisting the bar with one hand right at the stem.  You'll
see the same effect.   

    ian

One further thought.  The original question was how
to explain the gyroscopic effect in simple terms
without math.  I think I can do it without using
the words "angular momentum", "torque" and "gyroscopic precession". 
:-) 

Imagine someone fires a bullet horizontally past you from 
some distance away.  For all practical purposes the
bullet travels a sraight line toward you (ignore gravity for now).
As it passes you, you reach out and very quickly push down on
the bullet for a fraction of a second.  You've now deflected
the bullet onto a new course angled down toward the ground.
It was going horizontal and now it's angled down.  The change
in angle depends on how fast the bullet was travelling,
how heavy it is, and how much force you applied to it 
as it went by.   One straight line, with a bend in it
followed by another straight line.

Ok, so now someone ties a string to a second bullet and attaches
the other end of the string to a post with a pivot on it.
They stretch the string taut point the gun perpenticular to the
the string and fire fire the bullet and (again discounting gravity and drag)
the bullet travels horizontally.    But now because it attached
to the string it follows a curved path in a horizontal plane.  A 
circular path around the central post, round and round.    Again as the 
bullet passes by you, you reach out and push the bullet downward.   Again
it is deflected from its horizontal path toward the ground, but
the string still has a hold of it so it simple swoops down closer to
the ground reaching its lowest point one quarter turn (90 degrees)
from the place where you deflected it, and then continues on around
reaching its highest point 1/2 turn (180 degrees) beyond that
and when it returns back to the place where you deflected it
it occupies almost *exactly* the same space, just at a different
angle.   

If you continue to delfect it downward each time it comes
by the low side will get lower, the high side higher
and ultimately it continues to travel around and round the
post in a circle.   The plane of that circle is simply 
tipped more and more away from horizontal and where you 
deflect it, is actually "noticed" 90 degrees forward in 
rotation.     

THAT is gyroscopic precession.    

Any tipping force you apply to a spinning object is equivalent to 
deflecting the path of that bullet (every atom in the spinning object is
like the bullet on a string around the spinning axis), and it 
reacts to the force by tipping 90 degrees in rotation beyond 
the point you applied the force.   

For a simple home experiment, tie a tennis ball to a 2-3 foot
long string and then spin it around vertically.   Once you
have it going, stick your foot out and let it
deflect the ball a little bit (careful not to bean
yourself).   Notice which way the spinning circle
tilts.   If you spin the ball and string around
clockwise, and push outward with your foot at the
bottom of the circle, the circle will tilt out on the 
left, in on the right, instead of out at the bottom as 
you might originally expect.  

Try it.   Gyroscropic precession need not be so mysterious.

    ian

Wow folks, an impressive set of responses !

There was a Professor Laithwaite (at Manchester Uni ?) who was
intrigued by spinning objects. Despite being a prof., I don't think he
understood the gyroscope effect, as he was convinced that spinning an
object could cause anti-gravity (like the hanging bicycle wheel).

I had wondered about the gyroscope effect many years ago (when fixing
a puncture) - we never did this in school physics (seems like a major
omission). Since there were no teachers to ask, I imagined my own
explanation, but never followed it up till I came across this
fascinating Google answers site.

I tried the set of web sites from thx1138-ga. Some of the sites
explained it by saying it was because of "angular momentum" or
"gyroscopic inertia" - which isn't any kind of explanation ! But the
site at http://www.howstuffworks.com/gyroscope2.htm broke the wheel
down into moving segments and was just the kind of explanation I was
looking for.

daemon-ga gave a similar explanation to that site using a very nice
mind-illustration with bullets on pieces of string. Mention bullets
and you'll have the attention of schoolboys, so this could be very
good for the classroom ! Though flexible string does lose the forces
that operate among the opposite sides of a rim and the axle in a rigid
wheel.

I had been just thinking about the gyroscope effect, but then sdk-ga
and daemon-ga gave some fascinating (and new to me) insights into the
other effects that keep a bicycle upright. I had wondered how those
micro-scooters could possibly work with such small wheels ! I am
really amazed. I already had a lot of respect for the bicycle, but now
I'm thinking it may be one of the simplest, yet most sophisticated
machines ever invented. Excellent for individual and planetary health
into the bargain !

If it's OK, rather than select a "winner" I'd like to donate $15 to
the coming famine in southern Africa (probably thru Oxfam).

Thanks again for the great responses, 
Gareth

It's not a contest for your money.  thx1138 answered the question so
he deserves it,
as that's how this site works. I am not a Google researcher.  I just
provided my comments
because I gain some enjoyment out of explaining everyday physics in
everyday terms,
particularly the study of balancing  vehicles and dispelling the myth
of "gyroscopic action
holds you up directly" is a perpetual goal of mine.    
Spelling mistakes are left in as an exercise for the astute reader. 
:)

    ian

Can anyone explain why I am NOT able to ride a bike on a tredmill?  I
tried to do so and it is extremely wobbly.

Centripedal acceleration.   Or rather lack thereof.  
There's a simple equation which describes the relationship 
between your forward velocity, lean angle and the radius
of a turn.   It is.
lean angle = 90 - atan(v^2/r/g)
where
v = velocity
r = radius
g = gravitational constant
(angles in degrees not radians).

When a bike is going around a turn at a specific speed
it leans at a particular angle, but from the perspective of
the bike it is basically still balancing upright.  The forces you
feel as you turn are down straight through the bike rather than
sideways like in a car.    This relationship is all part of the balancing
act and is an important component of overall
stability.  The faster you go the more stable you become.
At a high speed, it takes a much greater lean
angle to induce a sharp turn than at a lower speed, 
and all the balancing effects described resist large changes
in lean, thus it helps overall stability.

Ok, all that said.. That equation only applies when you're
actually moving and have forward momentum.   When you're
riding on a treadmill (and they do make special bicycle
treadmills with two rollers in back and one in front), things
like gyroscopic precession, rake, trail etc, all still work, but
there's no forward momentum and thus no stability
to be had from the relationship described by the equation
above.   100% of your stability has to come from the front
tire driving under the CG, and there's no centripedal
acceleration to "lean against" as your speed increases.

This is also why riding a unicycle is so tricky.   At low speeds
it's a constant wobbly balancing act, but if you actually
get up some speed on a unicycle it'll become more stable, 
and it will lean into turns just like a bicycle (and numerous other craft,
including snowboards, mono-skis (snow and water), jet-skis etc, skateboards)

 
    ian

Thanks Ian, I think you may have uncovered the answer, but I disagree
with your particular explanation.  When riding on the treadmill, I do
have forward momentum relative to the treadmill.  Thus my lean angle =
90 - atan(v^2/r/g) should be correct relative to the surface of the
treadmill.  After all, suppose instead of a treadmill it was a flatbed
train moving in the opposite direction, etc.  Eventually it is easier
to visualize that the ground moving back one way is equal to the bike
moving forward the other way.

The key I think from looking at the equation and lies in the fact that
I have to start out slow on the treadmill because of the difficulty of
getting my bike on the treadmill and pushing the start button. 
Because I have to start out so slow, it is more difficult to balance
(lower v) just as it is on the road.  If I am brave enough to try, I
would imagine if I turn the speed up, I could ride relatively stable. 
The only problem is that on the road you have a few feet to the left
and right to play with where on a treadmill in my bedroom, there is
considerably less space.

Thanks for the equation and the time.

Nope, you're in the wrong frame of reference.  
Let's take your flatbed train, pretend it has a continuous line of cars to
ride over down its entire length.     If you were standing still on one
of the cars while it went around a sharp corner, you would have to lean into the
turn to remain balanced.    Same if you were sitting still on your bicycle
on top of the train car attempting to balance.   (growing up in a rural western 
state, I spent quite a bit of time riding in the back of pickup trucks and flatbed
trucks so it's something I can relate to quite easily).

Now let's say the train is going 20mph through a sharp turn,
and you are on the bike riding down the cars the opposite direction
also at 20mph.   You would remain stationary relative to
the ground and not have to lean the bike at all, as the train
passed under you.   If you tried to lean into the turn you would
simply ride off the edge of the car onto the ground.   
If you turned around and rode the same direction as the
train you'd have to lean even further as you and the train
both rounded the corner at the same time.   You'd have to
plug 40mph into the eq above instead of 20 or 0.  

The only momentum that matters for the lean angle is that 
relative to a fixed frame of reference (or as fixed as it can be)
which in this case is the earth.     

On the treadmill, the only thing that has any momentum
is the treadmill itself and the tires of the bike.    Yes you'd
have to lean a little to travel back and forth on the treadmill
but that's just a simple balancing act. There's no resistance to
falling over due to your foward momentum because you have none.

Too bad Merry-go-rounds have been banned from all playgrounds.
They were a fantastic tool for learning physics, including this little
particular problem.   (if you run in the opposite direction on a merry-go-ground
you don't lean in at all).  

   ian

Thanks for hanging in there with me Ian.  But I am still confused. 
What if my treadmill were a big concrete slab with trees and buildings
on it.  I cannot distinguish being on this treadmill (which also moves
all the air in the room with it) from riding on the street itself.  So
how does my momentum all of a sudden change?  Does the momentum of the
treadmill (which is now considerably greater) come into play? and if
so how?


Also, suppose I am in a train moving at a constant velocity (Constant
direction and magnitude).  If I am riding my bike in the opposite
direction at the same speed so that relative to the ground I am not
moving, it seems my turn angle would be the same as if I was riding on
the street at this same velocity.  How does this differ from the
situation on the tread mill?

If I could use the turn angle of my bike to calculate the velcoity of
the train relative to an external frame of reference, then doesn't
this violate what I learned as the first premise of relativity, namely
that the laws of physics are the same as observed in any
non-accelerating frame, i.e. you cannot conduct an experiment inside a
closed box that will tell you whether that box is moving at a constant
velocity relative to an outside frame of reference.  Please let me
know where I have all of this mixed up.

And again, thanks for your patience

The equation above to compute lean angle, has as one of its terms g,
the
acceleration due to gravity on earth.  In all of your examples, big
concrete
slab, moving train, and treadmill, they are all operating within a
much larger
overriding context, which is the earth and its gravity. In addition
the
velocity term is along a vector always perpendicular to that
gravitational acceleration.  They're tied together.

On the treadmill there is no velocity vector perpendicular to the
force of
gravity acting on the combined mass of the bike and yourself.  Only
the tire
rims are moving and that all motion cancels out relative to the
earth's
frame of reference because it's rotating.

If the slab/train/treadmill were all so massive that they generated
their own
gravity while floating freely in deep space then yes you could ride
across them
and compute a lean angle independent of what is observed riding over
the ground
on earth no matter how fast they were moving.  But then the constant
'g' in
the EQ would also have some other value, and the lean angle would 
be correspondingly different.  

You're right that in your black box you cannot detect that the box is
moving at
a constant velocity relative to an outside frame of reference from
inside the
box, but that's only if the box is free to move along a "true"
constant
velocity vector in deep space..  Velocity is a vector not just a
number
and it's only straight if it's free of external forces.  As soon as it
is acted upon by an external force (in this case gravity) you'll 
notice something in the box although it may take a while.  If the box
is
travelling over the ground then you feel the *external* 
force of gravity inside the box, while the ground is supporting it. 
Inside
your box you'd know immediately that the box was resting on a larger
object
with a gravitational pull thus you already know something about the
box's
frame of reference, which you didn't think you'd know.  

You say..  "fine, so perhaps it's orbiting around the earth in
freefall, then
you couldn't tell."  But if it's a box with rectangular dimensions,
tidal effects will eventually orient it a particular way, and you will
gradually
tend to float to one end of the box or the other over time, because in
an orbital context all the particles of the box and you inside *want*
to orbit at different velocities.  (lower orbits are faster, higher 
orbits slower).  They can't though because they're all connected to 
each other so stuff between the CG of the box and the gravitational 
source "falls" inward, and stuff outboard of the CG is "flung"
outward.
Some great Sci Fi based on this effect.    Check out Niven's short
stories.

Using relativity to describe macroscopic effects can be dangerous for
the above reasons.  "constant velocity" simply cannot apply to an
object
composed of more than one particle in a big gravitational well unless
it's
falling directly toward the gravitational source, and that's not the
case here..
Resting on the surface of that object pushes this all to the point of
being
a worthless comparison.  

But forget all that mumbo jumbo.  You *know* intuitively how this
works.  What
happens if you set a cup of coffee on the dash of your car and turn
the car
sharply? What if you hang a set of fuzzy dice from the mirror by a
string
and go around a corner? The angle that the dice hangs is the very same
computed lean angle as in the EQ above.  A bicycle riding in front of
the car would lean *exactly* the same amount as the fuzzy dice.  
(note
mass is irrelevant here). 

Just to bring this briefly back to why this is important to bicycles,
all the *other* stabilizing effects of a bicycle (gyroscopic
precession,
rake, trail, tire profile) all are more pronounced with increased 
or rapid changes in lean angle.   So if going faster around the
same turn produces a larger lean angle due to the EQ above, 
it means all the stabilization effects act with all the more
authority.   The bike is more stable when going fast, and less
so when going slow.  

This isn't rocket science.. Ok, well it is rocket science, but it's
not hard.  ;)

    ian

Oh crud.  That formatting is evil.  
It didn't look like that until after it was posted.  Sorry.
What an annoying glitch.  Wish I could clean it up.

You say that there is no velocity vecotr perpendicular to the
gravitational field on a treadmill.  If the veolcity vector of the
bike relative to the earth is the significant vector, then what
happens when I ride my bike (or walk for that matter) from my seat on
an airplane to the bathroom.  According to the equation and this
account of the appropriate velocity vector My lean angel would be
enormous.  Again, please help.
Thanks,
Buddy

When you walk from the front to the back of the airplane, you're
not even making a turn, at all. It's a straight line overlapping the
plane's own straight path of travel.