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Homework Help: Bike wheels forced precession

  1. Sep 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Bike has leaned, passing via corner.
    Centrifugal forces ( m*(V*V/R)*sin(alpha) ) are trying to upright the bike,
    gravity forces ( m*g*cos(alpha) ) to lean it further, so the balance lean angle comes
    tg(alpha) = V*V/R/g,
    where V is bike speed, R is curve radius, g is 9.81
    This math calculates the cog lean easily,
    but neglecting the gyro effects of rotating wheels.
    My question is how to assess this influence for a bike in corner?

    2. Relevant equations

    3. The attempt at a solution
    Let's assume bike is turning left. So the wheels, forced by the frame.
    Thus wheels try to precess, their angular momentum L which is pointing left,
    is forced to go up, due to the ext.momentum pointing up,
    this way wheel axle try to upright wheel, and bike respectively,
    creating frame reaction momentum M which counteracts,
    it's vector pointing backwards. This momentum can be
    calculated as cause of the forced precessing wheels to left following the corner.
    M= WxL , where W is angular speed via the corner : V/R,
    L is (wheel inertial moment)*V/(wheel radius),
    angle between them ( which sine will be used ) is 90+ alpha(lean angle)

    So, may I summarize that each wheel tries to help the bike go up,
    with momentum : (W/R)*((wheel inertial moment)*V/(wheel radius))*cos(alpha) ?
  2. jcsd
  3. Oct 3, 2009 #2
    if Mods consider my question posted into inappropriate section,
    please move it :smile:
  4. Oct 4, 2009 #3

    above to be read :
    ....with momentum : (V/R)*((wheel inertial moment)*V/(wheel radius))*cos(alpha) ,
    sorry for the typing mistake.
  5. Oct 6, 2009 #4


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    Hi twowheelsbg! Thanks for the PM. :smile:

    (have an omega: ω and an alpha: α :wink:)
    I don't think the rotational momentum or energy of the wheels has anything to do with the angle of the bicycle, nor does it provide any force.

    The equilibrium angle of the bicycle is solely determined by the equation you gave, tanα = v2/gr …

    it depends only on the speed of the bicycle, and the curve of its path, not on anything internal such as the rotation of the wheels.
  6. Oct 7, 2009 #5
    Thnks once for the reply,
    thnks twice for the greek characters :smile:

    While checking an gyroscope article,
    i read a paragraph with following:
    horizontal gyro is supported on a plate,
    his axle fixed only to spin by plate bearings.
    If we start to rotate the plate c.clockwise ( top view )
    this comes as external momentum M for the gyro, pointing up.
    If gyro rotates c.clockwise ( right view ) it's angular momentum L
    tries to go up, following it's change equal to the ext.momentum ( dL/dt=M )
    But the bearings restrict this axle movement, so the axle only pushes
    the bearings. This push produces reaction push from bearings to axle,
    which in turn ( pointing from our observer's view point through the diagram )
    acts as external moment again, attempting to rotate gyro axle c.clockwise
    (top view) as plate rotation dictates. In summary, the initial plate bearings push to gyro axle
    due to plate rotation ends with axle rotation same way. If we stop rotating the plate this push to axle disappears, so gyro keeps position and stops changing axle position.
    This was described as forced precession, and the bearings reactions were calculated as
    reason for this kind of precession, so momentum is
    M = R(reaction force at bearing ) * L ( lever or bearings distance )
    and ... here comes Rezal, Oiler and other wise guys which I admire :redface: ....
    M = ω1 ( plate rotation angular speed ) X { I ( gyro inertial moment ) * ω2 ( gyro angular speed ) } ... this vector multiplication same as scalar, vectors at 90 degrees.

    So same analogy I applied over the bike in the corner.
    The wheels are the gyros, bike is the plate, the only difference is
    that the gyros-wheels are not with horizontal axle.
    Wheels are changing axle direction while bike advances the corner ( plate rotates ),
    so following above logic wheels precess and there is momentum to frame,
    and momentum to wheels?

    Maybe, ( I am not sure ) those two moments are internal for the bike,
    and examining the outer influences they cancel each other out,
    and as you wrote, the bike balance is determined only by centrifugal and gravity forces.
    If we return back to plate-gyro example, gyro push over the plate would change it's position if plate is not well supported. I have felt this with working pc-hard drive in my hand.
    That's why maybe not, I am not sure :grumpy:
  7. Oct 7, 2009 #6


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    Hi twowheelsbg! :smile:
    Yes, I was only dealing with the equilibrium position …

    getting the bicycle into that position is quite complicated mathematically (and that's why learning to ride a bicycle is so difficult! :biggrin:) …

    as you say, it's exactly like trying to turn a gyro. :smile:

    For details of this "torque-induced precession", see …
    http://en.wikipedia.org/wiki/Precession#Torque-induced" :wink:
    Last edited by a moderator: Apr 24, 2017
  8. Oct 7, 2009 #7
    Hi, tiny-tim, thanks for the usefull web link!

    Sorry, but I can't see where exactly to place the 'separation' line making the difference :frown:

    Working pc hard drive in my hand pushes my hand transversely to my rotation,
    above plate receives reactions from the gyro and would incline if not well supported.
    Why then, the bike wheels reaction moments to frame, although small, wouldn't count?
    It seems rational to me these moments make small amendment to centrifugal one, this way the rider inclines further a little bit to compensate with increased gravity assistance to keep the blessed equilibrium.

    I humbly confess that trying to comprehend this phisics
    is far more difficult then negotiating corners :shy:
  9. Oct 7, 2009 #8


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    Yes, changing the angle of lean of the bicycle will produce/require an extra force, but if the angle of lean is constant, that doesn't apply, so in that sense the equilibrium angle of lean isn't affected.

    (except, I suppose, that even if the lean angle (roll angle) is constant, there will still be a very slight "yaw" rotation, since the whole bicycle is going in a circle … but that effect should be insignificant, and it won't affect the lean angle anyway)
  10. Oct 8, 2009 #9
    Passing a corner in physical context is very complicated of course,
    but I will try with a simplification. Lean angle changes almost all the time,
    and the balance condition is often very short time period,
    or fluctuations around such one happen.

    If, let's say approaching left turn having sufficient speed,
    we are obliged to lean into corner - otherwise the centrifugals will wipe us away from the curve. Leaning we use the gravity to balance. But how to lean? Pushing the bike to one side results in rider falling to the other, so we might lose control. That's why bikers use a technique, some call it 'counter steering'. We push the handlebar in opposite direction, right in our case. Bike 'feels' forced to go right, reacts with centrifugal impulse and starts leaning left. This is the moment when we return handlebar to neutral position, while bike continues to lean left under increasing gravity moment. If we take no further action, we would fall to left side, but approaching further into the left corner, we finally steer to left correcting the curve exact radius. This way we can increase centrifugal forces rapidly and to compensate the proportional advance of the gravity assisted lean. So the balance is felt somewhere ( thanks God not calculated :smile: ) and the curve negotiated. Of course we can adjust this balance going low ( speed decreased or handlebar adjustment in neutral direction widening the curve radius ) or going up ( speed increased or handlebar adjustment in more left tightening the curve radius ).

    Looking back at the boring calculations,
    it's obvious that if the coefficient of static friction is up to it's max ( road concrete - tire rubber ) - let's say 1.1,
    we can afford maximum lean 48 degrees of center of gravity (cog),
    side friction forces would be almost equal to bike's weight, and that would be their maximum assistance in compensating the side wiping centrifugal ones.
    Additionally when bike leans,
    ( especially a motor bike having sufficiently wider tires ),
    bike lean is roughly up to 5-6 degrees beyond cog lean,
    because when leaned, contact patch moves aside from tire center line.
    So adding this to perfect friction condition, bike leans up to 53-54 degrees.

    If I had figures for wheels inertial moments,
    I would calculate by the above formula this gyro moment -
    probably a small value, requiring degree or two additional gravity lean to compensate.
    ( I still experiment with practical ways to assess those wheel inertia moments ... sound ideas are welcome :blushing: ). I will update if any different results.

    So, probably, total bike lean calculable up to my knowledge/data tops to 55-56 degrees.
    In WSBK sport conditions, present lean max as far as I know is 71 degrees ....
    the only explanation in my mind would be that these guys don't use unknown physics,
    but use very sophisticated tires with a high-tec rubber-silicone compounds,
    granting them a coef. of static friction much over 1.1 ... amazing technology for me :bugeye:
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