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Why harder to turn motorbike handle at high speed?

  1. Mar 26, 2012 #1
    When riding a motorbike, at relatively high speeds the handle becomes harder to turn. Is this due to the conservation of angular momentum or is there another reason for it? I say another reason because the handle turns back if released ( I think). The conservation of the angular momentum doesn't justify the reaction.

    Note: Don't try this because it is extremely dangerous to turn the handle except for very small angles.
     
    Last edited: Mar 26, 2012
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  3. Mar 26, 2012 #2

    rcgldr

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    It's mostly due to angular momentum. There also is a caster effect, but I don't think that has as much effect as angular momentum. At moderate speeds, a motorcycle tends to straighten up due to steering geometry (trail), but as speed increases, the combined effect of angular momentum in the front and rear tires reduces the tendency to straighten up and instead a motorcycle tends to hold the current lean angle, taking as much effort to straighten up as it did to lean over.
     
  4. Mar 26, 2012 #3
    Thanks for the reply.

    The phenomenon seems complicated then. Lets calculate for a simple case when the wheel is not in touch with the road but it is rotating with the same angular velocity:

    Suppose the wheel is 5 kg. its radius is .4 m and is angular velocity is 69.5 rad/sec ( equivalent of 100 km/h)

    The magnitude of the angular momentum , L, becomes 55.6 Nm/s. Then we have T=dL/dt=55.6dθ/dt .

    50 N force in the handle would cause the torque of T=50*0.3 =15. ( the handle is assumed 0.6 m with the pivot in the center). Such a big force results in the rate of dθ/dt=15/56.5=0.265 rad/sec.
    This means we need about we need to apply such a force for about 3 seconds in order to turn the handle by 45 degrees. Quite slow for such a big force!

    For a running motorcycle, the wind force should be considered, I'm not sure how would it be on a turning wheel with thin spokes.


    Edit: 0.4 m for the radius of the wheel is a little high, 0.3 m is fine. The force becomes 3/4 of what we calculated then.
     
  5. Mar 26, 2012 #4

    jim hardy

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    I dont think it's gyroscopic at all.
    Wright Bros discovered a bicycle with caster in wrong direction(leading) is unridable.

    I'd think it's due to turning handlebar causes bike to turn sideways
    by an amount proportional to handlebar movement,
    changing direction of its momentum

    as forward component of v increases so does sideways component of v required to cause same direction change.

    so Δvsideways / Δhandlebar is in proportion to speed
    and trailing caster gives feedback in proportion to sideways force, masideways, a increasing with speed.

    sound plausible?

    Actually turning handlebars initially moves wheels out from under CG so bike leans into turn ,
    and that's why to start a turn you must initially pull bars opposite direction you wish to turn... but that's another story.
     
  6. Mar 26, 2012 #5
    Thanks jim,

    It is plausible but i am not sure yet. If its gyroscopic, the torque must be small for a very light wheel.
     
  7. Mar 26, 2012 #6

    rcgldr

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    Unridable would be a reference to stability, but the OP was asking about the effort to steer.

    This is one of the effects I mentioned. Both caster and angular momentum effects combine to increase the required torque to steer, but it's my opinion that angular momentum contibutes more, based on the fact that self-correction, which is related to caster and trail, essentially vanishes at high speeds (100 + mph, 160 + kph), and a motorcycle just tends to hold it's current lean angle (or the recovery is so slow that it's imperceptible at high speeds).
     
  8. Mar 26, 2012 #7

    jim hardy

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    My experience is not quite the same.
    Old British bikes of 1950's had an adjustable "Steering Damper" to prevent oscilations. The self-correcting negative feedback from geometry(caster, trail, and offset of forks from steering axis) combined with increased 'gain' ( if you will pardon the term) from increased forward speed, combined with inertia of whole front suspension about steering axis giving rise to closed loop oscillations that were divergent .

    The faster one went the more evident it was that one was approaching instability so he tightened up on the damper which was just a friction plate.
    Steering dampers disappeared from British bikes around 1970 and i wondered "How'd they do that?".

    If you know, please share ..

    Now that IS reminiscent of a gyroscope... are the moments in proper direction to explain it?
     
  9. Mar 26, 2012 #8

    K^2

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    It's both caster and gyroscopic. Gyroscopic effect is self-explanatory, but only applies to sharp movements. The reason caster return force increases is because the centrifugal reaction force on the wheel increases at higher speed, and due to caster angle, part of that force is transferred back as steering torque.
     
  10. Mar 26, 2012 #9

    rcgldr

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    It could be a combinations of too little trail, too much flex in the frame or swing arm (to rear tire), ... harmonics issue. The Royal Enfield Atlas had this problem. The 1970's Nortons didn't. Not sure about BSA or Triumph.

    The 1972 - 1975 Kawasaki Mach IV 750cc triple cylinder two strokes had this problem. I recall twin steering dampers, but don't recall if this was factory or aftermarket. If frame or swing-arm flex was the issue, it may not have completely cured the problem.

    I recall reading that the 1984 Suzuki GS1150ES could wobble the handlebars if you tried to ride hands free (sort of an old school test for wobble), but don't know if it needed a steering damper for racing.

    The 1992 to 1997 Honda CBR900RR motorcyles had a bit of wobble due to not enough trail (excessive triple clamp offset) in steering geometry. The 1998 and later versions fixed this by reducing the triple clamp offset by 5mm (0.2 in) (which increases trail). From wiki:

    In 1998, Honda continued subtle refinements in the CBR900RR's chassis. It saw frame stiffness closer to the original model's, revised suspension internals, and 5 mm (0.2 in) less triple clamp offset (an almost universal aftermarket upgrade to previous models)

    http://en.wikipedia.org/wiki/Honda_CBR900RR

    I don't recall any other motorcycles since the 1990's having speed wobble issues.
     
    Last edited: Mar 26, 2012
  11. Mar 27, 2012 #10

    jim hardy

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    My '56 Enfield twin( actually badged as Indian ) was textbook. Hands off a gentle tap on either bar would start a damped oscillation and the faster one went the more one had to tighten the damper to keep it from diverging. Damper was a prominent knob protruding up from center of steering head. It simply tightened a clutch to apply friction to turning of triplehead (thanks for that term) .

    Seems my friend's '65 BSA had a damper(memory fades) but the '72 Triumph i had for a while did not. That Triumph was amazingly stable and responsive. It was the bike made me aware of what they can do with geometry..

    Like you i've never felt it on any newer bike. Well, my last bike, 85 Sportster had some barely detectable but was well damped. Felt a lot like the 72 Triumph that's why i bought it..


    "Real motorcycles you kick-start."

    Thanks for sharing your expertise.

    old jim
     
  12. Mar 27, 2012 #11
    1, at high speeds you are more balenced due to air rushing around it
    2, as you turn there is a centrafugal force pushing you ouwards
     
  13. Mar 27, 2012 #12

    rcgldr

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    Mostly about being just old enough (started riding bikes in the late 1960's) to have heard the stories about the Atlas. I only saw a few of them. I think these had a second shift lever for instant shift into neutral? My only British bike was a Norton 850 (actually 828cc) Commando, back in the 1970's, right side shift with opposite pattern (one up, three down for gears 1->4).

    Back to the OP, the highest cornering speeds for motorcycles are around 200 mph, at Ilse of Mann TT (Tourist Trophy) track, which is a timed event as opposed to a normal race. The highest cornering speeds on a conventional track probably occur at the Daytona superbike race, which uses a banked turn that exits onto one of the tri-oval straights, and I recall that speeds reach 180mph by the exit of the turn, with the bikes well leaned over because of the banked turn. I recall a few articles and television commentary about how much steering effort it takes to straighten the bikes back up after exiting the banked turn.

    The mathematical models for bicycles and motorcycles indicate that there's a capsize speed where a bike will tend to slowly fall inwards with no steering inputs (hands free) once above a certain speed. Looking at the graphs, it's barely "unstable" so the rate of lean may be so slow that it's imperceptible and results in what I described as a tendency to hold a lean angle as opposed to straightening up. I only got to try 100 mph turns at a track one time, and was aware of this effect, and the bike did what I expected it to. Once leaned it stayed leaned until I countersteered to straighten it up; relaxing on the handlebars had no perceptible effect.

    Note self stability just means a tendency to straighten up, there's no direction stability as a lean distubance will change the direction of a bike, but it will recover to vertical unless at high speeds.

    At Delft University they did a study and experiments, but the model predicted capsize (very slow inwards lean instead of recovery) speed at 8 m/s while the actual bike was "very stable" at 30 kph = 8.33 m/s. The width / profile of the tires (maybe just the front) affects this, and I wonder if the mathematical model didn't take this into account, or if there was some other issue involved.

    Link to links to articles:
    http://home.tudelft.nl/index.php?id=13322&L=1 [Broken]

    video is the last one on the page, the 30kph run.
    http://bicycle.tudelft.nl/schwab/Bicycle/index.htm

    The point of all of this is that the angular momentum effects dominate over self-stability effects like trail at high speeds, and I would assume the increase in steering effort is because of the increasing angular momentum effects at higher speeds.
     
    Last edited by a moderator: May 5, 2017
  14. Mar 27, 2012 #13

    jim hardy

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    AHH Yes, the neutral finder.. with that awful clutch you needed one.
    BUt it was an interesting bike. Alternator coils under flywheel , ahead of the time for 1956. And magneto ignition. i didnt even have a battery. But the lighting i'm sure was inspiration for Mills Brothers' "Glow-worm" song. You know the old Brit bike jokes, " Prince of Darkness's family name is Lucas "

    And i was outraged when US Govt mandated gearshift on wrong side. I still sometimes tap rear brake instead of gearshift...

    That high speed uncorrected lean is interesting. Windage could come into play too i suppose, and the tire's point of contact moves off centerline of forks too.

    Last year EBAY had an Enfield interceptor in Oregon. Just a little out of reach for me, but i craved it.

    Much to learn. Thanks again !.
     
  15. Mar 27, 2012 #14
    It's because of conservation of momentum, but this resists the movement of the steering wheel in an indirect way.

    If you had a bike with the front wheel off the ground, and the bike clamped down, so you could only turn the front wheel, but no other movement of bike is possible, and you then you made the front wheel spin very quickly, you would be able to turn the handlebars with no resistance. If you would turn the handles to the right, there will be a torque that will want to make the bike fall to to the left, but this torque does not oppose the turning of the handles. Only if the bike can fall fall over to the left, that will in turn produce a torque that will make the handles turn to the left again. If the bike can't fall over to the left, there will be no torque to oppose your turning of the handles.

    If the bike is riding normally, a movement of the handles to the right, will make the bike lean to the left very quickly, partly because of conservation of angular momentum of the front wheel will produce a torque, but mainly because the friction with the ground will move the wheels to the right. This falling to the left will produce a torque that will make the front wheel turn to the left.
    If you quickly move the handles a bit to the right and then hold them still, the bike will continue falling over, and this will continue toproduce a torque that will make the handles turn back, until the bike has fallen completely. Note that the torque is not produced because the bike is leaning over, but because the lean angle increases.

    The torque that turns back the handlebars, is proportional to the angular momentum of the wheels, wich is proportional to the speed, and how fast the bike falls over for a certain steering angle is also proportional to speed, so torque pushing the handles back, depends on the square of the speed
     
  16. Mar 27, 2012 #15
    Thanks willem2, for analyzing the two cases separately and making it clear this way.

    First, I had overlooked the opposite torque on the body, but the torque is small for the total mass of the bike and the person riding it.

    Second, could you explain why " This falling to the left will produce a torque that will make the front wheel turn to the left"?

    also about the relation between the leaning angle and the produced torque.

    Thanks.
     
  17. Mar 28, 2012 #16

    rcgldr

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    It's due to angular momentum right hand rule. The angular momentum for the front wheel is a long vector to the left, the angular vector for a left "roll" (leaning to the left) is a short vector to the back. This results in an angular response or torque vector that is "up", which would tend to steer the front wheel to the left.

    At the rear tire, the angular momentum has the same vectors, with the rear tire trying to yaw the entire motorcycle to the left.
     
  18. Mar 28, 2012 #17

    jim hardy

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    That's interesting never thought of it.

    So a riderless bicycle should fall left more often than right?

    Now - right near the center of the machine is a massive flywheel spinning several times more rpm than the wheels, and it probably outweighs either wheel but not both. It's the cranksahft and counterweights of course.

    Does it play into left lean as well?
    I need a gyroscope.
     
  19. Mar 28, 2012 #18

    rcgldr

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    Using a vector to describe angular velocity, momentum, or torque is just a mathematical convention, in this case to indicate the front tire is spinning forwards. For mechanics, I don't see any reason that "left hand" rule couldn't be used instead, in which case the front wheel's angular momentum points to the right, a left roll points forward, still resulting in the wheel turning left if not opposed.

    If the bike leans right, the front tire turns right, using either right hand or left hand rule.

    The "long" vector part is only true at high speeds. If the wheel is spinning slower, the vector would be shorter.
     
  20. Mar 28, 2012 #19

    jim hardy

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    oops - thanks !
     
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