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I Way the cornering forces are felt when riding a motorcycle

  1. Dec 19, 2017 #26

    jbriggs444

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    60 degrees from the vertical? 30 degrees from the horizontal? That's an easy angle. 2 g's on the diagonal.

    You can easily see that with an equilateral triangle cut in half. The diagonal is twice the length of the half side that remains after the cut. The angle of an uncut corner is 60 degrees. The angle of the corner that was cut is 30 degrees. The half side represents gravity. The diagonal represents the diagonal acceleration. The length of the cut represents the centripetal acceleration.

    [Introduction to trigonometry]
     
  2. Dec 19, 2017 #27
    Well, how much force is that they feel when leaning 64 degrees which is what equals to 2g I think? Sorry I have no idea how to do the trigonometry bit
     
  3. Dec 19, 2017 #28

    jbriggs444

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    Trigonometry....

    You start with a right triangle. i.e. one with an angle that is at 90 degrees. That is the angle between gravity (vertical) and centripetal acceleration (horizontal).

    You pick one one of the two remaining angles. I am going to pick the 26 degree angle -- the angle with respect to the horizontal. [Darn, that's a heck of a lean]

    The "sine" of that angle is the ratio of the side opposite (gravity) to the hypotenuse (the diagonal -- or the total diagonal acceleration).

    We are after the inverse of that ratio -- the ratio of diagonal acceleration to gravity.

    So we pick up a calculator and ask for the sine of 26 degrees -- 0.438. And then take the reciprocal of that -- 2.28.

    Edit: you mention 2.0 g's being associated with 64 degrees. Could that be centripetal acceleration? Let's figure out the centripetal acceleration at 64 degrees.

    I'm still anchored to my 26 degree angle. Now we are after the ratio of the side adjacent (centripetal force) to the side opposite (gravity). That's the cotangent. My calculator does not have a cotangent button. So let's use the 64 degree angle instead and look for the ratio of the side opposite (centripetal force) to the side adjacent (gravity). That's the tangent -- 2.05

    You'd need 63.4 degrees to get a tangent of 2.00. On a calculator, the button for that operation is the "inverse tangent". The inverse tangent (or arc tangent) of 2.00 is 63.4 degrees.
     
    Last edited: Dec 19, 2017
  4. Dec 19, 2017 #29
    It must be incredible to lean 64 degrees and having over 2g playing tricks with the fluid in your ear and being out of sync with the visual orientation of the world whilst the road goes a few cm away from your head at god knows how many kilometers an hour must be unbelievably fun and terrifying
     
  5. Dec 19, 2017 #30
    Please, please, please keep such antics on official tracks. I've seen enough bikers scraped off roads to weep. One patch of mud, dung or gravel-wash, one slick of sloshed diesel, one dozy driver and they're gone.

    Where I used to live, we'd see a stream of bikers come off each IOM ferry after TT Week. They'd zoom across the big A59 junction's fly-over, lay into run-out's adverse-camber bend, straighten out and be waved down for doing 60+ in well-signed 30 zone. IIRC, record was 25 in an hour...
     
  6. Dec 19, 2017 #31
    Yes it was the centripetal acceleration I was talking about.

    Isn't the number 2.28 still correct for the amount of force the rider feels at 64 degrees?
     
  7. Dec 19, 2017 #32

    jbriggs444

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    2.28 g's at 64 degrees. Somewhat less at the 63.4 degrees calculated above.

    At 2.0 g's centripetal force, one can calculate the felt acceleration in either of two ways:

    $$a=\sqrt{a_{centripetal}^2+a_{gravity}^2}=\sqrt{2^2+1^2}=\sqrt{5}= 2.24$$
    $$a=\frac{1}{cos(arctan(\frac{a_{centripetal}}{a_{gravity}}))}=\frac{1}{cos({arctan(2)})}=2.24$$
     
    Last edited: Dec 19, 2017
  8. Dec 19, 2017 #33
    The motorcycle leans. The combined forces on the rider are normal to the seat.

    If course, if the rider hangs off of the bike, https://goo.gl/images/GYnzJn, the total center of gravity is shifted from that plane, the center of gravity of the rider is not aligned with the vehicle, the bike leans less, giving more clearance.
     
  9. Dec 19, 2017 #34

    berkeman

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  10. Dec 19, 2017 #35
    More interesting is what is called counter steering. The wheels, in particular, the front wheel, is a gyroscope. The angular momentum of the front wheel is significant. In order to turn left, the rider pushes forward on the left handle bar. The bike then leans to the left and the bike turns to the left. The effect is more pronounced as speed increases and there is a transition speed below which the steering is as expected and above which angular momentum dominates
     
  11. Dec 19, 2017 #36

    berkeman

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    You must have missed this recent (very long) thread: https://www.physicsforums.com/threads/motorcycle-physics-countersteering-and-bodysteering.927832/
     
  12. Dec 20, 2017 #37

    cjl

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    That's not a gyroscopic effect - rather, it comes from the fact that if you turn the wheel to the right (by pushing the left handlebar forward), the wheels will steer to the right, out from underneath the CG of the bike. Since the wheels are now to the right of the CG, the bike "falls" to the left, initiating the left lean needed for the left turn.
     
  13. Dec 20, 2017 #38

    rcgldr

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    The rider feels a net force vertical with respect to the lean angle of the bike (if the rider is not leaning inwards or outward relative to the bike). If the bike is leaning at 45 degrees (and in a coordinated turn), then the rider and bike are leaning at 45 degrees, and the rider feels a net force that is also at 45 degrees, so the rider doesn't feel any sideways forces (if the rider is not leaning inwards or outward relative to the bike). This is essentially the same as a coordinated banked turn in an aircraft.

    If the rider leans inwards on the bike, the rider would feel a torque and also a force that is not inline with the angle of the rider. The rider could lean to one side with the bike going straight (the bike leaned a bit the other way) and the feeling would be similar, just less force than if leaning inwards while cornering.
     
    Last edited: Dec 20, 2017
  14. Jan 2, 2018 #39
    The minimum force and acceleration is when the cycle is parked (or moving at a constant velocity). The acceleration due to gravity and the cooresponding weight.

    The total acceleration when the bike is cornering, accelerating or decelerating on a straight line will always be more, the vector sum of gravity and cornering.
     
  15. Jan 2, 2018 #40
    I believe that what they are referring to when they say the rider experiences 0.5g is that which is not due to gravity.
     
  16. Jan 2, 2018 #41
    https://en.m.wikipedia.org/wiki/Countersteering
     
  17. Jan 2, 2018 #42
    Oh, reading further in that link, "One effect of turning the front wheel is a roll moment caused by gyroscopic precession. The magnitude of this moment is proportional to the moment of inertia of the front wheel, its spin rate (forward motion), the rate that the rider turns the front wheel by applying a torque to the handlebars, and the cosine of the angle between the steering axis and the vertical.[10]".

    Note the term "of turning the front wheel is a roll moment caused by gyroscopic precession."
     
  18. Jan 2, 2018 #43

    berkeman

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    As the link shows, you can countersteer at speeds slow enough so that gyroscopic effects are minimal. cjl is correct that the main effect is to move the front tire's contact patch out from under the COM, which causes the bike to tip in. There is a small gyroscopic effect as well at higher speeds, but only adds a bit in my experience.
     
  19. Jan 2, 2018 #44
    Let's consider this intuitively. The classic presentation of the gyroscopic effect is a guy holding a spinning bicycle tire, sitting on a rotating chair. Compare the weigh of this bicycle tire and rotational velocity to the weight and rotational velocity of a motorcycle wheel at 40 mph. The bicycle pales by comparison. It can hardly be said that the precession of that motorcycle wheel is negligible. Clearly it is not.

    And, the phenomena does not occur at low speed. There is a crossover speed at which the steering changes from as expected to counter steering. As speed increases, the effect becomes more pronounced. What has changed? What has changed is the rotational velocity of the wheel.
     
  20. Jan 2, 2018 #45

    berkeman

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    Do you ride? What do you ride?
     
  21. Jan 2, 2018 #46

    rcgldr

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    The precession only exists while a torque is applied. Once the steering is completed, initially outwards to induce a lean, the precession stops since it's a reaction to torque, not angle.

    There are two primary speed ranges, one below the speed of self-stability, one in the range of self-stability. For a mathematical model using razor thin tires, there is a third speed range (capsize speed), above which a bike would tend to fall inwards at an extremely slow rate, but with real tires, this speed range doesn't exist or only exists well beyond any speed that a bike could achieve.

    Counter steering is in effect at all speeds. At slow speeds, it's much less noticeable as the steering input is very light. It's easier to see this effect in the case of unicycles, which don't have the self correction related to trail on a two wheeled bike.

     
  22. Jan 31, 2018 #47
    "Counter steering is in effect at all speeds."

    Yes. It is one of a number of effects that vary with things like wheel weight and radius, the bike's weight, angle of the front forks, distance from front to back wheels, and .... things I haven't considered. Some of these are in opposition to the counter steering affect, what we might call "normal steering.". It is apparent that the counter steering affect decreases as speed decreases, becoming negligible. My impression, from riding around a parking lot, is that at 20-25 mph, "normal steering" is predominant.

    It begs the question if why is there no apparent speed at which counter and normal steering component simple cancel out and make the bike simply unsteerable?
     
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