Wheelie

[Karen_Agusta]

Understanding the "Wheelie"

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First we can start with the picture I created using my image creation software ...

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A wheelie is in racing torque wasted lifting the front end - Rather than moving the vehicle forward - occuring due to

extreme torque being applied to the rear wheel. For Our Example - The distance between the centers of the wheels of a

motorcycle is 155 cm - The center of mass of the motorcycle, including the racer is 88.0 cm above the ground - slightly

forward of directly between the wheels - due to the weight / placement of the motorcycles engine - Thus our question is -

where does the horizontal acceleration of the motorcycle occur - causing the front wheel to rise off the ground?

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This can be descibed as \tau=rFsin\phi

Tau is amount of work done per unit of time

rFsin is the force against a rigid object against its displacement

phi is the angle against the axis of the object to which the force is being applied

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Thus we have amount of work done per unit of time - equals - the force against the rigid object - the power applied to the motocycles rear wheel per second - divided by the angle against the axis at which the force is being applied - the axis must be considered due to the following: Imagine the force of our equation were being applied downward - there would be no wheelie - were the angle of the force straight upward all force (100%) would be applied to lift the front wheel - were the angle of the force at forty five degrees (50%) of the force would be applied to lift the front wheel - thus the angle of the force - {phi} - has to be considered - Note: were the object not rigid - like a cooked spaghetti noodle - there would be no wheelie either - and that is why we divide the work done per unit of time against the rigid object by the angle of the force applied - NOTE: One other thing to consider here is that the tau (amount of work done per unit of time) has to deal with the issue of - I - which in physics is inertia or the tendency of a body to resist acceleration; the tendency of a body at rest to remain at rest - inertial is important due to the fact it is far easier to wheelie by red-lining the engine and dumpint the clutch - than it is to wheelie when already at a high rate of speed -

- because a wheelie is caused by excessive torque in relation to the movement of the bike ... the engine can generate far more excessive torque to a motorbike standing still ... than it can to one doing one hundred and forty miles per hour in top gear with its engine at red line ... because at that point the engine simple has no excessive torque to apply.

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Thus the Torque of the motor on the backwheel - applied via the chain - is what will lift the bike - However the torque of the motor also generates the - F = m*a (Force equals mass times acceleration) - of the overall system (bike/rider) - This has to be considered once again in racing as torque wasted moving the motorcycle forward - For if you were to lock the rear wheel into clamps so it could not rotate - all of the force applied to the rear wheel would lift the motorcycle - however in racing the rear wheel is not clamped so it can not move - the rear wheel generates forward acceleration which is the intent - So - This force which equals mass times acceleration - has to be greater than F = g (Force = Gravity) - we consider gravity because in all reality gravity - IS - the force which is being overcome when we wheelie the motorcycle - therefore - when force equals mass times acceleration is greater than the opposite force of gravity - causing the front wheel to be lifted - creating what we now call the wheelie.

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Because the back wheel is interacting with the body of the bike through the chain - an equal and opposite force - to the force of acceleration on the back wheel - is applied to the centre of mass - Essentially - Fa - (force equals accelaration) - is in the opposite direction as in the picture - of the bottom of the back wheel - WHICH IS - the pivot point - Therefore we may calculate the torque about that point with something like this:

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\sum\tau=0 F_{g}r\sin\theta=F_{a}r\sin\phi a=g\sin\phi/\sin\theta

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Sum divided by - tau - amount of work per unit of time - force gravity against a rigid object - sin - singularity or one object - divided by theta=F - theta - is the angle of displacement betweeen the amount of magnitude - magnitude is not merely weight it is the the extent of the length, breadth, and thickness of an object - Thus theta is the angle at which the force is being applied to the length of the object - the length must be considered for this reason - short wheelbase - motorcycle - easy wheelie - long wheelbase - rail dragster - - much more force to wheelie - phi - the polar angle of the z axis - z axis is One of three axes in a three-dimensional Cartesian coordinate system - Z being the vertical - x is horizontal - y is height - because we are talking about a wheelie we are concerned with the z or vertical axis - in our equation - \theta is the angle between Fg - force of gravity and the radius - and - \phi (the directional angle of the vertical axis) - is the angle between Fa - the force of acceleration - and the radius - thus we have gravity against a singular object and the angle of the force applied to it - We then consider the point on the ground that the CoM (constant of mass) is acting down on - as well as the CoM - and the height above the wheel that the bike has to put into the torque - that will accelerate the bike - you already have - the projection of the moment arms where the forces are perpendicular - To balance the moments then just before front wheel lifts.

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Because the back wheel is interacting with the body of the bike through the chain, an equal and opposite force to the force of acceleration on the back wheel is applied to the centre of mass. Set the bottom of the back wheel to be the pivot point and calculate the torque about that point with something like this: For the torque from the reaction force on the bottom of the back wheel...\tau_{R} = F_{a}R where Fa is that force and R is the radius of the wheel - \tau_{R} = MaR where M is the mass being accelerated - M is the mass of the bike - the torque of the weight at the centre of mass can be taken into account - \tau_{CM} = Mgl/2 with l being the length between wheels (were we to set the centre of mass directly between the two wheels) - \sum\tau = 0 so the torques have to be equal - MaR = Mgl/2 The masses cancel - a = gl/2R at the point the masses cancel if you apply force in excess of the point of the cancellation of mass ... you shall wheelie

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Or i suppose just buy a V Max and you to not have to worry at all about how to do a wheelie ... you just open the throttle

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http://www.youtube.com/watch?v=kIHxOOZ5B20

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Karen

[R6exy]

or get a propper bike lol

[Ryan09]

... If i wanted to wheelie i wouldnt buy a vmax.

[janed]

Well i think wheelie is good for the people who knows very well how to handle their bikes.Its not for me.I love to watch but wont try any time.