Motorcycle Q#1: Right Foot Peg Touching Road During Turn

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In a right turn on a motorcycle, if the right foot peg touches the road, it indicates the bike is leaning too much. Raising the body away from the turn may inadvertently increase the turning radius, leading to a potential off-road situation. To correct this, riders should either accelerate to lift the bike on its suspension or shift their weight to the right to maintain a tighter turning circle. Proper body positioning and throttle control are crucial for maintaining balance and avoiding accidents. Understanding these techniques can enhance riding safety during sharp turns.
Julian Solos
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Motorcycle Question #1:

Entering a sweeping right turn on a country road a little too hot, you have to lean your bike a lot to your right to negotiate the turn. During the turn you hear the right foot peg of your motorcycle start touching the road surface. You instinctively raise your body (to your left) away from the turn hoping to raise the bike away from the road surface. What will happen next?
 
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You push down on the bike to raise yourself up. You go up a little, the bike goes down a little. Peg digs a gouge in the road and you go flying.

Njorl
 
Originally posted by Julian Solos
Motorcycle Question #1:

Entering a sweeping right turn on a country road a little too hot, you have to lean your bike a lot to your right to negotiate the turn. During the turn you hear the right foot peg of your motorcycle start touching the road surface. You instinctively raise your body (to your left) away from the turn hoping to raise the bike away from the road surface. What will happen next?

You go off the road as your turning circle has increased!

You can do one of two things to rectify the situation:

1) Get the throttle on (if it isn't already) and if it is, give it some more. This lifts the bike up on the suspension and increases ground clearance.

2) Hang off the bike to the right, and move the bike up a little. Your overhanging mass keeps the turning circle tight, but removes the peg from the ground.

Hope this helps
Adrian Baker
250cc Classic Racing Motorcycle Club Champion 2003!
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

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