Motorcycle: Lean Angle, Speed, and Turn Radius

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The discussion focuses on the relationship between lean angle, speed, and turn radius for motorcycles negotiating curves. The lean angle θ is initially approximated by the formula tan(θ) = v²/(rg), where g is gravity and r is the curve radius. However, the conversation highlights the need to refine this relationship for motorcycles with wider tires, which can affect stability and handling. The impact of tire width is considered, with the assumption that the contact area is semicircular and the tire widths are uniform. Ultimately, the refined equation accounts for the height of the mass center and the geometry of the motorcycle during a lean.
Julian Solos
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Suppose you are riding a motorcycle at a constant speed v[/color] and about to enter a level, circular, curve of radius r[/color]. You intend to lean with the bike and want to make negotiating the curve as smooth and effortless as possible.

Then the lean angle (from the vertical) of the motorcycle \theta to which the motorcycle will settle into is approximated by the following relationship

tan (\theta) = \frac {v^2} {rg}

where g[/color] is the acceleration due to gravity.

If the widths of the tires of the motorcycle are narrow, like those of bicycle tires, the above relationship should give a good approximation. However, many of today's motorcycles have wide tires, i.e. 19 cm. Can we refine the above relationship with the addition of a variable or variables related to the widths of the tires of the motorcycle?

(For simplicity's sake, shall we assume the widths of the front and rear tires are the same so hat we need to use only one width and the sections of the tires which make contact with the ground are semicircles of the radius equal to the tire width and their shape remains the same throughout the range of possible lean angles?)
 
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Julian Solos said:
Suppose you are riding a motorcycle at a constant speed v and about to enter a level, circular, curve of radius r. You intend to lean with the bike and want to make negotiating the curve as smooth and effortless as possible.

Then the lean angle (from the vertical) of the motorcycle \theta to which the motorcycle will settle into is approximated by the following relationship

tan (\theta) = \frac {v^2} {rg}

where g is the acceleration due to gravity.

If the widths of the tires of the motorcycle are narrow, like those of bicycle tires, the above relationship should give a good approximation. However, many of today's motorcycles have wide tires, i.e. 19 cm. Can we refine the above relationship with the addition of a variable or variables related to the widths of the tires of the motorcycle?

(For simplicity's sake, shall we assume the widths of the front and rear tires are the same so hat we need to use only one width and the sections of the tires which make contact with the ground are semicircles of the radius equal to the tire width and their shape remains the same throughout the range of possible lean angles?)
Just need to consider the point of contact with the ground and the position of the nass centre of bike+rider. If the height of the mass centre is h when upright and the tyre radius is r then when the bike leans at angle θ the height is ##r+(h-r)\cos(\theta)## and the horizontal displacement of mass centre from point of contact is ##(h-r)\sin(\theta)##.
Going around a curve radius R at speed v we have ##\frac{v^2}{Rg}=\frac{(h-r)\sin(\theta)}{r+(h-r)\cos(\theta)}##.
Note that setting r=0 gives the simpler equation.
 
I built a device designed to brake angular velocity which seems to work based on below, i used a flexible shaft that could bow up and down so i could visually see what was happening for the prototypes. If you spin two wheels in opposite directions each with a magnitude of angular momentum L on a rigid shaft (equal magnitude opposite directions), then rotate the shaft at 90 degrees to the momentum vectors at constant angular velocity omega, then the resulting torques oppose each other...

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