Motorcycle: Lean Angle, Speed, and Turn Radius

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SUMMARY

The discussion focuses on the relationship between lean angle, speed, and turn radius for motorcycles, specifically using the formula tan(θ) = v²/(rg), where g is the acceleration due to gravity. It highlights the need to refine this relationship for motorcycles with wider tires, such as those measuring 19 cm. The conversation also introduces the concept of the motorcycle's mass center height and its impact on lean angle calculations, leading to a more complex equation that incorporates tire width and mass center displacement.

PREREQUISITES
  • Understanding of basic physics principles, particularly forces and motion.
  • Familiarity with trigonometric functions and their applications in real-world scenarios.
  • Knowledge of motorcycle dynamics and tire characteristics.
  • Basic grasp of circular motion concepts, including radius and speed relationships.
NEXT STEPS
  • Research the effects of tire width on motorcycle handling and stability.
  • Explore advanced motorcycle dynamics, focusing on lean angle calculations.
  • Study the physics of circular motion in relation to vehicle dynamics.
  • Learn about the role of the center of mass in vehicle performance during turns.
USEFUL FOR

Motorcycle enthusiasts, engineers, and riders looking to enhance their understanding of motorcycle dynamics, particularly in relation to lean angles and handling during turns.

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.
 

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