Bicycle Speed Bump


I am researching the force applied to a bicycle wheel when it hits a 2 5/8 inch high speed bump. The velocity of the bicycle is 15 mph. I am trying to solve for the impulse force in the vertical direction so I can select proper springs/shocks for the bike under 110 lbs of load. I have had a difficult time determining what methods and equations I should use. If you would have any information for me, I would greatly appreciate it. Thank-you.


I am making some minor changes and assumptions in the question. Let me know if this causes you a problem.

Consider a bicycle with 18 inch radius wheels. The weight of the bike and rider is 161 pounds. This weight is evenly distributed between the front and rear wheels. If the speed of the bicycle is 15 mph, what will be the force on the front suspension of the front wheel striking a 2 5/8 inch speed bump. Assume the speed bump has a rectangular cross section.

speed bump collisionThe diagram seen at the right depicts the situation. The wheel in blue rolling to the left strikes the speed bump in red. Assume for the moment that the collision is elastic and the wheel and speed bump are quite rigid. Immediately after impact the front axle follows the path shown in light green as it is carried around the point of contact at a distance of the radius of the wheel.

The force on the front suspension would be that sufficient to provide the necessary acceleration to deflect the path of the front axle from its horizontal path to follow the curved path shown in light green. The light blue arrows shows the front axle velocity prior to (Vi) and immediately after (Vf) the collision. The magenta arrow (DV) shows the change in velocity during the collision. The velocity immediately after the collision will be tangent to the light green path at its right end. This direction is perpendicular to the radius R since tangents are always perpendicular to a radius on a circle.

From the fact that lines Vi and (R-d) are perpendicular and lines Vf and R are perpendicular, the angle q is the same in both triangles. Since the change in velocity must be in the same direction as the acceleration and the acceleration must be in the same direction as the applied force and the applied force is transmitted from rim to axle along a radius, then the angle f is the same in both triangles. Therefore the velocity triangle is similar to the collision triangle and the magnitude of the change in velocity is the initial velocity times the sine of q, or |DV|=15*d/R=15*9.36/18=7.8mph. In feet per second that change in velocity is 11.44f/s

Now we have to determine how long was required for this velocity change in order to get the acceleration. If the wheel and speed bump were perfectly rigid, the time would be vanishingly small, requiring infinite acceleration. In fact the bicycle tire is going to compress and then expand during the collision. As soon as the tire starts compressing the wheel begins to accelerate, reducing the peak force required to achieve the new velocity. Without knowing a lot about the geometry of the tire, its inertia, its stiffness, its air pressure and so on, the best we can do is make some kind of reasonable guess about the effect of the tire on collision. Suppose for example that the tire extends the contact time of the collision to 100 milliseconds and cushions the force so that it is essentially flat over that time. Then the acceleration required to achieve the calculated change in velocity would be 11.44/0.1=114.4 f/s/s.

The force required to accelerate the front axle is its acceleration times its mass. The weight of the bike and rider was 161 pounds which is 5 slugs. Assuming that 1/2 the mass is coupled to the front axle, then the force will be 2.5*114.4=286 pounds force.