In the January 2007 issue of *Ride*, we ran a sequence of Ryan Sher doing the mother of all curved wallrides while on the Shadow Euro Tour (which will surely be in Shadow’s just-released video, *Into The Void*—appropriately titled for this trick). The wallride sent Ryan down a spiraled stair case and straight into an adjacent wall—although I always imagine that the wall just goes on forever and he’s stuck in an eternal-hell of a never-ending wallride, but whatever—it really caught people’s eye simply because how wild the setup was and shear wildness of ducking into a dark hallway. It sparked even more interest once we threw the animated sequence up on RideBMX.com…although it was forever lost and I had to re-make it for the purpose of showing you this: The Physics of Ryan Sher’s Curved Wallride, as written by Tom Schug.

You learn a lot of useless things in school, but applied science isn’t one of them. Read on, learn something about the “why” and “how” of what you do on a daily basis works, and geek out a bit. It’s okay sometimes. Be sure to leave any comments or questions, maybe Tom will chime in and elaborate.

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"What's In This Picture?"

Ryan Sher's Curved Wallride

By Tom Schug

The picture presented is of pro BMX rider and owner of Subrosa, Ryan Sher. This is a photo of him doing a curved wallride down a set of stairs in Lyon, France. The picture was taken by Keith Mulligan and printed in the January 2007 issue of RideBMX magazine (86-87). What makes this a great picture is that Ryan rides along a curved vertical wall without falling over as if he was on flat ground. There are many actions taking place here that can be explained by applying fundamental physics ideas. The question addressed here is how fast would Ryan have to be traveling in order to not fall over.

The first step in solving this physics problem is to draw a free body diagram, identify the forces and analyze the situation. (See Fig.1) Focusing on just the contact forces, the three forces acting in this situation are gravity, normal force, and a force of static friction. Because Ryan is moving with centripetal motion, the normal force is considered to be the force of the wall pushing against the tire towards the center of the curve. Also, because Ryan is not sliding up or down the wall, he is in vertical equilibrium. This means that the force of gravity pulling down on his tire is equal to the force of static friction between the wall and his tire keeping him from slipping downward. With these concepts and forces understood, some information can be derived in order to estimate his needed speed.

By the networking capabilities of Myspace, Ryan was contacted directly for the information pertaining to his bike and body weight. Ryan stated, "Well I weigh around 155 lbs., and my bike at the time was probably around 26 lbs." Another known piece of information is the coefficient of static friction between rubber and concrete. This was found in the textbook Physics For Scientists and Engineers A Strategic Approach by Randall D. Knight. Due to lack of definite certainty, the radius of the curve must be assumed.

•Mass of Ryan & bike = 181 lb. = 82.10 kg.

•Radius R = 7.5 ft. = 2.29 m.

•Force of gravity = 9.8 m/s

•Rubber on Concrete ?s = 1.00

In physics, the force of gravity is equal to the product of the mass and force of gravity acting on it . The force of gravity is always positive or negative 9.8 meters per second depending on the coordinate system used. In this case gravity is positive because it is pulling Ryan down at a positive rate towards the earth. The textbook describes the force of static friction as, "The value of depends on the force or forces that static friction has to balance to keep the object from moving. (133)" In this case then, is equal to the product of and the normal force. The normal force in this situation is a special kind called centripetal. Centripetal force occurs when an object moves at a constant speed in a circular motion. For simplicity of this problem, the fact that Ryan is not moving at a constant speed and that he is following the slope of the stairs downward on the wall will be discarded. It will be assumed he is in uniform circular motion. Newton's second law defines centripetal force as the product of mass and velocity squared all divided by the radius. The force of static friction is equal to the force of gravity as stated above because Ryan is in vertical equilibrium. Now, to solve the problem of how fast Ryan has to be going is simply just an equation with one unknown variable.

The calculations show that Ryan needs to be going approximately 10.5 miles per hour to make it around the wallride without falling over, which seems to be a logical speed. It must be realized though that this example only takes into account the exact point where Ryan's tire touches the wall. In reality Ryan's mass is not just all at his tire but in fact offset and at an angle to the wall. To account for this and adjust Ryan's speed, a different method can be used.

To compare speeds determined by the next method, two more variables and a new free body diagram will be needed. The first variable is an estimated angle of Ryan riding on the wall and the second is an estimate where his center of gravity is.

• Angle Ryan is riding = 40°

• Distance up from tire Ryan's center of gravity is = 4 ft. = 1.22 m.

This method uses the concepts of torque incorporated with centripetal motion. The diagram shows the torque, gravitational and centrifugal forces. (See Fig. 2) Centrifugal force is not actually a real force, but instead it is an imaginary force that is not shown on a free body diagrams. The centrifugal force in this picture is what Ryan feels pulling him towards the wall but in reality there is no force pulling him, hence imaginary force. For this picture though, the centrifugal force is defined by the equation for centripetal force. Torque is dependent on three factors; the magnitude of force, distance from the pivot point to the object, and the angle. All of these can be computed by using simple trigonometric functions. Since we will assume Ryan's angle is not changing, the torque pulling Ryan counterclockwise is equal to the torque pulling him clockwise (rotational equilibrium). If one torque were greater than the other, his angle would continue to either increase or decrease. Since one torque is dependent on velocity and both are equal to each other, a simple equation can be solved for velocity using algebra.

This method of finding Ryan's speed shows that he must be going 9.04 miles per hour or he will fall over. This number is slightly less than the previous calculation when only the points of contact were considered. This also seems to be a reasonable outcome. It may be even more reasonable than the first outcome because it places Ryan's weight at a more realistic position. The angle in the second method has a big factor in the speed needed to travel around the wall. Although many other combinations of riding angles or centers of gravity could be used to find different outcomes, it is reasonable to say that between the two methods used here, Ryan was riding around the wall at a rate of around 10 miles per hour.

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And le pièce de résistance courtesy of Mr. Keith Mulligan (click to get the image sequence):