Loop-the-Loop on a Motorcycle

The question was asked: “Can you give me a simple formula to determine what minimum speed is necessary for a motorcycle and rider of a given weight to be able to successfully ride loop-the-loop around a tube of a given diameter?”

Ostensibly this is simple; one just needs to calculate the minimum speed at which the outward

force exerted by the motorcycle at the top of the tube (we’ll call this the centrifugal force, for historical reasons) just balances the downward force, which of course is due to the acceleration of gravity g = 9.8 m/s/s or 32 f/s/s. A quick look at a textbook reveals that the centrifugal force F_c = mv²/r, where m is the mass of the rotating body, r is the radius at which it is orbiting, and v is its linear velocity. Thus we have

mv²/r = mg

This looks simple enough, but there are a few other considerations to be taken into account. Firstly, the above assumes a point mass at a given definite radius, but a motorcycle and rider subtend a rather large distance above the ground, especially compared to the likely radius of a loop-the-loop tube. In most physics problems it is sufficient to consider the motion of a body as being the same as an equivalent mass travelling at the point given by the body’s centre of mass. For a motorcycle, I understand that the centre of mass is usually approximately between the footpegs and the underside of the rear of the tank, so for the rest of this paper I’ll assume that this is at a height of 0.5 metre, and thus define an effective radius for the motion r_eff which is 0.5 m less than the radius of the tube being traversed.

All of the treatments I’ve seen so far assume that the motorcycle travels at a constant velocity v (this should of course be speed v, as the direction of motion is constantly changing). Consider the implications of this – it means that there is always an outward force of mv²/r_eff, which just happens to be matched at the zenith by the force of gravity mg. However, at the nadir of the loop, this force adds to gravity, meaning that the downward acceleration at that point is 2g – twice the gravitational field. We shall see that this is a best-case scenario, but the rider and the suspension should be ready for this overload!

Let’s have a quick look at some figures at this point. Assume the loop-the-loop is about two storeys high – 6 m. Then the radius is 3 m and my effective radius r_eff is 2.5 m. Noting that mass drops out of the equation above, v²/r_eff=g, or v=sqrt(g*r_eff). Therefore, v=Ö(9.8*2.5) » 5 m/s. Which is 18 km/h, or about 11 mph.

However, this isn’t the whole story. Think what it means to be travelling at constant speed around a vertical loop. Whilst ascending you are fighting gravity to some extent, and while descending gravity is assisting you. To maintain constant speed you therefore need to be accelerating upwards and decelerating downwards, in a sinusoidal fashion as it happens (as a function of angle from the vertical) – maximum acceleration needs to be 1 g at the 90-degree point upwards, and deceleration (i.e. braking) needs to be 1 g at the 90-degree point downwards. While a bike with good brakes can manage 1 g braking, 1 g acceleration is only in the domain of litre-bikes.

OK, so let’s look at another scenario, where the motorcycle itself applies no acceleration or deceleration. This corresponds to the ideal frictionless world, so in practice the rider should be prepared to give a little bit of throttle on the upward track, at least. Now, the difference in energy between a motorcycle at the bottom of the loop and the same bike at the top is called the potential energy for that distance and is simply the product of the mass, the height involved, and the acceleration due to gravity

E_p=mgh, or in this case E_p=mg*(2r_eff)

Assuming that the minimum speed v=sqrt(g*r_eff) is reached at the top of the loop, this much energy must be added to the bottom of the loop. In other words, the kinetic energy at the bottom of the loop, ½mv_b², must be the sum of the kinetic energy at the top, ½mv², and the potential energy lost (converted to kinetic energy) in falling from the top to the bottom

½ mv_b² = ½ mv² + mg*(2r_eff)

or, since m cancels throughout,

v_b = Ö(v² + g*(4r_eff))

Plugging in our example above, and approximating g=10, this gives v_b = Ö(25+100), or v_b » 11 m/s. This is more than twice the speed needed if a constant velocity could be achieved. However, look at its impact on the downward force at the bottom of the loop:

F = g + v_b²/r » 10 + 125/2.5 = 10 + 50 = 60 = 6 g!

This is obviously far higher than the human body can sustain without risking blackout and would also certainly lead to the suspension bottoming out. Strategies to minimise this would include actually performing some of the acceleration and braking mentioned above to try to keep the motorcycles speed closer to constant, or doing a one-time loop on a Mattel Hot-Wheels type ramp where the entrance and exit ramps actually meet the circle at a tangent some distance off the horizontal. The one bit of footage I recall – vaguely! – of a motorcycle doing a loop-the-loop, it was actually in a mesh sphere and IIRC the rider built up to the overhead loop by working his way higher and higher up the sphere. This “training” would seem preferable to attempting the difficult one-off loop from scratch!