Understanding which forces are at play when the axle rotates on the ground is a pesky problem of mine. The goal is to obtain a good understanding of the effect that different axle lengths have when turning (an effect that’s overstated, I feel, in the community). Physics was not my favorite subject in school. So, while the geometry of the truck is clear to me (and hopefully to you too), placing vectors of forces and calculating the torque that turns the axle when the rider tilts the deck was a puzzle for me. It turns out the problem is not so unattackable. In what follows, I report what I’ve found so far.

I’ve attached here my concept/model of the system rider-skateboard (please click on the picture). I built it just so that I can make sense of it myself, so it is as abstract as possible, but not more. Speaking of abstractions, I completely ignored the complexity of the forces placed upon the board by the rider and simply assumed the rider steps on one side of the board, as if the system was stationary (i.e., imagine the rider just steps on the rail – edge of deck- while the skateboard rests). At the moment, I don’t have the energy, capacity and the knowledge to tackle describing how a rider impacts a board when turning with any more complexity. My goal here is to digest how and why the axle turns.

Let me tell you how I arrived to that model, in small logical steps. The axis of rotation (for both the deck and the axle) is the pivot axis of the truck. The deck is free to rotate on a vertical plane, but not on the horizontal (as it is constrained by the other truck). Conversely, the axle of the truck is free to rotate on the horizontal plane (since the wheels, to simplify, rotate without friction) but not on the vertical (as they must remain on the ground). Therein lies the trick: since the deck cannot rotate on the horizontal plane, but the axle can, the torque that tends to rotate the deck on that plane is equal to the torque that rotates the axle instead (but in the opposite direction). (This is an intuitive step, which I believe is based on sound logic – the logic being that there can’t exist other forces in action that result in the torque, and that this torque must be solely responsible for turning the axle). So, the rotation (i.e., the sum of torques) can (I believe) be resolved on these two axii.

To put the above thought in a model, I constructed the two forces on the deck (rider weight B and spring/bushing force S) and resolved (decomposed) their vectors to get their projections (the red vectors) on the plane of the rotation (i.e. the plane perpendicular to the pivot axis). It was simple then to calculate the resulting torque that tends to rotate the deck and then resolve it into two component vectors on the appropriate plane (because, remember: the deck simply can’t rotate horizontally, while the wheels can). After this, it is also simple to calculate the moment of inertia of the wheels (i.e., two times the mass of one wheel times its squared distance from the pivot axis). Torque must always equal moment of inertia times angular velocity (i.e., the lighter the wheels, or the closer they are to the pivot axis, the faster they rotate).

Incidentally, please notice that after the rider has tilted the deck more than usual, torque starts reducing even though bushing resistance increases. That’s because the orientation of the forces of weight and bushing resistance favors the former. The trucks becoming too soft after we’ve turned them adequately, is something we must have all observed on our skateboards. Well, here’s the explanation!

The problem remains: which force does the rider actually effect on the deck? Definitely not only their weight, but also some configuration of a centrifugal force. I believe this might prove helpful. If only I could read such mathematics… Also, it seems correct to claim that front-offset trucks are harder to turn than back-offset trucks, all other things being equal. Intuitively. Why? It can’t only be that front-offset trucks have to carry more of the rider’s weight. To be continued…