This is an aside, intended for those who are interested in how one can and actually should investigate such systems:
The first analysis should be a
kinematic one, examining the motion of the system components without considering the forces or causes of the operation.
Let's consider a vehicle that in some specific time interval moves right by \$x\$ with respect to the ground. The large wheel rotates clockwise by \$\varphi = x / R\$ radians, when there is no slippage, because the surface of the wheel travels the same distance \$x = R \varphi\$. (Full turn is \$2 \pi\$ radians, and the length of the perimeter of a circle with radius \$R\$ is \$2 \pi R\$; similarly, the length of a circular arc with radius \$R\$ that distends an angle \$\varphi\$ in radians is \$R \varphi\$.)
In the same time interval, let's say the blue driven belt/surface moves left by \$y\$. This means that the surface of the small wheel travels a distance of \$x+y\$, and the small wheel turns by \$\theta = (x + y) / r\$, using the same logic as we used for the large wheel above.
The vehicle has a gearbox with gear ratio \$\lambda = \varphi / \theta\$, with \$\lambda \ne 0\$. That is, when the small wheel rotates one full turn, the large wheel rotates \$\lambda\$ turns. Here, let's say \$\lambda = 1:20 = 0.05\$, so that when the large wheel rotates one full turn, the small wheel rotates twenty turns. When \$\lambda \gt 0\$, the wheels turn in the same direction (as shown in the diagram), when \$\lambda \lt 0\$, the wheels rotate in the opposite directions.
We now have all equations we need to examine the kinematics of the system. (We don't know or care how much work is done to move the blue surface left by \$y\$, we just know it has done so.) We have:
$$\varphi = \frac{x}{R} \tag{1}\label{1}$$
$$\theta = \frac{x + y}{r} \tag{2}\label{2}$$
$$\varphi = \lambda \theta \tag{3}\label{3}$$
The key here is substituting the first two into the third,
$$\frac{x}{R} = \lambda \frac{x + y}{r}$$
which we can solve for \$x\$,
$$x = y \frac{\lambda R}{r - \lambda R} \tag{X}\label{X}$$
for \$y\$,
$$y = x \frac{r - \lambda R}{\lambda R} \tag{Y}\label{Y}$$
or for \$\lambda\$,
$$\lambda = \frac{r x}{R (x + y)}$$
We can immediately see that \$\frac{\lambda R}{r - \lambda R}\$ is the key term for considering how the vehicle moves. Let's use
$$x = \nu y \quad \iff \quad \nu = \frac{\lambda R}{r - \lambda R}$$
The only invalid gearing is when \$r = \lambda R\$, i.e. \$\lambda = \frac{r}{R}\$.
Examining the effect of \$\lambda\$ on \$\nu\$, we can make the following observations:
- When \$0 \lt \lambda \lt \frac{r}{2 R}\$, the vehicle moves as shown in the diagram, with \$0 \le \nu \le 1\$, i.e. \$x \lt y\$.
- When \$\lambda = \frac{r}{2 R}\$, the vehicle moves as shown in the diagram, with \$\nu = 1\$, i.e. \$x = y\$.
- When \$\frac{r}{2 R} \lt \lambda \lt \frac{r}{R}\$, the vehicle moves as shown in the diagram, with \$\nu \gt 1\$, i.e. \$x \gt y\$.
This means that the vehicle moves right faster than the blue surface moves left. - When \$\lambda \gt \frac{r}{R}\$, the vehicle moves left faster than the blue surface, with \$\nu \lt -1\$, i.e. \$-x = \lvert x \rvert \gt y\$.
This means that kinematically, with a gearbox where the wheels turn in the same direction (\$\lambda \gt 0\$), the vehicle can move right at any rate, but left only faster than the blue surface.
However, if we also consider gearboxes \$\lambda \lt 0\$, where the two wheels turn in different directions, we find that then \$\nu \lt 0\$ or \$\nu \gt 1\$, i.e. the vehicle can move left at any rate, but right only faster than the blue surface moves left.
That also means that varying the gearing ratio and direction \$\lambda\$, any constant velocity \$x\$ is possible for a given blue surface velocity \$y\$.
Note that a constant velocity kinematic analysis implies that all forces are in balance, since there is no acceleration. That includes both static and dynamic forces, and both linear and angular forces (torque). To determine acceleration, both linear and angular momentum would need to be considered also.
Simply put, examining a random subset of forces in a system does not tell you anything about the system. Either you do it fully, finding the rules for the vehicle behaviour, especially acceleration, or you do a kinematic analysis as I do in this post, and find out what kind of steady states are possible when losses due to friction etc. can be ignored.
The path of such analysis can be considered to start from kinematics (simplest, easy to teach to children; "when this moves
thus, how does
that have to move"), then rigid body kinetics (forces, torques –– perhaps via ragdoll physics), then to classical mechanics.
For real world models, friction is a major issue. Test vehicles tend to be light, and slip easily; this means it is
extremely important to use small gearing ratios, \$\lambda\$ close to zero. That is, you need a gearing where the driven/small wheel does many rotations per each large/ground/driving wheel rotation, regardless of whether they turn in the same or in the opposite direction.
I personally created a test vehicle, where the ground/driving wheel is connected to a vertical axis using Lego Technic worm screw, and a thread spooled to the vertical axis; see the attached image below. This gives \$\lambda = \pm 0.05\$ or so, but even then, adding more weight to the vehicle adds to its traction, and easier to observe behaviour. Depending on the spooling direction, \$\lambda\$ changes sign, so it is easy to test all the above scenarios. If you rename
trike.txt to
trike.ldr, you can open it in
LeoCAD.
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