Calculating Joint Forces

To properly size the articulation joint, I need to understand what forces it will experience. The first step is drawing a free body diagram of the full vehicle and solving for the wheel reaction forces.

Here is the full vehicle. The joint in the middle is represented by 3 ball joints forming the triangle. You can see the basic set-up of the joint here in this post.

The vertical forces all have to sum to zero so we can write,

$$W_r + W_f = F_r + F_f$$

And we can sum moments around any point to get a second equation. Let’s sum the moments around the Rear Half COG to get,

$$F_r \cdot X_r + W_f \cdot (X_f - X_r) = F_f \cdot (L - X_r)$$

Solving by substitution, the front wheel reaction is simply:

$$F_f = \frac{W_r X_r + W_f X_f}{L}$$

and the rear follows from the first equation:

$$F_r = W_r + W_f - F_f$$

This has a nice intuitive interpretation — each wheel reaction is just the sum of each weight multiplied by its distance from the opposite wheel, divided by the wheelbase.

Now we can cut the robot in half at the joint and just look at the rear of the robot, as shown below.

The top ball joint is connected by a link to the top joint on the front half of the robot. This link has ball joints at both ends and so can only transfer loads along its length, here denoted $F_L$. To balance the forces in the x direction, the bottom joint must see an equal and opposite x force, $F_{J,x}$. This bottom joint can apply a vertical force because it is rigidly connected to the rear chassis, just like how a trailer hitch has tongue weight that will compress your car’s suspension system. Here that vertical force is denoted $F_{J,y}$. $F_r$ is the rear wheel reaction force and is unchanged from the analysis above.

Balancing forces in x and y, and summing moments around the lower ball joint gives three equations. Solving each for the unknown joint force:

$$F_L = \frac{F_r \cdot L_r - W_r(L_r - X_r)}{h}$$

$$F_{J,x} = F_L$$

$$F_{J,y} = W_r - F_r$$

There is a nice special case worth noting. When $X_f = L$ the front section’s centre of gravity is directly above the front wheel — the term $W_f(L - X_f)$ in the full vehicle moment equation goes to zero. This means $F_r$ simplifies to:

$$F_r = \frac{W_r(L - X_r)}{L}$$

The front weight $W_f$ has dropped out entirely. Since $F_L$, $F_{J,x}$, and $F_{J,y}$ all depend only on $F_r$ and the rear section geometry, it follows that all joint forces become independent of the front weight when the front COG sits directly above the front axle. The front section is perfectly balanced on its own wheel and transfers nothing through the joint — it could be twice as heavy and the joint wouldn’t know.

Numerical Example

Plugging in the RoBug prototype dimensions with $W_r = 500$ N and the front COG over the front axle ($X_f = L$):