I don't think the micro servos are useless so much as you are over torquing them overly heavy loads at the end of overly long lever arms, and digital servos will try and fight harder to maintain position than analog servos will. They work fine on the tiny model aircraft they were meant for.
It appears you never considered how much torque was required before starting things, and just went for the cheapest servos.
Physics dictates that, the torque of a joint closer towards the base of a mult-jointed lever is always greater than the torque at all the joints further out towards the tip. This should be evident from mechanical intuition.
There are only penalties placing the same motor at every joint because the loads are never the same. It only ensures that motors are either underworked or overworked. Doing so only ensures that the motors further out have torque capability that can never be used (since whatever torque they exert is amplified and passed back towards the motors towards the base). And if you are working against gravity, then it's even worse because motors at further out are now heavier than they need to be with all that useless extra torque capability which means the motors towards the base now have even more of a load.
The minimum torque required by a motor is:
\$ \tau_{static} = \sum_{Joint}^{Tip} (Force \times Distance)\$
"Distance" is the distance through space from the joint of interest to the point where the force is applied in the worst case scenario. The worst case scenario is when the lever is fully extended as long as possible.
If this force is a weight, then the distance is from the joint of interest to the center-of-gravity of the component. For example, if this component is a segment of lever itself, you can probably assume a lever that is uniform along its length which means the COG is halfway along its length.
Also note that "Distance" is a distance through space. If the arm can extend to be completely straight then the distance follows along the arm linkages. But in the case that the arm cannot fully extend (maybe it can only ever fully extend 90 degrees), this distance is along line that cuts straight across space itself from the joint of interest to the point the force is acting at. It need not follow along the linkages of the arm.
Note that this is the minimum force required. It is not the force required to actually accelerate anything. In terms of lifting against gravity, this means this is the force required to hold something at a given position. Extra force is actually required to produce acceleration to get it to that position.
If you choose to account for this, there is an extra torque to add to the torque calculated above:
$$ \tau_{dynamic} = I \times \alpha $$
This formula is just the rotational equivalent of
$$F = M \times A$$
which is for a straight line.
\$\tau_{dynamic}\$ is the torque required to swing (or stop a swing) for everything under its own inertia under the worst case. Worst case being when everything further out is extended such that as much of the mass as possible is farthest away from the joint as possible.
\$\alpha\$ is the angular acceleration in \$radians/s^2\$. It is the rotation analog of linear acceleration.
\$I\$ is the "Moment of inertia" which is the rotational equivalent of mass and can be tricky to figure out if you have never seen it before.
To get moment of inertia, you first have to arrange everything past the joint into the worst case scenario and the calculate the center-of-gravity and moment of inertia of the entire assembly.
Worst case being when everything further out is extended such that as much of the mass as possible is farthest away from the joint as possible.
From there, it is most convenient to first turn every object in the arm into a point mass, rather than leaving them as the distributed masses they are. So place the mass of every component at it's center of gravity.
From there, the moment of inertia of the entire assembly is
\$I = \sum (m \times r^2)\$
Where:
\$m\$ is the mass of each object
\$r\$ is the distance from the joint to the center-of-gravity of each object
This is the simple way. You can technically model your components as simple geometric objects of uniform weight distribution (like beams, or cubes, or rectangular prisms) and look up the moment of inertia in a table. The table will provide the moment of inertia for the object spinning around it's center-of-gravity, so you have to use a simple equation called the "Parallel axis thereom" that translates that table's moment of inertia of the object about it's center-of-gravity to some arbitrary point in space (which will be your joint). Then you just add all the moment of inertias up.
For the record, I probably would not calculate moment of inertia and just pick a torque double the static torque.
Also...RC servo torque and speed ratings are not are not to be trusted. Take them with a grain of salt. They are almost certainly slower and weaker than what the manufacturer claims.
