#### What is in this article?:

- Linearized Model of a Hydraulic Motor
- Synchronizing the speed of two motors

## Synchronizing the speed of two motors

Hydraulic-system designers often connect two motors in series in an attempt to synchronize their speeds. In principle, this is a sound idea. In actuality, however, the degree of synchronizing is imperfect because of finite internal leakage resistances. The accompanying box illustrates a practical use of a mathematical model to quantify the degree of this nonequality of the two motor speeds.

Connecting two hydraulic motors in series in an attempt to synchronize their speeds is a sound idea. In reality, though, the synchronization is imperfect because of internal leakage resistances. We’ll now examine a scenario using a mathematical model to quantify the inequality of the two motor speeds.

Assume two hydraulic motors — each identical to that described previously — are to be connected in series and powered by a 60-in.^{3}/sec constant-flow source. As shown in Figure 3, the outlet port of the low-pressure motor is connected directly to tank, as are both case-drain ports. The high-pressure motor is connected to a 650-lb-in. load, but the shaft of the low-pressure motor is completely free. Both motors have a displacement of 1.88 in.^{3}/rev; leakage resistance from each motor port to case of 885 psi/(in.^{3}/sec); port-to-port leakage resistance of 696 psi/(in.^{3}/sec); and torque loss from friction and windage of 0.031 lb-in./rpm.

There are four unknowns: *P*_{1}, *P*_{2}, *N*_{1}, and *N*_{2}, so four equations will be written and solved simultaneously. Note from the illustration that *P*_{4} and *P*_{3} equal 0. Two node equations represent the summation of flows (*P*_{1} and *P*_{2} nodes) and two torque summation equations (*N*_{1} and *N*_{2}).

Flow summation at *P*_{1}:

Flow summation at *P*_{2}:

Torque summation at *N*_{1}:

Torque summation at *N*_{1}:

Substitution and linear algebra matrix are two common methods of solving four equations with four unknowns. However, the most practical method is by computer, and all popular spreadsheet programs have a simultaneous equation-solving capability. I solved these equations using the eQsolver capability in IDAS Engineering software. The results are:

*P*_{1} = 2,537 psig,

*P*_{2} = 186.6 psig,

*N*_{1} = 1,716 rpm, and

*N*_{2} = 1,802 rpm.

The solution to this problem demonstrates that there is nearly a 100-rpm difference between the two motor speeds. If we now solve the problem with the loads reversed (the upper motor is unloaded and the lower motor is loaded) we find that:

*P*_{1} = 2,521 psig,

*P*_{2} = 2,333 psig,

*N*_{1} = 1,815 rpm, and

*N*_{2} = 1,549 rpm.

This solution shows that there is nearly a 300-rpm change in the speed of the lower motor — a condition that certainly is less than ideal for the application, but without more specific information, judgments cannot be passed.

The point of this analysis is not to provide a means for achieving perfect motor speed synchronization. Rather, there is a more limited goal. First, using reasonable models of hydraulic machinery, it is possible to evaluate the consequences of implementing a given circuit concept before any hardware is even assembled. Second, circuit developers and designers can explore the endless “what ifs” that always occur at circuit design time.

The broader issue of perfect motor-speed synchronization requires closed-loop speed-control systems and will have to wait for some later discussion. Additionally, closed-loop control modeling must expand to include dynamic response because of the possibility of hunting and sustained oscillations.