SimMechanics

Checking Model Validity

Simulink can simulate a SimMechanics model only if it is valid. A model is valid if it satisfies the following rules:

• Every schematic in the model is topologically valid (see Checking Schematic Topology).
• The model contains at least one degree of freedom (see Counting Degrees of Freedom).

Checking Schematic Topology

To avoid simulation failures, you must ensure that the topology of your block diagram is valid. A block diagram is topologically valid if each schematic that it contains is valid. A schematic is valid if its spanning tree is valid. Thus to determine if your model is valid, first determine the spanning tree of each schematic that it contains and then the validity of the resulting spanning trees.

Determining a Schematic's Spanning Tree

To determine the spanning tree of a schematic, remove all blocks from the schematic except Body and Joint blocks and open every closed loop in the resulting reduced schematic. To open a closed loop, conceptually follow the loop-cutting rules in Cutting Closed Loops.

For example, here is a schematic with two closed loops.

Cutting the top loop at the Disassembled Prismatic and removing the Parallel Constraint block (thus simultaneously cutting the bottom loop) yields the schematic's spanning tree.

Determining the Validity of a Spanning Tree

To be valid, a spanning tree must meet the following requirements:

• The spanning tree must have at least one Ground block to serve as a reference to World.
• Every Joint block must be connected to exactly two Body blocks.
• Every non-Ground Body block must have a unique path to a Ground block. (This need not be true of the model schematic itself.) This ensures that, while each body moves via joints relative to other bodies, SimMechanics can resolve all bodies' motions into absolute motions with respect to World.
• Every non-Ground Body block at an end of a sequence of Bodies must have nonzero inertia (mass or inertial moment) associated with all joint primitives that can move. Each translational DoF must carry a nonzero mass, and each rotational DoF a nonzero inertial moment. This prevents infinite accelerations when forces are applied.

Examples of Invalid Schematic Topologies

Here are some examples of invalid topologies:

• This one-loop schematic lacks a Ground block.

  

• This open machine schematic has a dangling Joint block.

  

• Another open machine schematic features a zero-mass body at one end of a chain of bodies.

  

The last two invalid examples are dynamically (but not topologically) equivalent, because a zero-mass body is dynamically no body at all.

Counting Degrees of Freedom

Identifying and counting the independent degrees of freedom (DoFs) of a machine are important for trimming and linearizing SimMechanics models (see Trimming Mechanical Systems and Linearizing SimMechanics Models in the Running Mechanical Models chapter) and for correcting simulation errors (see Troubleshooting Simulation Errors in the Running Mechanical Models chapter).

Your SimMechanics model must have at least one DoF to be valid. A free physical body has six DoFs: three translational and three rotational. But in a machine, connections between bodies by joints, constraints, and drivers, and motion actuation by joint and body actuators reduce the machine's independent DoFs to a smaller number. You also reduce a body's DoFs if you confine the machine's motion to one or two spatial dimensions.

In SimMechanics, a Body block has no DoFs. Connecting Joints to a Body adds DoFs to the machine. The joint primitives represent the body's DoFs relative to other connected Bodies or Grounds. Connecting Constraint and Driver blocks to Bodies or motion-actuating joint primitives in Joints removes DoFs from the machine.

Finding Independent Degrees of Freedom

Here is the formula for determining the number of independent DoFs your model has:

# of independent DoFs = # of body DoFs + # of primitive DoFs -
# of motion restrictions

The following three steps define each term on the right-hand side:

1. Calculate the number of body DoFs from the number of Body and Joint blocks in your model:

    # of body DoFs = 6 * (number of Bodies - number of Joints)

1. If you have confined the machine to move in only two dimensions, replace the 6 by 3. If you have confined the machine to move in only one dimension, replace the 6 by 1.

2. Calculate the number of primitive DoFs by adding up the primitive DoFs from the Joint dialog boxes:
• Count one for each prismatic (P) or revolute (R) primitive.
• Count three for each spherical (S) primitive.
• Count zero for each weld (W) primitive.

1. Do not count a primitive DoF that is motion-actuated by a Joint Actuator.

3. Calculate the number of motion restrictions by adding up the motion restrictions of each Joint Actuator, Constraint, and Driver block. Different blocks from the Constraints & Drivers library impose different numbers of motion restrictions.
   
   

   Constraint Block	Restrictions	Driver Block	Restrictions
   Gear	One	Angle	One
   Parallel	Two	Distance	One
   Point-Curve	Two	Linear	One
   		Velocity	One

1. Be sure not to count redundant motion restrictions. These are restrictions that forbid the motion of joint primitives that could not move anyway even if the constraint were removed, because of how the joints are configured.

   Example: A body is connected to a ground by a single prismatic. You place a constraint on the body that prevents it from moving perpendicularly to the prismatic axis. The body could not move in that direction even if you removed the constraint. So the constraint is redundant, and you would not count it as a motion restriction.

DoF Example: Double Pendulum

The mech_dpen model from the Demos library represents planar double pendulum motion actuated by a Joint Actuator.

The double pendulum has two rigid bodies, such as two rods, confined to move in two dimensions. Ignoring the Joint Actuator temporarily, there are two bodies, two joints, and two revolute primitives, and thus 3 * (2 - 2) + 2 = 2 independent DoFs. There are many ways to represent these two DoFs, but the two revolute primitives are the simplest way.

Including the Joint Actuator in the DoF count removes the revolute primitive in the Revolute block as an independent DoF. So this model actually only has one independent DoF, the revolute primitive in the Revolute1 block.

DoF Example: Four Bar Mechanism

The four bar mechanism of A Four Bar Mechanism in the Learning Basic Procedures chapter has four revolutes. You can establish that only 3 * (3 - 4) + 4 = 1 of these DoFs is actually independent and arrive at the same result obtained in the Four Bar Mechanism example.

Modeling Sensors

Running Mechanical Models