Peter H. Werner
Senior Principal Engineer
Drive Systems Engineering
The goal of drive-system engineers is usually pretty straightforward. Specify the smallest motor and drive needed to accelerate and run a load at a required rate, and determine the braking forces necessary to stop the system in an acceptable period.
All too often drives engineers know the electrical details inside and out but fall short on the mechanical nuances. First, there seems to be a general lack of appreciation of the difference between weight and mass, particularly when using the imperial system of units. Convention is to consider weight as the force required to support a body, while mass is its resistance to acceleration.
Kilograms (mass) and pounds (force) are often rationalized incorrectly. And moving back and forth between imperial and metric engineering units regularly leads to confusion and faulty designs.
In many engineering disciplines motion is not a concern such as building a structure to support static loads and the issues relative to appreciating the relationships between mass and force, and what the term "weight" really means, do not come into play. It is only important when accelerating objects, which tends to be the realm of drive-system engineers.
Another issue today is that most engineers use sizing-software programs. These are valuable because no one wants to derive equations from the extreme basics just to size a motor. But in most cases, there are no references to how the equations were derived. So when engineering units differ from those in the software, there is little understanding as to how to establish new constants. And using canned equations without a grasp of the underlying fundamentals can unwittingly lead to gross errors. To help alleviate such problems, here's a look at the physical considerations that go into building a successful drive system.
Let's start with some basic concepts. While seemingly elementary, a failure to grasp these fundamental tenets is often a problem for drives engineers. Also, for simplicity, we'll address the subject in metric terms, although the same concepts apply to imperial units. (See Imperial information for additional details.)
Mass is the property of a body that quantifies its linear inertia and influences the force necessary to support it in a gravitational field. The term "weight," or influence, has no consistently defined engineering units, as do force and mass. The engineering units of mass are grams in the metric system and slugs in the imperial system. The engineering units of force, discussed later, are Newtons in the metric system and pounds in the imperial system.
In the metric system, weight (or influence) of a body is typically defined as its resistance to acceleration or its inertia. In the imperial system, weight is typically defined as the force, in pounds, to support a body in its gravitational field. Either is appropriate, but it is important to understand which engineering units are implied. Mass in kilograms is defined by a standard platinum-iridium cylinder kept at the International Bureau of Weights and Measures.
Force is the influence that, if applied to a free body, results chiefly in an acceleration of the body. The metric unit of force is the Newton. A force of 9.860665 N is developed by a 1-kg mass under a gravitational attraction of 9.80665 m/sec2,
Displacement is the difference between the initial location of a body and any later position. The meter is the basic linear metric unit of length. But in the rotational world, displacement is measured in radians the angle at the center of a circle subtended by an arc equal to the length of the radius. The radian is therefore the rotational displacement mechanism that aids in the transition between linear and rotational physics. The radian is unitless relative to distance, and therefore circumferentiallinear distance is established by the radius.
Circumference is geometrically defined as
Therefore, 1 radian
Rotational velocity is measured in terms of radians/sec, v , or revolutions per minute (rpm), S.
Torque is a force that produces rotation or torsion. It is the product of the force and the perpendicular distance from the line of action of the force to the axis of rotation,
Therefore, a 1-N force acting at a distance of 1 m produces torque of 1 Nm.
Linear mechanical power is the product of force and velocity,
By definition, a 1-N force moving a body at 1 m/sec requires 1 W of power, or
Rotational mechanical power is the product of torque and rotational velocity,
A torque of 1 Nm at 1 rad/sec, acting at a radius
of 1 m, requires 1 W of rotational power,
For rotational speed, S, in rpm, rotational power is
RADIUS OF GYRATION
When accelerating an object it is necessary to consider inertia, the tendency of a body to remain at rest, or in uniform motion, unless acted upon by an external force. Moment of inertia, a measure of a body's resistance to angular acceleration, equals the product of the body's mass and the square of its distance from the axis of rotation,
Moment of inertia, which depends on the shape of the object, is important when solving problems that have to do with how things rotate. Dividing by mass and taking the square root produces a variable with units of just distance, also known as the radius of gyration. Conceptually, the radius of gyration is the distance that, if the entire mass of the object were all concentrated at that radius, would give the same moment of inertia as the original object.
Because most motor loads are cylindrical (such as pumps and rollers) let's looks at the radius of gyration for a cylinder.
For a hollow cylinder, radius of gyration is
where r1 = radius to the cylinder ID and r2 = distance to the OD.
Note that for a cylindrical ring with all the mass concentrated at a single radius, r1 = r2 and K = r2.
For a solid cylinder, r1 = 0 and
or the rms value of the radius.
Determining the torque required for acceleration is necessary to properly size a motor. Torque is proportional to the product of inertia and angular acceleration, or
This can be restated in terms of speed, S, in rpm,
The torque required to accelerate a cylinder of known mass is
To determine the torque to accelerate a cylinder of known density, first determine the cylinder mass:
Torque is the product of inertia and angular acceleration,
where I = mK2.
Substituting for m and K,
There is also the need to determine the energy stored in the rotating mechanical system, to size components such as dynamic braking resistors, capacitor banks for power ride-through systems, and mechanical brakes.
Energy is defined as the product of power and time,
Peak power Ppis the product of torque and velocity,
Deceleration torque is
where is the deceleration rate in rad/sec2. Therefore,
Potential energy is
For a complete discussion of drive-system physics in terms of imperial units, including gravitational constants, imperial units of mass (slugs), and the like, contact the author at email@example.com.
For more information on Allen-Bradley drives: