Kinematics is defined as a branch of mechanics that describes the motion of bodies (objects) without considering the forces that caused the motion.

KINEMATIC ANALYSIS

Example of a kinematic analysis of soccer Kinematic analysis is the process of measuring kinematic quantities that are used to describe motion. Kinematic analysis is often used to study the motion of the body, limbs, and joints that occurs during various human movements, as well as the motion of external objects (e.g. sporting equipment or assistive devices) associated with human motion. This method of analysis can be used to evaluate functional performance under normal and abnormal conditions, which can serve as the basis for technique and training improvement, as well as injury prevention and rehabilitation (An, 1984).

KINEMATIC VARIABLES

There are five kinematic variables that are used to describe the spatial and temporal characteristics of human motion. When conducting a kinematic analysis any (or all) of these variables may be used. The five kinematic variables are:

Time
Position
Displacement
Velocity
Acceleration

The first two variables, time and position, are the most basic in that they do not depend on other variables. Displacement, velocity and acceleration are more complex and are built upon the variables of time and position. The variables of position, displacement, velocity and time can be discussed both in linear and angular terms. In the upcoming sections we will discuss each of these variables in detail.

Units

Before we cover the specific kinematic variables, it is important to note the units used to quantify these variables. The international system of units (SI units), which is the modern form of the metric system, should be the system of units used to express each of these variables. The subsequent description of each of these variables will include which SI unit to use and you can find information on how to convert units in the next section.

Converting Units

The following method uses a grid system in which units are cancelled on the diagonal, i.e. a unit in the numerator is cancelled by a unit in the denominator. We will start with the example of converting 10 miles per hour to the SI unit of m/s:

Start with 10 mi in the upper left corner of the grid and hr (because it is in the denominator) is in the bottom left of the grid. When there is no specific numerical value a 1 is included, so in this case we will include a 1 in the bottom left quadrant to go with the hr.

10 mi

1 hr

Using this grid system, we will add columns to the grid as needed and units will cancel on the diagonal by dividing units (grayed cells). We will start by cancelling the miles with the intent to get to the last top cell being meters (m). It is essential that we end with meters in the numerator (top) since that is where it is located in our desired unit of m/s. In this example we add a column and then fill in the conversion factor. Here we can use the direct relationship that 1 mi = 1609 m.

10 mi	1609 m
1 hr	1 mi

In this case, the miles have cancelled, so we’re left with m/hr for our unit. If this were our goal, we would multiply the top row to obtain the numerator and multiply the bottom row to obtain the denominator and then divide the numerator by the denominator, which would give us:

Numerator: 1609 * 10 = 16090

Denominator: 1 * 1 = 1

Result = 16090/1 = 16,090 m/hr

However, our goal is to convert to m/s, so next, let’s take care of the hours. Again, we are going to cancel on the diagonal, so in this case we’ll move to the top row and add a 3^rd column. The miles have already been converted to meters, so we’ll leave what we have so far and add another column to start cancelling the hours. We’re again working with the denominator, so to get rid of hours, so we’ll add another 1 hr to the top cell of the 3rd column. Then in the denominator we’ll include the conversion to minutes (1 hrs = 60 min):

10 mi	1609 m	1 hr
1 hr	1 mi	60 min

The hours would then cancel, so at this point we could multiply the top row to obtain the numerator and multiply the bottom row to obtain the denominator and then divide the numerator by the denominator, which would give us:

Numerator: 1609 * 10 * 1 = 16090

Denominator: 1 * 1 * 60= 60

Result: 16090/60 = 268.17 m/min

Again, our goal is m/s, so we now need to convert from minutes to seconds. Given that 60 min is in the denominator, we’ll add 1 min in the top cell of the 4^th column and then put the conversion from minutes to seconds in the bottom cell (1 min = 60 s). Then we'll check to make sure the units cancel on the diagonal. Miles cancel, leaving meters in the numerator. Hours and minutes cancel leaving seconds in the denominator:

10 mi	1609 m	1 hr	1 min
1 hr	1 mi	60 min	60 s

Then we multiply the values in the top row to obtain the numerator and multiply the values in the bottom row to obtain the denominator and then we divide the numerator by the denominator to obtain our final values of m/s.

Numerator: 1609 * 10 * 1 = 16090

Denominator: 1 * 1 * 60*60= 60

Result: 16090/3600 = 4.47 m/s

There are several different ways to convert units, so it is fine to use an alternative approach if you are used to something different. However, this method provides an organized approach that clarifies the conversion process.

TIME

Time is one of the most basics aspects of kinematic analysis and is essential for describing the temporal characteristics of human movement. There are several different ways of describing temporal characteristics. First, we can describe time both as relative or absolute. Relative time describes components of movement as a percentage of the movement time, whereas absolute is the actual time measurement (e.g. milliseconds, seconds, minutes, hours, days). While different scales of measurement can be used to describe absolute time, when using time in kinematic calculations it is important to use the SI unit of seconds (s).

Example of absolute (s) and relative time (%):

Absolute and relative time for a complex ski maneuver

In addition to describing motion using either absolute or relative time, often movements are broken down into discrete phases that are used to describe the temporal patterning of a movement. Phases are essentially discrete components of a particular movement. There is no hard and fast rule for how to break a movement into phases, but a good way to think of how to break apart the movement would be to think of how you would break it apart if you were teaching the movement. In this regard think of a balance of offering enough information while not overwhelming the learner with too much information. Additionally, key events, are often used to distinguish specific phases.

Examples of phases:

Example of phases of a basketball dunk

"Mansfield Womens Basketball Sequence " by Alex Hancook is licensed under CC BY-NC 2.0 . Phase titles added to original.

Example of phases of motion of a baseball pitch

"Baseball_pitching_motion_2004" by Rick Dikeman is licensed under CC BY-SA 3.0 . Phase titles added to original.

POSITION

Along with the variable time, position is one of the most basic kinematic variables. Position is described as the location in space relative to some reference point. To describe position typically a Cartesian coordinate system is used. The Cartesian coordinate system identifies a point in a plane by a set of numerical coordinates that are the signed distances to the point from two fixed perpendicular lines. Each of these fixed perpendicular lines is called an axis (plural axes) and the point where the lines meet is called the origin. The Cartesian coordinate system can be located anywhere, but usually it is oriented in a location that makes sense relative to the movement. For example, the vertical axis, anterior-posterior and medio-lateral directions. With the coordinate system centered at a location where force is measured (force plate). The SI unit used for linear position is the meter (m) and position can be described as two-dimensional (2D) or three dimensional (3D). For simplicity, biomechanists often analyze planar motion separately. For example, it is common to examine only the sagittal plane in running or walking.

2D Cartesian Coordinate System:

2D Cartesian coordinate system

Cartesian Coordinate System by Kbolino [Public domain]

3D Cartesian Coordinate System:

Cartesian coordinate system

Cartesian coordinate system for 3D by Jorge Stolfi [Public domain]

Alternatively, angular position is the angle through which a point or line has been rotated about a specified axis on the cartesian coordinate. The SI unit used for angular position is the radian (rad). However, degrees (^o) are often used to describe motion because they are used more frequently in everyday life and hence tend to make more intuitive sense to many individuals. It is important to note that when making many calculations it is essential to use radians, so it is good to get in the habit of converting to this unit.

Position can be a bit tricky to conceptualize, because it is a component that is used to describe movement, rather than it being a measure of movement itself. A good way to avoid this confusion is to think of position as being a snapshot in time during a movement.

360 degree rotations expressed in radian and degrees

30 degree rotations expressed in radian measure by Adrignola [CC0]

Sample Problems:

1. Here we have a reflective marker as in those that are tracked by motion capture systems. At position A the marker is 1.8 m from the origin (0 m) and at position B the marker is 3.2 m from the origin:

Linear position

2. Here we have a knee extension movement. The lower leg extends from position A = 22^o to position B = 3^o.

Angular position example

Adapted from "Body Movements" by Tonye Ogele CNX is licensed under CC BY-SA 3.0

DISPLACEMENT & DISTANCE

Displacement and distance are measures of how far an object has moved. Displacement and distance can be expressed in both linear and angular terms. Angular displacement/distance is a measurement of movement around the origin, whereas linear displacement/distance is movement anywhere within the Cartesian coordinate system. With displacement and distance, we are now moving into the kinematic variables that are built upon the variables of time and position.

Distance Displacement by User:Stannered licensed under CC BY-SA 3.0

Displacement is defined as the final change in position and is calculated as follows:

Linear displacement: d = Δx = x_f– x_i

Where d = linear displacement (m), Δ = change in, x = position (m; i = initial, f = final). The subscript of i after the x (as in x_i) indicates that the position value is the initial position value and a subscript of f (as in x_f) indicates that the position value is the final position value.

Angular displacement: θ= Δx = x_f – x_i

Where θ = angular displacement (rad), Δ = change in, x = position (rad; i = initial, f = final). The subscript of i after the x (as in x_i) indicates that the position value is the initial position value and a subscript of f (as in x_f) indicates that the position value is the final position value.

Distance is defined as the sum of the changes in position and is calculated by adding how far an object has moved in all directions.

Sample Problems:

1. Continuing with the example of the reflective marker that is tracked by motion capture systems. At position A the marker is 1.8 m from the origin (0 m) and at position B the marker is 3.2 m from the origin, so the linear displacement is calculated as:

d = x_f– x_i

d =3.2 m – 1.8 m = 1.4 m

Linear position

2. Continuing with the example of the knee extension movement. The lower leg extends from position A = 22^o to position B = 3^o. First we need to convert to radians:

3^o	2 𝛑 rad
360^o

(3^o * 2 𝛑 rad)/ 360^o = 0.052 rad

3^o	2 𝛑 rad
360^o

(22^o * 2 𝛑 rad)/ 360^o = 0.38 rad

The we can solve for displacement as the change in position:

θ = x_f – x_i

θ = 3^o – 22^o = -19^o

or

θ = 0.052 rad – 0.38 rad = -0.328 rad

Note: positive and negative signs indicate opposing directions. In this case the negative value indicates a counterclockwise movement (extension) and positive would indicate a clockwise movement (flexion).

Angular position example

Adapted from "Body Movements" by Tonye Ogele CNX is licensed under CC BY-SA 3.0

Scalars vs. Vectors

The main difference between the measures of displacement and distance is that displacement is a vector quantity and distance is a scalar quantity. Before diving further into a discussion of displacement and distance it is important to cover the difference between scalar and vector quantities.

Variables that are used to describe motion can be divided into two categories: scalars and vectors.

Scalars quantities are described by magnitude alone.
Vectors quantities are described by both magnitude and direction.

Let’s use linear displacement and distance to illustrate the difference between scalars and vectors. An easy way to think of the difference between displacement and distance is that distance would be what your car odometer would read and displacement would represent the distance as the crow flies (start to end of a movement).

Distance Displacement by User:Stannered is licensed under CC BY-SA 3.0

VELOCITY & SPEED

Velocity and speed are measures of how fast an object is moving. Velocity is the vector version of this quantity and speed is the scalar quantity. Velocity and speed can be expressed in both linear and angular terms, with the only difference being that with angular velocity/speed the movement is a rotation instead of along a linear path. Velocity is calculated as the displacement of an object divided by the amount of time that it took for the displacement to occur. The equation for velocity are as follows:

Linear velocity: v = d/Δt

Where v = linear velocity (m/s), d = linear displacement (m), Δ = change in and t = time (s)

Angular velocity: 𝜔 = θ/Δt

Where 𝜔 = angular velocity (rad/s), θ = angular displacement (rad) Δ = change in and t = time (s)

Speed is calculated as the distance an object has moved divided by the corresponding change in time.

Sample Problems:

1. Continuing with the example of the reflective marker that is tracked by motion capture systems. We calculated the displacement as 1.4 m. If this displacement took 0.75 s the velocity would be calculated as follows:

d = x_f– x_i

d =3.2 m – 1.8 m = 1.4 m

v = d/Δt

v = 1.4 m / 0.75 s

v = 2.27 m/s

Linear position

2. Continuing with the example of the knee extension movement. The displacement was calculated as -0.328 rad, assuming this occurs in 0.2 s the velocity would be calculated as:

θ = 0.052 rad – 0.38 rad = -0.328 rad

𝜔 = θ/Δt

𝜔 = -0.328 rad/0.2 s

𝜔 = -1.64 rad/s

Note: positive and negative signs indicate opposing directions. In this case the negative value indicates a counterclockwise movement (extension) and positive would indicate a clockwise movement (flexion).

Angular position example

Adapted from "Body Movements" by Tonye Ogele CNX is licensed under CC BY-SA 3.0

ACCELERATION

Acceleration is a measure of how quickly velocity is changing and is calculated as the change in velocity divided by the corresponding change in time. Similar to the other kinematic variables, acceleration can be expressed in both linear and angular terms, with the only difference being that angular acceleration is a rotation and linear acceleration is along a linear path. Acceleration is a vector quantity meaning that it has both a magnitude and direction, and there is no scalar equivalent of this variable. In reality in many cases acceleration is continuously changing, but often these small changes in acceleration are ignored and for simplicity we assume that acceleration is constant. The equations for acceleration are as follows:

Linear acceleration: a = Δv/Δt = (v_f – v_i)/Δt

Where a = linear acceleration (m/s²), v = linear velocity (m/s; i = initial, f = final), Δ = change in and t = time (s)

Angular acceleration: 𝛼 =Δω/Δt = (ω_f – ω_i)/Δt

Where 𝛼 = angular acceleration (rad/s²), ω = angular velocity (rad/s; i = initial, f = final), Δ = change in and t = time (s)

Sample Problems:

1. Continuing with the example of the reflective marker that is tracked by motion capture systems. We initially calculated the velocity of the marker as 2.27 m/s. If the marker accelerated to a velocity of 4.13 m/s in 0.8 s, the acceleration would be calculated as follows:

a = (v_f – v_i)/Δt

a = (4.13 m/s – 2.27 m/s)/ 0.8 s

a = 2.54 m/s

2. Continuing with the example of the knee extension movement. The velocity for the section of the movement that we observed was calculated as -1.64 rad/s. As discussed previously, positive and negative signs indicate changes in the direction of displacement or velocity. In this case, the angular velocity of -1.64 rad/s indicated an extensor movement. However, when discussing acceleration positive values indicate increasing speed and negative values indicate decreasing speed. Therefore, we'll ignore the sign for this calculation. Let's assume that the leg started from a stationary flexed position and then accelerated to the velocity of 1.64 rad/s in 0.3 s, so the acceleration of this movement would be calculated as follows:

𝛼 = (ω_f – ω_i)/Δt

𝛼 = (1.64 rad/s – 0 rad/s) / 0.3 s

𝛼 = 5.47 rad/s²

Angular acceleration

Adapted from "Body Movements" by Tonye Ogele CNX is licensed under CC BY-SA 3.0

References:

An, K. N. (1984). Kinematic analysis of human movement. Annals of Biomedical Engineering, 12(6), 585–597. doi: 10.1007/bf02371451

When no attribution is noted, images are from the Public Domain or the work of the author.