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Science has always been used to improve every aspect of our lives. That's why it isn't surprising to see, that the science of biomechanics is being used to improve and fine-tune the performance of athletes and players in the world of major sports as well.

There are billions of dollars at stake in the sports industry and the latest cutting-edge technology is being used to take the game(s) to a whole new level. There is a lot of money involved in a game such as Golf and here too the science of biomechanics is applied to perfect the sport of top professionals.

For those not in the know, Biomechanics revolves around the mechanical analysis of the body in motion. It uses Motion Capture Technology (MOCAP) or high-speed cameras to take shots and of a player's body in action and record the data. This data is then analyzed by Biomechanics to ascertain the best possible movement of the body to in order to generate maximum results.

The biomechanics of a golf swing can be divided into six main stages. These are:

1. Address
2. Backswing
3. Transition
4. Downswing
5. Impact/Contact
6. Follow-through
7. Finish

At the address stage, a golfer places their body in a comfortable, relaxed and balanced stance to begin the swing. This is a specific anatomical position which also includes getting the right grip. For a golfer to achieve this correct starting position, they must try to consistently maintain it from swing to swing.

The backswing is that part of the golf swing where the body is in the right position to begin the downswing. During this process, the torque or potential energy is created and stored in the body by rotation of the knees, hips, spine and shoulders. In order to achieve the right position here, a golfer must rotate his body parts around an imaginary axis of the body.

The transition stage is defined as the point where the backswing ends and the downswing begins. At this point the golfer begins forward movement of the swing. Studies have shown that this is where additional elastic energy is stored within the body which is later transferred to the golf ball at the point of impact.

After the execution of the transition stage, the downswing of the club into impact commences. Here the transformation of energy from potential energy in the body to kinetic energy in the club head occurs. The downswing ends at the point when impact with the golf ball takes place.

At the impact stage, the club head makes contact with the golf ball for approximately half a millisecond and sheer kinetic force is transferred to the golf ball to hit it in the desired direction.

The follow-through is basically the deceleration of the body and club after the impact stage is completed. The deceleration happens as a result of the reverse-absorption of the excessive kinetic energy which was not absorbed by the ball into the body of the golfer. Here, the body continues to rotate around its axis to a completion point where the club head stops its movement behind the golfer. The end position where the follow-through is completed is defined as the Finish point.

These are the stages of the bio mechanics of a golf swing and if all of these stages are executed correctly, a golfer can lift their game and lower their score.

Quickly improve your golf game with proven golf swing tips that can help lower your golf score by visiting http://www.golf-swing-improvement.com, a popular golfing website that provides tips, advice and resources to include information on golf swing aids, golf putting tips and the top selling golf ebooks that will improve your golf game.

Dynamic Analysis of Stepper Motor Mechanism

A force of one pound will accelerate a mass of one slug at one foot per second squared. The same relationship holds between the force, mass, time and distance units of the other measurement systems. Most people prefer to measure angles in degrees, and the common engineering practice of specifying mass in pounds or force in kilograms will not yield correct results in the formulas given here! Care must be taken to convert such irregular units to one of the standard systems outlined above before applying the formulas given here!

 

Statics

For a motor that turns S radians per step, the plot of torque versus angular position for the rotor relative to some initial equilibrium position will generally approximate a sinusoid. The actual shape of the curve depends on the pole geometry of both rotor and stator, and neither this curve nor the geometry information is given in the motor data sheets I've seen! For permanent magnet and hybrid motors, the actual curve usually looks sinusoidal, but looks can be misleading. For variable reluctance motors, the curve rarely even looks sinusoidal; trapezoidal and even assymetrical sawtooth curves are not uncommon.

 

For a three-winding variable reluctance or permanent magnet motors with S radians per step, the period of the torque versus position curve will be 3S; for a 5-phase permanent magnet motor, the period will be 5S. For a two-winding permanent magnet or hybrid motor, the most common type, the period will be 4S, as illustrated in Figure 2.1:

 

Figure 2.1

Again, for an ideal 2 winding permanent magnet motor, this can be mathematically expressed as:

T = -h sin( ((/2) / S)  )

Where:

T -- torque

h -- holding torque

S -- step angle, in radians

 = shaft angle, in radians

But remember, subtle departures from the ideal sinusoid described here are very common.

The single-winding holding torque of a stepping motor is the peak value of the torque versus position curve when the maximum allowed current is flowing through one motor winding. If you attempt to apply a torque greater than this to the motor rotor while maintaining power to one winding, it will rotate freely.

 

It is sometimes useful to distinguish between the electrical shaft angle and the mechanical shaft angle. In the mechanical frame of reference, 2 radians is defined as one full revolution. In the electrical frame of reference, a revolution is defined as one period of the torque versus shaft angle curve. Throughout this tutorial,  refers to the mechanical shaft angle, and ((/2)/S) gives the electrical angle for a motor with 4 steps per cycle of the torque curve.

 

Assuming that the torque versus angular position curve is a good approximation of a sinusoid, as long as the torque remains below the holding torque of the motor, the rotor will remain within 1/4 period of the equilibrium position. For a two-winding permanent magnet or hybrid motor, this means the rotor will remain within one step of the equilibrium position.

 

With no power to any of the motor windings, the torque does not always fall to zero! In variable reluctance stepping motors, residual magnetization in the magnetic circuits of the motor may lead to a small residual torque, and in permanent magnet and hybrid stepping motors, the combination of pole geometry and the permanently magnetized rotor may lead to significant torque with no applied power.

 

The residual torque in a permanent magnet or hybrid stepping motor is frequently referred to as the cogging torque or detent torque of the motor because a naive observer will frequently guess that there is a detent mechanism of some kind inside the motor. The most common motor designs yield a detent torque that varies sinusoidally with rotor angle, with an equilibrium position at every step and an amplitude of roughly 10% of the rated holding torque of the motor, but a quick survey of motors from one manufacturer (Phytron) shows values as high as 23% for one very small motor to a low of 2.6% for one mid-sized motor.

 

 

Half-Stepping and Micro stepping

 

So long as no part of the magnetic circuit saturates, powering two motor windings simultaneously will produce a torque versus position curve that is the sum of the torque versus position curves for the two motor windings taken in isolation. For a two-winding permanent magnet or hybrid motor, the two curves will be S radians out of phase, and if the currents in the two windings are equal, the peaks and valleys of the sum will be displaced S/2 radians from the peaks of the original curves, as shown in Figure 2.2:

 

Figure 2.2

This is the basis of half-stepping. The two-winding holding torque is the peak of the composite torque curve when two windings are carrying their maximum rated current. For common two-winding permanent magnet or hybrid stepping motors, the two-winding holding torque will be:

h2 = 20.5 h1

where:

h1 -- single-winding holding torque

h2 -- two-winding holding torque

 

This assumes that no part of the magnetic circuit is saturated and that the torque versus position curve for each winding is an ideal sinusoid.

 

Most permanent-magnet and variable-reluctance stepping motor data sheets quote the two-winding holding torque and not the single-winding figure; in part, this is because it is larger, and in part, it is because the most common full-step controllers always apply power to two windings at once.

 

If any part of the motor's magnetic circuits is saturated, the two torque curves will not add linearly. As a result, the composite torque will be less than the sum of the component torques and the equilibrium position of the composite may not be exactly S/2 radians from the equilibria of the original.

 

Microstepping allows even smaller steps by using different currents through the two motor windings, as shown in Figure 2.3:

 

Figure 2.3

For a two-winding variable reluctance or permanent magnet motor, assuming nonsaturating magnetic circuits, and assuming perfectly sinusoidal torque versus position curves for each motor winding, the following formula gives the key characteristics of the composite torque curve:

h = ( a2 + b2 )0.5

x = ( S / (/2) ) arctan( b / a )

Where:

a -- torque applied by winding with equilibrium at 0 radians.

b -- torque applied by winding with equilibrium at S radians.

h -- holding torque of composite.

x -- equilibrium position, in radians.

S -- step angle, in radians.

 

In the absence of saturation, the torques a and b are directly proportional to the currents through the corresponding windings. It is quite common to work with normalized currents and torques, so that the single-winding holding torque or the maximum current allowed in one motor winding is 1.0.

 

Friction and the Dead Zone

 

The torque versus position curve shown in Figure 2.1 does not take into account the torque the motor must exert to overcome friction! Note that frictional forces may be divided into two large categories, static or sliding friction, which requires a constant torque to overcome, regardless of velocity, and dynamic friction or viscous drag, which offers a resistance that varies with velocity. Here, we are concerned with the impact of static friction. Suppose the torque needed to overcome the static friction on the driven system is 1/2 the peak torque of the motor, as illustrated in Figure 2.4.

 

Figure 2.4

The dotted lines in Figure 2.4 show the torque needed to overcome friction; only that part of the torque curve outside the dotted lines is available to move the rotor. The curve showing the available torque as a function of shaft angle is the difference between these curves, as shown in Figure 2.5:

Figure 2.5

Note that the consequences of static friction are twofold. First, the total torque available to move the load is reduced, and second, there is a dead zone about each of the equilibria of the ideal motor. If the motor rotor is positioned anywhere within the dead zone for the current equilibrium position, the frictional torque will exceed the torque applied by the motor windings, and the rotor will not move. Assuming an ideal sinusoidal torque versus position curve in the absence of friction, the angular width of these dead zones will be:

d = 2 ( S / (/2) ) arcsin( f / h ) = ( S / (/4) ) arcsin( f / h )

where:

d -- width of dead zone, in radians

S -- step angle, in radians

f -- torque needed to overcome static friction

h -- holding torque

 

The important thing to note about the dead zone is that it limits the ultimate positioning accuracy! For the example, where the static friction is 1/2 the peak torque, a 90° per step motor will have dead-zones 60° wide! That means that successive steps may be as large as 150° and as small as 30°, depending on where in the dead zone the rotor stops after each step!

 

The presence of a dead zone has a significant impact on the utility of microstepping! If the dead zone is x° wide, then microstepping with a step size smaller than x° may not move the rotor at all. Thus, for systems intended to use high resolution microstepping, it is very important to minimize static friction.

 

Dynamics

Each time you step the motor, you electronically move the equilibrium position S radians. This moves the entire curve illustrated in Figure 2.1 a distance of S radians, as shown in Figure 2.6:

 

Figure 2.6

The first thing to note about the process of taking one step is that the maximum available torque is at a minimum when the rotor is halfway from one step to the next. This minimum determines the running torque, the maximum torque the motor can drive as it steps slowly forward. For common two-winding permanent magnet motors with ideal sinusoidal torque versus position curves and holding torque h, this will be h/(20.5). If the motor is stepped by powering two windings at a time, the running torque of an ideal two-winding permanent magnet motor will be the same as the single-winding holding torque.

It shoud be noted that at higher stepping speeds, the running torque is sometimes defined as the pull-out torque. That is, it is the maximum frictional torque the motor can overcome on a rotating load before the load is pulled out of step by the friction. Some motor data sheets define a second torque figure, the pull-in torque. This is the maximum frictional torque that the motor can overcome to accelerate a stopped load to synchronous speed. The pull-in torques documented on stepping motor data sheets are of questionable value because the pull-in torque depends on the moment of inertia of the load used when they were measured, and few motor data sheets document this!

 

In practice, there is always some friction, so after the equilibrium position moves one step, the rotor is likely to oscillate briefly about the new equilibrium position. The resulting trajectory may resemble the one shown in Figure 2.7:

 

 

 

 

Figure 2.7

 

Here, the trajectory of the equilibrium position is shown as a dotted line, while the solid curve shows the trajectory of the motor rotor.

 

Resonance

 

The resonant frequency of the motor rotor depends on the amplitude of the oscillation; but as the amplitude decreases, the resonant frequency rises to a well-defined small-amplitude frequency. This frequency depends on the step angle and on the ratio of the holding torque to the moment of inertia of the rotor. Either a higher torque or a lower moment will increase the frequency!

 

Formally, the small-amplitude resonance can be computed as follows: First, recall Newton's law for angular acceleration:

 

T = µ A

Where:

T -- torque applied to rotor

µ -- moment of inertia of rotor and load

A -- angular acceleration, in radians per second per second

We assume that, for small amplitudes, the torque on the rotor can be approximated as a linear function of the displacement from the equilibrium position. Therefore, Hooke's law applies:

T = -k 

where:

k -- the "spring constant" of the system, in torque units per radian

 -- angular position of rotor, in radians

We can equate the two formulas for the torque to get:

µ A = -k 

Note that acceleration is the second derivitive of position with respect to time:

A = d2/dt2

so we can rewrite this the above in differential equation form:

d2/dt2 = -(k/µ) 

To solve this, recall that, for:

f( t ) = a sin bt

The derivitives are:

df( t )/dt = ab cos bt

d2f( t )/dt2 = -ab2 sin bt = -b2 f(t)

Note that, throughout this discussion, we assumed that the rotor is resonating. Therefore, it has an equation of motion something like:

 = a sin (2 f t)

a = angular amplitude of resonance

f = resonant frequency

This is an admissable solution to the above differential equation if we agree that:

b = 2 f

b2 = k/µ

Solving for the resonant frequency f as a function of k and µ, we get:

f = ( k/µ )0.5 / 2

 

It is crucial to note that it is the moment of inertia of the rotor plus any coupled load that matters. The moment of the rotor, in isolation, is irrelevant! Some motor data sheets include information on resonance, but if any load is coupled to the rotor, the resonant frequency will change!

In practice, this oscillation can cause significant problems when the stepping rate is anywhere near a resonant frequency of the system; the result frequently appears as random and uncontrollable motion.

 

Resonance and the Ideal Motor

 

Up to this point, we have dealt only with the small-angle spring constant k for the system. This can be measured experimentally, but if the motor's torque versus position curve is sinusoidal, it is also a simple function of the motor's holding torque. Recall that:

 

T = -h sin( ((/2)/S)  )

The small angle spring constant k is the negative derivitive of T at the origin.

k = -dT / d = - (- h ((/2)/S) cos( 0 ) ) = (/2)(h / S)

Substituting this into the formula for frequency, we get:

f = ( (/2)(h / S) / µ )0.5 / 2 = ( h / ( 8 µ S ) )0.5

Given that the holding torque and resonant frequency of the system are easily measured, the easiest way to determine the moment of inertia of the moving parts in a system driven by a stepping motor is indirectly from the above relationship!

µ = h / ( 8 f2 S )

For practical purposes, it is usually not the torque or the moment of inertia that matters, but rather, the maximum sustainable acceleration that matters! Conveniently, this is a simple function of the resonant frequency! Starting with the Newton's law for angular acceleration:

A = T / µ

We can substitute the above formula for the moment of inertia as a function of resonant frequency, and then substitute the maximum sustainable running torque as a function of the holding torque to get:

A = ( h / ( 20.5 ) ) / ( h / ( 8 f2 S ) ) = 8 S f2 / (20.5)

Measuring acceleration in steps per second squared instead of in radians per second squared, this simplifies to:

Asteps = A / S = 8 f2 / (20.5)

Thus, for an ideal motor with a sinusoidal torque versus rotor position function, the maximum acceleration in steps per second squared is a trivial function of the resonant frequency of the motor and rigidly coupled load!

For a two-winding permanent-magnet or variable-reluctance motor, with an ideal sinusoidal torque-versus-position characteristic, the two-winding holding torque is a simple function of the single-winding holding torque:

 

h2 = 20.5 h1

Where:

h1 -- single-winding holding torque

h2 -- two-winding holding torque

Substituting this into the formula for resonant frequency, we can find the ratios of the resonant frequencies in these two operating modes:

f1 = ( h1 / ... )0.5

f2 = ( h2 / ... )0.5 = ( 20.5 h1 / ... )0.5 = 20.25 ( h1 / ... )0.5 = 20.25 f1 = 1.189... f1

This relationship only holds if the torque provided by the motor does not vary appreciably as the stepping rate varies between these two frequencies.

In general, as will be discussed later, the available torque will tend to remain relatively constant up until some cutoff stepping rate, and then it will fall. Therefore, this relationship only holds if the resonant frequencies are below this cutoff stepping rate. At stepping rates above the cutoff rate, the two frequencies will be closer to each other!

 

 

Living with Resonance

 

If a rigidly mounted stepping motor is rigidly coupled to a frictionless load and then stepped at a frequency near the resonant frequency, energy will be pumped into the resonant system, and the result of this is that the motor will literally lose control. There are three basic ways to deal with this problem:

 

Controlling resonance in the mechanism

 

Use of elastomeric motor mounts or elastomeric couplings between motor and load can drain energy out of the resonant system, preventing energy from accumulating to the extent that it allows the motor rotor to escape from control. Or, viscous damping can be used. Here, the damping will not only draw energy out of the resonant modes of the system, but it will also subtract from the total torque available at higher speeds. Magnetic eddy current damping is equivalent to viscous damping for our purposes.

 

Figure 2.8 illustrates the use of elastomeric couplings and viscous damping in two typical stepping motor applications, one using a lead screw to drive a load, and the other using a tendon drive:

 

Figure 2.8

In Figure 2.8, elastomeric moter mounts are shown at a and elastomeric couplings between the motor and load are shown at b and c. The end bearing for the lead screw or tendon, at d, offers an opportunity for viscous damping, as do the ways on which the load slides, at e. Even the friction found in sealed ball bearings or Teflon on steel ways can provide enough damping to prevent resonance problems.

 

Controlling resonance in the low-level drive circuitry

 

A resonating motor rotor will induce an alternating current voltage in the motor windings. If some motor winding is not currently being driven, shorting this winding will impose a drag on the motor rotor that is exactly equivalent to using a magnetic eddy current damper.

 

If some motor winding is currently being driven, the AC voltage induced by the resonance will tend to modulate the current through the winding. Clamping the motor current with an external inductor will counteract the resonance. Schemes based on this idea are incorporated into some of the drive circuits illustrated in later sections of this tutorial.

 

Controlling resonance in the high-level control system

 

The high level control system can avoid driving the motor at known resonant frequencies, accelerating and decelerating through these frequencies and never attempting sustained rotation at these speeds.

 

Recall that the resonant frequency of a motor in half-stepped mode will vary by up to 20% from one half-step to the next. As a result, half-stepping pumps energy into the resonant system less efficiently than full stepping. Furthermore, when operating near these resonant frequencies, the motor control system may preferentially use only the two-winding half steps when operating near the single-winding resonant frequency, and only the single-winding half steps when operating near the two-winding resonant frequency. Figure 2.9 illustrates this:

 

Figure 2.9

 

The darkened curve in Figure 2.9 shows the operating torque achieved by a simple control scheme that delivers useful torque over a wide range of speeds despite the fact that the available torque drops to zero at each resonance in the system. This solution is particularly effective if the resonant frequencies are sharply defined and well separated. This will be the case in minimally damped systems operating well below the cutoff speed defined in the next section.

 

Torque versus Speed

 

An important consideration in designing high-speed stepping motor controllers is the effect of the inductance of the motor windings. As with the torque versus angular position information, this is frequently poorly documented in motor data sheets, and indeed, for variable reluctance stepping motors, it is not a constant! The inductance of the motor winding determines the rise and fall time of the current through the windings. While we might hope for a square-wave plot of current versus time, the inductance forces an exponential, as illustrated in Figure 2.10:

 

Figure 2.10

               

The details of the current-versus-time function through each winding depend as much on the drive circuitry as they do on the motor itself! It is quite common for the time constants of these exponentials to differ. The rise time is determined by the drive voltage and drive circuitry, while the fall time depends on the circuitry used to dissipate the stored energy in the motor winding.

At low stepping rates, the rise and fall times of the current through the motor windings has little effect on the motor's performance, but at higher speeds, the effect of the inductance of the motor windings is to reduce the available torque, as shown in Figure 2.11:

 

Figure 2.11

The motor's maximum speed is defined as the speed at which the available torque falls to zero. Measuring maximum speed can be difficult when there are resonance problems, because these cause the torque to drop to zero prematurely. The cutoff speed is the speed above which the torque begins to fall. When the motor is operating below its cutoff speed, the rise and fall times of the current through the motor windings occupy an insignificant fraction of each step, while at the cutoff speed, the step duration is comparable to the sum of the rise and fall times. Note that a sharp cutoff is rare, and therefore, statements of a motor's cutoff speed are, of necessity, approximate.

The details of the torque versus speed relationship depend on the details of the rise and fall times in the motor windings, and these depend on the motor control system as well as the motor. Therefore, the cutoff speed and maximum speed for any particular motor depend, in part, on the control system! The torque versus speed curves published in motor data sheets occasionally come with documentation of the motor controller used to obtain that curve, but this is far from universal practice!

 

Similarly, the resonant speed depends on the moment of inertia of the entire rotating system, not just the motor rotor, and the extent to which the torque drops at resonance depends on the presence of mechanical damping and on the nature of the control system. Some published torque versus speed curves show very clear resonances without documenting the moment of inertia of the hardware that may have been attached to the motor shaft in order to make torque measurements.

 

The torque versus speed curve shown in Figure 2.11 is typical of the simplest of control systems. More complex control systems sometimes introduce electronic resonances that act to increase the available torque above the motor's low-speed torque. A common result of this is a peak in the available torque near the cutoff speed.

 

About the Author

Assistant professor in lord venkateswara engineering college.I am doing phd in sathyabama university, Tamil Nadu,India.

Help with this D'Alembert's Principle Question!?

1) Compare and contrast the use of D'Alembert's Principle' with the Principle of Convservation of energy when solving: -

The winding drum of a hoist has a mass of 160 Kg, an outer diameter of 600 mm and a radius of gyration 200 mm. A light cable wound around the drum carries a load of 110 kg. This is accelerated from an initial velocity of 1.5 m/s to 3 m/s, whilst travelling upwards through a distance of 5m. The frictional resistance to linear motion of the load is 150N and the friction torque in the bearings of the drum is 5Nm. Determine the work done, the input torque applied and the maximum input power delivered by the driving motor.

Given ( l = mK(2) ) g =( 9.81 m/s(2) )

the (2) means squared like mK (squared)

2) Evaluate the methods that might be used to determine the density of a solid material and the density of a liquid.

If someone answers this correctly THANKYOU SOOO MUCH!!!!

D'Alembert's Principle is a principle which there is very little point to using. You don't even need to know D'Alembert's name. It is only algebra from Newton's 2nd law. D'Alembert may not like that we are belittling him, but he doesn't matter.

I feel it is a very poor choice of method, because it makes you think that "inertia is a force", when it is really simply a property of matter. It is very confusing for them to make you use D'Alembert's principle.

All it does is tell you that you can treat inertial effects as forces or torques, acting in the opposite direction of acceleration or angular acceleration. Why not just use Newton's 2nd law directly instead?

I suppose it is set up for engineers used to statics class, where all forces add up to zero on bodies which do not accelerate. And then if bodies do accelerate, just use D'Alembert's principle...WHAT! Why not just learn properly Newton's 2nd law?

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