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Appendix 7

THE FORCES BEHIND

ROTATIONAL ENERGY

NATURE'S MOST POWERFUL ENERGY - FORM

This appendix will show many that even the most hideous mathematical equations and expressions can have simple explanations and be solved with a minimum of fuss.

Nature's apparent complexity comes from many interacting processes and mechanisms, so many the complexity of the mathematics needed to explain simple things becomes overly complex and seemingly convoluted.

There are so many interactions that it is necessary to consider just one atom. But when it comes to rotational energy, the important word is sequence. Nature's being refers to sequence, from the most active, but when the most active is satisfied, the process involves the current most active. Eventually, the initial absolute least active reaction becomes the most active and in the correct environment, will occur.

From First principles, the aim is to calculate the rotational parameters of a disk and a sphere. In so doing. the methods described do not require mathematical tables, or angular conversions, because everything works on revolutions per second. Believe it or not, this is a pure geometric exercise, not a mathematical treatise. The simple equations expressed in chapter 4 can be extrapolated to give some general approximations. To recapitulate, the following symbols are used :

density ( Note: Rho),
pi ,
radius r,
thickness t,
mass m,
velocity v,
Momentum M_m,
and Energy E.

As only those points of equal radius will have the same rotational attributes due to the distribution of matter across the rotating disk, then each region with common radius (a concentric ring, an annulus) can be used to determine the overall strength of a rotational effect. The following mathematical expressions appear awesome, although they are very simple .

THE ANNULUS (is a ring) with an outer radius r_o has an inner radius r_i,
Given the thickness 'T' and density of the annulus, all the parameters of motion can be computed.

As the Area of a disk is r²
then the Annulus has two areas to consider making the mathematics relatively simple, subtracting the area of a hypothetical inner disk, r_i from the outer radius disk, r_o
So the area of an annulus is A @ r_o - A @ r_i. Filling in the gaps, the area is given as
A = r_o² - r _i² or A = (r_o²- r_i ² )

So as the Volume equals ( r _o ² - r_i ² ) t ,

then the Mass in that volume is ( r_o ² - r_i ² ) t .

From the circumference of a disk, and the revolutions per second, the tip speed is given as,

( 2 r ) rps .

To determine the momentum of the annulus may be seen as a bit tricky, for the mean velocity needed in the calculations, is somewhere between the tip speed of the outer surface and the inner surface. Although the angular rotation is a constant, the speeds are considerably different. No matter how thin the annulus, the inside surface speed must travel slowly at

2 r_i (rps)

compared to the more rapidly moving outside surface, where the outside surface tip speed is

2 r_o rps

The outer tip speed can be determined from the difference D between the surface speeds of the inner r _i and outer r_o as

2 r_o rps = 2 ( r_i + ( r_o - r_i )) rps

or

2 r_o rps = 2 ( r_i + D) rps

To toy with this for a moment, the radius where the mean velocity exists is somewhere nearer the outer edge, perhaps around the point of equal areas, where the masses are equal, at a radius close to r_i + 0.707 ( D ) .

This approximation is an approximation. For the time being, the mean velocity to determine the momentum is

v = 2 ( r_i + 0.707 ( r_o - r_i )) . rps

However, since Momentum is given as M_m= m v then,

M_m = t ( r_o ² - r_i² ) . 2 ( r_i + 0.707 ( r_o - r_i )) . rps

Having determined the momentum, the energy of the annulus at this same 0.707 radius ring can be determined.

E » 2 t ³ ( r_o ² - r_i ² ) . ( r_i + 0.707 ( r_o - r_i )) . ( r_i + 0.707 ( r_o - r_i )) . rps ²

But, equal masses or proportionally equal momentums do not mean the application or distribution of equal forces or equal energies. The effect on the total system is dependent on both the tip speed and the mass distribution. Energy is a function of velocity squared! Effectively, an increase in tip speed is produced with any increase in the length of the circumference, so the outer annulus is capable of storing and releasing enormous amounts of energy compared to the inner volume. Similarly, the mass is a function of volume (a cubic value) and the density. As an example, a disk whose radius is 10 units, can be divided into equal masses at a radius of 7.0707 units. The outer annulus, only 2.93 units wide contains half the mass. However, as energy is a function of velocity squared, the closer the mass is to the circumference, the greater the energy. This disk would have its mean energy distribution occurring around 8.5 units!

Even though the equations above (in bold type) are multiplying two equations together, the final equations are seen to be growing in complexity. Mathematics is a language, and as such, to describe anything in detail requires some complexity. Contemplate for a moment, the sentence, "The sky is blue." It is quite possible to write a 10,000 word dissertation, a novel or a treatise concerning the blue sky. Depending on the writer's ability, the 10,000 words essay may appeal to the reader, or be rejected as bald and boring. Many people see a huge mathematical expression and immediately go into "shut-down" mode, where upon, they reject the rest of the mathematical essay as too hard. At fault is not the reader, rather it is the reader's educators that have introduced the problem. It is extremely difficult to find a mathematics teacher capable of teaching mathematics, because many have been corrupted by boring teachers.

As mathematics is a language, it is not impossible to write works of fiction, where the mathematics can tell a fascinating story without any foundation or basis in Nature. One must carefully examine what some educators claim to be unequivocal and perfect mathematics used to describe unobserved-untested events. One must look passed the logic and the mathematics to see a solid foundation, for just as the work, Alice in Wonderland, is a literary masterpiece, with superb logical development, it is a work of fiction, as is Oppenheimer's mathematics used to describe the blackhole. Today, that mathematical treatise could be parallelled to "The never-ending-story", since everyone else has added their bit to the story, increasing the fractle convolutions that have brought such a fantasy into what some believe to be science fact, an actual object that must exist, when there is no proof or basis in Science for such an object. Santa Clause has more credibility.

An exciting literary work can be turned into a grinding chore, if one does not know where to put the correct intonation, expression or feeling, as each word carries a certain meaning. Equally, mathematics requires the same intonation, expression and feeling, else it becomes either terrifying or boring. Educators have in the past, caused so many problems, because the education system throws people head-long into their role without adequate training. Teachers need to be able to communicate. In order to count, one must know how to count. To count sequentially, from 1 to 12,187 is a task that few would take-on, because many cannot count to 78 without losing count. There are simple techniques which can make counting enjoyable and extremely accurate. The fastest method is the grouping technique, where bundles of fives are checked and grouped into bundle of twenty, then into bundles of 25, then grouped into bundles of 100, then 500, and what is left over, is really, all that is counted, a pile of 30, a pile of 5, with two loose pieces, equalling a total of 12,187. A simple technique, where the highest number physically counted is just 5. This technique can be used to count the number of coffee beans in a bag, or grains of salt in a canister, while holding a conversation about history, times and ages.

Teachers must know how to communicate their thoughts, to cater for the needs of the student, to provide security so that confidence is encouraged, not discouraged, and that one's enthusiasm is productively brought to fruition. Some teachers of mathematics show-off their mathematical ability by solving two absolutely different equations, explaining both at the same time while using both hands to write the methods on the blackboard, killing the student's understanding by saturating the students coping mechanism. Script writing, public speaking, psychology and the use of the audio-visual aids, should be necessary pre-requisites before any potential teacher conducts a class. The teacher must be made to realise that people's mind's are at risk, and the damage that idle snide comments and discrimination make. Often, it will be seen that un-necessary complexity is introduced through short-cut approximations and short-hand notations, where the explanations of the procedures and symbols are vague or written in jargon.

To deprive any student from knowledge, for what-ever reason is a crime against that person. To kill someone's enthusiasm through a boring presentation, through economic rationalism, perhaps by reading verbatum from a text without consideration, (something that the student could read on their own,) deprives the student. Unfortunately, many mathematics teachers have been thrown-in at the deep-end without sufficient training, so it is that many readers will flick through the next few pages only to put the book down, stating with the emphatic, "I cannot understand the mathematics.", even when the mathematics are childish or simple. The fault is not with the author or the person, rather it should be laid at the feet of that person's educators for promoting such attitudes.

Throughout Science, upper and lower case letters of the Greek alphabet are used because there are not enough letters in the English alphabet to cater for all mathematical requirements. Certain letters have been reserved, such as pi, , having a recognised value. As on page one of this appendix, a superscript note ( i.e. Note: pie) is used for the pronunciation.

Over the course of the next few pages, the complexity of the mathematics is going to increase, but when one studies the process, it is all very simple. It is no more difficult than counting to 12,187. All it takes is a simple procedure to read the mathematics, putting the intonation, expression and feeling back into the written word. To approximate the forces more accurately for a disk or sphere, all the individual momentums and energies from the centre to the circumference need to be summed. As each annulus involves one calculation, many calculations are involved. As this is completed, the mass of the disk or sphere can be determined through the same process of addition. If compared with the geometrically calculated volume, the results can present a statement of accuracy, highlighting any errors introduced by cumulative calculations.

At a moment in time, the momentum around the axis ( at a specific radius) could be considered as linear momentum since the effective force is maximized at a contact point on the circumference producing a force in the same direction as the rotation (on the tangent). This is noticed with a working power grinder where the sparks fly off at a tangent to and with the rotational direction, where the energy exerted is seen with the quantity of sparks. The subscript "point" in the following calculations refers to the contact point where energy is released, as if all the energy in the disk is applied at that point. This simply says that if all the mass in the disk is exerted at this point at a radius "r" from the axis it would be travelling at a certain velocity and would have a certain amount of energy behind it.

Unless otherwise stated, multiplication is assumed, so that something like " r_o² t " is actually entered on a calculator as " 3.14159 x the values for r_o x r_o x t ". Removing the "x" or multiplier notation "times" means that the letter "x" can be used as a variable "X" The following equations look ominous only because the Mass is calculated from the density "r" where mass is volume time density. .

mass x velocity

M_{m (point)} = mv » ( r ². t ) . ( ( 2 r ) . rps )

or approximately 2 ² r ³ t . rps

The symbol " » " above is taken as "maybe-equal-to" as there are other considerations to examine. Again, the next equation appears hideously complex, when it is no more than ½ M_m.v or ½ mv² . When all the brackets and the constants are removed, the equation appears far simpler, but still rather messy.

E_(point) = ½ mv ² » ½ ( 2 ² r ³ t rps ) (( 2 r) rps )

» 2 ³ r ⁴ t rps ²

The use of the subscript (point) in the above equations will be used soon to show something that needs greater explanation. Normally, the mass of the disk is found by multiplying the area, given by A = r ² with the thickness "t" and the material's density "r" .

In the following, the Greek symbol "" (sigma) states that the mathematical expression after the symbol is actually part of a much larger expression where the result for each value is added many times to give a grand total figure. The number six could be represented as ₃ (1+1) when the expression 1+1=2 is added to itself three times. To give the subscript "3", states that the expression (1+1) is treated three times, so 2+2+2 = 6. If written as ₂₅ (x) , the sigma notation means that the value of x is summed "twenty five" times.

"On the night of my birthday, the my close friends gathered in total darkness to surprise me. When the lights came on I was shocked to see the of them, friends and foes all together. The evening was catered for by Sigma feast foods. The of the Sigma feast bill was paid for by my best friend who then split the bill's evenly amongst all the others. It was a great night, but then I paid the price on the next day for the alcohol consumed."

Many subscripts are used in sigma notation, such as _n where the "n" is read as "the sum any number of times". Obviously, an expression like _co when infinitely summed will be infinitely large. With respect to _n (figure 4-7)

the purpose is to determine the volume of the sphere which is normally given as

V= 4/3 r ³

Here the method will be to (all the annulli volumes) or ( x ³ ) for how many calculations it takes. The value of "t" is going to be as small as possible, meaning that each calculation run will be in the terms of several thousand for even a small sphere. Each expression is applicable within a defined range from the centre of the sphere to and including the surface. A sphere should have a mass of 4/3 r ³ which should equal (of all the annulus masses). Of course, there is going to be a difference between the two mass calculations. This is caused by quantization, where the addition of each annulus can lead to an approximation that can add a little more or take away small amounts from the cumulative total, hence, as mentioned earlier, the difference between the two will give a statement showing the accuracy of the procedure. That is to say,

Error = (4/3 r ³ ) - _n ( f (x) )

where ( f (x) ) is the mathematical function used to determine the volume.

This is a statement of the quantization error, an error caused by the n (process). For the disk, each annulus (and there will be "n" of them) will have a volume and mass given by the following equations. The range is from the internal radius r_i (or zero) to the outside radius r_o . To simplify the mathematics, each annulus must have a constant thickness, as well as a constant change in radius.

r_i ---> r_o Vol.r_o - Vol.r_i

As Volume is V = _n ( t ( r_o ² - r_i ² ))

Then mass is m = _n ( t ( r_o ² - r_i ² ) )

(volume x density)

The momentum of each annulus is mass times velocity or mass times rps times circumference which in the description of a disk was previously shown as an implied multiplication

Momentum _(point) M_m = m v = ( r ² t ) ((2 r) rps) .

However there is a problem as each annulus has an inner and an outer diameter. This merely makes the mathematical equations appear more daunting only by the nature of the equation's length when the changes made are relatively minor. Some things are easier to say in English. The modification takes into consideration the fact that the point where the mean momentum is realized should be inside each ring and just below the outer surface. This could arbitrarily treated as 0.707 of the ring's width so a correction factor is needed. In this case, the momentum is determined for each annulus as

M_m = _n ( ( r_o ² - r_i ² ) t ) ( 2 ( r_i + 0.707 t ) rps)

So then M_m = _n ( (r_o² - r_i ² ) t ( 2 ( r_i + 0.707 t ) rps )

Since Linear Kinetic Energy "E" equals ½ m v ² then the equation of energy in the disk is equal to the sum of energy in every annulus as given by simply multiplying the momentum by the velocity again and dividing by two. Again the length of the equation is going to appear daunting, but there is really nothing new added to the expression, rather it is refined. The equation for energy is adequately solved by:-

E = _n ( ½ ( r_o² - r_i ² ) t ) ( 2 ( r_i + 0.707 t ) rps ) ( 2 ( r_i + 0.707 t ) rps )

then E = _n ( ½ ( ² ( r_o ² - r_i ² ) t ( 2 ( r_i + 0.707 t ) rps ) ²))

So then E = _n ( 2 ³ ( r_o² - r_i ² ) t rps ² ( r_i + 0.707 t ) ² )

To reduce the effects of the constants each equation can be treated with its variables handled as a summarised constant In the above equation all the working functions which are stabilized once the calculations start, that is the value of

2 ³ t rps ² can be considered as the constant "K". So then

E = _n (( r_o ² - r_i ² ) ( r_i + 0.707 t) ² K )

Note Well:- The Constant 0.707 will now be challenged, since it is incorrect.

The problem with the expression _n is the number of calculations required to reach the bottom line answer. As the calculations are so involved, the best solution to much mundane work is to take an easy approach, using a programmable calculator or a home computer. But then, the computer of choice must be one that can multiply and add up simple numbers without gross errors being introduced deliberately by inferior software manufacturer, to otherwise force the need to purchase another expensive software package. By far the best computers that perform the calculations accurately are the 48K Sinclair Spectrum, the JVC MSX , the 8 bit Atari and 16 bit Amega computers. These computers may be slow, but at least they give reliable results, compared to MSDOS based systems.

The orbital rotational momentum must also be calculated as a segment of a ring from the gravitational centre of the Sun. To do this manually is far too much work, however a computer can slave over the calculations for several days. The simplest method is to use those expressions with the subscript "point", previously ignored.

M_{m (point)}= mv = ( r ² t ) . ( 2 r) . rps

and E_(point) = ½ mv ² = ½ . ( r ² t ) . (( 2 r) . rps ) ²

= 2 ³ r ⁴ t rps ²

As the disk rotates, the balance point for both momentum and energy exist elsewhere. This fact allowed steam engineers to design an engine for an application, be it efficiency, speed or power merely by altering the eccentric rod's coupling point to the driven wheels.

Balancing annulli for a rotating disk

Equal Volumes and Mass 0.707 r

Equal Momentum 0.7935 r

Equal Energy 0.8405 r

Very different is the sphere. The sphere stores a great deal more rotational energy which it can release in dramatic ways since it is more difficult to effectively transfer energy. There are devices which use and tap into this energy by surface pressure couplings, where pressure rollers run on the surface on a specific latitude band. Although each latitude slice has three different balancing rings, the locus of each balancing annulus gives the spheroid some amazing properties. The program ROTNERGY.BAS cannot determine the points of balance for a sphere, requiring a completely different approach. The alternate method considers the sphere as built up from concentric tubes of differing heights. The telescope maker's sagitta equation is used to estimate the height "H" of the cylinder, where the thickness is the wall thickness (O_R-I_R). This procedure dramatically reduces the number of computations to form a complete calculation. With fewer instructions, far higher precision can be obtained in the same time frame.

ROTNERGY.BAS uses two nested loops which means that one loop determines the radius from the centre along the axis as the second determines the sum of each annulus from the axis to the surface. As an example, the solid cylinder or piston with a height of 20 units and a radius of 10 units would require 100 calculations when a sample thickness of 1 unit is used, from its equator or midsection. Using the concentric tube method and the same number of calculations, the sample thickness ten times finer at 0.1 units. Both methods give identical bottom line results when the same resolution is used, however the annulus method takes a great deal longer to complete. Each has specific benefits which means that they must be used independently.

seemingly endless lines of code for a second time, it is faster to modify ROTNERGY.BAS into the second program. "ERGY.BAS" is therefore only partially listed in Appendix 8. It is virtually the same program with the main loop altered. Since there are obviously fewer BASIC instructions in ERGY.BAS, a double scan has been introduced to discover the particular annulus balance zones, revealing the following.

The biggest problem with ERGY.BAS is again, resolution where overlie thick samples produce zones that occur at unrealistic radial positions. For instance, with a resolution of 20 tubes, the sphere has balance points of

To improve the resolution, adjust the value of "TH" in the first 10 lines of code. There is a "How the program works" in Appendix 8 which should be most helpful to the experimenters and sceptics.

The question being asked is since the moment of inertia is given by I = ½ M R ² determine the kinetic energy of the pulley and its radius of gyration. It is now time to introduce the methods used in conventional Physics, since the short-cuts used in Physics can be helpful and faster. But these short-cuts involve using angular measurements, where the angles are expressed in radians.

As the Kinetic Energy is given as = ½ I ²

Note: Epsilon, Omega

Since I = ½ M R ²

Then = ¼ M R ^{2 2}

= ¼ x 600 x 82 x 362

= 1.27 x 107 ergs = 1.27 Joules = 1.27 nt-m

The computer program derives the following answers.

Tip Speed 288.021cms 5.73 rps = 343.8 RPM

Volume 201.0618 cc

Mass 600.0001gm

Momentum 119,229.9 gm/cm/s

Kinetic Energy 1.32181 E+07 gm/cm/s^2

No. of Calculations = 8

A 4% error is found (not disclosed by the program) when comparing the two methods. The physicist derives a value of 1.27 nt-m as against 1.32 nt-m. As this is explorative research, placing any conclusion based on just 8 calculations is unscientific. This illustrates a quantization error in the outer annulus. Low resolution scans produce high quantization errors when the algorithm is not "idealized". As the measurements are defined in steps, the zones between the steps become less defined. To measure something somewhere between the level of one step and the next, results in an error. Hence the results can be poor especially when the wrong resolution is chosen. For instance, in cash transactions the quantized scale's lowest level of resolution is five cents, meaning that prices are either rounded up, or rounded down to the next price divisible by five cents. If twenty million transaction occur each day, the quantization error amounts to a gain or loss of a million dollars. Such quantization errors have meant financial windfalls for some gamblers and gaol sentences for some white collar criminals.

The gravitational balance point (0.707 r ) is where equal volumes and masses exist both inside and outside this ring. Astronomers often use this ring to support telescope mirrors, but fail to consider that more mass now exists outside the ring because the concave mirror making process has removed a greater proportion of the mirror from the centre. When the telescope is in use, it is extremely rare for the telescope to point vertically, so the exercise seems almost pointless. Just the thickness and gravitational pull will deform the mirror. Many people do not understand that this 0.707 r zone may not be the point of balance.

Using a similar method of latitude slices, the Earth can be divided into disks only a centimetre thick, without relating to any 9999 figure Cosine tables to obtain the desired accuracy. In line 230 of the program (Appendix 8) is the expression for the radius of a disk whose depth is known, which conforms to a fixed radius of curvature. Once this radius is determined, the period of rotation of the Earth (around the axis and around the Sun) can be used to show the changing speed of the point on the circumference. At this point one must take into consideration the Earth's inclination to the orbit and the interaction between the latitude slice. This can all be solved through simple geometry where the equations involve a square root.

are used in different ways. To make sense of the geometry is not mind staggering, it is simple.

The radius of gyration "k" is often expressed in terms of the moment of inertia as I = M k ²
which since I = ½ M R ² states that k ² = (R²)/₂, so in this situation k = R * 0.707.
As the text book question involved a disk of 8 cm, then the answer would be 5.6 cm.
The moment of inertia " I "is subject to the distribution of mass.
In the case of a sphere I = ²/₅ M R ² whereas the cylinder can have three possible versions depending on the nature of the cylinder. As a solid I = ½ M R ², however as a hollow cylinder

In many ways, the truck is much like an inertial guidance system where the engine, flywheel, gearbox, and transmission links are in line. At the differential, the crown wheel (perhaps heavier than the flywheel) transfers the rotational energy through ninety degrees to the wheels, preventing the dual in line gyroscope effect. The truck's types can have a diameter of a metre and a half, perhaps weighing from 75 to 100 Kg, where this weight is mainly in the rubber outer radius of the wheel. As the truck is travelling at 110 Km/hour then the mathematics of the prime mover can be determined. The forces involved are extraordinary. Needless to say, many fellow motorists do not realise how much room a truck needs to stop and then (if they survived) wonder why the truck squashed their car.

To enter the parameters of just one wheel as a solid disk into ROTNERGY.BAS using the GOSUB 690 instruction. Then simply multiply the results by the number of wheels on the truck. A bogie drive prime move has 10 wheels and the tri-axle trailer has another 12 wheels, a total of 22 wheels. Each (including the hub) has a mass of 150Kg. One does not need to be that accurate for this is just to give an appreciation of a real world situation. Remember that nasty function with rps squared multiplied by the radius to the third power due to the distribution of mass! If any of these tyres breaks free, a great deal of energy can be released on impact. There is a great deal of energy stored in the wheels of the moving truck; so much so that with improper design, the prime mover can be a death trap to anyone using the roads. When the same prime mover is moving slowly and runs over a wooden plank, very little happens. However, if travelling at 110 Km/h, the same truck running over the same plank, transfers so much rotational energy to the plank that it is lifted from the ground and thrown into the air to become every motorist's nightmare. It is thrown with such energy it actually travels faster than the tip speed of the wheel, and may be thrown 50 metres vertically into the air, travelling some 150 metres down the road before striking the ground. Similarly, when the steel belt peels off the tyre at that speed, it can be thrown over 75 metres vertically into the air. Simple logic says that such objects should be absolutely stationary to the ground, however the real world situation and mathematics prove a disaster waiting to happen, for what goes up, comes down.

Many of these sights look most spectacular when observed, but a tragic accident if it lands on a another vehicle. At that moment, the relative velocities produce relative energies, where the victim's velocity adds to the impaling effect. People have been decapitated by such accidents, so it is very wise to avoid following any speeding truck. Then again, how many windscreens have been smashed by tiny stones picked up by the tyres of the car in front? Have you ever tried to break a windscreen? It takes tremendous energy, however a 5 gram pebble thrown from the road surface, can take out the windscreen and embeds itself in a person's skull with the impact force of a bullet. Although traffic authorities warn motorists about safe driving distance between moving vehicles, often expressing distances in terms of a car lengths for each ten Km/h, the tail-gating motorist is not made aware of the dangers, for injury and death can be less than a split second away. Forget the 10Km/h concept, for it is far too short. It is much safer to simply triple the recommended distances for each 10 Km/h, especially if following a truck. Travelling at 60Km/h, means giving away about 9 car lengths. Such a distance will give the motorist time to savour the moment and take the necessary accident prevention measures.

There is nothing difficult about the mathematics illustrated throughout this appendix, for simple and logical mathematics works throughout the Universe, as addition and subtraction. Every multiplication is just another form of addition. To use Sigma notation again, as 4 x 4 = 16 , so does ₄ ( 4) = 16.

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