mr.bill project

Scientific Organization – RUMAG -.

Analysis - Calculus Mr. William Frances Neil project .

General Mathematics

by Dr. Zbigniew Malczewski .

Static Analysis

by Jan Rutkowski .

To analyze the whole concept of Mr. Bill project we use a few different technics .

Number one .

We use building statics to establish if this arrangement is able to move or collapse .

Or maybe is solid and is not moving . Instead of the whole wheels we use only wheels diameter which are nothing else but simple building beams . The top beam (a2) (a10 ) is support in the centre and that beam is able to move in both directions . In centre that beam has small wheel that is permanently fixed to that beam .

The second (wheel) diameter beem is (x) (x1). That beem is support in point (x) .

Point (x1) is able to move freely down .

The second method is matchematics to establish if point (x) is able to move up .

Dr Malczewski at this stage situated point (x) as a solid fixed point . In his opinion this is mechanism that it will fall down until chains and beems allowed to and then it will stop .

The role of second small wheel is to be connected by belt with top small wheel .

The most important point in this project is point (x ) .

If beam (x) (x1) is able to free fall and at this same time point (x) is attached to the belt than that end of the beam will rise up and then such beam become rotational .

This is very interesting case . The centre of gravity in that beam will establish it self with a speed of rotation .

Second drawings by Dr Malczewski is reviling that mathematics is giving a surprise answer .

At this moment a few words about Gravity Technology . RUMAG is an Scientifc Organisation looking in to scientific methods for any arrangements that is giving motion . As you can see project have been analyze by the best of the best . Polish Mathematicians . History is telling us that Poland and the super brains matchematicians brake the secret code of “enigma “ . Every message that go through Hitler machine was program differently but knowing the system they conquer every one of them .

Gravity Technology have such people .They giving their time and talents just to proof what is working and what is not .

Gravity Technology possess answers to “”free energy”” to global warming , to fossil fuels crisis . Thanks to Gravity we have energy .

To find out how - simply look in to projects . If you have one - simply contact us . If we can help we will .

Most scientists are ignorants to that elementary energy and will dismiss possibility of gravity as a field full of energy .

To educate everyone we produce short films of our experiments . Simplicity is paramount and the most important fact , all of them work . You can try yourself on a kitchen table .

Look @ google – video-Technologia Grawitacji ( the last name is in Polish) but picture speaks every language . Enjoy and use it .

This is basic drawing of Mr . Bill project .

Black thick lines are diameters of the wheels to simplify calculation .

g1

g2 G

Analyzis for Mr. William Frances Neil - project .

By Dr. Zbigniew Malczewski .

Page 1 .

a2 a10

dx a

dx a12

P x x1 dx P/2

x12 P/2

Układ belek jest mechanizmem. Obciążenie (masa) belki dolnej x, x1 rozkłada się po połowie (Arrangements of this beams work as a mechanism ) . ( Mass of the beam x,x1 is divided in half ) z jednej strony na punkt podparcia (x) z drugiej na łańcuch (x1). (on one side on point x and on the another side x1 ).

Obciążenie to powoduje przemieszczenie łańcucha o dx jak na rysunku. ( This loading will produce motion to the chain -dx- as on drawing ) .

Prawa część układu opada a lewa unosi się. Punkt podparcia (x) nie zmienia się.

(The right side is moving down and the left up . Point x at this stage is stationery ).

Nie rozumiem, do czego służą koła złączone pasem. Moim zdaniem po opadnięciu belki dolnej układ przestanie działać.

(Don't know way we need two small wheels connect by belt . In my opinion this mechanism will work until beam x-x1 is free to fall and then stop ).

Using static knowledge to determine motion we use this drawing

+

a

_

b

b1

a1

+ c

x

x1

c1

a-a1 jest to belka luzno zawieszona w srodku jej ciezkosci . ( a – a1 Beam suport in the centre of it's gravity ) x-x1 jest to belka zamocowana tylko w punkcie x albo c1 . Punkt x1 jest luzno opadajacy . (x- x1 Beam attached or hang in point x . Point x1 is freefall) Belka ta moze byc dowolnego ciezaru oraz dlugosci . ( this beam can have any length or weight )

.

c-c1 krotka belka statycznie zamocowana do stalej konstrukcji w srodku jej ciezkosci i sprzezona z belka b-b1 ktora jest wyznaczona jako czesc belki a-a1 .

W obecnym rozwarzaniu b-b1 i c-c1 sa to male kola o tej samej srednicy sprzezone paskiem klinowym . ( c-c1 and b-b1 small wheels with this same diameter connect by belt ) ( b-b1 permanently fixed to beam a-a1 ) Konce belek a-a1 i x-x1 sa polaczone zwisajacym lancuchem takiej samej dlugosci i wagi .

(The end of beams a-a1 and x-x1 are connected by chain with this same length and weight ) Kolor niebieski symbolizuje opadanie (sily dzialajace do dolu oznaczone znakiem) “_ “ . ( colour blue symbolised “fall” and mark as “ _ “ ) Kolor czerwony – (rozowy) symbolizuje podnoszenie ( sily dzialajace do gory ) “+“. ( colour -red- symbolised force “ up” and is mark as “ +” ) Analizujac taka konstrukcje od strony statyki budowlanej punkt a1 i punkt x1 sa punktami opadajacymi i takie pozostana w tej aranzacji .

Punkt “a” bedac punktem belki sztywnej (stalej) polaczony z punktem “x” lub “c1” przyjmie kierunk odwrotny i bedzie ulegal podnoszeniu , znak “+”.

Przykladem sytuacyjnym takiego zachowania ruchu jest zabawka “YOYO”.

Rozwazmy matematycznie , Czy punkt “c1” i “x” beda podlegac unoszeniu czyli “+” ? . ( To analyze this construction in the knowledge of building statics - point -a1 and point - x1 are always on the “ fall” and they stay that way in this arrangement .

Point “a”is sytuated on a solid beam “a-a1” and connected by chain with point “x” will move in opposite direction which is “up” mark as “+” ).

(As an example of such motion and the closely I can think of is a “YoYo” .

Let's consider by matchematic's , possibility of raise up point “x” mark as “+” ).?

Projekt Pana Billa .

Opinia Pana Dr. Zbigniewa Malczewskiego .

Schemat obliczeniowy – etap pierwszy ( CALCULUS – STAGE ONE )

Vx

Vx*sinα

P

α

x1 o

x l/2

Vx1 l/2

Pierwsze równanie równowagi ( The first equation for equilibrium) Σ Mx1 = 0 Suma momentów względem punktu x1 jest równa zero ( total momentum in relation to point x1 is equal zero ) Vx*sinα*l – P*l/2 = 0 Vx = P/(2*sinα) Drugie równanie równowagi ( The second equilibrium equation ) Σ Px = 0 Suma rzutów sił na oś pionową jest równa zero ( Total force projection on vertical axle is equal zero ) Vx*sinα + Vx1 – P = 0 P/(2*sinα)*sinα + Vx1 – P = 0 Vx1 – P/2 = 0 Vx1 = P/2 Na przykład dla α = 30, sinα = 1/2 ( for example angle Alfa = 30 , sin ά= ½ ) szukane siły: ( search for force ) Vx = P Vx1 = P/2

Siła Vx przekazuje się na koło dolne w punkcie x, co daje moment siły: ( force Vx is transfer on bottom wheel at point x and that give it's momentum force ) M = (d/2) * Vx * cos(β-90) gdzie: ( where: ) d- średnica koła dolnego ( d- diameter of the bottom wheel ) β- kąt uwzględniający brak prostopadłości siły Vx do ramienia (średnicy koła) ( β – angle to determine perpendicular force Vx to the arm ( diameter of the wheel ) ) Schemat obliczeniowy – etap drugi ( CALCULUS - STAGE TWO )

d/2

a

a10 Vx l/2 Vx

P

x x1

Vx1 Suma momentów względem środka koła górnego daje równanie: ( Total momentum in relation to the centre of the top wheel is shown in equation : ) Σ M = (d/2) * Vx * cos(β-90) – Vx1 * l/2 Σ M = (d/2) * P/(2*sinα) * cos(β-90) – (P/2) * l/2

dla α = 30 otrzymamy: ( For ά = 30 we recive ) Σ M = (d/2) * P * cos(β-90) – P * l/4 Mogą wystąpić trzy przypadki ( We are facing three natural cases ) 1 . Σ M = 0, układ jest w równowadze Case 1 ( The whole mechanism is in balance ) 2.

3.

Σ M > 0, punkt x opada a x1 unosi się Case 2 ( point x go down and x1 rise up ) Σ M < 0, punkt x unosi się a x1 opada Case 3 ( point x will rise up and x1 will fall )

Co się stanie? ( So what will happen to this mechanism ? ) Zależy to od doboru geometrii układu. ( That depend on well-matched geometry of the whole mechanism and it's arrangement) Uwaga! Założyłem, że łańcuch pionowy przechodzący przez punkt x jest przelotowy. ( Attention ! assumption was made that chain go right through by point x ).

Dr. Zbigniew Malczewski .

As you can see we have answers . Your project will balance ; or it will rotate . Everything must have specific proportion. . As you notice the most important point x is going to rise only under very specific gentle angle . That angle is the answer to you project . If we abuse that angle , mechanism will stop working under force of gravity .

To give you some idea what we talking about – look in short films Technologia Grawitacji .google - video . http://video.google.com/ThumbnailServer2?app=vss&contentid=9a8caf1e02e55219&offsetm s=25000 http://video.google.com.au/videoplay?docid=-7811657463322217816&pr=googsl It was a pleasure to work on yours project .

Jan Rutkowski .

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