__A Simplified
Study of GIT__

Created on
08-16-98 - **JLN Labs*** ***-
**Last update 08-19-98

Let’s denote **A**
(rear part) the point where the track is widest (and where the
engine is), and **B** (front part) the opposite point where
the track is narrowest. Due to this variable width, the ball is
continuously exchanging angular momentum against linear momentum.
At point **B** its linear speed is maximum et its spinning RPM
is low, while at opposite point **A**, it is the other way
round : translation speed is low with an important spin, still
enhanced by the motor. It appears immediately that such a
movement of the ball is producing on the frame two effects
opposite to one another. Namely :

• **an effect due to
centrifugal force**. As the average speed of the ball during
the half turn **B** is much higher than during half turn **A**
it clearly generates a **forward** force

• **an effect due to
accelerations et decelerations of the ball (tangentiel effect)**.
When the shrinking track forces the ball to gain linear speed
thus loosing angular momentum, the ball is taking support from
the track, which generates a rearward force equal to the product
of ball mass by the acceleration it undergoes. Clearly, during
the path from **A** to **B**, such a force, always tangent
to the trajectory, is **rearward**. During the return path
from **B **to **A**, there is a braking of the ball
supported by the track, thus generating anew a force tangent to
the trajectory **rearward** as well**.** The resulting
effect is then the difference between these two effects. Using a
few simplifying assumptions, the calculation is easy. We then
asume :

• that the system is kept motionless

• that the trajectory of the ball is a circle (nearly the case)

• that the track is
made in such a way that the ball speed varies **linearly**
from **A** to **B** and conversely from **B** to **A**
(this is always possible) and that there is no slipping of the
ball on track.

• that there is no
friction, which allows to neglect the engine. If the frame is
kept motionless, clearly the return path **BA** will be
exactly symmetrical of path **AB, **thus generating the same
forces and allowing a calculation on the first half only.

Notations

**R** radius of
(circular) path of the center of the ball

**h** radius of the ball

**r** instantaneous
(variable) roliling radius of the ball

**a** angle defining ball position,
starting from origin **A** anticlockwise

**v** linear speed
(variable) of the ball.

**V** **maximum**
linear speed of the balle at point **B**

**r**** **ratio** **of** minimum **linear
speed of the ball at point **A **to the** maximum speed **at
point** B **

**Dynamics of a ball
rolling on a variable width track**

Kinetic energy of the ball
has two components : translation energy and spin the sum
remaining constant if friction is neglected. If **m** is the
mass of the ball,** I** its inertia and** w**
its spinning speed, we then have

**mv**^{2}** + I****w**^{2} = Constant

but if **r** is the
(variable) rolling radius at a given time, we have **w****
= v/r**, so

**(m + I/r**^{2}**)v**^{2}** = **Constant

On the other hand, as the
ball is hollow, its thickness can be neglected, so its inertia **I**
is equal to **8****p****/3.h**^{4}, **h** being the radius of the
ball. Expressed as a function of the ball mass **m =** **4****p****h****2** such inertia writes **2/3.mh**^{2} , so that, suppressing m, we can
write :

**(1+ 2/3.h**^{2}**/r**^{2}**)v**^{2} = Constant

In our case, with a track
width of 35 mm in **A**, the rolling radius** rA** of the
ball is 9,7 mm and reaches in **B **a value **rB** = 19,6
mm for a track width of 8 mm. It is then easy to get the ratio **v**^{2}**/V**^{2} which is :

**(1+ 2/3.h**^{2}**/rB**^{2}**)/(1+ 2/3.h**^{2}**/rA**^{2}**)**

the end result being **v**^{2}**/V**^{2}** **= 0,442, that is **v/V** = **r**
= 0,665

**Computation of
centrifugal force**

The linear speed **v**
varies from **V****r** (mini) to **V** (maxi) according
to the formula

**v = V****r****
+ ****a****/****p**.**(V - V****r****)
= V [****r**** + ****a****/****p****.(1 - ****r****)]**

which genetates a centrifugal force equal to (ball mass is taken as equal to unity)

**v**^{2}**/R = V**^{2}**/R.[****r****
+ ****a****/****p****.(1 - ****r****)]**^{2}

which must be multiplied by
**cos ****a **to get**
**the** **projection
on axis AB

**Computation of track
reaction**

Ball movement being
uniformly accelerated from **V1** in **A **to** V2** en **B**,
the tangential (constant) acceleration during this course of a
length of **p****R** equals **(V**^{2}-**V**^{2}**r**^{2}**)/2****p****R **and its component on axis **AB**
is

**V**^{2}**(1**-**r**^{2}**)/2****p****R.sin ****a**

**Summing up the two
forces**

it may be noted that the
radius of the ball does not appear in any of these two
calculations. The resulting thrust on a half turn as a function
of **a**, then writes :

**V**^{2}**/R.{[****r****
+ ****a****/****p****.(1 - ****r****)]**^{2}**
**.**cos** **a **+
**(1**-**r**^{2}**)/2****p****.sin ****a****}**

Giving to **r**
the value calculated above, this expression becomes :

**V**^{2}**/R.{[**0,665**
+ ****a****/****p****.**0,335**]**^{2}**
**.**cos** **a **+
0,558**/2****p****.sin ****a****}**

or, giving its value to **p**

**V**^{2}**/R.{[**0,665**
+ **0,106.**a****]**^{2}** **.**cos** **a **+
0,089**.sin ****a****}**

The curve below (continuous
line. Vertical scale has been inverted to get positive values for
forward thrusts) represents the thrust as a function of **a**..
The area of the right part being greater than the area of the
left part, this could be taken as a proof of a positive resulting
thrust, but, as the orbital velocity is not constant the
integration should not be made with respect of the angle **a**
but rather, as pointed out by Cyril Smith, with respect of time.
Before integration, he last expression must then be divided by **d****a****/dt**
= **(**0,665** + **0,106.**a****).
**It becomes

**- (**0,665**
+ **0,106.**a****)**** **.**cos** **a **+
0,089**.sin ****a / ****(**0,665** + **0,106.**a****)**

giving the curve in dotted line whose total resultant area is equal to zero, the right part being equal to the left part, at least to the accuracy of the Excel spreadsheet. The rearward thrust being applied during a longer period of time than the forward thrust, it can be understood that the end result amounts to zero.

**Conclusion**

This confirms the conclusions of Hans Holzherr and Charles North that can be found at

http://www.geocities.com/ResearchTriangle/1986/GIT/GITconclusion.html

Quoting them, I would say contrary to my first conclusion :

**If the GIT works it does
so because of some non-Newtonian principle which requires
demonstrating and formulating.**

Email
to Jean de Lagarde**: ****jlagarde@cyberaccess.fr**

Sujet
: **Excelent math for dumb weight slingers, now let's try to
analyze the TORQUE components**

Date : 18/08/1998 21:18:26

From: davidc@open.org (David E. Cowlishaw)

To: jlagarde@cyberaccess.fr

Jean de Lagarde;

Thanks for your interest and efforts on behalf of the GIT! I
reviewed

both your and Charles North (with Hans Holzherr)'s math
derivations, and

quite agree with you that the sum of the "tug of war"
created by the

stronger centrifugal balanced by the two opposing tangential
accelerations

will be zero.

That's as it should be, and several now have lent their own
proofs to

that integration over time.

However, as of today, few have considered the torque components

generated by the orbitals on the total sum of forces, perhaps
thinking that

the interaction and shift of linearly expressed momentum (as
centrifugal

force) into the NON LINEARLY considered spin accelerations of the
orbitals,

has no bearing on reality. Assumptions that are fatal to an
accurate

examination!

The torque generated on the orbital as the inverse acceleration

exchange is taking place, has a counter torque on the system of
course, and

here is where the "magic" translation of angular
acceleration to linear

takes place!

The tangential accelerations of the orbital NOW pair off a
portion of

their momentum exchange with the race as your analysis details,
and a

portion of that equally expressive radial momentum
"disappears" from the

center of system mass (which includes the linearly considered
mass of that

orbital, since spinning or non-spinning, it's linear mass doesn't
change),

around the axis of the orbital. Convieniently, it apparently
happens to be

pairing off with the tangential forces (also angular in two
opposing

halves) giving us momentum that is stored "in another
dimension".

Try to think of it this way, although the outside of the orbital
(and

it's center of mass) is decelerating and pushing back on the
race, the

wheel is accelerating the inside of the orbital (it's center of
turning

sees more wheel surface pass when it slows it's forward
velocity), "pushing

off" in the forward direction from ANGULAR mass, and on the
return cycle,

the unwinding spin also pushes forward on the center wheel, while
the net

zero sum of tangential and centrifugal is tilted in centrifugal's
faver IN

THIS DIMENSION (linear acceleration), because we can only have so
much

energy "combatting" or pairing off with the
centrifugal. The torque

coupling of the two bilatterally opposed torque expressions
(angular

orbital spin torque, pairing with orbital or tangential torque)
of

neccessity "steals" the linearly expressable momentum,
"swallowing" it as

spin momentum and releasing it in a (mostly ;) single direction
as the

cycle completes.

Find the vectors of the orbital spin torque acceleration's input,

realize that the momentum in the spin is subrtacted from
SOMEWHERE as a

moving mass interaction, and a more accurate math portrayal of MY
device,

rather than a dumb mass whirler will be achieved.

ANYway, thanks again for your efforts, the zero sum of
centrifugal and

tangential is quite needed to give us a baseline proof, then it
can be

shown that the actual centrifugal force imbalance derives from
the

"external mass" interaction with the (seperate or
outside) angular mass

property of the orbitals, since any momentum the orbital spin
axis contains

is removed (and returned in a reinforcing direction) from your
linearly

considered system mass, and the time and position in the cycle
that the

momentum is NOT linearly interacting but winding around a
traveling,

spinning orbital, is time the positive portions of the cycle can
gain

advantage spacially.

Follow the momentum rather than the mass, you'll find that it
does

indeed result in an exchange of angular acceleration for linear

acceleration! The total sums may be zero and your proof valid,
lacking

only the break out of the angularly force-paired portion of the
tangential

interaction stored as spin accelerations, which would show the
now

unpaired (linearly) direction of the centrifugally expressed
thrust so many

are finding in reality, contrary to current science dogma.

Remember also that centrifugal force is the square of velocity,
and

though extra time is spent on the race at lower velocities at the
tail,

time is merely ADDITIVE, and the sum total of centrifugal only,
all the way

around the race, is still tugging forward, the zero sum
difference is the

orbital's acceleration and deceleration against the center of
system mass,

known as the tangential force opposed pair (our magic
"hermaphrodite" force

expression mediator, being both linear AND angular!), that
completes the

"dumb" or non-spinning orbit mass variable orbit
velocity machine, and

results in a net zero for that machine (not the GIT).

Just a few thoughts here as I psych myself up to do the next
update

containing the Mars Con info, gotta go, talk with you later!

18 August 1998 - CC: Matheads Group; Jean-Louis Naudin

Reference: http://ourworld.compuserve.com/homepages/jlnaudin/html/GITanal.htm

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