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### Author Topic: Free energy from gravitation using Newtonian Physic  (Read 82417 times)

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #75 on: December 21, 2007, 10:10:05 PM »
Hello pequaide,,

it seems you have a different understanding of the term momentum.

According to kinematics and SI-Units  :
Momentum = Impuls = force x time = mass x velocity exp2  [kg m sec exp2]

This would give all three masses at speed 1.8   momentum of 9.79 ( see formula 3) and not 5.4

m3 alone with speed 1.8 ( spinning off the system ) will have 1/3 of this = 1kg x ( 1,8 * 1. = 3.24
leaving the rest of the momentum = 6.55 to the balanced masse m1 and m2 at speed 1.8.

There is no more momentum available  than 6.55 which you can transfer to m3 which will end up by then having the momentum 9.79

Simply repeating wrong data or calculations does not have any effect on physical facts.

Regards

Kator

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #75 on: December 21, 2007, 10:10:05 PM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #76 on: December 21, 2007, 11:39:27 PM »
Last I knew (mv) 3 kg * 1.808 m/sec = 5.425 units of momentum.  It is not three times the acceleration it is three times the velocity. Or how do you come up with 9.79? Try this next example.

A 4 kg mass that is dropped from .051 m will have a velocity of 1.00 m/sec, if it is dropped as a pendulum bob it will be moving 1.00 m/sec horizontally. If it is caught into a horizontal circle it will continue this 1.00 m/sec velocity around the circumference of the circle. If you then release the mass from that circle and catch it in a pendulum it will rise (as a pendulum bob).051m.

A four kilogram mass moving 1m/sec in a circle can give all of its motion to one of its four kilograms, as demonstrated by the cylinder and spheres experiment. A good scientist should ask themselves, what must be the velocity of that one kilogram when it has all the motion? This transfer of motion will leave 3 kilograms at rest, and Newton?s Three Laws of Motion requires that the momentum remain constant. Some complain that the experiment is moving in a circle; but that is a ridiculous complaint because ballistic pendulums move in circles and they are used to establish Newtonian Physics. Others sight direction of motion as a problem, but Atwood?s machines are used to prove F=ma and nearly all the mass in motion of an Atwood is moving in opposite directions (almost half the mass is going up and a little over half is going down).

When a one kilogram mass moving 4 m/sec strikes a 3 kilogram mass at rest the new velocity for the 4 kilogram combination must be 1 m/sec to comply with Newtonian Physics.

Shouldn?t we be able to return to the point from which we came? Which was 4 kilogram moving 1.00 m/sec. When the four kilograms gives all its motion to the one kilogram and then the one kilogram is directed to slide back into the 3 kilograms at rest, shouldn?t that return us to the original motion. The only velocity that could do that is 4 m/sec for the one kilogram mass when it has all the motion.

Mathematically we could state it like this:  4 kg * 1.00 m/sec = 1 kg * 4 m/sec = 4 kg * 1.00 m/sec.

Only 4 m/sec velocity for the one kilogram will bring you back to the point of origin, which was 4 kg moving 1.00 m/sec.

Four kilograms moving 1 m/sec has 2 joules of energy. 1/2mv?

One kilogram moving 4 m/sec has 8 joules of energy. The one kilogram moving 4 m/sec can rise to .8155 meters, four times higher than the total height of the four kilograms at .051 meters. .051 m * 4 = .204 m

Note that I have not even mentioned radius, I don?t need to, velocity is independent of radius. Ballistic pendulums always conserve linear momentum no mater what the length of the pendulum.  Galileo proved that velocity and radius are independent of each other.

Formulas are valueless when they require a quantity that we don?t even need to know. The angular momentum formula requires radius and we don?t even need to know radius.

Energy can be made in the laboratory for \$25, which is the cost of a cylinder and spheres machine.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #77 on: December 22, 2007, 03:32:48 AM »
We have a pendulum where I work that swings for the better part of an hour. The energy transfer in the cylinder and spheres experiment occurs in less than a half second. Put these two facts together and bearing friction percentage becomes minimal. We are looking at about 300% increases in energy in a 3 or 4 second cycle.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #77 on: December 22, 2007, 03:32:48 AM »

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #78 on: December 24, 2007, 02:07:45 AM »
Hallo pequaide,

I apologize. I was totally on the wrong track. I have done myself what I was suggesting to you and looked in my physics-books.
I had confusion with english physical terms because momentum is impulse in german language.I understand now what you found. I had another concept in the back of my mind ( Alans discovery m*v exp2) and was mixing up the formulas.

Now, the correct way to calculate the final energy of m3 ( after m1+m2 fully stopped) must me done via the momentum-change

F*t = m * Delta V ( change of velocity of mass m3 ) and not by direct energy-calculation since this is what actually happens :

- momentum- ( or impuls- ) transfer.

This formula is the correct link of calculating the final energy of m3. The mistake lies in simply splitting the energy of mass-system m1+m2+m3 into m1+m2 and m3.

Now the momentum of m1+m2 after m3 spun off is 3.6.
If this system is stopped lets say in 1 second the m1+m2 * Delta Velocity is - 3.6 adding as a
+3.6 to mass m3 resulting in v = 5.4 m/sec thus delivering energy = 14.58.

As I said earlier : I think the key lies in changing the direction of momentum-vector of spinning m3 by 90 deg to m1+m2-system. When this is done there cannot be transfered any information from m3-system back to m1+m2-system if
m3 is moving to a bigger radius. The momentum-vectors of both systems are made independant.Now then you can slow down m3-angular-velocity in order to transfer momentum
of m1+m2-system.

The challange lies in finding a practical and simple system. I think of water running down on an pelton-like turbine where
the water leaves at 6 pm entering another horizontal turbine-system and flowing then out radially ( both system impulswise coupled for momentum-transfer )

This is hard engineering-work and it takes some time. Alan is a good person for enginnering-ideas. Write him about this.

If you have systems like a pendulum you mentioned why not show a picture including the principle here ?

Have a look to this pendulum Alan describes  here : http://www.unifiedtheory.org.uk/

( I am sorry the hyperlink-button does not work with my online-watch-system)

at diagramm 16 : A PERPETUALLY RESONANT DAMPED AND FORCED SPRING

Kator

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #79 on: December 26, 2007, 04:31:23 PM »
Quote from P-Motion  It would need to be shown how 3 weights falling could increase the potential of an opposing mass above the potential of the falling weights. A math calculation does not show how the increase occurs.
From the math above, the 3 weights could only cause the opposing weight to rise about .5 meters.

All three one kilogram masses will eventually be raised to .5 meters, but they were only dropped .1666 meters

Velocity of an object dropped equals the square root of the product of 2 times acceleration times the distance dropped. In freefall that would be

V = sqrt of 2 * 9.81 m/sec * .1666 m  = 1.808 m/sec

This (1.808 m/sec) is the velocity of each of the three one kilogram objects that was dropped (by using a pendulum) .1666 m. The total momentum of the three is (3 kg * 1.808 m/sec) = 5.425 units.

If this momentum is then transferred to one (using the cylinder and spheres device) of these three kilograms the one kilogram will have to have 5.425 m/sec velocity. This will conserve the momentum of the three. This will leave two kilograms at rest. Rest means they have no ability to rise and therefore are at zero elevation. On the other hand the one kilogram is able to rise 1.50 meters.  This places the center of mass of the three objects at .50 meters. d = ? v?/a

The center of mass of the three one kilogram objects started at .1666 meters elevation and ends with the center of mass of all three at .50 m elevation. An energy increase to 300%

You could also use a 2 kg balance wheel (with the mass concentrated on the rim of the wheel) to achieve 5.425 units of momentum. Place 1 kg on the edge of the 2 kg rim. This will give an acceleration of 9.81/3 = 3.27 m/sec. The center of mass of the mounted rim will not change upon acceleration; only the elevation of the overbalanced 1 kg will change.

After dropping the overbalanced 1 kg mass .5 m the three kilograms will have 1.808 m/sec velocity each.  So an elevation of .5 m for all three can be achieved by dropping only one .5 m.

You will have to keep in mind whether you are starting by dropping three kg .1666 m in a pendulum or by dropping one kilogram in a 3 kg (total mass) overbalanced wheel.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #79 on: December 26, 2007, 04:31:23 PM »

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #80 on: December 27, 2007, 02:23:01 PM »
Hello P-Motion,

yes, I understand the problem you describe. The main problems are techical ones. Once the single mass ( in our discussion m3 ) leaves the balanced system at 6 pm at v = 1.808 m/sec no momentum can be transfered by the balanced-weight momentum because all masses have the same speed.

But before I get deeper in technical discussions I would like to find in this discussion a simple way to proof the principle. I am still sceptic about the formulas although the conservation of momentum is clear.Some of the  formula-stuff is directing our attention to wrong interconnections But, as you described it : How do we transfer momentum back to m3 ?

One idea came up but I am not sure it works : lets assume m3-momentum ( including m1m2 -momentum ) will hit a spring at 6 pm in full elastic collision, compress it and then stick to it by some sort of mechanism ( detached from m1,m2 ) while the spring is locked in compressed state by another mechanism. We then have all the time to set up a test-rig where we can release the spring and measure m3 climbing the rig or directly measuring the velocity of m3 and have not to deal with more kinematic problems

The question is : does this setup deliver the answer to the momentum-transfer ? Or do we need free moving masses to do so ?

@pequaide : what you you think ?

Kator

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #81 on: December 27, 2007, 09:02:50 PM »
A spring is unlike gravity. Gravity has a uniform quantity of force along the entire freefall (and rise) distance; this gives you uniform acceleration (and deceleration).

In a spring: if you double the distance of compression you also double the force applied to the object performing the compression.

If you then allow the spring to unload on a less massive object, the acceleration would be very large at first. This would move the lighter object out of the compression range in a very short period of time.

Since momentum is a function of time; F = ma, a = Ã¢Ë†â€ v/Ã¢Ë†â€ t, the quantity of momentum used to load the spring would be greater than the quantity of momentum delivered when the spring unloads.

It would be the opposite of the overbalanced wheel, because the overbalanced wheel takes longer to load than to reload (the time it takes to cast the overbalance mass back up in reverse freefall).

I think the key is not the force itself but the time over which the force acts.

In a 10kg (total mass with 1 kg overbalance and 9 kg balanced) wheel with 1kg of overbalance it takes 1.43 seconds for the one kilogram to drop one meter. In reverse freefall the 1 kg can rise 1 meter (back to the place from which it was dropped) in only .45 seconds.

This 1.4278/.4515 difference in time also gives you a proportional momentum difference of 14.007/4.429. And the energy increase is the square of this difference.  14.007/4.429 = 3.16;    3.16? = 10.00

Here is how it works: Transfer the 14.007 units of momentum that is developed by the overbalanced wheel back to the one kilogram of overbalance and it will rise 10.0 meters.  And it had only been dropped one meter.  Formula used d = ? v?/a;   v = Ã¢Ë†Å¡ (2*a*d);   d = ? at? or t = Ã¢Ë†Å¡ (2 * d / a) (these three are the same formula of course because t = v/a), F = ma was used to obtain acceleration, mv, and 1/2mv? or 1/2mt?a? because v? = a?t?.

I think the vertical and horizontal wheels would be a good choice of designs, I think this was post #77 this site (masses changed).

In the example you have 1 kg on each end (2 kg) of a one meter vertically mounted bar (the bar is very light weight, this could also be a wheel with evenly distributed mass) with a high quality center bearing.  Add an extra 1 kg to one of the end masses so that the overbalance 1 kg can rotates one half meter to the bottom. That will mean the extra mass will move from 90? to 180? (or from 3 o?clock to 6 o?clock). The acceleration rate should be one third that of gravitation 9.81* (1/3) = 3.27 m/sec/sec. This is from F = ma, a = F/m   9.81 newtons / 3 kg = 3.27 m/sec/sec

At the end of the .5 meter drop (of the extra 1 kg) all 3 kg will be moving 1.808 m/sec.  v = Ã¢Ë†Å¡ (2* .5m * 3.27 m/sec/sec) = 1.808 m/sec

As the extra (overbalanced) mass reaches the bottom release it into a light weight, .1 m radius, horizontally mounted wheel with a high quality bearing point. Now we have 2 kg moving 1.808 m/sec in a vertically mounted balanced wheel and 1 kg moving 1.808 m/sec in a horizontally mounted wheel.  Connect a string from the vertically mounted wheel to the horizontally mounted wheel so that the string is winding up on the vertical wheel and unwinding from the horizontal wheel. Now release the overbalanced mass from the horizontal wheel but keep it attached to another string that has been wrapped around the horizontal wheel (the cylinder and spheres experiment). While the overbalanced mass unwraps from the horizontal wheel it will absorb the momentum of the two 1 kg masses on the vertical wheel. This is the same phenomenon as the cylinder and spheres experiment.

Newtonian Physics predicts that it will now be moving 1.808 * 3 = 5.42 m/sec and it will rise 1.5 m. d = ? v?/a. It was dropped only .5 m.

Add the extra mass back to the top at 90? after you have transfer 2/3 of it energy to another system. You can still start over because you have three times the original energy to work with.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #81 on: December 27, 2007, 09:02:50 PM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #82 on: December 28, 2007, 03:45:08 PM »
Here is my definition of an overbalanced wheel.

Take a bicycle and turn it upside down, onto its seat and handle bars. The front tire of the bike is now a vertically mounted balanced wheel. Duct tape the padlock to the tire and now you have an overbalanced wheel.

If the vertically mounted balanced wheel has a mass of 9 kg and the out of balance or overbalanced mass is one kilogram then you have a 10 kilogram overbalanced wheel with an overbalance of one kilogram. The radius of rotation of all ten kilograms would be (roughly) the same. This would give you (F = ma) 9.81 newtons of force applied to 10 kilograms for a .981 m/sec/sec acceleration.  There is no need to consider lever arm length because they are all the same.

Thank you for allowing me to clarify this.

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #83 on: December 28, 2007, 08:40:12 PM »
Hello P-Motion,

searching for a technical solution I found the following website - unfortunately in german language.

http://doerler.gmxhome.de/

I find it helpful to visualize necessary kinematic basics in order to locate the technical problem with this approach.

Please go on the left navi-area to Impulserhaltung ( conservation of momentum ) and then to

Newtons's Wiege ( se-saw ). Here you can simulate momentum-exchange with 1,.. to 4 balls. Just change the number in the small window.
Insert number 3 and watch what happens. In this setup it is not possible to stopp all 3 masses and accelerate the left single mass to the same momentum of the tree masses swinging in from the right side.. So this is the problem to solve.

But this setup solves another problem you brought to our attention in one of your posts :
Quote
If the one weight goes around in a circle and then returns to the pendulum, it will be on the side opposite of the other 2 weights.
How would the pendulum be set up to catch it ? And how would it know when to release it ?

Choose number 1(ball) this will give you an idea how you can transfer the momentum of mass m3 leaving the wheel at 6 pm at velocity 1.8 m/s. By this you can put the horizontal wheel to a given distance away from the 6 pm - position.

The second topic is momentum-change

Left navi-area go to : Kraftstoss and then to Schau's dir an

Here you see the simulation of a car decellerated with different forces.

1) big force - very short time ( to get the car to stop )

to

3) very small force - big time  ( to get the car to stop )

I hope this helps to understand some of the basics here.

Regards

Kator

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #83 on: December 28, 2007, 08:40:12 PM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #84 on: December 28, 2007, 11:37:56 PM »
Pictures: note the slit and the extention of the string. The spheres have most or all the motion, they are much lighter than the cylinder.

#### supersam

• Sr. Member
• Posts: 475
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #85 on: December 29, 2007, 04:02:59 AM »
@ all,

i think i know for sure what is cleared up.  the fact that pequaide has the ability to post a video of his experiment!

lol
sam

#### Free Energy | searching for free energy and discussing free energy

##### Re: Free energy from gravitation using Newtonian Physic
« Reply #85 on: December 29, 2007, 04:02:59 AM »

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #86 on: December 29, 2007, 04:17:42 AM »
P-Motion Please be more specific as to what you think doesn?t work; I thought everything works just fine.

Both the last two pictures are of the same cylinder and spheres devices in freefall. It was spun by hand and released when the spheres were seated in the holes. The spheres unwind and swing out on the end of the strings, when they swing out the cylinder stops after about a quarter turn. The string is seen on both sides of the slit showing that the cylinder is not moving and that the spheres are. Video tapes can break it down into 1/30 of a second, the cylinder stops.

The mass of this particular cylinder is about 630g; the spheres have a mass of 67g each. 630+134 = 764, 764/134 = 5.7. This would be like 5.7 kg being stopped by transferring the motion to 1 kg.

#### pequaide

• Full Member
• Posts: 139
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #87 on: December 30, 2007, 12:52:23 AM »
P-Motion: You asked if it is overunity.

Answer; You start with 1 kg at .5 m elevation and end with 1 kg at 1.5 m elevation. I assume that this is one definition of overunity.

P-Motion:  What I wonder about is the ability of the spheres to transfer force to the cylinder or the other sphere.

Answer: The strings have tension in them, which means force is in the string, Newton?s Third Law states that the force in the string will be equal in both directions, and the Second Law states that the momentum change will be equal on both ends of the string.

P-Motion:  It seems the set up transfers force/potential from the cylinder to the spheres.

Answer: The setup transfers momentum from the cylinder to the spheres.

P-Motion:  For the spheres to be able to effect each other, they would have to be on an arm attached to the cylinder.

Answer: Are we concerned if the spheres affect each other. The spheres remove momentum from the cylinder, I am unconcerned if they affect each other.

P-Motion:  And the cylinder would have to be designed to allow for spin/velocity to be increased by gravity.

Answer: The vertical overbalance wheel is ?designed to allow for spin/velocity to be increased by gravity?. The cylinder and spheres portion is mounted horizontally and is unaffected by gravity.

P-Motion:  In other words, it would have to allow for linear momentum to be converted to angular momentum.

Answer: Motion changes from linear to circular easily; often; and is 100% efficient in both directions.

P-Motion:  But with the 2 spheres being attached by a string, the spin necessary for the cylinder to create enough inertia would be high.

Answer: Inertia occurs in the lowest degrees of slowness. When released the spheres attempt to maintain a straight line motion, whether they are moving slow or fast. The cylinder pulls the spheres from this straight line motion and the force in the string begins to increase, at any velocity.

P-Motion:  Since any overunity design is dependent upon 9.8m/s/s, and with a changing of the mechanics of a device, its' spin will be limited.

Answer: The momentum transfer of the cylinder and spheres device is not limited by any spin rate that I know of. The effect is the same; slow or fast. The overbalanced wheel portion is of course limited by the acceleration caused by gravity.

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #88 on: December 31, 2007, 05:53:54 PM »
Hello all,

I wonder if anybody has looked to my post #110, page 8, because I try to find analogies of impuls-transfer either in nature or  in man-made system we are not aware of.

P-Motion, I also like this thread because it is a challenge to figure out a way to transfer impuls in a way nature does not show us - or should I say scientific or technical education we went through has been conducted in a tunnel-view manner.

I looked at the pendulums ( animation I was posting in my reply #110, and to the formulas again and again. I asked myself why the heck is the conservation-process of momentum-transfer from two 1kg-  masses hitting a single 1kg - mass happening only in a manner focussed on the target-mass and not
on the velocity ? More clearly expressed : Why does the process not solely increase the velocity of the one kg-mass and stop the 2 kg masses in order to fullfill the equilibrium in this equation p = 2 kg * V1 = 1kg * 2*v1 ?

The answer is of course : intertia of masses prevail. The inertia of a 1kg mass cannot stop a 2 kg-mass.

I have not yet figure out how to technically do this besides what pequaide has presented here. I only have grasped this one idea of spin-inertia because here spin-inertia of a 2 kg mass on a wheel can be made equal in value by placing the reaction-partner-mass of 1kg on a wheel with double the radius of the two kg-mass-wheel. Now both systems having the same spin-intertia might provide a chance that  velocity of target mass is increased in a way the 2 kg mass is stopped.

Maybe that you all here have already mentally worked this trough. I myself concentrate on different way to do this momentum-transfer.

I hope I have been able to bring across my point here. As an enginneer in applies car-construction I have learned  tunnel-view realities and I always was struggeling to overcome this.

Another idea came to my mind : a flowing water-mass beeing stopped suddenly by a valve - a technique which is state of the art since almost 200 years - by name of
hydraulic ram pump

principle : Water-hammer-shock-wave

have a look here

http://virtual.clemson.edu/groups/irrig/Equip/ram.htm

Can this technique be changed so it fits in in this momentum-transfer-problem ?

The hydraulic-ramp-pump-principe increases pressure in a tank.Gas is compressible, water is almost incomressible and thus suitable for the elastical-bounce-case.
This pump does not transfer the momentum of a very long slow moving water-column beeing stopped->  to a small water-mass at high velocity.

What do you think ?

Best wishes for year 2008

Kator

#### Kator01

• Hero Member
• Posts: 857
##### Re: Free energy from gravitation using Newtonian Physic
« Reply #89 on: January 01, 2008, 07:33:27 PM »
Hello P-Motion,

I have a few questions. I simply do miss some steps in your description.

- two weights : weigth at left is in fiixed position ? I assume both masses are of the same weight.
- why does the center of the rotating arm has two positions ? Is the center ( axis ) not fixed ?
Does it wobble ?

I simply do not understand the whole process because I do not understand the basic setup especially this :

Quote
In the schematic, when the arm has rotated to the desired angle, if the over balanced weight changes its' course, it can return to a balanced position.

- ... What is the desired angle and at which step in the process ?
- ... Changes its course at what position or angle ?
- ... It can return to a balanced position at what angle ?

Can you please give me a more specific description if you have the time ?

Kator