IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

Floor · August 10, 2021, 04:59:28 AM

@Georege1

I started out to look at this design and diagram the forces involved there in.

But no, I really don't see any potential for "perpetual motion"or a " reactionless drive"
here at all.

There are some cancellations of momentum and forces by virtue of opposed
actions and reactions there, but again, I really don't see any novelty or point to
the device at all.

The incessant repetitions of these questions and even their answers seem to be
leading to nothing and to no where.

bah

George1 · August 10, 2021, 05:21:05 AM

To Floor.
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THE SIMPLE OBVIOUS FACT, that you keep constantly avoiding to answer my two simple questions 16 TIMES IN A ROW (!), unambiguously shows that you are either a stubborn ignoramus and/or a paid agent of the official science mafia, who makes some money by trying to manipulate the audience in a clumsy and unskillful manner. In any case it is more than evident for all honest members of good will of this forum, that you are simply a professional (but clumsy) cheater and an unworthy person! How much do they pay you? Shame on you!
========================
ANSWER THE TWO SIMPLE QUESTIONS BELOW!
LOOKING FORWARD TO YOUR TWO ANSWERS FOR THE 17TH TIME!
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Consider carefully and thoroughly again (and many times, if necessary!) the link https://www.youtube.com/watch?v=xX14NK8GrDY
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A key component for a proper understanding of the zigzag mechanical concept is the SIMPLE OBVIOUS FACT, that the zigzags generate a mechanical effect, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. And this SIMPLE OBVIOUS FACT is clearly explained and described in PART 1 and in PART 2.
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And now focus on PART 3 where:
Ma = 1 kg
Mb = 4 kg. (The value of Mb can be either increased or decreased as many times as you want.)
V1 = 1m/s = const
Ffr. = force of friction inside the zigzag channels = 0.0000001 N. (The latter can be further decreased as many times as you want.)
N = number of zigzags = 10. (The value of N can be either increased or decreased as many times as you want.)
Shapes of the zigzags = sinusoids. (The latter can replaced by any other curve patterns.)
QUESTION 1: V2 = ? (How many meters per second is V2 equal to?)
QUESTION 2: V3 = ? (How many meters per second is V3 equal to?)
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ANSWER THE ABOVE TWO SIMPLE QUESTIONS!
LOOKING FORWARD TO YOUR TWO ANSWERS FOR THE 17TH TIME!
==========================
Your masters will beat you! You have to run quickly and hide somewhere!

Floor · August 10, 2021, 05:50:39 PM

Quote from: Jerry Volland on August 10, 2021, 10:13:24 AM

So how many Coulomb's does a Joule have in zero seconds?
There is a Time factor. A Newton manifests during a duration of one second.

The joule is the SI unit of energy.

work = force x displacement
or
joules of work (energy) = newtons of force x meters of displacement, in the direction of the applied force.

The work done in displacing an object 1 meter against a force of 1 newton = 1 joule of energy. This is also referred to as the energy expended.

A 1 kilogram mass (equal to 1000 grams) exerts , 9.8066500286389 newtons of force down
(in earth's standard gravity).

1000 grams = 9.8066500286389 newtons
and
9.8066500286389 / 1 newton = 0.10197162099999948436449906616771 grams
there fore
0.10197162129779282425700927431896 grams or approximately 102 grams exerts
1 newton of force down in standard gravity.

A mass of about 102 grams exerts 1 newton of force down in standard gravity.

If we lift a 102 gram object 1 meter, we do about 1 joule of work upon that object.

Power (watts) is equal to the joules expended per second of time.

1 joule per second = 1 watt of power
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There is an inverse relationship between the force of gravity and the distance between the CENTERS of the two attracted objects.

If the distance is doubled, the gravitational force is decreased by a factor of 4. This is because the square of 2 is 2 x 2, which equals 4. If the distance between two objects is tripled, the force of gravity is decreased by a factor of 9. In this case, it is because the square of 3 is 3 x 3, which equals 9. This relationship is known as an inverse square relationship.

However, on the scale of a base ball in attraction to the earth, the distance between the CENTER of and the surface of, that base ball becomes insignificant within the calculation.

Similarly, a distance of 1000 meters above Earth's surface becomes insignificant in proportion to the distance from Earth's surface to its center.

There fore, a base ball weighs ALMOST exactly the same, whether it is 1 meter above, or 100 meters above the Earth's surface.
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Coulomb force is the force due to electric charge. It is the repelling force between two electrons but also the attracting force between an electron and a proton. Both of which are due to electric charge.
Like unto a gravitational force, coulomb force also diminishes by the inverse of the square of the distance between the CENTERS of two particles (point sources).

However.....

Unlike the Earth, Sub atomic particles ( protons and electrons) are very very small and there fore, very small changes in the distance between the center of one particle and the center of another particle, causes a large change in the force present.

And unlike gravity, electric charge has two polarities.

The magnitude of the electric force between two "electrons" is directly proportional to the amount of one electric charge, q1, multiplied by the other electric charge, q2, and inversely proportional to the square of the distance r between their centers.

The fixed numerical value of the elementary charge e (of 1 electron) is 1.602176634×10−19 coulomb
and
One coulomb is the charge of 6241509074460762607.776 elementary charges (electrons)
and
The numerical value of theses two quantities are the multiplicative inverses of each other.
Like this... The coulomb is exactly 1/(1.602176634×10−19) which is approximately 6.2415090744×1018, elementary - charges.

The charge of 6241509074460762607.776 protons is a + charge.
The same number of electrons has the same magnitude but opposite sign of charge.
That is − charge. 1 coulomb is 6241509074460762607.776 − charges.

The force from electric charge has other considerations as well.
example...
In calculating the force between two charged and macro world objects (for example two electrically charged plates), one must also consider the the area of the surfaces of those plates.
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Quote from Wikipedia

"Until 2019, the International System of Standards (SI) defined the ampere as follows:

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one meter apart in vacuum, would produce between these conductors a force equal to 2×10−7 newtons per meter of length ."

End of Wikipedia quote
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That force of 2×10−7 newtons, per meter of length is the result of the magnetic field surrounding the two conductors.

That repulsion force is magnetic and is in due to, both the electric charge (coulomb charge) and the motion of the electrons along the conductor.

note...

This is not the coulomb force present as the repulsion between the electric charges.
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The present (SI) quantification of the ampere (since May 2019).....
The AMPERE was then defined as one coulomb of charge flow per second. In SI, the unit of charge, the coulomb, is defined as the charge carried by one ampere during one second. However, this definition although not the SI standard until 2019, was in use earlier, within the Centimeters Grams Seconds (CGS) system (prior to 2019).

A current of one ampere is defined generally as one coulomb of −charge (electrons) going past a given point (generally in a conductor) per second, but strictly speaking, this could also be a current of + charge protons passing a given point in 1 second of time.
and
It seems as though there is no simple and straight forward way, to exactly correlate the force and displacement elements of mechanical work ( joule), to the process of arriving at its electrical energy equivalent ( joules).

FORCE
Voltage is a force, and that force is referred to as electromotive force. In SI terms the derived unit of electromotive FORCE is the volt. The volt is a unit of the electric potential between two points.

MASS or CHARGE per unit of time
The ampere unit, definition, INCLUDES A TIME ELEMENT (the second).
It is a unit of a quantity per second. It is a current, a flow of electrons.
The Ampere is a time based unit of measurement.

ENERGY
The joule unit of measurement of energy
The Joule is not a time based unit.

The coulomb may be thought of as either, a quantity of 6.2415090744×1018 negative charges or as a quantity of 6.2415090744×1018 electrons.
Coulombs per second = amps...... Amps x volts = watts.
A force of 1 volt will move 1 coulomb of electrons through a resistance of 1 ohm in 1 second of time.
But...
In SI units, ELECTRIC WORK is stated as joules of energy per coulomb, where 1 volt = 1 joule (of work) per 1 coulomb (of charge or electrons)....
1 volt = 1 joule / 1 coulomb
where as
Electric power (1Watt) is 1 Volt of force x 1 coulomb of electrons passing a point in a conductor in one second.
Volt x Coulomb = Joule but Volt x Coulomb / Second = Power as Watts.
however....
There is no length of displacement specified, as is specified, in the force times displacement equation which defines mechanical work.
There is only the movement of a specific quantity of electrons (one coulomb) through a point. The coulomb unit is substituted for the displacement unit (meters).
That which is analogous to a mechanical reactive force (equal to and opposite force), is the electrical resistance (stated in units of ohms of resistance) opposing the voltage.
One coulomb of electrons passing a point in a conductor is one joule of work (no time element).
Force as newtons x displacement as meters = joules.
Force as volts x quantity as coulombs = joules.
One coulomb of electrons passing a point in a conductor in one second, is a flow RATE of 1 ampere (a current). Like unto gallons per minute. Time element.
The ampere is not a unit of some quantity of electrons, it is a unit of a rate of flow (1 coulomb of electrons per second).
mechanical.... force x displacement = joules or newtons x meters = joules electric.... force x quantity = joules or volts x coulombs = joules
mechanical..... force x displacement / time = watts or newtons x meters = joules and 1 joule / 1 seconds = 1 watt .... 1 joule per second = 1 watt electric..... force x quantity / time = watts or 1 volt x 1 coulomb = 1 joule and 1 joule / 1 second = 1 watt. But also, 1 coulomb / 1 second = 1 amperes and so 1 volt x 1 ampere = 1 watt.
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1 coulomb is a QUANTITY of 6.2415090744×1018 negative charges or 6.2415090744×1018 electrons.

1 coulomb PASSING A POINT in a conductor IN 1 SECOND OF TIME is 1 ampere of
current FLOW.

1 ampere of current will flow through 1 ohm of resistance when an electromotive force potential
difference of 1 volt is present.

1 ampere times 1 volt is equal to 1 watt of electrical power, and / or 1 joule per second.

Floor · August 10, 2021, 06:39:02 PM

THERE IS NO TIME PERIOD DURING WHICH ENERGY
MUST BE TRANSFERRED IN ORDER TO QUANTIFY the JOULE OF ENERGY

Neither is there a rate of acceleration which must occur in order that joules of energy can act or be defined. Such a stipulation would only apply to the defining of the JOULE PER SECOND.
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Time is intrinsic within and unalienable from the newton of force and its definition. This is because the newton is defined in terms of acceleration against the reactive force caused by
the inertia of a mass. That reactive force of inertia exists only during an acceleration. Of
course, acceleration is described and defined in terms of a displacement over an interval of
time per another interval of time, as in m/s2 (meters per second per second).
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But there is no time period which must be stipulated in order to quantify the joule of energy.
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If we transfer 2 joules of energy, in a manner such that it will cause the uniform acceleration of a 1 Kg mass, that energy acts as a force.
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If we transfer 2 joules of energy to cause the uniform acceleration of a 1 Kg mass (against the inertia of that mass solely)
and
that 1 Kg mass accelerates to a velocity of 2 meters per second in 1 second (2 m/s/s).

The object will have displaced only 1 meter during that second.

2 newtons of force must have acted upon the mass, because the 1kg mass accelerated at 2 m/s/s.
because
1 N by definition is = 1kg • (1 m/s/s) .... therefore 1kg • (2 m/s/s) = 2 N

If the force applied is 2 newtons and the displacement is 1 meter the energy expended is 2 joules.

1kg • (1 m/s/s) = 1N ... and 1N • a displacement of 1 meter = 1 joule ....
therefore ....
1 kg • 2 m/s/s = 2 N ... 2 N • displacement of 1meter = 2 joules

But it is not necessary for that acceleration to occur in one second of time, in order that 2 joules of energy are expended.

When we transfer joules of energy to cause the acceleration of a 1 Kg mass (against the inertia of that mass solely), the time required for said 1 Kg mass to reach a velocity of 2m/s, will be dependent upon the rate at which the joules of energy are transferred to the 1 Kg mass. The joules per second.
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When we transfer 2 joules of energy to cause the acceleration of a 1 Kg mass (against the inertia of that mass solely), that 1 Kg mass WILL ACCELERATE from rest, at the rate of 2 meter per second per that second, ONLY WHEN the energy is transferred AT A RATE OF 2 joules per second.

This is the amount of energy PER SECOND (2 joules per second) (as force in newtons) required to accelerate a 1 kilogram mass, from rest, to a velocity of 2 m/s, IN ONE SECOND of time. (against the inertia of that mass solely).

Redundantly stated, this is also the amount of energy required, to cause an object to displace 1 meter against a force of 2 newtons.

Also redundantly stated, it will not matter if the transfer of those joules of energy has occurred over some period of time other than 1 second. It still remains that it is just 2 joules of energy.
The energy transferred in accelerating a 1 kilogram mass (against the inertia of that mass solely) to some specified velocity, is the same amount whether the mass is accelerated to that specified velocity, rapidly or slowly.
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a first example

If it takes 1/2 second of time for a 1 kilogram mass to accelerate from rest, to a final velocity of 2 meters per second, the energy was transferred at a rate of 2 joules per 1/2 second.

2 m/s in 1/2 second is an acceleration at the rate of 4 m/s in 1 second or 4 m/s/s.

The average velocity of a constant or uniform acceleration is 1/2 of the final velocity.
The average velocity of the 2 m/s final velocity is 1 m/s.
The duration of the acceleration is 1/2 second.

The displacement that occurred during the 1/2 second is the average velocity times the duration of the acceleration. This is 1 m/s times 1/2 second or 1/2 meter of displacement.

We have a 1kg mass accelerating at a rate of 4 m/s/s. This is equal to 4 newtons of force.
4 newtons of force times a displacement of 1/2 meter is 2 joules per that 1/2 second.
and
This is 2 joules per that 1/2 second, 2 joules have given rise to the acceleration.
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a second example

If it takes 20 seconds of time for a 1 kilogram mass to ACCELERATE to a 2 meters per second final velocity.
then
The 1 kilogram mass accelerated to a velocity of 2/20th of a meter per second in the first second.

The mass accelerated an additional 2/20th of a meter per second, during second number 2, for a total velocity of 4/20th of a meter per second by the end of second number two.

A total velocity of 6/20th of a meter per second was reached by the end of second number 3, and so on.
and
At the end of 20 seconds the mass will have reached a total velocity of 40/20th of a meter per second, or in other words a final velocity of 2 meters per second.
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2 m/s in 20 seconds is an acceleration at the rate of 0.1 meter in 1 second or 0.1 m/s/s.

The average velocity of a constant or uniform acceleration is 1/2 of the final velocity. The average velocity during this acceleration (to the final velocity of 2 m/s) is 1 m/s.

The duration of the acceleration is 20 seconds.

The displacement that occurred during the 20 seconds is the average velocity times the duration of the acceleration. This is 1m/s times 20 seconds or 20 meters of displacement.

We have a 1kg mass accelerating at a rate of 0.1 m/s/s. This is equal to 0.1 newton of force.
0.1 newton of force times a displacement of 20 meter equals 2 joules.

That's 2 joules per 20 seconds, this is 2 joules of work or 2 joules of energy transferred to accelerate the 1 kg mass to a velocity of 2 m/s.
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Whether the velocity of 2 m/s is reached via an acceleration rate of 4m/s/s during 1/2 second or the velocity of 2 m/s is reached via an acceleration rate of 0.1 m/s/s during 20 seconds ....

It requires the same amount of energy to accelerate the 1 kilogram mass to the velocity of 2 m/s. It requires the same amount of energy whether the acceleration is a rapid one or a slow one.
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The force applied and / or the JOULES PER SECOND transferred are different in the two scenarios.
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lancaIV · August 10, 2021, 07:05:25 PM

1000 RPM generator , 1 000 W rated output

1 year = 8766 hours

50% periodic full load = (4383 hours x 1 000 RPM) = 4 383 000 RPM in one year with 4383 KWh conversion

100% permanent 50% load = (8766 hours x 500 RPM) = 4 383 000 RPM in one year with 1095,75 KWh conversion

same RPM per annum,but different force conversion gain !

https://www.raeng.org.uk/publications/other/23-wind-turbine

important for each kind of rotoric permanent/periodic fixed/variable conversion devices !

IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

Floor

George1

Floor

Floor

lancaIV