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

## Mechanical free energy devices => mechanic => Topic started by: kmarinas86 on January 07, 2009, 11:38:30 PM

Title: Critical Response to Tom Napier's (1999) Critical Analysis of Newman Claims
Post by: kmarinas86 on January 07, 2009, 11:38:30 PM
http://www.phact.org/e/z/newmann.htm

PART ONE AND TWO

Quote from: Tom Napier
Newman -- point by point

Part 1, Currents and field strengths   http://www.phact.org/e/skeptic/newman.htm

Introduction:

This document is one of a series which address specific errors made by Joseph Newman in his book, "The Energy Machine of Joseph Newman."
This section examines the connection between the current flowing in a solenoid, its number of turns and the resulting magnetic field and power consumption.  It relates to the section of Newman's book which runs from page 297 to page 302 and, in particular, to the table on page 299.  (Page numbers refer to the 8th edition.)

Nothing to dispute here.

Quote from: Tom Napier
Newman claims:

1)  That the current required to generate a given magnetic field can be reduced by increasing the number of turns in the solenoid.

This is of course true in that even a small current MAY be set to produce a magnetic field with greater overall potential energy AND strength.

Quote from: Tom Napier
2)  That the power required to generate a given magnetic field falls in proportion to the number of turns and tends to zero as the number of turns increases.

This is true for the amount of potential energy in the field, but not for the DENSITY of it. Over time, a small amount of input power will result in a large amount of input energy. No matter how much you change the coil design, the amount of input power has inherent limitations in one's ability to make the field intensity of the coil greater. HOWEVER, do not confuse field intensity with force, since field intensity is proportional to the square root of the potential energy density of the field, which in turn is proportional to the pressure of the field. That means to increase field intensity by x times increases the pressure by x^2, and on top of that, it fails to account for area to derive the force from this pressure, as well as the distance over which the force is applied.

Quote from: Tom Napier
3)  That this proves that the magnetic field "emanates from the atoms of the conductor and not from the current."

1 and 2 are not proof or even evidence of such an origin. It simply is coherent with the idea that the magnetic field potential energy must be derived from an interaction which can be assisted or even caused by the presence of charged particles (in this case atoms). Newman has claimed in his patent submission that charge flow through resistance determines the magnetic flux generated and that magnetic flux through resistance produces charge flow, which is not wrong at all, although his spoken demonstrations (notably as of late) implies that he ignores the influence of current.

Quote from: Tom Napier
My response:

Conventional physics states that the magnetic field intensity generated in the center of a hollow cylinder of a given length is a linear function of the total current flowing round the cylinder.  It is not a function of the diameter of the cylinder provided its length is many times its diameter.

IT IS TRUE that current is a linear function with respect to field intensity in such a given coil. HOWEVER, it is not true that this is linearly proportional to either power consumption or energy stored in the field.  That is because power relies on voltage not just current while the energy stored in the field relies not only proportionally to the square of the field intensity but also the volume, since the field intensity is proportional to the square root of the field's potential energy density.

Quote from: Tom Napier
Physics also states that it is immaterial how the current flowing in the cylinder is generated.

However, what is not immaterial is the WAVEFORM of the current itself, which has dependency on how the current is generated in the first place.

Quote from: Tom Napier
The total can consist of a high current flowing round the cylinder once or a small current flowing round the cylinder many times. Which is preferred is a practical matter, not a physical one.  This has been known to be true for about 180 years.  Thus Claim 1) is true and is in complete accordance with conventional physics.

Actually, by definition, preference entails a intention that may be aligned or not aligned with other people's generally accepted practices.

The person deciding what is to be preferred may do so for the sake of experimental research rather meeting the demands of an organization.

Quote from: Tom Napier
No power is required merely to maintain a static magnetic field.  To establish a field in the first place requires a momentary input of power.  This power is stored in the field and can, in principle, be recovered when the current is turned off.  Until then, power is only needed to force the required current through the winding resistance of the solenoid.  This power generates heat in the solenoid but does nothing else useful.

A static magnetic field is produced by BOUND CURRENTS that arise from quantum mechanical principles. Although they have potential energy surrounding them based on a field, it cannot be a source of power. The reason why engineers design motors that incorporate induction between electromagnets and permanent magnets is that permanent magnets have higher magnetization. Magnetization is the amount of magnetic moment per unit volume. By crossing a field from a solenoid with a strongly magnetized object, such as a neodymium magnet, A TORQUE is produced according to:

1) the square of the field of strength of the solenoid
2) the shape and size of the magnet vs. the shape and size of the solenoid interior
3) the magnetization of the magnetic object

Quote from: Tom Napier
It can be shown, see below, that this power does not depend on the number of turns of wire but only on the dimensions of the solenoid.  It can also be shown that there is an minimum power required for any given field, no matter how big the solenoid is made.  This shows that Newman's Claims 2) and 3) are incorrect.

You have no idea what the power output would be. In your analysis:
1) You have not defined the torque of the system.
2) You have not defined how quickly such torque can be produced and destroyed.

Quote from: Tom Napier
The analysis:

To make life simpler I'm going to assume square section wire of side "T" inches.  The results can easily be corrected for round wire if one wants.  (I am using "*" as a multiplication sign and, to avoid superscripts, I am representing exponentiation as repeated multiplication.)

Let's call the inside radius of the solenoid "P" and the outside radius "Q."  Its height is "H."  The cross-section of one side of the solenoid is thus a rectangle H by (Q - P) inches.  Since the wire has an area of T*T the number of turns "N" in the solenoid is H*(Q - P)/(T*T).
The volume "V" of the solenoid is pi*H*(Q*Q - P*P).  Its mass is this number times the density of copper which is 0.320 lb/cuin.  Since the area of the wire is T*T its length "L," assuming no wasted space, is V/(T*T).  This length of wire has a resistance "R" given by L/(T*T) times the resistivity of copper which is 0.63 micro-ohms per inch.  The power needed to drive a given current "I" through the solenoid is its resistance multiplied by I*I.
Provided the diameter of the solenoid is smaller than its height the field strength at its center is proportional to N*I/H.

This is pretty straight forward as there is no disagreement with the scenario itself. However the conclusions lack precision, as follows:

Quote from: Tom Napier
From these results one can show, see the appendix, that the power required to maintain a field strength of F Webers is:

W =  808*F*F*H*(Q + P)     Watts.
-----------------
(Q - P)

This formula only shows the "field strength" in "Webers" when in fact the unit for field strength in Telsas is "webers/meter^2". Also, comparing power input to power output makes no sense without having knowledge of both. Once someone defines the actual force and velocities then can one calculate the RATE of work being done.

The Weber is an SI unit, and in here it stands specifically for MAGNETIC FLUX. The magnetic flux accounts for the area in which magnetic field intensity exists. While it requires the incorporation of area, it fails to account for the fact that the pressure in the field is proportional to the square of the field intensity. Thus, Webers must be squared first of all, thereby squaring both the differential area as well as the field intensity. But then you must consider the fact that once you have determined the field strength (measured in the SI unit TESLAS) and squared it, you must multiply it by (1/2)*volume/magnetic constant to get the amount of potential energy stored in the field.

Quote from: Tom Napier
That is, the power required to maintain the given field is a function only of the field strength and the dimensions of the solenoid.

This says nothing about the fact that you must change the field rapidly to get any work done rapidly. Thus the electric field must be variable with time and space, and quickly so, to get much work done in a short amount of time. This need for swiftness is especially true for smaller electric fields which are typical of smaller motors. Keep in mind that with the magnetic field, both its flux and its strength are quantified using different SI units. The same separation of flux and strength applies for the electric field. The SI unit of electric flux is the call "volt-meter".

Turning on the coil increases its own potential energy. In fact, it increases the potential energy in the entire system EXCEPT FOR THE SOURCE OF ENERGY since the potential energy is defined between interacting particles. It becomes clear that electrical potential energy with respect to SUBATOMIC levels is DILUTED while the potential energy of the MACROSCOPIC level becomes more EVIDENT, though the force per particle significantly affected becomes much weaker as the particles significantly affected increase.

The relationship between power input through electrical activity versus power output that is useful for motive power involves many physics parameters, and often it comes from energy that is not evident at the macroscopic level. After all, energy sources were understood after WORK was defined, as a result of the need to understand the nature of atoms, particularly in chemistry and nuclear physics. Electricity is not a root source of energy but is simply the common language by which all potential bound energy is converted into motive power. Electrical potential energy does not self-multiply and is a constant of the universe.

Quote from: Tom Napier
It is completely independent of the number of turns.  As Q becomes much greater than P, (Q + P)/(Q - P) tends to unity.  That is, for a given height of solenoid the power tends to a constant no matter how much copper one adds.  Adding to the height increases the power needed.  This directly contradicts Newman's Claim 2).

Increases in the number of turns for a given mass, shape, and volume of a coil consisting of the same conducting material increases in the inductance. To achieve this requires that the resistance to increase proportionally.

When given the same allocation for coil volume, both resistance and inductance scale proportional to the SQUARE of turns, n.

In either case, a DC current will take nearly the same amount of time to reach some percentage of its final value, but this value for maximum current (voltage/resistance) will be lower for the high inductance, high resistance circuit. Given the same time frame (i.e. same frequency), it will consume n^2 times less power to generate n^2 times less torque.

As long as the person switches to higher gauge wire to fit in the exact same space and mass as the original coil, one will be decreasing the power density of the system IF DC VOLTAGE IS KEPT THE SAME.

However, if one were to achieve the same amp turns of the system, one would have to increase voltages by the amount, n*(original power factor/final power factor) while currents would increase by n.  The first reason is that between the two first cases, the second case requires that the current would fall by n^2 while the cross-section of each turn falls by n only. The second reason is that as one increases the DC voltage, the amount of time the motor spends on each rotation is reduced, requiring a briefer switching interval, potentially reducing the power factor.  As a result, with the higher turns, it is possible to manage the same current density as the original coil with a voltage that is increased by a factor (n+Îµ) greater than the relative increase in turns (n). Under certain conditions, such a coil of the same volume can withstand higher power transmission with the same current densities.

By increasing resistance and inductance, the heat losses for the same current increase relatively by n^2. The benefit is that n^2 times as much work can be done per revolution (because of n^2 higher torque). Heat losses are also n^2 times as much per second in the same volume.

HOWEVER, because the voltage is increased by an amount greater than n, RPM MAY INCREASE. Keeping this in mind, the increase of the energy lost per revolution may be just a factor less than n^2. Thus, heat losses per revolution MAY fall faster than work done per revolution, as would be expected from reducing the power factor. Explained further...

In cases where input torque climbs faster than the shaft torque, the % of free-load rpm dropped under load will fall. As a result, such systems which already possess a lower power factor will be able to reduce the power factor faster than the voltages increase. The fact is that this will increase rpm slightly beyond a linear relationship will voltage, even under load, but only so slight.

In other words, altering the number of windings in an electric motor have many non-linear influences of performance, even when ignoring their orientation.

Quote from: Tom Napier
A worked example:

One can increase the number of turns in a solenoid in two different ways.  One can use thinner wire so that the new solenoid has the same dimensions and mass as the old one.  Alternatively. one can add more turns of the same wire.  This gives a solenoid having a much larger diameter and mass than before.

That is the scenario, so we don't question this.

Quote from: Tom Napier
In both cases the current can be reduced in proportion to the increased number of turns and the magnetic field will remain the same.

Your continued reference to the magnetic field without distinguishing magnetic flux vs. magnetic field strength undermines your arguments for those who have received accurate education of the workings of electromagnetism, especially for those with an extensive background in modern physics.

Quote from: Tom Napier
Let us suppose that all solenoids are 10 inches high and have an internal diameter of 2 inches.  That is, H = 10 and P = 1.  We'll start with a reference solenoid which has an external diameter of 3 inches (Q = 1.5 inches) and which is wound with 0.1 inch wire.  (Roughly equivalent to 9 AWG wire.)  Obviously it has five layers of 100 turns each.  That is, there are 500 turns.  We are going to put 2 amps through it to get a field of 1000 A.turns.  (Just for the record, the central field intensity will be 49.5 gauss or 0.00495 Webers.)
The volume of the solenoid is 39.270 cuin and its mass is 12.57 lbs.  The length of the wire is 3927 inches and its resistance is 0.2474 ohms.  Putting 2 amps through the wire requires 4 x 0.2474 Watts, that is 0.9896 W.  That's our reference figure.

Case 1)
Keep the same solenoid dimensions and use thinner wire.
Let's use 0.025 inch square wire.  (Roughly equivalent to 21 AWG wire.)  Its side is a quarter of the previous wire so its area is a sixteenth.  Since the dimensions of the solenoid are the same the number of turns has gone up by a factor of 16, that is we now have 8000 turns.  To get the same magnetic field we only need a sixteenth the current or 0.125 Amps.  The volume of the wire hasn't changed so its length must have gone up by 16.  We now have 62832 inches of wire or almost exactly a mile.
The length has increased by 16 and the area has decreased by 16 so the resistance has increased by 256.  We now have 63.335 ohms with 0.125 amps flowing through it.  That's still 0.9896 watts or exactly as much power as before.
That is, changing the wire size has reduced the necessary current but has made no difference at all to the power.  What has changed is the voltage required.  In the original solenoid we needed less than half a volt to put enough current through the coil.  The new coil takes a sixteenth the current and hence sixteen times the voltage or 7.9 V to be exact.  In practice it is more convenient to drive a solenoid with 8 volts at 0.125 amps than it is to use a 2 A, 0.5 V supply.

As I explained before, this does not deal with the issue of output power at all.

Quote from: Tom Napier
Note particularly that Newman's prediction that using more turns would reduce the required power did not hold up.

His view (Newman's view) of the magnetic field is that of a potential energy source. Your view of it so far lacks any sense of:

1) how much torque there ought to exist between magnetic components relative to the power being delivered to the device
2) how the speed of the motor influences the amount of power drawn as well the subsequent torque

Quote from: Tom Napier
In his table he assumes that the voltage driving the coil is a constant 10 V and derives the power from 10 times the current.  As can be seen above, this is absurd.  If you applied 10 V to the reference solenoid it would pass about 40 amps and probably melt.

It follows from your belief about this motor that current would have enough time to reach its full value. The truth is that with a large multi-layer solenoid of great mass, one would have a high ratio between of inductance over resistance. The result is a large charging time for the coil, which according to the patent, allows for delays in the circuit as much as TWO SECONDS said to be proven when a light bulb was connected in SERIES with the device. Even then, the resistive voltage drop of the light bulb was comparatively insignificant relative to that of the coil, much less its inductive voltage drop! This means that the current in a very massive inductor must be delayed significantly before it can reach the 40 amps you speak of!

Quote from: Tom Napier
Case 2)
Let's keep the wire size the same but reduce the current to 0.125 A.  We still need 8000 turns to get the same magnetic field.  That is, at 100 turns per layer we need 80 layers of wire.  That's 8 inches worth, the solenoid is now 18 inches in diameter.  (2 inch central hole plus two 8 inch windings.)  Its volume is now 2513.27 cuin and its weight a whopping 804 lbs, 64 times the mass of the reference solenoid.  This is beginning to sound like a real Newman coil.
The wire is now 251327 inches long, that's almost 4 miles.  It has a resistance of 15.834 ohms.  If we put 0.125 A through it, it will dissipate 0.2474 Watts.  This is a quarter of the dissipation of the reference coil.

You have the same problem here, but with a coil better resembling the properties of a Newman motor. Yet you made no significant improvement in your arguments here.

Quote from: Tom Napier
A summary:
In case 1) where we went from 500 to 8000 turns but used the same mass of copper the power required was identical.  In case 2) we went from 500 to 8000 turns.  We used 64 times as much copper and reduced the power to a quarter.  This obviously contradicts Newman's contention that the power required for a given field falls directly in proportion to the number of turns or to the mass of copper used.  Thus both his Claim 2) and Claim 3) are incorrect.

Your measurements of the field are characterized by your reference to Webers as a measurement of "magnetic field intensity" instead of properly attributing it to "magnetic flux". Also, by not calculating output power, you make no attempt thus far to calculate the efficiency.

Quote from: Tom Napier
For completeness I should point out that a solenoid 18 inches in diameter requires a somewhat larger current to achieve the same magnetic field intensity as would one only 3 inches in diameter unless the length of the solenoid is increased in proportion.

Doing a final step like this is not a means of correcting errors since it is treated simply as a "cherry on the top". This is supposed to go with the rest of your calculations in a very rigorous manner.

Quote from: Tom Napier
A final confirmation:

I predicted that the power needed for a given field is:

808*F*F*H*(Q + P)     Watts.
-----------------
(Q - P)

I claimed that the field was 0.004949 Weber in a solenoid 10 inches high with an outside radius of 1.5 inches and an inside radius of 1 inch.  The above equation gives 0.990 Watts which agrees with the figure found above for both the 500 turn and the 8000 turn solenoids.

In the second case H and P remained at 10 and 1 inches but Q became 9 inches.  We calculate the power to be 0.247 Watts which also agrees with the result above.  If P is made negligibly small compared to Q then the minimum power required is 0.198 Watts, not zero as predicted by Newman.

Again, and again, the field you speak of has no mention of how much potential energy is stored or how quickly it is done so.

Quote from: Tom Napier
Appendix:

1] Volume of copper, V = H*pi*(Q*Q - P*P)  cubic inches

2] Length of wire, L = H*pi*(Q*Q - P*P)/T*T  inches

3] Resistance, R = 0.63*H*pi*(Q*Q - P*P)     ohms
---------------------
T*T*T*T*1,000,000

4] Number of turns, N = H*(Q - P)/T*T

5] Field, F = 0.00004949*N*I/H   Webers

You call this a formula for Webers, but this is for units of Teslas. To get Webers you have to integrate Telsas with respect to differential area. I'm not sure about the use of "0.00004949" either, which seems to stem from Maxwell's equations but appears simplified to make it understandable to college undergraduates.

Quote from: Tom Napier
combining 4] and 5] we get

6] Field, F = 0.00004949*I*(Q - P)/T*T   Webers

7] Current, I = F*T*T/0.00004949*(Q - P)  Amps

8] Power = I*I*R Watts

=                F*F*T*T*T*T*0.63*H*pi*(Q*Q - P*P)
-------------------------------------------------------
0.00004949*0.00004949*(Q - P)*(Q - P)*T*T*T*T*1,000,000

9] This reduces to:     808*F*F*H*(Q + P)     Watts
-----------------
(Q - P)

By adding a source of bound current (i.e. permanent magnets), the value for F would increase. From there it could be maintained without any additional power. Of course, like I have said many times to myself and others, "Magnets do not do any work.! The field must change over time!"

Quote from: Tom Napier
Newman -- point by point

Part 2, Commutators and coils

Introduction:

This document is one of a series which address specific errors in the book, "The Energy Machine of Joseph Newman."
This section examines the construction of Newman's "Energy Machine" and attempts to make sense of his performance figures.  It relates to the section of Newman's book which runs from page 60 to page 70.  (Page numbers refer to the 8th edition.)

The early models of the Newman Energy Machine consist of a rotating magnet placed near or inside an air-cored solenoid.  A commutator attached to the magnet shaft switches the current from the battery through the windings of the solenoid.
The principle of the motor is that the interaction of the field generated by the current flowing through the coil and the field of the permanent magnet causes the rotating part to make almost a half turn.  By reversing the direction of the current flow at the end of the first half turn one can cause the rotor to make a second half turn.  The current is then switched back to its original direction for the third half turn, and so on.  The commutator and brushes act as a reversing switch to change the current direction twice per turn.  This automatically causes a series of impulses which tend to turn the rotor in the same direction, resulting in continuous rotation.
This form of motor dates back to the earliest days of the electric motor.  It is still used today as a laboratory demonstration or a toy, though more commonly in the form in which a coil rotates within a fixed magnet.
The advantage of using a rotating coil is 1) the moving part can be much lighter since the heavy permanent magnet is stationary and 2) fewer brushes are needed since the coil being driven rotates with the commutator and can be directly wired to it.  The chief disadvantage of this simple motor is that its torque varies considerably throughout the cycle and drops to zero twice per cycle.  Should a motor stop near this dead-point it will not start again without being pushed.
For these reasons practical motors are built with many more windings on the armature and hence many more commutator segments.  This means that the active winding is always the one with the most torque being exerted on it.  This gives both a smoother and a stronger torque characteristic for a given input current.  Multi-pole motors always self-start.
One interesting feature of the motor is that it has a well defined maximum speed.  As the commutator switches on the battery voltage to the coil the current starts to rise at a rate which is inversely proportional to the coil's inductance.

This is the section where you mention inductance.

Quote from: Tom Napier
However, as the current increases the drop across the coil's resistance reduces the effective voltage and hence the rate of rise of the current.  As a result the current rises more slowly as time passes and, given long enough, reaches a fixed value V/R.  Normal motors have low inductance windings so the current only reaches the resistance limited value at very low motor speeds.  This is why DC motors take a large current when they start or are stalled.
However, at high motor speeds a second effect takes over.  The rotating magnet induces a voltage in the coil which opposes the input voltage.  At a high enough motor speed this induced voltage cancels out the driving voltage and the coil current drops to zero.  This is a good time to break the commutator connection.  Firstly, since the coil current is zero, no large voltage spikes will be induced.  Secondly, it stops the induced voltage driving a reverse current through the coil and slowing down the motor.

Newman's motor has a single stationary winding and a rotating two-pole magnet.  Since he uses very big and heavy coils this makes sense.  Because the coil is stationary his commutator has four brushes, two to carry the current in from the battery and two more to carry the current out to the coil.  His motor has two other odd features.  One is that the two halves of the commutator are not continuous, each is cut into 10 sets of three segments.  During each half turn of the magnet the coil is in turn powered, unpowered and short-circuited.  The other strange feature is that the coil has a much higher resistance and inductance than is commonly found in electric motors.  If the concepts of resistance and inductance are unfamiliar to you see the Appendix.

This section is a very good analysis thus far.

Quote from: Tom Napier
To make a magnetic field change rapidly, as you must do to reverse the field as the rotor turns, you must either apply a very high voltage or use a low inductance coil.  Newman has chosen to use a high voltage, everyone else wants motors which run from 12 volts or 110 volts so they use a lower inductance.
When you apply a voltage to a coil the current through it starts to rise.

It is also true that the density of charge in the conductor will rise with the increase in voltage. Capacitance in a wire of a given conducting substance also adds the same way as resistance. Two wires connected in series have more capacitance than two wires connected in parallel. Unlike electrical devices which contain a high capacitance per volume, wires capacitance differs by being relatively thin as opposed to flat. The capacitance is very dependent on the distribution of charges.

Quote from: Tom Napier
If neither the coil or the power source had any resistance the current would rise for as long as you applied the voltage.  In practice both the supply and the coil have a considerable resistance.  The rate of rise of the current, initially high, falls exponentially with time.  It eventually settles down at a constant value controlled only by the total resistance and the voltage.  If you cut off the voltage soon enough only the initial rise will occur.

No dispute is called for here.

Quote from: Tom Napier
Appendix 1.    Resistance:

All normal conductors have some resistance, that is, when current passes through them some electrical energy is converted into heat energy.  Except in special cases, such as water heaters, resistance is a bad thing.  When one wants simply to generate a magnetic field by passing a current through a coil of wire its resistance is a nuisance.  Without it no energy would be needed to maintain the field.

That that is one of the many purposes for using permanent magnets.

Quote from: Tom Napier
The power lost to heating the wire can be calculated from I*I*R watts where I is the current in amps and R is the resistance of the wire in ohms.  However, when the current is varying with time, as it is in a Newman motor, measuring or computing the total power lost becomes rather complicated.
A conventional current meter will only give the correct value for the current when the current is constant.  If the current is pulsing on and off the meter is probably going to give an incorrect value.  However, even if the meter does read the mean current correctly despite the pulses, this does not allow one to calculate the mean power!  Because the peak power depends on the square of the current the ratio of the mean power to the mean current depends on the length and shape of the pulses.

And it depends on resistance, which in this case is variable as a result of mechanical switching! This is not a friendly scenario for someone needing testing prior to licensed certification to manfacture motors!

Quote from: Tom Napier
As a simple example, suppose 1 amp is passing continuously through a 1 ohm resistor.  The mean current is 1 amp and the mean power is 1 watt.  Now pulse the current so that 10 amps flows for one tenth of the time and no current flows the rest of the time.  The mean current remains 1 amp.  However, during the pulse the peak power is 100 watts.  The other nine tenths of the time the power is zero.  That is, the mean power is 10 watts.  Even though the mean current is the same the input power has risen by a factor of ten.  The power being dissipated in the resistor has also gone up by a factor of ten, possibly making it dangerously hot.  Thus, the mean power cannot be calculated from the mean current.  This, unfortunately is what Newman and Hastings do regularly.

Tough luck!

Quote from: Tom Napier
Appendix 2.    Inductance:

When you change the current flowing in a coil of wire you also change the magnetic field surrounding it.  This changing field induces a voltage across the coil which acts against the source providing the input current.  The faster the current changes the higher the voltage must be applied to make it change.  The extent to which a coil resists a change in current is its inductance.  Inductance is simply defined as the ratio between the rate of current input change and the voltage required to make the current change.  If a coil's current increases at one amp per second when one volt is connected to it, it has an inductance of one henry.

Perfectly clear to my ear, go on.

Quote from: Tom Napier
Note that a pure inductor does not limit the current.  If the source could supply a constant one volt for any output current the current could increase at an amp per second for ever.  All the power supplied by the source would be stored in the magnetic field.  This power can be recovered when the field is reduced to zero again.
Of course in real life the resistance of the coil would limit the current to one volt divided by the coil resistance.  If this was 0.1 ohms no more than 10 amps would flow into the coil and the magnetic field would stop increasing at that point.  The stored energy is I*I*L/2 joules.  The coil had an inductance of 1 henry so it would be storing 50 joules or enough energy to light a 1 watt bulb for 50 seconds.
Unfortunately, input energy is needed to keep the current flowing through the resistance of the coil.  This energy would be I*I*R or 10 watts in this case.  That is, the resistance would be wasting as much energy as the inductance has stored every five seconds.  That's why inductors are not generally used to store energy except for very short periods.  For example, if we tried to keep the current flowing by shorting the ends of the coil all the stored energy would be turned into heat in tens of seconds.
If you suddenly stop the current flowing, for example, by opening a switch, the magnetic field collapses very quickly.  This generates a very large voltage across the ends of the coil.  This voltage can be high enough to create sparks across the switch contacts.  Not only is this damaging, it also wastes the energy which was stored in the coil.

Clearly so.

CONTINUED ***
Title: Re: Critical Response to Tom Napier's (1999) Critical Analysis of Newman Claims
Post by: kmarinas86 on January 08, 2009, 12:22:27 AM
http://www.phact.org/e/z/newmann.htm

PART THREE AND FOUR

Quote from: Tom Napier
Newman -- point by point

Part 3, Roger Hastings' errors

Introduction:

This document is one of a series which address specific errors in the book, "The Energy Machine of Joseph Newman."
This section examines the arguments used by Dr Roger Hastings to demonstrate an over-unity motor performance.  It relates to Chapter 5 of Newman's book which runs from page 22 to page 35.  (Page numbers refer to the 8th edition.)

In this chapter, written in 1982, Dr Roger Hastings attempts to show from measurements and calculations that Newman's motor has an efficiency much greater than unity.  He makes several elementary mistakes in his reasoning.

Driving the oil pump

In the first section Hastings compares the performance of a normal DC electric motor driving an oil pump with that of the Newman motor driving the same pump.  We are not told much about the normal motor except that it is claimed to have 80% efficiency and that, when driving the pump at 1 Hz, it consumed 2 amps at 12 volts.  From this Hastings deduces that it takes 19 watts (12 V times 2 A times 0.8) to drive the pump and therefore, if Newman's motor can drive it with less input power, its efficiency must be greater than 100%.
This calculation is completely incorrect.  The efficiency of an electric motor does not have a fixed value, it varies from its peak value to zero as the motor speed decreases.  A typical 12 volt motor would exhibit its rated efficiency at a speed in the 3000 to 8000 rpm range.  Unless the DC motor was geared down to drive the low speed load, and Hastings makes no suggestion that it was, it was running virtually stalled and had almost zero efficiency when running at 1 Hz (60 rpm).  Almost all of the 24 watt input would have been heating the motor windings.

Clearly!

Quote from: Tom Napier
From this we can deduce that it takes very little power to drive the oil pump at 1 Hz.  Newman's motor is inherently a low speed, high torque device.  It would have little difficulty in running at 60 rpm when driving a pump.  Based only on the battery life, Hastings estimates that the Newman motor is consuming 2.4 watts.  This is consistent with an efficiency of well under 100%.

All of this is normal of course.

Quote from: Tom Napier
Power to torque ratio

Hastings then reports on a comparison of the static torques of the two motors and uses this to compare the ratios for torque per watt of the two motors.  This comparison, taking the ratio of input power to output force, is dimensionally nonsensical.

Definitely, as far as a measurement of efficiency it is.

Quote from: Tom Napier
One can easily generate a static torque from two permanent magnets (or a spring) without using any input power at all.  Does this make the magnets infinitely efficient?
As mentioned above, Newman's motor generates a substantial torque when stationary.  Normal electric motors do not generate as much torque but they can continue to supply it when rotating at high speed.  Since output power is torque times rotation rate, a normal electric motor has a substantial power output at high speeds.  Newman's motor has a higher initial torque but is incapable of high speeds; it can manage perhaps five turns per second.  Thus the power output of the conventional motor is much higher than the power output of Newman's motor.  The faster a conventional motor runs the less input current it takes thus, as mentioned above, its efficiency rises with speed.

All motors run more efficiently at certain speeds. It's no question that motor as simple as Newman design would become more efficient at higher speeds. The speed in a Newman motor is limited by back-emf (BEMF) problems. The speed of conventional motors it limited by coil overheating. If these were not issues, then one would continue increasing the efficiency until they did so. This puts everyone on the same boat.

Quote from: Tom Napier
Starting power equals running power?

In his section C) Hastings makes a massive blunder.  He computes the energy required to spin up the Newman motor from the moment of inertia of the rotor.  This calculation is correct as far as it goes.  It shows that the motor requires an input of 13 watts to accelerate it from 0 to 6 Hz in 21 seconds.  Hastings then states, "This yields a minimum energy to keep the rotor spinning at 6 Hz of 13 watts.  Therefore the batteries must be supplying at least 13/70 = 190 mamps."

Which is complete bullshit, especially when coming from someone purported to be recognized as "the former senior physicist at Sperry-Univac".

Quote from: Tom Napier
This again is total nonsense.  All he has demonstrated is that the batteries had to supply 190 milliamps during the 21 seconds it took to spin up the motor.  Once the motor is spinning the power needed to keep it spinning could be quite low.

Like all motors would.

Quote from: Tom Napier
Hastings supposes that the 21 second initial battery drain continued during the entire ten hour test run.  He then expresses amazement that the batteries have lasted longer than the two hours predicted from the 190 milliamp current.  Thus is like wondering why your car battery doesn't run down during a 100 mile trip since the starter takes 100 amps.

Well said.

Quote from: Tom Napier
Motors and currents

In any normal DC motor the input current starts off high.  It is limited only by the resistance of the windings which is made low to avoid losses.  However, the current drain drops enormously as the motor gets up to speed because of the back EMF generated by its motion.  For example, a quick test on a Radio Shack 12 V motor running on 5 V (to avoid damage) showed a drain of 1.7 A when stationary, falling to 0.27 A when running with no load.  The induced EMF in the motor coils subtracts from the applied voltage, leaving less voltage to drive the current.  Of course, reducing the current also reduces the field and hence the motor torque.

Either way, to get to that speed in the first place, input voltage was required.

Quote from: Tom Napier
Since the current through each winding is being turned on and off by the commutator the inductance of the winding contributes to the reduced current drain.  This inductance makes the coil current take a finite time to rise to its static value.  If the commutator switched the current on and off rapidly enough the current would always remain low.  In practice the winding inductance is low enough for this not to be a significant effect.

First of all, inductance increases the amount of torque you have per current. If your motor demands a torque that increases with the square of the rpm, then doubling the current, which quadruples the torque, will allow you to double the rpm, increasing power eightfold. The counter-emf will increase proportional to the rpm and thus, the induction reduces the current by a proportional amount. Inductance is not so much a bad thing as long as higher voltages can be handled (see a good example of this at powertecmotors.com).

Quote from: Tom Napier
For example, a three pole motor from Radio Shack which is rated at 8300 rpm at 3 V has a coil inductance of 275 uH and a resistance of 0.77 ohms.  At 8300 rpm each pole is active for 2.4 mS.  However, 3 V applied to 275 uH gives a current rise rate of 11 amps per mS.  Even though the effective coil voltage is less than 3 V, due to the back EMF, it is apparent that it is the coil resistance which dominates the current.

How Newman's motor differs

Experiment with a small scale Newman-style motor showed that if the supply voltage is applied to the coil for a complete half rotation the current flowing is initially limited by the coil inductance.  At low speeds the coil resistance eventually limits the current to a fixed value.  However, as the speed increases the voltage induced in the coil by the rotating magnet becomes more important.  This voltage drives down the input current.  The resultant is a current pulse which rises slowly because of the coil inductance then falls slowly because of the induced voltage.  The motor settles down to a natural speed at which the coil current is zero when the commutator breaks it.  If you try to make the motor go faster, e.g. by changing the commutator phase and length, the induced coil current actually reverses, slowing the rotor.

Exactly the same thing I witness with the motor I designed.

Quote from: Tom Napier
Incidently, even when the coil current is quite small the voltage pulse across it as the commutator breaks the current can be hundreds of volts.  This pulse oscillates at the self-resonant frequency of the coil which is some 40 kHz in my case.

The commutator is critical. It is a makes or breaks the situation. That plus my inept construction skills is why I made the commutator on my motor easy to adjust since it has no base except a battery holder and connecting wire. This made it relatively quicker, yet still massively time consuming to find optimal modes of operation. But the improvements possible by this ad hoc setup prove that many who are keeping the commutator relatively stable can have losses which are orders of magnitude higher than need be simply due to one's inflexibility to change it!

Quote from: Tom Napier
In the Newman motors the commutator is segmented and it has coil shorting segments.  The latter allow the induced voltage to pass a current through the coil resistance, heating it and wasting input power.  This also acts as a brake on the rotor.
The time available to charge the coil is much smaller than the time constant given by its resistance and inductance so the mean running current is lower than the static current.  Twenty times per revolution the segmented commutator allows the field to build up then it shorts the coil so that the field collapses .  This stops the current and field ever reaching their static values but is wasteful since all the energy which was just put into the field is lost to heat.

Which is simply not true. If you put potential energy in the field, do you really think all of that will go to waste? Actually, the idea that the energy even develops in the field at all is a bit misguided. Each magnetic field line is actually a differential element of a conic surface that emanates from magnetic poles. Each of these conic surfaces correspond to different voltages, electrical potential energy per charge, which are connected by means mutual bridging (where band-like surfaces share the same potential energy per charge as the magnetic field cones they are fused to). Because magnetic field lines cannot cross each other, just as a point cannot have more than one level of electrical potential energy per charge, these cones are impossible to puncture and must exist for all of time.

In two-dimensional representations, it is hard to imagine how you get coupling between millions of magnets suspended in three-dimensional space to be mutually inter-connected with every magnet connected to every other magnet. However, the laws of physics require this to be so. The mutual bridging in question consists of differential areas located between infinitesimally seperated differential line elements each representing a certain degree of rotational displacement. If one particle rotates, then it affects the rotational displacement of the field lines of other particles. Individual bridges have no beginning or end as it is impossible for magnetic field lines to converge unless magnetic monopoles exist, the existence which is doubted by most.

Quote from: Tom Napier
Just before the coil is shorted there is a blank section on the commutator.  On reaching this section the field starts to collapse, generating a very high voltage across the coil.  This voltage generates sparks, Newman sometimes uses it to light fluorescent lamps.  However, all that is happening is that some of the energy which the battery supplied at a lower voltage is being returned at a much higher voltage.  (This effect is used in the fly-back transformer used in every TV set.)  The rest of the energy goes into heating the coil when the coil is shorted.  The whole process then begins again.
By the way, the fact that the magnetic field drops during rotation makes Hastings' computations of the motor torque misleading.

Agreed.

Quote from: Tom Napier
Some real numbers

To quantify this effect, assume the motor is turning at 300 rpm.  There are twenty commutator segments per turn, each consisting of 50% on, 30% open circuit and 20% short circuit.  Ignoring the width of the brushes, each segment makes contact for 5 mS.  Taking Hastings figures for the coil resistance (13 ohms), inductance (23 H) and applied voltage (200 V), we have an initial current rise of 8.7 amps per second.  Thus in 5 mS the current will rise to 43 mA.  The mean current will be a quarter of this.
Under static conditions the coil would pass some 15 amps if 200 V were applied from a low impedance source.  That is, it would be dissipating over 3 kilowatts.  Presumably  the battery resistance is limiting the voltage to some more reasonable value but this makes one doubt Hastings' figures for the power consumed during start-up.

Astute observation.

Quote from: Tom Napier
If one assumes the battery resistance is about 0.5 ohm per cell a 200 V battery would have a resistance of around 70 ohms, much higher than the coil resistance of 13 ohms.  Thus initially much more power is being dissipated inside the battery than is going into the motor.  (Total resistance 83 ohms, therefore current = 2.4 amps.  Battery dissipation = 406 watts, motor dissipation = 75 watts.  Motor efficiency < 15.7%.)

This says nothing about output power.

Quote from: Tom Napier
Running on empty

Hastings makes one curious statement, that "the old batteries have worn down to a point at which they will not even run a 1 1/2 V small toy motor."  It is not clear whether he applied the entire battery to the motor or just one or two cells.  Small toy motors take an amp or two to get started and old batteries just cannot supply that much current, their internal resistance is too high.  However it is not surprising that they can still supply the 100 milliamps or so needed to spin up Newman's high impedance motor.

Hastings makes a complicated estimate of the inherent motor efficiency and demonstrates that it cannot be higher than 56%.  He then says, "It is clear that measured efficencies for the Newman motor are far in excess of predicted efficiencies."  Why, then did he trouble to prove that its efficiency could not exceed 56%.  Does he not believe his own calculations?

Whose or what formulas did he implement? That's the real question.

Quote from: Tom Napier
Newman -- point by point

Part 4, Roger Hastings makes more errors

Introduction:

This document is one of a series which address specific errors in the book, "The Energy Machine of Joseph Newman."
This section examines further arguments used by Dr Roger Hastings to demonstrate an over-unity motor performance.  It relates to Chapter 6 of Newman's book which runs from page 36 to page 39.  (Page numbers refer to the 8th edition.)
In this chapter, written in 1984, Dr Roger Hastings attempts to show from measurements and calculations that a small version of Newman's motor has an efficiency much greater than unity.  He once again makes several elementary mistakes in his reasoning.

The performance

Hastings claims that the motor has an output "in excess of 10 watts" for an input of less than 0.5 watts.  If true this would be a remarkable achievement.  However there are extensive errors in the way inputs and outputs are measured and compared.  In particular, Hastings seems to be unaware of how powers and currents must be computed in pulse circuits.  His 10 watt output is not the mechanical output of the motor he is testing.  It is actually his erroneous estimate of the energy being wasted in the resistance of the solenoid.

Hastings is incompetent for his inability to realize that the measuring equipment as well as some "physical laws" are not valid in the cases in question. For example, Ohm's law (which requires a power factor of unity), among others are claimed to be laws of physics. Many of the ones taught in undergrad school are just approximation tools for engineers who either:

1) Lack full understanding Maxwell's equations, or
2) Lack time to apply them properly.

Physics courses in college are specifically tailored for engineers who deal with electrical equipment whose design and standardization is managed by government regulation. Phasors are not needed in Maxwell's theory, yet they are used in power engineering precisely because AC power is used in nearly all power utility lines. The limitation of knowledge to that required in practice limits our ability to understand the actually theory behind the laws themselves.

Hastings has abandoned Newman nonetheless, perhaps by realizing the inconvenience of doing the measurements right! You would think this would have got his name removed from the retirement fund records at Sperry-Univac (now Sperry Corporation)! People have asked again and again for verification of his former employment and failed to get it.

Quote from: Tom Napier
The motor configuration

The motor tested by Hastings consists of a solenoid about 13 inches high by 11 inches diameter.  This is wound with 30 gauge wire and is claimed to weigh 145 pounds.  Its resistance is claimed to be 50 kilohms and its inductance is claimed to be 16,000 H.  Sitting alongside this is a cylindrical magnet mounted on a bearing.  It has a commutator attached to its shaft which switches the current being delivered to the solenoid.  There is about a 11.5 inch separation between the center of the solenoid and the magnet bearing.  From the scale on the photograph the magnet appears to be 4 inches in diameter and 8 inches long.  It is alleged to weigh 14 pounds.  Elsewhere in Newman's book (page 65) there is a further reference to this magnet as weighing 14 pounds and having a 4 inch diameter.  There is a discrepancy here since a solid cylindrical magnet of these dimensions would weigh over 28 pounds, not the 14 pounds reported.

Agreed.

Do the math, and you find that if the magnet is 28 pounds, then its density should be 28 pounds / (pi (2 inches)^2 * 8 inches). That gives us in metric 7.71 grams per cubic centimeter. That sounds like iron which has about the same density as a ferrite magnet.

Quote from: Tom Napier
The motor was driven by a 304 V DC supply.  A passing reference makes it clear that this came from 9 V transistor radio batteries connected in series.  (In 1984 these would have been carbon-zinc batteries considerably inferior to today's alkaline batteries.)
The commutator is apparently 5 inches in diameter and must generate considerable friction.  It differs considerably from the simple two pole design which would normally be used in a rotating magnet motor to reverse the current flow twice per rotation.  In Newman's design the commutator contains twenty sections each comprising a through connection, an open circuit and a short across the coil.  This has several consequences, not the least of which is that it makes the motor's operation difficult to analyze!  The main consequence is that the current into the coil is interrupted before it has risen to its full value.  During the open-circuit section of the commutator the collapsing field generates a large voltage spike across the coil.  This is followed by a current pulse as the coil is short-circuited.  This series of operations can be categorized as energy input, high voltage output and energy dissipation.

You forgot state the obvious part: The rotor itself!

Quote from: Tom Napier
Each through section is effective for 0.0156 of a revolution, the open-circuit for 0.0094 and the short-circuit for 0.0063 of a revolution.  At the 136 turns per minute quoted by Hastings this corresponds to about 7.5 mS, 4.5 mS and 3 mS respectively.
It should also be noted that, as the brushes pass from the shorting position to the through position, a short-circuit path exists across the battery.  In the photographs on page 64 there appear to be gaps or insulating sections in the commutator at this point but it is not clear whether these are wider than the brushes, as they would have to be to avoid a short.  If there is a short-circuit condition a high current will flow briefly through the battery twenty times per turn.  This is not a high efficiency design.
Taking a somewhat naive approach it can be argued that the commutator applies the 304 V supply to the claimed 16000 H inductor for 7.5 mS twenty times per revolution and that the shorting sector resets the solenoid field to zero at the same rate.  The inductive rate of rise of the current would be 19 microamps per millisecond.  At the end of each through sector of the commutator the peak current would be 143 uA.  The mean current would be 71 uA for 31.25 % of each rotation, giving a mean current of 22 microamps.  This is significantly smaller than either the 1.2 milliamps or 14 milliamps  claimed by Hastings.

Hastings has it wrong, which is probably due to haste - iron - ically.

Quote from: Tom Napier
Hastings' test results

The photograph on page 37 shows the current waveform allegedly measured with a 1 ohm shunt in series with the solenoid while the commutator was turning.  On the timescale of the photo, 2 mS per division, the risetimes of the stepped waveform are invisible.  The fall times have slopes of 0.5 amps per mS.  The peak current is 1.5 A and the entire pulse occupies about 5 mS.
As they say in the puzzle books, "What is wrong with this picture?"  For a start, it cannot possibly represent the current flowing through an inductor.

Unless it has a capacitance.

Try to show me a material with any of the following:
1) Zero resistance
2) Zero inductance
3) Zero capacitance

Every material in existence has RESISTANCE, INDUCTANCE, AND CAPACITANCE.

Otherwise applying voltage would allow for:
1) Infinite velocity of charge
2) Infinite acceleration of charge
3) Infinite charge

Never mind the fact that DIVISION BY ZERO IS UNDEFINED!

All three are physical IMPOSSIBILITIES.

Quote from: Tom Napier
300 volts applied to 16000 H would cause the current to rise from zero at about 19 mA per second.  Neither the apparent virtually instantaneous rise nor the 500 amps/S fall can occur in a circuit containing a 16000 H inductor.  A substantial capacitance across the inductor would be required to pass currents having these rates of change.

The wire has a built in capacitance, while simultaneously having a inherent limit to the amount of current that can exist for a given voltage, and when wound into a coil, its inductance becomes "statistically significant". All of this shows the fallacy of "physical laws" which rely on the concept of an "ideal" circuit which does not exist in Maxwell's equations.

Quote from: Tom Napier
It is much more likely that the scope picture arises from the common-mode voltage generated across the commutator than from the differential-mode voltage across the 1 ohm shunt.

It arises from both with absolute certainty. What is not certain for me is the contribution of each as well as that of the interaction between them.

Quote from: Tom Napier
The schematic on page 37 shows shunts on both sides of the commutator with connections from both to the scope.  Unless a scope with a true differential input was being used meaningless results would likely arise with this configuration when a 300 V supply is being switched.

One odd aspect of Hastings reasoning is that he attempts to show that the battery is supplying a large current.  For example, he claims that the circuit breaker in an ammeter opened when the ammeter measured the input current.  This is supposed to confirm that large current pulses were flowing.  Since ammeters, if they have circuit breakers at all, tend to drop out only if an overload current persists for a second or so it is difficult to see what was happening here.  Of course, as mentioned above, the commutator may be applying a short-circuit to the battery.  The point of this test appears to be to show that at least 150 mA is flowing for long enough to trip the meter.
Hastings then measures the input current by measuring its heating effect in a 500 ohm resistor.  He measures a 1 degree C temperature change in 15 minutes and concludes that a mean power of 0.1 W is being dissipated.

What the hell?

Quote from: Tom Napier
I doubt if any competent scientist would have drawn such a conclusion from such a small temperature change unless he were using a very elaborate calorimeter but let that pass.

Even Blacklight Power has 1000x more competence than that.

Quote from: Tom Napier
Hastings then commits a blunder which crops up again and again in his writing.  He assumes that you can derive the average current in a resistor from its average power dissipation and thus derives an average input current of 14 mA.  This result is just not true.

The resistance of the system is variable, however, the power dissipated in the resistor is the power dissipated in the resistor. That does not mean all power that enters the resistor gets dissipated!

Quote from: Tom Napier
Let's take an example.  Suppose that the current through the resistor is flowing 30% of the time, not untypical of the current into a Newman motor.  If the resistor's average dissipation is 0.1 W it is dissipating 0.333 W when the current is on and, of course, 0 W the other 70% of the time.  To dissipate 0.333 W in a 500 ohm resistor requires a current of almost 26 mA.  Since this is on for 30% of the time the average current is 7.75 mA, much lower than the 14 mA claimed by Hastings.

The low power factor of the circuit makes it utterly vain to measure power output in a "resistive load". Anyone who is a master in electrical engineering should be able to quickly understand why.

The reason for the low power factor is that the positive (inductive) reactance and negative (capacitive) reactance. When the resistance increases due to the air gap opening, the air gap takes the majority of resistive voltage drop. The rest of the circuit loses its share of the applied voltage. Not only that, input apparent power goes down as quickly as the current does. The quick fall in current requires a negative inductive voltage drop. When the current becomes negative, only the capacitive voltage drop remains postive, but it has to be in excess of the applied voltage. That ratio of charge / capacitance must drastically increase. It is clear that it will, because applying a voltage to an open circuit can hardly increase the charge of the circuit, instead, the charge, if it can escape the battery, simply leaks slowly into the surrounding air.

The fact in the end is that the capacitive voltage drop IS the applied voltage. The proof in this is that capacitors can serve as batteries for a circuit. The voltage is like that in a battery, the only difference is the lack of chemical reactions to produce electrons, and you don't need chemical reactions to have an applied voltage!. It would have a negative reactance compared to inductance simply because it is on the other side of the equation. V_C = V_L + V_R ...

Quote from: Tom Napier
Just to complete the picture let's look at the average voltage across the resistor.  To pass 25.82 mA through 500 ohms takes 12.91 V.  Since this is only on for 30 % of the time the average voltage is 3.87 V.  If we applied DC we would have needed 14.14 mA at 7.07 V to get 100 mW.  To calculate the true average power we have to multiply the peak current by the peak voltage and then average over time.  Multiplying the average voltage by the average current makes it look as if we got 100 mW out for only 30 mW in.  Making this error has led a number of people to think they have got power for nothing.

It took me a few months to learn this stuff through the internet and experimentation. Too few people are willing to admit they made a mistake, and far less have the ability to validly debunk diluted teachings sold to poorly performing students as "laws of physics" that are in truth approximations and then only valid in limited circumstances.

Quote from: Tom Napier
Hastings then measures the DC input as 1.2 mA using a meter.  You would think that getting such values as 1.5 amps, 14 mA and 1.2 mA by different measurement methods would have led him to think again.  He then argues that the mean current must be lower than 14 mA since there is no apparent battery voltage drop after four hours; what then was the point of the measurement with the resistor which, erroneously, showed that the input current was 14 mA?

Hastings mentions that he has observed negative current spikes and that these cannot come from the battery.  Naturally, when you short out a huge inductor you get negative current spikes.  Why is Hastings surprised?  In any case, rotating a large magnet near a solenoid induces a reverse voltage and current.  The spikes disappear when the magnet is mounted on top of the coil instead of alongside it.  Again, why is Hastings surprised?

Not all engineers have skills created equally. That's absolutely certain.

Quote from: Tom Napier
The bottom line

Now comes the number juggling.  The 14 mA which was incorrectly computed as being the battery input current now magically becomes the current flowing in the coil.  The lowest of all the currents measured, 1.2 mA, is picked as the battery input current.  Since both numbers were mismeasured, why not use them the other way around?  Because that doesn't give the answer Hastings wants.
He argues that 1.2 mA times 304 V gives an input of 360 mW.  Luckily, since the battery voltage is fixed, multiplying the mean current by the voltage gives the correct answer in this case, assuming that the measurement of 1.2 mA means anything at all.
An average of 14 mA is flowing into the coil, says Hastings, and it has a 50 kilohm resistance.  That means that it is dissipating 10 watts.  Wow!  27 times the input power!  So what went wrong?
For a start, we are using a 304 V input.  As Hastings has already computed, at most that can drive 6 mA through the coil.  Where did that 14 mA come from?  Then, as I mention above, the battery is being applied only briefly to the coil.  Because of its huge inductance the current never rises to anything like 6 mA.  The peak current is 143 microamps, a factor of 10,000 lower than the peak current Hastings claims.  The power being dissipated is miniscule.

Power input = power dissipated + undissipated power

What exactly is the undissipated power then? Do you even grasp that hole in your reasoning?

Quote from: Tom Napier
The real efficiency

Of course there is another factor which Hastings and Newman completely overlook.  The efficiency of a motor is measured as the ratio between its output power in mechanical form to its electrical input power.

Energy is simultaneously mechanical and electric. Electrical potential energy consumption simultaneously becomes magnetic potential energy and kinetic energy, and that's probably because they are the same thing! Magnetic potential energy logically extends from Special Relativity and the same is true for Relativistically-corrected Kinetic Energy. Coincidence? I think not.

Sound (longitudinal waves) are based on the same thing. Rarefaction is conversion of stored energy to kinetic energy, and compression is conversion of kinetic energy to stored energy. The same thing when you give a blow to someone's chin, when somebody knocks on the door, and other kinds of things like explosions - even supernovae. It's all the same physics.

Quote from: Tom Napier
Ignoring the mechanical output and counting the resistive losses in the windings as "output" is not playing fair.  Still, if the loses truly were greater than the input power this would be remarkable in its own right.  One wonders why Hastings didn't measure the temperature rise of the solenoid to confirm that excess power really was being generated.  After all, 10 W input to 145 pounds of copper will heat it at almost 1.5 degrees C per hour.  After the four hour run mentioned in the report the temperature rise would have been significant.  How odd Hastings didn't notice it.

"How few are the people who understand the origin of energy. Alas, there is no such thing, except itself!" ....

Quote from: Tom Napier
# There will still be lots of scientists and fringe inventors still looking for cold fusion or over unity machine or space energy and the following is a great summary of current free energy claims. Before you get too excited about any of them, consider that none have made a serious effort to go for my \$10000 prize for proof of a free energy machine.
#

Another account: For those who believe Joe Newman has invented a free energy machine, let me give you the conclusion I came to after spending hundreds of hours experimenting with his ideas and studying his books. I was once very impressed after hearing about the motors Joe Newman claimed he had been building. Joe claimed his motors were able to produce more energy output than input. Most of the people I knew trained in physics laughed at me for believing that such a motor could be built, but having an open mind, and little previous knowledge on the subject of electric motors, I was determined to see if I could build a motor as Joe described. After reading his book over and over, and watching some of Joes video tapes, I could see Joes basic idea for building better electric motors was quite simple. One point Joe use to emphasize was that in his opinion, motor manufacturers have been misled into believing the electro magnetic fields from a coil of wire came from the amount of current passing through the coil, and that the wire it self had nothing to do with the strength of the field produced. and as a result, Joe claimed they have been building motors improperly with a minimal amount of wire, and much potential energy gets wasted because of this, Joes belief is that the magnetic fields produced in a coil of wire can be made much more efficient by increasing the amount of wire used in his stator coils, Ideally, Joe says the stator coil should have many miles of wire in them so as to create the maximum amount of atom alignment with in the coil when electricity is applied, and for the least amount of input current. In one of Joes video he does a demonstration which shows the difference in magnetic field strengths between a small coil and a large one to prove his point, and it is clearly evident there is some truth to what he says, but if you lack a firm understanding regarding the use of electro magnetism in motors, you will not realize there is a catch which he does not mention. I'm referring to the problems you encounter when you try and make use of this larger magnetic field in a motor. It is called counter electromotive force, or EMF for short.

It just means that your inductance will increase as a result of the interaction between magnetically interacting components. The problem is that this still reduces torque, but not always by the amount people assume. The current falls and so the fall of squared current is even greater. Unless the inductance here rises faster than the current falls, then this will reduce rather than increase the efficiency of your motor. HOWEVER the SINGLE way that counter-emf (CEMF) can improve efficiency is for magnetic flux to rise as the current falls, which can only occur in a motor that LACKS ANGULARLY SYMMETRIC INPUT while sufficiently incorporating BOUND CURRENTS.

Quote from: Tom Napier
I never understood what that term meant until I built my first scaled down model of Joes motor. What I discovered was that Joe was correct about more wire in a stator coil being an advantage over a small amount of wire, but only to a point, what Joe did not mention was the fact that as you increase the size of your coil, there comes a point where the addition of extra wire does not help produce greater power, not only because of the extra resistance from the wire, or the fact that as your coil becomes larger your wire becomes further away from the rotating magnetic armature, but because of an internal counter force which develops in a coil as the armature turns, and it works against your armature turning beyond a limited speed.

If you cannot increase the voltage, if friction and drag is not the issue, then you are stuck there and can't improve it. That's is the fact one has to put up with when building any motor.

Quote from: Tom Napier
I did not understand why it was when I added additional wire to my coil my motor kept turning slower. This went against everything Joe had taught in his video tape, and I was at a loss to understand what was happening. A gentleman with a good understanding in motor design explained this problem to me, He said this counter force I observed is a characteristic of any motor, and especially one with a large coil like mine. and this counter force would prevent me from ever building a motor which was able to produce more energy output than input, he told me because my motor was working much like a generator, as soon as the generated voltage reached the same level as my power supply voltage, this counter force would all ways keep me from seeing more energy output than input. This is called Lenz Law, a well known principle to motor builders

If Lenz's law didn't exist, all voltages would be infinite, which is obviously not the case. Because of Lenz's law, the values for voltages actually have meaning, but that does not mean you understand it by being told what it is!

Quote from: Tom Napier
...but I was still convinced Joe had something, so I built a special commutator which was shown in Joes book. It was a tough one to build on account of the fact that it was built in a way which could reverse the current to my coil every 180 degrees of rotation, and it also had to be made with special contact segments which when rotating would send short pulses of voltage from my batteries to my coil as the commutator turned, and in between each input pulse to my coil, one more segment was needed to short out my coil. Everything was made to very tight tolerances,

People always assume that this will help. Not in this case! Irregularity is your friend. That doesn't mean that motor should be full of dirt and crap. It helps to experiment without making the project such a deterministically achieved failure.

Quote from: Tom Napier
, but I still had no luck when it came to getting more energy out of my motor than I put in to it,

Energy is conserved in the universe, not power. Saying that power is a universal constant basically says that entropy in the universe increases at a constant rate. Every cosmologist worthy of a PhD can know constancy of rate change of entropy does not exist by looking at the evolution of galaxies! The same principle applies for anything that has the property of hysteresis. And that includes EVERYTHING!

Quote from: Tom Napier
and the short segment of my commutator made the motor run even slower. What really surprised me was when I attached a few more very strong rare earth magnets to my armature expecting it to turn much faster, and it slowed down again.

Nothing that is complete is clear cut. Ever. People who accept the fact a day not learning is a day wasted in futility may systemically overcome such traps, but not all of them. Such is everything in life!

Quote from: Tom Napier

"The proud and free always lend an ear to honesty. But something things just have to rest." ....

END OF THIS CRITIQUE ***
Title: Re: Critical Response to Tom Napier's (1999) Critical Analysis of Newman Claims
Post by: JustAnElectrician on January 31, 2009, 05:04:33 AM
;D
Kmarinas86, as I have not critically (or totally) examined your response to Napier for exact scientific scrutiny, (nor am I qualified to do so) I perceive
that you are more than a 23 year old "kid" (assuming your birth year is 1986) AND YOU HAVE SOME GOOD SCHOOLING!!! Remembering that
"classical EM theory" has many shortcomings (Heavisides equations integrate OUT any ZPE interaction, just to make the math work with closed systems)
you seem to know your stuff! I responded to your youtube vids, but you did not reply, (hard to filter out the haters!!!) so I availed myself to this opportunity. GREAT TO SEE that youth is still wide-eyed for truth! Keep it up, and realize that NEWMAN's theory only really works with MILES long coils or hyper-fast switching, which you are the best U-Tuber (longest coils) to try but you seem slightly jaded on his theory. You have seen his high-TORQUE low-amp input, that alone proves something in my humble opinion, and I have the classical training (EM) to validate it. KEEP FIGHTING the good fight, there is NO WAY we know everything about electromagnetics, peeps like you and me can uncover MORE!!!

-ROB GreenPower