I said in Patent No. 8606 of my own inventing that when the polar number ratio of stator to rotor is made to be 1 to 2 and the direction of electric force line against the rotation movement of rotor is made to be 90˚, the electrostatic induction generator can generate electricity without energy being supplied because the electric force does not prevent rotor from rotating (For more detailed explanation, refer to Patent No. 8606.). But later on the result of my further research and experiment on it showed that the direction of electric force line changes according to whether its electric resistance is strong or weak. In other words, when electricity flows well by making the electric resistance null or extremely weak, the direction of electric force line(Electric force lines are marked with two straight lines and one of them is named F.) becomes 90˚as illustrated in Figure 1. On the other hand, when electricity becomes difficult to flow by making the electric resistance strong, it causes an amount of positive(+) electricity which has not become electric current to remain on the polar plate(b), and then the attraction between the electricity and negative(-) electricity on the polar plate(B) makes the direction of F go over 90˚ as illustrated in Figure 2. The result is that the horizontal component(Fx) playing a role as a cause of input gets strong. Accordingly, in order to get larger output than input and generate electricity simultaneously with providing some of output to input, or in order to generate electricity without energy being provided from outside, it is necessary to make the direction of F close to 90˚ even in case of a strong resistance.

[Basic Principle]
  The basic principle of this invention is illustrated in Figure 3. It shows that the polar plate a and b in both sides of insulator(Y) are connected with resistance line R1 and the polar plate a' and b' are connected with resistance line R2. This is especially a complement to the basic principle of Patent No. 8606 by doing it twofold.


illustrated in Figure 3, when electricity is stored in the polar plate A and B by a D.C high voltage generator, (-) and (+) electricity are produced in a and b by electrostatic induction action. In that case, because A and B are completely insulated, it is assumed that (+) and (-) electricity of A and B once stored remain kept for a long time, and (+) and (-) electric force of A and B as marked with dotted line operate on a, a' and b, b' as long as A and B. If, like Figure 4, A and B in Figure 3 are pushed or pulled into the direction of → , (-) electricity will be produced in a because there will appear a part escaping from (+) electric attraction of A and (+) electricity will be produced in b because there will appear a part escaping from (-) electric attraction of B, while (-) and (+) electricity flow into R1 owing to their own attractions as soon as they escape from (+) and (-) electric attraction of A and B. In that case, it is ceratin that (+) electricity of A does not have its attraction action on the outside part of dotted line in a which (-) electricity isn`t left and (-) electricity of B also does not have its attraction action on the outside part of dotted line in b which (+) electricity isn`t left. And because there appear parts taken by (+) and (-) electric attraction of A and B in a' and b', (-) and (+) electricity are produced in a' , and (+) and (-) electricity are produced in b' by making electrostatic induction occur there. In that case, while (-) electricity of a' and (+) electricity of b' remain left as a result of being attracted by (+) and (-) electricity of A and B, (+) electricity of a' and (-) electricity of b' not affected by (+) and (-) electric attraction of A and B, (+) electricity of a' and (-) electricity of b' not influenced by (+) and (-) electric attraction of A and B flow into R2 owing to their own attractions. Accordingly, it is certain that (+) electricity of A does not have its counter-attraction action on the outside part of dotted line in a' which (+) electricity isn`t left and (-) electricity of B does not have its counter-attraction action on the outside part of dotted line in b' which (-) electricity isn`t left. Therefore, because the electric force line against the horizontal movement of A and B is toward 90˚,neither toward a and b nor toward a' and b', the following expressions are resulted.
        Fx = Fcos90˚= 0
         Fy = Fsin90˚= F                (cos90˚= 0 , sin90˚= 1)

   These expressions are the same as Patent No. 8606(Refer to pp.121~127 in Patent Official Report No. 483). But the expressions in Patent 8606 are made only when the resistance is null or weak, while the revised expressions are made even when the resistance is strong. In other words, like Figure 1, electricity flows well when the resistance is null or weak, whereas in case of Figure 3 electricity flows with its own electric attraction even when the resistance is strong. And because the electric current flowing into R1 and R2 does not arise from magnetic field, it does not disturb the horizontal movement of A and B. Therefore, as A and B need not work for electric force and magnetic force, the electrostatic induction generator operated by this principle can generate electricity without energy being provided.
But as a matter of fact, electric attraction gets weak due to the thickness of insulator(Y) and dielectric constant. As a result, the electricity which has not become electric current comes to be left and it makes the direction of F escape from 90˚. Accordingly, the thinner the thickness gets and the lower the dielectric constant gets, the better Y is. That is, as the thickness of Y is thinner and the dielectric constant of Y is lower, the direction of F gets closer to 90˚.
 
  Now, a formula for calculating output will be described.

   1.Formula for Calculating the Electrostatic Capacity of Electrostatic Induction Generator
Wherever B of Figure 1 and A and B of Figure 3 are located, this can be thought of as a condenser. So in order to calculate the capacity of electrostatic induction generator, the formula to calculate the capacity of condenser can directly be used. If B of Figure 1 is a stator, it is said to be a simple-system stator(Patent No. 8606) and if A and B of Figure 3 are stators, they are said to be plural-system stators(Patent applied for). But for the purpose of calculating the capacity, the simple system need not be distinguished from the plural system because both of them are the same. So by means of a easier simple system, B of Figure 1 which acts as a stator will be explained like Figure 5, and b and b' which act as rotors will be explained like Figure 6.


                                                                                                                                                                                                                                                                                                                                                                                 

         The effective area of rotor comes out by subtracting π r12and the area of insulated line(11) from πr22, while, as illustrated in Figure 5 and 6, the area of stator should be half as much as the area of rotor because it is half as much as the area of rotor. And if the number of rotor is n and then the capacity is calculated by the method to calculate the capacity of condenser, both sides might be considered to be 2n. But because the effective area of stator is a half of the effective area of rotor, the calculation should be n. And also if the polar number(It will be explained later on.) of stator is E, the number of insulated line(11) is 2E(The number of insulated line is the same as the polar number of rotor.). Thus, after the units are defined as follows,
     ① r2[cm] - distance from the center of rotor to external length of circle
     ② r1[cm] - distance from the center of rotor to internal length of circle
     ③ d [cm] - distance of stator and rotor
     ④ n - the number of rotor
     ⑤ 2E - the number of insulated line
     ⑥ α[cm] - width of insulated line
the capacity will be calculated by using the formula to calculate the capacity of condenser which is

C =1 ×S[F]
9×10114πd
In that case, the effective area should first be obtained. If the whole area of a rotor is S2 and the ineffective area is S1, the effective area S is as follows.

           S = S2 - S1 And,

           S1 = πr1 2 + 2E(r2 -r1)α

           S2 = πr22

So, the effective area of a rotor S is as follows.

          S = πr22 - {πr12 +2E(r2 -r1)α}

           = πr22 -πr12 - 2E(r2 - r1)α

           = π(r22 -r12) - 2E(r2 - r1)α

           = (r2 -r1) {π(r2 +r1)- 2Eα }

And also, because the number of rotor is n, the whole effective area of rotors(S) is as follows.

           nS = n(r2 - r1){π(r2 + r1)- 2Eα }


Therefore, the formula to calculate the capacity of electrostatic induction generator is as follows.

C = 1 × n(r2 - r 1) {π(r2 +r 1)- 2Eα } [F]
9×1011 4πd

2. The Rotation Movement of Rotor and Charge * Discharge of Electric Energy

    For the sake of convenience, after it was assumed that A and B in Figure 3 do horizontal movements, the principle was explained. But, in fact, because they do rotation movements, A and B are treated as stators, and a' and b' as rotors like Figure 7~11. And though slip rings are shown to be four different-sized concentric circles, R1 is expressed to seem connected directly with a, b and R2 connected directly with a', b' by omitting slip rings and brushes in Figure 8~11. And in case A and B are equal to a and b like Figure 7, the capacity is expressed as Cab, and in case A and B are equal to a' and b' like Figure 9, the capacity is expressed as Ca'b'. When (+) and (-) electricity are stored in A and B with D.C. high voltage generator, (-) and (+) electricity are produced in a and b owing to the action of electrostatic induction. And because A and B are completely insulated, it is assumed that (+) and (-) electricity once stored here are not changed and are preserved forever.

By means of the formula,

W    =    1CV2 [J]
2

electric energy of

Wab    =    1 CabV2 [J]
2

is stored in Figure 7. In that case, in order to make the voltage of electric energy
           V(Wab = 1/2 CabV2 [J]but the original voltage is V.), the voltage of D.C. high voltage generator which is authorized by A and B must be made to be 2V because Figure 3 is the same that two in Figure 1 are connected in series(Refer to series connection method of condenser) .

    Like in Figure 8, when a rotor is rotated to the direction of → , (-) electricity is produced in a escaping from (+) electric attraction of A and (+) electricity is produced in b escaping from (-) electric attraction of B. But these electricities flow into R1 by their own attraction as soon as they escape from (+) and (-) electric attraction of A and B. And because the part which took (+) electric attraction of A appears in a', electrostatic induction action takes place, and then (-) and (+) electricity are produced. And because the part which took (-) electric attraction of B appears in b', electrostatic induction operation occurs, and then (+) and (-) electricity are produced. In that case, (-) electric attraction of a' and (+) electric attraction of b' remain kept because they are attracted by (+) and (-) electric attraction of A and B, while (+) electric attraction of a' and (-) electric attraction of b' flow into R2 by their own attractions. Accordingly, like in Figure 9, the electric energy of

Wab = 1 CabV2 [J]
2

first stored during rotation until a' and b' are equal to A and B, and the electric energy of

Wa'b' = 1 Ca'b'V2 [J]
2

not taken by the electric attraction of A and B among electric energy newly produced by electrostatic induction action are stored. Therefore, the amount of electric energy flows to R1 and R2 while a rotor is half rotated is two times as much as the amount of electric energy first stored.

&When a rotor is rotated like in Figure 10, contrary to Figure 8, (-) electricity appears in a' escaping from (+) electric attraction of A and (+) electricity appears in b' escaping from (-) electric attraction of B. But these electricities flow into R2 by their own attraction as soon as they escape from (+) and (-) electric attraction of A and B. And because the part which took (+) electric attraction of A appears in a, electrostatic induction action occurs, and then (-) and (+) electricity are produced. And because the part which took (-) electric attraction of B appears in b, electrostatic induction action occurs, and then (+) and (-) electricity are produced. In that case, (-) electric attraction of a and (+) electric attraction of b remain kept because they are attracted by (+) and (-) electric attraction of A and B, while (+) electric attraction of a and (-) electric attraction of b flow into R1 by their own attractions. But in that case, because the pole is changed like in Figure 8, a and b' are (-) poles and b and a' are (+) poles. On the other hand, because a' and b are (-) poles, and b' and a are (+) poles like in Figure 10, the direction of electric current is opposite to Figure 8. Accordingly, the electric current of electrostatic induction generator is an alternating current.

When a rotor reaches the state like in Figure 11, the electric energy of

Wa'b' = 1 Ca'b'V2 [J]
2

in Figure 9 stored and the electric energy of

Wab = 1 CabV2 [J]
2

not taken by the electric attraction of A and B among electric energy produced newly by electrostatic induction action flow into R1 and R2, but the electric energy of

Wab = 1 CabV2 [J]
2

not taken by the electric attraction of A and B is stored. Therefore, the amount of electric energy which flows to R1 and R2 while a rotor is once rotated is

                                          2(Wab + Wa'b') = 2(1/2 CabV2 +1/2 Ca'b'V2 ) [J]

And for the sake of convenience, Cab and Ca'b' are distinguished, but because they are actually the same capacity, if Cab and Ca'b' are C,

                 2(Wab + Wa'b') = 2(1/2 CabV2 +1/2 Ca'b'V2 )

                                        = 2(1/2 CV2 +1/2 CV2 )

                                        = 2CV2 [J]

And because the amount of electric energy flowing into R1 and R2 is proportioned to the number of rotations, if the number of rotations per t second is N, the following expression becomes made.

                                            W = 2 CV2 Nt [J]

Like this, whenever a rotor is half rotated, a half of the electric energy newly produced by the stored electric energy and electrostatic induction action flows into external circuit, and the other half is repeatedly stored and at the same time electricity comes to flow, the direction of electric current changing. In that case, the type of wave motion gets rectangular. In other words, electric current is proportioned to the area of rotor, whenever a rotor is half rotated, the direction of electric current is changed. And also, because the area of rotor is linearly constant whenever changed again, it gets developed into a rectangle-wave alternating current.

But this is not a good form of wave. In order to get an applicable big output, the voltage should be strengthened to generate electricity. But it cannot be used as it is. So the voltage should be moderately lowered by using a transformer, in that case sine wave is the most efficient. Electric current is proportioned to the changing area of rotor. Like in Figure 12, either the pole of electric connector may be made rectangular(dotted line) and the pole of rotor may be made a form of sine, or the polar number may be heightened to the maximum because the higher the polar number gets, the closer it gets to sine wave. Accordingly, because the polar number is heightened to the maximum to make output high, the pole of rotor necessarily need not be made to be a form of sine.

3. Relation between Polar Number, Output and Electric Current

Polar number would mean that of stator and the rate of stator to rotor is 1 to 2.

   When a stator with one pole like in Figure 5 is combined with a rotor with two poles like in Figure 6, a simple rotor like in Figure 13 is made.

Let`s suppose that when a stator is rotated in the direction of → and for t second with a certain velocity, the changed areas become a[㎠] and a'[㎠](The unit of [㎠] will below be omitted.). And then when a stator of two poles is combined with a rotor with four poles after they made like Figure 14 with the same sizes as Figure 5 and 6, a simple generator like in Figure 15 is made.

And when a rotor is rotated in the direction of → and for t second at the same velocity with Figure 13, the changed areas become b, b', c and c'. Accordingly,

            a=a'=b=b'=c=c'

            2(a+a') = (b+b') + (c+c')

Like this, they are the same sizes, but when polar number becomes two, the changed area also becomes double and when polar number becomes three, the changed area also becomes three times ..... and when polar number becomes n, the changed area also becomes n. That is to say, the changing area of rotor becomes proportioned to polar number. Thus, because the electricity stored in the whole area of rotor does not flow into external circuit at a time, but whenever a rotor is rotated, the electricity of changing area flows into external circuit, electric current becomes increased with a certain size, proportioned to polar number. In other words, when polar number is heightened two, three, ...... n times, the capacity is increased two, three, ...... n times, or the velocity of rotation becomes two, three, ...... n times fast. Therefore, if the former formula of

          W = 2 CV2 Nt [J]

is multiplied by polar number(E),

the amount of electric energy flowing to outside while rotor is rotated for t second becomes

         W = 2 CV2 NEt [J]

When this is expressed by electric power(P),

P =W [W]
t

p= 2 CV2 NEt = 2 CV2 NE [W]
t

This is a basic formula to calculate the output of electrostatic induction generator.
   And this formula of output is broken down into electric voltage and electric current.

Because

            P=Vⅰ [W]

can be broken down into

            P=Vⅰ [W]

So when

            P = 2 CV2 NE [W]

is broken down into

            P = Vㆍ2 CVNE [W] ,

V is electric voltage and

            ⅰ = 2 CVNE [A]

is electric current.

Like this, it can be found that while sizes are the same, output and electric current are proportioned to polar number. This is already made public as No. 91-13655.

And because frequency (f) is the product of rotation no.(N) multiplied by polar no.(E), it becomes

f = NE / sec [Hz]

So the frequency of electrostatic induction generator is much higher than that of common alternating current(50/60[HZ]) because the number of rotations and that of poles are heightened to the maximum to heighten output.

But polar number should not be heightened without limit. Because when polar number is heightened, the number of insulated line(11) proportioned to polar number is also increased, and it means that the effective area is decreased as much. Thus, polar number should be fixed moderately.

  Now, the making of stator and rotor, and the way both of them are combined will be explained.

1.Stator

   A stator makes a rotor produce electrostatic induction, keeping a strong electric connection owing to the high voltage of D.C. high voltage generator. When electricity is generated into just one of (+) and (-) poles in D.C high voltage like B in Figure 1, it is called a simple-system stator(Patent No. 8606). And when electricity is generated into both of (+) and (-) poles like A and B of Figure 3, it is called a plural-system stator.

    After many insulated plates with 2mm thick are made quadrilateral, they are cut in half and the opposite angles of coupling holes(7, 8) are also cut to decompose and compose them freely. And after posting thin aluminum paper on it and with razor blade drawing lines on electrode(1), coupling wires(3, 4) and another coupling wires(5, 6) connecting all electrodes with coupling holes(7, 8) like a form of Figure 16, detach the aluminum paper from other parts except electrode(1), coupling wires(3, 4, 5, 6) and coupling holes(7, 8). And electrode, coupling wire and coupling hole are made on the back side just like on the front side, but all electrodes(1) on both front and back side should match each other and the same. And also, as high voltage is allowed in stator, insulation is very important. So an insulated tape was attached four times and after cutting it, a strong adhesive was posted on the edge.

But I made this easily. If a stator is practically made, this insulation alone is not enough. As long as there isn`t a perfect insulator, electricity flows to places(2) other than electrode(1) under high voltage. So the result is that the voltage of electrode becomes nearly the same as that of non-electrode and the output decreases much. So in order not to transmit electricity to non-electrode, as illustrated in Figure 16, it should be detached(2; dotted line part) or it should be sunken in.

2. Rotor

   A rotor makes electricity flow into outside, doing electrostatic induction action and rotation movement owing to a strong electric force of a stator.

    In order to make the electricity of a rotor flow into outside, the basic principle of Patent No. 8606(Figure 1) was supplemented like Figure 3, but this alone is not enough. Because electric attraction arising from both sides of a rotor gets weak and electricity does not flow well when the insulated material is thick and the dielectric constant is high. Thus, the thinner insulated material and the higher dielectric constant are good.

    Many epoxy plates 1[mm] thick and both sides coated with copper plate are made circular like Figure 17, and then axis holes(17) are drilled to make the axis of a rotor go through in the middle of them and many bolt holes(18) are drilled to make thin and long bolt go through. And then transparent tape is attached on them. After drawing lines with razor blade to make a form of electrode(d, e, f, g, h, I, j, k) like Figure 17, detach tape from non-electrode and draw lines with black enamel on places to become coupling wires(12, 13, 14a, 14a'), circular circuits(15, 16) and integrated electrodes(a, a'). And connect electrodes(d, f, h, j) with integrated electrode(a) and electrodes(e, g, l, k) with integrated electrode(a'). And use the same method on back side as front side. Then electrodes(d', f', h', j') are connected with integrated electrode(b) and electrodes (e', g', l', k') with integrated electrode(b'). In that case, like a rotor, d and d', e and e' ..... k and k' should match each other and the same.

                                                                                                                                                                                                                                                                              And small conducting holes(19a, 19a', 20b, 20b') are drilled in all integrated electrodes(a, a', b, b'), so it is easy to connect them with conducting wires. And when these rotors are put in solvent with chloride of second iron, copper is left in place with tape and enamel posted but the other copper are dissolved and the rotors like Figure 17 and 18 are made. And in order to prevent short-circuit, the surface of a rotor is thinly painted with clear rocker.

    After making many rotors in this way and dividing them into two, One is marked with C and the other with D. And then integrated electrodes of Group C are marked with ac, b'c, a'c, bc like Figure 19 and integrated electrodes of Group D are marked with aD, b'D, a'D, bD like Figure 20. After that, put small circular plates(26) to adjust space between C and D like Figure 21, and pass the axises of rotor(27) and long bolts(30) through axis holes(17) and many bolt holes(18) in order of C, D, ..... And at the same time after passing conducting wire through conducting holes(19a, 19a', 20b, 20b') of the same integrated electrodes(ac-ac, a'c-a'c ..... b'D-b'D), solder integrated electrode and conducting wire. In that case, the number of integrated electrodes, or conducting wires is eight(ac, b'c, a'c, bc, aD, b'D, a'D, bD) but they should be divided into two. As it is shown in the development figure of stator and rotor in Figure 22, bD, ac, b'c and a'D are (-) poles and aD, bc, b'D and a'c are (+) poles(It is A.C but is classified into (+) and (-) like D.C because it is changed into the same pole.). So when they are made into one, it leads to two. And finally after drawing these two conducting wires out of the last small circular plates(26') and connecting them with two slip rings(33, 34), make the insulated lines(11, 11') of rotor in a row and fasten both sides of the axises of rotor(27) and long bolts(30) with nuts(28, 29, 31). Now, the assembling of a rotor like Figure 21 is completed.

3. The Combination of Stator and Rotor

After assembling a rotor and a stator in this way, when both of them are combined, it becomes an electrostatic induction generator. Now, after assembling a stator, the way to combine it with a rotor will be described.

First, divide stators made like Figure 16 into two(A and B) and mark the bottom stator of A as A1, the top stator of A as A2, the bottom stator of B as B1, and the top stator of B as B2. And, as illustrated in Figure 23, fasten front and back supporters(39, 39') on prop(38) with acryl sticks(41, 41') 3[mm] thick and then fasten stators with adhesive in order of A1, B1, A1,.... on acryl sticks.

                                                                                                                           the assembled stator and fasten the bearing And then put the rotor on (43) of the axis of the rotor on supporters(39, 39') with nuts to make the rotor rotate well. And fasten acryl stick(42') standing on both sides with adhesive, pushing A2 into the grooves(9,10) of A1 and B2 into B1. And fasten the stator with acryl sticks diagonally so as to be crushed(Treat back side with the same way as front side). And as illustrated in Figure 24, pass conducting wire(44A1) through the coupling hole(8) of A1, conducting wire(44A2) through the coupling hole(7) of A2, conducting wire(44B1) through the coupling hole(8) of B1 and conducting wire(44B2) through the coupling hole(7) of B2, and connect 44A1 with 44A2 and 44B1 with 44B2. If so, all stators are connected with two conducting wires. When generating electricity, connect these two conducting wires with (+) and (-) poles of D.C. high voltage generating device.(A simple-system stator is one that connects all of the stators with one conducting wire and then connects the conducting wire with either of (+) and (-) poles of D.C. high voltage generating device.). And finally after contacting brush(40) with slip rings(33, 34), fasten it with supporter(39). Now, an electrostatic induction generator like Figure 23 is completed.

As mentioned above, I explained the making of an electrostatic induction generator. But this is not practical but is made for the purpose of experimenting. So in order to put it into practice, the selection of materials and the method of making need doing differently.



This is a picture shot to experiment in lighting an incandescent electric lamp with a generator of a simple-system stator of my own making. This experiment only is to prove the basic principle. When occasions come, I will do further experiment.


   

       This is the explanation of the experimental processes of Perpetual Mobile (Electrostatic Induction Generator) and their result.
      I myself made a generator on these conditions for the purpose of experimenting the Electostatic Induction Generator.

    ① R2 - - - - - - - - 12.6[cm]

    ② R1 - - - - - - - -   8.4[cm]

    ③ d - - - - - - - - 0.094[cm]

    ④ n - - - - - - - - 47 sheets

    ⑤ 2E - - - - - - - - 48poles

    ⑥ a- - - - - - - - 0.1[cm]

  When the capacity for it is calculated,                          

C = 1.13 × 10-8[F]

comes out (See the calculation formula of electrostatic capacity in the Electrostatic Induction Generator.)
   When the weather is dry(Humidity: 50% below), raise ome A.C. generator to about 20,000[V] and rectified voltage to the stator for 0.3 to 0.5 seconds and then detach the diode from the stator.
In that case, the voltage falls sharply ; nonetheless, a relative high voltage remains in the stator for quite a long time. The experiments are at the level of a degee to prove the pinciple of the Perpetual Mobile because I couldn't generate a very high voltage.

▒▒ TEST 1 ▒▒
InputOutputRemarks
Voltage28.7[V]Voltage First Electric Current 76[mA]
Electric Current1.91[A]Electric Current The Number of Rotation 25/sec
Electric Power54.8[W]Electric Power  


▒▒ TEST 2 ▒▒
InputOutputRemarks
Voltage25.6[V]Voltage47.1[V]First Electric Current54[mA]
Electric Current3.08[A]Electric Current0.56[A]The Number of Rotation 20.6/sec
Electric Power78.8[W]Electric Power26.6[W] 

▒▒ TEST 3 ▒▒
InputOutputRemarks
Voltage25[V] Voltage78.5[V]First Electric Current41[mA]
Electric Current3.4[A]Electric Current0.43[A] The Number of Rotation 19/sec
Electric Power85[W] Electric Power33.7[W] 

   Test 1 shows the result that high voltage is added to a stator. And the input is got by measuring the electric current and voltage of D.C. motor.
   Test 2 shows the result that the electric current and voltage of the second side was measured after supplying the first side of a strong transformer with the output(electric power) of generator. Owing to impedance(resistance) of transformer, while the first electric current and rotation no. was more decreased than Test 1, the input was increased.
   Test 3 is the result shown when an incandescence electric lamp is just turned on(Refer to photo.). Because of the resistance of transformer and the incandescence electric lamp, the first electric current was more decreased than Test 1 and 2 and the second electric current(0.43[A]) was also more decreased than the second electric current(0.56[A]) of Test 2, while the input was more increased than Test 1 and 2.

   Thus, Judging from Test 1, 2 and 3, it is found that when electric current increases, the input decreases and when electric current decreases, the input increases. It means that input changes in accordance with the increase and decrease of electric current and it is right opposite to the case of electromagnetic induction generator. The cause is that when electric current is increased, electricity as little as the increase of electric current is left in rotor and thus horizontal component to cause input becomes weak, on the contrary, when electric current is decreased, electricity as much as the decrease of electric current is left in rotor and thus the electricity produces a strong horizontal component(Fx of Figure 2).

   Accordingly, in order to more output than input, the amount of electricity left in rotor should be diminished to the minimum by increasing electric current to the maximum in case of strong resistance. So a plural-system stator is needed to solve the problem.    

According to the experiment result above, it seems that the principle of the Perpetual Mobile has proved. The reason the output was small is that it was generated by the remaiing voltage. In other words, I couldn't generate enough voltage to make a large amount of output.
If one wants to get more than tens of theousands of volt of voltage one will practically have to give the stator a high voltage continuously with D.C. high voltage generating device. Thus, let's suppose that the output is calculated on the assumption that the stator is given the voltage of 104, 2 × 104[V] and 3 × 104.
    And suppose that the capacity is C = 10-8 [F] to make the calculation simple although the original capacity is C = 1.13 × 10-8[F]. And then, put the voltage of 104[V], 2 × 104[V] and 3 × 104[V] on

    ①C = 10-8 [F]

    ② N = 5 × 10/sec (at a vacuum sate)

    ③E = 102 pole ( I got 24 pole because I did't make a precise device. I thinks that more than 102 can be acquired if a previse device is made.), ad one gets the output of

                          P = 10[KW], 40[KW] and 90[KW]

through the output formula of

                          P = 2CV2NE[W].

And also, when voltage can be raised even to hundreds of thousands of voit, one can guess how much the output of the Electostatic Induction Generator will be and how much a deree the voltage in output will take.
         Like this, one can get a very high output by raising voltage. In that case, in order to raise voltage, electric power in principle isn't expended. In fact, a small amount of electric power leaks out. But such electric power loss can be ignored because a high output is produced.
    When one generates electricity with this high voltage, one may doubt that it is dangerous. But there is no worry about it. After voltage is raised to generate electricity, the raised voltage can be lowered by transformer.

Common Use and Its Pervasive Effect

So far, I have explained the basic principle of the Electrostatic Induction Generator, and then I will talk about the common use of the Electrostatic Induction Generator and its pervasive effect. We are now faced with critical crises throughout the world, such as energy problem, air pollution, radioactive contamination, noise pollution, acid rain, and unusual climate. In order to work out these problems, scientists and inventors all over the world have been engaged in research on nuclear fusion, solar heat, wind force, etc. But they didn`t get satisfactory accomplishments yet. In fact, the electronic induction generator requires a large amount of energy. It is too difficult to put nuclear fusion into common use, and the natural energy like solar heat and wind force has a low density and it is very changeable according to season, and day and night. But, the Electrostatic Induction Generator unlike the electronic induction generator can produce a very strong output without energy being provided from outside. So, I think it can be a alternative to the present problems humanity is faced with. Now, I will be more specific about it by taking a car into an example. All kinds of pollutants which cars emit and their noise are very harmful to living things including people. So, scientists invented electric cars to solve such problems, but they didn`t come into common use yet. Electricity should be charged into storage battery to operate them. But, the driving distance by one charge is about 100 to 120km, and also it takes about five hours to charge it. In addition to that, it is necessary to build many power plants and thus they require a lot of energy. However, the Electrostatic Induction Generator can address such problems as those of cars very simply. What`s better, it can be applied to airplane, ship, power plant, etc. Therefore, when it comes into common use, its pervasive effect will be enormous. Besides that, whatever has a good insulation and doesn`t produce gas in vacuum can be used as material of generator. And, it is easy to make it because its structure is simple. Even when a short circuit or leaking occurs, there is a very little chance of bringing about a fire and elctrocution. Because the electric current of the Electrostatic Induction Generator is limited to

i = 2CVNE [A]

so that it doesn`t increase any longer even in case of a short circuit or leaking. In that sense, when the Electrostatic Induction Generator is practically used, it is clear that its pervasive effect will be tremendous. By the way, the inventor said that its material must not produce gas in vacuum. Why is that? The electric current of the Electrostatic Induction Generator is relatively weak, and also the weak electric current flows into each pole. So, the force of the electric current is close nearly to zero. Using such a character, one has to generate electricity with the Electrostatic Induction Generator. I will explain it more specifically by taking a generator of my own making into an example. The electrode of one side of generator is 48, so that the electrode of both sides is 96. And the number of rotors is 47. So, the total number of poles in rotor is

4,512(96×47)

And also, the first electric current(the electric current of generator) in Experiment 1 is 76[mA]. Finally,

76[mA]÷4,512 = 0.0168[mA].

Thus, the electric current flowing in one pole of rotor becomes nearly zero. And the resistance of electrode is also almost zero because the electrode of rotor is thin and not long, and it is a wide and flat board. In that case, heat cannot be produced. Even though the Electrostatic Induction Generator is made precisely and then voltage is raised, the electric current won`t increase as much as a strong heat will be produced owing to the features of it. Such a weak heat will naturally be cooled without using a special cooling device. That`s why the Electrostatic Induction Generator is generated in vacuum. The occurrence of gas in vacuum decreases the degree of vacuum, so that the material without having any gas in vacuum is necessary. and As the air resistance is almost null under the state of vacuum, the input decreases to a minimum and the output increases owing to the number of increased rotations. Thus, Operating the Electrostatic Induction Generator under the state of vacuum is very effective. The better the state of vacuum becomes, the higher the effect becomes.

   As mentioned above, I, both the inventor and applicant, explained an electrostatic induction generator. I announced in the application that electrostatic induction generator can generate electricity without energy being provided from outside because larger output than input is obtained and thus some of the output is provided to input. I don`t think most of people agree with me because it contradicts the law of physics. But the supply of energy in electrostatic induction generator only is concerned with the direction of electric force line and it has nothing to do with anything else. So I wish people will think about whether the basic principle that the direction of F becomes 90˚ in case of following Figure 1 and 3 is valid or not, prior to their objection to my view.