rexresearch1.com
Jean LAFFORGUE
Field Propulsion Thruster
Field Propulsion Thruster
https://mail.linas.org/mirrors/jnaudin.free.fr/2002.06.03/lfpt/html/lelpv1.htm
The Lafforgue's Electrostatic Pressure Experiment
by Jean-Louis Naudin
On January 23, 2002, I have tested successfully the Jean-Claude Lafforgue's basic experiment about the Electrostatic Pressure. An asymmetrical electrostatic pressure is the main principle used in his patent for producing a thrust Vs external referential ( the Universe ). Lafforgue has called this the " Action Force " or the " Expansion Force ". The setup of this electrostatic pressure experiment is fully described in his patent FR2651388 page 35...
http://jnaudin.free.fr/lfpt/html/lfptv1.htm
The Lafforgue's Field Propulsion Thruster LFPT v1.0
Built and tested by Jean-Louis Naudin
On January 20, 2002, I have replicated and tested successfully the Basic Lafforgue's Field Propulsion Thruster fully described in his patent FR2651388 ( see the photo below ).
Tested apparatus description :
I have used the same design as described in the Lafforgue's patent at the fig7 page 48. The asymmetrical capacitor is built with 3 aluminum armatures ( 0.5 mm thick ). These armatures have the exact shape of the Lafforgue's capacitor described in the Fig 7 of his patent. The entire capacitor is fully enclosed with epoxy resin and sandwiched between two mylar sheets so that there is no contact with the surrounding medium ( the air )...
http://linas.org/mirrors/jnaudin.free.fr/2002.06.03/lfpt/html/lfptorg.htm
The Lafforgue's Field Propulsion Thruster Calculator
By Jean-Louis Naudin
The electrostatic forces are "generated" by the interaction of polarised atoms (q) on an electric vector field (E), i.e. F=q.E, positive or negative according to the sign of (q). By suitable geometry of the polarised armatures, and/or by the use of electric charges induced by induction effect, and/or by the use of suitably arranged dielectrics, it becomes possible to break the symmetry of the expansion forces on one of the axes of the three-axis reference system and thus to obtain a non-zero resultant force (Fn). The isolated system is then propelled, and it draws the charges (q) and (E) along with it, thus F=q.E remains constant. This force can be used as a means of propulsion for any vehicle or even as motive force for producing energy.
Machine Translation :
Electrostatic forces are "generated" by interaction of polarized atoms (q) on an electric vector field (E), i.e. F=q.
E positive or negative depending on the sign of (q).
By an adapted geometry of the polarized armatures, and/or by the use of electric charges induced by influence effect, and/or by the use of suitably arranged dielectrics, it becomes possible to break the symmetry of the expansion forces on one of the axes of the reference trihedron and thus obtain a non-zero resulting force (F). The isolated system is then propelled and it carries with it the charges (q) and (E), therefore F=q. E remains constant. This force can be used as a means of propulsion for any machine or as a driving force to produce energy.
Electrostatics, in general education, is often studied in a summary manner, as a preamble to the much more important and detailed courses which relate to electrodynamics.
In fact, electrostatics is an extremely complex science which calls upon the laws of classical physics (macroscopic domain), quantum mechanics (electron, hole, orbital, etc.) and relativistic physics since any vector field has the property of distorting the space-time continuum in which it circulates. Until now, electrostatics has been considered as a science relating to polarized conductors which are in equilibrium as a result of the symmetry of action-reaction forces. The author of this patent of invention will endeavor to demonstrate that the free equilibrium between polarized material subsystems, constituting an isolated system, is in fact only a particular case. j)5 The present patent proposes some examples of realization, non-limiting, which make it possible to break this symmetry by: - an adapted geometry of the polarized surfaces, - the use of suitably arranged dielectrics, - the introduction of induced charges, 20 -a judicious distribution of the potential between (î-tf-t-A) polarized armatures.
All the proposals made are carried out in a vacuum, requiring almost perfect machining of the armatures and the use of flawless dielectrics. 25 2To differentiate the text and the symbols, the latter will be indicated in parentheses when the clarity of the summary makes it necessary. The text is written using verbs in the first person plural, as is customary in any mathematical demonstration. 30The chosen observers are equipped with a clock which measures time (tjja vector space, a Euclidean affine space and a frame of reference. The frame of reference linked to the space-time continuum will be considered Galilean and will be called ( f\, ) We will use the following symbols: 55 - £: coefficient of absolute permittivity - £,: coefficient of relative permittivity of the vacuum -: coefficient of relative permittivity of a dielectricelectric vector field (or electrostatic)-AV-or -P '—^jpotential potential difference electric charge surface volume volume density of charge surface density of charge capacity of a capacitor energy, work pressure (energy per unit of volume) variation between the initial (<*•) and final state ((3) unit vector on the normal raised on the surface element unit vectors of the frame of referenceoc, ^ force A an element (armature, dielectric, continuum.. .) will be attached a frame of reference (p.). A force which acts between 2 elements will then be represented by the sym= "bole f (fVR-tO which means that (£) is exerted between the frames of reference (^and îS- i»), and that it is oriented from the first () towards the second (R^) Thusf (6URk) s(>1 KO With the symbol (F) we will associate the indices (i) and (j) which correspond= pond to the unit vectors of the axes (ox) and (oy) according to the direction of (f) relative to the Galilean frame of reference () attached to the space-time continuum, i.e. for example f^^jCLO where the force (f) is supported by the axis (ox) of unit vector (i) and it is directed by (R-«. ) to (R-t).
A force which acts on 2 material frames of reference (C^l^t) will be called a connecting force. A force which acts between a material frame of reference (R-^) and the frame of reference linked to the space-time continuum (R,) will be called an “action” force. 5/; The author considers that all the laws of electrostatics are known to the reader. It will also limit recalls to the force experienced by a driver. The energy of a distribution of charges is equal to V r ^ g qy^ where the charges ele narrow s ( ^ ) are brought to the potential that of a capacitor is equal to Wii-3-where. (G) measures capacity.
The pressure (P^ in the thermodynamic sense of the term, is equal to the energy per unit of volume, i.e. (-with (e Jdistance between the surface almatuies S) with c * £ S, (plane capacitor) ande-S . *L* ^'•^ . <rV-^ with ^<. ç; ( planou- capacitor, - ^ therefore "Pc <r, f err " z We then have fP.S. = £ê'S and, taking the unit vector at the -»normal (n) raised on the surface (s), we have (refer to figure 1) It is extremely important to note that this force (called expansion force) is of the action force type. , i.e. ^(R«..(o) since it is exerted between an armature (du,) and the space-time continuum (R.,,) which contains It is also very important to note that the vector ( p) is always supported by the normal (n) to the polarized surface element.
If a dielectric, with a relative parttivitity coefficient (£t) touches the conductor, the force is multiplied by () if the potential (y) is kept constant. 4; Without involving the depolarizing field which reigns within the die, we can say schematically that a dielectric, for what interests us: 25-which has an external vector field (g), collinear with the surface of separation between the dielectric (&£*.) and the vacuum (/o) (which must also be considered as a dielectric) undergoes a surface force (see figure 2) 50oriented along the normal (n) to the element (d) and directed towards the vacuum, that is to say towards the lowest coefficient of permittivity. -which undergoes a gradient of ( <? ), generates within it a volume force (see figure 3)35 which can be reduced to a surface force (it being understood that J fsoitzT '"" which we reduce to (ct-S) where the field electrical is constant.
Electrostatics presents a double interest, because of the expansion forces which are of the "action force" type (since resulting from the in= teraction of polarized scalar quantities on a vector field) and because it allows man to model these forces as he sees fit, both in module and in direction. In mechanics, the forces which act between 2 material systems (V) and (?
>k), are connecting forces which act between 2 centers of gravity. (see figure 4). 10The forces '? (&«. IfU) and f^tlRo-) are supported by the same line have the same module and an opposite direction, therefore In electrostatics on the other hand, the expansion forces: -result from an interaction between the charges, electric (<r) por * 15ted by the surface element) (see figure 5) and the electric vector field (), i.e. F* I. «"«k, it is therefore an acting force, - are always carried by the normal (n) to the surface element (is) - As soon as the field line is not rectilinear but curved, it appears that the resulting force is different from zero, as shown schematically in Figure 5. Finally, it should be noted, contrary to popular belief, that these forces of expansion can reach astronomical figures. Here are some examples of implementation. They are not restrictive. 25 Their only interest is to show different models making it possible to break the symmetry, that is to say to obtain a resulting expansion force different from zero. g.
Refer to figure 6 50 6-1 /]^ £3cri£tion: Consider a parallelepiped (marker 1) in which we fix: -on the lateral faces, 2 metal frames (marker 2) connected to the negative pole of a generator, - on the lower face, a central metal armature (mark $>) 55 connected to the positive pole of the same generator. We give to the upper end of the positive armature, a profile called "constant field", that is to say that the vector field t = — is identical at every point of the armature, whether in its part flat or curved. This profile is obtained using the theory of conformal transformations.
The parametric equation of this profile is: =£i*\ (9) The positive reinforcement (reference 3) constitutes a "thick blade" with a thickness of 1' spacing between the central reinforcement (positive) and each lateral reinforcement ( negative) being equal to (), with (e*-k'5.) since10Naturally we must not give the same profile to the upper part and the lower part of the central frame, otherwise the resulting force would be zero.
It is appropriate to adopt a geometry which generates a different force at these two ends. 15There are several solutions for this. Let's examine the one that consists of giving (see figure 6): - at the upper part of the lateral frames, a quarter-circle shape (convex part), - at the lower part of each side of the central frame, a 20in shape quarter circle (concave part), while maintaining the spacing (). The available volume between the armatures is filled with a solid, homogeneous dielectric having a high (). If the parallelepiped is made with the same dielectric, we obtain a block of dielectric 25 in which the reinforcements form inclusions. Let's charge the capacitor then isolate the generator armatures. To facilitate the identification of the forces, we will take a reference = 0; ) of unit vectors (i) and (j) and we will divide the isolated system into 5 zones comprising (see figure 7): 30-(A), the upper curved part. -(B), the flat part of the central frame -(G), the lower curved part •6-2/Id£ntification_àes_forces^ of expansion6-2-1 /_E3a35Let us consider (refer to figure 8) a surface element ( S„ ) of the central frame, small enough to assimilate it to a pottion of plan.
Let us project ( -£, ) onto the axis (ox ), i.e. S, Let us consider the electrostatic pressure (?
) which is exerted on (S), Let us express the product?«. Sk Let f. . S«*? f S, Cto» 0 = Cf. o>8) ^, f^ Sso ÇyS, s Tc.S^ Likewise?,. S^ = ? o C SSi. *) --S*k *} S?«. Sso*o .^a. * ^'^OOr f-« ^. St= ? ^. So^o-^^. tfc ^s^T.-S-10Thus, knowing (6) (let's not forget that the electric field is constant) we know that (then (? „*£»£,. ^) and (S»^) is none other than the master couple of the central armature, i.e. (2.e-), therefore —» - -As for the component ("^V) it has the same module but a opposite direction on the 2 opposite faces of the positive armature, therefore their resultant is zero. The field lines are always emitted perpendicular to the surface element. The expansion forces which are generated on the negative reinforcements (reference 2) are therefore in the plane (ox) and they have the same module 20 but in an opposite direction on each negative reinforcement, therefore their resultant is zero (see figure 7) . Thus ^=^=^ •£*£*&,6-2-2/ En_zone ^B^s 25All the forces are contained in the plane (ox), they have the same modulus but are in opposite directions between the positive and negative, so their resultant is zero. é_2-3/_In zone_(ç)_j 30on each side of the isolated system, the positive and negative armatures form a quarter of a cylindrical capacitor.
Let's see right away that we can give the curvature we want (see figures 9 and 10) to the profile of the reinforcements. Let us call (^^) the interior radius of the negative reinforcement and (R) the exterior radius of the positive reinforcement. We always have &p - P* •+--The capacity of each quarter of a cylindrical capacitor is equal to The electric charge (q) located on each armature is equal to c]= c.v But we immediately guess that the density of the surface charges( •*• ) will be greater on the smallest surface, that is to say that (tf^) will be greater than (o-xh i.e. The expansion forces will then be equal to: -on the negative reinforcement : -on the positive armature:The expansion force which is exerted on the small radius (rt^) is therefore greater than that which is exerted on the large radius (6. ^ ). These forces have two components (Fj) and (fy). The components (F^) cancel out 2 to 2 as a result of the inverse curvature of the quarters of a circle between the 2 sides of the isolated system.
"" Only the vertical components (F^) remain which we know are equal to: ^^£,£,. TT £.(3- hence a resultant force: and since Q=C.Vwith c^ iC^^^j / iig d' ^^a) and since there are "2 sides in a quarter circle," we let's get 6-2-4/ Resulting force of expn: It is equal to F^ (zone A) + F(zone C) i.e.: Given that (^p) is greater than (£<. ), since ( RR^e) the expansion force in zone (A) is directed towards (+y) and that of zone (C) is directed towards (-y).
In conclusion, we can say that the different geometry of zones (A) and (C) has the effect of breaking the symmetry of the expansion forces which act on the isolated system.
6-3/ Example: 6-3-1/ £a]_cul_d£ J_a force Jlé£uJ_tantej_ Let's orient the central frame downwards (-y) and the quarter circles upwards (+y). The capacity of a quarter of a cylindrical capacitor is equal to: ,3-, Or for the 2 quarters of a circle Furthermore, the expansion force at the top of the positive armature is equal to âavec2o, or [F"k 4-^^^ ! Cw> ^f,. ^.< 5-6 36^9^%^ hence 6-3-2/ Se£sjd£Sj§vol_uti£ns_: n Tz.
or the tables: - considering (e) constant and varying (&-t) ft*^c m *.^'oiî'oi'-% .63 SX-és«r ^*f* -^« -3 <J3?ft*^c m *.^'oiî'oi'-% .63 SX-és«r ^*f* -^« -3<J3?=similarly keeping (£.<>) constant and varying (e)fi.i'SUi? -^3?
We have every interest/a radius (&.) and a spacing of the reinforcements •"'(e) as small as possible. But we're limited with (e) due to the stiffness of the dielectric, since more (e) decreases and more (E) increases, for a given voltage. To increase the expansion force acting on the isolated system, consider mounting (N) isolated systems in parallel (see fig 11). With a value of (£-. '5' HT), of (e^Pî- U>), a system occupies a width of Afi-^^-i-e. +MConsider an assembly constituting a square of side 72.83 m, that is to say that (N= 1000), the resulting expansion force is then equal to:E "(ï.' orThis assembly could then lift from earth a mass equal to: â- And since each isolated system does not disturb its environment (except at the 2 lateral ends where the electric field leaves the armatures and on which it generates forces whose resultant is zero because they are opposite 2 to 2) we can stack (N'i together, which increases the resulting force accordingly.
Be^crbjgtion j^ 5Let us keep the central "constant field" armature and the lateral armatures of the previous example, we will make the following modifications to the lower part (zone C) of the isolated system (see figure 12). Let's take a ( ^ ) greater than the thickness of a negative reinforcement and 10 let's take a rounding in the shape of 1/2 circle, both for the negative and positive reinforcement ( P.* ) . At the end of this rounding, let us give the reinforcements another profile also called "constant field", again obtained by the theory of conformal transformations, and whose parametric equation is equal to: The spacing between the reinforcements being always equal to (e), the thickness of each reinforcement is equal to (ô/i). This profile, directed towards (+y) makes it possible to obtain a constant electric field, both in the rectilinear central part (i.e. S^) and in the profiled part at the ends.
The expansion force which acts at these 2 ends of the reinforcements is equal to:25As for the components (f-) they have the same module but in an opposite direction, their resultant is therefore zero. Concerning the dielectric, we have no interest in letting it leave the isolated system, otherwise it would undergo a volume expansion force ( |f f-ôûJÊ^), 1^ would be oriented towards the () maximum, it that is to say that 50 the dielectric would experience a force directed towardsyThe dielectric is therefore cut at the exit of the 1/2 circle (see figure 12). The field lines are then collinear with the separation surface between the solid dielectric and the vacuum (£) (which must also be considered as a dielectric).
by mii-te of the difference ' 2oa co<£friciontB dt pexaittivitrf antr » the dlilectiique «t the vldo, we can say that it o'exercsToroa oriented with the video (co^fficiont, therefore prenoica, more weak.), <SgB.l« t 7-2/ Srom£l£^_ria-lisa.tionj_ Let us keep the values of the previous example, namely: In zone (i.):.
C Rj [e. * «WK) - » ^ » t. ? * I - 5636, ^ Eef ST. in zone (3): —».
in zone (c): -I^15• fKle.}- ftî.
...It should be noted that ljJCfujp nais and links are in the same direction by hearing the geometry of zone (c). since the resulting force is oriented towards (-y), we have every interest in extending the dielectric in zone (d).
The force acting on the surface of the dielectric, collinear with the vector (2) no longer exists.
By center there is exerted a voluaic force -.f.^aj £ within the dielectric in the region where aJsls reigns, we overflow the dielectric so that it-captures the 2 ends 15 of the arnatuxas (refer to figure IJ }. •^ îAll then have: -at the 2 ends of the araatures, a force _ -in the dielectric (the vol-unique force being mathematically reduced to L 20 "a surface force) hence zero Resulting force eh - "each ZOrie-^O}.
What remains then is 7-3/_SS.5 ⣠i. ^^Whatever the spacing (£), or the value of (S.* J, the resulting force is always zero in aone (jj). Seste then/ * ter -3 with ^ tV>/ '.^rjo-WAî-ï «f^tffii tfVt ïtoavec ^ tV>/ '.^rjo-WAî-ï «f^tffii tfVt ïtoavec ( (LS-O^ -)|l(7 -3 _z.
^3.3«t53-»t»ylZ. 832-(? ) is at.'-airtant greater than C ^i )"( «- ')small. . However, the scalability is relatively limited with (A.*), which is not necessary because we must take into account the possible need to reduce the bulk. Pax against (F.”) *•larger variations with). 3>=s when-iine ((lj) is smaller than 2 times the thickness of the negative arma-tare, we can adopt the construction of figure 74, where the,/o" circle is made up of two parts of a circle, separated by a rectilinear portion (within which the resulting force is zero). We are also limited by the rigidity of the electrical system. At this level, we can play on pressure, and there. iUMuual temperatures the dielectric will be subjected, so that we will always have to stick to an electric field which does not generate an electric arc. - the chosen example is a compromise between (} and () small, which generates m electric field equal to ou3.lo * / volts/centimeter, which does not pose a problem.
We see that when (Fy) and (e) vary in the ratio of 10~* to 1 that the value of (F^) varies respectively by Si63y «pf*£4 s 4'. The preponderance of (e) is explained by variations in the electric vector field. We also note that the variations of (£.1) play on the resulting force of the zone (C), i.e. the 1/2 circles, while the variations of (e) play on the force which acts in the area (A). 8ie adapted to agnatureg and This third example is intended to show that there exists a large number of solutions, since the geometry makes it possible to generate a break in symmetry. Let us describe a device which directly generates a motor torque. 8-1/ £escn£tion_ j_ 5Let (see figure 15) be a planar capacitor (zone A) whose plates have a spacing (e) and a thickness (e/z) - Let us fix a quarter of a capacitor at each end cylindrical (zone B) oriented in the opposite direction, and whose interior radii are (fty) and (£p) with) - Let us give the ends of the reinforcements a "constant field" profile 40 (zone C) whose parametric equation is equal to: Let us place between the armatures a solid and homogeneous dielectric having a high (£>(,).
Two cases can be considered; either the dielectric is cut before the constant field profiles (fig 15), or it leaves the capacitor and traps the constant field profiles (fig 16). 8-2/ Determination of the motor torque: Let be a reference frame (0; XY) of unit vectors (i) and (j). -e£ £one_(A)j. ÎJ»The forces S^and 5 are both in the same plane, have the same module and an opposite direction, their sum is therefore zero. Z.Jl.5. Jl_A__ We have (refer to § 6 - page 7- line 10): ----- sot*" ri - ----- — W= _--- i.e. . r^ = —— . - in ^ £-n ft.p> -in zone (C): The forces Fet F, are both in the same plane, have the same module §0 and an opposite direction, their sum is therefore zero (taking into account the smallness of (e) the deviation of the lever arms is considered negligible). An equal expansion force (Ç, + F) is exerted at the end of the reinforcements. f <. -v * A force is exerted on the dielectric directed outwards and equal to: If the dielectric leaves the capacitor (until E =0), a force is exerted at the end of the armatures ( Ç, +)=but there is also a volume force (f^.
= P. grad E) which we can reduce to a surface force (Maxwell tensor) and which is oriented towards (E) maximum, therefore towards ( +x), and which is equal to ^< + -££, £ therefore the sum (-F^-F^1=0 Let us consider the projections of the forces on the axes (ox) and (oy) and ^0 let us determine the respective lever arms. Let us call (£) the length of zone A.
[0020]
p___> _d. ££^ - v. - • “2.
a^Let us specify that the sign (-) is raised in front of the product (F^.) such that it is positive, taking into account the sign taken for (* w) The sign of the lever arm (*) is: . positive if flC^/J t- ê.
P-p.
< ^.^ot.negative sif <^ /'^ /* ^'^^The motor torque is then equal to: -without dielectric outside: which can be summarized, with R^v, ^ol e. ^ .^ < y),^^^ -with dielectric on the outside: which can be summarized, with (Ir > Jl^T-a ^c,'^ /^g < /jV^v^. and since' there are 2 ends which have the same torques, you must multiply these results by 2 to know the motor torque; 8-3/Exempl£:_ •R^s i.ii?--vr=fo«£'* height armatures=l m Consider 2 capacitors arranged in the same plane and oriented at 90° to each other on the same axis of rotation. Everything is placed in a composite material box in which the air has been evacuated (so as not to slow down the rotation by aerodynamic drag). The dielector traps the ends.
With (R= 5.10" }, we have, therefore L,is (-) and the product-^'^is (-).
We then obtain: SOlt "W« Ît.-Ms -<5,3^?-W. "HeffaWo.+ r 5As in i 6-3-Z page 9, the power increases if (e) and (R^decrease . "*us-z But we have a (E) equal to E ==10 / 1.57.10 =63.694 volt/meter. We can further increase it subject to respecting the conditions defined in § 16 g/: Let two meta.lUq.ties bodies (a) and (b) not polarized and rigidly connected together by a structure in insulating material ( see figure 17). They are placed in a vacuum. Let's connect (a) to the terminal (+) of a generator and the body (ï) to the home 15 (-) of the same generator, which delivers a potential difference (^V). Charges (<]) will appear on the entire surface of the bodies (i.) and (B), but almost all of these charges will be located on the surfaces of (a) and (b) which are facing each other. 'one from the other. Indeed, with “Tî. C-f, - -Y. £„ * -the most important () is located in the area where the 2 bodies (A) and (B) are closest to each other.
Oss 2 bodies constitute in some way a capacitor. 25On the rest of the surface, ( "") is infinitely small, because the field lines start from the bodies (a) and (b) and close after having traveled a great distance, which leads to a ( ^ui Y) low. Consider the presentation in Figure 17 of the polarized sub-layers, excessively enlarged to facilitate schematization. 50Let’s call () and ( <\J ) these initial charges. Now let us isolate (a) and (b) from the generator. Let us approach and then establish contact between (a) and a metallic body (c), at the same potential as (a) and carrying a charge (<£ ). Refer to figure 18. 55These charges ( <{* ) will pass into the body (a) because they spontaneously tend to extend over the maximum surface (minimum pressure). But instead of spreading over the entire surface of (A+C), almost all of the charges (4Î) will still be located in the sub- or polarized layer where the (gradE) is maximum, that is. that is to say in the zone the 2 surfaces of (a) and (b) are facing each other.
We could also say that the charges ( ^ ) cannot occupy a quantum state already occupied on (a) (Pauli exclusion principle) must then necessarily go into the polarized sublayer where 5() is maximum, this -which allows them to occupy an immediately higher free quantum state. Refer to figure 19. This increase in charges (£. Cj= <{) leads to an increase in the flux ( $- J ) and, since the flux is conserved, a charge ( ^ ) must necessarily appear on the body (l). 10II then appears on (b) next to (a) of the charges (X ^ - ^ -). We call these charges ( ^ ) charges induced by influence. They correspond to a supply of electrons. By virtue of the principle of conservation of "electric charge", the electrons (^) will give way, on the opposite face of (b) to "holes" having a charge (). 15It is in our interest to move this exterior surface as far as possible from (b) (see figure 20) so that the minimum number of field lines starting from (b) closes on (a).
The field lines of () of (b) then close on what we call a pseudo-armature (walls of the laboratory, particles suspended in the cosmos) which can be located at infinity. 20on the surfaces facing (a) and (b), we have: -so' ^? 25-etkV*6i? c-so-with?, “Ç” since referred to a capacitor. But there remain the charges (t/-) on (b), which have no symmetry on (a). The balance is therefore broken. 30Note that the body (b) is at the potential (- V) on the face opposite (a). Since a metallic body must necessarily be at the same potential throughout its mass, we have charges (q^") at the potential (-V). The charges (and ^) in (b) are separated by a "neutral line".
Example_of realization J Consider 2 planar capacitors (see figure 2.1) which we will identify by the indices (o) and (l).
Let's connect ( ^e and) in parallel. Initially we will take identical capacitors (C/, ) and (^), i.e. S- 100 m" and & » 1 cm« «-o* m and we will take a supply voltage equal to /o' volts, then a dielectric at £, /o. So.V j10Let's give the marks (see figure) ' - (A) to the armature (C) - (G) " "l of ()15The 2 capacitors being charged, let's cut the electrical connections with the generatorex but let's keep the connection between the 2 positive reinforcements. Let's separate the negative reinforcement (d) from the positive reinforcement (c). Refer to figure 22. As (d) deviates from (c) the potential difference between these 2 armatures varies. This potential variation is transmitted to the armature (b), which causes the potential of (a), but the ^V » (. V. -•remains constant. In fact, passing a certain distance between (c) and (d) the vector field 5., decreases, which has the effect of causing a transfer of charges () from (c) to (b), where the field electric is big.
These charges (^) will be added to the charges () and charges (4~) will appear on the interior face of (a) by effect of influence. By virtue of the principle of conservation of electric charge, charges ( ^* ) will appear on the exterior face of (a). The expansion force which will be exerted on this face of the reinforcement 5° (A), which is facing the void, will then be equal to: We can give the thickness that we desire to the reinforcements, this which simplifies their industrial production. The reinforcement (d) being sufficiently far away, we have two solutions, either keep it as is, then the loads (^) will be distributed over the entire surface of the reinforcement which results in a zero resulting force as a result of the symmetry of (cr), is to remove the charges ("^). Most of the field lines starting from (^), on the exterior surface of (A)t will close infinitely on a pseudo-armature (dust suspended in the cosmos).
The field lines which start from () from (to) and which close on the exterior face of (b) will generate charges ( ^' ), therefore () on the interior face of (b), which results in () on the interior face of (a) (the flux is conserved) and consequently new ones on the exterior face of (a) (the electric charge is conserved). Refer to figure 23. The flow then tends to self-amplify. In fact it will stabilize for a given flow. And since we have a £«]* equal to ^q' on the external faces of (a) and (§), this "looping generates a zero resulting force. Do not remain. that the expansion force linked to the initial () of (a). It is naturally possible to take the surface(s) we desire and place the capacitors (£«) and (<m) one above the other, (&*) and (C) being back to back , so as to reduce clutter. 20With a surface corresponding to a square of sides, the force of expansion would allow the ascent of a mass equal to ton s, which corresponds to a traction of lf-5*/.tf?
C Kgf/m. 10; 10-1 / Description j_ 25i.e. 2 flat cddensators placed in a vacuum and whose armatures are respectively identified by (a" and B*) then (C* and d"). Refer to figure 24. Let's connect (&*" and C*) together, just like (A" and D"). It is considered that the reinforcements are held together by rigid and insulating supports. A dielectric in is placed between the armatures 50 (a" and B"") then (C*et iT(le.fcEPlets (A' and l") at the negative pole, then (B^et G*") at the positive pole d 'a generator. Let's charge the capacitors then remove the connections (A~ and) as well as their connection with the negative pole of the generator.
We then have ^=cv andS be ?
We also have: Now let's remove the armature (d " ). Refer to figure 25. It is easily demonstrated that the energy spent to move this armature away is equal to 2! 2651388the electrical energy which was initially stored in the capacitor, i.e. V/rV < •*. resume*. So far this installation is similar to that of the previous example. But now we are going to change procedure. 5Indeed, we will prevent the transfer of charges (^) from (c) to (b). To do this, we connect (a" «t B ^ ) to the "terminals (-) and (+) of a generator whose ddp we increase so that the potential of (b) is always greater than that of (c ). It will then appear some charges (^^which will generate whatever charges (.^) by influence (the flow is conserved). Oes some charges (^ ^will make room for ( ^on the exterior face of (c) they will reinforce the ( ^^.initials. But this small (/Center (b^ and C) is mainly intended to force the loads (to remain on the external face of the reinforcement (c).
A force results: and the symmetry is broken due to the small -M/» (^-10-2/_Examplej^ Consider a circular capacitor with a radius «1.5 m (which corresponds approximately to the master torque of the nacelle of a reactor). Let "1-, w"V /m- Furthermore f,«8s /0"%V*t-^Hence:: C - 4 £7. -2.9 The importance of this force is explained by the large value of (^), which is due (in part) to the () between the reinforcements, and to the fact that (? ) is exerted on the surface of the dielectric which is held on the armature 50 (c) facing the vacuum. If such a propellant should not pose big problems in a vacuum, it risks behaving in the atmosphere like a dust vacuum cleaner since the dust will be polarized by the field lines and irresistibly attracted by the polarized armature (c). . It will then be necessary to provide a device to chase them away.
We defined in paragraph 2, two types of forces: -noted ¥( ft,, | HjJ, for a "connecting" force, in the sense that it acts simultaneously on 2 material systems.
5Usually, the direction is given by the line which connects the 2 frames of reference (do and R) and the direction is given from the first frame of reference (£,„) towards the second frame of reference (fl.). -noted ffe-o |&), for an "action" force, in the sense that it only acts on a single material system (R-ft), the second frame of reference (&„) corresponding to = 10 weighting to the continuum of space-time which contains- ;. These definitions being given, let us consider a negatively polarized armature for example and corresponding to the frame of reference (P^iv). We know that the polarity is given by the sign of the electric charge carriers, i.e. electrons (-^éo-Mo" c) in the present example. 15Let us examine what happens in the thickness (a) of the polarized sublayer whose thickness is approximately equal to 1 Angstrom (). Consider an element of volume (dv), surface () and height (“-). The polarized charges are located in (a). Let (x) be the distance from a point (p) to the interior surface of the polarized sublayer (see Figure 26).
^ The electric field (£) in (p) is equal to: and the electric field (g^) at the free surface is equal to:25The quantity of charges contained in a layer (x, x+dx) is equal to The force which is exerted on is equal to 50.so-P oW- - ^ ^^ .that is ûjl *.and J e<? te * jf^j'^ou () is the surface electric field. The force exerted on all the charges contained in the polarized sublayer of thickness (a) is equal to: .soitdf - £ ^ df .and sinceg «. ^.therefore the pressureP- _^_ * — This pressure is directed towards the outside of the free surface, and it is this which maintains the electric charges () in the polarized sublayer in the immediately higher quantum state liter. What must be understood is that the electric vector field which manifests itself on the surface () and outside of the armatures in the frame of reference (H) of the space-time continuum, in fact extends in the 15 polarized sublayer, in the same direction, and this results in a force of interaction between the electric vector field and the polarized solar quantities (p).
It is therefore a so-called "action" force of the fce-ftl^ type. We can simplify by saying that the vector field; acts on the scale quantities (a-_), hence "generated" force f-(cr^&,(nm (or 20 m is the unit vector of the high normal surd*). Polarized charges generate a vector field (^) which, in return, acts on them. It is the same with the vector field of gravitation which is generated by matter) and which acts, in return, on it, either? =™..t-25The only notable difference lies in the direction of these interaction forces. The force of gravitation tends to compress the material while the force of electrostatic expansion tends to diffuse it, to suck it outwards, in order to reduce the electrostatic pressure. The electric particles () being "sucked" by the electric field = 50 tric (g"), with a force to the armature the force exerted on them.
We know that the chemical potential of a particle (^) in the 24 phase (k) is noted: with ^ = tlieiraodîmainiqTae potential of the phase (k) containing (Nj-) particles of species (^). (mJ'J is a partial derivative considering that the other parameters (temperature, pressure, number of particles of species (£)) remain const= anti. We also know that the electrochemical potential is noted: corresponding to the chemical potential plus the electrical term C^r^) ^cule part () is at the potential V^ IWe also know, as Sommerfeld and Brillouin demonstrated that - electrons free are distributed in a conductor entered the potential energy (-1^) and the first Fermi level (^) (both 1 5 referred to the same origin). The Fermi energy 4 ^ ^F 'C'^J^^ reassures the thickness of the condugtion band envisaged (see figure 27). On the macroscopic level, we fix an average energy Uc^-êtf oàfc (£) is the charge of an electron and f the Gaîvani potential) 20We see that the chemical potential () of an electron corresponds to the energy difference between the level (f) and (^) and or the electrochemical potential-fIn fact, on the surface of the metal the electrons, driven by a certain speed (kinetic energy) spend a certain part of their time in the vacuum, at the exterior of the metal.
The result is, on the surface of the metal, a double layer of inverse polarity (over a Debye thickness) and an external potential (see Figure 28) negative compared to the internal potential. The electrons are then trapped, or trapped , in a potential tank of 50 Tolta. By applying a potential difference between 2.armatu±es (A and B) we measure using a voltmeter the *<? c ^C/j-'/O-The iW(vi)-(yi) = ddp of Volta, governs phenomena in a vacuum (.electric field). 55We can then say that: -the polarized charges (M) undergo a volume force c^adP, which attracts them towards the region where the electric field is greatest, namely on the face of the armatures, -the charges (^) do not can emerge from the metal because they are trapped in a potential bowl, - as a result, the action force >(JL|(Oïf-becomes a bonding force ^(^tj/fcftlbetween the electric charge(( «\ ) and the rest of the frame (a).
As a result, the entire armature is attracted by the electric field, and a total force is exerted: This force is called "expansion force", and it results from an interaction between the electric vector field (e), located in the frame of reference (du) of the space-time continuum, and the scalar quantities ( ^" ), which are sensitive to the vector field in question. The expansion force is therefore indeed an “action force”.
Consider an electric charge ( ^j ) and a line of the electric vector field ( E ) which passes through (). An "action" force is then "generated" and its value is (tf* ^.) where (n) corresponds to the tang= 20 between the field line at (^). Now we know that the electric vector field (s) derives from a potential £ ^ - ^t-d-V = We also know that the potential varies along a field line and that the vectors (E) are always orthogonal to the lines which connect the points which are at the same potential (equipotential) The field lines are therefore always orthogonal to the surface element (<^6) of a polarized armature, the latter constituting an equipoten=; tielle. But the field lines which remain orthogonal between 2 armatures, 50 or "the equipotential surfaces which remain collinear with the surface of the armatures" only constitute a very particular case. This is the case in the volume located inside the plates of a plane capacitor.
Most often we find field lines which diverge, which we can trace, just like equipotential surfaces, using 55 calculations which are not very complicated but which nevertheless require computer tools due to the large number of calculations to be carried out. perform. Consider a curved field line (see figure 5) which starts from a charge (^*) (marker i) and arrives at a charge ()(marker2).
Let's define a reference (0; X,Y ).
The force acting on the positive charge is equal to f(a,,|R^rfS and the force acting on the negative charge is equal to F^i [<?.,,)We can say that the principle of action-reaction is respected 5since these 2 forces (and) are equal in module and of opposite sign, i.e. iF(.**|(l. t Rol^) = o However, this principle, in the framework of classical physics, is based on the fact that these 2 forces - action-reaction - are supported by the same "line". It is then obvious that their sum is zero. 10But in the field of electrostatics, the field line which supports the two forces - action-reaction - is most often a "short" one. The geometry intervenes at the level of the reciprocal positioning of the charges ('j), at the level of the flow path (|), and of course the positioning, the shape and the orientation of the polarized surface elements 15 (clS). As a result, the sum of these action-reaction forces is no longer zero and we can even, by an adapted geometry of the surface elements (ol5) and (els_) arrange so that the sum of l one of the components is zero and the other components add up.
20This is the case for the field lines which start from the top of the central reinforcement in the examples of ^ 6 "and 7. Thus, electrostatics, which some consider to be a very old science, where there was nothing more to discover, or which did not present any industrial interest, is in fact the only one where man can model to his liking. convenience the curvature of the space-time continuum in which the field lines (E) circulate. Thus the principle of action-reaction remains true when a field line connects 2 particles of opposite sign, but their sum is only zero if the field line is a straight line, or if the polarized surface elements -30 are oriented along 2 normals having the same direction and an opposite direction. The examples given in £9 and 10, where the field lines start from an armature of the isolated system and close on a pseudo-armature located at infinity, constitute an extreme case. This pseudo-armature can be constituted by all the material particles which are suspended in the cosmos and where the field lines circulate.
The electric field of a charged particle is equal to £=^ (with (n) unit vector in the chosen direction). The electric field varies as (), so it drops quickly. 40We can, consequently, imagine that a field line starts from a polarized armature, moves away on its geodesic, is attenuated in (j) then, encountering no scalar quantity that it can polarize, that it becomes equal to zero. From then on, we have a force of action without a force of reaction. This is the extreme case, but possible. Most often, a reaction force will be exerted on dust, an asteroid, a planet, etc. But if the action force is equal to ? o**i, the reaction force, oriented in any direction, will only be equal to^ ^i with E,;^:-. 10All this shows that the norm of classical physics relating to the zero resultant of action-reaction forces deserves to receive a new, more explicit formulation.
Not only must the curvature of the field line be taken into account, but also the fact that these forces are not of the "bonding force" type? Ce.l^), but that each of them is of the "force of action" type, that is to say? (ft,|t,) « KM**) +^C»l0 Take 2 sheets of paper. Draw a curved field line that starts from one leaf and ends on the other. Position a charge (<^) at one end and a charge (“j~) at the other end. 20Now spread the 2 sheets of paper apart. Oe diagram to show (or rather image) that each force is "generated", each on its own, by "interaction" of an electric vector field (e) and a scalar quantity. The opposite signs of the two forces result from the interaction of the same curvature of the space-time continuum by the electric field, on scalar quantities of inverse polarity. It is the same in the universe at the level of gravitatian forces, where) is replaced by (wl) and where (e) is replaced by (5-).
1550The expression JU^F-cif naturally assumes that the 2 terms (and () are measured in the same reference frame. Now, let's take the example of ^ 6 where the force exerted on the central armature (figure 7) is internal to the isolated system. To evaluate the expansion force, the observer will have to move 55 with his frame of reference (R») in the polarized sublayer'. ^ Indeed (f*-1-£), which means that this action force ffR* 1&«J, to exist, requires a fulcrum (^.) and a vector field). As soon as we leave the last polarized atomic layer, only (E) remains in (^o). ^^ 40The same observer cannot both measure ( "r ) and (), although he does it indirectly by measuring the acceleration of the entire isolated system, unless he measures the brake force (is external to the isolated system) if the displacement is not accelerated. So we formulate the following statement by saying that the energy apparently involved corresponds to the displacement) of a force (?
) which we can measure through its effects on a frame of reference external to the isolated system. But this definition does not seem satisfactory because the driving force is internal to the isolated system. 10Let us then give an explanation based on the laws of thermodynamics. The pressure (P) corresponds to the energy per unit volume, or W =. ( ?. if ) or (), it is | say, in the case of a planar capacitor (easiest example) that; 15. <?,J1_In reality, the energy of the capacitor is located in the polarized sublayer, but we can consider that it is located in the volume of the continuum. of space-time located between the frames where the field lines circulate. We then find the famous expressions of thermodynamics. We can also, by spending mechanical energy to move one armature relative to the other, write that: WP-iri -SeWP-iri -SeSince the charges ( ^ ) and the coefficient () remains constant, whatever the value of the spacing (e-).
50In the thermodynamic sense, we can say that the energy stored in a charged capacitor is equal to: which corresponds to -a work of "transfer". It is the same for penetrating a body of volume () within a liquid at pressure (p)./ jr-t-If we let a loaded armature/get closer to the other loaded armature 5 (“ " ), the variation in energy will be equal to: which corresponds to “relaxation” work. It is the same when an immersed body of volume (v) rises towards the liter surface of the liquid. 10This connection between a charged armature and a body immersed in a liquid comes from the fact that the cost? rps inuiie±gé undergoes the Archimedean thrust force ( | î.-4Ûï3f ), just as the polarized sublayer undergoes a volume force (£«^[f ). Similarity, except that: 15-with the submerged body, the energy recovered from the displacement (iursPiv) is limited by the dimension of the container since (d-»-) corresponds to the volume displaced during the movement. -with condensate, the energy recovered from the movement of the reinforcements (dw* Wv) is limited by the spacing between the reinforcements.
20But let us first consider the case where. the field lines start from a polarized and charged armature, and close infinitely on a pseudo-armature, as indicated in the examples of £ 9 and 10, and scb.éma= tized figures 22 and 25. The vector ?- Jj^-s./^orthogonal to the surface just like the electric field vector e=r _-The expansion force which is exerted in the polarized sublayer |a/uid well as that which is exerts only on the surface loads and absolutely does not depend on the distance which separates the 2 reinforcements. The second armature, of different polarity from the first, only intervenes 50 to ensure a (cù^d V) and therefore an electric field (e) since the field derives from a potential. In the case of a single armature, the potential will vary in (and the electric field in (Ve") >'electrostatic pressure since it is proportional to (E55We can then say that: -the expansion force is exerted only in the polarized sublayer, or on the surface of the armature, since it is "generated" by interaction between () or () and (e), - the field lines always start perpendicular to the surface element, just like the force vector which is collinear with it, -the field lines can then remain parallel, diverge or converge, this does not change the module and the meaning of (ï), 5-it should then be considered that (p) is supported on ( E), therefore on the space-time continuum.
The expansion force is of the "action force" type and is written ^(R-fl/fto) It is based on the space-time continuum alone. It is the same for all the armatures, whatever their orientation and their polarization. The principle of action-reaction consists of saying that the force of expansion has the opposite direction to (tg [R-o) But it is very important to note that all these forces are based on the curvature of the space-time continuum. Then, whether the reinforcements are stationary or in motion, this does not change anything in the values of (f), of (<r), of (^), of (e) and therefore of (*? ). Thus, under the action of the resulting expansion force the isolated system moves, carrying with it ( <r ), therefore ( e ) and consequently ( ]? ). Our isolated systems are therefore “self-accelerated”. Since all values (cr, oj, ë) remain constant, it should be recognized that the internal energy of the isolated system remains constant over time.
The expression then corresponds to a potential function which relates to ç. a frame of reference external to the isolated system. The work of relaxation (cfàsffiv) is just as infinite as is the 25 () of the continuum. This amounts to assimilating the continuum to a fluid, in which a force (.f^ <ûcJ P) is exerted, due to an anisotropy of its curvature, located in the (). We then broke the symmetry. ^^ Finally, at the energy level, any "external balance" means nothing, the only significant balance is located inside the system 30 isolated by "the maintenance of the electrostatic pressure", which is exerted continuously on all polarized surfaces; In fact, we should speak of "electrostatic depression" to bring it closer to the concepts of thermodynamics, since this force of expansion if directed towards the outside of the metal and tends to increase its volume . Let's carry out an extremely simple experiment, which consists of placing a cylinder vertically, with the upper end closed and a piston able to slide at the lower end.
Refer to figure 29. In the initial state the piston is located at the top of the cylinder. Let us then exert a vertical force (+? ) on the piston and a force (-f) at the top of the 4 cylinder.
The work provided is equal to (*= ^= P-S 4 = P. AIT ) calling (p) the atmospheric pressure which remains constant.
Let's tie one. beggar to the piston, of a weight .-^ =-t1<j =-^ .An observer= external observer who does not see the person holding the piston, will wonder by what miracle this beggar can hold in the air ! We are there with our isolated electrostatic system which, undergoing a force of vertical expansion) would remain immobile in the air, in equilibrium. In fact, the person who keeps the cylinder immobile expends energy (she felt in his tense muscles). This corresponds either to the work required to keep the pig immobile above the ground (this is the apparent side of the experience), or to maintain the vacuum in the cylinder (this is is the invisible side of the experience). Let us then consider that this person, being tired by this expenditure of energy, attaches the cable connected to the cylinder to a hook fixed to the structure 15 of the bracket. The piston then remains in its position and the bitch remains suspended in the air, without us expending the slightest energy. What happened? Looking more closely at the hook attached to the metal structure, we see that the stresses (force f = ^ c|) have deformed the molecular structure of the metal. The sum of the intermolecular forces balances the force (F). Naturally the force (f) travels to the ground, but what is interesting is to note that the deformation of the atomic bonds generates a force (f) which balances the external stress, and this 25 indefinitely without the expenditure of the force (f). less energy by man. The deformations in our electrostatic systems are located in the polarized sublayer. The energy required to maintain the electrons and holes in their higher quantum state is taken from the electric field, which is due to the polarized charges. The system is self-maintaining 30Consider an electrostatic system stationary and suspended in the gravitational vector field. We then have 2 forces of the same module but of opposite direction and which both result from the interaction of a vector field on scalar quantities (i.e. -m and g then q • and E j. 35There is equilibrium between these 2 action forces ^OaIO. This balance is broken when only (f* ^ Ê) remains. The isolated system then stabilizes under a constant acceleration () with and the energy apparently involved corresponds to the work of relaxation dv-f.Jir, We could also say that: JlvTs f.J s?• Vjlr* f.yfcJK'VItalîdt' Let V^c^lll^c ill" 2.*12-11 expression qM does not involve a frame of reference exterior since I no longer have the term (dî). 5However, considering (W) and (M) as invariants, the terms (f) and (i") vary depending on the frame of reference where we place ourselves, as demonstrated relativistic mechanics. By leaving mathematical formulations, our mind experiences some difficulty in conceiving the deformations of the space-time continuum. 10The concept linked to energy, which seemed to be solid as a rock, also comes to vacillate because it is based on the concept of “binding force” (cause external to the isolated system). We can more easily conceive the phenomena highlighted, by saying that: 15-the internal energy of the isolated system remains constant over time, -the potential function-? therefore evaluated from the outside, corresponds to the work of relaxation (^), where the space-time continuum is assimilated to a fluid. We can then compare our isolated system to an oil within which we have generated anisotropy. In the same way that an air bubble rises towards the free surface of a li” which, this oil moves according to the greatest value of (“jôàlLP”). And since this bubble carries (a) and (É*), it moves on its equipotential. 25As for the energy (-5—L), it corresponds to the maintenance over time (t) of a force (p) which is generated by interaction of (^ J and (£), which are independent of time The free interior energy (w, is transformed into exterior energy (oW* f.Jlr) recoverable through the interaction of the anisotropic continuum 30 of the bubble on our isotropic continuum. The transition from one to the other is done via a potential function. Finally; the most important term of this potential function would remain "time $". 3514/: Let us consider a charged plane capacitor placed vertically on a table provided with a plate made of insulating material. Refer to figure 37. On the exterior faces of the frames, we fix a hook to which is attached a cable, which passes through a pulley connected to the table, and at the end of which is fixed a mass (m) . 5Let's remove the fixings which held the frames immobile. Taking as the value of the mass (m), that which allows the equality < Expansion force) = Gravity force (r*'vncL ) > - we then see that the two reinforcements remain in the position they had, under the action oftwo action forces ^r&M^oJ etnf£e), both being independent of time.
10Let's move the positive armature (A) by a length (AtO), giving it a slight impulse. We then obtain an energy gain on the mass which rises (&» i ^-/m-O-Ai>) and we know that there is a loss of energy within the capacitor as a result of the reduction in volume (i -Mr»^^ or â^ -r). Now let's move the negative armature (B) by the same (>•*- ). We see 15 -then the mass (m) decreases, which corresponds to a loss of energy (A^V*but we also know that the energy of the capacitor increases as a result of the increase in its volume (AWr ^F -tt- or-MAY/). It is therefore a "transfer" of energy, from the capacitor to the mass of (A) and transfer of energy from the mass (B) to the capacitor. The energy balance is therefore generally zero. If we move the two armatures simultaneously, nothing changes, we simultaneously have the ("t-rtrt^C "'*$'ûv)" les (4-fû-ïet les).
But let's focus our attention on the positive armature (A), where the energy gain occurs. The energy of the capacitor is equal to (vT-i- 2. )• In its displacement (âX)> the energy gain on the armature (A), i.e. (j^Wicorresponds to a transfer of energy from the continuum located between the armatures' towards the armature (A) then the mass (m), and the energy of the capacitor decreases by (! _. ). It is very important to remember that there is a transfer of energy from the continuum to the moving armature. Let us then imagine that we only have one polarized armature (A) on one of these faces (refer to § 9).35In this configuration, we can always say that the energy gain on the armature comes from of an energy transfer coming from the continuum curved by the electric vector field, i.e. (? C^[^o)û-v sWe usually say in electrostatics that the field lines close to infinity on a "pseudo-armature".
Then, the (—&V) generated by the displacement of the positive armature (A), leads to a C+AV) equivalent of the pseudo armature, and this without expending the slightest energy since the vector field there is zero and that there is no matter, and therefore no polarized charges. This observation is the first lesson to be learned from this observation. Let us draw the graph representing pressure versus distance (d). The pressure decreases in C) since it is proportional to the square of the vector field. 10In contact with the reinforcement, the pressure is represented by the segment (AB), and it is zero at (E)(refer to figure 38). Let's move the armature by one (A^) which is represented by the segment (AD). Before this movement, the pressure at point (D) was equal to (DG), then it changes to the value (DC =AB). 15If the expansion work (recovered on the displacement of the armature (A) can be schematized by the surface (ABCD), the energy corresponding to the increase in pressure in.
(D), corresponds to a thermodynamic work of "transfer" (V *fP ~-) <ïwhich can be schematized by the surface (BCGÏÏ), in an operation, isothermal. 20By taking a point where. the pressure is initially zero, we would have equality between the work of expansion and the work of transfer. But, because there is a colossal “but”, this transfer work corresponds to the transition from the planar continuum to a curved continuum, and this is produced by the effect of the electric vector field__which is free and independent of time. The transfer work is therefore free for us. This is the second important lesson brought by this observation. We owe this energy gain to the fact that the structure of the space-time continuum is modified when a vector field circulates there, and this curvature is maintained without expenditure of energy by the independent vector fields of the time . But we also owe it to the fact that the structure of the space-time continuum has what I will call an "attenuation coefficient". in the sense that it decreases this curvature with distance.
35If the structure of the space-time continuum did not have this property, the slightest gravitational vector field would swallow up all the matter in the universe.
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IS/; If I put myself in thought in the place of the person studying this document, I suppose that he can admit the curvature of the continuum by curved vector field lines, then its anisotropy, due to the grad E. 5But can Does she have difficulty admitting the non-existence of "reaction" forces and the existence of only "action" forces from the continuum.
Let us then recall 2 very well-known experiments in electrostatics. Let us first consider a point in the universe where there is no matter and no parasitic vector fields. Let's place a container containing a current-conducting liquid at this location and immerse a hydrometer into it, it will sink partially. If we polarize the liquid, we then see that the hydrometer rises. Refer to figure 30. If we place a sphere made of flexible and conductive material in the same place. current, we see, once the sphere is polarized, that it grows. ^The phenomena observed are proof that the electro=static expansion forces are generated by the interaction “electric vector field/A polarized charges”, therefore that these are indeed action forces which are exerted from the continuum, which move with the electric charges on which they act and which generate the electric vector field.16industrial applications, linked to the use of the expansion force generated within an isolated system similar to the One of those which have been described above is very numerous, but there are two which immediately come to mind.
25- IS-l/ Po£ce_ de_pM]3ulsi°n î Let us first of all exMdjies the means required to ensure the propulsion (and/or the sustentation in any vector field of gravitation) of any land, naval, aerial and of course vehicle of any spacecraft (fr'tfirf6les .various devices schematized in figures 7-11-12-13-14-21 and 22, 24 and iû25 can be used for this purpose. Several of these devices can be associated on the same machine, by orienting them in various directions then regulating the various resulting forces* generated by modulating the potential difference (ddp) applied to each of them, so as to obtain the total resultant force (f. T ) deared. ,. Let, for example, as diagram.tL»é «.gu» 31, six xsole systems (SE), schematized by parallelepipeds, oriented according to the trihedron (Ojx.y.z) in: + (ox) by the (SE) deiepèie (F) - (ox) ""(E) + (oy. ) ' "( D ) - (oy) ""( B ) + (oz) » "( A ) - (oz) ""( C ) Activation of only (Si)reference (a) will result in the machine towards (+ oz), the activation of the only (SE) of pepère (d) will lead the machine towards (+ oy), ete... and the activation of the (SE) of markers (A) and for example will cause the machine in the direction of the total resulting force (of the two: components (ra) and (V).i> generated by the (SE) of marks (a) and (l).
All connections remain possible between the various isolated systems, activated or not, in a more or less intensive manner, thus making it possible to obtain a total resulting force centered in (o), and whose vector is defined in 15. spherical coordinates, thus giving the machine the possibility “to move in all directions (figure 32). It is naturally possible to climb as much (if) as one wants. This arrangement allows any more equipped machine, and especially aeronautics and space, to be able to change direction extremely quickly, to stop and then start again. reverse direction promptly. Military interest is therefore immense. 16-2/this driving force: Let us then examine the use of these isolated systems (se) as a generator of a motive force intended to drive for example an alternator, an air compressor, or any other generator. 25Let the solution schematized in Figure 53 include: -an axis of rotation (oz) equipped with 2 bearings whose outer cage is linked to a rigid structure, -m radius (OA), collinear with (oy), fixed to the axis ( oz) eti (0) et.d'end (a) -tm isolated system, schematized by m parallelepiped, which generates a resultant force coplanar to the plane (xoz) and collinear to the axis (ox).
, The resulting force then plays the role of mofecice force and the axis (oz) is driven in rotation.
Let (1) be the modulus of the radius (Ôl), ( ^c. ) the resulting force generated by one of the isolated systems shown schematically in Figure 7-11-12-15-14-21 and 22-2ft-et 25, the available power is then equal to VI. «>, with “> -rotation speed. It can be as large as we want and it only stabilizes when the braking force (counter electromotive force for example of an electric generator which is regulated by the excitation current) reaches the same value, but in opposite direction, that the driving force (F*)- Several spokes can be mounted on the axis (oz, in the same plane and several planes of (SI) can be stacked on top of each other on axis (oz). as shown schematically in Figure 34. Let us take the (SI) described in paragraph 6 and schematize figures 6 and 35. Let us consider that the depth is 5 meters and its length is 22 meters. That is to say a lever arm of 100 meters. The eastern circumference at i* R. = 628.31 m, we can therefore arrange 628.31: 22=28.55, i.e. 28 (SI) - nlan Let us consider 2 planes of rotation (which can be contrarotated on the same plane .
Let us consider * (jx^ua to remove the cut-off effect if this device is on a machine) we then pose a power by taking “. *-* .. "-30 revolutions per second):" - These devices can be mounted, either in a fixed installation, or in any machine, propelled and stabilized by the means cited in §15-1. then uniting the energy required for the operation of offshore easements. To reduce losses due to aerodynamic drag, the (SI) will be enclosed in a tunnel, which will take the form of a torus, no longer offering any grip on aerodynamic drag, apart from the lever arms (Ai caret of the torus and are fixed to the axis of rotation. Their drag can be reduced by using lever arms on both sides, discs whose diameter is equal to the internal diameter of the torus.....,. „ We can also mount one or more of (SI) schematized in Figure.7-11-12-13-14-22 and 25, on the stator of a linear electric motor whose rotor has the shape of a circumference.
By placing, as explained above, several (SI) one after the other on as many stators of linear electric motors, and enclosing them in a tunnel, which takes the form of a torus, we can recover a equally unlimited energy.- The connections between the linear electric motors and the external electrical network can be made by means of mobile brushes, fixed to the stators and which come to rest on a collector, with several tracks. fixed on "pji interior of the torus. The connections with the outside are made or- oar waterproof connections.._,i.„nfact- the air vacuum inside the torus, It being understood that we have made the vioe u a-u.
nothing limits the speed of movement of the stators relative to the rotor; and ^ the energy recovered <&•=is just as unlimited as is the term V. The power delivered can be quantified in mega or gigawatts, it becomes possible to power powerful lasers, thus making it possible to produce powerfully armed machines, well beyond the concepts currently accepted in the Strategic Defense Initiative program.: Refer to figure 35 - Or the isolated system described in § 6.
The side (2) and central (3) frames are made of current-conducting material. These metal parts do not undergo friction, but must have: - on the external, non-polarized faces, a <machining finish by simple millingfw - on the polarized internal faces, a perfect machining finish (wjt-) supplemented by electrolytic deposit (ifâ-e) • •15'Lvolume located between the polarized armatures is filled with Titanane • Barium. It is not necessary to machine it in a single block, but it can be made up of a stack of BaTiO plates. 11 must be machined by superfinishing after polishing (wv) and contain no homogeneity defects. The Barium Titanante crystals must be oriented in the ^7 electric field following the mesh of highest dielectric constancy. The assembly formed by the armatures and the dielectric forms a parallel = epiped which is enclosed in a box, made of insulating material, and the highest vacuum is carried out inside the box (4) To facilitate its production, the box can come to rest on the external faces of the reinforcements and the dielectric.
It not only allows the vacuum to be maintained inside, but it must maintain the device in compression and contribute to the rigidity of the whole by bonding the lateral reinforcements. lowerssopen'et/fes - In operationTV. of the ddp (therefore of the generated vector field) used, it will be necessary to maintain the temperature of the dielectric at 120C, which allows a (£„,) of almost 7000 .while it is only 1000 at 20C. It is also possible to use as a dielectric KPO^Ha from the monopotassium phosphate group, whose (£,.) can reach 32000 around 110K. a^-The parametric equation of the top of the central frame is:. St'wcfr c -As a result, the opposite faces of the positive and negative armatures are only parallel and spaced (e) apart at infinity.
•39will admit a deviation of.
1.10 m between the value of (y) and the axis (ox), tangent to the central armature at infinity. This value corresponds to (deducing that the point (x) is located at -9.2103 m. Let us take: this value for the dimensions CA) and (B) of figure 35. While retaining the values of the example in § 6-3, i.e. (e)= 1.57.10, ftor= 5.10 R.sft^-i-e, =20.7.10" nuC*""* avo* *UT,(i«^*n.iff"*C«p*"»"«'The value of (C) is therefore equal to (R^t*+" H), with H=thickness of the end side of the central frame, i.e. H=l.93.10 m, therefore (C)=4.10 m. In the horizontal plane, the dimensions are as follows: -thickness of the side reinforcements =1.10T) -distance between the 2 side reinforcements=.((l.57 x 2) + (1.57 x 2)).10 = 6 .28. KfV Let's take 1 meter as depth, which leads to ( fft. )=^^o^5"Newtons. If the depth is (M) meters, the ( ?«. ) is multiplied by (M). Given the low value of (e), the dimensions are not respected in Figure 35, in order to make it more explicit.
Thee force which acts between (2N) material systems is called in this patent "connection force" and it is denoted f^R.^ l'V) The outfits (P-) and (r.a) correspond to the associated references to the material systems ((( ) and (f^) and the vector ( Facts in the direction from ( ^ ) to (ftp ).
The opposite direction vector is denoted ^L^-M^rfl) It is admitted and demonstrated that the resulting force of the internal forces of an isolated system is zero, because an action force is associated with a reaction force of the same module and opposite direction. These forces correspond to "connection forces", i.e. Yf^-I^j) + ^(^el^^)10But there are forces which are "generated" by interaction between a vector field and scalar quantities ( material) which are sensitive to this vector field. This is how it is, between the gravitational vector field (g) and the -a. ^>-* . matter (m), i.e. F « m g, or between the electric vectosie-1 field CE and an electric charge (q), i.e. F = q E. This type of force is called in this patent "action force" and it is denoted ? (&*.where (R^) corresponds to the material frame of reference which carries the electric charges and (&„) corresponds to the frame of reference linked to the space-time continuum in which the electric vector field circulates.
The “action” force F = q. E is also called "expansion force". It is always emitted perpendicular to the surface element which contains the electric charges (q). The expansion force is therefore -> collinear with the vector E, at the level of the fulcrum.aphysics We commonly say that 1*electrostatics is the domain/of. the polarized bodies remain in equilibrium under the action of the various expansion forces. Indeed, an expansion force is generated at the two ends of the same field line and, as a result of the reverse polarization of the charges (q), the two vectors (F) are in opposite directions, so their result is zero. 30The first claim concerns the fact that these 2 forces are not linked together by the principle of action-reaction as happens with "linking forces" but that it is indeed 2 dp forces type “action force”.
By an appropriate geometry it then becomes possible to curve the 35 field lines and adapt to suit the direction, the direction and the module of the expansion forces, so as to obtain a resulting force different from ero.
The expansion force acts as an "action" force by interaction of the electric charges and the electric field E within the polarized sublayer of the armatures, transmits this action force to the polarized atoms.
The atoms of the metal frames being linked together, the 5d "action force is transformed into a "bonding" force, and the action force is transmitted to the entire metal structure of the isolated system. The electric charges which undergo the action force remain on their metal where they are held by the electric field, and do not leave the metallic structure because they are trapped in a potential bowl.
Device and method characterized in that a dielectric which undergoes a pressure (p) and an electric field gradient (grad E) is the seat of a volume force (f»P grad E), oriented in the direction of the field maximum electric. 15O this force is also an "action" force which is generated by interaction of the dielectric dipoles on the electric vector field. By associating this volume force with the surface forces of claim 1, it becomes even easier to break the symmetry of the action forces and thereby obtain a resultant force different from zero* 3,. /Device and method according to claim 1 characterized in that the expansion force (F= q. E ^, depends only on the value of the electric sector field (e) at the level of the charge (q). Whether the vector field (e) is then rectilinear or curved, constant or not, this changes nothing to (f). 25We can conceive that the field lines, which start from a polarized surface and on which the action force is exerted (F=q.
E), then diverge in the universe, and the electric field decreases with the distance (e) and tends towards zero according to the expression S s -£^- » -p- Then, in this extreme case, there does not exist only one action force, at only one end of the field line. The key word of these inventions is "breaking symmetry" and the various examples presented below are of no interest other than to offer non-limiting solutions from which it is possible to calculate the value of the resulting force. ,all the more easily as the chosen surfaces do not have ambiguous singular points.
Device and method according to claims 1 and 2 characterized in that we consider two armatures at the potential (- T), collinear with (oy), of a frame of reference (0; X,Y). They are considered to be long towards (+ oy) and they end towards (- oy) with 2, convex quarter circles of radius (*v). Refer to figure 7. We place at the cup part and between these 2 negative armatures, a central armature at potential (+ V), the upper end of which has a so-called '* constant field' profile (i.e. the electric field is 5 the mên" in the flat and short part of the reinforcement) and whose parametric equation is *•«• L^ ««CP^ * -f) J This reinforcement has a thickness (2e) and 1 • spacing between the central reinforcement and each of the lateral reinforcements is equal to (e). The central reinforcement ends laterally towards (-y) by the concave face 10 of 2 quarters of a circle of radius (ft*), while maintaining a spacing (e) between the negative and positive reinforcements.
The fltix (4*'S) s® conserves between the surfaces (% l** ) and (S) but as a result of the difference of the surfaces (Sp>S<-) it appears that (e>S,) therefore ( ^>fn. ) hence the appearance of a resultant force oriented towards (-y). 15 As a result of the geometry adopted, the components on (ox) of the ex-• pansion forces generated are opposed to 2, only the components on (oy) remain, i.e. the resulting force: where (E, and ^) are the coefficients of interttirity of the vacuum and the dielectric 20 which can be placed between the armatures, where (v) is the potential difference and (L) the depth of the armatures.
Hspositive and process according to claims 1-2-5 and 5 characterized in that the system isolated above is modified as follows. Laterally, the quarter circles are replaced by semi-circles of radius (<R) for the negative reinforcement and ((^) for the positive reinforcement. Refer to figure 12.
At the end of the semi-circle, the reinforcements are extended towards (+y) by a rectilinear portion then they end with another profile called "constant electric field" whose parametric equation is £«i2Uijfc»|-û »yWe place a dielectric between the armatures and we consider that the dielectric traps the ends of the lateral armatures then that it ". extends towards (+y) until the electric field is zero. the armatures: • lateral have a thickness (e.) and their spacing is always equal to (e). The dielectric located at the end of the lateral reinforcements undergoes a 3.5 volume force (^«VT^Tê) which is directed towards the zone where the maximum electric field.
_» We demonstrate that the action force ^R^o) which acts between the dielectric (Ri) and the continuum (“-“) is equal in module and of opposite sign to the action force? (X\M «acts at the ends of the lateral reinforcements 40, i.e.f - i f,e. LLa. Toroe recoil, directed vtrs (-y) is then equal to j W^V" ' -"* é/Device and method according to claims 1-2 «t J cainctéxized in that the central part is constituted by a plane capacitor, extended 5 on either side by a quarter-circle shaped capacitor but in opposite direction, then giving the ends of the 2 armatures a constant field profile whose parametric equation is ("«^ <« £ - < *» f • and jp- <jP + &* V ). Refer to figure 16. the dielectric placed between the armatures traps the two ends 10 at constant field and extends outside until ( ^ud'O ). By calculating the value of the lever arms on (ox) and (oy), calling (£) the 1/2 length of the central rectilinear part, and carrying out the product (F.for each of the components, taking ftp^ .4*o"î- e. and Ra^λ** we obtain a motor torque equal to: By placing at the center of gravity an axis of rotation on which a generator is engaged, we can collect a power equal to â ( ÎsCéj), with ( û> ) = rotation speed.
This power can be as great as we want, since this device is enclosed in an enclosure of composite material within which the air has been evacuated, so as not to be slowed down by aerodynamic drag, f /Device and method according to claims 1-2 and 4 characterized in that two surface plane capacitors (s) have armatures which. carry the marks (A,B and C,D). A dielectric is placed between the armatures. 25 the armatures (a) and (k) are connected to the negative pole and the armatures (b) and (c) are connected to the positive pole of a generator. After having charged the 2 capacitors, we isolate them from the generator, we • maintain the electrical connection between (3.)' and (c) then we remove the armature (d). Refer to figures 21 and 22. .50The electric charges (^~) will be distributed over the entire surface of the armatures (c) and (X) in order to undergo a minimum pressure then they will move towards the polarized sub-layer of the armature (B.) where the electric field is maximum, and they will reinforce the initial charges (^), which increases the electric field (£).
55The loads (will then generate by influence loads (<Ç) on the reinforcement. (A) because the flow ( | ) is conserved between (a) and (B).
As a result of the principle of conservation of electric charge, the appearance of chargesa () interior surface of (A) will lead to the appearance of charges ( ^*) on the exterior face of (fi).
The few electrical charges which could remain on the armature-5 structure ec) generate a zero resulting force since these charges are distributed over the entire surface of the armature. The expansion forces generated on the interior faces of the reinforcements (a) and (b) cancel each other 2-to-2 since they have a tiny module but an opposite direction. All that remains is the expansion force which is exerted on the exterior face of the magnet (A), i.e. fc ^/C^^^a.5)ELspositive and method according to claims 1-2 and 4 characterized in that that 2 surface plane coadensators (s) have annatures which carry the marks (A,B and C,D). We consider that the 4 magnets are stacked, one on top of the other. A dielectric is placed between them.
The armatures (A and L) are connected to the negative pole and the armatures (B and are connected to the positive pole of a generator.
We charge the 2 capacitors (A/B) then (C/D). The electrical connections between the armatures (c) and (l) and the generator are removed. We remove the armature (]>) **and we regulate the generator voltage so that the voltage of (b) is always greater than the potential of (c). Refer to figures 24 and 25- The electrical charges ()•*«*«*of the armature (c) where they were initially because the potential of (b) is greater than that of (c).
The electric field established between (b) and (c) is not necessarily important, but its existence is sufficient to give the meaning of (), consequently to maintain the charges (C) on the exterior face of ( c) from which the (AY) continues by a circulation of the field lines towards the infinity of the continuum.
^ The expansion force and t^JR,) cancel out at 2, it is the same for 1(R>c) and l(U^), ** then remains when T^ïO is frffe S C sidielectric remains glued to the frame G). Devices and methods according to claims 1-2-5-4-5-6-7-8- •* 9 characterized in this goe the polarized surfaces undergo an electrostatic pressure of which the resulting value (p) can be calculated. When moving the isolated system, the energy involved, and measured from an external reference, is equal to where (5, ) corresponds to the main torque resulting from the polarized surfaces measured in the direction of movement. This work V«?.corresponds to a work of relaxationIn the course of movement the axmatuies preserve the cats ( tfl ), therefore (e) remains constant, just like ( ^ ). IF the isolated system is not subjected to a braking force, it acquires an acceleration, where. (m) corresponds to the mass of the isolated system. We can then write: Thus, the energy collected wtlf.a'wou W«it_L is none other than the energy which should be spent to maintain the electrostatic pressure (?
) or (^ ). Now these two values depend on (e) being independent of time. So the 10 most important term becomes time (t). Ifl/devices and methods according to claims 1-2-3-4-5-6-7-8-9 and 10 characterized in that the isolated systems described above generate a resulting expansion force (). This can be used either to ensure propulsion and/or lift in any gravitational vector field, of any terrestrial, naval, air and/or space, as shown schematically in Figure 36. Several isolated systems can be mounted on the same machine, by orienting them in various directions and activating them more or less intensely by varying the ddp applied to each of them, which gives the machine unequaled maneuverability, as shown schematically. figure 51. This expansion force can also be used as a driving force. By fixing the isolated system at the end of a lever lever (l), itself being fixed to an axis of rotation which drives a generator; (figure 33), the expansion force () propels the isolated system, this which causes the rotation of the axis and drives the generator (alternator for example). the power collected is equal to D=fJ. «J „ It can be as large as we want since it is directly proportional to ( û> ).
Tfo stabilized speed will be reached when the counter electromotive force of the driven alternator (fig 35) will be equal in module and opposite in direction to 30 the driving force (î^)