THE FNAL NVESTGATON ON TORS EXPERMENT N AQNO S SET P n the llwing invetigatin, we ae ging t exaine the equatin Syte G, accding t Pe Aquin clai. THE EQATONS FOR THE TORS EXPERMENT ARE THE FOLLOW: Velcity EM Wave in n Pwde v µ σ 4π µσ Wking Fequency the Syte Magnetic Peeability n Pwde Cnductivity n Pwde Radiatin Reitance Syte G( n Tu i nt ued a we can ee ) 1 3 3 R π µ σ z 3 z Length the Antenna ( ) The the Paaete ae Knwn the Velcity in n Pwde. ELF Enegy Abbed by an at and calculatin the Reactive ndex in Tu Shield n R η Sa Stu n c c, whee v v 4π µσ peed light η eiciency abbptin S, S ae the Suace the n At and the Tu Suace a tu v, µ, σ ae the Shield n Tu Cuent thugh Antenna Abbed Enegy by An n at n eactin index in Shield n Tu R Radiatin Reitance the Antenna
Thickne calculatin Ttal Enegy Abbtin δ π 1 µ σ, 5 δ, we have ttal abptin by the Shield n Tu Enegy and Pwe Denity Requied Gavity Shielding an n At in Tu D Gavity Shielding Gavity Shielding DGavity Shielding n Gavity Shielding 5 n S in at c Gavity Shielding a Pwe Denity thugh an n At Gavity Shielding eactin index in n Tu Enegy equied Gavity Shielding an n At Radiated Pwe Requied the Gavity Shielding the Tu.Calculatin can be pvided by tw independent equatin. Gavity Shielding P D S Tu Gavity Shielding Gavity Shielding tu P R Tu Gavity Shielding The Cuent Peak Sinu Cuent Gavity Shielding Gavity Shielding R Radiatin Reitance the Antenna Calculatin the a the Shield Tu n V π R and V π R ttal ext tu epty inne tu V V V hield n ttal epty d V hield n in hield n Nw we will calculate the eult the given value paaete: µ µ µ π σ 10 S/ 7 75 ο 75 4 10 H/ 60 z 1 Hz
η 1 Sa Stu c, 46 10 0,374 8 3 10 19 µ µ µ π 7 5000 ο 5000 4 10 H / 7 1, 03 10 S/ σ ( ) 7 6 in at 55,85 1,66 10 Kg 9, 7 10 Kg Kg din 7840 3 0, 03175 and R 0,3 ext V inne 0,031 5 10 hield n hield n 1, 96 5 3 Kg tu Reult the equatin abve value paaete v 3,8 10 / ec R 115 OHM R 0,115 OHM v λ 47 z 1 1 λ 47 4 Macni Antenna in the n Pwde ppagating ediu v 3 48 10 / n 6 10 7 3 3 δ 0,134 10 and 5δ 0,67 10 5δ Thickne 0,67 tu
Gavity Shielding 147 10 Jule an At n 6 10 and µ 5000µ ο Tu 0 7 D Gavity Shielding 358,53 an At P 358,53 0,374, then Tu Gavity Shielding P 134,09 Tu Gavity Shielding Atic Abbtin g ( Tu Shield ) i ( Tu Shield ) i ( Tu Shield ) 1+ n 1 i( n At) c { } 7 4 1,96 Kg 1,96Kg 1+, 1 10 1 gtu ( ) 0 The Weight Tu i nulled : 0 48,76 A 34,58A P R, whee R 0,115 OHM, then P 136, 70 That ean u calculatin ae cect, becaue bee we calculated that: P 358,53 0,374, then Tu Gavity Shielding P 134,09 Tu Gavity Shielding We have a dieence,61, becaue the cuent wa calculated by SyG.xl Gaph.S thi dieence i an expected eult.me accuate the cuent, then cle t the exact pwe 134,09. S 0 144,95A ( ) 33,04Kg g Tu That ean: 33, 04Kg + 1,96Kg 35Kg
Accding t Aquin the Syte G ha an nitial Weight 35 Kg.We und that 144,95A the Syte G weight i nulled.t i clealy that all the jb i dne by the tu weight 1,96Kg. When the tu weight bece negative enugh, it nulliie the added weight the cntuctin which i 33,04 Kg (weight the n pwde,cable, e.t.c). S the Antigavity eect (Negative Ma Tu ) i epnible the abve eult the nulling the whle weight the Syte G. Then the weight 33,04Kg we can cnide it a a Weight Lad.Then the equatin ay be e-witten a llw: { } 7 4 + + 1,96Kg 1,96 Kg 1+,1 10 1 g ( Tu) Lad Lad 0 The Equatin (J.X Equatin) Aquin Syte G i the Fllwing: Kg Kg 7 4 g 35 1,96 1+, 1 10 ( ) 0 1 Syte G Then again 0 144,95A G,i nulled. the Gavitatinal Ma the Syte The equied Pwe t achieve the abve ( Null the weight Syte G and nt t Gavity Shield it) eult i: P R, whee R 0,115 OHM, then P 108,10 Nw t null a Syte G with weight 100 Kg( with added lad65kg ) in 60 Hz with ae Tu Weight and Suace, the equied cuent and pwe ae: 37,38 A and P 340
CALCLATON FOR THE SED TRANSFORMER Pe Aquin, ued the llwing tue paaete: n1 1 P iay tun n Secnday tun 1 Z1 Z 11500VA Z n1 1 0V and 36,6V n 6 and 11500VA Z ƛ 0,115OHM Z That ean the Piay t Secnday tun the Tane and 11500VA, give the ight edance ( Matching ) the ecnday, t have the axiu deliveing pwe t the Lad ( ELF Tu Antenna ). By the baic Electnic, we knw the axiu Pwe, whee can be deliveed by a Tane t a Lad, i when we have atching cnditin, between the Secnday and the Lad.That ean: Zec RLad Z R ƛ 0,115OHM 115OHM Then the Maxiu Pwe which can be deliveed by the Tane t the Antenna i: 36,6 158, 44 A 158, 44 ƛ 3, 4A 0, 31 Z + R Maxiu Pwe, whee can be deliveed t the Antenna: P R 158, 44 0,115 ƛ 886 then P ƛ 886 Thi pwe can null a Syte G with ttal weight 88,14Kg. A we have een, the equied pwe t null the weight 35Kg the Syte G, i 108,10.Thi ean, we have a technical ubject hee. 108,10 k 100 41,86% 886 ued Pwe
A we knw in Electnic, when we want t ake a Pwe Supply, we che a Tane with the Duble Deliveing Pwe. Exaple:F a Lad 100W, we che a Tane which can delive in atching Cnditin 00.Thi happen, any ean.se ae:nt velad the Tane Oveheating it in high Cuent and becaue the divegence ( +-5% ) the 0V the Netwk.Me becaue l Pwe in the ce and t the nn Main Lad cpnent. S by thi pint view, Pe ade a technical ptin, which pint in vey all pecent, that the expeient eally tk place. Cautin:The Abve invetigatin wa ceated, by cnideing that the Releative Magnetic Peeability i 5000 Tu Shield and i cntant v Cuent, Tepeatue and the act.we knw that in pactice thi cannt be happened, becaue the Peeability change v Cuent, Tepeatue and t e the Fact. S in the next Dcuent, we will ty t appach the pble, e pactical, that ean cnideing that the Peeability change v e act, which thi will help t ee the natual cnditin the expeient and the tuth abut it. Sinceely Jhn Xydu Electnic Enginee Geece