Newman的能源机-翻译版-第6章-更小装置的描述.doc

Chapter 6 DESCRIPTION OF SMALLER UNIT第6章 更小装置的描述WITH AN AFFIDAVIT BY DR. ROGER HASTINGS 带有黑斯廷斯博士的宣誓书The following is a smaller unit (see photograph 15-C2 below) composed of 30-gauge, insulated, copper wire weighing approximately 145 lbs. (atoms) and having a rotating magnet of 14 lbs.(atoms). This portable unit, with very little current input, clearly demonstrates an energy output which is greater than the external energy input. With 300 volts input of pressure, only 1.5 milliamps of current (volume of gyroscopic particles) went into the copper coil (of atoms), which is less than 0.5 watt input for an energy output in excess of I 0 watts.Photograph 15-C2:下面是一个小装置（看图15-C2），由30 gauge的绝缘铜线组成，重大约145磅（原子），有一个14磅（原子）的可转动磁铁。这个便携装置输入很小的电流可以明显的看到输出的能量多于输入。用300伏的输入电压，只有1.5毫安的电流（大量陀螺子）输入到铜线圈（原子），这小于1.5瓦的输入产生多于10瓦的输出。See below copy of a test conducted by Dr. Roger Hastings utilizing the 15-C2 unit.看下面Dr. Roger Hastings复制的15-C2装置的测试。AFFIDAVITJune 17, 1984TO WHOM IT MAY CONCERN :On June 16 and 17, 1984 I ran a series of tests on Joseph Newman s 145 lb . motor with 14 lb . rotary . These tests show that power is generated by the motor which greatly exceeds the battery input power . The results are summarized briefly below :1884年六月16和17号，我做了一系列关于纽曼的有14磅转子145磅发电机的测试。这些测试说明发电机产生的能量多于电池的输入能量。结果总结如下：1. Demonstration of Large Current Spikes Produced by the Motor A . Oscilloscope ReadingsThe oscilloscope showed large (1 Amp ) staircase current spikes of significant time duration , which were initiated when the commutator switched , and flowed both in the coil and battery portions of the circuit . A picture of this spike taken on the coil side of the commutator is attached . A block diagram of the circuit is shown below .1. 证明发电机产生了有高峰值的大电流A. 示波器读数示波器显示了重要的持久的大的（大于1安）梯形电流峰，当换向器切换时从新开始，流向电路中的线圈和电池。一张图版，在换向器在线圈一侧时拍摄，附在下面。电路的图表在下面展示。B . Circuit Breaker TestsAn ammeter which has a built in circuit breaker was placed in the circuit . When the meter was placed on the 100 ma scale, the breaker opened, both on the battery and coil side of the commutator .The breaker did not open when the meter was placed on the one amp scale, however, it was verified that a current input of more than 1.5 times the full scale deflection did not open the breaker.B. 电流短路测试一个有电路开关的电流表加入到电路中。当电流表放到100 毫安的偏移时,开关打开,换向器在电池和开关的一侧.当电流表电流超过1安时电路断开,然而,证明多于满刻度1.5倍的输入电流没有打开开关.C . Temperature RiseA five hundred ohm risistor was placed in series with the battery . The resistor was water-proofed and placed in a small thermos container with a precision thermometer. A temperature rise of approximately one degree Centigrade was observed in a period of fifteen minutes. To raise the temperature of the 21 grams of water by 1 degree in fifteen minutes requires at least an average power of:C. 温度升高一个500欧的薄膜晶体管与电池串连。晶体管是防水的，放到一个有温度计的热水瓶中。在15分钟内观察到大约升高了1度。在15分钟内升高这21克水1度所需求最少的平均能量为：Since the power supplied by the current flowing in the resistor is I2R, where I is the average current and R = 500, it follows that a current of at least 14 ma on the average must flow in the circuit . This result was verified experimentally by supplying 14 ma to the 500 resistor via a battery and series resistors .If the current contained in the spikes (attached photo) is averaged over the cycle time, the result is consistent with an average current of 14 ma.因为能量由电流流过晶体管产生，符合I2R，I是平均电流，R = 500，也就是说电路必需流过最少14毫安的电流。这个结果被实验证明，通过一个电池提供14毫安的电流流过500的薄膜晶体管。如果电流，包含峰值（附图中），是周期内的平均值，结果是和14毫安的平均电流等同。注：C是用电路加热水的方式来计算系统的输出能量。2. Demonstration that Large Current Spikes are not Produced by the Battery.2.证明大电流峰不是由电池产生。A.Current ReadingsWhen a Simpson amp meter is placed in series with the batteries, a d.c . input current of 1.2 ma is registered. The battery input current is therefore 1.2 ma .当一个Simpson电流计和电流串连，一个12毫安的直流输入电流被检测到。电池输入因此是1.2毫安。B.Expected Input CurrentWhen the rotor is stopped, the input current from the batteries is measured to be 6 ma (this is in agreement with 304 volts and 50 K/L coil resistance).The coil inductance , as calculated from the number of wire turns and the geometry, is 16,000 Henries. At the operating speed of 136 r.p.m., the inductive reactance of the coil is 230 K/L., which is much large than the coil resistance . The expected battery input current is 304 V/230 K/L= 1.3 ma, in good agreement with the measured input of 1.2 ma.B. 期望的输入电流当轮子停止，来自电池的输入电流为6毫安（这符合204伏和50 K/L的线圈电阻）。线圈电感，从线圈圈数和形状计算出来，是16,000 H。在136圈每分钟的速度，线圈的感抗是230 K/L，远远大于线圈电阻。期望的电流输入是304 V/230 K/L= 1.3 ma，非常好的符合测试到的输入1.2毫安。C.Constant Battery VoltageDuring four hours of continuous running of the motor, the voltage remained constant at 304 volts.If the 15 ma average current contained in the spikes came from the batteries, they would drain down significantly in the four hour period.By draining 14 ma from a fresh 9 volt transistor battery identical to those on the motor it was found that the 14 ma drain causes the voltage to drop by 2% per hour. Thus if the 14 ma were originating at the battery, the battery voltage would drop by 24 volts in four hours . No drop was observed .C.恒定的电池电压在发电机连续运行的四个小时中，电压保持在304 v的恒定水平。如果15毫安的平均电流包含来自电池的峰值，它们将在四小时内耗尽。通过从一个9v的晶体管电池释放14毫安电流，和电动机一致，发现14毫安的消耗会引起电压每小时下降2%。因些如果14毫安完全由电池产生，电池电压将将在四小时内下降24伏。但没有发现电压下降。注：C在证明能量不全来自于电池。D.Larger Current Spikes on Coil SideThe current spikes, as recorded on the scope, were larger on the coil side of the commutator than on the battery side.This indicates that the spikes originate at the coil, with some loss occuring at the commutator .D. 线圈一侧的大电流峰电流峰，如记录的幅度，换向器线圈一侧大于电池一侧。这指出峰值产生于线圈，在换向器处有一些的损耗。E.Negative CurrentA significant portion of the spike in the battery circuit is negative (opposing the battery voltage). The battery cannot generate such a negative current.E. 反向电流在电池电路峰值一个重要的部分是反向电流（和电池电压相反）。电池不能产生这样一个反向电流。F.Dependence Upon Rotary PositionThe intensity of the spikes varies greatly with the placement of the rotary. For example, when the rotor is on the side (outside) of the coil the spikes are large . They virtually disappear when the rotary is placed on top of the coil .F.依赖于旋转位置峰值的强度变化非常依赖于旋转的位置。例如，当转子在线圈一侧（外侧）峰值最大。当旋转到线圈顶部时消失。3. Power and Useful Output3.功率和有用的输出A . Output verses Input PowerSince an average of 14 ma flows through the 50 K/L coil, the heat dissipated in the coil is ten watts . The battery input is 1.2 ma times 304 volts, or 0.36 watts.The heat generated in the coil is 27 times the input power . Note that if the ten watts were delivered by the batteries, they would drain down very quickly. These batteries have been used in frequent demonstrations for long durations by Mr. Newman over the past several months . As mentioned above , four hours of motor operation during these tests did not measureably lower the battery voltage.A. 输出和输入的