AndrewH
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In what regard? I go into technology that could of been used at the time from the simple; using sirens to create a thermal field bending light rays to create a mirage effect, to other possible methods. I go into Einsteins Unified Feild theory, explain it, and how it was said to be used in relation to the PX, other technologies that could be employed to create optical invisibility... I go into detail analysis of Bieleks description of the equipment and how since he was an electrical engineer it simply just does not add up for example; (I hope this shows I did do some "due diligence upon the technology mentioned" 
Can a Destroyer Escort power all the equipment Al Bielek’s states was used in the experiment?
A massive power deficiency exists based on the documentation of Bielek’s lectures. Considering that Bielek claims to have a PhD in some area of science, he should have caught the error. The following analysis may negate the implementation of the Eldridge system.
“..and the two generators remained essentially the same, 75 KVA each.” – Al Bielek
The total power available is:
Pmax = VA x pf watts.
Where P is the available power in watts, VA is the Volts x Amps at the generator output and pf is the power factor.
A multi-phase generator is assumed. Whenever a multi-phase generator is connected to a load and modulation of the load is present (It was stated that the transmitters were modulated), the efficiency of power transfer is decreased by the power factor. The pf can be from 0.9 to 0.5 for a modulated transmitter load. Assume a pf of 0.9. The total power available from the two generators is:
Pmax = 2 x 75,000 x 0.9 = 135,000 watts.
Apparently, there were two loads, The first:
“The final version some 3,000 ‘6L6’ tubes were used to drive the field coils of the two generators, and that was an accurate rendition, by the way, perhaps not to the exact position, but actually in the sense that there were a large number of tubes, about 3,000 by count.” – Al Bielek
The primary power source is presumed to be the generators, though the rhetoric appears contradictory. The load is presumed to be the 4 deck mounted coils (not the generators) which were driven by 3,000 6L6’s. The cathode power required for the 6L6 is 6.3vac at 0.9a or 5.67 watts. A transformer converts the primary voltage to the cathode heater voltage level. Transformer efficiency is about 0.9. The generator power required for the cathodes is:
Pcathode = 3,000 x 5.67 / 0.9 = 18,900 watts.
The maximum 6L6 power output (Class A operation) is 17.5 watts. The 6L6 Plate Efficiency is assumed to be the maximum available, 30%. A power supply is required to convert the ac primary power to dc for the plate. Power supply efficiency is about 80%. The generator power required to drive the amplifiers is:
Pplate = 3,000 x 17.5/(0.3 x 0.8) = 218,750 watts.
The total 6L6 power required is:
P6L6 = Pcathode + Pplate = 18,900 + 218,750
= 237,650 watts.
“So various subsystems were tested; the generators, the RF transmitters. Tesla used three, Von Neumann went to four and he finally decided the power of the transmitters selected by Tesla, which were General Electric. The 500 kilowatt CW was not sufficient. He put boosters on them to raise each one to 2 megawatt CW and the two generators remained essentially the same, 75 KVA each. Fairly low frequency, motor-driven, special synchronizing circuits to make sure the two generators were in absolute sync, otherwise, it wouldn’t work. A special generator system was built with another very exotic device– it was inherited directly from Tesla– and that was the Zero Time Reference Generator.” – Al Bielek
Normally, transmitter power is given as the Effective Radiated Power (ERP) which includes antenna gain. The photos of the mast mounted antennas do not show apertures that could exhibit any gain, but for the sake of argument, I will assume a 10 dB antenna gain for each transmitter. A 2 megawatt CW transmitter is quite unusual. Most transmitters of this power level are pulsed with duty cycles less than 10%.
However, CW was stated, so the total CW ERP from the 4 antennas is then 8106 watts. Removing the antenna gain, the antenna input or transmitter output power is then 8105 watts. There are two components to a transmitter: The power amplifier and the modulator.
The efficiency of the power amplifier is approximately 30% (given the benefit of the doubt). This means that the dc input power required to drive the transmitters is 8×105/0.3 or 2.4×106 watts. The modulator (amplitude or phase modulation) has about the same efficiency as the transmitter and takes approximately 1/100 of the power of the amplifier or 0.024×106 watts. The dc power supplies that would drive the transmitter and modulator are about 80% efficient in converting primary power to dc power. The primary generator power required to drive the transmitters and their modulators is therefore:
Pt = (2.4×106 + 0.024×106)/0.8 or 3.03 megawatts.
There is a significant discrepancy in available power:
CW Power shortage = Pt + P6L6 – Pmax watts
CW Power shortage = 3,030,000 + 237,650 – 135,000
= 3,132,650 watts
If the transmitters were pulse modulated at 1% duty cycle, the average power required for the transmitters and modulators would be about 33,000 watts. There is still a power shortage:
Pulse Power Shortage = 135,000 – 33,000 – 237,650
= -135,650 watts.
The above figures do not include the losses incurred from the use of the motor-driven synchronizing devices for the generators.
Where did the missing power come from if the experiment existed as outlined by Bielek?
Can a Destroyer Escort power all the equipment Al Bielek’s states was used in the experiment?
A massive power deficiency exists based on the documentation of Bielek’s lectures. Considering that Bielek claims to have a PhD in some area of science, he should have caught the error. The following analysis may negate the implementation of the Eldridge system.
“..and the two generators remained essentially the same, 75 KVA each.” – Al Bielek
The total power available is:
Pmax = VA x pf watts.
Where P is the available power in watts, VA is the Volts x Amps at the generator output and pf is the power factor.
A multi-phase generator is assumed. Whenever a multi-phase generator is connected to a load and modulation of the load is present (It was stated that the transmitters were modulated), the efficiency of power transfer is decreased by the power factor. The pf can be from 0.9 to 0.5 for a modulated transmitter load. Assume a pf of 0.9. The total power available from the two generators is:
Pmax = 2 x 75,000 x 0.9 = 135,000 watts.
Apparently, there were two loads, The first:
“The final version some 3,000 ‘6L6’ tubes were used to drive the field coils of the two generators, and that was an accurate rendition, by the way, perhaps not to the exact position, but actually in the sense that there were a large number of tubes, about 3,000 by count.” – Al Bielek
The primary power source is presumed to be the generators, though the rhetoric appears contradictory. The load is presumed to be the 4 deck mounted coils (not the generators) which were driven by 3,000 6L6’s. The cathode power required for the 6L6 is 6.3vac at 0.9a or 5.67 watts. A transformer converts the primary voltage to the cathode heater voltage level. Transformer efficiency is about 0.9. The generator power required for the cathodes is:
Pcathode = 3,000 x 5.67 / 0.9 = 18,900 watts.
The maximum 6L6 power output (Class A operation) is 17.5 watts. The 6L6 Plate Efficiency is assumed to be the maximum available, 30%. A power supply is required to convert the ac primary power to dc for the plate. Power supply efficiency is about 80%. The generator power required to drive the amplifiers is:
Pplate = 3,000 x 17.5/(0.3 x 0.8) = 218,750 watts.
The total 6L6 power required is:
P6L6 = Pcathode + Pplate = 18,900 + 218,750
= 237,650 watts.
“So various subsystems were tested; the generators, the RF transmitters. Tesla used three, Von Neumann went to four and he finally decided the power of the transmitters selected by Tesla, which were General Electric. The 500 kilowatt CW was not sufficient. He put boosters on them to raise each one to 2 megawatt CW and the two generators remained essentially the same, 75 KVA each. Fairly low frequency, motor-driven, special synchronizing circuits to make sure the two generators were in absolute sync, otherwise, it wouldn’t work. A special generator system was built with another very exotic device– it was inherited directly from Tesla– and that was the Zero Time Reference Generator.” – Al Bielek
Normally, transmitter power is given as the Effective Radiated Power (ERP) which includes antenna gain. The photos of the mast mounted antennas do not show apertures that could exhibit any gain, but for the sake of argument, I will assume a 10 dB antenna gain for each transmitter. A 2 megawatt CW transmitter is quite unusual. Most transmitters of this power level are pulsed with duty cycles less than 10%.
However, CW was stated, so the total CW ERP from the 4 antennas is then 8106 watts. Removing the antenna gain, the antenna input or transmitter output power is then 8105 watts. There are two components to a transmitter: The power amplifier and the modulator.
The efficiency of the power amplifier is approximately 30% (given the benefit of the doubt). This means that the dc input power required to drive the transmitters is 8×105/0.3 or 2.4×106 watts. The modulator (amplitude or phase modulation) has about the same efficiency as the transmitter and takes approximately 1/100 of the power of the amplifier or 0.024×106 watts. The dc power supplies that would drive the transmitter and modulator are about 80% efficient in converting primary power to dc power. The primary generator power required to drive the transmitters and their modulators is therefore:
Pt = (2.4×106 + 0.024×106)/0.8 or 3.03 megawatts.
There is a significant discrepancy in available power:
CW Power shortage = Pt + P6L6 – Pmax watts
CW Power shortage = 3,030,000 + 237,650 – 135,000
= 3,132,650 watts
If the transmitters were pulse modulated at 1% duty cycle, the average power required for the transmitters and modulators would be about 33,000 watts. There is still a power shortage:
Pulse Power Shortage = 135,000 – 33,000 – 237,650
= -135,650 watts.
The above figures do not include the losses incurred from the use of the motor-driven synchronizing devices for the generators.
Where did the missing power come from if the experiment existed as outlined by Bielek?
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