http://www.youtube.com/watch?v=n5RyNKNjzbU

**Stay tuned for project updates!**

1. Assume motor turns at 7200 rpm = 120 revolutions/second

2. Assume there are 8 contacts/revolution

3. Therefore there are 120 * 8 = 960 contacts/second

This means that during 1 cycle of 60 Hz AC input waveform,
there will be 16 spark gap firings, or 8 spark gap firings in each 120
Hz half cycle (positive or negative).

Since the voltage peaks every 120 Hz, there will be a
predominance of this frequency in the transmitted modulated note if the
rotary spark gap is synchronized to the AC input waveform (multiples of
120 Hz). This is too low a frequency to have a piercing transmitted
signal. Therefore, the motor should be run asynchronously (actually
the better term in free running, as some AC induction motors are asynchronous
but still follow the AC frequency with a fixed delay, or phase shift).
The best is a universal AC/DC series connected (armature and field windings),
and run at a significantly different rate, such as 7000 rpm or 7500 rpm
in this example. This will make the 933 - 1000 Hz tone be predominant
in the transmitted modulated note. (The alternative is to use a 500
- 1000 Hz alternator as the primary supply fequency and a high frequency
synchronous motor = very expensive solution!)

4. Before each spark gap firing, the capacitor charges up: Energy = 1/2 * Capacitance * Voltage squared, or E=1/2CV^2

5. For C = 0.01 uF, V = 14,100 volts peak AC, E = 1 Joule

6. For a resonant operating frequency in the primary tank, or "oscillation transformer", of 1500 kHz, period = 0.7 microseconds

7. For a primary LOADED tank circuit Q=7 (set by the primary capacitor and spark gap resistance), there are approximately 10 damped oscillation cycles

8. Therefore each spark discharge will create RF energy for 10 * 0.7 = 7 microseconds

9. Power = Energy / Time

10. Therefore the PEAK power transmitted in each 7 microsecond discharge period = 1 Joule / 7*10^-06 = 143 kilowatts!

11. The AVERAGE power is 143 kW * 7 microseconds * 960 repeats per second = 960 watts

The hot discussion of the day in 1912-1920 was what transmitted modulated note was best, higher or lower tone. With everything equal, a lower tone or repetition rated made the average transmitted power proportionately lower. However, this gave more time for the primary capacitor to charge up fully to whatever the value of the AC input voltage was at the time during the input AC cycle, so one could end up with higher peak and therefore average transmitted power. Too low a note though, and intelligibility suffered because of noise in the receiver.

In a Tesla coil design, the load is a high Q secondary coil, not an antenna, and since average power is not the important goal, but peak power to create the highest terminal voltage, the design criteria are slightly different. Here synchronous rotary gaps and 120 Hz repetition will give the highest peak power. Note that a "CW" transmitter, while giving much higher average power for communications, will be the WORST type of Tesla coil driver by orders of magnitude, simply because of this peak/average ratio difference!

Since the peak powers of even a modest spark transmitter are 10's or even 100's of kilowatts, the antenna system must be extremely well insulated with 12" insulators, and at night, glowing bluish-purple "creepy crawlers" or corona will be seen on the antenna wires!

To keep the bandwith as narrow as possible, the oscillation transformer used "loose coupling" to the antenna resonant circut, with this circuit having a Q as high as 35. With a secondary Q of 35, the -3 dB power bandwidth points at 1500 kHz was +/- 21.4 kHz. This is an amazingly respectable figure for a 1912 transmitter. A proper station was therefore NOT as wide as a barn door as common gossip seems to indicate.

Long Live King Spark!

http://www.youtube.com/watch?v=Uua5cO3_26c

An important equation and design notes:

e^-dn = 0.01

where d is the logarithmic decrement of a damped wave

n are the number of waves or cycles

0.01 represents 1/100 of the original amplitude of the wave, a useful
"fully decayed" approximation, since this represents 1/10000 of the original
power

(-40 dB).

The Radio Act of 1912 stated that the logarithmic decrement of the wave had to be 0.2 maximum, thus ensuring at a minimum, n = 23 waves.

Since d=pi/Q

Q = 16 was the minimum acceptable system Q, including the antenna.

**NOTE:**

All transmitting antennas, unlike receiving antennas, need to be resonated
at the operating frequency so that they present a relatively low and resistive
impedance to the transmitter for maximum power transfer. At low frequencies,
to avoid using loading coils with attendant ohmic losses, it is important
to use large capacitive loading giving rise to the ubiquitous multiple
wire flat top antennas. The radiation resistance is determined by
the vertical wire length and the amount of top capacitive loading, since
this loading forces the current distribution along the vertical wire to
become completely uniform rather than linearly decreasing to zero at the
top. Thus the top loading wires do not contribute much to the radiation
of the electromagnetic waves, and serve only to tune the vertical part
of the antenna to a lower resonant frequency and to raise its radiation
resistance. This increase in radiation resistance of the antenna
also improves the efficiency of the antenna because any ohmic and ground
losses that are present are then smaller in proportion.

The Q of the antenna is calculated from the equivalent series Rradiation,
L, and C lumped components centered at the operating frequency.

Q = Xc/Rradiation.

The capacitance C, in picofarads, of a flat top antenna is:

C = 1.1124 (36 * SQRT (A) + 7.97 * A / h)

where A = area of the flat top, in square meters

h = height of the antenna above ground, in meters

Example:

Inverted L flat top antenna:

horizontal length = 120 feet

height = 40 feet

flat top comprised of 4 wires spaced within 10 feet width

Resonant wavelength = 4* (40 + 120 feet) * 0.3048 meters/feet = 195 meters, or 1500 kHz.

A = 120 * 10 = 1200 sq. ft. = 111.5 sq. meters

h = 40 feet = 12.2 meters

C = 504 pF

Xc = 1/(2*pi*f*C) = 210 ohms at 1500 kHz

This antenna can be considered to be a Marconi vertical of 40 feet, with top capacitive loading. For this short monopole over ground, with capacitive loading which results in an almost UNIFORM current distribution throughout the 40 foot length, the formula for Rradiation is:

Rradiation = 160 * (PI)^2 * h^2 / wavelength^2

where h = 40 feet = 12.2 meters, and wavelength = 195 meters

Therefore Rradiation = 6.2 ohms

Therefore in this example, Q = 210/6.2 = 33.8, or a decrement of 0.09, a truly excellent figure.

The loaded Q thus calculated is the one to use in the above design equations starting at step no. 7 concerning the peak power calculation, rather than the oscillation transformer's primary Q which is only 7. The energy will be spread out over a longer time in the actual antenna wires, and so the power calculation applies there.

Titanic CQD/SOS distress call using a spark transmitter