wrote in message
ups.com...

My question is this: so long as final output to the antenna is within
the requirement of part 15 unlicensed operation, is part 15 unlicensed
operation allowed within a band normally governed under part 97?
Part 15 operation would easily cover a 30 mile radius on 80M.

http://www.access.gpo.gov/nara/cfr/w...7cfr15_06.html

Part 15.223 says in part...

"Subpart C_Intentional Radiators
Sec. 15.223 Operation in the band 1.705-10 MHz.

(a) The field strength of any emission within the band 1.705-10.0
MHz shall not exceed 100 microvolts/meter at a distance of 30 meters.
However, if the bandwidth of the emission is less than 10% of the center
frequency, the field strength shall not exceed 15 microvolts/meter or
(the bandwidth of the device in kHz) divided by (the center frequency of
the device in MHz) microvolts/meter at a distance of 30 meters,
whichever is the higher level."

For CW operation, your bandwidth is zero for all practical purposes. You
will therefore be allowed 15 uV/m at 30 meters.

From this I made a very crude estimate of your allowed transmitter output
power. Assuming your transmitter antenna is 100% efficient and radiates
hemispherically, your transmitter output power is allowed to be no more than
about 1.7 nanowatts (1.7e-9 watts). For ideal circumstances (0 dB receiver
antenna gain, 100% efficient receiver antenna), you would get less than 5 uV
at a receiver 400 meters away. I stress that this is for ideal
circumstances. If your transmitting antenna is less efficient, you can run
more power, but less of your power gets launched. How efficient is that
antenna, anyway? On the other hand, your antenna may have some directivity
which would further limit your power.

This is making my head hurt.

All that can really be said about this estimate is that, to be safe, you
should not have much more than about 2 nanowatts output power unless you
have the ability to measure the field strength according to regulations.

Good luck with your project.

73,
John

John,

So, say I had a CW oscillator on the colorburst frequency. I'll shunt
the output
with a 50 ohm resistor to ground. I'll put a random wire at the top of
the
resistor.

Basic equations yield P = (EE) / R

If R is 50 Ohms and maximum output power is 1.7e-9 watts, then E
is 4.1e-5 volts.

So, would a measurement of 40 microvolts or less across the resistor
be satisfactory?

Thanks,

The Eternal Squire

On Jan 25, 9:44 pm, "John" wrote:
wrote in oglegroups.com...

My question is this: so long as final output to the antenna is within
the requirement of part 15 unlicensed operation, is part 15 unlicensed
operation allowed within a band normally governed under part 97?
Part 15 operation would easily cover a 30 mile radius on 80M. http://www.access.gpo.gov/nara/cfr/w...7cfr15_06.html

Part 15.223 says in part...

"Subpart C_Intentional Radiators
Sec. 15.223 Operation in the band 1.705-10 MHz.

(a) The field strength of any emission within the band 1.705-10.0
MHz shall not exceed 100 microvolts/meter at a distance of 30 meters.
However, if the bandwidth of the emission is less than 10% of the center
frequency, the field strength shall not exceed 15 microvolts/meter or
(the bandwidth of the device in kHz) divided by (the center frequency of
the device in MHz) microvolts/meter at a distance of 30 meters,
whichever is the higher level."

For CW operation, your bandwidth is zero for all practical purposes. You
will therefore be allowed 15 uV/m at 30 meters.

From this I made a very crude estimate of your allowed transmitter output
power. Assuming your transmitter antenna is 100% efficient and radiates
hemispherically, your transmitter output power is allowed to be no more than
about 1.7 nanowatts (1.7e-9 watts). For ideal circumstances (0 dB receiver
antenna gain, 100% efficient receiver antenna), you would get less than 5 uV
at a receiver 400 meters away. I stress that this is for ideal
circumstances. If your transmitting antenna is less efficient, you can run
more power, but less of your power gets launched. How efficient is that
antenna, anyway? On the other hand, your antenna may have some directivity
which would further limit your power.

This is making my head hurt.

All that can really be said about this estimate is that, to be safe, you
should not have much more than about 2 nanowatts output power unless you
have the ability to measure the field strength according to regulations.

Good luck with your project.

73,
John

wrote:
John,

So, say I had a CW oscillator on the colorburst frequency. I'll shunt
the output
with a 50 ohm resistor to ground. I'll put a random wire at the top of
the
resistor.

Basic equations yield P = (EE) / R

If R is 50 Ohms and maximum output power is 1.7e-9 watts, then E
is 4.1e-5 volts.

So, would a measurement of 40 microvolts or less across the resistor
be satisfactory?

Thanks,

The Eternal Squire


Good question. And, although I support the effort, which was stated, in
helping the youngsters--you also must recognize I would be slow to give
advice. I don't wish to invoke the pointy-headed attorneys when I have
no horse in the race ... grin

Proceed at your own peril ...

Warmest regards,
JS

John Smith I wrote:

But .0000000017 watts?

I thought we were talking a full watt!

(more of a pirate in me than some grin)

JS

John Smith I wrote:

Anyway, I would pursue this much differently.

Indian reservations are exempt from MANY of the rules, regulations and
laws the rest of us non-indigenous peoples are subject too.

I'd see if the above were not a factor in all this.

JS

John Smith I wrote:
John Smith I wrote:

Anyway, I would pursue this much differently.

Indian reservations are exempt from MANY of the rules, regulations and
laws the rest of us non-indigenous peoples are subject too.

I'd see if the above were not a factor in all this.

JS

Just think! Instead of the Indians having a monopoly on all gambling in
California (yes, I know, it is really vegas using the Indians), they
could now go ahead and get a monopoly on all broadcasting!

JS

John Smith I wrote:
John Smith I wrote:

John Smith I wrote:

Anyway, I would pursue this much differently.

Indian reservations are exempt from MANY of the rules, regulations and
laws the rest of us non-indigenous peoples are subject too.

I'd see if the above were not a factor in all this.

JS


Just think! Instead of the Indians having a monopoly on all gambling in
California (yes, I know, it is really vegas using the Indians), they
could now go ahead and get a monopoly on all broadcasting!

JS

Myself and a colleague discussed the number of permits and variances
needed to launch real, orbital-type rockets and wondered how much would
be required if you launched from a Native reservation.

wrote in message
oups.com...
John,

So, say I had a CW oscillator on the colorburst frequency. I'll shunt
the output
with a 50 ohm resistor to ground. I'll put a random wire at the top of
the
resistor.

Basic equations yield P = (EE) / R

If R is 50 Ohms and maximum output power is 1.7e-9 watts, then E
is 4.1e-5 volts.

So, would a measurement of 40 microvolts or less across the resistor
be satisfactory?

Thanks,

The Eternal Squire

I think it would be about 300 uV across 50 ohms ( I think you are off by
about an order of magnitude). Yes, I think that would be okay. A good place
to understand this stuff is on the ARRL site. Look into the BPL and RFI
things there. Read everything you can find about this subject. If you
research this thoroughly, you will find allies in the field, I'm sure.

My intent here is not to discourage you in your project, but to encourage
you to think critically. Always ask yourself why this has not been done
before. Most of the time there is a reason, but sometimes (a small
percentage of the time) it is because nobody has had an interest in the
subject.

But, you didn't ask for my philosophy. I sometimes get carried away. I
apologize for that.

Cheers and 73,
John

If (mind you, IF) I was going to make a device that would bootleg a signal
over a 30 mile range, the LAST frequency I would pick is one that had a real
good chance of messing up somebody's color TV set in that radius.

Google on Mouser, crystal, and you will get HUNDREDS of cheap crystals that
aren't going to be messing up anybody's home entertainment devices. You
might also google for "emergency frequency" to keep away from the Coast
Guard's "hit the red button" frequencies.

If it twer me, I'd probably find a quiet spot on 80 meters for the little
rugrats.

Jim



wrote in message
oups.com...
John,

So, say I had a CW oscillator on the colorburst frequency.

Okay, suppose I make a colorburst CW oscillator and shunt it with a 50
ohm resistor.
I'd put a random wire at the top of the resistor.

So from basic equations P = (EE)/R, with a power limit of 1.7 nanovolt
this gives a voltage limit of 0.6 microvolt.

I'm afraid I don't have a voltmeter or scope that goes that low.

The Eternal Squire

On Jan 25, 9:44 pm, "John" wrote:
wrote in oglegroups.com...

My question is this: so long as final output to the antenna is within
the requirement of part 15 unlicensed operation, is part 15 unlicensed
operation allowed within a band normally governed under part 97?
Part 15 operation would easily cover a 30 mile radius on 80M. http://www.access.gpo.gov/nara/cfr/w...7cfr15_06.html

Part 15.223 says in part...

"Subpart C_Intentional Radiators
Sec. 15.223 Operation in the band 1.705-10 MHz.

(a) The field strength of any emission within the band 1.705-10.0
MHz shall not exceed 100 microvolts/meter at a distance of 30 meters.
However, if the bandwidth of the emission is less than 10% of the center
frequency, the field strength shall not exceed 15 microvolts/meter or
(the bandwidth of the device in kHz) divided by (the center frequency of
the device in MHz) microvolts/meter at a distance of 30 meters,
whichever is the higher level."

For CW operation, your bandwidth is zero for all practical purposes. You
will therefore be allowed 15 uV/m at 30 meters.

From this I made a very crude estimate of your allowed transmitter output
power. Assuming your transmitter antenna is 100% efficient and radiates
hemispherically, your transmitter output power is allowed to be no more than
about 1.7 nanowatts (1.7e-9 watts). For ideal circumstances (0 dB receiver
antenna gain, 100% efficient receiver antenna), you would get less than 5 uV
at a receiver 400 meters away. I stress that this is for ideal
circumstances. If your transmitting antenna is less efficient, you can run
more power, but less of your power gets launched. How efficient is that
antenna, anyway? On the other hand, your antenna may have some directivity
which would further limit your power.

This is making my head hurt.

All that can really be said about this estimate is that, to be safe, you
should not have much more than about 2 nanowatts output power unless you
have the ability to measure the field strength according to regulations.

Good luck with your project.

73,
John