Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.

Richard wrote:

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is
only an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything
other than what he has written.
...

Actually, he's probably low. That would be true only if the efficiency of
the device is 100% and most of them are probably no where near that
efficient.

So explain how he's wrong, other than that?

Richard wrote:

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is
only an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything
other than what he has written.
...

Actually, he's probably low. That would be true only if the efficiency of
the device is 100% and most of them are probably no where near that
efficient.

So explain how he's wrong, other than that?

In New Youirk -- We Say "Forget A Bout It." Why talk to a concrete wall.

If you try to correct all the false stuff on the radio groups or anywhere
else -- you would have to hire a full time team of a dozen people (at
least).


Caveat Lector -- Let the reader beware and leave it go at that.

Regarding the power calculation -- see URL:
http://www.ameritron.com/ameritron/p...rodid=ALS-500M

This is a 400 watt CW (500W PEP) mobile amplifier and it "requires 14 VDC
at 80 amperes peak current for PA transistors and separate line for 14 - 16
VDC at 4 amperes for control and bias circuits.That ought to help. Tis a
matter of efficiency

You do the math

Elmer E Ing

---------------------------------------------------

"Richard" anom@anom wrote in message ...
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.

In New Youirk -- We Say "Forget A Bout It." Why talk to a concrete wall.

If you try to correct all the false stuff on the radio groups or anywhere
else -- you would have to hire a full time team of a dozen people (at
least).


Caveat Lector -- Let the reader beware and leave it go at that.

Regarding the power calculation -- see URL:
http://www.ameritron.com/ameritron/p...rodid=ALS-500M

This is a 400 watt CW (500W PEP) mobile amplifier and it "requires 14 VDC
at 80 amperes peak current for PA transistors and separate line for 14 - 16
VDC at 4 amperes for control and bias circuits.That ought to help. Tis a
matter of efficiency

You do the math

Elmer E Ing

---------------------------------------------------

"Richard" anom@anom wrote in message ...
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.

"Z.Z." wrote in message
...
Richard wrote:

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is
only an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything
other than what he has written.
...

Actually, he's probably low. That would be true only if the efficiency of
the device is 100% and most of them are probably no where near that
efficient.

So explain how he's wrong, other than that?

He's mixing RF power with line power.
As I tried to explain it to him, if you were to put a voltmeter at the RF
connector and keyed down, you would not see 12 volts at 100 amps as he
claims.
In order to do that, you'd need an antenna cable an inch thick for the
center wire.
Plus the fact, that his theory suggests the antenna is receiving that much
juice, would ultimately fry the antenna every time you keyed down.
Standard RF coax such as RG8 does not handle more than a few amps at the
most simply for the fact of it's size.
Plus the fact that internal wiring of the amplifier would also have to be
capable of handling the extreme amperage.
I don't think circuit boards can handle it.
Let alone any transistors, resistors, or capacitors.
Do they even make a 1200 watt resistor?

I also tried to compare his theory with that of an inverter.
The rated output power is by no means any where near what the rated input
power is.
Obviously, to many people just do not understand the bare basics of
electronics and radios.

"Z.Z." wrote in message
...
Richard wrote:

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is
only an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything
other than what he has written.
...

Actually, he's probably low. That would be true only if the efficiency of
the device is 100% and most of them are probably no where near that
efficient.

So explain how he's wrong, other than that?

He's mixing RF power with line power.
As I tried to explain it to him, if you were to put a voltmeter at the RF
connector and keyed down, you would not see 12 volts at 100 amps as he
claims.
In order to do that, you'd need an antenna cable an inch thick for the
center wire.
Plus the fact, that his theory suggests the antenna is receiving that much
juice, would ultimately fry the antenna every time you keyed down.
Standard RF coax such as RG8 does not handle more than a few amps at the
most simply for the fact of it's size.
Plus the fact that internal wiring of the amplifier would also have to be
capable of handling the extreme amperage.
I don't think circuit boards can handle it.
Let alone any transistors, resistors, or capacitors.
Do they even make a 1200 watt resistor?

I also tried to compare his theory with that of an inverter.
The rated output power is by no means any where near what the rated input
power is.
Obviously, to many people just do not understand the bare basics of
electronics and radios.

Yon knucklehead is assuming 100% efficiency or doesn't know the meaning of
the term.

So ask Yon knucklehead how one builds an amplifier that yields 1200 Watts
of DC Power into 1200 Watts of RF power ??
Lots of folks would like to know how to do that.

Power out divided by power in times 100 = % efficiency

"Richard" anom@anom wrote in message ...
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.

Yon knucklehead is assuming 100% efficiency or doesn't know the meaning of
the term.

So ask Yon knucklehead how one builds an amplifier that yields 1200 Watts
of DC Power into 1200 Watts of RF power ??
Lots of folks would like to know how to do that.

Power out divided by power in times 100 = % efficiency

"Richard" anom@anom wrote in message ...
Some knucklehead in misc.transport.trucking is showing off what a real ass
he is in the way of knowledge.



That should take you to the proper thread. If not, look for the post from
"mad dog" posted at 9:35 am.

I kid you not, these are his words:

quote
Keep the P=IxE formula in mind when choosing a mobile amplifier.
Power = Current x Voltage
or Current = Power / Voltage
P=Power or Wattage
I =Current draw
E=Energy or Voltage
Example: Tommy truck driver wants to install a 1200 watt spurious
emissions generator on his Kenworth with a 12VDC electrical
system. We just learned that Current draw is calculated by dividing
the Voltage into the Wattage so if the 1200 watt cataract producer has
a Input Voltage of 12 Volts then the Current draw will
be...............................
One Hundred Amps (100). Yes folks you read that right,
that is more current draw than most alternators can produce.So please
be realistic when adding accessories to your already marginal electrical
system or you may find yourself playing the "Prince of Darkness"
on a very wet and rainy night.

--
714 Sandpile,
The Mad Dog wavin' good bye

/quote

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is only
an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything other
than what he has written.
So if one of you could come in to the thread and explain how damn wrong he
is, I would appreciate it.

thanks for the time.

Oh boy -- lets assume an antenna is totally resistive -- 50 ohms or better
still a dummy load of 50 ohms that exhibits no inductance or capacitance.

Now since Power in watts = I squared R where I is the current and R is a
pure 50 ohm resistance
transpose and solve for I = square root of P over R

I get about 4.9 amperes RF current

now

E=I times R so 4.9 times 50 = 244 volts RF volts

That's what you should see at the antenna. Try that on old knucklehead.

With any inductive reactance or capacitive reactance --- different ball
game.

Gurus check my math please




Elmer E Ing


"Richard" anom@anom wrote in message
...

"Z.Z." wrote in message
...
Richard wrote:

He is wrongfully assuming that RF power output is equal to saying
12volts*100amps.
Assuming the rated output power is 1200 watts.
As I have dealt with radios for years, I know for a fact that this is
only an idiot's explanation.
He's big bad mister trucker trying to show off how smart he is.

I have tried to explailn it to him and he will not believe anything
other than what he has written.
...

Actually, he's probably low. That would be true only if the efficiency
of
the device is 100% and most of them are probably no where near that
efficient.

So explain how he's wrong, other than that?

He's mixing RF power with line power.
As I tried to explain it to him, if you were to put a voltmeter at the RF
connector and keyed down, you would not see 12 volts at 100 amps as he
claims.
In order to do that, you'd need an antenna cable an inch thick for the
center wire.
Plus the fact, that his theory suggests the antenna is receiving that much
juice, would ultimately fry the antenna every time you keyed down.
Standard RF coax such as RG8 does not handle more than a few amps at the
most simply for the fact of it's size.
Plus the fact that internal wiring of the amplifier would also have to be
capable of handling the extreme amperage.
I don't think circuit boards can handle it.
Let alone any transistors, resistors, or capacitors.
Do they even make a 1200 watt resistor?

I also tried to compare his theory with that of an inverter.
The rated output power is by no means any where near what the rated input
power is.
Obviously, to many people just do not understand the bare basics of
electronics and radios.