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.

"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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

RF power is not electrical power.
As audio power is not electrical power.

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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

RF power is not electrical power.
As audio power is not electrical power.

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

Richard wrote:

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


RF power is not electrical power.
As audio power is not electrical power.

????????????? What kind of power is RF? Magical? At which frequency does
the electric current stop following the laws of physics?

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

Are you serious? And you wanted *us* to get on that newsgroup and make
asses of ourselves like you just did?

Better check your math and your understanding of the Ohm's law, the
Power equation and the efficiency of the AB class amplifiers.

U= I*R, P= U*I = P= I^2*R or P= U^2/R

= I= SQR(P/R) and *not* I= SQR(P)/R like you mistakenly claim

Elmer's calculation is 100% correct. The trucker is right. You are wrong.



73 .... WA7AA




--

Anti-spam measu look me up on qrz.com if you need to reply directly

Richard wrote:

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


RF power is not electrical power.
As audio power is not electrical power.

????????????? What kind of power is RF? Magical? At which frequency does
the electric current stop following the laws of physics?

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

Are you serious? And you wanted *us* to get on that newsgroup and make
asses of ourselves like you just did?

Better check your math and your understanding of the Ohm's law, the
Power equation and the efficiency of the AB class amplifiers.

U= I*R, P= U*I = P= I^2*R or P= U^2/R

= I= SQR(P/R) and *not* I= SQR(P)/R like you mistakenly claim

Elmer's calculation is 100% correct. The trucker is right. You are wrong.



73 .... WA7AA




--

Anti-spam measu look me up on qrz.com if you need to reply directly

I is NOT the the square root of power. It is the square root of (power
divided by R) You first have to divide the power by the resistance THEN take
the square root.

And E=I x R not I x I x R

Back to Ohms Law --see URL:
http://www.angelfire.com/pa/baconbacon/page2.html

and AC RMS power is the same as DC power. Provided the circuit has no
inductance and capacitance and RMS values are used -- see OHMS LAW URL above
and URL:
http://www1.jaycar.com.au/images_uploaded/ohmpower.pdf

About now I think you are putting me on, so the old Elmer is exit stage
left.

Elmer E Ing
------------------------------------------------

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

"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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

RF power is not electrical power.
As audio power is not electrical power.

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

I is NOT the the square root of power. It is the square root of (power
divided by R) You first have to divide the power by the resistance THEN take
the square root.

And E=I x R not I x I x R

Back to Ohms Law --see URL:
http://www.angelfire.com/pa/baconbacon/page2.html

and AC RMS power is the same as DC power. Provided the circuit has no
inductance and capacitance and RMS values are used -- see OHMS LAW URL above
and URL:
http://www1.jaycar.com.au/images_uploaded/ohmpower.pdf

About now I think you are putting me on, so the old Elmer is exit stage
left.

Elmer E Ing
------------------------------------------------

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

"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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

RF power is not electrical power.
As audio power is not electrical power.

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.

Your assuming the input and the ouput power are the same, not so. For most
amps the effiiciency is about 50 to 60% of the input. For solid state
figure 50%. So in order to figure the current needed to run the amp you need
to assume a 50% rule. he was only partly right. So you take his answer and
divide it by 0.5 and you get 200Amps. That's right folks.

73 Dan N0FQN
"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.

Your assuming the input and the ouput power are the same, not so. For most
amps the effiiciency is about 50 to 60% of the input. For solid state
figure 50%. So in order to figure the current needed to run the amp you need
to assume a 50% rule. he was only partly right. So you take his answer and
divide it by 0.5 and you get 200Amps. That's right folks.

73 Dan N0FQN
"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.

On Sun, 27 Jul 2003 22:16:15 -0500, "Richard" anom@anom wrote:


"Elmer E Ing" wrote in message
news:VM_Ua.12050$ff.5596@fed1read01...
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

RF power is not electrical power.
As audio power is not electrical power.

Square root of P or 1200 watts in our case = 34.641
34.641/50 = 0.6928 amps.

E=IIR

(0.6928*0.6928)*50 = 23.99 volts.

Therfor, the wire can easily handle the power.
Use of a calculator helps.
Your theory was correct. The decimal point was not.




Elmer's calculations were correct.

Power(P) = voltage (E) x current (I), and
voltage (E) = current (I) x resistance (R)

so... P = I * I * R

and I = square root( P / R )
= sqr( 1200 / 50 )
= sqr( 14 )
~= 4.898979 Amps

and E = I * R = 4.9 * 50 ~= 245 volts

RG-58C/U (Belden 8262) is rated at 1,400 volts RMS. The 20 AWG center
conductor of RG-58C/U (Belden 8262) is good for about 6 amps RMS.

Ref:
http://ecom.belden.com/static/ZZBLDN...TA.HTM?P0=8262


73 de Leigh W3NLB