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?

"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.

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.

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.

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

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