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Absolute nonsense. ...as to be expected from dumbass Myers.On Sunday, April 8, 2012 9:57:56 AM UTC-6, k...@att.bizzzzzzzzzzzz wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned. Go back to your 555s.
You're as dumb as DimBulb.
If you use the term "watts" in connection with AC, any engineer (as opposed to engineering student) will understand that you are talking about true power, which has been corrected for the phase difference or "power factor." Watts IS watts; no engineer worth the title will confuse "watts" with "volt-amps."
I * V does equal W in AC, since in AC work that HAS to be a vector calculation.
---John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
---if you look at this formula.
P = I+V*Cos(x), you'll notice that "I" is used as "Amperes" here.
This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.
Moron, reading between the lines, anyone with as much as half a brain, wouldOn Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
What a dumbass.But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
What a stupid twit. "<>" == NOT EQUAL TOAnd, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
Like I said, you're as dumb as AlwaysWrong. Keep proving it.I guess you never made it that far, though...
Like AlwaysWrong, you insist on proving that you're *always* wrong. I have---
Go back to your 555s.
---
Interesting that those of you who haven't been able to get a handle on
how to use a 555 efficaciously, for any purpose, try to use your
ignorance to make us, who use them with delight, seem inferior.
AlwaysWrong.---
You're as dumb as DimBulb.
---
Perhaps, but it seems you still have a long way to go before you get
here.
He didn't. You know, that pot and kettle diversion.On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
No, he was talking about volts and amps. You can't get to watts from there.On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
That implies a load with a negative impedance.On Sat, 07 Apr 2012 19:56:22 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson"
wolftracks@invalid.com> wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
1000 watts at 120VAC is about 8.3 amps.
1000 watts at12VDC is about 83 amps. <--typo? Shouldn't it still be 8.3?
If you have an efficient switching regulator, as many DC/AC converters
are, and you connect them to a constant load, they will indeed exhibit
a negative input impedance. As you increase the DC input voltage, the
input current will drop, such as to maintain about constant input
power.
---
But, in the poster's context, that's all irrelevant since what he was
asking was that for constant power into a load, as _output_ voltage
decreases, must _output_ current not increase.
The original post didn't specify if the measurements were on the inputIndeed it must and - for a perfect supply with a constant voltage
input - the input current will change not one whit.
---On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
No, when you se "VA" you simply don't know "W"; not enough information given.John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
What? Did you really mean to write that nonsense?PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
You've just specified the PF by stating a resistive load, so no, you're stillHaving AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
Your "COS(X)" *is* the power factor, which is only true for sine waves.if you look at this formula.
P = I+V*Cos(x),
???you'll notice that "I" is used as "Amperes" here.
WTF, is the COS(X) term, if not to cover reactive power? Jamie, go back toThis is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.
Then your comments about what first-year engineering students wouldOn Sun, 08 Apr 2012 13:40:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
---
None.
I'm largely self-taught.
On Sun, 08 Apr 2012 16:03:02 -0400, Jamie
jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:
John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
---
There's no such thing as "RMS power."
Oh really..On Sun, 08 Apr 2012 16:03:02 -0400, Jamie
jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:
John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
---
There's no such thing as "RMS power."
---
think I intended it to be that way ? If so, you are naive.if you look at this formula.
P = I+V*Cos(x), you'll notice that "I" is used as "Amperes" here.
This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.
---
You don't know what you're talking about.
Well excuse me, I slipped with the keyboard. I hope you really don't
And if you want to start punching at the bit, from what I can see withIn the first place, it's not I+V*cos(x), it's I*V*cos(x) and, in the
second place, the cos(x) term is used to determine the actual power
dissipated.
It shows.On Sun, 08 Apr 2012 13:40:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
---
None.
I'm largely self-taught.
I first encountered it in college in calc class and then in second semesterOn Sun, 08 Apr 2012 15:53:20 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 13:40:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
---
None.
I'm largely self-taught.
Then your comments about what first-year engineering students would
have been taught, and people not making it that far, are absurd.
I don't recall learning about reactive power first-year; that was more
like year 2 or 3, when I took the electric machinery courses.
We barely touched on circuit theory in physics class, just a veryOn Sun, 08 Apr 2012 14:10:19 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 08 Apr 2012 15:53:20 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 13:40:56 -0700, John Larkin
jjlarkin@highNOTlandTHIStechnologyPART.com> wrote:
On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
I guess you never made it that far, though...
What engineering school did you graduate from?
---
None.
I'm largely self-taught.
Then your comments about what first-year engineering students would
have been taught, and people not making it that far, are absurd.
I don't recall learning about reactive power first-year; that was more
like year 2 or 3, when I took the electric machinery courses.
I first encountered it in college in calc class and then in second semester
physics. ...both first year.
Joining the AlwaysWrong brigade?krw@att.bizzzzzzzzzzzz wrote:
On Sun, 08 Apr 2012 16:03:02 -0400, Jamie
jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:
John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
No, when you se "VA" you simply don't know "W"; not enough information given.
With an inverter you can *bet* the power factor is not unity.
PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
What? Did you really mean to write that nonsense?
Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
You've just specified the PF by stating a resistive load, so no, you're still
wrong. Without the PF explicitly stated you *cannot* get there from here.
if you look at this formula.
P = I+V*Cos(x),
Your "COS(X)" *is* the power factor, which is only true for sine waves.
you'll notice that "I" is used as "Amperes" here.
???
This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.
WTF, is the COS(X) term, if not to cover reactive power? Jamie, go back to
your ham shack.
How lonely you must be.
You don't have to tell us about your Saturday night. Really.Isn't it a bitch when sitting there late at night scratching your
balls because you, just don't get what you though you did.
Sorry, but you failed EE-101, as usual, Maynard.Sorry, But you failed, again.
You *certainly* don't show it here. You're wrong about just about everything.As for Ham radio, that is just a Hobby and social hour get-together, I
actually do work in the Electrical/Electronics field and I do very well
at it. I take that opinion from bonuses, raises and royalties earned.
"Engineering level work" tells it all. Like Fields (and DimBulb), you *aren't*I've never had a problem being employed and been so for many years
performing engineering level work. I guess If "I" was so miss led as you
seem to think, our company that I work for would have been in trouble
years ago.
You even sound like AlwaysWrong; "I worked for". "I worked on". "I workedOh btw, I used to work for a company that made electronic components
for the patriot missiles. Yes, the actual electronics involved in the
missiles and transport. I bet that really scares you now !
I believe you! Really, I do!I must say, ignorance is bliss.
I don't drink, dimwit. You really are slow.Put the beer down, it's impairing your judgment.
How lonely you must be.On Sun, 08 Apr 2012 16:03:02 -0400, Jamie
jamie_ka1lpa_not_valid_after_ka1lpa_@charter.net> wrote:
John Fields wrote:
On Sun, 08 Apr 2012 09:39:37 -0700, Fred Abse
excretatauris@invalid.invalid> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields wrote:
Watts is watts...
But ain't always Volt-Amperes.
---
Of course, but since the OP couched his problem in terms of watts,
then VA is irrelevant.
When I see the term "VA", I know we're dealing with "REACTIVE" power.
No, when you se "VA" you simply don't know "W"; not enough information given.
With an inverter you can *bet* the power factor is not unity.
PF (Power Factors) denotes the difference between "REACTIVE" and
"RESISTIVE (True power)" So, using the term VA is assumed power.
What? Did you really mean to write that nonsense?
Having AC in the equation has nothing to do with it actually, I can
put AC into a purely non reactive load and it would simply power.. There
difference being is, you need to take measurements along the vectors to
come with a sum of power with in a time frame. Normally, with a clean
sinusoidal wave, we just assume RMS power.
You've just specified the PF by stating a resistive load, so no, you're still
wrong. Without the PF explicitly stated you *cannot* get there from here.
if you look at this formula.
P = I+V*Cos(x),
Your "COS(X)" *is* the power factor, which is only true for sine waves.
you'll notice that "I" is used as "Amperes" here.
???
This is a AC power formula but you don't see any distinction here with
the use of "VA" as would be in case of "REACTIVE" power.
WTF, is the COS(X) term, if not to cover reactive power? Jamie, go back to
your ham shack.
---On Sun, 08 Apr 2012 14:43:57 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 11:57:56 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 10:18:16 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sun, 08 Apr 2012 09:53:38 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sun, 08 Apr 2012 03:59:19 -0500, John Fields
jfields@austininstruments.com> wrote:
On Sat, 07 Apr 2012 20:18:26 -0400, "krw@att.bizzzzzzzzzzzz"
krw@att.bizzzzzzzzzzzz> wrote:
On Sat, 07 Apr 2012 17:02:40 -0700, "W. eWatson" <wolftracks@invalid.com
wrote:
Could the following be a typo. Written by someone to me on inverters (DC
to AC).
Volts x Amps = Watts so as the voltage goes down, the amperage goes up
to maintain the same number of watts.
This isn't strictly true for AC.
---
P
How does I = --- manage to not work out for AC?
E
PF, dumbass.
---
Watts is watts...
But, as any first-year engineering student knows, I * V <> W, where AC is
concerned.
---
If you had perspicacity, and could read between the lines, you'd have
noticed that the OP couched his query in terms of watts, implying the
load was resistive.
Moron, reading between the lines, anyone with as much as half a brain, would
understand that he was talking about an INVERTER, which is *NOT* resistive.
---But, since you don't, you missed that the cosine of the phase angle
between voltage and current - in the resistive load he alluded to -
would be 1, and volt-amperes would be precisely equal to watts.
What a dumbass.
---And, by the way, any first-year engineering student would have been
taught that, in a reactive circuit, your: "I * V <> W" is nonsense
since volt-amperes can be greater than - but never less than - watts.
What a stupid twit. "<>" == NOT EQUAL TO
---I guess you never made it that far, though...
Like I said, you're as dumb as AlwaysWrong. Keep proving it.
---Go back to your 555s.
---
Interesting that those of you who haven't been able to get a handle on
how to use a 555 efficaciously, for any purpose, try to use your
ignorance to make us, who use them with delight, seem inferior.
Like AlwaysWrong, you insist on proving that you're *always* wrong.
---I have
nothing against the 555, just one-trick-ponies, like you.
---You're as dumb as DimBulb.
---
Perhaps, but it seems you still have a long way to go before you get
here.
AlwaysWrong.
I first encountered it in college in calc class and then in second
semester physics. ...both first year.
There is nothing to be ashamed of in being self-taught. Lots of highlyI'm largely self-taught.
It shows.
As I try to follow along through this thread, I find myself having toThere's no such thing as "RMS power." ---