On Tuesday, July 4, 2023 at 12:36:10&#8239;PM UTC-4, TTman wrote:
> >>>> That's a surprise... 
> >>> 
> >>> That's the point. You don't understand this and seem to be unwilling to learn anything new. You say you don't understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem. 
> >>> 
> >> Yes, I see it, but it defies apparent simplistic logic. 
> > 
> > Yes, that's what we've tried to tell you. You can't just say, double resistance, so half power in the pot. 
> > 
> > 
> >> I see it's V xV 
> >> or IxI for power, hence the quarter/sq root. No wonder nothing I 
> >> designed/built worked... 
> > 
> > Yeah, if you ignore the math, things don't go well. 
> > 
> > What made the lightbulb come on? 
> >
> I went back to basics. I admit I just didn't think about this properly 
> at all. I'm 72 and clearly my brain is screwed and incapable of clear 
> thinking. Never mind, I won't be long for this world and I won't have to 
> think about things like this.

So, that is what I have to look forward to in three years?  Crap! 

I nearly stopped working for some four or five years, and I the next time I had a serious task, I could tell I had lost something.  But then I got a sizeable job and had to work intensely for months.  It pretty much all came back.  I'm not as quick with numbers, not being able to do as much in my head now.  But the logic is all there.  

One of the things I do to get to sleep is to recite the alphabet backwards.  So, I'm ready for the intoxication tests now.  

-- 

  Rick C.

  +-++ Get 1,000 miles of free Supercharging
  +-++ Tesla referral code - https://ts.la/richard11209

>>>> That's a surprise...
>>>
>>> That's the point. You don't understand this and seem to be unwilling to learn anything new. You say you don't understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.
>>>
>> Yes, I see it, but it defies apparent simplistic logic.
> 
> Yes, that's what we've tried to tell you.  You can't just say, double resistance, so half power in the pot.
> 
> 
>> I see it's V xV
>> or IxI for power, hence the quarter/sq root. No wonder nothing I
>> designed/built worked...
> 
> Yeah, if you ignore the math, things don't go well.
> 
> What made the lightbulb come on?
> 
I went back to basics. I admit I just didn't think about this properly 
at all. I'm 72 and clearly my brain is screwed and incapable of clear 
thinking. Never mind, I won't be long for this world and I won't have to 
think about things like this.

-- 
This email has been checked for viruses by Avast antivirus software.
www.avast.com

On Tuesday, July 4, 2023 at 7:02:34&#8239;AM UTC-4, TTman wrote:
> On 04/07/2023 01:05, Ricky wrote: 
> > On Monday, July 3, 2023 at 5:40:51&#8239;PM UTC-4, TTman wrote: 
> >>> No. 
> >>> ...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> >>>> making Rt - 40 Ohms. 
> >>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. 
> >>>> 
> >>> 
> >>> Correct. Now for goodness sake do the next step in the math. 
> >>> 
> >>> 125 volts across 20 ohms produces 781.25 watts in the pot 
> >>> heater and 781.25 watts in the added resistor. 
> >>> But you said you want 1500 watts in the pot heater. 
> >>> 
> >>> Ed 
> >> That's a surprise... 
> > 
> > That's the point. You don't understand this and seem to be unwilling to learn anything new. You say you don't understand what others post, then repeat the wrongthink that you have in your head. Maybe now you will pay a bit more attention to what others have posted and learn something about the problem. 
> >
> Yes, I see it, but it defies apparent simplistic logic.

Yes, that's what we've tried to tell you.  You can't just say, double resistance, so half power in the pot.  


> I see it's V xV 
> or IxI for power, hence the quarter/sq root. No wonder nothing I 
> designed/built worked...

Yeah, if you ignore the math, things don't go well. 

What made the lightbulb come on? 

-- 

  Rick C.

  +-+- Get 1,000 miles of free Supercharging
  +-+- Tesla referral code - https://ts.la/richard11209

On 04/07/2023 01:05, Ricky wrote:
> On Monday, July 3, 2023 at 5:40:51&#8239;PM UTC-4, TTman wrote:
>>> No.
>>> ...HR= 20 Ohms. Add another resistor in series = 20 Ohms
>>>> making Rt - 40 Ohms.
>>>>    If I apply 250 V across 40 Ohms, I get 125V across each resistor.
>>>>
>>>
>>> Correct.  Now for goodness sake do the next step in the math.
>>>
>>> 125 volts across 20 ohms produces 781.25 watts in the pot
>>> heater and 781.25 watts in the added resistor.
>>> But you said you want 1500 watts in the pot heater.
>>>
>>> Ed
>> That's a surprise...
> 
> That's the point.  You don't understand this and seem to be unwilling to learn anything new.  You say you don't understand what others post, then repeat the wrongthink that you have in your head.  Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.
> 
Yes, I see it, but it defies apparent simplistic logic.I see it's V xV 
or IxI for power, hence the quarter/sq root. No wonder nothing I 
designed/built worked...

-- 
This email has been checked for viruses by Avast antivirus software.
www.avast.com

Ed wrote:
> Don wrote:

<snip>

>> "If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant" (as
>> stated above) is my premise. This 250VAC is treated as Vmains by me.
>>      My mind mechanically treats Vmains as a constant - because it is.
>> Most electric outlets in my world source about 120 VAC regardless of the
>> load connected.
>>      When Vmains is held constant, a doubling of Rmains halves Pmains.
>
> Correct. He'll get half power out of the combination of HR (the heater
> resistance) plus AR (the added resistance). The original power was 3000
> watts. Half power is 1500 watts. So he'll get 750 watts out of AR and
> 750 watts out of HR.
>
> BUT THE OP WANTS 1500 WATTS OUT OF HR.
>
> Your math solves the wrong problem.
> From the op: "As part of an experiment with solar, I want to drive my
> 220V 3kW immersion heater with a suitable diode so that the effective
> power is ~1.5kW..."

There's actually a couple of problems. One problem is people who try to
shout down wrong thought of how doubling a resistor halves the power. My 
math solves this problem by illustrating an exception.
    If you keep reading my previous followup (the part you snipped)
you'll discover how my math eventually illustrates a solution to the 
OP's 1500 W problem. But, you need to read my followup in its entirety
to see it.

In the end, sloppy nomenclature caused this thread's miscommunication.

    "When I use [the word volt]," Humpty Dumpty said in rather a
    scornful tone, "it means just what I choose it to mean -
    neither more nor less."

Danke,

-- 
Don, KB7RPU, https://www.qsl.net/kb7rpu
There was a young lady named Bright Whose speed was far faster than light;
She set out one day In a relative way And returned on the previous night.

On Monday, July 3, 2023 at 6:54:37&#8239;PM UTC-4, ehsjr wrote:
> On 7/2/2023 7:35 PM, Don wrote: 
> > Ed wrote: 
> >> Don wrote: 
> >>> Ricky wrote: 
> >>>> Don wrote: 
> >>>>> Ricky wrote: 
> >>>>>> TTman wrote: 
> >>>>>>> Ed wrote: 
> >>>>>>>> 
> >>>>>>>> The communication hasn't worked so far, perhaps it's been a bit 
> >>>>>>>> unclear. I'll try a bit differently. The key point is that doubling 
> >>>>>>>> the resistance does NOT halve the power. 
> >>>>>>>> 
> >>>>>>>> Take the resistance of your heater (HR), which is about 20.8 ohms, 
> >>>>>>>> and figure what voltage needs to be applied to it to run it at 1500 
> >>>>>>>> watts. Here's the math: 
> >>>>>>>> Solve for heater resistance (HR): 
> >>>>>>>> P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83 
> >>>>>>>> 
> >>>>>>>> Now what voltage must be applied to HR to equal 1500 watts? 
> >>>>>>>> 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so 
> >>>>>>>> E = sqroot 31250 so E = 176.77volts 
> >>>>>>>> 
> >>>>>>>> Ok, once you know that voltage you can solve for the rest. 
> >>>>>>>> We need to add a series resistance (AR) to drop the voltage 
> >>>>>>>> across HR from 250 to 176.77volts 
> >>>>>>>> 
> >>>>>>>> Compute the current through the series string of HR plus AR 
> >>>>>>>> 
> >>>>>>>> Solve for I through HR at 1500 watts: 
> >>>>>>>> P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps 
> >>>>>>>> 
> >>>>>>>> The total voltage (250) minus VHR (176.77) equals VAR, the 
> >>>>>>>> voltage across the (AddedResistance) AR 
> >>>>>>>> 
> >>>>>>>> 250-176.77 = 73.223volts 
> >>>>>>>> 
> >>>>>>>> Solve for power dissipated in AR 
> >>>>>>>> 
> >>>>>>>> P=I*E= 8.486*73.223=621.32 watts in AR 
> >>>>>>>> 
> >>>>>>>> I hope this makes it a bit clearer. 
> >>>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> >>>>>>> making Rt - 40 Ohms. 
> >>>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. 
> >>>>>> 
> >>>>>> and... you get half the current, so each resistor will dissipate a quarte 
> >>>>>>>> I hope this makes it a bit clearer. 
> >>>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> >>>>>>> making Rt - 40 Ohms. 
> >>>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor. 
> >>>>>> 
> >>>>>> and... you get half the current, so each resistor will dissipate a quarter 
> >>>>>> of the original power. The total power will be half of the original power, 
> >>>>>> but if I understand your limitations on the design, the series resistor can 
> >>>>>> not be used to heat the pot. 
> >>>>>> 
> >>>>>> I'm not going to continue to beat you over the heat to get you to pay 
> >>>>>> attentn to the equations. But here are equations for power, one more time. 
> >>>>>> 
> >>>>>> P = E * I 
> >>>>>> P = E^2 / R 
> >>>>>> P = I^2 * R 
> >>>>>> 
> >>>>>> Apply them to the appropriate devices, rather than just saying, "twice the 
> >>>>>> resistance, half the power". Also, try solving for the actual current and 
> >>>>>> the voltage on each device. This isn't hard, but sometimes it takes a bit 
> >>>>>> of work to get used to it. 
> >>>>> 
> >>>>> Thanks guys. You helped me put it all together. If we keep 250VAC 
> >>>>> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind 
> >>>>> misrepresents the V^2 constant as a slope in a linear equation. 
> >>>>> But V^2 is not a slope because P=V^2/R is not linear. The 
> >>>>> multiplicative inverse term, 1/R, makes it non-linear. 
> >>>>> Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it 
> >>>>> looks similar to a 1/x plot. 
> >>>>> 
> >>>>> This (so far unanswered) question was previously asked by me: 
> >>>>> 
> >>>>> Does the -3 dB half power point play a role with kW (eg not kWh)? 
> >>>>> 
> >>>>> Ordinarily the half power point pertains to bandwidth. It's a little 
> >>>>> eccentric to apply it to a constant frequency. But it can be done. By 
> >>>>> inspection, the half power point of 3000 W yields 40 Ohms. 
> >>>>> 
> >>>>> Note. 
> >>>>> 
> >>>>> [1] octave code: 
> >>>>> 
> >>>>> R=linspace(10,50,40); 
> >>>>> P=250^2 ./ R; 
> >>>>> plot(R, P); 
> >>>>> xlabel('R (Ohms)'); 
> >>>>> ylabel('P (Watts)'); 
> > 
> > <snipped> 
> > 
> >>>> You are solving an irrelevant problem. No one cares about reducing the 
> >>>> total power to half. That's the mistake that TTman is making. I 
> >>>> originally said he could add an equivalent heater in series with the one 
> >>>> he's already using and the total power would be half. He said that won't 
> >>>> work because only the pot heater is producing useful heat. 
> >>>> 
> >>>> If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts, 
> >>>> how much resistance do you need to add in series to make the pot heater 
> >>>> dissipate 1,500 watts? 
> >>> 
> >>> By inspection, from my plot, you need to add 0 Ohms in series in order 
> >>> to dissipate 3000 W from 250 VAC. On the other hand, when you add 20 
> >>> additional Ohms, in series to the original 20 Ohms, you obtain 1500 W. 
> >>> Again, by inspection. 
> >>> As stated above, 250 VAC^2 is a constant. 
> >> 
> >> No. You make a resistive voltage divider when you add a series 
> >> resistor. The voltage across the original resistor is reduced 
> >> (except if the added R is 0 ohms). Whatever leads one to 
> >> think the voltage is constant is erroneous or not properly 
> >> understood. 
> > 
> > "If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant" (as 
> > stated above) is my premise. This 250VAC is treated as Vmains by me. 
> > My mind mechanically treats Vmains as a constant - because it is. 
> > Most electric outlets in my world source about 120 VAC regardless of the 
> > load connected. 
> > When Vmains is held constant, a doubling of Rmains halves Pmains.
> Correct. He'll get half power out of the combination of HR (the heater 
> resistance) plus AR (the added resistance). The original power was 3000 
> watts. Half power is 1500 watts. So he'll get 750 watts out of AR and 
> 750 watts out of HR. 
> 
> BUT THE OP WANTS 1500 WATTS OUT OF HR. 
> 
> Your math solves the wrong problem. 
> From the op: "As part of an experiment with solar, I want to drive my 
> 220V 3kW immersion heater with a suitable diode so that the effective 
> power is ~1.5kW..." 

The OP's problem has been solved many times in this thread, in different ways.  But the OP doesn't understand any of them.  Now, people are just trying to educate the OP.  

-- 

  Rick C.

  +--+ Get 1,000 miles of free Supercharging
  +--+ Tesla referral code - https://ts.la/richard11209

On Monday, July 3, 2023 at 5:40:51&#8239;PM UTC-4, TTman wrote:
> > No. 
> > ...HR= 20 Ohms. Add another resistor in series = 20 Ohms 
> >> making Rt - 40 Ohms. 
> >>   If I apply 250 V across 40 Ohms, I get 125V across each resistor. 
> >> 
> > 
> > Correct.  Now for goodness sake do the next step in the math. 
> > 
> > 125 volts across 20 ohms produces 781.25 watts in the pot 
> > heater and 781.25 watts in the added resistor. 
> > But you said you want 1500 watts in the pot heater. 
> > 
> > Ed
> That's a surprise...

That's the point.  You don't understand this and seem to be unwilling to learn anything new.  You say you don't understand what others post, then repeat the wrongthink that you have in your head.  Maybe now you will pay a bit more attention to what others have posted and learn something about the problem.

-- 

  Rick C.

  +--- Get 1,000 miles of free Supercharging
  +--- Tesla referral code - https://ts.la/richard11209

On 7/2/2023 7:35 PM, Don wrote:
> Ed wrote:
>> Don wrote:
>>> Ricky wrote:
>>>> Don wrote:
>>>>> Ricky wrote:
>>>>>> TTman wrote:
>>>>>>> Ed wrote:
>>>>>>>>
>>>>>>>> The communication hasn't worked so far, perhaps it's been a bit
>>>>>>>> unclear. I'll try a bit differently. The key point is that doubling
>>>>>>>> the resistance does NOT halve the power.
>>>>>>>>
>>>>>>>> Take the resistance of your heater (HR), which is about 20.8 ohms,
>>>>>>>> and figure what voltage needs to be applied to it to run it at 1500
>>>>>>>> watts. Here's the math:
>>>>>>>> Solve for heater resistance (HR):
>>>>>>>> P=E^2/HR so HR=E^2/P so HR=250*250/3000 so HR=62500/3000 = 20.83
>>>>>>>>
>>>>>>>> Now what voltage must be applied to HR to equal 1500 watts?
>>>>>>>> 1500 = E*E/20.83 so 20.83 * 1500 = E^E so 31250 = E*E so
>>>>>>>> E = sqroot 31250 so E = 176.77volts
>>>>>>>>
>>>>>>>> Ok, once you know that voltage you can solve for the rest.
>>>>>>>> We need to add a series resistance (AR) to drop the voltage
>>>>>>>> across HR from 250 to 176.77volts
>>>>>>>>
>>>>>>>> Compute the current through the series string of HR plus AR
>>>>>>>>
>>>>>>>> Solve for I through HR at 1500 watts:
>>>>>>>> P=E*I so 1500 = 176.77 * I so I = 1500/176.77 = 8.485amps
>>>>>>>>
>>>>>>>> The total voltage (250) minus VHR (176.77) equals VAR, the
>>>>>>>> voltage across the (AddedResistance) AR
>>>>>>>>
>>>>>>>> 250-176.77 = 73.223volts
>>>>>>>>
>>>>>>>> Solve for power dissipated in AR
>>>>>>>>
>>>>>>>> P=I*E= 8.486*73.223=621.32 watts in AR
>>>>>>>>
>>>>>>>> I hope this makes it a bit clearer.
>>>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
>>>>>>> making Rt - 40 Ohms.
>>>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor.
>>>>>>
>>>>>> and... you get half the current, so each resistor will dissipate a quarte
>>>>>>>> I hope this makes it a bit clearer.
>>>>>>> Too complicated...HR= 20 Ohms. Add another resistor in series = 20 Ohms
>>>>>>> making Rt - 40 Ohms.
>>>>>>> If I apply 250 V across 40 Ohms, I get 125V across each resistor.
>>>>>>
>>>>>> and... you get half the current, so each resistor will dissipate a quarter
>>>>>> of the original power. The total power will be half of the original power,
>>>>>> but if I understand your limitations on the design, the series resistor can
>>>>>> not be used to heat the pot.
>>>>>>
>>>>>> I'm not going to continue to beat you over the heat to get you to pay
>>>>>> attentn to the equations. But here are equations for power, one more time.
>>>>>>
>>>>>> P = E * I
>>>>>> P = E^2 / R
>>>>>> P = I^2 * R
>>>>>>
>>>>>> Apply them to the appropriate devices, rather than just saying, "twice the
>>>>>> resistance, half the power". Also, try solving for the actual current and
>>>>>> the voltage on each device. This isn't hard, but sometimes it takes a bit
>>>>>> of work to get used to it.
>>>>>
>>>>> Thanks guys. You helped me put it all together. If we keep 250VAC
>>>>> constant then V^2 (eg 250VAC^2) is a constant. Initially, my mind
>>>>> misrepresents the V^2 constant as a slope in a linear equation.
>>>>>      But V^2 is not a slope because P=V^2/R is not linear. The
>>>>> multiplicative inverse term, 1/R, makes it non-linear.
>>>>>      Here's a plot: <https://crcomp.net/misc/pr.png> [1]. As expected, it
>>>>> looks similar to a 1/x plot.
>>>>>
>>>>> This (so far unanswered) question was previously asked by me:
>>>>>
>>>>>      Does the -3 dB half power point play a role with kW (eg not kWh)?
>>>>>
>>>>> Ordinarily the half power point pertains to bandwidth. It's a little
>>>>> eccentric to apply it to a constant frequency. But it can be done. By
>>>>> inspection, the half power point of 3000 W yields 40 Ohms.
>>>>>
>>>>> Note.
>>>>>
>>>>> [1] octave code:
>>>>>
>>>>>      R=linspace(10,50,40);
>>>>>      P=250^2 ./ R;
>>>>>      plot(R, P);
>>>>>      xlabel('R (Ohms)');
>>>>>      ylabel('P (Watts)');
> 
> <snipped>
> 
>>>> You are solving an irrelevant problem.  No one cares about reducing the
>>>> total power to half.  That's the mistake that TTman is making.  I
>>>> originally said he could add an equivalent heater in series with the one
>>>> he's already using and the total power would be half.  He said that won't
>>>> work because only the pot heater is producing useful heat.
>>>>
>>>> If the pot heater (at 18 or 20 ohms, pick one) is dissipating 3,000 watts,
>>>> how much resistance do you need to add in series to make the pot heater
>>>> dissipate 1,500 watts?
>>>
>>> By inspection, from my plot, you need to add 0 Ohms in series in order
>>> to dissipate 3000 W from 250 VAC. On the other hand, when you add 20
>>> additional Ohms, in series to the original 20 Ohms, you obtain 1500 W.
>>> Again, by inspection.
>>>       As stated above, 250 VAC^2 is a constant.
>>
>> No. You make a resistive voltage divider when you add a series
>> resistor. The voltage across the original resistor is reduced
>> (except if the added R is 0 ohms).  Whatever leads one to
>> think the voltage is constant is erroneous or not properly
>> understood.
> 
> "If we keep 250VAC constant then V^2 (eg 250VAC^2) is a constant" (as
> stated above) is my premise. This 250VAC is treated as Vmains by me.
>      My mind mechanically treats Vmains as a constant - because it is.
> Most electric outlets in my world source about 120 VAC regardless of the
> load connected.
>      When Vmains is held constant, a doubling of Rmains halves Pmains.

Correct. He'll get half power out of the combination of HR (the heater
resistance) plus AR (the added resistance). The original power was 3000
watts. Half power is 1500 watts. So he'll get 750 watts out of AR and 
750 watts out of HR.

BUT THE OP WANTS 1500 WATTS OUT OF HR.

Your math solves the wrong problem.
 From the op: "As part of an experiment with solar, I want to drive my 
220V 3kW immersion heater with a suitable diode so that the effective 
power is ~1.5kW..."

Ed

<snip>