Quote from: Rosemary Ainslie on May 09, 2012, 12:45:04 PM
Actually TK you are STILL wrong.  You are talking about the POWER that is being dissipated at the element resistor.  You CANNOT separate that POWER from its context.  You CLAIM that there is an instantaneous power dissipation based on 20 watts determined by the current flow and the voltage - V x I.

NOW.  Any determination of the ACTUAL wattage delivered or dissipated is a determinant of the POWER.  And POWER IS DEFINED AS DETAILED IN THE DOWNLOAD FROM WIKI. THAT sum requires that average over time - and THAT'S the ONLY WAY TO DETERMINE WATTAGE. 

Rosie Pose

YOU FOOL. You don't even know what that equation means. What do those little triangles mean, Ainslie? You have no clue.
What does the plain symbol "W" mean and what does it mean with the subscript "avg"?
Instantaneous power and average power: two different things sharing the same unit, the Watt.

Quote from: TinselKoala on May 09, 2012, 12:40:41 PM
What, then, is the power during that time, unequivocally?

What are you arguing about? PW and I are talking, and have been talking, about the ON time exclusively. And that is because we are examining the veracity of a claim YOU MADE ABOUT THE ON TIME CURRENT in those blog posts above.

A claim that you have just now admitted is wrong.


320 mA at 62 volts is NOT "next to nothing".

It is certainly NOT indicative of the power delivered by the battery EITHER.  It is only indicative of the amount of power that is delivered during 1/6th of a duty cycle.

Quote from: Rosemary Ainslie on May 09, 2012, 12:50:14 PM
AGAIN.  You are trying to IMPLY that watts can be separated from POWER.  IF you are trying to determine the amount of energy dissipated at the load resistor then you are REFERENCING A POWER MEASUREMENT.  This is ALWAYS BASED ON WATTS and those WATTS NEED TO BE DETERMINED OVER TIME.  AGAIN.  Here's that equation.

Rosie Pose

NO AINSLIE YOU LIE AND MISREPRESENT YET AGAIN. The Watt is a measure of INSTANTANEOUS POWER. How am I trying to "separate watts from power"? Like many  things you say, this makes no sense and is a gross mischaracterisation of what I AM trying to do. In other words you LIE again.

You are confounding Joules, instantaneous power in Watts, and average power in Watts.

You don't understand this stuff, you can't provide anybody or any reference that actually supports your position and you do not understand even the simple equation from WIKI that you keep parroting.

But all that is moot, since you have AGREED that 320 mA flows during the ON time, and this ALONE refutes your entire claims in 117 and 118, where you say that there is only a tiny current. 320 mA at 62 volts is not tiny and represents significant power.

Look.

I travel at 10 miles per hour for half an hour.
Then I speed up and travel at 20 miles per hour for another half an hour.

How far did I go in that hour? What was my AVERAGE SPEED? What was my speed at the 15 minute mark?

I went (10 miles/hour) x ( 1/2 hour) = 5 miles. Note that the units cancel and are correct.
I also went (20 miles/hour) x (1/2 hour) = 10 miles. Note that the units cancel and are correct.

Adding these, I get 15 miles travelled, in one hour. Taking one hour as the interval and 15 miles travelled, I AVERAGED 15 miles PER hour. But my speed at the 15 minute mark was 10 miles per hour.

This is the difference between an INSTANTANEOUS RATE and an AVERAGE RATE, Ainslie.

A Watt is like a Mile Per Hour. A Joule is like a Mile.  A mile is NOT the same thing as a mile per hour, and is NOT the same thing as an hour per mile.


And this, by the way, is what is meant by "SHOWING" a computational result. I SHOW the numbers, the units, the operations, the results and the interpretation. All you do is waggle your gums.

Quote from: TinselKoala on May 09, 2012, 12:59:49 PM
NO AINSLIE YOU LIE AND MISREPRESENT YET AGAIN. The Watt is a measure of INSTANTANEOUS POWER. How am I trying to "separate watts from power"? Like many  things you say, this makes no sense and is a gross mischaracterisation of what I AM trying to do. In other words you LIE again.

You are confounding Joules, instantaneous power in Watts, and average power in Watts.

You don't understand this stuff, you can't provide anybody or any reference that actually supports your position and you do not understand even the simple equation from WIKI that you keep parroting.

But all that is moot, since you have AGREED that 320 mA flows during the ON time, and this ALONE refutes your entire claims in 117 and 118, where you say that there is only a tiny current. 320 mA at 62 volts is not tiny and represents significant power.

I'm actually heartily sick of this.  You are now AGAIN trying to infer that there is POWER delivered by the battery based on a figure that is NOT representative of WATTAGE.  The wattage delivered by the battery is 3.33 watts.  The power delivered by the battery CAN ONLY BE BASED ON THAT FIGURE.  And for you to state that I don't know what those 'little triangles' represent?  I keep telling you.  I have a functional intelligence.  Why would I NOT know? 

You can try and argue otherwise TK.  But you are UTTERLY WRONG.  And I'm rather shocked that picowatt is endorsing this value.  He at least KNOWS who reads these threads.  Your analysis is QUITE SIMPLY and COMPLETELY WRONG.  AT NO POINT IS THERE the delivery of 20 watts.  There is only the measure of 20 watts over 1/6th of each duty cycle - making it a total of 3.33 watts - AT BEST.

Rosemary