ROTFLMFAO hysterically ... OHHH MY GAWWWWD ... this is GREAT ...
Posplayer, do you realize what you've just done ... you've in your own words said that you are incompetent and unable to solve even a (in your own words "EASY") power calculation correctly.
And you did it with such an arrogant tone, implying that _WE_ were not smart enough to do it.
OH THE IRONY ... the sweet sweet irony ...
HAHAHAHAHAHAHAHAHAHA ... I still can't help laughing ...
this is not intended as a sophomore course in electrical engineering but since graham "called me out" (i guess that is a challenge to produce actual equations) and nobody else seems to know how to calculate the power in a dummy load, i have summarized the calculations for a 10 ohm power resistor below.
the equations are a simple and straight forward application of balanced, non-reactive, 3 phase power equations. This is typically covered in a 1st semester junior year in a ee curriculum (i probably did it in an undergraduate lab back in the late 70's). It is probably covered much earlier in a trade school. There are many readily available "how to" tutorials all over the web for this stuff. If you don't know how to do it, maybe you should go pick up a reference and figure it out.
I absolutely agree, anyone who wants to claim competence in electronics should be able to figure this out, and do it correctly.
Its pretty basic stuff, pretty much EE 101 level, not advanced at all.
there is nothing particularly complicated and with a minimum sense of mathematical and technical awareness of electrical circuits should know how to apply the equations in the referenced link.
You preach it brother, you would have to be an incompetent idiot to screw it up.
it would be very unusual for someone with an ee degree to not remember at least doing these calculation and they would be fully expected to be able to figure it out even if they forgot something from many years ago. It certainly would not even cross my mind to suggest someone with a ee could not go back and figure out these equations. So when someone suggests i don't know how to do it, i take it as an insult.
Take it however you want posplayer, but Graham was absolutely right,
YOU DONT KNOW HOW TO DO IT, you try below and you screw it up. I will point out how in exact detail.
It would be equivalent of me claiming that graham doesn't know how to change the brake pads on a bike.
I absolutely agree. I love your analogy, it fits perfectly ... And below you show yourself to be the electronics equivalent of a mechanic that can't even change the brake pads on a bike.
It's your analogy, but I love it, I'm going to have to borrow it from you and use it every time it applies ... Based on your past performance, I have a feeling you're going to hear it a lot.
Please note that it is you who have said what an easy problem this is, and how it SHOULD be no problem for a competent engineer or technician to figure out.
That would be a pretty ludicrous statement, because assuming he does work as a mechanic (not sure why he would lie about that), he is bound to have found himself in a situation where he needed to change pads. It is expected.
this thread makes me question why i even bother to post analysis of gs electrical systems and trouble shooting when what happens is that it draws fire from apparently insecure idiots hiding behind the internet (i'm specifically referring to balzar and graham at the moment).
Please stop posting "analysis" ... you almost always screw it up, leading to nothing but confusion ... you understand the simple stuff, but anytime you delve into something beyond the most basic, all you do is come up with baseless theories that have little to no basis in reality, don't help people, and cause confusion or actual harm.
I've told you before, I'm not particularly out to get you, if you post sensible stuff I have no issue with you (I've never come down on you for advocating thoroughly cleaning connections for example, or you're method of tracking down high resistance connections ... these are sensible ideas, please continue with them)
But on the other hand, you do also post a lot of stupid stuff. These two threads about the loaded stator tests are a prime example. When you advocate stupidity I'll let you know.
i try to keep the material relevant at a minimum technical level to convey the issues for generally non technical members. I'm not paid to do this, so it is more of a way of "paying it forward" for the help i have received.
for those individuals that can't figure out these simple calculations , but nevertheless claim that they are wrong, for that i will bestow the highest honor i can; display it with pride :
Oh yes posplayer, please edjumacate me, I thirst for knowledge.
Stator resistance :
R_stator - measured in ohms
R_stator _leg = 0.5 (spec is 0.4-0.5 ohms)
Phase Resistor:
R_dummy leg - stator leg resistor value in ohms
R_dummy_leg selected for a 100 watt thermal constraint
R_dummy_leg = 10 ohms
Phase to Phase Voltage:
VAC_pp - ac voltage 0-peak phase to phase
VAC_pp - 80 VAC at 5000 RPM (measured)
This is potentially where you make your first mistake. You notation seems to indicate that this is Phase to Phase peak voltage. It is not. IT IS PHASE to PHASE
RMS AC voltage.
NOTE the RMS, it is important !
Do you understand the difference between RMS and peak, and how they are measured. You might consider it insulting to ask, but I suspect you are going to try to claim this is peak, so I'm going to head that off before you even start.
Please note that where the Suzuki manuals mention measuring the AC unloaded voltage they use a Volt meter; unless using a specialized meter which indicates otherwise, these always measure it as RMS voltage.
To confirm this, I just a little while ago went and measured the phase to phase voltage on my 650 at 5000 rpm: about 92 or 93 VAC RMS (which means that the peak voltage is ~ 130, so higher definately does not help your case ... but we'll let you stay with 80 VAC RMS for the purpose of these calculations)
If you want to dispute that the correct voltage is 80 VAC RMS you better have a
darn good case ...
(Are you getting a sinking feeling in the pit of your stomach yet ... )
Phase to Neutral Voltage:
VAC_pn - ac voltage 0-peak phase to neutral
VAC_pn = VAC_pp/sqrt(3) = 46.2 VAC 0-peak
When you apply the 1/sqrt(3) conversion factor to go from phase-phase to phase-neutral, your units stay the same ... so if you start with peak you get peak, and as in this case, when you start with RMS you get RMS ...
I'll assume you realize this ...
so what you have is 46.2 volts
RMS phase to neutral.
Power _Total - total power dissipated in resistive loads (ignores reactance)
Power_Total = 3/2*VAC_pn^2/(R_stator_leg+R_dummy_leg/2)
Power_Total = 312.2 watts
And this is where you go off the rails ... the voltage in this calculation needs the
PEAK phase-neutral value (check your wiki link from the first post if its unclear)
The 46.2 volts above is RMS.
To convert the RMS voltage to Peak voltage (for sines) you multiply by 1.414 and get 65.34 volts peak phase-to-neutral ... thats the voltage you plug into 3/2*VAC_pn^2/(R_stator_leg+R_dummy_leg/2)
You missed the step of converting the RMS to peak.
Any competent EE would know the units he is actually working with, and use the correct ones that apply ... this really is EE 101 level stuff ... and you hosed it.
And because this is the real world, you don't just fail a test, you potentially set someones bike on fire. So, yeah, when you screw it up it matters enough that you need to hear about it
... and if you hadn't taken the condescending tone you have here I wouldn't even laugh at you, but you did, so eat crow buddy ... is it tasty?
By your own definition you can't correctly do the equivalent of safely installing brake pads.
Using the correct value of 65.34 volts peak p-n you get a total power of 609.90 watts ... double your incorrectly calculated value ...
Power per resistor - power per dummy leg
Power per resistor - Power Total/3
Power per resistor = 104.1 Watts
Using the correct total you really get 203 watts/resistor.
That is for a stator with 80 VAC RMS ph-ph ... Since mine (and presumably others as well) run at 90+ VAC RMS ph-ph its worth running those numbers as well, since this is a worse case than the 80
at 90 VAC the power per resistor comes out to 257 watts/resistor
(and a minor note ... you really get ~5% less since that gets lost in the stator ... but close enough)
Amps per Phase - current through each leg
Amps per Phase - 2*Power per resistor/VAC_pn
Amps per phase - 4.5 Amp 0-peak
Thank you for your post posplayer, you have proven that you cannot do a simple power calculation that is covered in the most basic of EE101 courses ...
You have unequivocally shown your level of competence
Cogratulations.