From: jlittler on
On Feb 9, 7:49 pm, atec <"atec77 "@hotmail.com> wrote:
> jlitt...(a)my-deja.com wrote:
> > On Feb 9, 2:51 pm, atec <"atec77 "@hotmail.com> wrote:
> >> jlitt...(a)my-deja.com wrote:
> >>> On Feb 9, 12:58 am, atec <"atec77 "@hotmail.com> wrote:
> >>>> Andrew McKenna wrote:
> >>>>> sharkey wrote:
> >>>>>> Andrew McKenna <NOcmorSPAM3...(a)NObigpond.SPAMnet.au> wrote:
> >>>>>>> I think your critics are thinking of their bicycles with dynamo
> >>>>>>> powered headlights :-) More electrical load might make you discover
> >>>>>>> that you need to push harder to achieve the same results but there's
> >>>>>>> no way the dynamo itself gets harder to spin.
> >>>>>> What? You need to push harder to spin it but it doesn't get harder to
> >>>>>> spin?
> >>>>>> -----sharks
> >>>>> No, you need to push harder to get the result if you add electrical
> >>>>> load. It cannot possibly get harder to spin.
> >>>> Now thats wrong in so many ways .
> >>> Actually he's dead right if you read it carefully - typical engineer's
> >>> wording though. They never speak english.
> >>> The MECHANICAL load is unchanged IE the effort to physically spin the
> >>> metallicy bits around will be unchanged regardless of electrical load.
> >> The way I understand it that statement is wrong , mechanical input
> >> required to generate electrical energy bears relationships to each
> >> other, more electrical energy required means more drive is required to
> >> overcome the resistance to turning .A change in turning momentum is
> >> proportional to energy required.
>
> > And your statement is indeed correct - there's a difference between
> > the mechanical input and the mechanical load. Pure semantics of
> > course. mechanical load (as torque) plus electrical load(as torque)
> > equals mechanical input required(as torque). The mechanical load is a
> > constant (ceteris paribus), the electrical load changes with, well,
> > the electrical load <grin> (1). To be more accurate the torque/turning
> > force that you have to provide to generate a current equal to the
> > current being drawn is increased as the current required increases(2)
>
> > JL
> > (1) there obviously being more than one meaning of the word load in
> > this context
>
> I will never agree to that ( I stick to what I know and have been taught)

I think that translates as I'm too stupid to think for myself, n'est
ce pas ?

> - one being current drawn, the other being turning force> required
>
> load is load , as in work to be done /required so whether explained as
> turning force or current drawn they bear a direct relationship of watts
> and where's consumed

Ya gotta love a guy who makes a statement to which the only answer is
"yeah but"...<pause> actually no you don't.

Yes, load is load, it's measured in work done, hence why I put the
bits in brackets about how it was measured (that'd be that torque
thingy). But Andrew seemed to be differentiating between two different
types of work (he later said things that make me doubt he had a clue,
but that's a different discussion) the actual statement he made is
(pedantically) correct.

There's work to turn the metal things in the alternator around- that
load is a constant, it doesn't change regardless of the electrickery.

Then there's the electrickery, the EMF is going to create a different
load depending on the watts being consumed by dint of how much
electrical load is attached.

Add the two together and you get the total one is variable, one is
constant. Not difficult really.

>
> (2) still not sure I'm explaining that particularly well
>
> No you didn't , I suspect it's some bullshite machination dreamed up
> to split hairs buy a drunken masturbaterpoofta ( tell me if I am wrong)

Well i wasn't drunk but I don't know what the latter bit means so I
can't comment, it's obviously intended to be pejorative so I guess the
obvious reply is "gefuggedyastupidkuntgetabigblackdogupya"

JL

From: Nev.. on
Peter Cremasco wrote:
> On Fri, 9 Feb 2007 08:22:27 +1100, "Dale Porter" <daleaporter(a)gmail.com>
> wrote:
>
>> "Peter Cremasco" <FirstName.LastName(a)bigpond.com> wrote in message news:ms4ns2duvu4b7tvncuicr1vedevs58cnjj(a)4ax.com...
>>
>>> Strangely enough, the Camira engine speed increases when the a/c is
>>> switched on. I think that's the ECU overcompensating for the extra load,
>>> though.
>>>
>> That would be due to the air-con compressor not being driven by the engine, more the other way around. It starts up and runs at a
>> higher speed than the engine does at idle.
>
> The compressor is driven by the engine, via two belts.

Nah mate, it's a Camira :)

Nev.. (I think you were whooshed)
'04 CBR1100XX
From: Knobdoodle on

<jlittler(a)my-deja.com> wrote:
> And your statement is indeed correct - there's a difference between
> the mechanical input and the mechanical load. Pure semantics of
> course. mechanical load (as torque) plus electrical load(as torque)
> equals mechanical input required(as torque). The mechanical load is a
> constant (ceteris paribus), the electrical load changes with, well,
> the electrical load <grin> (1). To be more accurate the torque/turning
> force that you have to provide to generate a current equal to the
> current being drawn is increased as the current required increases(2)
>
I'm not surprised that YOU are having this discussion John, but I'm
absolutely STUNNED that you're having it with a poster who display about
half the mental capacity of an overripe marrow!

C'mon; do you reeeeeealy reckon he's following you here?
--
Clem
[No; I'm not either]


From: Hammo on



On 9/2/07 9:23 AM, in article
1170973410.349356.238620(a)m58g2000cwm.googlegroups.com,
"jlittler(a)my-deja.com" <jlittler(a)my-deja.com> wrote:

> On Feb 9, 12:56 am, Hammo <hbaj2...(a)aapt.net.au> wrote:
>> On 9/2/07 12:34 AM, in article
>> YFFyh.4682$sd2....(a)news-server.bigpond.net.au, "Knobdoodle"
>>
>>
>>
>>
>>
>> <knobdoo...(a)hotmail.com> wrote:
>>
>>> "Nev.." <i...(a)mindless.com> wrote:
>>>> I think you've been reading too many physics books and you've lost sight
>>>> of reality. Are you saying that if I measure something once per second
>>>> and then multiply that by 3600 my result is not an accurate measure of an
>>>> hourly rate? Do you think the computer controlling the fuel rate just
>>>> guesses?
>>
>>> No; it actuates the injector the exact amount that it's been told to for the
>>> conditions it's measured.
>>> It then displays the exact mpg (L/Hr, Km per kilojoule or whatever) that's
>>> it's been told to display too.
>>
>>> But it doesn't have any idea what a litre actually is and it certainly
>>> doean't have any ability to actually measure one!
>>
>> Eh?
>>
>> It's measured, but it can't measure?
>
> No, it's calculated based on what actually is "measured" (or more
> accurately specified by the efi).
>
> The efi opens the injector for a period according to what it's lookup
> table tells it is the right amount given the parameters it has sensors
> measuring. It ASSUMES the fuel pressure (and hence fuel flow) is
> correct (unless there has been some new fangled advances in EFI I'm
> not aware of). I know of no cars that measure actual fuel flow, they
> usually calculate your fuel consumption based on the amount of time
> they've opened the injectors, and the number of KM's travelled. Those
> two parameters are based on a number of implicit assumptions:
> - the fuel pressure is correct
> - there are no blockages restricting fuel flow
> - the size of your tyres are as specified and hence the number of K's
> calculated is correct
>
> etc etc
>
> it's a reasonably accurate estimate and more than good enough for the
> purposes for which it is used. That doesn't mean the actual fuel flow
> is measured.

Eh?

It's measured but it doesn't measure?

Hammo

From: Knobdoodle on

"Theo Bekkers" <tbekkers(a)bekkers.com.au> wrote:
> jlittler(a)my-deja.com wrote:
>> (2) still not sure I'm explaining that particularly well
>
> Those of us that agreed with you understood it.
>
Errmmm..... [shuffles guiltily]
--
Clem


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