From: "Tom Del Rosso"
References: <0001HW.B9C26E7400A7161B165FEAC0@news.covad.net> <7tGLdeAS6fn9EwBv@jmwa.demon.co.uk>
Subject: Re: Best way to power array of LEDs?
X-Newsreader: Microsoft Outlook Express 6.00.2600.0000
NNTP-Posting-Date: Tue, 05 Nov 2002 05:35:22 GMT
Organization: AT&T Worldnet
Date: Tue, 05 Nov 2002 05:35:22 GMT
"Bob Wilson" wrote in
> Turning on the radio does not magically make the engine more efficient
> clearly what you are doing is to add load to the alternator (in order
> power the radio), and hence you the engine has to work a little harder
> consume more gas) to do the extra work.
> What if your alternator was capable of somehow outputting a megawatt?
> Suddenly switching on a load that is using FAR more energy than it
> drive the car, would be an extreme example of this. If the answer to
> question was that you are dissipating less heat and burning no more
> then your engine would be running so efficiently that it would be
> very cold, VERY fast!
If the alternator could supply a megawatt, then I'd expect it to be a
bigger mechanical load than the drive train even with no electrical
load. When *less* current is drawn, does that mechanical load change
proportionately, or does the alternator, to some extent at least,
convert some of its unused capacity to more internal heat?
> Bottom line: you don't get something for nothing.
Well I wasn't thinking of getting something for nothing. I was thinking
that it sounds too ideal for an increased load on the alternator to
convert into an exactly proportionate load increase on the engine. If I
draw less power than the alternator can supply, isn't it just wasting
some of that power because of the non-ideal properties of the
alternator? In other words does a generator precisely convert a higher
load current into more mechanical resistance?