[DML] Re: How to measure HP (WAS: "Fastest" DeLoreans)
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[DML] Re: How to measure HP (WAS: "Fastest" DeLoreans)


I like this topic too.  It's way too easy for someone unscrupulous to
inflate the power ratings of his engine(s).  You are correct.  Neither
the "constant loss" nor the "percentage loss" numbers for drive train
loss match.  Neither can be correct (friction losses have at least a
linear, a quadratic, and a cubic component, all variable, even for a
given drivetrain).  If your VQ35 were mine, I would use a percentage
based on the best available data, in this case 26%.  It is a plausible
figure for drivetrain loss.  35% is less plausible, I don't buy that
number.  So your VQ35 might be 210 + 26% = 265 horsepower at the
engine.  That's quite a lot.

To editorialize:  engine measurements taken without the ancillaries
that don't DISCLOSE they were taken without the ancillaries are
somewhat deceptive in my opinion.  

Also, engines that have been modified to the point where flooring them
instantly subjects the drivetrain to catastrophic stresses don't
impress me much.  There is at least one publicized "world's fastest
DeLorean" that fits into this category.  I don't consider a car so
powerful that when you step on the gas, you don't go anywhere because
you just broke it, to be that interesting, but given that I own the
"World's Most Overheating-prone DeLorean", I probably shouldn't talk.  

Rick Gendreau

--- In dmcnews@xxxxxxxxxxxxxxx, Marc Levy <malevy_nj@...> wrote:

> My question-
> If you are saying we should all speak in terms of
> engine horsepower, what loss factor should we
> calculate for the drive train?  ... My VQ35 is 210 WHP.  When
> someone walks up to me at a show and asks me "How much
> power is it?"  what is the correct answer?
> 210 + 35  = 245HP ?
> 210 + 26% = 265HP ?
> 210 + 67  = 277HP ?
> 210 + 35% = 283HP ?

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