formulas to relate diameter of rotating mass to HP

[mark kibort <mkibort@xxxxxxxxxxxxxx>]

Message text written by INTERNET:bmw@domain.elided
>Any formulas for how to relate the mass of those "rotating thingy"s to t=
he
power drain??  If I reduce that mass by 5%, can I assume a 5% lower drain=
?
<

Ok, if you have a 1 foot diameter flywheel and it weighs 4 lbs and you ra=
ce
somone>>>>>>>>
the acceleration from 60 to 100 takes about 9 seconds in a 200 hp 2600 lb=
s
car.>>>>>>>>>
each 1000 rpm segment is between 3 and 5 seconds depending on the HP to
weight of the car>>>>>>>.
3 seconds to accelerate 4 lbs, one foot in diameter, to a difference of
speed of 1000 rpm, is less that>>>>>>>
.5 hp.   (more like a .25 hp depending on the shape of the pulley of
course)>>>>>>>>

It is not even a factor , so dont waste any money on lighter flyweels.  =

Nascars do it to drop total weight, and increase>>>>>>>>
reving responsiveness and shiftability as  they dont use clutches all the=

time and the engine speeds change very quickly.>>>>>>>

(ie if you rev in idle from 1500 to 7000rpm within a 1/2 second, that
becomes big hp. )>>>>>>


Anyway, chassis dynos cannot measure a gain because it acts like your car=

is on the road.    typical dyno runs last>>>>>>>
10 seconds over a speed range of 3000 to 6000 rpm.    do you thing it cou=
ld
measure , or is there a diff????????

ALSO,  reducing the pulley  sizes  for reducing accesory speeds is also a=

waste.   alternator will produe the same amps that>>>>> needs to produce =
at
any speed , thus pulling the same HP.   Water pumps could gain and so cou=
ld
driven fans,>>>>>>>
maybe PS pumps.  A/C is usually off anyway and is free wheeling>>>

Mark Kibort>>>>>

how does the text look now>>>>>>    except for the>>>>>>>>(temp fix i
think)>>>>>>>

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