# Rotating Mass, Available Horsepower, and Acceleration

 Rotating Mass, Available Horsepower, and Acceleration

We see or experience centrifugal force and
inertia every day of our lives, and intuitively know how it works. One of the
best examples is twirling a rope with a weight on the end. We know the faster
and further out the weight is, and the heavier the weight, the further the
weight will fly when we let go. This is because we store more energy in the
weight as we move it faster, swing it in a wider circle, or use a heavier
weight. The energy isn’t really wasted, it is stored and released later.

## Common Questions

There are several
common questions
A few
of them are:

If I use a lighter
crankshaft, how much
power is gained?

If cluster gear
weight of a
transmission is
reduced, how much
will acceleration
improve?

How much will a
light-weight
aluminum flywheel
improve
acceleration?

How much ET can be
gained from using lighter wheels?

Does an aluminum driveshaft speed up my car?

These
questions can be
the weight change,
the distance out
from the center the
weight change occurs
at, the speed (RPM),
and the time period
over which the RPM
change occurs. Even without exact calculations, we can get a feel for how things
change. This will help us budget our money and make better choices, or at least
have a feeling for what we are accomplishing.

on-line calculators, we can
understand how changes in rotating mass will affect available horsepower in a
vehicle.

## What does a rotating mass actually do?

A rotating mass
does not really
consume or dissipate energy. A rotating
mass stores energy. The rotating mass
eventually either returns energy to
the system in a useful way, or
something converts the stored energy to some
other form of unwanted energy. The conversion might be with a friction,
converting to heat. The
energy stored might
be helpful, like the smoothing of cylinder pulses in an engine flywheel. The
energy stored also might not do
anything at all, or the stored energy
can even be harmful, reducing acceleration or braking.

Accelerating an unnecessary rotating mass requires energy, and
the acceleration process saps some of the horsepower we have available to
accelerate our vehicles. Reducing
available horsepower affects acceleration in a very predictable manner, and the
horsepower amount needed to spin something up gives us some feel for how important a part change might be.

Four things
determine the effect
of rotating mass. Every one
of these things is important:

1. How quickly and
often a
rotating mass
speeds up or slows
down. Every time it is forced to speed up or slow down, it takes or releases
energy

2. How heavy the
rotating mass is. More weight (with no other changes) stores or releases more
energy

3. The rotating
weight’s distance
outwards from the
centerline. The further out, the more energy pushed in and out of a given
weight

4. How fast the weight
spins, or the speed the weight travels in a given circle diameter. The higher
the RPM, the more energy stored

Here are how these things work:

1. If we push energy
into the rotating
mass and pull
energy out several
times, we move more power
around than if we
make a slow, smooth,
change in speed. It takes much more effort to repeatedly speed and slow
something in a short period of time than to gradually speed it or slow it
2. The amount of weight is the least important
thing! If we double the weight (with no other changes) we only double the
stored energy
3. Weight distance
from the center
line is very
important, because
it determines the
weight’s circular velocity
(speed)! Stored
energy goes up by
the SQUARE of the
we replace a 4-inch
diameter hollow
driveshaft with an
8-inch
diameter tube of
exactly the same weight,
it is not just double. It is twice the size squared, or four times
the stored energy when it weighs the same!
4. The faster we spin the weight, the more energy it stores. If we double
RPM, we
multiply stored
energy four times.
Again it is a
square of the
change, just like
weight distance from centerline is a
square.

The above is very important. If
we double the effective “circle size” the weight is rotating at, we get
four times the stored energy. If we simply double the weight without
changing the spinning radius, we just double stored energy:

• If we reduce
mass from twenty
pounds to ten pounds,
keeping the same distance
out and same peak RPM, we
reduce stored
energy to half the
original amount.
Reducing weight is a
one-for-one change.
• If we cut diameter in half while
keeping the same weight
and RPM, stored energy
will be 1/4 the original
stored energy. This
change is a square.
Twice is a “four times”
effect. 2*2=4. Four
times is a sixteen
time effect on
stored energy.
4*4=16
• If we cut
RPM in half, we
would reduce stored
energy to 1/4 the
original amount.
Once again this is a
squared change.
Change RPM three
times, and the
stored energy
changes nine times.
3*3=9

We should
carefully think
means when we
change things.
Some changes are
worthwhile, some are
not. We also cannot use carte blanche rules, like the silly rumor that reducing
a rotating weight is like dropping the vehicle weight four times that amount. As
a matter of fact, it is probably never four times. It is more likely closer to
one, and might even be less than one!

Wheel Changes

Let’s assume, just as an example, all
of a wheel’s weight is at the
outer edge and
remains at the outer
edge. If
we reduce a wheel’s
diameter but keep
the overall weight
the same, the wheel
is a spinning ring
with smaller
diameter. The
smaller diameter
increases the wheel’s RPM at
the same vehicle speed.
The smaller diameter
also moves the
spinning weight
closer to the
center.

Let’s say we cut
diameter in half.
Now think about how fast the
wheel spins. RPM
will be twice
what it was at the
same speed. The
half size diameter
reduction spins the
wheel twice as fast,
and that would
increase stored
energy to four times the
original amount if
the weight was the
same distance out. But the weight isn’t the
same distance out.
The spinning weight
is now half size.
This 1/2 size
reduction decreases
stored energy by
four times!

If we did not change the weight or weight
distribution, and we reduced a wheel and tire diameter by half but drove the
same speed, nothing would change. It would be a major change that just broke even. Moving the
weight closer to
the rotation center reduced stored energy, but the increased RPM to maintain the same speed
increased stored energy the same amount.
One cancelled the other, and stored energy did
not change!

If we change tire and wheel diameter without
changing weight distribution and weight in the tire and wheel, we don’t change a
thing.
In this example, we gained
nothing from a significant physical change.
We
also lost nothing.

Lightening the tire or wheel some distance out from the hub reduces
stored energy. This is especially true if
the weight reduction is far out
from the center. If we change the weight one-pound fourteen inches out, it is
like changing weight four-pounds seven inches out.
Which brings up an
important point we
almost never hear
mentioned, a lower
weight part might
not be lighter at
the outside edge. It
might be lighter in
the center, where
the weight reduction
doesn’t mean much.

It is more important to make
something as light
as possible on the
OUTER edge, rather than near
the (wheel) center.
Spending money on
smaller or lighter
rotors to save
rotating weight should be down the list, because the
rotating weight is
closer to the wheel hub. Unless the
rotors are huge and
we take weight out of the
rotor’s outer areas, things
will not change much.
(A light rotor and wheel is good for
reducing un-sprung
weight, and that
helps keep
our tires in
contact with the
reduces vehicle
weight. But this is a
different problem.
Here we are talking
the bounce inertia

If we spent money on
the same weight
reduction in the wheel,
reducing weight out
a little further away from the
center, we would do
much better.
We would be removing weight
further out from the
center, where it
does the most good.

If we spent
our money on a
lighter tire we
would be
getting the very most
return for the
weight change. The
tire’s weight change
is mostly outside
between the rim edge
get maximum effect
from the weight change!

a lighter tire, we
know for sure the
weight comes off the
most critical area.
rotor, it is close
to the center and,
for the same weight
change, the return is
much less.

The wheels also
speed up and slow
an 11-second car, we
have 11-seconds to
speed the wheel up.
Most of the
horsepower pushed
into the wheel and stored is
pushed in near the
end, when
acceleration is
least. Since we have
more time to push
the bigger amount of
energy into the
wheel, it takes less
horsepower than we
might expect.

Drive Shaft Example

Now let’s think
The driveshaft is a
fairly thin hollow tube.
Nearly all drive shaft
weight is
at the outside,
since it is (of
course) hollow.
The shaft also turns
at the same RPM no
matter what the
driveshaft diameter, because
the RPM is set by
the rear end ratio, tire
diameter, and
vehicle speed.
If we make a
driveshaft lighter
and keep everything
else the same, the
vehicle acceleration change is
often
insignificant.

Why would it be insignificant in
most cases?

In the
first place, the
drive shaft is small
in diameter. With a
small diameter, less
energy is stored for
a given weight. In
the second place, a
driveshaft is
really not that
heavy. A steel
Mustang driveshaft weighs
somewhere around 30 pounds, so
we just can’t take that
much weight out.

Also, the
driveshaft
and smoothly
over a long period
of time. It
accelerates fastest
at slowest speeds,
and that is when it
needs the least
energy to spin up. Because it
has a long time to
spin up, is a small
diameter, and because it does not weigh much,
the driveshaft does not remove
very much horsepower
at any instant of
time. Despite
what we are told, a change
in driveshaft weight has, at
best, a very
small effect on
acceleration. Likely
any change is
immeasurable in a
street/strip car.

Now a lighter shaft
certainly can help
in a very light
vehicle. It can also help in a road
race car (as will a light crank and flywheel), because road racing requires
instantly changing from acceleration to deceleration.

A light driveshaft won’t change
anything significant or measureable in a 3000-pound
11-second car,
except how fast
dollars leave your
wallet!

Another worry is driveshaft diameter. If we go from
a 30-pound 3-inch steel driveshaft to a 30-pound 3.5-inch aluminum shaft, we
move the weight out 3.5/3 = 1.167 times. That increases stored energy 1.167^2
times, or 1.36 times. If we store 0.3 horsepower in the shaft, changing the
diameter will increase that to 0.4 horsepower. We would have to reduce weight
14.3% to 25.7 pounds just to break even with the diameter increase.

The worst thing about a driveshaft is the diameter
is so small, and the acceleration time is so long, there just isn’t much
horsepower being sapped from the system. A typical steel driveshaft in a typical
12 or 13 second car only stores an average of about 1/4 horsepower. If we got
100% of that back with a zero weight shaft, we would never notice it.

Good reasons to change a driveshaft are to get rid
of vibration and harmonic resonances in the shaft, to make it stronger, or to
simplify a two-piece driveshaft system. The silliest reason is to speed the car
up. Even if we only pay \$100 for a shaft, it would typically be much less than
1/4-horsepower average gain. That would be paying much more than 100/.25 = \$400
per horsepower.  Paying a lot more than \$400 per horsepower is not a good
investment.

Flywheel Change Horsepower

A flywheel can be
fairly heavy, and
the weight is a good
distance out from
the center. It spins
at crankshaft speed,
and it has to abruptly change speed (slow down) at every
up-shift.

While a driveshaft stores around 1/4 horsepower
spread over 12-13 seconds of time, the flywheel is entirely different. The
stored energy in a 25-pound 13.5 inch diameter flywheel at 6000 RPM is something
around 32,916 joule-seconds. This is 44.12 horsepower-seconds of power.

Looking at a 13.5 inch diameter 25 pound steel
flywheel at 6000 RPM, we have:

A 25 pound (400 oz)  flywheel has
about 32,916 joules of energy. Since each joule is one watt/second, and since
746 watts equal one horsepower, we have 32,916/746 = 44.12 horsepower-seconds
stored. This would be one horsepower applied over 44 seconds, or 88 horsepower
over 1/2-second, to reach 6000 RPM from zero.

An aluminum 12.5 pound flywheel would have half that
energy, because weight is a direct one-for-one change in energy. If we ran it on
a calculator we would see:

This is about 16,485 joules, or 22
horsepower-seconds of power. Before we run off thinking we will gain 22
horsepower by swapping flywheels, we have to realize this is horsepower-seconds
. Also, the flywheel is NOT starting from zero speed!

If we launch our car at 4000 RPM, the flywheel
starts at 44.4% of the 6000 RPM energy. (square of 4000/6000 times 6K RPM power)
This is 19.6 horsepower-seconds with a steel wheel, and 9.8 horsepower-seconds
with an aluminum wheel, of initial launch stored energy. We only have to add 12
horsepower-seconds to the aluminum wheel, and 24 horsepower-seconds with the
steel wheel.

If we have two seconds to spin up to 6000, the
engine will push either 6 horsepower average with aluminum, or 12 horsepower
average with steel, into the flywheel. If we have 6 seconds to spool up, average
horsepower is either 2 or 4 horsepower.

Also, not all of that stored energy is wasted. When
we shift to second gear, the extra energy is returned on the shift. The engine
gets a “boost” as the flywheel power is returned to the slower-turning
transmission input shaft.

A car with street tires and traction issues will
benefit from a heavier flywheel on launch, and be hurt more on the shift.

A car with good traction will benefit on the launch
with the heavier wheel, but could loose a little power as it runs up through low
gear.

Either way, we are not talking much power, and the
results are highly dependent on the vehicle. We might gain ET or lose ET
depending on many factors.

There are two general ways the flywheel affects
acceleration, although this can vary. In a
light car with very
fast 60-foot times,
a lighter wheel can
slightly improve
60-foot times. This
is because the
launch is often at
full throttle, the
car generally has a
steep gear, and we
want to plant the
tires hard into the
track without
encouraging spin.
The tires hook hard,
and usually have a
very soft sidewall
that absorbs shock.
We want the engine
to quickly match the
RPM needed to move
the rear wheels, and
not overpower the
available traction.
It is a wide open
throttle high-RPM
launch.

A typical
street-strip car is
different. Generally
we can’t launch at
wide open throttle,
the tires are
stiffer walled, the
suspension is
heavier, and things
just don’t hit as
hard. We actually
want a heavy wheel
(and a heavy crank)
to smooth out the
power. This lets us
have a much more
controlled launch,
and smoothes out any
sudden application
of throttle. An
aluminum wheel,
especially when the
car is severely
traction limited and
heavy, can really
hurt 60-foot times.
A light aluminum
wheel not only makes
a street car hard to
drive, it usually hurts at
the track. It is
a heavy street
machine.

Now that we have seen the flywheel in detail, we can
compare it to the driveshaft. The flywheel is typically around 20-40
horsepower-seconds of energy, and has a somewhat small effect on overall power.
The driveshaft has about 5.5 horsepower-seconds of energy, and has the full
length of the track to spin up.

In a 13 second car, the driveshaft consumes about
1.5 horsepower maximum over the first sixty feet.  If we cut the driveshaft
weight in half with no increase in diameter, we would pick up 0.75 horsepower.
If we increase diameter from 3 to 3.5 inches and reduce weight by 30%, we would
just barely change the driveshaft’s stored energy. The added diameter would
increase stored energy 1.36 times, and the reduced weight would reduce stored
energy by 30%. We would gain about 0.25 horsepower-seconds, or 0.02 horsepower
average power gain over the track length with a 13 second car.

Crankshaft

A crankshaft is a
bit worse than
a drive shaft.
A crankshaft
accelerates and
changes speeds through
every gear, so it is
constantly storing
and returning energy
to the system. In
low gears it spins
up pretty fast,
spinning up from
“launch” RPM to
shift RPM. This
spin up repeats at
every shift. The
crank also
has to be heavy to
support the pounding
and tugging of the
pistons and rods as
they accelerate and
decelerate, so we
are dealing with
some weight.

Fortunately the
crank diameter is
small. A 3-inch stroke
requires only a 1.5-inch throw radius. Unless we make a
huge change in
OUTSIDE weight in
the counterweights,
in most engines making the crank lighter
makes very little sense. The
dumbest thing to do
is hollow out the
crankshaft center
because it is the
smallest rotating diameter
area. Don’t believe
this?
following technical
paper from the

Scat Crankshaft
website
.

Lightweight
Crankshafts-
Performance or
Deception

Scat has it 100%
correct. Many bench
racers, and even
some crankshaft
manufacturers,
exaggerate a good
bit! They
remove weight where
it makes little
difference in stored
energy, but might
make a difference in
strength. Some
transmission experts
wrong thing also. If
outside edge weight
of the largest-diameter fastest-spinning parts that
speed up and slow
down at every shift,
we are worrying
parts. If we worry
speed up at the rate
of the driveshaft,
we would be wasting
our efforts.

The purpose of the
examples was to give
you a feel for what
to look at first.
Any weight reduction
is good for
horsepower to weight
ratio, but some
weight reduction has
a bigger payback.
Things that change
speed often, change
speed rapidly,
and/or are heavy at
a large distance out
from the
center…make the
most difference.
Look there first.

The last “things” to
small diameter
“things” that change
speed a smaller
amount, change speed
over a longer time,
and change speed
less often. They
will have much less
stored energy. If we
want to reduce
rotating mass we
should look at the
heaviest things that
speed up and slow
down most often,
spin the fastest,
and are large in
diameter with most
of the weight at the
outside edge.

# Why do things work this way?

First we have to
understand what
power and energy
are,
and what rotating
mass does with that
power or energy.
Contrary to popular
belief, rotating
mass does not
consume energy. A
rotating (or moving)
mass stores
energy. This effect
is very much the
same as pouring
energy in a bucket,
much like
charging a capacitor
in an electronics
circuit. Virtually
all of the stored
energy, except for
that lost by
conversion to heat,
is still there and
available to do work
at some time in the
future. That future
where energy is
returned might be
milliseconds later
and help us out, or
it could be some
considerable time
later and waste
energy. This is why
time is very
important.

One example of
useful energy
storage is the
flywheel and
crankshaft of a car.
The force on the
crankshaft is in
pulses. A common
four cycle V8 has four power
cycles per
crankshaft
revolution, and there
are 100 turns of the
crank per second.
At 6000 RPM an
8-cylinder 4-cycle
has 400 power pulses
per second. The flywheel (along
with the harmonic
dampener and weight
of the rotating assembly)
smoothes these pulses
out by storing and
releasing the pulsed energy
from the explosions
in the cylinders.
The result is a
smooth rotation that
will not tear gears
up, vibrate the car, or
beat on
bearings.

We should always
remember rotation,
or movement of a
mass, does not
actually destroy
energy. If it did,
the earth would have
stopped spinning
millions of years
ago! The key to
understanding how
weight changes
affect performance
is to understand
some very simple
basic energy flow in
the system.

Definitions:

Energy

Energy is
the
capacity
of a
physical
system to
perform
work.
Energy
exists in
many forms
like heat,
mechanical,
electrical,
and
others.
According
to the law
of
conservation
of energy,

the total
energy of
a system
remains
constant
.
Energy may
be
transformed
into
another
form, but
it is
constant
within a
system.

For
example,
we all
know two
pool balls
eventually
come to
rest after
colliding.
They stop
moving
only
because
the

applied
energy

(from
moving the
cue stick)
is
eventually
converted
to heat
(from
friction
with air
and the
table) and
sound
(which is
not very
much of
the energy
loss).
The ball
movement
along the
table’s
felt
surface
and
through
the air
transfers
energy
outside
the two
moving
balls to
the air
and
environment
around the
table and
into the
table
itself.
The
temperature
of the
table and
air rises
ever so
slightly,
because
the
applied
energy
moves
outside
the system
we “see”!
Since the
heat
energy is
around in
a very
large
area, we
don’t
notice the
temperature
rise. We
just
notice the
balls
quickly
quit
moving.

Another
example is
our car’s
brakes.
The energy
stored in
the moving
weight of
the car is
converted
to heat by
friction
of brake
rubbing
against
metal
rotors
attached
to the
rotating
wheels.
This
converts
stored
energy
(the
engine put
into the
weight of
the
vehicle)
into heat,
and the
heat
(containing
all of
that
energy)
out into
the air.
Most of
what we
actually
do in a
car is
move heat
around.

Newton’s first law

A mass continues
in its state of
rest, or continues
uniform motion in
a straight line,
unless it is
compelled to
change that state
by forces
impressed upon it.

Old guys like
lot of time on
their hands to
things, but they
got it right. A
rocket coasting
through outer
space is a good
example. It will
go on forever in a
straight line
unless it hits
something, or
unless gravity or
some other  force
pulls it in a new
direction.
The earth
wants to
move in a straight
line, except
gravitational
attraction to the
sun bends its path
constantly.
A bullet
reacts the same
way, except
friction with air
and gravity
changes the
direction and
over distance.

Newton’s second
law

The acceleration
produced by a
particular force
acting on a body
is directly
proportional to
the magnitude of
the force and
inversely
proportional to
the mass of the
body.

We push harder
and/or longer, and
something moves
faster. If it is
heavier, we need
to push longer or
harder (or both)
to obtain the same
speed. It takes
more energy to
accelerate a
heavier object to
the same speed as
we might move a
lighter object to
that same speed.
We can either
apply more force
or apply the same
force over a
longer time to
make something
move faster. It is
times the POWER, or the amount of TIME an amount of POWER is
applied. This is
why those big
showoffs can
eventually move a
large boat, a
an airplane. All
it takes is low
friction and
enough time and
someone who can’t
move a Volkswagen
with two flat
tires can roll a
car.

## Acceleration, by definition, is a change in direction or speed. If we slow something down it is acceleration, just in a negative direction. If we turn a vehicle or any other mass in a new direction, it is really acceleration at a new angle or in a new direction. This is why we can compare or define braking and cornering in G-force (g’s), just as we do with “taking off” acceleration.

We apply force (and
this means we apply
energy) over time (force applied over time is power) to
accelerate an
object. If we want
to spin a top, we
apply force
off-center from the
axis and at right
angles to the axis.
The top stores the
energy we apply, and
continues to rotate.
Over time the stored
force is converted
to heat from
friction and the top
until it finally
stops.

Force is pressure or
energy. The product
of the time we
apply the force

and the amount of
force
is the
power. Power over
time is a very
useful thing to us because it means we can do work with it. Power alone, without
time it is applied,
is not not so useful. Let me give
some examples:

“Watts” are a
measure of power,
much like
horsepower. “Watts”
alone are not
speed, because a
watt does not
include a defined application time. A watt
is only power level, or work level, of energy over an undefined time.

If we include one
hour’s time we would
have a watt-hour.
Kilowatt-hours,
watt-seconds,
watt-hours, and
other combinations
of power level and
time define
electrical energy or
work. This is why we
billed for
kilowatt-hours at
our homes! If we
were billed for
plain old “watts”,
it would not tell
anyone how much
“work” we bought.
Watts are a true
scalar (single
dimension) measure of ability to do work,
just as horsepower
is. Both indicate a
force or the ability
to do work, but both
lack any inclusion
of work time, so we have
no idea how much
work was done, or
could be done.

Horsepower is a
function of RPM and
torque, just like
watts are volts
times amperes.
Horsepower is an
ability to do useful work,
but doing actual
work requires
time
. Torque is
pressure, and since
it does not include
speed it is not a
very useful measure
of system power or
the ability to
accelerate or move
weight. Despite what
we hear,

crankshaft

torque is not
directly related to
moving something off
the line or pulling
the engine, it is
horsepower. The
horsepower (torque
at a certain RPM) is
eventually converted
through gears and
other mechanical
devices to a new
torque value at a
different RPM.
Eventually all we
rotational pressure
on the contact patch
of our tires that
thrusts our car
forward. A 800 lb/ft
torque at 2000 RPM
engine does not
accelerate a vehicle
as well as a 400
lb/ft engine at 5000
RPM, because
horsepower is a
product of torque
and RPM. The higher
RPM engine can be
geared to provide
more forward
pressure at the
wheels. The
higher RPM engine, with less torque, has
more horsepower.

If you notice, ET
calculators don’t
is because torque
does not quantify
the ability to do
work. ET calculators
because horsepower
clearly defines an
ability to do work.

Joules are another
common measure of
ability to do work. A joule
includes both time
and force
(pressure). A single
joule is one
watt-second, or the
equivalent of one
watt applied for one
second. A single
joule could be 10
watts applied for
1/10th of a second
(10*1/10 = 1), the
product of time and
force only has to be
ONE watt-second to
make one joule. If
we applied TWO watts
for 1/2 second, we
have the same work.
Two watts for 1/2
second is one joule
(2*1/2=1).

Horsepower can also
be stated in
kilowatts. One
horsepower is
approximately 0.7457
kilowatts, or 745.7
watts (the exact
value is

0.745699872
kilowatts).

This means 746 watts
for one second is
746 joules and that
is one
horsepower-second!
One kilowatt is
1.341 horsepower.

Many European
engines are rated in
horsepower, you’ve
probably seen that.
A 300-horsepower
engine would be
kilowatts. Your
house probably
consumes between 2
to 5 kilowatts of
average power,
depending on how
large it is and how
you heat or cool.
This is somewhere
between 2-1/2 to 7
horsepower of
average power. Think
of what would happen
to the power grid if
we converted all our
cars and trucks,
like the Greenies
want, to run on
electricity! We
would run out of
electricity very
quickly.

How many joules are
in 1492 watts when
applied for 1/2
second? 1/2 times
1492 or 746 joules!
746 joules is one
horsepower-second.
We could rate our
engines in joules if
we needed to include
both power and time.

## Horsepower and Acceleration

We know horsepower
alone is not a
measure of useful work results,
we must know the
time a certain
horsepower is
applied (or removed)
to know how it
affects
acceleration.
Fortunately there
are horsepower
calculators that
predict ET for a
given power. These
calculators work
because they know
the distance, they
know the applied
horsepower (they
assume it is
constant), and from
that they can
calculate speed and
elapsed time. They
do this because they
assume the power is
applied constantly
and they calculate
the speed change
over time. From the
speed and time, they
get the distance.
When they see 1/4
mile (or 1/8th mile)
they stop
calculating and
display the speed
and the time taken
to reach that speed
and distance.

Now here is an
interesting thing.
It takes a certain
number of
horsepower-seconds
(certain energy
applied) to reach a
certain speed for a
given weight. If we
make the vehicle
twice as heavy, it
takes twice as many
horsepower-seconds
(twice as much
energy) to go the
very same speed.

For example, go to

Now let’s apply 100
HP to go 1/4 mile in
a 1000 pound
vehicle. We went
108.6 MPH in 12.55
seconds. Now let’s
say we have a 2000
pound car. To have
the same speed and
time, we have to
also double the
applied force. If we
apply 200 HP in our
2000 pound car we
have exactly the
same ET and MPH! Now
we know why
insurance companies,
in the late 60’s,
often limited
insurance to a car
with 10:1 weight to
horsepower ratio or
more. They didn’t
care if it was a
4,400 pound Super
Bee Dodge with a 425
HP hemi or a 315 HP
3200 pound Hurst
Rambler Scrambler,
the insurance
companies wanted
weight to power over
10:1 ratio or you
insurance.  10:1
weight-horsepower is
at very best a 108.6
MPH at 12.55 seconds
car! My American
Motors 10:1
Weight-HP Hurst S/C
Rambler, as a documented fact,
set a new national
ET record of 12.54
seconds in the 1/4
mile back around
1970.

Rotating Mass

Let’s say we want to
change the drive
shaft rotating mass
to improve power
available to the
rear wheels. We all
know most of the
weight in a
driveshaft is at the
outer edge. It is a
hollow tube. Let’s
say the original
shaft weighed 30
pounds, and we want
to change it to a 15
pound aluminum
shaft.  The drive
shaft is 3.5 inches
in diameter.

We can go to another
calculator to find
the joules stored in
the driveshaft! When
we know the joules,
we know the
horsepower-seconds
sapped from moving
the car. Let’s say
the engine peaks at
6000 RPM at the end
of the 1/4 mile, and
that took 13
seconds.

Go to this
calculator:

The original
driveshaft weighed
to spin it to 6000
RPM. If we input
that, we see it
consumed (and
stored) 5310 joules.
480 ounces in a 3.5
inch diameter RING
(hollow center) and
6000 RPM.

That is 5310/746 =
7.12
horsepower-seconds
to spin the shaft to
6000.  Since the
time was 13 seconds,
the shaft soaked up
0.548 horsepower
distributed over
that 13 seconds.

Now we change to the
aluminum shaft.
Everything is the
same except the
weight, it is now 15
pounds or 240
ounces. Using that
flywheel calculator
we find we used 2655
joules. This is
2655/746 = 3.56
horsepower-seconds.
Over 13 seconds, we
“stored” .274
horsepower. The net
gain in available
energy over 13
1/4 horsepower.

Here is the real
rule of how this
works….

If we are spinning
up a very large
diameter mass, or a
very heavy mass, and
we do it rapidly, we
sacrifice a lot of
available power. If
we are spinning up a
very small diameter
mass, especially
over a longer period
of time, we give up
less power at any
instant.

The change from an
aluminum flywheel to
a steel flywheel is
much more pronounced
than the change of
the same weight in a
driveshaft because
the aluminum wheel
is much larger in
diameter. We also
speed and slow the
flywheel as we
accelerate and
smoothly spinning
the thing up like a
driveshaft.

The truth is for
drag racing, unless
we have a God-awful
race car where we
have to instantly
change power, an
aluminum wheel
barely makes a
perceptible change
over a steel
flywheel. The
aluminum wheel can
actually be slower
in a drag car,
because the applied
power is not as
smooth. It is harder
to get a light
aluminum flywheel
out of the hole, and
that can easily
offset any small
“available power”
change.

## Summary

This is an
approximation
designed to give you
a reasonable feel
for how a change in
rotating mass
affects
acceleration. We can
see the power
extracted to spin a
weight up is not
very much if we do
not spin it up too
quickly, or if what
we spin is not very
heavy and/or very
large in diameter.
The “feeling” most
people cling to (and
parrot) is that
“heavier rotating
mass kills
acceleration”.  This
is generally not
true at all for big heavy cars, although it can be true. Most
make no appreciable
difference in the
grand scheme of
things. I would
never bother
changing from steel
to an aluminum
driveshaft in my
car, because my car
takes 11 seconds to
go 1/4 mile. The car
weighs 3000 pounds,
and this means I
might save 20 pounds
of weight and 1/2
horsepower lost to
spinning that weight
over the length of
the track. \$400 is
not a good
investment at all
for 1/2 horsepower
over the length of
the track, or the
extra 1/2 horsepower
applied for 11
seconds I have to
extract at the end
and convert back to
heat with my brakes.

I don’t really have
fast things spin up
at this point. I
don’t care if the
crank is 12 pounds
lighter out of 50
pounds. I don’t care
if the driveshaft is
15 pounds lighter
out of 30 pounds! Right now
that \$400 to \$1000
would go a lot
20 more engine
horsepower, or
removed 60 pounds of
static weight. When
I start running out
of easy power, then
I will spend money
making expensive
things lighter. The
big problem right
now is traction, so
right now I want to
smooth the power
out. The last thing
I need is to make
the car more
critical for launch
RPM by using a
lighter flywheel or
shock the tires more
by using a lighter
driveshaft. The
first major weight
reduction will be
the front K members,
because that would
remove weight from
the front and
larger percentage of
weight to the rear
wheels! The last
weight reduction
for my car will be an aluminum
flywheel or
driveshaft.