|
Tech Article Title |
Author |
Date |
|
Effect of Wheel Mass on Acceleration |
2001 |
What is the effect of a change in wheel/tire
mass? Several people have quoted factors of x6 or x8 to determine the equivalent
non-rotating mass variation. That is, an increase of 10lbs/wheel could be as
detrimental to performance as carrying Arnold Schwarzenegger and Scooby Doo as passengers.
I decided to refresh my Physics and came up with the following conclusion: from
the point of view of acceleration, an increase of X in wheel or tire mass is no
worse than an increase of 2X in passenger mass. Not 6x, not
8x, just 2x worst case. This is why.
At any given speed/gear combination there is maximum torque T available at the
wheels. The torque does two things: (1) opposes the rolling resistance and the
aerodynamic drag and (2) accelerates the car. The equation that describes the
equilibrium is:
T = I*u + M*r*a + D*r
*: denotes multiplication
T: torque at the wheels
I: total moment of inertia of the rotating parts
u: angular acceleration of the rotating parts
M: total mass of the vehicle
r: external radius of the tires
a: linear acceleration
D: total drag and rolling resistance
The first term, I*u, is the torque used in making the wheels rotate faster. The
second, M*r*a, is the torque used to accelerate everything (wheels included) in
the direction of movement. The third is the torque used to cancel the drag and
rolling resistance.
Another way of writing the equation above is:
T/r = I*u/r + M*a + D
Now the terms are forces. The left side is the force available at the contact
patch.
If the tires are not slipping, the angular acceleration and the linear
acceleration are related in the following way:
a = u*r
Replacing in the equation and moving things around, we get
T/r - D = (I/r^2 + M)*a
^: denotes exponentiation (r^2 means "r squared")
We can say that the left side is the force available for acceleration. Such
force accelerates an "non-rotating equivalent mass" E,
E = I/r^2 + M
Now suppose that we increase the mass of the wheels and tires by an amount X.
Both the total mass and the moment of inertia will increase; let's call the new
mass M' and the new moment of inertia I'. Obviously,
M' = M + X
What about I'? Well, the moment of inertia of a "punctual mass" [m]
(a mass concentrated in a point) at a distance [r] from the axis of rotation is
[m*r^2]. That is,
the moment of inertia depends critically on the distance between the mass and
the axis of rotation.
In a real wheel+tire combination the mass is
distributed in different amounts at different
distances from the center. In order to compute the total moment of inertia we
would need to know the mass distribution and use integral calculus. We can do a
simpler thing though. We can make the pessimistic assumption that all of the
mass increment is
located on the periphery of the tire, that is, at a distance [r] from the
center. This assumption is pessimistic because in a real wheel some of the mass
will be located closer that [r] and will contribute less to the total momentum
(it is not too pessimistic though: most of the mass is located pretty far from
the center, if not at the periphery). So now we can compute I',
I' = I + X*r^2
The new "equivalent mass" is,
E' = I'/r^2 + M' = I/r^2 + M + 2*X = E + 2*X
In other words, from the acceleration point of view, the equivalent
non-rotating mass increment corresponding to an increment X in rotating mass is
- at worst - 2X.
NOTE: After doing some approximations and assumptions about mass distribution
in a typical wheel+tire combo, I believe that 1.7X is
a better approximation. A 10lb/wheel mass increase would not hurt acceleration
worse than carrying RinTinTin.
|
Message Title: Two aspects I know ... |
|
Posted by:
ceric on 2003-02-25 at |
Message:
Roughly, we need to consider
tires/wheels, rotors/brake/pads/suspension elements separately. In combination,
they are "unsprung weight" (they move when
your car hits a dip on the road - spring, unspring).
Lighter tires/wheels combo enables faster acceleration due to smaller inertia,
more fuel efficiency as hcat7 said. Such lighter tires/wheels combo also
contributes to lighter "unsprung weight".
To understand "unsprung weight", consider
two metal balls (1 small, 1 large) attached together by a spring. The smaller
being the unsprung weight, the larger one being the
rest of the weight excluding the springs/struts(i.e.
"spring"). When the car hits a dip(potholes,
etc) it is like someone pulls the smaller ball while holding the large ball and
then let it go. The vibration begins to affect the larger ball. The smaller the
"small ball", the less effect it has on the larger ball. Therefore,
you feel more "attached" to the road surface. That is why alloy
wheels and aluminum suspension components are popular for sports cars. Fuel
saving and quicker acceleration are icing on the cake.
Nowest trend is that "ceramic rotors" are
gaining popularity (used on some Porsches). It is only 1/3 of the weight of a
steel rotors and fade-resistent. It saves unsprung weight. More attached to the
road surface.
That is my understanding. Please correct if I am wrong here.
P.S. I don't think it affect "vibration frequency". That is body
structure related. It does affect how much weight hits the body shell when your
car hit a pothole... smae frequency of vibration but
hit a lot harder.
|
Message Title: Sit back and relax... its Spring Time! |
|
Posted by:
hcat7 on 2003-02-25 at |
Message:
Wheels and tires are unsprung weight. They sit without any spring to hold them
in place. (rotors etc.... but let's not get crazy yet)
Now the rest of the car (I know.. not rotors and ...)
sits in place being held up by springs. This is sprung weight. Think of your
feet as unsprung weight. The rest of you is sprung. If you loose weight (like your beer gut)then you can move faster. Same with cars.
Now imagine you are wearing Army boots. They are heavy (don't give me a ration
about nylon/kevlar boots either). Same
with huge gumbo tires and steel wheels. Your feet don't want to start
running as quickly as when you have your Nike Air Force Ones. Your car will
start to roll quicker with light alloy wheels and lighter thinner ply tires.
Next, think about when you kick with those Army boots on... It is hard to
start, but once your big foot is moving you can't stop it very quickly. Same with wheels/tires. A light combination will react
quicker and quit bouncing quicker. However, kick a trash can with your Army
boots and it won't hurt your foot. Now do it with your shower shoes on. Ouch!
You just bent your 19" lightweight rims (shower shoe wheels) on a pot
hole. Sometimes a little weight (don't give me garbage about wheel strength on
alloys, this is an illustration) can mean strength. So what we have is that:
A light unsprung wheel/tire will react and start to
roll quicker
A sprung weight (most of the car) helps your performance if you can lose a bit
of that weight (think of car dieting for health)
This is mostly just offered under the influence of Tylenol 3 for my
toothache.....
It is hard to reduce sprung
weight on a BMW. Think of how you are going to remove 40 pounds. Passenger seat? Rear seat? Glass? It is easiest to remove the weight on your wheels by
going to lighter alloy and lighter tires. Unless you are going to go carbon
fiber body panels, start with wheel/tires.
Hey Malachi,
I saw your question about unsprung weight and decided
to email you since you
may not look down the page. :)
It goes like this. Unsprung weight is defined as any
weight (or vehicle
mass) not carried by the springs. This includes the wheels and tires,
brakes, 1/2 of the suspension arms (effective mass) and not too much else.
Sprung weight is everything else such as the body, exhaust, engine etc. The
reduction of unsprung weight is generally considered
more important than the
reduction of sprung weight as the reduction of unsprung
weight allows the
tire to remain in contact with the road better and upsets the sprung mass
less (increasing grip on bumpy roads). Consider a body of infinite mass with
a very light suspension. As this suspension hits bumps it is able to follow
the road very well since it has very low inertia (momentum to carry it over
the downside of the bump, the tire pretty much has to be in contact on the
upside....) and hence will remain in contact with the road better as it can
change direction faster. Now consider a body of relatively small mass with a
very heavy suspension. As this one hits the bump the wheel will have a lot
of upward momentum which will carry it over the downside of the bump and it
will not remain in contact. In addition the body will be upset quite a bit
as a comparatively large force is now coming into the spring mount as a
function of the wheel's displacement. The other consideration here is that
the heavy body will require comparatively high spring rates as well to keep
the nose off the pavement under baking combined with bumps.
This is why heavy cars ride a hell of a lot nicer than light ones. The force
required to restore the wheel to equilibrium may be the same in both cases
(given the same mass suspension and same size bump) however the heavy car
has a lot more inertia than the light car and is upset a lot less. Although
is a very simplistic view. Of course heavy cars are not fast and this is not
the best thing for handling either due to tire considerations.
Now since somebody mentioned vibration of the suspension here is how that
works. If we consider the suspension as a 1 degree of freedom linear mass
spring damper system the natural frequency is the square root of (the spring
rate divided by the mass). So as the mass goes up the natural frequency goes
down and as the spring rate goes up the natural frequency goes up. Damping
does effect this but unless you are mathematically
inclined or really
interested I won't bother going into it.
Another consideration already mentioned is the rotational inertia of some of
the unsprung mass. If the unsprung
mass rotates (wheels etc) then it must
not only be accelereated with the car but must also
be accelerated angularly
(increased rpm). You can think of this as the difference between a light
flywheel and a very heavy flywheel on the engine. Its
really the same thing.
The heavy flywheel will make the engine rev much slower than the light one
and effect acceleration. Don't downplay this effect. I was on a tuning
project where some larger "sport" wheels were proposed for a sport
truck by
marketing. Even though the overall size of the tire was the same (no effect
on gearing) the vehicle was slower to 60 and in the quarter mile. This was
partly due to the increase in mass but also very much due to the fact that
the mass of the wheel and tire was concentrated farther out from center (on
a larger diameter).
Carrol Smith's Tune to Win or, if you are mathmatically inclined, Race Car
Vehicle Dynamics by Milliken and Milliken are the two best books out there
to explain a lot of this. Tune to Win is an excellent primer and very
practical. Stay away from any books by Gillespie (SAE loves him and he may
know something but his book sucks) and any book called 'How to Make your
****** handle'.
Let me know if I can help with anything else. Or if this is
really unclear.
Steve