# I Equivalent mass of rotating parts

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1. Jan 3, 2018

### feak1

Hi everyone !!
i want to find the equivalent mass of a car rotating parts and i came across this site which provides some cars formula : http://www.thecartech.com/subjects/auto_eng/car_performance_formulas.htm

So far i understand almost everything until i get to the part where inertia resistance is calculated as an equivalent mass of rotating parts. Here is the quote :

meq = equivalent mass of rotating parts [kg]

= [ Iw (1/rw)2 + Ip hf (if /rw)2 + Ie ht (if ig / rw)2]

where:

Iw = polar moment of inertia of wheels and axles ≈ 2.7 [kg m2]

Ip = polar moment of inertia of propeller shaft ≈ 0.05 [kg m2]

Ie = polar moment of inertia of engine ≈ 0.2 [kg/m2] + polar moment of inertia of flywheel and clutch ≈ 0.5 [kg m2]

hf = mechanical efficiency of final drive

ht = mechanical efficiency of transmission system (hg x hf)

ig = gearbox reduction ratio [ig1 or ig2 or ………….]

if = final drive reduction ratio

Now what i dont understand is the first part : Iw (1/rw)2 ???
Why do we divide 1 by the wheel radius?? is this a mistyping mistake??
Can someone explain this to me please?

Thanks !!

2. Jan 3, 2018

### BvU

Hello Feak1,

Well, dimensionally, it fits !
We have Newton $F = ma$ and that is called the inertia resistance.

For the wheels you have something extra: you don't just have to make them go faster but hey also have to rotate faster.

That is an additional $\tau = I\alpha$ with
$\tau$ the torque,
$I$ the moment of inertia ($={1\over 2} mr^2$ for a simple cylinder) and
$\alpha$ the angular acceleration.​

With $\ a = \alpha r \$ for a rolling cylinder and $\ \tau = r \times F$,
you can work $\ \tau = I\alpha\$ around to $\ r \times F = I {\displaystyle {a\over r }}\$ and hence $\ F = {\displaystyle {I\over r^2}}$

3. Jan 4, 2018

### A.T.

You divide by the wheel radius squared in all 3 terms, for the reason BvU gave. For the engine and drive train terms you also have the gearing ratios if and ig, but the wheel is accelerated directly by the road with rw as moment arm, so the gearing ratio is 1.

Last edited: Jan 4, 2018
4. Jan 4, 2018

### Dr.D

The easy way to get an equivalent mass for most situations is to write the kinetic energy in terms of a single generalized velocity, say q-dot. Then
T = kinetic energy
= (q-dot)^2/2 (sum of inertia terms)
That sum of inertia terms is the generalized inertia.

5. Jan 4, 2018

6. Jan 4, 2018

### Dr.D

The kinematic analysis posted by Ferroo (#5) is such a painful way to do this problem. Instead, consider the two scalar loop equations
r * cos(theta)+L * cos(phi) - x = 0
r * sin(theta) - L * sin(phi)=0
where
L = connecting rod length
x = slider position from crank axis
theta = crank angle, zero with the slider at TDC
phi = connecting rod obliquity angle

The two look equations are easily solved for phi and x for any assigned value of theta. Consider that done.

The two loop equations can be differentiated with respect to time to relate theta-dot (crank speed) to x-dot (slider speed) and phi-dot (obliquity angular speed). These two differentiated equations can be solved to produce x-dot = f(theta-dot) and phi-dot = g(theta-dot).

At this point, you can write the entire kinetic energy of the slider-crank assembly in terms of theta-dot. From there on, simply factor out (1/2) theta-dot^2 and what is left is the generalized inertia.

Using vectors simply makes an easy problem much more difficult.

7. Jan 8, 2018

### feak1

wow thanks a lot for the answers !!!

Now if i understand correctly and make an example of this :

using this formula : Iw (1/rw)2

wheelMass = 20 kg

rw = 0.330 Meters

Iw = polar moment of inertia of wheel = 1/2 * wheelMass * rw2 = 1.093 [kg m2]

equivalent mass of rotating wheel = 1.093 * 9.143 = 9.999 Kg

So now an extra 9.999kg will be added up to the initial wheel weight , making the wheel feel like 29.999 kg , am i wrong to say that?

Now i have another doubt in my mind since it seems that the formula calculates only one wheel, should i multiplicate the results by the number of driven wheels or by 4 ???

I m asking all this because i m a mechanic and im doing a basic drag race simulator program which i like to implement wheel weight as part of a tuning process.
So far i m able to make the car accelerate using these formulas which work OK but changing wheelMass won't affect acceleration as much as i thought it would.
(about 0.03 sec or less on a 1/4 mile for 10 kg difference)

Everything worked great till i implemented the equivalent mass of rotating parts as part of the accelerating formula cause i wanted to make the acceleration more realistic.

Here s the formulas i use to make the car accelerate maybe you guys can help me understand what i am doing wrong :

total resistance = air resistance + rolling resistance.
total Mass = car Mass + equivalent mass of rotating parts.
car acceleration = (car tractive effort - total resistance) / total mass.

Am i missing something here?

8. Jan 9, 2018

### A.T.

In your original post you define:

Iw = polar moment of inertia of wheels and axles

This sounds like all wheels and axles are accounted for.

9. Jan 9, 2018

### BvU

Seems OK to me: for a simple cylinder (as in post #2) you get an extra $F = {1\over 2} ma$.

Note: do use the superscript buttons; they make a big difference in legibility​

Yes that is for 1 wheel of 20 kg. So multiply by the number of wheels -- if they are all the same size (Are they in drag racers ?). Otherwise you should add up the contributions per wheel.

Compare the extra 10 kg with the total mass. If it is 0.5% and the acceleration is around 5 m/s2 you indeed expect 0.03 " difference at 400 m ( $d = {1\over 2} a t^2 \Rightarrow t = \sqrt {2d\over a}$ )
Must be a calculation error - check the various terms (all units OK ?) . As you've seen one whole wheel is only a small correction.

10. Jan 9, 2018

### feak1

Sorry for that i didn't know how to use them.

Lets say i have a rwd car (rear wheel drive) , do i multiply by 2 cause they are the ones directly affected or do i really just need to multiply by 4 no matter what wheel drive type i m using?

I am not sure i understand this part correctly or where to add this extra value , If i use the previous numbers : m = 20kg and a = 5m/s2

F = 0.5 x 20kg x 5m/s2

F = 50 Kg m2

so when accelerating at 5m/s2 , my wheel would feel like 20 kg + 10kg + 50kg = 80kg?
And when cruising at constant speed, my wheel would feel like 20kg + 10kg = 30kg?

11. Jan 9, 2018

### BvU

Has nothing to do with two or four wheel drive. All of these wheels have to move faster (requiring $F = ma$) and they have to rotate faster (requiring $F = {1\over 2 } m a$ ). That is really all there's to it.

Define 'feel' !
Where on earth do you let the 50 kg come from ?

You are making a mess of things when you write
This is painful to my eyes.

The force to accelerate a 20 kg wheel at 5 m/s2 on a perfectly smooth non-friction surface is equal to ($F_1 = ma$) 100 kg m/s2 or 100 N.

On a road and with no slipping the torque needed for the corresponding angular acceleration -- when translated to a force -- is $F_2 = {1\over 2} ma \$ as we discussed in post #2. So 50 N.

Cruising at constant speed means there is no acceleration ($a = 0$). What cartech calls 'Inertia resistance' doesn't enter. And 'feeling' isn't playing a role either.

That is all.

12. Jan 9, 2018

### feak1

well some people learn by making mistakes , i guess i am one of those .

*F = 50N