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(OP)
Good day.

Is there any simple method to approximate a radius of gyration of a car?

I have found on a recent thread the next statement:
useful term that represents the departure from one (1.00) of the yaw radius of gyration [in meters]

Can someone explain more what does it mean 1.00 in relation with CoG?

Iam trying to make front heavy Audi RS2 Avant Quattro with I5 engine be more agile.
Is it right in a real world trying to tune response on a production wagon by playing with PMOI and weight?

Chassis of the subject

P.S. Sorry for the grammar, English isn't my native language.

I'd think make and model of tires and roll bar stiffness and even tire pressure could have a pretty significant effect there.

How do your current tires rate on the Tire Rack test reviews ( not surveys )?
What are you tire pressures?

A buddy did REAL well running his stock MGA in gymkhanas running Sears snow tires, just on the back.

I remember when the longitudinal front engine Renault Fuego first came out. For about 45 minutes it had about the quickest Road and Track dry slalom time.
It was physics versus Fast power steering, carefully balanced sway bars, and aided or perhaps abetted by those curious and unique Michelin 365-mm (14.4-inch) TRX

2
G'day Boris,

I think, you may mixing/confusing two different terms here. yaw radius of gyration (often labeled k) and a measure, which is known as the Dynamic Index in yaw (DI).
The often as "rule of thumb" cited DI=1 for a full sized sedan, is not the same as a yaw radius of gyration of 1 m^2.
Yaw radius of gyration = SQRT(MoI_z/Mass) MoI_z = Moment of Inertia aroud the (vertical) z- axis a.k.a. Moment of Inertia in Yaw.
To estimate MoI_z, (and this assumes a DI=1), you need the wheelbase of the car, the mass of the car, and the location of the CG - which is easy enough to measure, with a set of scales.
Then you do the following calculation MoI_z=a*b*Mass ( a= distance of the front axis to the CG location, b= distance of the rear axis of the CG location)

Let's use an example which is close to the car you ask about.
I will use a Audi quattro 4000 (in Europe better known as a Audi 90 quattro from 1984, the model before the B4)
The wheelbase of the car, is given as 2.520m, the distance a = 1.124m, from this is follows that the distance b = 1.396m (the car is "front heavy, and uses the same base engine I5 as the car you question about)
the mass is given with 1240 kg (empty).

Now, if we use this data (which come from the NHTSA database, and you will see soon, why I use them), we can extimate the MoI in yaw as.
MoI_z = 1.124m * 1.396m * 1240kg = 1945.7 kgm^2
which gives as a k^2 value of 1.5691 m^2 [k^2 = MoI_z/mass]
and if we take the square root of this k^2 value we get our radius of gyration in yaw
k = SQRT(k^2) = SQRT(1.5691)
k = 1.2526 m ---> as you see, the radius if gyration is not 1 m, but the DI=1

so what is the DI? --> it's k^2/(a*b)

But the estimate of DI=1 is not always true, it just happened to be a reasonable "rule of thumb" for a "standard (front engine, rear wheel drive) full size sedan, something like a BMW 7 series etc.
The measured data for the Audi 4000 quattro, show a MoI of 2352 kgm^2 [compare to our estimated 1945.7]
Therefore the radius of gyration for this car is:
k = SQRT(MoI_z/mass) = SQRT(2352/1240) = SQRT(1.8968)
k = 1.3772 m [ compare to our estimated 1.2526m]

the resaon that k is larger then the estimate is, because this car has a DI of >1 [k^2/(a+b)] = 1.2088

There are other, (better?) , rules of thumb to estimate MoI which take the overall length of the car, and the ration overall legth to wheelbase into account, which gives a indication of the "overhangs" and thereby a better "guess" for the mass distribution, but I can't remember the formula out of the top of my head. But I'm sure someone on here, will know it.

So, all things considered, you may use a DI=1.2, measure the wheelbase and the mass of your car in question, and that should give you a good enough estimate for your radius of gyratio (k) - IMHO

On the other hand, without relocating heavy items &/or altering the wheelbase (front overhang) it will be difficult - IMHO - to change this value by much. Not sure if this is a daily driver, or a track/race car.
The only item, which I can see, that you could reasonable easy relocate is the battery, if it is not allready at the back.
See, what Audi did with their Audi Sport quattro S1 group B rally car, to try and change this --> they placed the radiators at the rear, but this is quite a heavy modification for little gain.
But this is up to you, if the battery is still at the front, you can try to move it more backwards (boot) and then you can try to move the front wheels a bit more forward, to get the engine a bit more "over the front axle" [see similar issue with BMW 3 series 4 vs. 6 cylinder cars].
If this is a track/race car, and cost and effort is not a limiting factor, you should consider moving the complete engine gearbox backwards, but this is a "major project", which needs a modification of the front fire wall, a shorter propshaft, and maybe longer front drive shafts and other mods. Not sure this is worse it. But that's fr you to decide/figure out.

best regards & good luck
TC3000

O.K. - I just looked up the other "rule of thumb" formula

MoI_z = 0.1296*mass[kg]*wheelbase[m]*overall length[m] ---> the 0.1296 is a regression coefficient (fudge factor) to fit a curve onto the measured data of 56 German production cars ( Rau et.all), but this approach is a bit "dated" (1982) as well, and the correlation with modern cars (SUV's, general higher weight etc.), is maybe not always good.

in the above example, we would get
MoI_z = 0.1296*1240*2.520*4.465
MoI_z = 1808 kgm^2
k= 1.195 m, which is less, then the DI=1 estimate

I think, this estimate is more suited for "compact" cars, with short overhangs and a DI<1.

(OP)

Yes I was confused by DI (I've never read in books about it and never heard in the university)

It's a weekend car with some track day action.

My plan is:
Fiber hood - 10kg save 1.3m from CG = 10x1.69 = 16.9 kgm^2
AC Remove - 20kg save from 1.6m = 20x2.56 = 51.2 kgm^2
Electric PAS - 10kg save from 1.6m = 10x2.56 = 25.6 kgm^2
Alloy Rad - 10kg save from 1.4m = 10x1.96 = 19.6 kgm^2
Remove spare wheel - 20 kg save from -1.6m = 20x 2.56 = 51.2 kgm^2
Light battery - 10 kg save from -1.8 = 10x3.24 = 32.4 kgm^2
Remove tow bar - 10 kg save from -2 = 10x4 = 40 kgm^2

Total = 236.9 = -10% from NHTSA 2352 kgm^2 and it's nearly 1 DI

Am I correct in my approximation and in DI understanding?

I need it only to realize - will it be worthwhile work due cost and effort?
What DI is a ballpark for weekend track day car? Or it would be wise to keep 1 DI and tune other factors of the transient response?

P.S. There is a big gap between books,sandbox and the real world - called experience. I don't want to do any work only by my guess.

"will it be worthwhile work due cost and effort?"

Aye there's the rub, as the famous Danish race engineer Hamlet said. If your calculation is correct then you've changed the polar moment by 10% (which is pretty good), and reduced the mass by 100 kg.

The latter is worth it anyway, tho I'd leave the EPAS in, and the alloy rad sounds like an extravagance.

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

(OP)
Thanks!

There are 2 heavy rads 20kg total that I want to change with 5 kg alu one.

And there is a tandem pas with hydro assisted brakes line.I want to change it with an epas and move it near the rear axle.

The quattro wagon for a track sounds like an extravagance to me.But I want to learn how it works and possibly make it better for my purpose.

I recognize the original statement, seeing that I wrote it. It is based on real measurements of hundreds of vehicles: both inertias and total masses AND the analysis of theoretical vehicle dynamics differential equations.. The relationship I mentioned is a large factor in yaw and sideslip dynamics because the balance between the SUM of side forces and their DIFFERENCE places demands on tire and suspension characteristics in order to meet handling goals. What is evident to those willing to construct the coupled differential equations that govern these motions is that the inertia to mass ratio is a strong player in the foundation of the plant dynamics (denominators) for yaw velocity and sideslip angle motions. To simplify this relationship, this ratio is often reduced to an index form, which suggests that the vehicle is only a bar bell. Unfortunately, vehicles are not actually bar bells. There a few point masses. The big players (engine and transmission for example) have their own inherent principle inertias so just adding a motor as a point mass doesn't work very well in getting total vehicle inertias from the placement of (point) masses. BTW: the powertrain principle inertias seldom line up with the vehicle body inertias.

The factor you mention is a big player when it's time to synthesis vehicle dynamic characteristics (gains and bandwidths/response times). Then it is evident that an 'inboard' mass distribution puts lower demands on things like tire cornering stiffness. (Remember the gains and bandwidths are targeted by design, not settled for). The bottom line is that a vehicle requires lower cornering stiffnesses at each axle in order to meet its handling goals if this index is lower.

So now for your specific question. Moving BIG parts inward is good. Small motors are good, manual transmissions are good. Light glass is good. Fuel tank inward is good. Passengers inward are good. Spare tire left in the garage is good. Heavy bumpers are bad.

Note that it is the transient response traits that mainly benefit from this effort. For the same wheel weights, tires needing to generate lower yaw moments will ride better and ultimately deliver better fuel economy via lower rolling resistance.

TC3000. Moving the battery to the boot will be great for F-R weight distribution, but will probably be further from CG (on this front-heavy car) and thus increase MOI. (Great post on the whole - please excuse the nit-pick)

Agility is a function of (mainly) MOI(Z) and wheelbase. Moving the front wheels forwards 25mm is a 1% improvement and changes weight distribution 2%.

Torque vectoring has a huge effect on agility. Add brake drag to the inside wheels in proportion to steering angle and throttle %.

je suis charlie

(OP)
More I try to understand than more I dont and more I like it :)

My starting point is a Millikens formula:
ω2 = CfxCr/m2 * l2/k2 x (1+KV2/V2)

CfCr - will be tuned on the last stage by dampers, chassis and alignment
m2 - I will reduce it by my plan above
k2 - It will reduce with mass
K - stability factor will be slightly negative value due slightly oversteer
V2 - will be as max as possible in each particular situation

Then Ill try to open the next can of worms and to dig in overshoot and yaw damping.

Is my platform correct? Or Milliken is outdated on that subject?

(OP)
Ok, thanks! Useful hint.

@ GG
No Problem with being precise, you are correct, that moving the battery all the way to the rear, may doesn't help with MoI or has the potential to make it worth. In terms of reduction in k, Boris should move it to the CG position, so that the mass*r^2 contribution vanishes, and only the MoI of the battery around its own CG remains as contribution to the overall MoI of the car/vehicle.

@Boris
please keep in mind, pay attention to what cibachrome said, you have only part of the k^2 contribution considered in your calculation.
A real physiacl object is more then a "point mass", because it will have a MoI around it's own CG. This MoI will not vanish, even if you place the object closer to the CG of the car.
Maybe look up some infos (Google or textbook) about the "parallel axis theorem" [in some countries known as "Steiner Rule"], to get a better understanding on the subject.
In short the contribution of a object, (let´s say the engine for sake of an example), to the Overall MoI of the body (car) is governed by.

I = I_cm + mr^2 --> I_cm = the MoI of the object (engine) around it's own mass center (CG) + m = mass of object * r^2 = square of the distance/radius of the CG (of the object/engine) to the point of Rotation of the overall system (car).
As you can see, so far, you have only considered the second term.

Now, as we go a bit "off topic" , you have to distinguish between the "transient" part of the response, and here MoI Features quite prominently, and the steady state part. Which in short means how well are your tyres matched to the weight distribution of your car. A front &/or rear heavy car doesnt has to be US/OS in itself, you just need to find the "right" tyre for this.
So, this can turn a bit into a "Chicken & Egg" Problem, do you have a car, and "tune" (find) the "correct" tyre for it, or do you have a given tyre, and try to "fit your car".
Your initial question, about DI and the Ratio between mass distribution and MoI (radius of gyration), has a large influence about "how fast" and "in which direction" (initial at least) the slipangles of your rear tyres build up.
You will may find only examples which deal with Baseball/cricket bats &/or Tennis racks, but let this not distract you.
Remember, that you can see a steering Input at the front wheels/axle as a "Impulse" to the System (sprung mass), and that this "Impulse" will have to cause a change in "momentum" (linear & angular) of the system (car /sprung mass).
If the linear or angular response to this Impulse dominates, is a function of DI, and this will define how and in which direction your rear slipangles build up.

If you have understood this for the simplified "plane" (2D)Motion (yaw only), you can think about a real car, and how it will respond to this "Impulse" at the front axle. It will become a 3D "problem", where the "Body" (sprungmass of the car) will respond with more then one rotation (combined response of yaw and roll), to the Impulse, and doing so simulatanously. Therefore, it's not only MoI in yaw (z-axis) which enters the equation here, but MoI in roll (x-axis) as well. The instant motion axis can (and often will be) inclined in space, which you Need to consider when using the MoI values measured around the "principal" axis.
But for a start, you may focus on a simpler "plane" (yaw only) Response model, and play/experiment around with this a bit.

Reducing mass, and changing the CG position, as well as trying to lower the CG height, has it's own benefits, and is worth some effort - no question. But you may don't manage to Change your k (or k^2) value significantly with this.
So while lowering m (mass), will change MoI, which has it´s own benefits, it may does not change k at the same time (at least not automatically, it can, but it doesn't has to).

I don't have my copy of Milliken here with me, and I may don't understand correctly what you mean with the CfCr terms, you use, so appologise if I get it wrong. In case they stand for the tyre Cornering Stiffness front and rear, I don't think, that you can alter These values much with "dampers" - alignment (camber,toe) yes
A other "cheap" and simple way worth - IMHO - to consider, is the use of different wheel/rim width, for a given/Chosen tyre.
I'm not talking about "Boy racers" stretching, but you will find, that for a given tyre size, you have usually a 1-1.5" wheel width "window", which you can use to your Advantage, to fine tune handling/Balance, with comparable little effort - but that's all up to you.
If you learn/understand how DI affects your rear tyre slipangle build up, I think you are on your way, and you will know, what you would like to do, and where you want to go with it.

Good luck - it's an interesting, but not "easy" topic.

P.S.: It seems, that you can't change older posts on here, but maybe one of the mods/admins could be kind enough to correct a typo in the above post. In the lower part, of my first post it reads:

#### Quote (because this car has a DI of >1 [k^2/(a+b))

= 1.2088]

DI = k^2/(a*b)

Thanks
TC3000

OK, I just reviewed this thread again and want to add a more practical suggestion. I'm reading that you want to make your own personal car more "agile". Let's say by this you mean more "nimble" or more "responsive", or more "maneuverable".

Forget the weight stuff, and all that MOI theory. Keep the car saleable. Instead, there are some simple and easy changes you can make that can be reversed when you've had enough.

To make the car more "nimble", put a faster steering gear in it. Instead of the say 16.5:1 or 17.5:1 gear, put a 13 or 14:1 into it from another model. Maybe aftermarket, maybe production VW something. This will raise your steering efforts and may stall your power steering pump, so at the next step, put a higher flow rate pump into it. Putting the higher output pump into it will nimblize the car all by itself. (Ff its electric power steering this will need some controller help and maybe some even bigger primary wires.

If you can't or won't change the gear stuff, consider putting a shorter steering arm length into it. Either after-market or junkyard knuckle stuff. Try to maintain the outer tire-rod ball end height, otherwise you will screw up the ride and roll steer parameters these cars have.

These are things any proper boy-racer can do to make their car "more agile". After all, it is an AWD "Station Wagon" according to your picture. Next time purchase a separate car to run on sunny days and keep the wagon for a safe family trip with all your luggage on board and the kiddies strapped in.

"Its a weekend car with some track day action"
Better tyres will sure help with agility but steering ratio won't improve times on the track unless there is a problem with crossed arms or ability to turn the wheel fast enough.

je suis charlie

I don't believe he's on a "track". He's driving a 'Classic' 5 door station wagon on curvy roads and fast highways. The car is a powerhouse with numb steering and "wooden" brakes. Agility is maneuverability and "fun to drive". Where is 'track' ever mentioned??

I'm surprised you didn't tell him to chop a 30.cm front-to-back section out of the center of the car. That will surely give him a new "track" position. (Hey, that's a good one !!)

(OP)

You are right about my target.
Twisty roads and some track action for fun not for a competition.
I love this car and want to make it adequate to handle with 380hp. I don`t want to make it Miata or Mr-2.
It is more like a drag car now that can drive by the straight line only.

Your post helps me to clarify my options and plan.

But it has 14:1 rack, A048 Yokohama 225x40x18 tires, 8,5J mag wheels and powerful PAS pump already.

If I understand you right, you suggest to work out with other variables from this eq.

* Maximize Cf,Cr
* Balance the car
* Reduce stability factor (1+KV^2)/V^2
* And try to reduce time for rear wheels slip angle build up
* Leave k and mass as is

Am I right?

Another thing that I want to try is to increase front dampers low speed compression force. Main idea that it will increase lateral force on the fornt outer wheel so yaw will build up quicker.
Can it help in my situation? Or there is a mistake in understanding?

@cibachrome "Where is 'track' ever mentioned??"

Post 5 line 3.

je suis charlie

But he (and I) meant track FUN, not track serious competition. The fun factor stuff should be limited to reversible changes.

"Another thing that I want to try is to increase front dampers low speed compression force. Main idea that it will increase lateral force on the fornt outer wheel so yaw will build up quicker.
Can it help in my situation? Or there is a mistake in understanding?"

That will probably reduce agility by increasing front weight transfer on turn-in. If the car is not already picking up the inside rear, you may be able to improve turn-in by doing the opposite - increasing rear damping.

je suis charlie

This may be a dumb question, but:
Generally, when the car turns, where's the center of rotation?
Would it (ever?) be at the CG? (unless all four tires are sliding)
Without significant rear steering, wouldn't it usually be somewhere near the center of the rear axle?
Then the important MOI for each component would be its mass and its distance from the real center of rotation to the CG of that component, wouldn't it?
regards
Jay

Jay Maechtlen
http://www.laserpubs.com/techcomm

The instantaneous center of rotation is at the intersection of the perpendiculars to the velocity vector at each axle, ie somewhere a long way from the car.

That's pretty funny and fundamental, need to have a think about that.

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

When its going in a straight line the axis of rotation is a long way of to the side from the car(infinity in fact)and and moves in closer as you turn tighter.

je suis charlie

Relocate battery to the trunk if you haven't already. Some of those euro cars are coming with them already mounted in the trunk now, like my volvo S60 T5 that I had. It was a complete electro mechanical blunder of an automobile, but they got a few simple things right.

"Formal education is a weapon, whose effect depends on who holds it in his hands and at whom it is aimed." ~ Joseph Stalin

Moving the battery to the boot will be great for F-R weight distribution, but will probably be further from CG (on this front-heavy car) and thus increase MOI.

de ja vu all over again

je suis charlie

If you really want lower MOI, do this!

"Formal education is a weapon, whose effect depends on who holds it in his hands and at whom it is aimed." ~ Joseph Stalin

Does the "too light to hook up " at ~ 4;50 really make sense?

I'm guessing the front/rear weight bias may be pretty bad possibly worsened by rear roll stiffness, or the max camber adjustment is now insufficient, or soomething.

If your issue is mostly turn-in, check tire temps to see if they’re pushing or not hooking up. Either Ackerman or anti-Ackerman –the latter is theoretically preferable- can help as well as relative lower front roll stiffness. Torque vectoring would really help but I have no idea as to its availability in your case.

In my opinion you’d be much better rewarded to work on dial in variables before resorting to the heroics you’re considering.

If I take the average in my database of all cars that I have had measured the average radius of giration of a car turns out to be 45% of it's wheelbase. Note however that the "extremes" go from 40% to 50%.

Cheers

dynatune4xl

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