It has always bothered me how a car like the 911 can still manage to be so competitive with an absolutely horrendous weight distribution.

I decided to delve into the matter and calculate just what makes the 911 so unique, and what forces allow it to keep that heavy rear end in check. To do this analysis some justice we must look at the car in a few critical scenarios. First we will look at launch characteristics, then we will work through a full throttle high speed cornering scenario, and then a no throttle cornering scenario, and finally we will do a braking scenario.

Before you continue reading, if you havenâ€™t yet I strongly recommend you read through some of my other articles. To get a better idea of what I am doing.

Tyres Part 1

http://www.autospies.com/news/Tyres-Tirâ€¦

Tyres Part 2

http://www.autospies.com/news/Tyres-Tirâ€¦

Tyres Part 3

http://www.autospies.com/news/Tyres-Tirâ€¦

Static Weight Distribution

http://www.autospies.com/news/Static-Weâ€¦

Dynamic Weight Distribution

http://www.autospies.com/news/Dynamic-wâ€¦

Weight Distribution and Turning Moments

http://www.autospies.com/news/The-Effecâ€¦

First, I set out and researched some numbers to help with the calculations. I have all of the numbers I used listed below for the 911 GT2. Itâ€™s important to note at this point that this is a simplification used to understand how the car works, not a super accurate depiction of exactly what goes on in the car. As a disclaimer, there are far more variables in real life, these variables will all change to some degree, and all calculations are of course done with enough assumptions to show a point without the use of a super computer and high end simulation software.

Launch Torque of 500lb-ft at 2200rpm (assumed value to include drive train losses)

Torque of 440lb-ft at 5950rpm in 4th gear at 125mph (assumed value to include drive train losses)

1st gear ratio of 3.15

4th gear ratio of 1.09

Final drive ratio of 3.44

Weight 3175lbs

Front 37%

Rear 63%

Wheelbase 89.4 inches

Width ~70 inches (lowered a bit to accommodate for center of tyre patch)

Tyre Diameter 22"

Longitudinal tyre coefficient of friction of 1.2

Lateral tyre coefficient of friction of 1.0

COG height (approximately 1/3rd of vehicle height, so approximately 16.7")

Front down force at 190mph 120pounds

Rear down force at 190mph 306pounds

First, letâ€™s determine the static weight distribution and distances of the COG.

3175lbs*0.37=1175lbs on the front wheels

3175lbs*0.63=2000lbs on the rear wheels

89.4*0.37=33.1 inches ahead of rear wheels

89.4*0.63=56.3 inches behind front wheels

With this data we can start finding the dynamic weight distributions of the 911gt2 under launch scenarios.

To see how much tractive force is available to the wheels in first gear we must first find the overall gear ratio.

3.17*3.44=10.8

Then multiply the torque by this number and divide by the wheel diameter (remembering to convert the wheel diameter to feet!)

10.8*500lbft/(22in/(12in/ft))=2955pounds

So we have 2955 pounds of forward thrust available to the car in this scenario, the next question is whether the vehicle will be traction limited. To answer this, we multiply the coefficient of friction by the weight on the driving wheels.

1.2*2000=2400 pounds

Since 2400

We want to see what this 2400 pound force will do to the overall weight distribution of the vehicle. By multiplying it by the distance to the COG we find the forces turning moment (i.e. the torque lifting the front of the car up and squatting the rear end). Dividing that moment by the wheelbase will give us the weight transfer from the front to the rear of the car.

2400lbs*(16.7in/(12in/ft))/(89.4in/(12in/ft))=447lbs

So initially under traction limited acceleration 447lbs will be transferred from the front wheels to the rear. But with this added traction in the back from the weight transfer our tractive force will increase.

2000lbs+447lbs=2447lbs

2447lbs*1.2=2937lbs of tractive force

Now if we re-run the calculation with 2937lbs of rear traction we will get 548lbs of weight transfer leading to a 2548lbs of tractive force. Repeat this enough times and the answer will steady out at 576lbs of weight transfer. Leaving us with 599lbs over the front wheels and 2576lbs over the rear wheels; a 23%front 77% rear weight distribution. It is important to note now that the time that this weight transfer will take place depends on the polar moment of the car the damping characteristics of the car and the spring rate. In other words, when the moment the moment torque is applied, the car will effectively have 2447lbs of force on the driving wheels and that number increases to 2576lbs as the car settles out moments later.

What does this mean when driving the car? Simply, under hard acceleration as much as 77% of the vehicles weight is over the drive wheels. Allowing it to accelerate much harder than it would if it had a 50/50 weight distribution.

We will now look at the full throttle corner. For this we will assume the car is in 4th gear at 125mph cornering at about 1.0g.

Like the previous scenario we wish to find the available torque and compare it to the available traction.

1.09*3.44=3.75 overall gear ratio

3.75*440lbft/(22in/(12in/ft))=900lbs of engine force at the wheel

900lbs is far less than the 2400lbs of tractive force for the static weight distribution. Therefore we will assume that we are not traction limited in this case. Next, we shall apply the 900lbs to the vehicle just like we did for the previous case. I will skip showing these calculations since they are repetitive and donâ€™t add any value to this already lengthy post.

At full throttle in 4th gear at 125mph we have a settled out weight distribution of 2202lbs on the rear and 973lbs in the front. Next we shall apply the cornering forces by multiplying the vehicles weight by the height of the COG to find the torque that is making the car roll.

3175lbs*(16.7in/(12in/ft))=4410lbft of roll inducing torque

Dividing this torque by the distance between the wheels will yield the overall lateral weight transfer.

4410lbft/(70in/(12in/ft))=756lbs of weight transfer

Next, we wish to determine the down force over the front and rear wheels in this scenario. Since we have the down force at 190mph, we will linearly interpolate to find the approximate force at 125mph. It is important to note that down force is not a linear scale, we are assuming it to be linear due to a lack of data and the fact that in the majority of cases less than 200mph it is fairly linear.

((125mph)/(190mph))*120lbs = 80 pounds of front down force

((125mph)/(190mph))*306lbs = 201 pounds of rear down force

Finally, we adding together all of these individual weight transfers to find the overall weight distribution. To distribute the amount of roll seen between the front and rear wheels, we will use a ratio using the location of the COG. To simplify the weight transfer and since I couldnâ€™t find sufficient data we will ignore the difference in roll bar stiffnessâ€™s front/rear spring rates. But we can make some inferences as to how this assumption will affect our result. Namely, the 911 uses stiffer rear suspension components then front, this allows for less weight transfer in the rear and more in the front.

Front inside wheel

(1175lbs/2)-(202lbs/2)-(756lbs*0.37)+80lbs=286.8lbs

Front outside wheel

(1175lbs/2)-(202lbs/2)+(756lbs*0.37)+80lbs=846.2lbs

Rear inside wheel

(1175lbs/2)+(202lbs/2)-(756lbs*0.37)+201lbs=609.8lbs

Rear outside wheel

(1175lbs/2)+(202lbs/2)+(756lbs*0.37)+201lbs=1169.2lbs

Well, now with about 4 times more weight on the rear outside wheel we are starting to get a glimpse of what exactly what keeps that rear end in line. But before we finalize any conclusions letâ€™s have a look at how much traction is left over in each tyre. For this we will assume linear coefficients of frictions to be conservative and to help with the calculations. In reality, the more a tyre is loaded, the lower its coefficient of friction will be which is why itâ€™s generally a good idea to minimize weight transfer in a race car.

Force Applied To Front Wheels

1175lbs laterally

Force Applied To Rear Wheels

2000lbs laterally

900lbs longitudinally

Sqrt(2000lbs^2+900^2)=2193lbs net force

Sqrt((1.2*(900/2193))^2+(1.0*(2000/2193))^2)=1.04 effective coefficient of friction for rear wheels

Leftover Traction at Front Wheels

1.0*(286.8lbs+846.2lbs)-1175lbs=-42.0lbs

Leftover Traction at Rear Wheels (100% throttle)

1.04*(609.8lbs+1169.2lbs)-2193lbs=-342.8lbs

Looking at the leftover traction we can see that at full throttle the rear end is going into an over steer condition. Next we shall look at a no throttle condition to see whether the car will over or under steer.

Front inside wheel

(1175lbs/2)-(756lbs*0.37)+80lbs=387.8lbs

Front outside wheel

(1175lbs/2)+(756lbs*0.37)+80lbs=947.2lbs

Rear inside wheel

(1175lbs/2)-(756lbs*0.37)+201lbs=508.8lbs

Rear outside wheel

(1175lbs/2)+(756lbs*0.37)+201lbs=1068.2lbs

Force Applied To Rear Wheels

2000lbs laterally

Leftover Traction at Front Wheels (No Throttle)

1.0*(387.8lbs+947.2lbs)-1175lbs=160.0lbs

Leftover Traction at Rear Wheels (No Throttle)

1.0*(508.8lbs+1068.2lbs)-2000lbs=-423.0lbs

What is interesting to note from these calculations is that the car exhibits increased stability (reduced oversteer) when on the throttle due to the balance of the car going more over the rear wheels. This is rather counter intuitive since most rwd cars will become more unstable as the throttle is pushed. This key piece of information is what we need in figuring out why the 911 feels so stable in a corner. Instead of losing rear traction when you hit the gas pedal you actually gain a little bit! This works of course until you exceed the limits of your tyres and the massive drop in the coefficient of friction isnâ€™t able to counter this effect.

Finally, letâ€™s take a simple look at braking performance. Under full braking the weight will transfer forward evening out the vehicle. Since braking is a traction limited action the full braking force on the vehicle will beâ€¦

3175lbs*1.2=3810lbs

Applying this braking force to the cog yieldsâ€¦

3810lbs*(16.7in/(12in/ft))/(89.4in/(12in/ft))=711.7lbs of weight transfer.

1175+711.7=1886.7lbs

2000-711.7=1288.3lbs

This means about 68% of the weight is over the front wheels and 32% of the weight is over the rear wheels under peak braking. If the 911 were more evenly balanced the front wheels would see an even greater load on the front wheels, and due to non-linear tyres this would minimize braking distance.

So what have we learned?

-The rear weight distribution is good for rwd launches and hard braking.

-The car becomes more stable under power so long as the limits of the tyres arenâ€™t being exceeded. This effect gives the driver extra confidence as the car will want to go straighter under gas and will want to turn in more when off the gas.

-The excessive weight difference between the rear outside wheel and the front inside wheel limits peak cornering capability due to the non-linearity of tyres.