Why do so many magazines try to sell g-force as absolute grip? The highest G-force attainable in a car without down force is less then 1.00G. When they test on a track they have to drive at high speeds or on a banked turn in order to get the numbers that you see in tests. In order to test the true absolute grip, you would have to drive a car at relatively low speeds and corner hard until the wheels brake loose. Without down force or a banked turn it is physically impossible to go beyond 1.00G. The weight of the car Horizontally cannot exceed the total weight of the car. So whenever you see claims of G-forces in excess of 1.00g the results are already out of the realm of reality. Manufactures will claim High G's but it is mostly a marketing ploy. If you put the same tires on a Camry and recreate the same conditions (speed, degree of bank Etc.) you will likely get the same result with the g-force. The true test of a cars handling performance is its ability to keep as much rubber meeting the road at all time, such as when traversing through quick transitions while going over sharp bumps such as a bridge expansions. G-force alone will NOT tell you the measure of a cars handling performance, just the tires lateral grip.

 

I thought that they used a flat 200 ft diamter concrete circle?

 

It's been a while since Newtonian vector physics, but I remember a little.

I am no expert in g-force tests on auto, but I do not think they do it on bank curves .. I was alway under the impression it was done weaving through cones on a flat road.

A level car turning (with, of course the proper suspension) can take more than 1G ... I think I know what you were considering when you said that it was physically impossible for a car to take over 1g : 1g = the wieght of the vehicle .... so if i apply an ounce over that of force perpendicular to the car's orientation to the ground it should move, right? That would be the case if there were no resistant factors: friction of tire, etc ...

In considering lateral g-forces in a car, you:

1. Cannot get away from creating additional "downward" forces when corning in a car ... this is what the design of the suspension system does ... it takes and allocates tranferring weight in such a manner to produce downforce on the tires.

2. The faster you go, the greater the aerodynamic downforce is applied. Considering the tires (friction cooefficients, etc), the track, blah, blah ... there exist an optimal speed and angle of attack into a corner to get the maximum g-force that it can produce without breaking.

You are correct that a banking would allow for a greater lateral g-force reading .. but this would be only from an orientation from level ground .. not from a perpendicular orientation from the degree of the bank ..

Either way, I think they do do the ratings on level ground. Any know for sure??

 

Every test I've ever seen was done on a 200ft. diameter skid pad. All level. The laws of physics of course cannot be broken but besides the suspension geometry, you also have to consider the tire compounds and mating surfaces. Of course if the skid pad were made of perfectly smooth glass or stainless steel, your 1G limit theory would be absolute. The fact of the matter is that the rubber conforms to the rough surface and basically creates thousands of little fingers that are grabbing and holding on to the surface. Much like a gecko climbs up a seemingly flat wall. The gecko obviously has more stick power than 1G, but would probably be close to that figure if he was trying to climb a smooth stainless steel surface. Otherwise he wouldn't be able to climb on anything greater than a 45 degree angle.

It is because of the steering and suspension geometry as well as the roll characteristics of the vehicle that make the tires distort and/or not meet the pavement perfectly flat. This actually takes away from it's theoretical 1G limit. It's the stickiness of the tires mating with the pavement that makes up for it. I'm not exactly an expert on this subject so I'm sure there are others that could explain it better and I'm sure they'll pipe in.

 

What you are forgetting is the 1G down force (Acceleration of Gravity)created by gravity is also acting on the car in addition to the frictional component of the tires. The force vector of 1 G lateral and 1 G vertical has a resultant force of the square root of 2G (about1.2G) at a 45degree angle down from level. The key is actually with tire adhesion. With tire adhesion of zero any lateral accelleration (force) would displace the body laterally. With infinite adhesion (think Super Superglue), no amount of lateral force could displace the body laterally. The greater the tire contact with the road, the greater the down force(weight of car+any aerodynamic downforce), and the more conforming (sticky) the tire compound, the greater the lateral Gs possible. 1G is not the theoretical maximum one might suppose (actually there is no theoretical limit, only practical ones) and I would venture that Formula 1 cars routinely excede 1G.

 

hmm 1.4G s here on hard braking ( enough to make you )from 135 mph down to 60 mph to make a 180* right hand corner on VIR #1&2 then 1G accelorating out of #2

 

Because it get people to buy the magazines!

This was believed to be true in the 1950s, but tire technology has displaced this 'impediment'. The tire in contact with the road is not sliding on a microscopically smooth surface, but a surface where the rubber moelcules in the tirea bond to the road and then release in the microseconds the contact patch runs over the road.

A) racing slicks have a coefficient of friction the 1.6 Gs range (or higher).

B) most sports cars can corner at higher G loads when accelerating rather than at constant velocity around a skid path. That is why the same cars that get 0.95 to 1.0 Gs in the skid pan are often measured at 1.15 to 1.22 Gs in operation on actual race tracks.

C) dragsters leave the line at around 5Gs and this is before the aerodynamics have any chance in creating downforce.

 

Ive been able to pull 1.5 in this sunday at the AutoX.. Neck snappingly fast transition.. sickening to my ridealong

 

You my friend are sadly mistaken. You would be correct if the only force at work keeping the car on it's path was friction. But thankfully tires also use a wonderful property called adhesion. Get a tire up to temp and it literally glues itself to the road, allowing the vehicle to sustain more than 1 g of lateral acceleration, no downforce required.

 

Even if there is an ounce of reality to claims above 1.00g, then they are selling the tires, not the car... if you read my email, I said people should not use it to determine a cars handling performance, just the lateral grip of the tire.

 

I believe the racing slick 1.6g is not rated lateral, but in the foward axis. this takes into account tire flex and other factors, such as the effect of a lauch when a car tranfers weight to the rear(positive g). I am completly aware that a car can reach in excess of 1.00g, but I do not believe it will do so at slow speeds on the horizontal axis. There have been several experiments done on cars with ridiculously low centers of gravity (some students at stanford tried this) and excessivly wide tires and they were not able to break 0.95g(this may not be exact, I only rememeber that it was less then 1.00g).

 

When you lower the speed you're lowering the acceleration measured from the change in direction. So at low speed you will have lower g measurements, that's obvious isn't it? Stopped you'd have no lateral g rating. Makes sense right? You use a constant radius turn and you keep that radius but you increase the speed until you break traction. Your idea of lowering the speed of the test would automatically lower the G measured unless you modified the radius to compensate. They use a constant radius on purpose. So you must increase the speed to break point. Another way to look at it would be the speed at which the car lost traction on a flat 200' radius skidpad. Then you'd know your velocity for a constant radius 200' turn at a track you like. Or an exit ramp.

This takes out the instant spikes and gives you a smooth speed where your traction would gradually break away in a smooth turn. I find it useful. Knowing the physics you can calculate the car's breakaway velocity point reasonably accurately for any flat constant radius turn. How can you find that not useful?? It definately tells you good information about how a car turns. You just need to look at the information available differantly.

Look at it this way, enter your favorite turn while carving through that radius and then vary the speed. Slightly faster and you'll widen. Slightly off throttle and you'll tighten and point your nose into the turn. The perfect balance point for off or on throttle oversteer. That's the most enjoyable part of track days for me. Balancing the car there sideways in turns.

It's dancing.

-mikey

 

Neko,

So when they "test" to find out the g-force rating, they simply drive on a flat surface in a circle (200' radius ??)? Then, of course, at the max point before it breaks this is the rating?

Thanks

 

You go as fast as you can without sliding outside of the line. Then they time one lap and can calculate mathematically what the g-force is.

 

cool thanks!

 

Here's an equation:

Lateral Frictional Force = Coefficient of Friction x Normal Force

The only thing keeping a car from sliding around is the friction that the tires provide. 1 G would require the coefficient to be 1. Coefficients can easily exceed 1.0, meaning that it is very possible for a car to exceed 1 G laterally without breaking traction.

 

If a car is driving around a 200' radius, it would need to travel at approximately 55mph to create a 1 G lateral acceleration.

 

No need to time a lap. Max velocity going in a circle of a known radius is easily converted to a lateral acceleration. Travel in a circle of any radius at any velocity and you will want to continue in a straight line. As the car turns and you want to go straight you push agains the edge of the seat. Your feeling the acceleration they are measuring if that's an easy way to picture it. Just as when you brake you feel as if your heavy against the seat belt or when you accelerate you feel your back in the seat. When you turn you can easily measure the force pushing you into the sides of the seat with the radius of the turn and the speed you make the turn at. At any instant you just need those two to calculate instantaneous lateral G.

The skid pad just is a ideal test so we all do it the same way just like a 1/4 mile run. We're setting up with the same conditions so we can compare and recreate.

Nice things about the skidpad:
it's flat. It's got rings of known radius.

With those two things you can slowly approach the limits of traction and gracefully exceed them to know exactly where smooth driving will exceed your grip limits in an ideal situation. Knowing that you can kind of compensate for lots of other conditions knowing something about the way the car will lose grip. On the skidpad you can also see just how the car will lose grip with certain tires. Will it lose it smoothly or will it just all of a sudden GO.

The originator made the point that it just measures tires. That's only half true. It measures the tire car combination.
Unless otherwise noted for a test they are using the stock tires sold on the car so if you use those tires too you will have the same performance at the same tempeture in general.

Our corvettes, I'm picking mine up at the museum May3!, come with tires very tuned to them. Changing tires will drastically change behaviour thoughout the cars envelope.

-mikey

 

The skidpad is measured as an average of clockwise and counterclockwise circles. The driver makes sure to align the driver's side tire with the dotted line of the path going both directions. Since clockwise represents a slightly smaller circumference and counter-clockwise represents a slightly larger circumference, the combination is an accurate average. The laps are timed because it is impossible to get an accurate average speed around the entire circle without timing it. If the distance is known and the time is known, the speed is known. The calculation is v=2?r/t or speed=circumference/time the acceleration is a=vˆ2/r. If you combine all this with 32ft./second/second for 1 g and a 100ft. radius circle, you get a formula of (628/t)ˆ2/3200. Therefore, if you do the circle in 12 seconds average, you are pulling .8558 g's. If you do it in 11 seconds, you are pulling 1.01855 g's.

I really didn't mean to confuse everybody but here it is.

that question mark in the first formula is supposed to be pi but I guess the html doesn't translate that from my computer.

 

ahhh, so they measure the entire lap and are therefore getting the average G for the entire circle not the exact traction loss point? So basically it's as fast as the driver can do the circle without drifting the wheels off the line?


-mikey