Bravo Ilmor and thanks for sharing this short video, shows what a great engineering house they are. Maybe we don’t have to trash the gas engines and go to ??? solar powered bikes and lithium powered environmental disasters…‘Breaking News’ Co2 is being absorbed by the high nitrogen levels in the earth’s atmosphere…slowing down/reducing global warming, gosh darn it all.
Best, JW
So Bernie, as you opened this topic and others are contributing…Is it Torque (specific term) or Power?(seems like an overbroad term) that was causing Mike Roddy’s TWR Spa winning XJ-S to lose traction at 190mph+ on a straight away…He had calm air conditions, no crosswinds, was not drafting anyone else, nor close to any group that had just gone by, when this thing occurred? Gummy Race tires and he had been running in a straight line accelerating. Puzzled, best, JW
Kirbert, Totally agree. Maximising useful area under the power curve (between shift points) is the target.
Equivalent to your wife talking non-stop and getting more words in, while you shout loudly in frustration, but can’t string a coherent sentence together …
Jon, don’t know the circumstances, but it’s torque that breaks the traction, just as it’s torque that loosens the nut. The amount of torque you can withstand is dependent on the stickiness of the tyres and the weight on them. At 190mph, I would guess there was a lot of lift (!) which would reduce the available traction.
It was interesting to read Nissan’s engineers defend their decision to make their car heavier (to improve acceleration) because they wanted to maximise the mechanical grip on the rear tyres …
Um…no.
No, it isnt.
MARK that makes excellent sense, well said!
Well, they can defend it, but, then…there’s physics!
“Friction is the resistance to motion of objects in contact with each other. The standard friction equation detemines the resistive force of sliding friction for hard surfaces, when you know the normal force pushing the two surfaces together and the coefficient of friction for the two surfaces.
When applied to sliding friction of hard surfaces, the equation implies that friction is independent of the area of the surfaces in contact.
This equation can also apply to soft surfaces, rolling friction and fluid friction, but the coefficient of friction may depend on area, shape and viscosity factors.
Questions you may have include:
When a force is applied to an object, the resistive force of friction acts in the opposite direction, parallel to the surfaces.
The standard equation for determining the resistive force of friction when trying to slide two solid objects together states that the force of friction equals the coefficient friction times the normal force pushing the two objects together. This equation is written as
Fr = μN
where:
Fr and N are measured in units of force, which are pounds or newtons. μ is a number between 0 (zero) and ∞ (infinity).
Applies to static and kinetic
This equation applies to both static and kinetic sliding friction. Static friction is the friction before an object starts to slide. Kinetic friction is the friction when the object is actually moving or sliding.
Static friction and kinetic friction have different coefficient of friction values.
Independent of area for sliding hard surfaces
An interesting result of this equation is that in the case of sliding friction of hard surfaces, the friction is independent of the area of the surfaces. In other words, it is just as difficult to move a 1 square-cm object as a 1 square-meter object, if they both are pressed to the surface with the same amount of force.
This is not intuitive. You would think that there is more friction when the surfaces are larger, but the friction equation states otherwise. You can verify this fact with experiments.
Soft, adhesion, rolling and fluid
In situations where the surfaces deform or there is molecular adhesion, the friction is not independent of the areas in contact. In these cases surface area usually comes into play. This is also true for rolling and fluid friction.
When solid surfaces are soft and deform or when one material is a fluid, the shape of the solid object may be a factor.
Although the standard friction equation still holds, the coefficient of friction may have area, shape and other factors included in it.
OK
So, what this says is that the greater the force pushing down, which is mass times G, gravity the greater the force needed to move the surfaces past each other.
So adding weight does increase friction and therby traction. But as you notice once the system becomes kinetic (and in the case of cars, rolling) the equations change.
Keep in mind also that this is for determining friction between two surfaces. It has nothing to do with the inertia of the objects. Inertia comes into play also in this complex situation. You need force to overcome inertia, but the application of that force depends on the success of the friction between the tires and the road. If you make it to heavy you will not be able to move it because there will not be enough friction to overcome the inertia. This is what happens when your tires just spin on pavement, providing the engine has sufficient power to overcome enough of the friction to spin the tires.
So, there is a balance between weight, friction, inertia, and power. It will vary for every vehicle.
While adding some weight does work, a better approach would be to improve the coefficient of friction for the tires. This way you get better traction without the extra weight. Bottom line, invest in the best snow tires you can get and lighten the load. Provided you can afford it. Otherwise throw a bag of salt in the trunk and when you get stuck you can throw it under the wheels to try and increase the friction.”
Yes, Kazutoshi Mizuno was talking about a traction limited situation.
“What is a supercar’s ideal weight?” he asks. Someone attempts to answer him along the lines of the lighter, the better. “You are wrong. Ideal weight is 1,740kg. Why?” At this point he grabs hold of a chair and moves it along the carpet with ease. “When I put weight on, it sticks to the surface. Same with cars - you must have weight to gain traction, which is why the GT-R is the world’s fastest accelerating four-seat production car.”
“So, how do you make a car? I can show you a very easy example. Imagine a high-speed corner in an F1 car, and it is using the best tyres in the world. An F1 car weighs 560kg, more than 600kg with the driver. How much downforce does an F1 car generate? Currently, it is around 1,300kg. So what is the total weight? 1,860kg [about the same as a GT-R with the driver on board, coincidentally]. A GT1 racing car weighs between 1,200 and 1,300kg. Plus downforce of 600kg, the actual weight on the car is 1,800kg… you see, very easy!”
Sorry Bernard, I didn’t mean to butt in.
Worth a read.
https://www.topgear.com/car-news/top-gear-magazine/top-gear-chats-mizuno-san-mr-gt-r
"Tyre grip is better than maximum engine torque. All customers can enjoy this supercar drive! All customers can easily enjoy 500bhp and accelerate to 62mph in three seconds! Anyone, anywhere, anytime can enjoy this. Other cars, only a special development engineer can enjoy them…”
Very impressive, Paul. I am surprised–not with your knowledge, but with your willingness to enter the realm of long nerdy discourse.
I only add that almost all basic physics actually fails when it comes to dry driving conditions encountered by cars (particularly sports cars) and tyres, particularly modern ones.
As you state, the common idea of friction is that it depends on the material and the downward force. That BTW is not physics (derivable from physical laws) but rather the definition of a “constant,” called coefficient of friction: mu = frictional force f / downforce or weight mg. Empirically, in simple lab experiments, mu can’t be more than 1 ( f < F) and as you mention, mu is independent of the area of contact. But with tyres that intercalate into the pavement, f can be greater than corner weight mg, and area of contact matters greatly. And mu is not a constant, but a function of corner weight (weight on that tyre). That mu depends on weight isn’t surprising, because the simple definition of mu as a constant is approximately true and convenient to use, but not absolutely true. Electrical resistance, R, is a similar empirical “constant.” Some things have fairly constant R (like resistors) but most do not (like light bulbs, toasters, etc).
A significant point that AFAIK is always true is that mu depends systematically on corner weight (the weigh applied to the tyre). Specifically, it increases with decreasing weight–being maximum with the tyre just sitting on the pavement, with no car attached. As weight increases, f increases but in less than direct proportion because mu is decreasing. The importance of f (or grip as it’s called) is that it determines maximum acceleration, braking (deceleration), and lateral acceleration (g’s on skidpad) before traction is lost. Real physics now comes into play–the relationship between acceleration and f is f = m x a where a is the acceleration. As Paul mentioned, the weight is m x g. So m contributes identically to corner weight and to lateral inertial forces and doesn’t need to be considered at all to a first approximation! BUT, grip itself decreases with corner weight.
As a result, (1) A lighter car (on tyres and pavement) will accelerate, brake, and corner better than a heavier car. If via weight transfer you increase one corner weight 100 lbs and decrease another 100 lbs, the total grip decreases–the corner with more weight loses more grip than the corner with less weight gains it. If you increase downforce without adding weight (fan, wing), then you get somewhere because, even though friction doesn’t increase in proportion to the added downforce, you’ve not added any mass, so acceleration and cornering can increase. IMHO.
Your elucidation reminds me—wonderfully so!!— of when my Dad first explained this to me: unfortunately, being 10 limited my ability to grok it!
I called a physics guru friend of mine, just after posting the quoted article, and he told me exactly what you posited, above.
Ill end with a real world example of how faulty the assumption can be, that adding weight increases tractive effort: semi-trucks.
In the mountains, these 80,000 pound monsters are helpless, without chaining up.
More than likely it was aerodynamics, specifically lack of downforce
Ah… so the Mark X would be the fastest Jaguar off the line?
Snicker giggle snort…
There’s some figurin’ missing from the pitcha’…
Interesting fact that the coefficient of friction reduces with increased contact pressure…but it is another interesting fact that the effective coefficient of friction INCREASES with tyre slip. I had to research this because my son was charged with “loss of traction” by the police when they saw him spin one of his back wheels. The coefficient of friction increases to a maximum at about 15-20% slip and returns to about the same level as the static coefficient at about 100% slip. This is why drag racers dial in the right amount of wheel spin to maximise acceleration.
So…highly likely the Nissan Engineer is aware of this fact and actually knows how the load and slip coefficients vary when his magnificent GTR attacks the tarmac!!!
…and…happy ending…the police dropped the charges !!!
OK. I’ll bite. I think we are in danger of practicing pseudo science.
I grant that the Nissan Engineer raised howls of laughter and derisive grins. However, he clearly achieved the desired effect. So a rational observer might think the we are missing a critical detail and not be so quick to laugh.
We know you can increase traction by adding weight (or longitudinal load transfer). (Practical observation)
We know you can increase braking by forcing load transfer prior to maximum braking effort.
We can explain how you can increase acceleration by adding weight, but it appears that you can.
So a closer look to the fine print.
From Wikipedia …
"Traction
Load transfer causes the available traction at all four wheels to vary as the car brakes, accelerates, or turns. This bias to one pair of tires doing more “work” than the other pair results in a net loss of total available traction. The net loss can be attributed to the phenomenon known as tire load sensitivity.
An exception is during positive acceleration when the engine power is driving two or fewer wheels. In this situation where all the tires are not being utilized load transfer can be advantageous."
And from Wikipedia “Load Sensitivity”
"The load sensitivity of most real tires in their typical operating range is such that the coefficient of friction decreases as the vertical load, Fz, increases. The maximum lateral force that can be developed does increase as the vertical load increases, but at a diminishing rate.
Or as Robert pointed out
“As weight increases, f increases but in less than direct proportion because mu is decreasing. The importance of f (or grip as it’s called) is that it determines maximum acceleration, braking (deceleration), and lateral acceleration (g’s on skidpad) before traction is lost.”
However it appears that his conclusion must be incorrect. ie That a lighter car will accelerate better than a heavier car.
We can easily disprove this statement by referencing the acceleration rate of an Austin Maxi and a Mk X Jag.
Mk X Jag : 1778 kg, 0-100 9.6 s
Austin Maxi : 979 kg, 0-100 13.2 s
We can conclude that this statement would only be true given all other aspects of the vehicle were identical.
I would suggest that, if there were sufficient torque available (including gearing and final drive ratio), a heavier vehicle could indeed out accelerate a lighter vehicle because it could potentially achieve more traction (mechanical grip) …
We could also conclude that if we added too much weight (or too much load transfer), that the gains from increased traction would be offset by the added inertia of the vehicle.
From the equations everyone has quoted of Fr=uN, so long as doubling N doesn’t reduce u by more than 1/2, your net Fr goes up. The smart money would presumably be in encouraging longitudinal weight transfer (adding effective weight on rear wheels without adding to vehicle mass). In a 4WD it is probably more complex and you’d want to keep weight on your front tyres.
I have no idea if that number comes out to 1800kg on a 4WD vehicle with the dimensions of the GTR, as is suggest by Mr Mizuno, but it does appear that he knew what he was talking about. It would be an interesting exercise to crunch the numbers.
An interesting post, Mark. I agree with most of it but not all of it.
One good point you make is that when only one or two wheels are being “driven” it is indeed desirable to transfer weight to those wheels. Placing a sandbag in the trunk is sometimes helpful, on ice or similar, because the weight (and concomitant frictional force increase) is over only those wheels. But even then, you’re trying to prevent wheel spin, not maximize acceleration. Keeping the weight of the total car constant but shifting it to the driving wheels is good…you’re decreasing total grip of all four wheels, but grip at the front doesn’t matter if they’re not driven. The torque reaction when accelerating does this for RWD–ideally you lift the front off the ground. For FWD it’s the opposite; you’re transferring weight to the undriven rear wheels. IMO, this is one reason why RWD on dry ground beats FWD, but not AWD.
OTOH, I would argue that best braking is when all wheels support the same weight. Not sure what you mean by “forcing weight transfer prior to braking.” Typically, braking torque transfers weight to the front, opposite of acceleration. Assuming the brakes are adequate on all four wheels, that weight transfer slows braking.
Yes, I certainly agree, and didn’t mean to imply that light cars with small engines perform better than heavy ones with big engines.
Here’s where I think I see an error in your logic, Mark. I’ll repeat from my earlier post: If you assume that u is constant and you double N, then you double Fr as you say. But what does that get you? Depends on how you double N. If you do it with a ground effects fan, or with aerodynamic ground force, you get more grip. But if you do it by increasing the weight, then you’re increasing the mass exactly in proportion. That mass causes inertia that resists acceleration, braking, and cornering. So you have no loss (because Fr has doubled as well) but no gain either–doesn’t matter how massive the car is, because change in mass causes change in downforce, plus change in inertia by same amount. But, now consider the “second order” effect of u decreasing with increased weight–that upsets the balance and the lighter car wins. Doesn’t have to be 50% decrease for X2 increase in weight…only has to be 1% because weight (mg) and inertia (ma) are precisely balanced–one of the most important and consequential physical laws.
YOOUWZZA<< has this turned into mess of numbers , and NO winners!
back many years ago the FAMOUS booze runners , mountains of Tennesse ,Caroliners, Kentucky ETC.
they built some 1939/1940 V8 Fords and found using REALITY that by adding a big heavy tank of HOOCH, White lightning, the cars handle better and road better with heavy duty built up springs.
cars where heavy ,but with engine tricks , left many a Revenurs in the dust on rough bumpy back woods horse paths.
there is a great song about them guys ,that later in life Made the NASCAR races famous also!
and think that most couldnt read or write, much less do any usable math, EXCEPT COUNT THE MONEY!!
and you wouldnt want to race them for money etheir , with anything built in GB at that time!.. LOL, gotta love it?
RON
may as well add this ;;.
1967 Mustang electric car , looks nice , goes fast , 0-60 in 1.94 seconds, measured at TEXAS mile at 174 MPH. torque estimated at 1800 lb.FT.
his old records , using two double overdrive gears , 2.20 rear gears!
is building a new electric drive train and updated motors, plus new tech batteries, and lots of reinforcements!
he lives about one mile from me in AUSTIN Texas, home of COTA F1 track.
Well, shitfar, I cooda come and seen you, too when I was down!
I passed the exit—on some highway— for COTA.