Question about vehicle dynamics Pop quiz

"DC3 LS, will be perpetually replacing cars until the end of time" (dc3ls-)
10/30/2017 at 10:11 • Filed to: None

Update: So I made this post last night and scheduled it. Later I looked it up anyway and I kind of get. At least as far as I think someone without an engineering degree can.

So now it’s a pop quiz to see who can get it right!

Pic (I think) unrelated

So if you have two cars that can pull the same amount of maximum lateral g’s, but weight differently. The lighter one will be able to corner faster, because the heavier one will reach the same amount of lateral g’s sooner because it has more weight.

Edit: This is wrong. Both cars will be going the same speed.

But if for example you’re trail braking. You’re transferring weight from the rear to the front. This causes the back end to have less grip than if there was no weight transfer.

But from the first example. Shouldn’t the back end have more grip since there’s less weight on it? And shouldn’t the extra weight on the front cause understeer? Obviously understeer will happen if you do it too hard, because you’ll overload the front, either with too much weight or too much braking and cornering.

DISCUSSION (24)

PartyPooper2012 > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:16

So car is going around a curve. Lighter one allegedly will corner at higher speed.

The braking is a whole other beast. What are we trying to solve for here?

Are you trying to drift in the corner? Are you trying to stop while in curve?

If you pull handbrake in the curve that locks up rear wheels, the wheels will kick out towards the outside of the turn and you may spin out depending if you are on the throttle or not.

beardsbynelly - Rikerbeard > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:20

That extra weight on the front is like downforce. It can improve grip up until it lets go and understeers.

You’re fighting having enough grip on the rear to stop oversteer and having more weight on the front to inprove turn in

Drive a mid or rear engine car and you’ll notice it more.

nermal > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:21

Transferring weight to the front increases friction on the front tires, providing more grip for braking and steering. Up until a point when they get overloaded, and you understeer.

Transferring weight off of the rear end helps it to rotate. Up until a point where you spin.

Basically, trail braking makes you faster, but there’s a fine line between as fast as possible and crashing. Proper brake and steering modulation from turn-in to apex is key.

Party-vi > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:27

Remembering nothing of my certainly average performance in college physics, lateral g forces measure acceleration, so two cars with the same lateral g’s should accelerate from a corner at the same rate of speed.

A Lamborghini Murcielago LP640 weighs 3,726lb, and can complete Motortrend’s 200ft figure eight track in the same time as a Viper SRT10 with a weight of 3,450. In this instance both cars pull 1.05 lateral g, but one weighs 10% more, and they both have equal performance around the corners.

Can someone correct my C- physics comment?

DC3 LS, will be perpetually replacing cars until the end of time > PartyPooper2012
10/30/2017 at 10:40

Just weight transfer. Doesn’t have to be braking. It could just be transfer from the inside to the outside wheels.

interstate366, now In The Industry > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:40

Doesn’t matter until we know the driven wheels of each car. Don’t you know that drive wheels are the only thing that matters around here? /s

DC3 LS, will be perpetually replacing cars until the end of time > beardsbynelly - Rikerbeard
10/30/2017 at 10:42

Pretty much spot on, but you didn’t use the fancy words.

You can tell a Finn but you can't tell him much > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:47

This part of your statement is incorrect. Lateral acceleration is dependent on the speed of the vehicle squared and the radius of the curve (Ac = S^2/R). If both vehicles in question can generate the same lateral acceleration then they can both corner at the same speed, their weights do not matter.

DC3 LS, will be perpetually replacing cars until the end of time > Party-vi
10/30/2017 at 10:48

I’ll have to double check the book. But I think your wrong.

DC3 LS, will be perpetually replacing cars until the end of time > Party-vi
10/30/2017 at 10:58

Nvm. You’re right. See “You can tell a Finn...” for details.

BREADwagon > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:59

You’re almost there, but you have a couple of things wrong. The same lateral G for a car, regardless of weight, equals the same acceleration....so if they are both going around the same radius turn at the same G, each car would have the same cornering speed. The major difference would be the force required from the tires to attain that acceleration.

Acceleration is Force divided by Mass. If you want the same acceleration, but you increase the mass (weight) of the car, you need to increase the force directly proportional to the increase in mass. So, if the car weights more, the tires need to apply more force to achieve the same cornering speed around the same radius (to achieve the same acceleration).

You’re getting force and acceleration mixed up.

Now, the maximum force a tire can ‘give’ is a function of the tire/surface friction force and the Normal force on it (which, when overly simplified, we can call weight). Assuming the tire/surface friction coefficient remains equal all around, we can say that the maximum force a tire can apply is directly proportional to the weight applied to it...ie: more weight on tire equals more cornering force the tire can give. However, that cornering force is negated by the added mass of the car...unless you keep the weight of the car the same and increase the normal force on the tire via Aerodynamics downforce...

In your example of trail braking, the weight transfers forward, increasing the maximum force the front tires can apply. However, you are using all of that force to accelerate backwards (decelerate), and can’t use that force to turn. If you try to turn while using all of the available traction (friction force) on that tire, the tire will slide and you would understeer.

Look up a ‘Traction Circle’ for more details.

Party-vi > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 10:59

My professor would be so proud

random001 > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 11:02

Nope, Party-man is damned near spot on. You can usually figure these problems out by looking at the units. Small trick from a Physics major. Lateral-G’s, being G’s, being acceleration, have units of m/s^2. Nowhere in there is mass a factor.

Now, that’s the simple answer. You need to trace that way the hell back, though, for the most complete answer, to include weight, static and kinetic friction, polar moment, etc. etc. to get the really complicated answer, which is that lateral-g stats can not tell you the answer to your question. If the heavier car is more advanced, and pulls in some traction control wizardry, it might be able to pull 1.05 LG’s at higher speeds than a much lighter, simpler car like a lotus 7. It’s never that simple.

MasterMario - Keeper of the V8s > Party-vi
10/30/2017 at 11:05

Your C- physics did not fail you.

notsomethingstructural > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 11:22

lateral G’s are a ratio of the inward centripetal acceleration compared to gravity. so the centripetal acceleration equals the square of the tangential velocity divided by the radius of the curve (v^2/r), then divide that by the constant for g and presto. So in metric, if your radial acceleration is 10.29 m/s^2 that’s 1.05g (since g = 9.8 m/s^2). You’ll notice that mass (or weight) does not come up in those formulas, since for racecar physics, the centripetal acceleration is a measure of how the car is actually moving, not the mechanics of how the car is doing so.

the mechanics of “how does the car turn” are a function of grip. so step 1: force / mass = acceleration. we have the acceleration from above. so now - yes, the mass of the car comes into play, and lower mass means that lower force from the tires is required to make the turn. now you have a lateral force.

step 2: the lateral force the tires can resist is a product of the total downward force at each tire and the friction coefficient between the rubber and the asphalt (there’s more to this, as you could guess, but the analogy serves the purpose.) when you trail brake, the center of gravity of the car - which is elevated above the axles - produces what amounts to a torque that pitches the car forward (it pitches to the outside of the turn, too, but let’s ignore that for now.) it’s resisted by the front tires, which is why your front springs compress under heavy braking. since the weight of the car is the weight of the car - it doesn’t change - the increase in force on the front tires is similarly reduced on the rear tires.

(as an aside, imagine you took a 2' long 2x4, stood it on end and put lego wheels on the corners. if you try to brake, the 2x4 is going flying right over the front axle. the fact that the wheelbase on cars is way longer than the distance from the center of gravity to the axles is what stops them from flipping over the front. this torque is also why you don’t ever lock the front wheel up on a motorcycle).

so since the front tires are the ones doing the steering, this is good - because as explained, the force the tires can resist is a product of the TOTAL downward force, which includes things like the weight of the car, the pitching due to braking, downforce, all sorts of crap. the more you load up the front tires, the better you grip.

so trail braking allows you to turn better because it concentrates more force on the front tires, allowing them to produce more grip. it comes at the expense of the grip on the rear tires (since there is now less downward force on those tires).

hope this helps.

WRXforScience > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 12:08

Friction is determined by only two factors: coefficient of friction (how rough/smooth the surface boundaries are), and the Normal Force (how hard the surfaces are pressed together). The coefficient of friction is determined by the tires and the surface, while the Normal Force is determined by the weight of the object and any other vertical forces (like down-force or lift) acting on the object.

The maximum tangential velocity an object moving in a circular path can attain is equal to the square root of the product of its coefficient of friction, the radius of the curve, and the acceleration due to gravity. (Or, the smallest radius curve is equal to the speed squared divided by both the coefficient of friction and the acceleration due to gravity).

The mass of the objects will factor into the force required to achieve the same acceleration. Dynamic load (weight transfer) can increase the Normal Force acting on individual tires, which increases the overall friction of that particular tire at the expense of grip at another tire. Down-force will increase the overall friction by increasing the Normal Force without increasing the mass of the object (thereby granting a larger centripetal acceleration).

BiTurbo228 - Dr Frankenstein of Spitfires > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 12:36

Another interesting quirk of physics is that more weight on a wheel provides it with more grip, not less. So, in your example where both of them dive forwards, the front end will get more grip from the weight transfer and the rear end will get less (the sensation of ‘going light’ when you’re driving).

So, heavier cars actually produce more grip, all else being equal.

The reason that lighter cars tend to be able to corner quicker is more due to lower inertia and lower weight transfer (so although the heavier car has overall higher peak grip it also throws its weight around more so there’s a greater disparity in grip between tyres causing one end or other to lose grip).

There’s a whole ton of other confounding variables in that equation though, but if all else was equal then that’s pretty much how it goes. Heavier car creates more peak grip, but through a host of side-effects from its increased weight winds up being slower through the corners.

Spoon II > DC3 LS, will be perpetually replacing cars until the end of time
10/30/2017 at 12:43

To answer your original question, two cars pulling the same lateral g’s are simply going around a corner of the same radius at the same speed. They rely on the equation A = V^2/R (where A is acceleration, V is velocity, and R is radius). But, we can delve a little deeper into your mass question with another formula (F=MA where F is force, M is mass, and A is acceleration). We rearrange that to A=F/M and substitute it into A=V^2/R, resulting in F/M=V^2/R, and we solve for F. Thus, F = (M*V^2)/R. From this, we can see that as we increase mass, we increase the amount of force required. That’s why a heavier car will require bigger tires, more downforce, or some combination of the two to round corners the same as a lighter car. More massive objects have more momentum, and resist change to a greater extent. Essentially, the problem of greater mass is solved by Clarkson’s favourite saying “More Power!!!”.

uofime-2 > BREADwagon
10/30/2017 at 12:50

Just to pile on to what is really well stated here. it is worth noting that mu of a tire is not constant, it is a function of loading (as well as a bunch of other things). This means that grip is not 100% proportional to normal force. This fact is why tire sizing and weight transfer are important.

Spanfeller is a twat > DC3 LS, will be perpetually replacing cars until the end of time
11/01/2017 at 02:22

Is the COG (Well, centre of mass) really also the rotational axis of the car as the wheels turn?

Shouldn’t the axis be closer to the back? as only the front wheels provide rotation under normal circumstances

And about grip, grip in its basic form is friction, and for tires not to loose it they must remain within the bounds of the friction. tires rotate, so no slip means the tire remains within the bounds of static friction.

Meaning that all the forces applied to the tire must be below the static friction which is the coeficient times forces applied.

If a car is nose heavy, its easier for the forces to be larger because more mass is outside the place where friction is managed and torque increases, but its only handled by two wheels. this doesn’t apply to mid engine cars because while the rotational torque increases as the large masses leave the place where friction is applied, this is handled by four tires.

If the back of a vehicle is lighter, it means that the tires cannot provoke as much friction as one would like because of weight. friction is weight times the coefficient.

In the end, grip is a function in the z axis and displacement is a function in the X and Y axis, but the efficiency of displacement is linked to the grip . so really sometimes the forces are similar but end up being applied very differently.

I’m in first year physics for Mechatronics, so take what I say with a large grain of salt.

handyjoe > Party-vi
11/01/2017 at 10:05

The problem is that the acceleration you’re measuring in lateral g is towards the middle of the corner. It’s a function of grip (much like any measure of acceleration). Once they get out of the corner we’re back to measuring horsepower.

Party-vi > handyjoe
11/01/2017 at 10:09

But we don’t care about “out of the corner”, just their speed in cornering :P

handyjoe > DC3 LS, will be perpetually replacing cars until the end of time
11/01/2017 at 10:10

The back end will have less grip because you’ve reduced the normal force. If you reduce the amount you’re pressing down, it won’t grip as hard. Try lightly dragging your hand across the table vs. doing it while pressing down.

The extra weight on the front can cause understeer if you overload the tires, but if you don’t overload them you’ve applied more normal force, providing more grip on the wheels doing the steering. So it will have more grip, until it doesn’t.

handyjoe > Party-vi
11/01/2017 at 12:16

Well...yes, you’re right. It just takes more magic with more mass.