Tires & Acceleration: Newsletter Challenge (04/11/06)

I think that the question here revolves, if I may be so crude, about the equations of centripetal force:

ar = v2 / r

And from F=ma we can derive…

F = ma = mv2/r

Where v2 is really V squared or V*V or angular velocity squared.

Since both cars start and accelerate at an equal rate, then the smaller diameter wheel must rotate faster. All things being equal, which they are not, the V squared component may rule the day for the smaller wheel/tire combination.

The point is that there is more angular inertia required for the smaller wheel, which amounts to torque output, which, as Newton noted, creates an opposite and equal reaction in the system, so the nose of the car raises more with a smaller wheel.

Which is why, in part, Drag Race vehicles have tall rear tires as well as wide.

Yes. Another is heat. More rubber takes longer to heat up and more chance to cool. The other is obviously contact patch size to improve traction.

The key element is that the mass of the car "system" is the same. The car with the larger, more massive tires must have a lower mass than the other car (excluding tires). The force causing the nose to rise is the torque, resisted by the gravitational pull on the centre of gravity. As the mass of the larger wheel car is lower, there is less gravitational resitance for the same torque.

An actual drag racing question! The two cars have identical mass, but the one with big tires is stated to have more mass in the tires. More rotating mass means slower acceleration, but both reach 60 mph at the same time. So the small tire car would start fast and run out of power, while the big tire car would start slow and build speed. Additionally, they leave you in the cold as to what "big" verses "small" wheels means. Are they the same height and different widths? Or different heights and the same widths? Or something in between. If the big tire is a taller tire, then that means the identical center of gravity is relatively lower on the big tire car. That means that the big tire would be less likely to lift the front end, which would mean less weight transfer onto the back tires at the hit of the throttle. Door cars that drag race try to transfer all of the weight to the back tires on initial acceleration to keep the tires from spinning. That's why door cars are often lifted in the front and have the rear wheels moved forward (an "Altered"). In this case the small (short) tire car would have better geometry for doing this. So scenario 1 says the small tire car leaves harder, thus using acceleration to shift more weight back, and assuming small means short, the second scenario means the small tire car has better geometry for shifting weight rearward. Sister is correct.

I'm confused, what idiot is scoring this thing. Eric answers one way, Jeff answers the opposite, and both are scored 1. I like Randall's answer best, indeterminate, too many unknowns.

The Score of 1 is automatic for anyone logged in; as is the 0 for ACs. A Score of 1 doesn't actually count for anything...unless you lose it.

Goto the FAQ section and look up scoring for more info.

And get registered and log in; it's nice to know who you're talking to! :o)

Since both cars are accelerating at the same rate and have the same mass and height of c.o.g, they will both nose up by the same angle. Samsook

i don't think the answers so far get to the point (s). whether either car "noses up" or not cannot be determined unless we know that there is some degree of elasticity in the tires, and how much. neither one necessarily would, if neither yielded to a downward force. once that is resolved, it would seem that, if the center of mass is at equal height, a force applied below that would create a tendency for the entire mass to spin around the center of mass, which is rearwards at the top and downwards at the rear of the car. The further the force is below the center of mass, the greater is that tendency and hence the downward force on the rear, which of course translates into an upward force on the front. So if the tires of both vehicles have an equal degree of resistance to the downward force, the smaller tires would produce a greater force and more deflection of the nose upwards. I'm not an advanced physicist, so i won't give equations, but i think the analysis is sound.

Rotational inertia? Heating the rubber? This is a much simpler engineering analysis. It's not the tires, per se, it's the location of the where the force is applied to the vehicle: the axle. Since the two cars accelerate at the same rate, the forward force applied to them is obviously the same. The smaller (I assume diameter) tire sets the axle closer to the ground. Since the center of mass of the cars are at the same height, the forward force is applied to this car at a point that is a greater distance below the center of mass as compared to the tall tired car. The longer radius from the center of mass to the line of force provides a greater torque around the center of mass. more torque = more nose lift.

Indeterminate. The question places a lot of emphasis on the tires being more massive: not the axle being further from the ground. Most posts (except the first perhaps) so far have missed the equal and opposite torque on the car from the rear axle. All other things being equal the car with the more "massive" rear tires will nose up more. The higher the axle on the bigger wheeled car the more likely the sister is to be right. The centre of mass (of the car)'s position front to back is also significant.

The answer is simple. It has nothing to do with torque, gravity or the location of the centre of mass. It's about power and energy. The car with the big wheels must have more kinetic energy when it is going at 60 mph because the heavy wheels have greater rotational kinetic energy. As both reach 60 mph in the same time, the engine of the heavy-wheeled car must have produced this extra energy during this period, and therefore must be more powerful. As the cars go faster, eventually they reach their top speed when air and rolling resistance match the forward force generated by their engines. Assuming the rolling and air resistance is the same in each case - we have no reason to assume it is not despite the heavier wheels - the heavy-wheeled car with its more powerful engine will eventually noes ahead.

The one with the looser suspention.

Let's keep it simple... 1. The same force is required at the contact point of the Wheel Tire combination with the ground to accelerate both cars identically. 2. The required torque would be greater for the larger wheeled car because of both the larger tire mass and the larger assumed tire radius. More required torque = more apt to want to raise the front of the car. The reason for the bigger tires when dragging has to do with Raising the Speed at high RPM and controling the Force at the contact point for sticktion and lowing the center of gravity in relation to the centerline of the axel so accleration in the forward direction foces the nose down rather than up, but is relativly insignificant compared with the torque created at the contact point.... or maybe I'm just spinning my wheels.

When I consider the circumstances of this case, the answer becomes fairly clear, given a few assumptions.
When you consider that the point of force (at and parallel to the road behind the center of gravity) each car has the same moment arm at the same acceleration pushing it up, however if you incorporate the additional force required to rotate the wheels the answer becomes apparent.
If we assume that the 'big' wheels have more mass to them at a greater radius than the 'small' wheeled car, the added force required to accelerate the rotational speed of the wheels will create an additional tortional force that will cause the front end to pull up more.
However, if we also consider the center of gravity of the vehicles are both the same and don't assume that the two sizes of wheels are the same weight, then the heavier wheeld car will have to have more of the body weight distributed further forward, which will change the dynamic of the equation.
If we assume the weight of both sizes of wheels to be equal, then the answer is simply that the added torque required to rotate the wheels at the same rate as the body of the car, then the larger wheeled vehicle will rise more.

Not replying to any particular posting, but I think most are off track (pun intended!) and get involved in unnecessary detail - road friction, heat etc. Noseup is greater with the bigger tyres. For same total mass and same acceleration, car with bigger/heavier tyres needs greater force, as all the mass is accelerated linearly, but additional rotational energy is required for the bigger wheels. This needs higher torque to the wheels,and the reaction to this is increased noseup. Force between tyres and road is unchanged. In addition, the net mass of the car (total mass minus tyre mass) is what keeps the nose down. (assuming the C of G is same distance along car, as well as same height, question doesn't say) Both these effects contribute to higher noseup with bigger tyres. So the son is right. The question doesn't say tyres are same OD, but above holds if "bigger" tyres are same OD as smaller, or more (reasonable assumption). Only case noseup is less for bigger tyres is if "bigger" tyres have smaller OD, OD ratio at crossover depending on mass ratio of tyres (and on radius of gyration, if we want to be pedantic). Finally, a help request - how do you insert paragraph breaks? When I hit return a break appears, but it isn't there on Preview or transmitted response. It's clearly possible, as on response from Anonymous Coward on Tuesday April 11, @08:09AM (#4075), so what am I missing?

You are missing the HML coding for the paragraph breaks. What you need to do is place the characters >P> where you want the break, but you must reverse the left greater than character for it to work.

I had to do that so that they printed!!! Otherwise you will not see the character combination.

Make sure you have HTML Formatted selected on the pull down menu immediately to the right of the SUBMIT button.

Good Luck!

I can't give a fancy equasion to prove the fact that a smaller tire will lift the front end of a drag racer, but I've been around them for quite some time. I think the main reason that larger tires are used has to do with dispersing heat through i'm sure with large rear tires and smaller front tires, this increases the amount of travel the car must go before the wind catches the underside of the chasis and cockpit. I kindof think of it as arm wrestling. Less force is needed for the person who is winning because gravity is pulling down as well. The same case with the wind resistance. With larger tires, it positions the vehicle at a downward angle, so the force of the wind is supose to take care of some of the lift of the vehicle caused by the heavy acceleration. Morris Rules!

Just tried the para break and it worked - that's great, thanks.

Back to the noseup question, ignore the last bit of my previous one about noseup being less for bigger tyres under some conditions. Got the maths wrong, I used wheel radius at a place where it should have been height of C of G. Doing it right (assuming I have now) noseup less for "bigger" wheels only happens at impractical values of diameter and mass for the 2 types i.e. "bigger" wheels only slightly heavier and much smaller dia than smaller.

Cheers

After reflecting on what you stated, I think that the tire/wheel mass may not be significant. Here is my rational.

1) The ratio of mass between the rear wheels with their tires is less than the mass of the remainder of the vehicle by a very significant amount. NHRA dragsters with a 300" wheelbase are in the realm of 2200 lbs!

2) I believe that: The moment of inertia for the wheel is less than the force required to propel the vehicle forward by a significant amount. In other words, if you measure the force required at the axel to accelerate the wheels in free space to achieve, say a rate of 60 mph, and then measure the force required at the same axel to propel the car and wheels to the same speed, most of the force required to achieve the target speed is used to accelerate the total vehicle mass down the drag strip.

3) Based on points 1 and 2, I propose that the differences in mass and size between the large and small wheels play a very small part in the equation and can be essentially ignored. Some or most of the wheels' impact will be neutralized anyway since the angular force is a product of the mass times the square of the angular velocity of the wheel divided by the radius of the wheel ( F = [m * v * v] / r ). The larger wheels' angular velocity is slower than a smaller wheel and the delta is further exasperated when you square those speeds and divide them by their radius.

4) What I believe is a significant factor is the distance between the drive axle and the vehicle's center of gravity. If the center of gravity is extremely close to the road surface and far from the wheel's axel, more torque is required to rotate the vehicle about that axel. If the center of gravity is at or god forbid, above the axel (such as it would be with a small wheel), the force required to rotate the vehicle body would be less.

5) If you ever watch a dragster launch, the greatest body lift occurs at launch, when the angular speed of the rear wheels is much slower than at mid point of the run. Most of the power applied to the wheels and the contact patch where tire meets road surface is consumed in accelerating the vehicle forward.

Since the center of gravity is equal for both cars relative to the road surface, the torque advantage goes to the vehicle with the smaller wheels as far as vehicle lift goes. If both cars require the same amount of power to accelerate to 60 mph, the moment of inertia of the vehicle's body is greater for the larger wheels because the distance between the axel and the center of gravity is longer than that for the vehicle with the smaller tires. The smaller inertial moment for the small wheeled vehicle yields a greater potential for that vehicle's nose to lift.

OK after much restraint I will try to make this very complex situation as simple as possible. First and formost the whole situation revolves around getting the tires into their effective temperature range so as to allow the maximum adhesion. If one does the pure science it doesn't work unless many variables are taken into consideration. some of these are tire compounding, tire sidewall design, tire underlayment design, track surface, chassis rigidity, engine and suspension design,(squat or anti squat), engine torque charateristics, center of gravity height, driver location, front to rear pitch and dive centers, transmission type if any, flywheel mass, engine rotational mass, air density, clutch slippage rate and the control therof. It all is reduced to getting a balanced setup. When determining the amount of antidive induced into the vehicle all of the aforementioned variables are considered and the optimum is usually unobtainable but closely approximated using a large database that balances everything into a harmonic that will allow for the tires to maintain maximum adhesion as well as growth rate. maximum nose up can be achieved through the correct weight transfer and done so by matching the engines torque, the amount of clutch slippage, engine fueling, and intake air pressure utilizing blower characteristics in conjunction with all the other variables. in AAfuel or any other rear suspensionless car the tires are of a smalloer diameter and when leaving the starting line the weignt transfer causes the sidewalls to wrinkle therby further inducing a reduction in effective tire diamerer and lowering the effective gear ratio and inducing greater downforce. as the vehicle accelerates the tires adhesion is affected by track surface and aerodynamics as air can become trapped under the tires as well as the molten tire compound which induces additional slip but as the car ends the race the tires grow tremendously in diameter and narrow in section and as they do they lift the vehicle which induces furtner downforce If it all goes well this is balanced and the tires are kept at optimum and the nose stays at a fairly constant height (lower is best as the body styles are designed to act as airdams (Diplanes) so as to minimize lift and the destructive blowovers.(flying cars) Now if the cars have rear suspension the whole thing changes as the antisdquat setting can be adjusted so as to utilize the rear axles rotational torque to lift the vehicle in harmony with the mass and center of gravity height which is constantly changine with the amount of front and rear lift as well as the front to rear bias of mass. One thing that has been kept quiet is that the tires are really springs and their natual frequency is changed by rotational speed and due to this effect the vehicles springing must be matched to the natural frequency of the unit as a whole along with the tire's ever changing harmonic. Add the fact that the valving of the shock absorbers and elastomeric frequency of the bushings or metallic attachments for the suspension linkage must all be in resonance harmonically with everything else then the whole thing gets really involved. So in a brief answer either vehicle can be tuned to have the exact same amount of lift. After all this very brief synopsis of some of the factors one must consider that a larger diameter tire will have more contact area then a smaller tire of the same width but the smaller tire will change the amount of downforce due to the center of mass being located further rearward and when launching will end up with the same effective contact patch If the tire pressures are equal. Additional factors are the larger tire's sidewall's cooling effect of the tread due to more exposure to ambient air which would possibly reduce the tires temp below optimum and on the other hand the smaller tires cooling characteristics may push it into an overheat situation. Essentially this has been argued over the ages by many folks some even used to claim that the ultimate top speed was limited to 200 mph in the 1/4. If you think that it's as simple as tire size just look at the early AAfuelers with narrow tires that would wheelie back in the 50's and compare the state of the art today. get it wrong and you are either on your top or go up in smoke. It really is a good question and anyone who really thinks that there is an answer needs some application experience and would benefit by attending a few races. By the way there is a segment of the sport now that is dedicated just to lifting the front wheels and keeping them up. This may seem like it is really long but in fact it is only a micron of what we consider when in the planning stages of a new vehicle, and if it were I would be making a lot more racing these things. Good thing this wasn's about cornering force or I would ge kicked off of this site for using up too much band width Sorry about the length but this is really much more complex than most folks realize CG

Awesome post!

Yes, it is amazing the amount of details that go into racing. Drag racing seams kind of simple if you are like me and unfamiliar with the sport. However, it is obviously very complex and I would imagine the complexity is due to the highly competitive nature of the sport and the need to push technology to the limit.

Thanks for the lengthly description. Truly a fascinating insight!

The car with the larger tires will require more torque to wheels in order for it to start moving, The additional torque required will cause the nose of lift into the air. The car with the smaller tires will require less torque to the wheels in order for it to start moving..

The key is that the angle between a line connecting the center of mass of the vehicle and the center of mass of the tires is greater with respect to the angle of force/angle of motion for the car with the smaller tires. Therefore a greater percentage of the force vector will applie to the vertical direction or the "lift" of the front of the car. The daughter was correct and the son was wrong. The car with the smaller tires will have more lift.

Well I'm just a country EE, but I do know that dragsters have tall drive wheels for gearing purposes, avoiding the 25k+ RPM required to achieve 240MPH. The reason why they must do this with torque as opposed to HP is also linked to the top RPM, which goes down as stroke is increased. Further, the tires are designed to expand outwardly almost 2:1 their profile height at rest, greatly increasing their O.D. as the vehicle accelerates, which extends the operating range of the power band of the engine in terms of RPM by way of increased MPH (brilliant). The problem stipulates there is no CG delta while the tires are different. The gentleman who indicated the vector math model is partially correct IMHO, but misses the main point and, given a dragster's extended body length, the vector math falls off into the noise. So here it is: The torque applied through the axle to the road by the engine drive line is divided by the tire O.D., while the inertia of the vehicle is multiplied by the tire O.D.. If the body weight integrated over the body length exceeds the maximum available torque applied to the road, the nose will never lift off the ground (though some lift will be effected causing a loss of potential steering traction). Note that if stiction gives way to the translated power (where the tire "meats" the road), then a burn-out goes into effect, which serves as a waste gate for those who do have extraneous torque applied to the tire stiction as compared to the body's counter torsion, preventing a wheely from occurring, the lost energy going into the tire as heat, and the impact is a reduction in acceleration and therefore reduced terminal velocity. However, if the stiction can handle the torque applied, and that torque is greater than the counter-torsion of the body, the front wheels will pull off the road, since the body will begin to rotate opposite to the forward moving tire direction. Once the wheely bars hit the ground, again, a self-limiting burn-out occurs as a result of vehicle wheight being shared by the wheely bars, reducing stiction of the drive wheels. MOST SIMPLY: Assuming constant torque, CG and axle offset (wheelbase), and assuming enough tire stiction to maintain traction, as the drive tire diameter is increased, the ratio between the torque applicable to the road versus the counter-torque exerted to the body decreases, eventually causing the body to rotate counter to the tire direction lifting the nose off the ground (in the fwd direction only, assuming rear wheel drive). The dragster rail's extended body length proves this: It is solely to enable the body's integral weight across that length to overcome the potential counter torque applied by the driveline at the axle. Smaller wheels will also avoid nose lift, but they will burn up immediately, because the torque is too great with a short radius to the ground, compared to the max stiction maintaned by the contact patch of the tire -- even if the contact patch were the same in both cases, the litle wheels sit and spin with no appreciable nose lift. I'll be hear all week -- thank you! You are all much too kind... Thank you all... (I really hope I'm right about this -- I'm not a coward for nothing you know).

One assumption: The centers of mass must be the same distance from the Back Axle on each car. The distance measured at a line verticle from the center of the axle to the center of the mass. These conditions met: the smaller wheeled car would have the best leverage. The other condition that would make it a wash, is if each set of tires on each car had the same coefficient of traction, then it would be a wash. The Big tired car would have better grip on the roadway, and may equal the same lift because of the increased torque of the Big tire axel shaft. The small tired car would have more wheel spin, and easier lift conditions. The big tired car would have harder lift, but the axel shaft should have more torque from the greater grip of the wider tires. It should be close to a wash. Depending on the moisture in the air. Thanks, Fris.

There are lots more things to be considered here than tire size. Given that the bigger tires are heavier that means the lighter car can add weight anywhere as long as the center of mass and total mass remain the same. This allows final gear changes, lower first gear ratio's, hi stall speed torque converters. Wider tires, stickier tires 90/10 shocks on the front & 50/50 shocks on the back etc. All of these could potentially affect the launch a lot more than tire size.

its all about finishing the race in top gear, at the end of the engines rev range, so diff gears are modified as well as tyres to achieve the outcome, which is for all that dont listen, finishing race in top gear at the end of the engines rev range. so if both cars are identical except for wheels, the smaller wheels will lead at start of race with the bigger wheels winning at end....simple