The effect of a flywheel on a car's launch

[Emre]

Hello,I have been searching for a good answer but I couldn't find any.

Let's say a car's engine is producing 200 Nm of torque throughout its entire range.since gear ratios and tire radius are fixed,it must provide the same amount of wheel force in a gear.So,theoretically,whatever RPM you launch at,the wheel force should be the same.

But in real world,tire slip becomes too much when we try to launch at high RPM.This is true for any car I have driven so far.Even though the torque at 6000 RPM is lower than torque at 3000 RPM,the car far easily lose traction if we try to launch at 6000 RPM.

Is it so because if we launch at high RPM,the rapid deceleration of flywheel provides extra torque to the crankshaft in addition to torque produced by the engine?I have done some calculations,assuming the flywheel is 28 cm in diameter,ansd has a mass of 10 kg.If we dump the clutch and the RPM rapidly drops from 6000 to 3000,I found that not huge amounts of torque is produced at all.

Is the idea true?Does dumping the clutch at high RPM provides huge amount of extra torque?Is it the reason even low torque cars like s2000 launch really fast?

 

[ChemAir]

Edit: All the below assumes drag type tires and a track with available traction...

Yes, the flywheel mass has an effect, but compared to car weight and horsepower that is small. Generally, the fastest acceleration will be when the maximum power is applied, and that is typically close to maximum RPM/torque. Starting a race/acceleration with the engine already at peak power means you will be at or close to peak power for the entire run.

What you find in drag racing, is that when all engine/car combinations have very similar power and weight, things like rotating mass start to be the differences between the fastest and the slowest. The faster cars will optimize their weights to maximize acceleration. Flywheel mass to hold engine rpm high while the car starts moving is part of this equation. Other ways include very low friction bearings, light weight tires/wheels/brakes, low friction brakes, etc...

I believe the S2000 has the advantage of reasonable power and pretty low weight.

 

[Spinnor]

If we dump the clutch and the RPM rapidly drops from 6000 to 3000,I found that not huge amounts of torque is produced at all.

That does not seem right to me, you have a lot of rotating mass in the engine and it all decelerates from, your example, 6000 to 3000 RPM in a second or two. I think that equates to a lot of torque.

Your basic idea is correct I think, if you want your car to accelerate as fast as possible from a start you rev the engine before you "pop" the clutch. I think all drag racers do this.

 

[Rive]

Is the idea true?Does dumping the clutch at high RPM provides huge amount of extra torque?

No. The RPM got high only because before dumping the clutch the only load what balances the inflow of fuel is only the loss of the engine (what's higher at high RPM). After the clutch dumped, it's the loss at low RPM and the acceleration together.

The engine has a delay between giving in more fuel and getting the matching power. To keep the RPM high before start is to eliminate this delay => faster, more responsive start.

 

[Spinnor]

Using your numbers for the flywheel and assuming the all the mass is at the circumference of the flywheel and using a 6000 to 3000 RPM drop in say 2 seconds I got a torque close to 200Nm. The angular deceleration of the rotating mass helps accelerate the car for a short period but then as the engine tries to speed back up the same rotating mass tends to slow the angular acceleration of the engine.

 

[Emre]

Using your numbers for the flywheel and assuming the all the mass is at the circumference of the flywheel and using a 6000 to 3000 RPM drop in say 2 seconds I got a torque close to 200Nm. The angular deceleration of the rotating mass helps accelerate the car for a short period but then as the engine tries to speed back up the same rotating mass tends to slow the angular acceleration of the engine.

Could you please show your calculations here?I also think just as like you,because even small cars with engines 1.0-1.2 or so can produce tire spin at high RPM easily.Moreover,when you shift up to second gear a bit faster,you can still hear a tiny hiccup.So they must be providing a huge amount of torque just momentarily,for 1 second or so?

 

[Emre]

So this is exactly a good illustration of what I say.This car produces 320 Nm between 1750-2500 or so.But it launches at a very high RPM,therefore tires slip too much.You can also hear the hiccup when the driver shifts to second gear.

 

[Spinnor]

Torque = Iα = inertia x angular acceleration

For your example if the RPM goes from 6000 to 3000 in 2 seconds what is the angular deceleration in radians/sec?

What is the inertia if the 10Kg mass is considered all located at the circumference of your example, flywheel diameter = .28m?

Multiply the two numbers to get the torque.

 

[Emre]

Torque = Iα = inertia x angular acceleration

For your example if the RPM goes from 6000 to 3000 in 2 seconds what is the angular deceleration in radians/sec?

What is the inertia if the 10Kg mass is considered all located at the circumference of your example, flywheel diameter = .28m?

Multiply the two numbers to get the torque.

The formula is mr^2/2 for the disc.It gives me 0.098 kgm^2.Multiplying this with the acceleration doesn't yield a good torque value.

 

[ChemAir]

That does not seem right to me, you have a lot of rotating mass in the engine and it all decelerates from, your example, 6000 to 3000 RPM in a second or two. I think that equates to a lot of torque.

6000-->3000 doesn't really happen in practice for a well set up car. That would be seen as a bog, and the car would slow.

It varies, but launch RPM for HP limited cars typically is close to a high torque point, moving to peak power in a few tenths of a second. Ideally, the tire slip (by wheelie bars, clutch, power application, available traction, etc.) is all carefully controlled to keep the engine at the best RPM possible, optimizing wheel speed and acceleration. The data acquisition used even by sportsman racers is extensive.

 

[jack action]

I think the time factor is underestimated here. The relationship between force $F$, energy $E$ and time $t$ is:
$$F = \frac{dE}{dt}$$
First thing to note is that whatever energy you have at 3000 rpm, you must have 4 times the energy at 6000 rpm, thus 4 times the force.

But one also should note that if $dt=0$, then the force is infinite. So anything is possible, no matter the amount of energy stored.

In theory, $dt$ is zero. In reality, it is not. That is because every component deforms. How they deform depends on their stiffness. Hooke's law ($F=kx$) demonstrate that 4 times the force will yield 4 times the displacement. Again that is in a perfect world. Real springs are not infinite in length and they will collapse to a «solid height» where the spring stiffness will dramatically increase.

So it is possible that, at 3000 rpm, the deformation of the components is high enough to absorb the shock between the engine and the tire-road contact patch. Then, once at 6000 rpm, the components are 1) deforming a lot faster and 2) may reach their deformation limit - increasing the effective stiffness of the system - thus shortening drastically the absorption time $dt$ even more. At that point, something has to give (slip or break).

The thing to remember is that evaluating that $dt$ might be a lot more complicated than one thinks and there is no real limit on how low it can be.

 

[Spinnor]

The formula is mr^2/2 for the disc.It gives me 0.098 kgm^2.Multiplying this with the acceleration doesn't yield a good torque value.

Looks like we both might have made a mistake. 10x.14x.14=.196 (edit I assumed all the mass on the circumference)

What number did you use for angular deceleration, even if 2 seconds is not a realistic number? I had something like 157 rad/s^2.

 

[Emre]

Looks like we both might have made a mistake. 10x.14x.14=.196

What number did you use for angular deceleration, even if 2 seconds is not a realistic number? I had something like 157 rad/s^2.

I actually used 1 second :) .But isn't there a dividor mr^2/2 in the formula for the disc?

 

[Spinnor]

But isn't there a dividor mr^2/2 in the formula for the disc?

Right, I did not assume a disc but a rim.

 

[Spinnor]

I actually used 1 second

So what number did you get for angular deceleration?

 

[Emre]

So what number did you get for angular deceleration?

314 rad/sec^2

 

[Emre]

I have been doing research still.I realized that I hadn't taken into account the effect of crankshaft.After all,it decelerates just as flywheel does.Assuming 2 kgm^2 for inertia gives really really huge amount of torque.If we add the flywheel effect to this,I think we can understand why high RPM launch is very important no matter how much torque you have.

 

[Spinnor]

314 rad/sec^2

That looks right.

 

[Spinnor]

I think we can understand why high RPM launch is very important no matter how much torque you have.

High RPM equates with the most power, you want power to accelerate fast?

From, https://www.quora.com/Why-is-it-tha...hing-a-peak-even-though-the-RPM-is-increasing

From, https://www.google.com/search?safe=...j35i39k1j0i30k1j0i8i30k1j0i24k1.0.mqikaVr3JjI

Also you don't want your wheels to chatter, why do they do that?

 

[Emre]

Crankshaft's inertia is not even compareble to that of flywheel,actually.

High RPM equates with the most power, you want power to accelerate fast?

View attachment 223049

From, https://www.quora.com/Why-is-it-tha...hing-a-peak-even-though-the-RPM-is-increasing

From, https://www.google.com/search?safe=...j35i39k1j0i30k1j0i8i30k1j0i24k1.0.mqikaVr3JjI

Also you don't want your wheels to chatter, why do they do that?

In a given gear,the car accelerates the fastest at torque peak moment.However,you can always do better by downshifting,bringing the RPM to max power.There is only one gear we aren't able to do this,the first gear.So yes,if you downshift or upshift,we want power.

 

[Emre]

I want to ask another question.The formula dL/dT=Torque doesn't account for the doubling effect of speed?I mean,if RPM doubles,the energy must also double so torque should be four times more?

 

[Mohankpvk]

The maximum acceleration that could be given to a vehicle by rotating the wheels ( by the engine) depends on the friction between the tires and the road.By acceleration I mean the quickness with which you can increase your vehicles speed from zero to a desired value(whilelaunching the vehicle).If you accelerate more than the maximum that could be offered by the friction then the tire slips.

 

[OldYat47]

I didn't see this mentioned, but sitting at the lights the engine is producing zero net torque. The engine may be spinning quite fast, but all it's doing is keeping itself going. The instant the clutch begins to load up the engine begins to slow. The flywheel gives a stored amount of inertia to get the car going, giving the engine time to regain RPM and to start producing useable torque. There is a balance between the moment of inertia of the flywheel and optimum acceleration. A flywheel with a higher moment of inertia gives a good launch but is harder to spin back up. A flywheel with a lower moment of inertia is easier to spin up but has less inertia to aid launching the car.
The rotating mass of the crankshaft isn't any help since (virtually) all its inertia is used on the compression stroke.