racing tranny 2

[wangp1283]

It's often said that the moment of inertia of the transmission is very important to a light, racing car. So important that shaving a pound off the transmission rotating part is going to do more good than shaving a pound off the vehicle body. Can someone explain? Is it really significant? I think there is also a formula that relates all this, and it has a lot to do with the gear ratios.

Another thing I've noticed is that a lot of racing cars and motorcycles have gear ratios that are very closely spaced. For example, a lot of those Japanese road bikes may have a 5 speed transmission with a overall ratio span less than 3 or 2.5. (if you divide the tallest gear by the shortest, they are usually less than 3). Why is this?

Does anyone know the gear span (and ratios if possible) of a typical F1 car or Rally car?

Thanks

 

[GregLocock]

Reducing the weight of rotating components is almost always better than reducing the weight of static components, because you then have a smaller rotational and translational inertia to accelerate.

The formula is pretty obvious if you use an energy method.

You should be able to work out the ratios of an F1 box, and its span, if you can find a trace of RPM over a lap.

eg

http://members.optusnet.com.au/greglocock/cepstrum.htm

Williams and Renault may have better traces on their F1 sites.









Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[Metalguy]

"Another thing I've noticed is that a lot of racing cars and motorcycles have gear ratios that are very closely spaced. For example, a lot of those Japanese road bikes may have a 5 speed transmission with a overall ratio span less than 3 or 2.5. (if you divide the tallest gear by the shortest, they are usually less than 3). Why is this?"

With high power-to-weight ratios, they don't need real low gears (generally). But such engines usually don't have real flat torque curves, so having many closly-spaced gears helps keep the revs where they belong. Notice how "good" trans. have the higher gears much more closely spaced than the lower ones-a point usually ignored by Detroit for their "commoner" cars.

Also, while any weight reduction is beneficial, esp. rotating weight, the difference to the engine while accelerating in gear is just about zero.

 

[wangp1283]

"Also, while any weight reduction is beneficial, esp. rotating weight, the difference to the engine while accelerating in gear is just about zero."

Can you explain the last part further? You mean it's only beneficial while shifting?

 

[Metalguy]

When an engine is accelerating a car/bike, the rate of RPM increase is so slow (relatively) that the weight of flywheels and gears doesn't mean much. On dirt bikes where the rear wheel can easily spin and lose traction, a heavy flywheel can slow the spin-up and keep better traction.

The revs can change very fast during shifting, so gear/clutch-plate weight can have a big influence, but once the engine is hooked up to the load it's all slow-motion from a spin-up POV-especially in the higher gears.

 

[GregLocock]

Referred rotational inertia of the engine and driveline is about 25% of the vehicle's mass, in first gear.

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[patprimmer]

In my opinion, reducing rotational weight in the engine, especially the flywheel (because of it's large diameter), makes a significant difference. Further down the drive line, the emphasis reduces with gearing and therefore rate of speed change. Despite this, I also think that the wheels play a significant part as even though the rate of rpm change is at the slowest part of the drive train, the weight and diameter are significant.

A weight saving any were in the drive train still has a larger influence than a similar weight saving on the chassis.

Even the smallest saving should be exploited in racing, but if budget dictates, spend the available money on the drive train.

Regards

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Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[GregLocock]

Can't argue with that that... but bear in mind that it is an N^2 relationship, so once you are out of first gear then engine inertia matters less and less. Also, it is far easier to pull a couple of kg out of the body than 0.5 kg (or whatever) out of an engine.






Cheers

Greg Locock

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[wangp1283]

I'm trying to calculate the rotational inertia involved in a racing vehicle.

Can someone give me an idea of the diameter and width of the gears (an average) in a

F1 car
a Rally car

Thanks.

 

[evelrod]

Pat, I know this is not necessarily relevant to drag racing, but let's not forget that the ultra light drive train is felt mostly in top gear on 'deceleration'. An engine with a light flywheel will rev much quicker than an equal engine with a heavy flywheel under no load conditions, but the effect is rather diminished in top gear. Now, on the other hand, under heavy deceleration, the light flywheel engined vehicle is at advantage. I know I don't use the engine exclusively for braking, but it still is a factor, and oft times, a big factor...particularly when braking where no gearchange is needed.

Greg, I try to make the drive train a light as is practical because in most class racing, a vehicle weight limit is in effect. My problem is with the idiot that drives my car. He has gotten totally FAT in his old age. ;-)

Rod

Rod

 

[Metalguy]

Greg,
"Referred rotational inertia of the engine and driveline is about 25% of the vehicle's mass, in first gear."

Where did you get that from? I won't say it's impossible, but it's hardly a "normal" % with common 1st gear ratios and flywheel/gear train weights/inertias.

You'd need very heavy inertia and very low 1st gear to have such a high %.

 

[evelrod]

To the original poster---

http://www.taylor-race.com

You will get more info from these folks about racing gearboxes and differentials than just about anybody on the planet.

Rod

 

[GregLocock]

Metalguy

That's a rule of thumb number from many years ago when I was first working in the area of performance prediction.

Reasons why it might be right (1) the guy who told me to use it was a brilliant and thorough engineer (2) the extreme difficulty cars have in getting 0-60 times under 3 seconds. 3) full throttle no load accel time for an engine is 1 or 2 seconds

Let's do it the hard way. mass of crank is about 40 kg, arm is about 0.05 m, so I is 0.1 kg m^2

overall gear ratio is about 10, so that's 10 kg m^2 at the wheel. rolling radius is 0.3,so that's about 100 kg of referred inertia

I'm happy enough that it is in the ball park, given that I've ignored a lot of the other components, especially the flywheel and clutch.





Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[BillyShope]

Was searching for a specific inertia value and Google brought up this old thread. I realize this site is supposed to be for engineers, but I also realize that a lot of non-engineers come here for help, so, Greg, I hope you don't mind if I "translate" your calculations into something more recognizable to the American motorsports enthusiast.

Greg refers to an inertia of 0.1 Kg-m^2. If you multiply this by 3417, you get the inertia in lbm-in^2.

Multiplying a tire radius of 0.3 meters by 39.37 will give you the radius in inches.

Finally, that which is often called the "equivalent mass" is found by multiplying the inertia and the square of the overall gearing and then dividing by the square of the tire radius. Using Greg's figures, this would be:

(341.7)(10)^2/(11.81)^2 = 245 lbm

(Don't let the "lbm" throw you. The numerical value is the same as the "lbf" or, simply, weight.)

This equivalent mass is the extra mass (weight) equivalent of the effect of the rotating inertia.

I thought this translation might be helpful, Greg, since all that metric stuff just doesn't fly here. The government tried to force it on us a few years ago, but we're a rebellious lot. And, no, I'm not trying to "change" anybody. Years ago, I was one of those who thought it made much more sense to slash "O's" instead of "zeros," and I ended up on the losing side, so I've learned to live with such things.

 

[Greenlight]

I cut and pasted this from my Excel program. The columns don't exactly line up here, but you can see the difference in equiv. wts. by changing the flywheel wt. by one lb. (from 9 lbs. to 10 lbs.) and the effect in each gear.


Setup 1 Setup 2
Flywheel Weight (lb) Wf = 9 10
Flywheel Radius (in) Rf = 5 5
1 st Gear Ratio 1 = 3.19 3.19
2nd Gear Ratio 2 = 2.22 2.22
3rd Gear Ratio 3 = 1.6 1.6
4th Gear Ratio 4 = 1.29 1.29
5th Gear Ratio 5 = 1 1
Final Drive Ratio FD = 5.57 5.57
Tire Radius (in) Rt = 13 13

Equivalent Weight 1st gear (lb) We1 = 210 234
Equivalent Weight 2nd gear (lb) We2 = 102 113
Equivalent Weight 3rd gear (lb) We3 = 53 59
Equivalent Weight 4th gear (lb) We4 = 34 38
Equivalent Weight 5th gear (lb) We5 = 21 23


So, as you can see there is a distinct difference even in high gear.

I hope this helps.

Greenlight

 

[patprimmer]

One other point re inertia of rotating components.

It seems so simple that it seems offencive to mention it, but I haven't seen anyone take the normal or linear inertia of rotating components during acceleration as well the rotating inertia.

Regards

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Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[Metalguy]

Greenlight,
Try this. You're at Bonneville in your not-so-aerodynamic car/truck. Your little 2 liter engine can turn 12,000 RPM, and you're in high gear approaching top speed. It takes ~5 seconds to go from 11,000 to 12,000. The engine doesn't feel a 2 lb. wheel vs a 10 lb. one.

I have an old LP record of "Sounds of Bonneville". Blown hemi has a clutch problem. It holds until the torque reaches some level. You can hear the engine pulling the car slowly up to speed, and when the clutch slips the engine revs skyrocket instantly-far less than a second.

My AA/FC had a very heavy triple disk Crowerglide. Throttle response in neutral on a blown fuel engine is *immediate*, even with a heavy flywheel/clutch.

Now, where does your formula account for engine RPM rate of increase?

 

[BillyShope]

Working backwards from Greenlight's figures and assuming 350 foot-pounds available from an engine that is no longer attached to the rear wheels, that means an acceleration of around 13,000 rad/second². Yes, that's pretty instantaneous and it's exactly what Greenlight's equations would predict.

And, Pat, wouldn't the weight of the rotating components be included in the total car weight? Are you suggesting they should be separated? Why?

 

[patprimmer]

No

I am suggesting that it has a double effect.

A pound off the flywheel is a pound off the car, and has that benefit as well as a pound of the rotating mass, so it also has that benefit. a pound off the car is only an advantage in the first instant.

After reading the above sentence, I now realise why grammar was never a good subject for me.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[BillyShope]

Okay, Pat. I hadn't been thinking in terms of "deltas," so your comment threw me.

Just noticed that reference to "fora." Someone took high school Latin! Was watching a documentary the other day on college bands and their halftime performances. The documentary was probably made back in the fifties. They referred to "football stadia around the country." Had to laugh. I came out of retirement to teach at a private Christian high school for a year and the kids are so dumbed down now that only a very, VERY small percentage of their generation would ever understand that. I once made reference to turn "apices" in a hotrodding forum post and that resulted in some curious responses.