PTskater944

I noticed that the solutions to a lot of problems was to look at www.clarks-garage.com, so I decided to read through the 944 FAQ's. The 944 FAQ just has a link to another site, and on that site I saw something that caught my eye. I noticed it said was 15 psi was pretty much the limit if you want to stay safe. It said that beyond 15psi would require lots of modifications, including a different transmission to handle the power. I know a lot of people on these forums are running 18+ psi, have any of you had problems with the transmission? Does anybody know how much power they can take?

PS Heres the exact wording and the link:

"15 PSI or 1.1 bar of boost appears to be the maximum amount of boost you can run with a reasonable margin of safety. Going beyond this (such as the injectors maximum of 17 psi) would require extensive modification such as replacing the fuel pump, seals, transmission to handle the extra power, etc., etc."

http://www.connact.com/~kgross/FAQ/944faq13.html

Chris Prack

A car that I put together ran 18 psi for three seasons in SCCA with 315 or 335 Hoosier's on the back and today runs the 335's making up to 462 rwhp and we have not had a single problem with the stock 951 gear box.

BTW it's a std. turbo box.

rhesus

Lots of people run stock transmissions on more than 15psi... even with bigger turbos, etc.
I haven't heard many complaints about that area on our cars.

seb928s

Haven't heard of any problems from people running more hen 15psi. What I have heard and seen are the stock cv joints going out if you lunch it.

badass951

This is very true, however I have a feeling the CV joints are a weak link whether or not you launch the car.

Chris Prack

Haven't had any failure's there either. Only problem that has come up is the stub axle nuts kept coming loose even with green locktite on them. Trashed two bearings and one Big Red. Replaced both stub axles, nuts and used 964 cup car double nut. No further problems.

ehall

I read the above, so I have a question. I've never driven on the track. My assumption is that other than pit stops, you don't "launch" the car, like a street rod. Is that the case? If so, why the above warning. How much of the "stop light to stop light abuse" hysteria is real, and how much of it is hearsay? All I've ever heard is that you can burn cv's and clutches by stop light racing, and that sort of thing. Isn't most race drivng, okay virtually all, from a roll on situation? If so, why did the cup cars have hardened 1st and 2nd gears? or am I wrong about that?

hosrom_951

You could run whatever boost you like as long as you have the car tunes (Air/Fuel ratio is set and Timming, etc). However, even on a bone stock car, you could break the transmission, diff and/or CV's by shifting hard and being 'too' agressive on the car.

The launch is the key, you don't just redline on first grea and drop the clutch, it may survive for a while, but you could end up with a chewed up 1st and 2nd gears and a broken ring & pinion, not to mention CV's again. (This is on a STOCK car im talking about).

So even if you have 200whp or 600whp, the driving technic would differ slighlty, but the main issues (launching and shifting) should be smooth and not agressive.

Laust Pedersen

Chris P., That’s great information. Do you or anybody else know the torque limit for the OEM clutch, i.e. at what torque does it start to slip?

Spinning tires put an upper limit on the stress of components after the differential and just to get a sense of how much TQ (as measured on a wheel dyno) is needed to spin the tires in gears 1-5 I generated this simple formula which should be fairly accurate:

TQs = Wt*wr*mu*Rw/Gr/Dr

Where,
TQs is torque for slip (as measured on a wheel dyno)
Wt is the weight of the car (3000 lb)
Wr is the relative weight on the rear wheels, with weight transfer (75%)
mu is the wheel to ground friction coefficient (1.0)
Rw is the rolling radius of the wheel (1.025 feet)
Gr is the gear ratio (3.500, 2.059, 1.400, 1.034 and 0.829)
Dr is the differential gear ratio (3.375)

Plugging these numbers in gives the following torque requirement for wheel slip in each gear to be:

1st … > 195 ft-lb
2nd …> 332 ft-lb
3rd … > 488 ft-lb
4th … > 661 ft-lb
5th … > 824 ft-lb

So some of our cars should be able to spin the tires in 1st and 2nd gear on dry asphalt without "dropping the clutch".
If anybody sees a gross flaw in the formula or numerical assumptions, please let me know.

Laust

PS Edited 1-9-2005 to reflect more weight transfer.

PTskater944

I'm not sure, but I think the value you're using as your ground friction coefficient is too high. I remember a presentation that a police officer gave our science class about how the police use laws of physics to determine what happened in car crashes, like how fast the person was probably going and how much time he had to react...stuff like that. Anyways, I think he mentioned the friction coefficient on a normal road to be more around 0.7. I might be thinking about something else though, but I'm pretty sure its around 0.7 or 0.6.

I think in one of Johns racing videos he brings wheelspin in 3rd gear, which according to your formula would take 377 ft-lbs of torque. I don't think he has that much on a stock turbo. And another thing, I think the torque required to spin the wheels in each gear isn't a specific number. Doesn't it also depend on how gruadually or rapidly the torque is increased?

slim_boy_fat

Just some observations not 'having a go'

i assume this is for 1 wheel spinning? The weight of the car it spread across 2.
What about the LSD and the torque transfer across to the non spinning wheel?
I would have thought the weight transfer might be slightly higher.
Where does mu=1 come from ?
Contact friction is related to area of tyre in contact with the ground, the tyre/ground friction coefficient and the vertical load on the wheel.

Cheers.

Waterguy

On the other hand, I think your values are a bit low. Rage2 has posted his best 0-60' time on street tires of 2.3 seconds. This translates to an average acceleration of 0.7 g. To achieve this, he must have either had an 'effective' friction factor of greater than 1 or 70% weight on the rear wheels (i.e. more wieght transfer.) Rage has an LSD, so this does not address single-wheel spin.

Tires reach maximum traction at about 15% slip. They also grip the rough surface of the pavement, so the concept of a friction coefficient does not really hold.

I estimate that you require about 190 ft-lbs of torque at the rear wheels in first gear to achieve 0.7 g acceleration (2.3 second 0-60'.) In second, about 350 ft-lbs would required, in third about 540 ft-lbs.

azmi951

there is a break away Mu and a sliding Mu, also you neglect weight transfer which is substantial.

I dont think that a single equation has the capacity to calculate wheelspin TQ and static would do nothing, it would need to be a dynamic calculation.

Tomas L

I think Laust's assumptions are reasonable.
In this case you should use break away friction which for a wide performance roadtire is close to 1, it's definately higher than 0.7. The police probably must use a safe value so that they don't convict inocent people. 0.7 migth be correct for a sideways sliding grocery getter but not for longitudinal traction of a performance tire. But remeber that grip is affected by other parameters, for instance temperature, dirt on the surface, etc. A cold day with a slippery asphalt will yield different (lower) friction. On a good day, and on a good surface (for instance a dragstrip) it's possible to get higher friction than 1.
The fact that maximum coefficient of friction occurs at som slip doesn't really matter for this.

I also belive that weigth transfer may be a bit greater than 60% at the rear wheels.
This calculation is for two wheels and also works for non LSD cars. Since we don't have a solid rear axle, weigth and torque should be the same on both wheels when acceleration straigth forward. Yes, one wheel will normaly break loose earlier but the other will be on the edge of loosing traction.
The torque value calculated is rear wheel torque, add driveline loss to get engine torque.

The only dynamic event we need to include in this simple estimation is weigth transfer. There's no point in making this more complicated since friction will change to much depending on the conditions.

It's possible to break the tires loose with lower torque than this calculation gives (even when actual friction is used) by using energy stored in the driveline and the flexibility of the drivline.

Tomas

Laust Pedersen

I guess you are all right, since reality as usual is more complicated than simple models such as this wheel slip formula, but it is correct under the simple assumptions given.

Based on the feedback and thinking more about it, my proposed weight transfer of 40%/60% is probably underestimated and may even approach 25%/75% if you also take into consideration, that our transaxle converts an engine twisting moment into an attempt of the two drive-shafts to lift the engine (through the torque tube), thereby adding even more weight on the rear wheels than the traditional center of gravity based weight transfer would suggest.

So if you can find a level road with a static (as opposed to dynamic) tire-road friction coefficient of 1.0, gently engage the clutch at low rpm, step on the accelerator to WOT, then the wheels will brake loose at the listed (transmission loss compensated) torque levels.

Laust

PS I edited the previous post to reflect this increased weight transfer.