9five1

Mike asked me to see if anyone out there had a chart, or idea of how fast he would be going at a specific RPM in 5th gear for an 85.5 944na.

He is driving from Indiana with no speedo, so that would be helpful

Thanks for the help!
~Eric

theedge

Do either of you have the original manuals? Each manual has a chart for all models. Id check mine, but its at home.

Lawn Gnome

Wouldn't that speed change over the course of 20 years?

9five1

That should do it! Thanks for the info!

The only thing that should change the speed over time (i would think) is the tire size.

~Eric

Yabo

Eric, tell him to check if the plug just bounced out of the speedo sensor, its on the top of the flange for the half shaft on teh driver side of the transmission, it happens soemtimes. Unless he's already sure why it isn't working, it's worth a shot! Don't even need to jack up the car probably.

Rock

i had to do this because i had an early tranny in a later car... my only guide was that about 3000RPMs (give or take 100) youre going about 80mph

M758

132 mph at 5800 RPM
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Seriously what is the rear tire size? I can calculate it from that.

Scott at Team Harco

Is this a Relativity question?

Lawn Gnome

I was just thinking the engine would not be running the same as it did when it was brand new. I also know that my stock 944 may be slightly faster or slower than someone else's stock 944.

Yabo

It doesn't matter if the engine isn't running as well. unless your clutch is slipping.. it can't really change, short of smaller/larger tires.

3000 rpm is 3000 rpm no matter what. if the engine is running worse it may take more gas to get 3000 rpm, or more time, whatever, but 3000 rpm will always be 3000 revolutions per minute. And, that means 3000 revolutions of the drive shaft (or at least some constant multiple), and the tranny is connecting you to a hard gear, which means a certain amount of wheel rotation per minute... how could any of this change?

Scott at Team Harco

This equation should work for any vehicle. I have inserted stock ratios and tire sizes for the 85.5 in this example. Corrected on 9/6 to show US 5th gear ratio (not European).

Let:
v = vehicle speed in mph
s = engine speed in rpm
gr = selected gear ratio = 0.730 (was 0.829) for 5th gear (overdrive)
fdr = final drive ratio = 3.889 (9 : 35)
tc = tire circumference in feet = 6.583' for 215/60R15 tire
c = conversion factor constant = 60 min/hr/ 5280 ft/mile = 0.01136

We have now:

v = s x (tc x c)/(gr x fdr)
s x (6.583 x 0.01136)/(0.730 x 3.889) = s x (0.0748)/2.839 = s x 0.0263

Insert any engine speed for s and multiply by 0.0263 to get approximate vehicle speed. e.g.

s = 2000 rpm, v = 52.6 mph
2500 rpm, 65.8 mph
3000 rpm, 78.9 mph

Feel free to add your own factors - such as the 'Mike Factor' (m), the loaded tire radius factor (lr) or the traveling in Indiana factor (i).
As usual, Your Mileage May Vary (the YMMV factor).

silverBean

That's a cool formula Scott, but a bit off from what I get, 2k is 50, but at 3k I am at 80. It doesn't exactly add up.

centerpunch

Assuming your car has a manual tranny, your clutch is not slipping, and you are in the same gear, that means either your tach is wrong or your speedometer is wrong.

We're talking about gears here. If the engine turns the input shaft to the tranny 50% faster, the wheels have no choice but to go 50% faster.

Giantviper

hey wait a minute guys he is talking about an N/A. An N/A is not at 80 mph at 3000rpm in 5th gear maybe a turbo is though. My N/A at 3000rpm in 5th gear is about 62mph give or take a few mph. 80 mph is like 3700 or 3800rpms.

Scott at Team Harco

As mentioned already, this is for a specific model (actually over a few years) of 944 with specific gear ratios and tire size. The equation is not wrong - your variables (gear ratios, etc) MAY be different from the ones used in the example. All you need to do is insert the appropriate gear ratios and tire circumference for your application and you will get numbers accurate for your car.

If your speedometer and/or tach are in error then that is another issue and independent to the validity of the equation.