Project84

I was doodling with some numbers trying to figure the gear ratios and I don't know if I'm thinking about this right. I'll show you guys what I did, and you tell me if its wrong or right.

This is the transmission gear ratios for the manual tranny in the first gen.

1st 3.674
2nd 2.217
3rd 1.432
4th 1.000
5th 0.825

I decided to use an even RPM to make the math less complicated, so I went with 1,000rpms. OK, now if an engine is at 1,000rpms, clutch out, the input shaft on the tranny will rotate at the same 1,000rpms.

Now, if I select 1st gear, I will divide 1,000 by 3.674 since the transmission gear goes around 1 times every 3.674 times the input shaft goes around. The result is that for every 1000 rotations of the input shaft, the output shaft, coupled to first gear, will rotate 272.18 times.

Ok, so now that the output shaft goes around 272.18 times, that means that the pinion gear goes around the same. To figure the rear final drive ratio for a diff with 3.933 Ring and pinion gears, I again divide the 272.18 by 3.933 since the ring gear goes around 1 time for every 3.933 times the pinion goes around. My result from this as 69.20.

So for an engine at 1,000revs per minutes in first gear (ratio 3.674) the tranny output goes around at 272.18 per minute and the rear axles will go around 69.20 per minute.

If this is not correct, please correct me.

Next, I changed the ring and pinion gears and kept every thing else the same.

For a tranny output of 272.18 and a rear diff with 4.10 R&P gears, I divide the 272.18 by 4.10 instead of 3.933. Now, my final drive is 66.38.
If I divide by 4.33, my final drive is 62.85

So is it right to assume that 3 cars, all at 1,000rpms in first gear with the same 3.674 ratio, same tire and rim size, but one with a 3.933 rear one at 4.10, and one at 4.33 the one with 3.933 gears will go slightly further after one minute?
I'm trying to figure out why larger gear ratios seem to be more favored over smaller ratios, which appear to be more efficient.

All help is appreciated,
Thanks

Evil Aviator

Smaller gear ratios = faster acceleration, larger gear ratios = faster speed. This formula may give you more understandable numbers:

Speed in mph = (Rear tire circumference in inches * engine rpm) / (Transmisison gear ratio * Differential ratio * 1056)

If you don't want to bother jacking up your car and measuring your tire circumference with a tape measure, this formula will be fairly close, although manufacturing tolerances vary in the tire industry:

Circumference in inches = pi * (Wheel diameter + (2 * (Aspect ratio * Tread width in millimeters / 254)))

For example, given a 205/60 R15:
Wheel diameter = 15 in
Aspect ratio = 60
Tread width = 205 mm

Calculated circumference = 77.55 in

Project84

It works out to where the smaller gear ratio, (at any given RPM and tranny gear ratio) makes you go faster than the larger gear ratio under the same given conditions. But what does the 1056 represent in the equation?

Evil Aviator

Oops, I just realized that my non-automotive slang may be confusing people, as the terms "larger" and "smaller" are relative. I guess the more traditional automotive terms of "taller" and "shorter" would probably also be confusing. Just for an easier clarification:

Transmission:
1st gear 3.674 = better acceleration, less speed
2nd gear 2.217 = less acceleration, more speed
3rd gear 1.432 = even less acceleration, even more speed
... etc.

Differential:
4.330 = better acceleration, less speed
4.100 = less acceleration, more speed
3.909 = even less acceleration, even more speed
... etc.

Sorry about any confusion. If in doubt, just work the equation.

The 1056 is a conversion factor because the tire circumference is in inches, and the engine rpm is revolutions per minute, but you want your answer in miles per hour. So...

1 mile = 5280 feet, and 1 foot = 12 inches, so 1 mile = 63360 inches
1 hour = 60 minutes

63360 / 60 = 1056, which is your conversion factor.

RotorMotorDriver

iTrader: (1)

Wow, I just learned A LOT!! Thanks guys .

~T.J.

Project84

Ok, makes perfect sense now. 4.33 gives you better acceleration because it gives more torque than 3.909

Here are the formulas to figure torque. (thanks a bunch evil aviator)

Driveshaft torque = engine flywheel torque (at a given rpm) * transmission ratio(for the gear you are in)

Example: If flywheel torque is 100 ft/lbs while engine is at 2500rpms, and 1st gear is 3.674, driveshaft torque is 367.4 ft/lbs

Next, Multiply the driveshaft torque by the differential ratio to find the axle torque, so you can see why a 4.33 will have more torque than a 3.909

Example: Driveshaft torque was 367.4 @2500rpms.
367.4*3.909=1436.16 ft/lbs at rear axles, while
367.4*4.33=1590.842 ft/lbs at rear axles, all at 2500rpms.

So, using this info, you can figure out what rpm per each gear would be the optimal shift point for your engine, tranny and rear setup.

rotor vs. piston

http://www.geocities.com/z_design_st...ion_z28_6.html

This is a link to a GREAT transmission calculator. Just never mind the Z28, just enter your information and you can find out how fast you car will go at a given RPM and gear. Have fun.

Project84

If you have a dyno sheet for you car, you will see the max amount of rear wheel torque for your car. lets say 1590.842@2500rpms. To find the driveshaft torque, divide that by the diff. gear ratio(4.33)

1590.842/4.33=367.4ft/lbs @ 2500rpms driveshaft torque.

to find flywheel torque, you must divide the driveshaft torque by the tranny gear.

So 367.4/3.674=100 ft/lbs @2500.

Just a little inverted math. I don't have a running car to dyno, (well, got the '94 explorer and the '96 escort) so I don't know exactly how much info they show, but just knowing some of your gear ratios, you can find the engine torque by working backwards through the equation.

Evil Aviator

It doesn't quite work like that because of the various drivetrain losses, and you forgot about the last big gear in the system, otherwise known as the wheel/tire. Anyway, it's the torque at the wheels that matters, and that's what the chassis dyno reads. The only purpose of the "driveshaft torque" calculation is to determine shift ponts (one of many methods), and this only works because the differential ratio and wheel/tire diameter are constants (more or less) and can therefore be ignored for this calculation.

BTW, some dynos measure the drivetrain loss after your run when the drivetrain is decelerating, and then the dyno can estimate your flywheel horsepower. Otherwise, the 15% drivetrain loss rule of thumb works fairly well.

Evil Aviator

Correction:

Circumference in inches = pi * (Wheel diameter + (2 * (Aspect ratio * (Tread width in millimeters / 25.4))))

Also note that the Aspect Ratio is a decimal (ie 60AR = .60).

Sorry about that. These equations are probably in a book somewhere, but I just make them up as I go along and sometimes things get lost in the process.

FYI for those technical types following along, all I did was convert the tread width from mm to inches, multiply by the aspect ratio to find the tire wall height, multiplied that by two and added it to the rim diameter to get the overall diameter, and then multiplied that times pi to get the circumference. Make sense?