speedlever

I just picked up a '99 Civic LX which has an EX 5 spd tranny that replaced the automatic tranny it came with. The PO who did the swap did not replace the final drive. And per the Honda owners site, the manual tranny version came with a 4.06 final drive while the auto version had a 4.36 final drive.

To further confuse things, the OEM tire is a P185/65R14 while my new Civic has P195/60R15 tires on 15 inch wheels.

To the best of my (limited) knowledge, the tranny swap seems to have been done properly with the starter being changed, the tranny mount/bracket changed, new flywheel/clutch/pressure plate, new ECU, etc.

I expect my Civic will turn higher rpms for a given speed than would a stock 5 spd Civic due to the final drive being 4.36 rather than 4.06. But I don't know if the oversize tire size will offset that to any degree or not.

Anyone know?

ezone

Final drive ratio cannot change speedo error on that car. Speedo data is taken from the axle rotation, the VSS is driven at axle speed.

Tire size is the factor which will change your true road speed vs. any given needle indication, but the tire sizes you mention are fairly close.

Find an online tire size calculator (it's late and I'm not feeling it), compare the different sizes you mentioned above, the calculator program probably will tell you the % of speedo error to expect.

There's always GPS speed checks too.

speedlever

Thanks. I'm not really asking about speedo error here. I'm asking how the changes affect the apparent/effective final drive ratio. I would think the EX tranny with the Auto final drive (4.36) will turn higher rpms at a given speed as compared to a standard 5 spd gearbox and final drive (4.06). But I've seen different numbers for the final drive so am not sure if those numbers I got from Honda are accurate or not.
Final Drive Ratio (MT/AT): 4.06/4.36 from:
http://owners.honda.com/vehicles/information/1999/Civic-Sedan/specs#mid^EJ652XPBW

http://www.clubcivic.com/forum/threa...n-specs.71070/

The wiki gives 4.357 (4.36) for the auto final drive which agrees with Honda:
https://en.wikipedia.org/wiki/Honda_...xth_generation)

Combine the oversize tires (which I think would take the 4.36 to some lower number) and I'm wondering how much higher the effect final drive ratio would actually be? 4.25?

I roughly figure the OEM 185/65R14 is 23.5 dia with 884 revs/mile
The 195/60R15 is 24.2 dia 858 revs/mile

(using Tirerack.com and using General Altimax RT43 tires for both examples)

The calculator found here:
http://www.onlineconversion.com/bigger_tires.htm

shows a lower rather than a higher effective final drive which seems bassackwards to me. Or maybe my thinking is bassackwards. Won't larger tires make a final drive ratio *effectively* higher (lower numerically)?

speedlever

Ok, I found another site which produced results that I expected.

http://www.onallcylinders.com/2015/0...l-drive-ratio/

Using the formulas from this site, I figure an effective final drive of 4.23.

23.47/24.21 x 4.36 = 4.2267

Tire Diameter = 2 x (section width/25.4) x (aspect ratio/100) + rim diameter
2(195/25.4)(.6)+15 = 24.21 in dia

****************************From the linked site*****************
Please note that these formulas are for metric-rated passenger and light truck tires. Slicks, large off-road, and some larger, Pro Street style tires are already rated by diameter, so you don’t have to do any calculating for them.

We will be using a P235/60R-15 tire as our example: 235 is the section width, 60 the aspect ratio, and 15 the required wheel rim diameter. Here is the basic formula:

Tire Diameter = 2 x (section width/25.4) x (aspect ratio/100) + rim diameter

When you plug in the values from our sample tire, the formula looks like this:

Tire Diameter = 2 x (235/25.4) x (60/100) + 15

Now, let’s do the calculations:

Tire Diameter = 2 x (9.25 x .60) + 15
Tire Diameter = 2 x 5.55 + 15
Tire Diameter = 26.1 inches (round down to 26)

Calculating Effective Drive Ratio

Once you determine the diameter of your old tires and potential replacements, you can compare the effect of new tires on the final drive ratio. For example, if you have 26-inch tall tires with a 3.55 gear, and you want to go to 28-inch tall tires. To calculate the effective drive ratio of the new tires, multiply the diameters together, then multiply that figure by the gear ratio as follows:

26/28 = .92857142857

.92857142857 x 3.55 = 3.2964 or 3.30
************************************************** *************

This makes me think the original site has the new and old tire fields reversed. I get the new answer if I reverse the inputs in the new and old fields.

I don't know if I'm misreading/misunderstanding what the first site is trying to calculate or not, but the last site provided an answer closer to my expectations. Assuming my expectations were correct!

ezone

Lunch break, no time here...

Terminology can be a funny thing when people talk about gear ratios.
A "taller" ratio could refer to the numerical ratio, or it could refer to the effect of the ratio change. Confusing at best LOL


In drag racing....one of the quickest and easiest ways to gain the advantage of a taller (higher) numerical rear end gear ratio is to simply install shorter tires.
Quicker acceleration rate due to increased mechanical advantage, but with the trade off of reduced road speed at a given RPM or reduced top speed at maximum RPM.

The reverse is true if you switch to a taller tire.

Nobody cares about the cars speedo in drag racing except the kids that don't understand how this works. The only speed that matters there is what the time slip says you ran (ET).