Axle Gearing: How to choose the ratio

[Scott Brady]

For years, people (and many still do) chose axle gearing based on a simple chart. For a 33" tire, you install 4.10 gears, for 35" tires, you install 4.56 gears, etc.

Gearing selection is significantly more complex than that, and is based on engine peak torque (not HP), tire diameter, final drive ratio, desired performance attributes (economy or acceleration), etc.

So I thought is would be helpful to start a thread on exactly how to select an axle ratio.

Step 1: Determine peak torque RPM. For example, the 3.4L Tacoma V6 generates peak torque at 3,600 RPM.

Step 2: Determine tire diameter (32.6" or whatever).

Step 3: Determine overdrive ratio. For the Tacoma automatic, it is .705:1. The remainder is used, so for the calculation it would be 1-.705= .295. The calculation uses the multiplier 1.295

Step 4: Determine maximum desired cruising speed in MPH. This is not desired top speed, but cruising speed. For me, it is 80 mph

Step 5: Determine if you want power or economy. If you want power than multiply your peak torque RPM by .85. If you want economy, multiply your peak torque RPM by .75. (Example: Power= 3,600*.85= 3,060)

Step 6: Calculate required ratio:

Formula=

Gear Ratio = RPM x Tire Diameter x overdrive ratio remainder
MPH x 336

So for the examples above:

1. 3060 x 32.6= 99756

2. 80 x 336= 26880

3. 99756/26880= 3.711

4. 3.711 * 1.295= 4.81

4.81 is the desired axle ratio. 4.88 is a typical production ratio, and would be ideal.

Here is another example: peak torque= 3,200 rpm, tire diameter=30.8, overdrive=.80, desired cruising speed=75 and the owner prefers power over economy.

1. 2720 x 30.8= 83776
2. 75 x 336= 25200
3. 83776 / 25200 = 3.33
4. 3.33 * 1.20 = 3.98 ratio

Last example: My truck: peak torque= 3,600 rpm, tire diameter=33.2, overdrive=.705, desired cruising speed=75 and I prefer power over economy.

1. 3060 x 33.2= 101592
2. 75 x 336= 25200
3. 101592/25200= 4.03
4. 4.03 * 1.295 = 5.22 ratio

How does your vehicle equate?

 

[articulate]

Not to be a tweaker, but is the formula supposed to look like this?

So, if I'm using 4 beautiful BFG AT 265/75r 16 tires on my Frontier, I come up with the following ratios:
Power = 5.01
Economy = 4.42

(according to BFG, this tire measures a full 32"...)

4.88 would be nearly an exact meet-me-in-the-middle solution, as well as a popular axle ratio. Very interesting.

So, where's the formula that gives us $1200 for the re-gear? :sunflower

Cool stuff!

Mark

 

[Scenic WonderRunner]

I found this link below useful when I was upgrading from my stock 225 tires to my 31" / or. There were some other good calculator Links I used......when I find them I will add them here.

I have 31x10.50x15 tires (Cooper All Terrains....but hey!.....they were practically free!).....I have an Auto with Overdrive.....so at 65mph with O/D on.....I'm only turning 2,450 RPM.....with Factory Stock 4:10 Gears.

Scott..........I'm not saying your formulas are not perfect or correct............I'm sure they are dead on! I'm just sharing links that I had found (quick look) when I was trying to decide on tires but really didn't know anyone to ask. It would be interesting to compare your "Dead ON" Formulas to the chart Links.....and see if they really know what they are talking about.



Gear Ratio & Tire Size Chart<~this one is great because it helps you decide between Better Gas Mileage to More Power.


Axle Ratio/Tire Size Calculator 1.1

 

[Scott Brady]

Thanks Mark, that is exactly how the calculation should look. I will use your cool image from now on.

I put together this method a few months ago after doing some research. There were no good calculators for determining the correct gear ratio based upon engine torque and driving style. So, this was my attempt.

 

[Pskhaat]

Okay, so this might be an anomoly, but in my imaginary world I have a Toyota FJ60 that has a gasser 4.2 I6 (2F) engine in it that peaks torque at 1800 RPMs and I'm in it for power, if I want to run 35" tires @75mph and final drive ratio (tranny) in 4th is 1.0 I come up with a ratio of 2.125.

Is my maths correct? In the case for low torque engines like the Toyota 2F and diesels, there may need to be a sliding scale or factor applied.

Now this all said, that FJ60 performs best stock with 3.73s at 55mph and IMO simply wasn't made for comfortable highway speeds

 

[articulate]

Okay, so this might be an anomoly, but in my imaginary world I have a Toyota FJ60 that has a gasser 4.2 I6 (2F) engine in it that peaks torque at 1800 RPMs and I'm in it for power, if I want to run 35" tires @75mph and final drive ratio (tranny) in 4th is 1.0 I come up with a ratio of 2.125.

Is my maths correct? In the case for low torque engines like the Toyota 2F and diesels, there may need to be a sliding scale or factor applied.

Now this all said, that FJ60 performs best stock with 3.73s at 55mph and IMO simply wasn't made for comfortable highway speeds

Isn't the FJ60 an "older" one? Back in the day, weren't normal highway speeds 60 mph or so?

On one of Mark's (Senor Scenic) links, there is this simple formula, too:

This one assumes you want to have similar performace to factory specs, and is a simple proportional guage: if your tires go up so much, then your gearing goes down so much.

Scott's version - as we all expected - is far more advanced and likely produces a much lower gear ratio than in the simpler formula above; well, excepting pskhaat's machine. MAN! I could eat at Brady's table-o-knowledge any day of the week... :chowtime: The thinking never ends with this guy.

Anyway, I believe I've found another fine line between geeky and hardcore. Back when that FJ60 first came out, there was a vast difference between geeky and hardcore. Now these two are interchangeable - if not the same


Cheers,
Mark

 

[Scott Brady]

Okay, so this might be an anomoly, but in my imaginary world I have a Toyota FJ60 that has a gasser 4.2 I6 (2F) engine in it that peaks torque at 1800 RPMs and I'm in it for power, if I want to run 35" tires @75mph and final drive ratio (tranny) in 4th is 1.0 I come up with a ratio of 2.125.

Is my maths correct? In the case for low torque engines like the Toyota 2F and diesels, there may need to be a sliding scale or factor applied.

Now this all said, that FJ60 performs best stock with 3.73s at 55mph and IMO simply wasn't made for comfortable highway speeds

Your math is correct. The formula is intended to focus on the greatest efficiency of the motor. So, either you would have to sacrifice efficiency by reving the motor above peak torque, or 75mph is not really feasible with the combination you indicated. I do not remember ever driving at 75 mph in my 60. :elkgrin:

BUT, having said that, I will look at including the actual torque number, but then it would have to be adjusted by the curb weight (to factor a torque value per pound).

I have to do some work

 

[flyingwil]

There is also an article on this in the April edition of Four Wheeler Magazine. However, the above posts, have better information! Way to go guys!

 

[DaveInDenver]

Yeah, pskhaat, I get the similar results when I run this with my truck. The 22R-E torque peak is 140 lb-ft at 2800 RPM. I want power, 75 MPH is plenty for me , I have 0.85 overdrive in the W56 tranny and run 33x9.50 BFG (so I use 32.6").

The formula says I should use 3.56:1 ratios. No way a 22R-E is going to move a truck down the highway in 5th gear on 33" tires with 3.56 gears.

But if I use the HP peak of 115 HP at 4800 RPM, I get a 6.07:1 ratio. I went with 5.29 with my 33" tires and I think that's pretty decent. At least 5.29 does accomplish 75MPH without much fuss.

Since HP is calculated by:

HP = torque (lb-ft) x 2 x pi * RPM

Using HP is taking torque, but adding time. Since it's not instantaneous force (torque is just force through an angle), but the rate of sustained force that moves the vehicle, I think using HP can make more sense.

But also using torque and not HP deals only with the rolling resistance (which is basically constant at all speeds) of the truck, not the aerodynamic drag (which is exponential with speed). Starting a truck from a dead stop (or overcoming an obstacle) requires immediate torque, but keeping it moving at some speed is power (i.e. sustained force over time). Since you gear a truck to put the engine in the right RPMs to move the truck on the road, I'd think it's available HP that is going to determine how easily it is to move your your lifted brick moves down the road through the air.

Mixing in efficiency is whole new game. That point is probably no where near either the HP or torque peaks. My guess is that you want to pick some number between peak torque and peak HP to use as the engine's peak performance point. That's probably the skew that's needed, to determine how much to skew between the torque and power. What the number is going to be different for all engines, so the bias needs to be built in at steps 1 and 5 in the original calculations.

Sorry, just thinking out loud. Feel free to ignore if it's total bunk.

 

[Scott Brady]

Great post Dave. As the variables are considered, my formula appears to only work with a modern DOHC V6

So, it will be fun to consider other options of calculating the result.

 

[DaveInDenver]

Yeah, Scott, what's curious is that the R motors are SOHC, aluminum heads, etc. They are positively modern compared to the 2F in the FJ60, so why exactly the results for my '91 (with EFI, even) and pskhaat's Cruiser jived rather than with your 5VZ is a bit baffling. Most likely I think it's the overdrive ratio.

I'm not sure there's a perfect absolute formula here. At least, I think you need more variables to make it more universal, probably including more engine characteristics or different variable for stick and automatic. You need to use internal losses, which for the engine is going to be implicitly accounted for in the HP numbers anyway. I do think using HP is the key over torque, though. Just curious, why do you specifically use torque and not HP?

I was messing with your formula a little.

If you completely remove the 0.75 and 0.85 RPM multipliers for performance verses mileage, my results are 4.16 & 7.14. No help there.

If I use the mileage multiplier I get 3.12 and 5.36. That last one is interesting because it used 0.75 x HP peak and produced almost the exact ratio I ended up using. Not sure if that's not just coincidental, though. But still it sparked an idea...

My thought was those fraction of peak modifiers. Those are the key to the formula I think.

I wondered what would happen if you use 0.65 (MPG) & 0.75 (power) of peak HP? So, I did that.

I found that when I use 75MPH top speed, peak horsepower RPM, a 0.65 multiplier (for economy) and 29" tires that I get 4.13 ratio. Which is incidentally the ratio that Toyota put in my truck from the factory with the original tires (225/75R15 is about 28.5).

Feeling lucky, I put my current parameters through (32.6" tires, left top speed at 75MPH) and it gives me 4.64 for mileage and 5.36 for power. Again, pretty cool, since the charts all tell me to use 4.56 with 33" tires to retain stock gearing and I used 5.29 and feel like I'm driving a Japanese Power Wagon.

Finally, I put in your parameters (32.6", 0.705 OD, 80MPH top speed) and left the HP peak at 4800 (the 5VZ-FE had a peak of 190 HP at 4800 RPM) and it pops out 4.90 for mileage and 5.65 for power. The multipliers were still 0.65 and 0.75.

Tweaking the multipliers to 0.64/0.74 gives me the ratios of 4.57 and 5.28 for a 5-speed at 75MPH. If I use 0.60/0.70 for your truck with the 0.705 overdrive automatic, I get 4.82 and 5.63 for 75MPH and 4.52/5.28 for 80MPH.

Interesting, eh?

 

[Scott Brady]

I do think using HP is the key over torque, though. Just curious, why do you specifically use torque and not HP?

Typically, modern engines operate at their greatest efficiency at 80% of peak torque. Torque (rotating mass and force) has a greater influence on sustained speed than HP does. HP is an RPM tool (watts), especially in a DOHC gas motor, where the RPM band is much wider than a diesel.

So, I am of the opinion that torque is the most critical asset of a motor (for our purposes) as it is the turning force of the engine (ft-lb or N-m).

 

[Scott Brady]

But, yeah, that's my point. Dynamometers don't measure HP, they measure torque and RPM, leaving HP to be calculated from that. Calculated pwer is completely dependent on measured torque.

Well said, and the reason I choose not to refer to HP, as torque is the true unit of measure we are most interested in.

BTW, the definition of torque is not force through a rotating mass, but actually a force at a point through a lever arm (a vector and an angle). Power is produced by applying a torque through a rotating shaft. The rotational speed of the shaft is RPMs and that's where the time component comes in. That eventually translates to work by applying the power to the wheels and actually moving the truck.

Thank you for the additional detail. The rotating mass of the motor is what generates the torque, which is measured as force (which you indicate in the lever comment above). That is why I referred to the rotating mass, not as the unit of measure, but as the physical action.

As you mention, HP is a calculated measurement:

A foot pound of torque is defined as the twisting force necessary to support a one pound weight on a horizontal bar (which has no weight), one foot from the fulcrum.

So engineers determine how much power is required to do something and then design an engine or motor that generates enough torque at some desired operating RPM to do that. It's all interdependent.

I'm wondering where the 80% of torque peak comes from? Has this got to do with the volumetric efficiency at some RPM? I honestly don't know, just asking.

As you mention, a vehicle motor is designed for a specific task and operating range. So I took the liberty to assume p ) that peak torque would provide the greatest performance within that range (reasonable), but not necessarily the greatest efficiency (not all of the energy is required for cruising). Cruising at 80% (2,880) of peak torque (3,600 rpm) for my Tacoma 3.4L generates the greatest MPG, and leaves a range of about 1,000 rpm's for climbing and passing. After 1,000 rpms above 2,880 the torque band drops considerably, and I am only left with (time or RPM) in the power equation, but force dips...

 

[Pskhaat]

As to the HP vs Torque debate, a factor that can be generically used (and multiplied by something to scale out) is the stroke/bore ratio. An overstroked engine will live better on lower revs and generate lower torque curve vs an overbored engine which can pump the RPMs and have it's torque higher in the RPM range.

2F = 102mm/94mm = ~108.5%

Which could somehow be used as a multiplier to raise/lower the gear ratios. But we always have the anomolies like this:
[FONT=Arial, Helvetica, sans-serif]
22RE = 89mm/92mm = ~96.7%

Which means the gears would be even lower against intuition on the 22RE. Okay, let's keep on going. Maybe the rated torque/hp max:

2F = 3600rpm/1800rpm = 2
22RE = 4600rpm/3400rpm ~= 1.35

Combine them linearly and get a correction factor of

2F ~= 2.17
22RE ~= 1.31

For the orignal posts where the 2F was listed at 2.125 with the correction factor comes ~4.61 for the 2F (not bad). For the 22RE listed at 3.56 is ~4.66 (not bad).

Thoughts on the correction factor of:

(stoke * HP peak) / ( bore * Torque peak)


[/FONT]

 

[flywgn]

Here's a link to a site I've used for a couple of years.

http://www.4lo.com/calc/gearratio.htm

Allen