A few different people had been asking me how I calculated the rolling resistance for larger tires in order to get the most efficient gearing.

As some might recall I said that the "perfect" gearing (as close to stock) for 33's on a AX-15/4.0 XJ was around 3.5 to 3.8 (depending on the assumptions made)

SO, a few asked me to calculate this for them, but I didnt have anytime, so I finally found the calculations and I thought I would post them, if you have rudimentary math skills and a calculator you should be able to do calc.

This calculation has some conservative assumptions, flat road, highest Cd et cetera.

I did this calculation for my XJ that has the AX-15 (0.79 5th), 4.0 and 3.07 gears. Originally it had 215-75-15 (27.7in) tires.

I then calculated at 70mph, what the engine speed was = 2060.145 rev/min

the equation for rolling resistance is
Frr=Cr*M*g+Cd*Af*0.5*p*V^2

Cr=rolling resistance coefficient for tires (~0.01 to 0.015 for tires on concrete)
M= mass of vehicle (~1700kg)
g= gravitational constant (we do drive our Jeeps on earth, haha (9.807 m/s^2)
Cd= drag coefficient (~0.57 early 90s XJ, I have seen as low as 0.47 for later models, but you have to do some research)
Af= frontal area (width*[height-clearance]) 2.579m^2
p= air density at your altitude (~0.95 kg/m^3 here at ~2km)
V= velocity of the car (112.63 kph which is 70mph)

so if you plug everything in
Frr=933.819 N or 209.931 lbf resisting the Jeeps forward movement.

NOW
F= tangential force exerted by the tire on the road.
I used F=Frr to find the steady state (no acceleration)

so to find what the engine output is at F, you simply model the drive train

so F=933.819 N
Taxle=F*D/2 (this is the force at the tire contact patch multiplied by the radius of the tire)
Taxle=328.51 N*m

Raxle=3.07
Rtrans=0.079

Tflywheel= Taxle/(Raxle*Rtrans) you divide because the gear ration is input to output

Tflywheel,=135.451 N*m or 99.903 lbf*ft

So you need 99.903 lbf*ft output by the engine to maintain the 70mph

First you have to re-calculate the rolling resistance for the bigger tire (mass consideration) mine were about 100 pounds heavier all together and I fudged the Cd and Cr slightly

So for the larger tires the Frr=968 N
also which is not too huge a difference, the main culprit will be the higher "gearing" the new height of the tire will produce

so work it again for F=968 N and D=33in

now you want the Tflywheel to equal 135.451N*m, so solve for Raxle

the equations are Taxle=F*D/2 and Tflywheel=Taxle/(Raxle*Rtrans)

so solve for Raxle

Raxle=(F*D/2)/Tflywheel*Rtrans
Raxle=3.791

So there you go, 215-75-15 with 3.07 gearing will perform the same as 33x12.5x15 with 3.791 gearing (with the assumptions made)

 

might be fine for the street but lower is better on the trails put 4:56s in it and call it a day

 

x2. Why make it difficult. Lol

 

Do you by any chance happen to be an engineer? I found this quite informative.

 

I love the math enthusiast but I found a calc online http://www.grimmjeeper.com/gears.html

 

yes, Mechanical Engineer

 

that doesnt take into account drag from wind, or mass or tire rolling resistance, its just a gear calculator which can be done with simple multiplication

 

I think this is one for Mythbusters!

 

I used pretty conservative numbers for the coefficients, it could be argued that even less gear is required.

from a purely gear ratio stand point

the ratio required to have 215s=33s is 3.657 axle ratio (@70mph)

this will leave the engine speed and road speed at 70mph to be exactly the same for the 215s and 33s