A few questions about my 90 4wd 3.0L
 

I hope I make these questions as clear as possible & if not, let me know & I'll try to specify.

1. My transfer case is leaking the Dexron pretty badly. The whole backside of the transfer is covered in grease leading me to believe that the rear output seal is bad considering it's on the top half of the transfer. Am I right in this assumption & how much of a pain is it to fix this?

2. I've heard a few debates over which exhaust headers are the best one's for the 3.0 or even if it's practical to tackle that, but I was thinking there had to be something to improve that rediculously designed exhaust manifold setup. Here's my idea: Take the crossover pipe out completely. Plug the hole in the driver side where the crossover hooked up with an exhaust donut & a piece of steel cut to cover the hole. Now run a pipe out of the passenger manifold to meet with the factory exhaust pipe down low right before the cat. How does that sound? Is it feasible?

3. What have been the best size tires to use overall? With gas mileage, power, acceleration, and off-roading all taken into consideration.

Thanks to all who answer. I love this place!!!!!!

FRED!

i've never done it myself, but friend said it took him two hours to do.

everything i've read said messing with the exhaust is the least effective way of spending your money to increase power on the 3.0. i'd look elsewhere first

gas milage: stiffest you can get - most loss (with respect to the tires) comes from the energy wasted in "squishing" the tire

power: by definition power is conserved (minus frictional losses) in the drivetrain, tire size has no effect

acceleration: you'll get the most torque to the ground the quickest with smaller tires, but you'll reach a slower top speed

off-roading: lot's of choices, BFG KO's seem to be highly regarded as great DD plus off roading tires, BFG MT or Goodyear MT/R seem to get good marks in the serious off road use also with use as a DD. people Firestone Destination MT is a great and cheap mud/snow tire. personally i run Michelin LTX which are primarily a street tire. i want to switch to BFG KO's like my brother has, but i can't justify it until i wear out the Michelin's, but they seem to be lasting forever.

Good post. However, you will not necessarily reach a lower top speed. You will be in a different torque requrement with different tires at speed. I lost 20% in mpg going from stock to 31's on an '84 pu. You won't be able to redline in high gear in either case.

I've had BFG AT's, GY MT/R's, and BFG MT's. Given the Discount Tire warranty, the AT's are my favorite. The MT/R's are better than the MT's on rock in my experience, and a little quieter on the fwy. The AT's are best in snow, not bad on rock, absolutely quiet on the fwy - but they have tender sidewalls, hence the warranty requirement... I've also had a couple sets of LTX AT's - don't underestimate their ability - plus I got over 100 k miles on a set on an '85 4Runner.

hey mike.


Work is force over a distance, and Power is the time it takes to do the work. So a truck with smaller tires will go farther in a given time with the same force (engine) than a truck with larger tires. So... Larger tire = less power. Gearing can greatly increase power.

The idea you expressed is that the trucks energy is constant regardless of it's tire size. This is just a semantic thing really, lots of people treat power work energy as synonyms.

Fred, off road and gas mileage/acceleration are oxymorons really. The best compramise IMO is the BFG AT's or MT's on the agressive side. a 30-31inch should give you clearance without loosing tooooooo much gas.

thanks for the replies! I was leaning towards the BFG AT's before I posted. My truck came with the 31" tire package, so that's what I'll probably go with.

* anyone have any other opinions on the exhaust alteration?

Your first sentence is almost correct. Power is work divided by time. Energy equals work. However, *engines* make the power. Unless you change the engine you don't change the power. The people that treat power and work or energy as synonyms are wrong.

What happens when you change gears or tire size is that you are in a different part of the power and efficiency curves at a given speed, and with tire size changes you change the wind resistance as well.



Steve H

My first sentence is correct, that is as long as yours is because they are saying the same thing differently. I would like to dissagree with you on the idea engines make power. An egine only makes a force ( to the lager system, obviously the pistons are displaced etc ). Work = F*D*cos where F = force, d = displacement, and the angle (theta) is defined as the angle between the force and the displacement vector. Now I have seen in several physic books that they relate power to a car engine as quoted here "A car engine is an example of a machine which is given a power rating. The power rating relates to how rapidly the car can accelerate the car. Suppose that a 40-horsepower engine could accelerate the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this were the case, then a car with four times the horsepower could do the same amount of work in one-fourth the time. That is, a 160-horsepower engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point is that for the same amount of work, power and time are inversely proportional."

With that in mind Tire size and gearing obviously effect the time therefore inversely effecting the power Power= Force*Velocity or Power= Work/Time. One has to reason that more then the engine has a roll to play in the power of the car *as a system.

In respect of your exhaust question , bang for your buck and for modest gains, a 2 1/4" system from the cat back is most effective. A free flowing mandrel bent system from the cat back with a single muffler will be most effective. From everything I have read, a 2 1/2" or bigger system robs you of torque which is something we do not need :)

Also, a free flowing cat may help.

Jury is still out on headers with folks in both camps with most saying they are a major PITA to install plus by the time you have added a cross over or Y pipe to get our new headers to hook up to your exhaust, you cannot possibly get enough performance to justify the $700 you have probably spent by now.

Start with a 2 1/4 " mandrel bent cat back system and go from there.

David

Headers & a 2 1/4" exhaust, mandrel bent, with a free flowing cat and a quality aftermarket muffler is a solid option. However many claim the headers aren't worth the PITA they are to install. So I'd get a 2 1/4" mandrel bent system installed from the exhaust manifold back with a high flow cat and again quality muffler. A guy I know who has an 89 4runner with the 3.0 V6 did the 2 1/4" exhaust from the manifold back, high flow cat and high flow muffler and he says while you loose a tiny bit of low end torque, he says he can rev it to 5000+ rpm's much more easily and he says it just pulls solid right on to the redline. This is what I will be doing at some point.

ok, we're starting to get off topic here, and i apologize to those who really don't care about this stuff, but i want to make sure that the whole energy/work/power debate is cleared up (at least a little).

first - work and energy are the same: DeltaE=F*d
second - power is energy done per unit time (or work done per unit time)

so, a car engine does produce power. it does a certain amount of work in a certian amount of time, that is power. that's why engines are given horespower ratings!

now, back to my original statement that tire size doesn't affect power output. there have been some good comments in this thread about how tire size will affect where the engine is in the power band or the effect of windresistance because a of a taller tire and such. but i'm ignoring that right now, because i'm talking about given the same power output from the engine, the tire size will not affect the amount of power going to the road.

first we have power output from the engine that goes through the drivetrain and finally makes it's way to the rear axle. there will be power losses in the drivetrain, but those will be constant regardless of tiresize. so a certain power output from the engine will map directly to a certain amount of power at the rear axle and that comes from a certain torque applied by the rear axle per unit time. from now on, i'll just be talking about the power at the rear axle. since we're talking about things rotating we need to be talking about torques rather than forces:

W=T*theta (work is Torque applied times the angular displacement - analogous to F*d)

Power is still W/t (work per unit time)

P=T*theta/t (power is the amount Torque applied time the angular displacement divided by the time)

So how does that translate to work done to move the truck?

the distance the truck moves is d=theta*r where r is the radius of the tire, the Force that makes it to the road is F=T/r (remember that T=r x F - so if we assume that all the forces are perpendicular and we don't need to worry about the cross product, then F=T/r)

so the power that makes it to the road is F*d/t , but F=T/r and d=theta*r or (T/r)*(theta*r)/t, the r's cancel out and as expected, the Power at the rear axle is equal to the power applied to the road.

conceptually, it goes like this: a smaller wheel is easier to turn, but you go less distance per turn. the amount that it's easier to turn by is exactly equal to the less distance you go. In terms of power, given the same amount of power, you can turn a smaller wheel faster, but since you're going less distance per revolution, the end result is you go the same speed.

granted, this is the sperical cow example where we don't worry about the extra losses due to flexing of the tire or things like that. but to first order, the tire size will not have any effect on power output.

i've attached a pdf of my work. it's easier to see it written out than to try to follow equations typed in plain text.

Mike, text is much easier than your hand writting. I will spare everyone the headache of trying to read mine but not posting it. I follow your equations but time is left as a variable and if we test this using T*theta/t on a vehicles 0-60 or 1/4 mile time we will see that the time will increase with larger tires. We know this is true from real life exsperance and I do not beleive it is from increase in the tires mass alone.

Mike is correct. Gearing can make things equal, except the wind resistance (major) and tire weight (extremely minor so as to be negligible). The force is determined by the torque of the engine times the gearing.

you guys are good.

i understand what you are saying, but i honestly dont know how to respond...

but, you guys are good. :banger:

Okay, I'll bite.
I see the logical argument you present, but you seem to be conveniently leaving something out - drawing your conclusion prematurely. There must be something missing.
Because in the real world, not on paper but on pavement - there IS a Very Real loss of ability to do work, and a Very Real increase in fuel required to do this work, when larger tires are installed but gears are not changed. So, if the amount of Power Applied is unchanged, the method of implementation certainly matters. Finish your exercise for us, will ya?

I believe I know what's missing in your example. (well, lots of things - like air resistance, resistance to accellerate due to mass, etc. You example could only be true in a vacuum on level ground) Torque. Both horsepower and torque do the work.

A while back I wrote a description of why changing gears is much more effective than adding horsepower to compensate for bigger tires. The visual example I used was moving a rock with a lever. If we use that analogy, then you are saying:

Use a lever and fulcrum to move a rock. For our example we'll use an 9 foot lever with the fulcrum in the center. Lets say it takes one pound of pressure exerted for one second to move the lever through its arc and move the rock. During that one second I burn a certain number of calories.

Now, move the fulcrum closer to the rock, to 3 feet from the rock. Move the rock and it seems much easier due to the 2:1 leverage. Now it takes 1/2 pound of pressure, but 2 seconds to swing the lever though it's arc. In your example I would expend half as many calories per second but in those 2 seconds I burn the same total number of calories by the time the work is done, so the power applied calculates to be the same. At a steady cruise speed that is true, the same amount of power is being applied to move the rock at either fulcrum length.

As you suggest in your example we won't worry about overcoming air resistance (although it's a big deal!) or rolling resistance (which is minor) right now.

So, with either gear combo the same amount of power is propelling the car along at a steady 60 mph. But to accellerate or go uphill and the vehicle with the better leverage ratio will accellerate faster, making more use of the available power. To accellerate the car (overcoming resistance to change inertia) or go up a hill (overcoming gravity) you find that the application of torque matters a lot. Final gearing in the axle is still torque multiplication, just like the gears in the transmission. How much does the ratio matter? I believe the inverse-square law applies to the application of energy and angular momentum. That's what you are missing and that's why your example doesn't match the real world.

well much of the real world difference comes from what steveh said earlier in the thread, and that's different tires/gearing will put you in a different location in the powerband for the engine.

before i give an example, let me state a few assumptions that i think most people here will agree with:
1 - under normal driving conditions, the engine provides more torque at higher rpms than at lower ones (granted, at some high rpm the engine passes through the torque peak and starts to decline, but generally you stay below the torque peak in everyday driving)
2 - the horsepower output, which is a function of torque and rpm is a monotonically increasing function of rpm (again, under normal driving conditions) - horsepower is basically torque*rpm*(some other constant to make the units workout) here's a link for those who care: http://auto.howstuffworks.com/question622.htm

so lets say we're driving down the street at 3000 rpm in 4th gear (1:1 ratio) with a 4:10 rear end. that means that the rear wheels are turning around at 732 rpm. with 28" tires that's about 61 miles an hour. if we wanted to drive 61 miles an hour with 31" tires that's means the engine needs to turn at 2710 rpm. we already knew this: a bigger tire means the engine spins slower to go the same speed. but given my assumptions from above, the slower engine speed means that the engine is producing less power.

so, if you're drag racing, which was an example given earlier, as you accelerate, the engine attached to the smaller tires will rev up faster and generate more horsepower than the engine attached to the larger tires. so that's where the confusion is happening - the engine is generating different amounts of power in the two cases. if we could magically make the engines generate the same amount of power, you wouldn't notice this.

that's also why you get the increased fuel usage with bigger tires (aside from the energy loss from squishing the bigger tire), you need to work the engine harder in a less than optimal part of the powerband to travel the same speed. plus since you accelerate more slowly, you're running more time at full throttle and you burn more gas.

so yes, you do notice what feels like a decrease in power when you switch to bigger tires, but it's only because the engine is operating at different speeds and the different speeds mean the engine generates different amounts of power. if you had a magical engine that would generate the same amount of power at all rpms, you wouldn't notice the difference in tire size at all. (note: electric motors come close to this, and that's why they use them to power trains)

i hope i explained this well, it's so much easier for a nerd like me to explain this stuff on a chalkboard than in plain text in a forum :)

i remember reading that example when you first posted it, and i was going to search for it to help clarify what i was saying, but i forgot to. that example is a perfect analogy here.

Just touching on something that people seem to be ignoring, the fact that larger tires are heavier and have more mass at a farther distance from the center of the wheel. This causes them to be more difficult to start and stop spinning. Your practical real world example from childhood would be the merry-go-round device on the playground. Put 5 kids on it, and have one person start to spin it by running around. If the 5 kids are on the outside of the ring, it will be very difficult to get it going fast, if they are in the inside it will be much easier. Same mass, just different location. Get it spinning with the 5 kids on the outside, have them climb to the center and the thing will speed up, have them move out to the edge again and it will slow down. So yes, tire size and weight do become a factor to what happens to the power generated by your engines. ie, bigger(taller) heavier tires will generally slow down accelleration and braking(traction and friction being even, which it's probably not, but if it were)

Still love it here Fred ?? :D

For my situation where the truck is geared for and came with 31s, I will probably not go any bigger than 32's as this is a daily driver and I really cannot afford to re-gear and I do NOT need to further reduce the lack lustre acceleration.

David

Yep, very true. Another variable that for the sake of convenience we are not considering in our friendly discussion.

And don't forget money! Money removed from your wallet to buy gears lightens the load and makes the vehicle easier to accellerate.... :P