3:42 rears

@REALITY098765 /Troll

 

Sorry for bursting some bubbles but
Facts are facts.
Dispute them if you can so we all learn.

 

@REALITY098765
Many members on this forum(and this thread) have huge experience and knowledge in this industry. Like Merrill Lynch, when these people speak, everyone listens!!!
Prove you aren't a troll. Back up your words and prove them wrong!

 

Those are ALL copy and paste from the WWW.in post 39.Not my words directly but what I believe, just facts.
So I did back up my words with facts but I can see where you thought differently.
Prove THEM wrong with facts,

Facts. Not conjecture.

 

@REALITY098765
Your right, you win. I'm wrong, I lose. I'm sorry.
(This is very similar to "conversation " with the EX..)..

 

What you say here is 100 percent correct. The issue is that you're speaking in terms of torque applied to the axle shaft. Everyone else here is talking torque applied to the ground. We're talking the "effective" gear ratio, which is derived from a combination of the transmission gear ratio, any aux trans ratio if equipped, rear end ratio, and tire size.
The radius of the tire is like a bar on your ratchet tightening a nut. Larger the tire, larger the radius, larger the "bar" and the more torque you could apply to the nut with a given force. Simplified formula for torque is Force x radius. T=Fr We can use this simplified formula because we are assuming the force is perpendicular to the radius. Picture the radius extending straight down to the ground from the center of the axle shaft, and the force is acting on the tip of this line, at the surface of the road and parallel to the road. Basically this forms two lines in the shape of an L.
This force, F, is essentially what we feel when giving the truck throttle. So lets go back to our example where T is the torque at the axle shaft. Like you said before, without changing the gear ratio, this number will be fixed. T is a constant. Now we're going to change tire size, thus changing r. If we decrease r, then F must be increased in order to still equal the constant T. Remember T=Fr.
Throw some numbers in to visualise it, lets say T=100, F=25 and r=4 if we decrease r by half for example, r will equal 2. In order for T to remain at 100, F must double to 50.
So ultimately what we have proved is that putting a smaller diameter tire on a truck will effectively increase the torque that you can apply to the ground.

You keep asking everyone to answer "what do the numbers mean" but I have no idea what number you are talking about so I cannot answer that. My only guess is you are referring to the gear ratio numbers which are literally just that, a ratio of the number of times one gear turns in relation to the other. In the case of his 3.42:1 rear end gears, the pinion gear will be turning 3.42 complete revolutions for every one revolution of the ring gear.

 

Correct
And the number of the number of teeth on the crown gear divided by the number of teeth on the pinion gear is what creates that.

And every rotation of the ring gear equals 1 revolution of the tire.
And that ratio doesn't change with tire size.
Only the roll out of the tire changes with tire height.
If you measure the torque at the center of the axle how does tire size change that?
The way you explain it you are measuring resistance. Now maybe if you put the truck in reverse it would work.
That's the way I see it anyway maybe wrongly .

The issue seems to be where the torque comes from.
The relation of the crown to the pinion is where the torque comes from or the longer lever application is created ,not the end of a bar the diameter of the tire. You can measure torque there but not create it [without an outside force]
If we assume an engine produces 1000 lb ft and direct trans gear , that equals 1000 times 3.42 =3,420 lb ft.at the axle,
1000 times 4.10 =4,100 lb ft at the axle.
It's just Torque multiplication.

 

This whole thing is just going completely over your head, I literally said that torque at the axle would remain constant... force applied to the ground by the tire would be changing. I proved it to you mathematically.
Take that 4100 lbft of torque and put it into the formula that I gave you, then solve for F using r for a 22.5 tire and also r for a 24.5 tire. F is clearly not going to be the same for both instances...
Look at formulas for the the pulling force provided by a winch. This is the exact same idea. The winch motor provides a constant torque, but the pulling force varies based on the effective radius. The effective radius changes based on the number of layers of cable wrapped around the drum. You'll have more pulling power when only one layer is on the drum than you will when you have already wound up 5 layers.
Go get yourself a bicycle with the biggest diameter rear wheel you can find, probably a 29" would be my guess. Get on it and see how much effort it takes to pedal up a certain hill. Now swap that rear wheel out for the smallest possible one you can find, probably like a 12" off some kids bike. Make sure you've got the same gear ratio as before. Pedal up that same hill. You won't have to push nearly as hard on the pedals to get moving.
I don't think there exists any more ways to possibly explain this. Nobody is trying to say that changing tire size is going to change the ratio of the gears in the rear end. What everybody is saying is that changing tire size will effect the force that is transmitted to the ground.

 

Well said