Why 4.1 axle?
I tow an 8000 pound trailer with a GMC 2500 with 6.0l engine, 4 speed transmission, and 4.1 axle ratio. I rarely need 3rd gear on the highway; only on steep hills. For my next truck, why not get si...
Forum Discussion
Getting lost in the weeds of detail...PM asking for explanation because you guys have lost them in the weeds
HP = torque x RPM / 5252 is the formula to figure HP (and is how much work it can do)
Gear boxes...from the tranny connected to the ICE, to the transfer case (if there is one), to the differential all have gear ratios that play with each other...or modifies their input shaft torque/HP by the gear ratio and some times changed direction of that rotation
Example using an even number for discussion : 100 ft/lbs of torque at the ICE flywheel or flex plate
1st gear of any tranny is, say 6:1. Means that 100 ft/lbs is multiplied by 6, to have 600 ft/lbs at the tranny output shaft
To the transfer case, but let's say there isn't any for this discussion
To the drive shaft that connects to the diff input shaft with 600 ft/lbs of torque
The diff has a 3.73:1 gear ratio. Means that 600 ft/lbs is multiplied by 3.73, to have 2,238 ft/lbs as it's output, plus this gear box changed direction by 90* and splits it to two axles, both having the same torque...that will change if open or LS or locker, but forget that to keep it simple...
Each of these gear boxes also changes their input RPMs to the diff by their gear ratios. Therefore, an increase in HP, reference the formula posted above
Just using 1st gear. The other gears inside any of these gear boxes will have different ratios and might have an over drive, so the gear ratio might be 0.7 or something like. So multiplying the torque or RPM by that number will have less torque and less RPM at the output shaft
So what these guys are talking about are various tranny's (gear boxes) and their different gear ratios that then has different ft/Lb multiplications from their input shaft to output shaft.
Another point being discussed is the 'spread' of ratios from 1st to 2nd to 3rd...etc. This is to keep the ICE spinning in it's best RPM range (often referred to as 'power curve'). This varies from type of ICE, to the architecture of that ICE. Most ICE's for this discussion spin up to around 5K-7K, whereas an ICE with a Desmodromic valve architecture (no springs or rocker or push rod) can spin up to or well over 16,000 RPM. There are some very specialized ICEs with OHC that can spin up there, but they are F1 types, which cost in the $1Million per engine range
So in 1st...it has the highest multiplier ratio, but will run out of RPMs (red line) and then the tranny needs to shift into 2nd, a higher (lower numeric) ratio to continue the process...all the way up to the top gear (1:1), then into OD if your tranny has one or them...
HP = torque x RPM / 5252 is the formula to figure HP (and is how much work it can do)
Gear boxes...from the tranny connected to the ICE, to the transfer case (if there is one), to the differential all have gear ratios that play with each other...or modifies their input shaft torque/HP by the gear ratio and some times changed direction of that rotation
Example using an even number for discussion : 100 ft/lbs of torque at the ICE flywheel or flex plate
1st gear of any tranny is, say 6:1. Means that 100 ft/lbs is multiplied by 6, to have 600 ft/lbs at the tranny output shaft
To the transfer case, but let's say there isn't any for this discussion
To the drive shaft that connects to the diff input shaft with 600 ft/lbs of torque
The diff has a 3.73:1 gear ratio. Means that 600 ft/lbs is multiplied by 3.73, to have 2,238 ft/lbs as it's output, plus this gear box changed direction by 90* and splits it to two axles, both having the same torque...that will change if open or LS or locker, but forget that to keep it simple...
This is the OP's topic/question...why a higher numeric ratio...it provides more torque because it has a higher numeric multiplier 4.1:1, which will provide 2,460 ft/lbs
I'm planning to go to a 5.3:1 on when update my Sub, which will provide 3,228 in this exampleEach of these gear boxes also changes their input RPMs to the diff by their gear ratios. Therefore, an increase in HP, reference the formula posted above
Just using 1st gear. The other gears inside any of these gear boxes will have different ratios and might have an over drive, so the gear ratio might be 0.7 or something like. So multiplying the torque or RPM by that number will have less torque and less RPM at the output shaft
So what these guys are talking about are various tranny's (gear boxes) and their different gear ratios that then has different ft/Lb multiplications from their input shaft to output shaft.
Another point being discussed is the 'spread' of ratios from 1st to 2nd to 3rd...etc. This is to keep the ICE spinning in it's best RPM range (often referred to as 'power curve'). This varies from type of ICE, to the architecture of that ICE. Most ICE's for this discussion spin up to around 5K-7K, whereas an ICE with a Desmodromic valve architecture (no springs or rocker or push rod) can spin up to or well over 16,000 RPM. There are some very specialized ICEs with OHC that can spin up there, but they are F1 types, which cost in the $1Million per engine range
So in 1st...it has the highest multiplier ratio, but will run out of RPMs (red line) and then the tranny needs to shift into 2nd, a higher (lower numeric) ratio to continue the process...all the way up to the top gear (1:1), then into OD if your tranny has one or them...
