Changing from 3.73 to 4.10 gears

[oakmandan]

Hello all,
After towing our new Reflection 320MKS from Eastern Iowa to Estes Park Colorado and back, I need more power for the hills. I have a 2015 F-250 with the 6.2 Liter engine. I have 3.73 years. Can they be swapped out for 4.10 gears, and will it help ?
Thanks

[Turtle_n_Peeps]

Ben, here is where I'm at. These guys say it much better than I can in limited time.

Here is a calculator you can play with.

As Thomas Andrews would say: "tiz a mathmatical certainty."

If you play with this calculator for awhile you will know if you have an engine that makes 1000 ft/lbs at 1000 RPM you "will" make 190.4 HP. "Tiz a mathmatical certainty." If you jump the RPM to 2000 and keep the torque at 1000 ft/lbs tiz a mathmatical certainty you will get 380.8 HP.

I "know" 5252 is the point that torque and HP will cross on a chart. "tiz a mathmatical certainty."

All the above are mathmatical certaities just as one will never make power (read HP) with gears.

[rhagfo]

4x4ord wrote:


To pull your 20500 lb unit up a 7% grade at a steady 55 mph would require very close to 290 rear wheel horsepower. (Exactly 210 HP to raise the load up the elevation gain plus another roughly 80 HP to overcome wind and rolling resistance) That means something like 342 engine horsepower. So long as the vehicle is geared appropriately it makes no difference whether that 342 HP is being generated by a diesel engine running at 2500 rpm, a gasoline engine running 4500 rpm or an electric motor running 1750.


Edit: I corrected a calculation error but even if your truck is capable of putting 85% of its engine's power to the pavement it would only be capable of 50 mph pulling 20500 lb gcvw up a 7% grade... and this assumes only 65 hp is being used to overcome wind and rolling resistance. I'm thinking the hill you've measured your speed on is not marked accurately or is not a constant 7% grade.

I forgot to mention that at the base of this climb is a 45 mph corner that is truly 45 mph when towing so I also accelerated to 55 while going up this grade.

[tatest]

That is not enough of a change to be worthwhile, and if it won't stay in the highest overdrive gear on 3.73, it likely won't with a 4.10. With a six-speed automatic, you have a selection of overdrives, the automatic will figure out which one is needed for the load and speed at any given time.

Most of the time when manufacturers go for steeper final drive, it is to raise the GCWR, get the overall gearing low enough to meet the standards for starting from dead stop on a grade, taking the hit on MPG at highway speed unloaded.

[4x4ord]

rhagfo wrote:

Turtle n Peeps wrote:

parker.rowe wrote:

Huntindog wrote:
Ford 2015 towing guideShows that a 4.30 truck can tow 3000# more than a 3.73 truck.

Yeeeep. Guess the people who designed the truck say they make a difference.

Yes, that's because a short gear will give you a better duty cycle. That's why a 450, 550 with the same engine tranny combo will give you a more choices of shorter gears.

Here is a great article that tells you about the myth of torque vs HP. Read the article and then read #1. The guy is spot on if you do the math.

I have an issue with the last part of this statement.

Enough theory: Lets see an understandable example of torque and horsepower

Let’s say that you are loading the back of your pickup with rocks. The bed is two feet high, and you lift six fifty-pound rocks in one minute. You have done 2 feet x 300 lbs = 600 ft-lbs of work. Power is determined by how fast you do this work, so you have done 600 ft-lb of work in 60 seconds, requiring 600 ft-lbs / 60 seconds = 10 ft-lbs/sec of power. A horsepower is 550 ft-lbs/sec, so you have used 10 / 550, or .018 horsepower. So, now let’s say that you work twice as fast, and lift 12 rocks in a minute in place of the 6. You have now loaded your pickup using .036 horsepower. Note that in both cases, you have exerted the same force (torque) when lifting each rock of 2 ft x 50 lbs = 100 ft-lbs. In the second case, however, it required twice as much power with the same amount of torque.
This is because the torque was occurring at a higher rate of speed.


So it isn't the same amount of torque, 12 rocks at 50 lbs ea is 600#, 2' X 600 lbs = 1,200ft-lbs of work.

So 1,200 ft-lbs / 60 sec = 20 ft-lbs per sec of work, or .036 Hp

In the second case you did twice the work in the same time, requiring twice the HP.

This is great in Theory and math formula.

It doesn't seem to workout in real life, most HD Pickups gas engines High HP/Low Torque ratio top out at about 16,000# towing capacity with most needing 4.30 to 4.60 gears to get there at near 4,500 to 5,000 rpm.

Today's Diesels with low hp/Torque ratio are rated to to tow nearly twice that amount about 30,000 lbs with 4.10 gears and maybe 3,000 rpm.

I my personal case my 2001 Cummins factory rated at 235 HP and 410 ft-lbs of torque, is likely near 300 HP and 610 ft-lbs of torque. I have added a DS Power Puck (50 HP and 140 ft-lbs claimed increase) and a set of RV275 injectors. The math is 275 + 50 = 325, but in reality not likely cumulative increase so about 300 HP. Torque 410 + 140 = 550, have never seen a torque rating for RV275 injectors so 60 ft=-lbs is a guess. My TV has 3.55 gears currently the GCVW of my rig is 20,500 lbs, I can put that up a 7 percent grade at 55 mph in direct drive (4th gear) at about 2,500 RPM.

So while love Theory and Math, when it comes to the power difference between a gas engine and a diesel it doesn't always seem to work out.

To pull your 20500 lb unit up a 7% grade at a steady 55 mph would require very close to 290 rear wheel horsepower. (Exactly 210 HP to raise the load up the elevation gain plus another roughly 80 HP to overcome wind and rolling resistance) That means something like 342 engine horsepower. So long as the vehicle is geared appropriately it makes no difference whether that 342 HP is being generated by a diesel engine running at 2500 rpm, a gasoline engine running 4500 rpm or an electric motor running 1750.


Edit: I corrected a calculation error but even if your truck is capable of putting 85% of its engine's power to the pavement it would only be capable of 50 mph pulling 20500 lb gcvw up a 7% grade... and this assumes only 65 hp is being used to overcome wind and rolling resistance. I'm thinking the hill you've measured your speed on is not marked accurately or is not a constant 7% grade.

[BenK]

Spot on and add two more attributes

There is one more torque multiplier if there is an automatic in this food chain of torque from the ICE to pavement...TC's have a torque multiplication factor in the average range of 15%-20%, so your formula below needs to have that additional torque factor...until the TC locks up...then that slippage/multiplier goes away

Other is that today's computer control system has 'torque management' as one major control factor in how much torque is allowed to be delivered

Why with today's levels of power...smoking the tires is tough to near impossible to do

Huntindog wrote:
Torque matters.
All of the Diesel guys are debating when the TQ numbers will go over 1000, and which brand will do it first...

Let me show in simple math why the engine race isn't all it's cracked up to be.

Lets say your motor has 500 TQ.
And your 1st gear is 6 to 1.
You will have 3,000 TQ!, leaving the tranny.
If your differential ratio is 3 to 1, you will have 9,000 TQ!, to the rear wheels.

If the diff is 4 to 1, it will be 12,000 torque!!
These are BIG numbers, so big that it makes the manufacturers adding 20-50 #s to the motors, seem like small potatos.... Of course an extra 50 TQ would mean an extra 900-1,200 to the rear wheels.

That is the secret magic of gearing.

[rhagfo]

dodge guy wrote:
With a 3500lb CCC he will be hard pressed to get anywhere near that 14k lb rating!

True, But got the “We watch closely how we load it” statement! Why have a nice big trailer if you can’t carry the items you need to enjoy camping!
With a 10,500# Dry weight they should easily be at 11,500# ready to camp, and lock out 5th and 6th gear.

[Huntindog]

Torque matters.
All of the Diesel guys are debating when the TQ numbers will go over 1000, and which brand will do it first...

Let me show in simple math why the engine race isn't all it's cracked up to be.

Lets say your motor has 500 TQ.
And your 1st gear is 6 to 1.
You will have 3,000 TQ!, leaving the tranny.
If your differential ratio is 3 to 1, you will have 9,000 TQ!, to the rear wheels.

If the diff is 4 to 1, it will be 12,000 torque!!
These are BIG numbers, so big that it makes the manufacturers adding 20-50 #s to the motors, seem like small potatos.... Of course an extra 50 TQ would mean an extra 900-1,200 to the rear wheels.

That is the secret magic of gearing.

[BenK]

It would help if you rode a multi-speed bicycle...if not...either look at one or rent/borrow one

Then ride it in the lowest gear and highest gear....noting the dia/size of tge rear gear in which you are pedaling...forget anout the crank/pedal gears (they are called chain ring gears)...as they "seem" to work in the opposite manner (ask if you wish info)

All the while paying attention to the effort vs speed

Ben
Via smartphone...please excuse my fat finger typos

[time2roll]

oakmandan wrote:
Thanks for the input. I have mastered the 6 speed and lock out 6th and 5th most of the time. The truck is not an everyday driver. It is the 5th wheel puller and Mama's grocery getter. I am no where near my GVWR. We watch closely how we load it.

Considering it is not used for general commuting and you have the top two gears locked out.... I would go 4.56/4.88 and start using those top two gears. 4.30 minimum if spending money. If you get off pavement much also consider a pair of TrueTrac differentials at the same time.

[BenK]

Turtle is one of my reference guys...agree and kinda sorta disagree on this one...

Correct in that gear ratios do NOT change the input torque....it mulitiplies that input torque via the ratio between the input gear (pinion) and driven gear (ring gear).

Multiplication can be zero (1:1)...to less (2:1)...to more (1:3.73)...etc, etc.



Here is a site that explains it well

How do gears multiply torque?

quora.com wrote:




Click For Full-Size Image.



Assume the smaller gear is driven by a motor producing X units of power. This energy is used to rotate smaller gear. To turn the gear faster you'll need more power. Smaller gear in turn rotates the bigger (driven) gear.

If smaller gear rotates 4 times per second, bigger gear will rotate only once per second. {because, smaller gear (9 teeth) has to rotate 4 times to traverse all the 36 teeth of bigger gear.}


Now,

Power is constant (since energy is conserved)**
Bigger gear rotates 4 times slower than smaller gear {i.e. Angular speed is 1/4th the smaller gear}


We can easily see,

Power = Constant
Angular Speed1 = 4*Angular Speed2
Torque1*Angular Speed1 = Torque2*Angular Speed2
Hence, Torque2 = 4*Torque1 {i.e. Bigger gear produces 4 times bigger torque. Torque is multiplied.}


If you get an opportunity to observe gearbox of an automobile, do observe the sizes of 1st, 2nd and 3rd stage gears. You'll observe the 1st is the biggest gear and then size reduces successively. This is because initially car needs maximum torque and bigger gear would result is larger torque multiplication

Forget about the rear gear box...AKA differential, for a moment and look at the gear box ahead of the diff

It has an even higher numeric (AKA...lower gear ratio) than the rear gear box...why ?



The exact same ICE power source for this discussion and use 200 ft/lbs as an easy reference to noodle in this discussion


Say the gear box bolted to the ICE via the bell housing will have 200 ft/lbs on the input shaft.

Quick search got this:

• Ford F150 10 speed is 4.69
• RAM 8 speed is 4.71
• GM 6 speed is 4.03

So all about the same ratio's and if you look at their second gear box (diff) ratio's...most are in the 3.xx range

So they all have about the same gear box ratio from the transmission (gear box #1) to the differential (gear box #2)

Hold that and now to how a gear box ratio works...simply put...gears either or both change the direction of rotation to transmit torque...and multiplies that torque (minus losses)

Changes direction from straight on, in line with input shaft...to off set. Most common is 90*


So, with the exact same ICE in first...the torque multiplication in 1st has higher torque by multiplication...say 6 times the base input shaft torque. That is what gets it moving...but will run out of steam as the ICE reaches it's red line or whatever the computer says to shift into a higher gear


2nd would be that higher gear and has a smaller multiplication. So less torque, but with the ability to spin the output shaft faster, but with less torque multiplication

3rd, 4th, 5th, etc....are using less and less torque multiplication for more RPMs, but less torque...as the mass is no moving and does NOT need to get it moving from a standstill..

All of that is then modified by the rear gear box's (diff) ratio, but is singular gear ratio...

[dodge_guy]

With a 3500lb CCC he will be hard pressed to get anywhere near that 14k lb rating!

[goducks10]

enblethen wrote:
specs for 320MKS show GVW as 13,995. That is real close to 14,000

Ooops didn't scroll down far enough.

[oakmandan]

Thanks for the input. I have mastered the 6 speed and lock out 6th and 5th most of the time. The truck is not an everyday driver. It is the 5th wheel puller and Mama's grocery getter. I am no where near my GVWR. We watch closely how we load it.

[enblethen]

specs for 320MKS show GVW as 13,995. That is real close to 14,000

[goducks10]

Crabbypatty wrote:
I just looked up your fiver and its max is 14,000# thats places you 2000# over the limit of what you can tow. Find a scale and see what you weigh as you might be overloaded for the truck set up.

Where are you getting the 14,000 from?

https://www.kz-rv.com/2016-products/durango-1500-fifth-wheels/2016-D281RLT.html

Looks like a perfect match IMO for a 3/4 ton gasser.