It wasn't too many pages ago you were calling me a dodge lady. I hope you can see that I'm just trying to help everyone understand the truth, as well as I can percieve it anyway.

 

Being of a math/science background, I'm always interested in seeing equations, even theoretical/hypothetical ones, but I also believe that they need to be grounded in real-world application if they are going to be used to compare two or more systems. That being said, I see some potential weak points in the equation presented. For one, what is the stall speed of the converter in the Torqshift, and what is the flywheel torque of the PSD at that speed? In other words, you use 560 ft-lbs. of torque at take-off, but isn't that the peak torque (at a higher rpms), not the torque at take-off engine rpms? Second, why didn't you use stock rear-end numbers? I just went to the Ford build website, and from the factory, you can't get a 4.10 rear-end with a PSD and Torqshift, only 3.73. Let's compare what Dodge and Ford produce, not stock Dodge with modified Ford, otherwise we'll just degrade into "who can build the best custom vehicle" (or, "who has the most money").

On edit: Dodge offers both rear-end ratios with the Cummins manual, 3.73 AND 4.10.

Now, the rest I would like to see supported with real-world tests. The rear-wheel torque numbers you come up with seem out of sight. I don't believe that there is a driveline made for production vehicles that could withstand 10,000+ft-lbs of torque. The Torque Converter may multiply torque, but gearing (tranny and differential) does not. If it took that magnitude of torque to get a vehicle rolling, I submit that you would not be able to push a vehicle on flat ground. What we're after is the amount of horizontal force that can be applied by the tire against the ground. How much force does it take to push a vehicle (in neutral, of course)? 100 lbs? 200lbs? On a vertical incline, it would take a force greater than the weight of the vehicle plus any friction of internal components, so it would be in the neighborhood of 7500 lbs or so (depending on the vehicle). That number drops off drastically as you reduce the incline.

The bottom line is, we would need rear-wheel dynamometer results. Marine Ironman, I'm curious as to what discipline of engineer you are. And, before you start accusing me of being a Dodge troll, check my signature (I own 2 Ford trucks), and let me say that I also own a Dodge minivan, and have a bro-in-law who used to work on the Dodge Cummins (as a mech engineer, about 2 years ago), but my only brand loyalty is that I don't like Chevys (I like both Fords and Dodges for various reasons). I said that just to let you know that I am up-front and honest, not interested in "tweaked" or "skewed" numbers, and would seek to avoid doing that myself.

Jason

 

The torque converter 1.86 number is a specification given with the TorqShift transmission documentation. It is the stall torque ratio. Secondly, when using an automatic transmission for towing, the reason autos are recommended for towing is because the engine can rev up higher, while the transaxle is stationary or moving slower. In this case the PSD can easily go to 2000 rpm, it's max torque area.





On edit: Dodge offers both rear-end ratios with the Cummins manual, 3.73 AND 4.10.

Now, the rest I would like to see supported with real-world tests. The rear-wheel torque numbers you come up with seem out of sight. I don't believe that there is a driveline made for production vehicles that could withstand 10,000+ft-lbs of torque. The Torque Converter may multiply torque, but gearing (tranny and differential) does not. If it took that magnitude of torque to get a vehicle rolling, I submit that you would not be able to push a vehicle on flat ground. What we're after is the amount of horizontal force that can be applied by the tire against the ground. How much force does it take to push a vehicle (in neutral, of course)? 100 lbs? 200lbs? On a vertical incline, it would take a force greater than the weight of the vehicle plus any friction of internal components, so it would be in the neighborhood of 7500 lbs or so (depending on the vehicle). That number drops off drastically as you reduce the incline.

The bottom line is, we would need rear-wheel dynamometer results. Marine Ironman, I'm curious as to what discipline of engineer you are. And, before you start accusing me of being a Dodge troll, check my signature (I own 2 Ford trucks), and let me say that I also own a Dodge minivan, and have a bro-in-law who used to work on the Dodge Cummins (as a mech engineer, about 2 years ago), but my only brand loyalty is that I don't like Chevys (I like both Fords and Dodges for various reasons). I said that just to let you know that I am up-front and honest, not interested in "tweaked" or "skewed" numbers, and would seek to avoid doing that myself.

Jason[/QUOTE]

 

I guess you have never run a 4x4 then. Lets run the numbers.

Engine produces 560lbs@2000
First gear = 3.09 to 1
4lo = 2.72 to 1
Rear end = 3.73

560 X 3.09 = 1730
1730 X 2.72 = 4706
3460 X 3.73 = 17553

This is without any multiplication.

 

I got cut off on my previos post. here is the rest from Ironman:

IRONMAN: for purposes of this comparison, it doesn't really matter. It's the relative comparison that is the point. You can multiply both by 3.73. Also ... just so you know ... I had to pay extra to get a 3.73 on my excursion, and the 4.10 can be had. Check w/ a dealer. They options are 3.73, 3.73LS, 4.10 and 4.10LS (limited slip). I also own a Dodge, and I got it with the 4.10 rear end for towing purposes.

IRONMAN: Au contraire amigo. At each gear, the torque is stepped up and the RPM is stepped down (or vice versa) The product of the Torque and RPM remains the same (i.e. energy is conserved). For example, if an engine like cummins puts out 600 ft-lbs at 1600 rpm ... then ... goes through 5th gear (1.0 ratio) and then goes through a 4.0 rear differential, we then have 2400 ft-lbs being applied at 400 rpm. 400 x 2400 = 960,000 ft-lbs of work per minute. Also 600 x 1600 = 960,000 ft-lbs of work per minute. Energy is conserved. In reality there is some loss through the drive train, but energy is still conserved. It is still there in the form of heat from frictional loss.

IRONMAN: Of course it doesn't take that amount of torque to move just a vehicle. If the truck is empty the actual torque output will be much less, because the load on the engine is less. Those maximum torque numbers, from engine curves, tell you what the engine MAY put out under the highest load that it can take.

EXAMPLE: What would it take to "ellicit" 600 ft-lbs of torque at 1600 rpm? It would take 960,000 ft-lbs per minute above x 6.28 radians per revolution is 6,028,800 ft-lbs over that minute. At 1600 rpm, a Dodge/Cummins is going about 50 mph (my guess, but you can redo this with the exact number) that is about ... oh ... 4400 feet per minute. So this work does 6,028,800 ft-lbs over a minute and moves 4400 feet in that minute. Divide, and we see that the load for that was 1370 pounds. That 1370 pounds represents the TOTAL RESISTANCE of the entire vehicle to moving forward. Friction in the trailer bearins. Wind resistance. Etc. Depending on the coefficient of friction and wind loading, you can make your own guesses as to the total combined weight that can be pulled. Shape of the truck/trailer makes a big difference.

IRONMAN: Have you ever seen that guy that pulls train locomotives with his teeth? All this depends on the coefficient of friction.

IRONMAN: Yes and no. To do this analysis, we can take Dodge and Ford's claims for their torque-produced-under-load curves. Secondly, I hold my first degree in Naval Architecture (ship design) which covers engines, diesels, steam turbines, gearing, nuclear power, jet engines, nozzles, fluid dynamics, thermo, and etc. Second degree is a master's in System Engineering.

IRONMAN: Okay, good.

 

These calculations are all well and good for those among us that like theory
Has anyone actually dynoed a couple of Dodge Cummins Manuals vs. A torqshift PSD?
I think it would be interesting to know what is actually at the rear wheels.
Also overall tyre diameter will make a difference in the "effective torque" available.

 

Dynos? You may as well have the engine's horoscope read. No two dynos are the same. You don't have any clue between two different dyno's as to the load used, dynamics of the load, calibration of the load, and so forth.

Then most people who dyno their engines have modified them anyway.

Ironman

 

I think it's Truck Trend magazine that put a PSD/auto up against a Cummins/manual and the Dodge flat out got spanked. It should be on newstands now.

 

Likewise, no 2 engines are the same. I would bet few engines would produce the torques quoted above, some maybe more, some less, although production variation has improved over the years, there is still can be around a 5% margin (either +ve or -ve) between quoted figures, and actual figures measured off an engine dyno.
If it was done on the same dyno under the same conditions (ie. perfect scenario), with around 10 of both types of vehicles you could expect some good results, but I, like you doubt this would have ever been done. But it would be far more useful than multiplying torques, as there are so much room for error (have we accounted for driveline losses?)

 

So did the Duramax. And this was the 310/590 DMax and 325/600 Cummins that got whooped by a 325/560 PSD in 0-60, 1/4 mile, 1/8 mile, towing accel., and several other categories.

 

I think we all need to take a step back and remember 1 thing about the cummins motors.


It's won the grand national truck pulling compition for the last 10 years - how can you argue about that. The Dmax and current PSD are not even in the ballpark. It's an awsome low speed puller - BOTTOM LINE. If you disagree go to a pull - I've been to many.

 

DMAX-HD, I believe this is a stock comparison. The trucks you are comparing are far from stock.

 

If someone were looking to extremely modify a diesel truck engine, it looks like the Cummins is the clear choice then(just look at the banks sidewinder). Not related to this thread though... continue.

 

It's not the PSD's fault that Banks didn't decide to use the PSD and went with the Cummins. If Banks said that the reason was because a PSD wasn't capable, then their engine selection would be relevant, but it isn't, and again, this is far from stock, and it's not even doing what diesel engines are designed in principle to do.

 

who says you have to engage the cummins at 1000 RPM? with a manual trans mission, you can engage the cluthch at any rpm. so you can re calculate the dodge using the 600 ft/lbs. I have driven a manual for years, I have yet to engage any engine at or below 1000 rpm, gasoline or diesel, and I have drien both PSD and cummins trucks with a manual.