Automatic Transmission Calculations

[Archive]

This is my first post on this
message board but I've been a TDR
member since its inception. I have a question which probably stems from my total ignorance of auto transmissions and torque converters.

I've read numerous articles and even posts in this forum about a "torque multiplication effect" of the torque converter. Admittedly I'm a scientist and not an engineer, but I really don't understand this principle. Unless there's some "gear" reduction in the torque converter, you can't get a higher torque output than the input torque. That would violate laws of physics.

I'm interested in understanding this because I have done some calculations to determine the average torque output of my engine based on a number of 40-60 mph hand timed runs. If there really is some torque multiplication factor... that would foul up my torque calculations!

Can someone explain this principle?



[This message has been edited by wxman (edited 12-13-1999). ]

 

[Archive]

Yes the torque can increase, under the right conditions. With the TC clutch disengaged, the output RPMs are always lower than the input RPMs if power is being transferred from the engine to the wheels. The transmission output RPMs are higher than the engine RPMs if you are engine braking. When you are applying power from the engine, the power into the TC is input rpm times input torque. The power out of the TC is output RPM times output torque. The output power is less than the input power by the amount of the hydraulic losses in the TC. This hydrulic power loss is what causes the heat buildup in your transmission. Thus, cruising on a level road with the TC clutch unlocked you may have less torque and RPM than the input values. When climbing a hill with your fifth wheel in tow, the difference in the input and output RPMs increase and the output torque increases above the input torque. When the TC clutch disengages on a hill, the engine RPMs increase while the axle RPMs remain the same and you can feel the increased horsepower to the wheels (even including the increased HP loss to the transmission fluid and cooler). Also, torque converters are typically rated or promoted by the stall speed which is the maximum engine RPM (full throttle) with zero RPM output (you need good brakes for this test). Torque converters typically have a torque multiplication of 2 to 3 at stall and might have a torque multiplication of 0. 9 to 0. 95 while cruising on a level road. They put torque converter lockup clutches on in the 80's to get 1. 0 and increase gas mileage.

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'97 SLT 2500, Cummins 4x2 LB Club Cab, auto, 3. 54 LSD, white/driftwood, air dam, sliding rear window, towing/camper pkgs, Rhino liner, stainless bed caps, stake pocket tie downs, chrome exhaust turn down, stainless wheel well trim, Carr steps, Cobra 25W CB/Firestick, J-MAC louvered aluminum headache rack and fifth wheel tail gate, Reese 15k fifth wheel hitch, 29 ft Dreamer fifth wheel, fully loaded GCW=18,000 lbs. Kennewick, Washington.

 

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All I can say is incredible... A very indepth question and an equally deep response. Good info to know for us automatic owners. You guys get my vote to go on Jeopardy...

Pat Daniels
91 4x4

 

[Archive]

wxman,

For a simple explanation, think of the torque converter as 2 fans sitting face to face. One fan has the standard pitch(engine) and one fan has a deeper pitch(torque multiplication) set of blades(transmission). Turn on the (engine) fan and the transmission fan will start to turn with ever increasing speed. This represents the transmission starting to turn. Eventually, the two fans will turn at approximately the same speed(depending on the pitch of the blades).

This is oversimplification of the process, but hopefully, you can get the idea. By using torque multipication, the automatic transmissions can use fewer gears and wider "gear splits" or gaps between the gears.

I don't know what the torque ratio is on the Dodge transmission, but they are on the average of 2 to 1 at stall speed on the transmissions I am familar with(Allison, Borg-Warner, etc. used in medium, heavy trucks and off highway construction equipment).

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97 3500CC,Auto,4. 10LSD,All Options incl Leather,Jardine 4" Cat back Exhaust,The following options removed and in box waiting on the Y2K: Pyro,Boost,Trans. Temp Gauges, PacBrake, RPM 40Gal. Aux. Fuel Tank, Reese 20K Hitch, Highway Products Tool Box. 2000 3500QC 6 Speed,4. 10LSD,SLT+, Camper & Tow Pkgs on Order since 9-28-99, Built 12-01-99, On Train, Stopped at Alliance(8 miles from my house),TX, 12-7 2:24PM waiting on truck transport to deliver to dealer 20 miles away.



[This message has been edited by Bill Stockard (edited 12-14-1999). ]

 

[Archive]

Thanks guys for your response. So if I understand correctly, there is sort of a "variable gear reduction" within the torque converter depending on the rpm of the output turbine relative to the rpm of the input turbine. Is this correct?

My 89 has a three speed (727) non-lockup converter. I guess that means my "brilliant" idea of calculating my engine's torque from timed 40 to 60 mph runs (my transmission doesn't shift out of high gear over 35 mph) isn't valid. I was going to use:

(change in) v = 1/2 at^2

solving for a (delta v and time known/determined)

and then using

f=ma

to determine force requred to accelerate the mass of my truck (as weighed on truck scales) from 40 mph to 60 mph in the time obtained.

Of course, I would've had to use a correction factor to take into account the tire rpm compared to the engine rpm @ essentially no load (e. g. crusing rpm at 60 mph).

If there's an infinitely variable gear reduction in the torque converter, then I guess it couldn't even be dyno'd on a chassis dynamometer? Does anyone have any ideas on how I can figure my engine's torque without taking the engine out and putting it on an engine dynamometer?

I guess this method would be valid for a standard transmission, wouldn't it?

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1989 D250, 727 auto trans, 16 cm^2 turbo housing, Banks Powerpack, Turbo Shop Intercooler