DIY_EFI Digest V4 #85

[Raymond C Drouillard]

On Thu, 4 Feb 1999 15:53:48 -0600 (CST) Roger Heflin 
<rah at horizon.hit.net> writes:
>
>
>On Thu, 4 Feb 1999, Greg Hermann wrote:
>
>> >Re: TC's and manual trans (was: Re: Smooth strategy)
>> >
>> >
>> >
>> >
>> >>>No way--a two element torque converter is a violation of Newton's
laws!!>>
>> >
>> >>>Regards, Greg>>
>> >
>> >Greg
>> >I don't know much about Newton's laws however I have been under cars
for
>> >the last 45 years, working mostly on auto trans. so I know a little
bit
>> >about torque converters. To get torque multiplication in a 2 element
TC
>> >you need curved vanes plus a  different number of blades (vanes) in
the
>> >turbine and impeller. There was usually a split guide ring (doughnut)
as
>> >well. While this arrangement is not as effective as a TC with a
stator
>> >it nevertheless multiplies torque and is therfore correctly termed a
>> >Torque converter.
>> >
>> >Theo from downunder
>> 
>> Let's don't mistake slip--lock-up speed  characteristics for torque
>> multiplication. If you have torque A coming into the unit on the input
>> shaft, and torque B going out of it on the output shaft, For A to be
>> different from B, there has GOT to be a torque reaction against the
case.
>> No way out of it. Otherwise we are talking about a close kin of the
famous
>> (?) 300mpg carburettor!
>>
>
>Actaully I think this is how it works.  The reason it multiplies is
>much simpler, rpmin != rpmout.   The easy model is more like 
>rpmin*torquein = rpmout*torqueout*efficiency.   I don't believe there
>is a torque reaction against the case (except waht is actually
>decreasing efficiency).   Basically if rpmout is 10% less that rpmin,
>then you could boost torque by an appropiate amount to make the
>quation balance out.   Also generally this is probably a max torque
>multiplication which may complicate the model above more, but I have
>used this model in some program I have written and it is pretty
>accurate with street convertors, with the high slippage convertors the
>model above does not work exactly right.     It is really no different
>that having a gear that multiplies torque except it is less efficent
>and somewhat more variable on the multiplications.


Your equation is absolutely correct.  The point that Greg is making is
that any change in torque between input and output must be balanced by a
reaction torque against something else.

This is stated in first semester statics class as "The sum of the moments
about any point is zero".

torque in = reaction torque + torque out.

Note that the reaction torque can be either positive or negative (it is a
vector quantity).

If the torque at the crankshaft is 250 Ft-Lbs and the output is 350
Ft-Lbs, the reaction torque is 100 Ft Lbs in the opposite direction of
the output torque.

This works for torque converters, gears, or whatever.  In the case of
gears, you have the force of the shaft against the case.

Ray Drouillard

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