At 12:52 PM 8/12/99 EDT, Jetmugg@aol.com wrote:
>In a message dated 8/12/99 8:57:54 AM Pacific Daylight Time,
>MATT_BARRET@EARTHTECH.COM writes:
>
><< Subj: RE: DML: RE: Ring Gear changes: 3.55 vs. 3.90
> Date: 8/12/99 8:57:54 AM Pacific Daylight Time
> From: MATT_BARRET@EARTHTECH.COM (Barret, Matt)
> Sender: owner-dakota-truck@buffnet4.buffnet.net
> Reply-to: dakota-truck@buffnet.net
> To: dakota-truck@buffnet.net ('dakota-truck@buffnet.net')
>
> HEY BOB, WHAT ARE YOU THINK'N, IT A NO BRAINER?? THE BIGGER THE TIRE THE
> BIGGER THE DIAMETER, THUS THE MORE TIMES THE DRIVE SHAFT WILL TURN FOR EVERY
> REVOLUTION OF THE TIRE.
>
>No offense intended - but the statement made above is 100% incorrect. One
>revolution is one revolution, whether the tire is 10 inches diameter or 100
>feet in diameter. What will change is the distance traveled in one tire
>revolution. Obviously, the bigger tire will travel a longer distance
through
>one complete revolution. However, the drive shaft will turn the exact same
>number of turns in one revolution of the axle shaft, regardless of what
>diameter the tire is. Believe it - it is a fact.
Thank you, Jetmugg@aol.com, not so much for your well-stated input (of which I
agree) but the opportunity to correct an error in finding the 'effective gear
ratio' example that I used and put out on the DML.
The example should have said:
Take the mounted stock tire, mark it and where this marker meets the ground,
move the truck one complete tire rotation and mark the ground where the tire
marker is ... this lays out the circumference of the tire (so far, no change
in the example). Now, mount the different diameter tire, mark it and where
it meets the ground. (Here's where I erred). From this ground marker,
measure
out the distance that the circumference was with the stock tire and mark
the ground.
Set the different diameter tire so that the tire marker is lined up on top of
one of the markers on the ground. Move the truck (and forget the marker on
the
tire) until an imaginary radius from the end of the axle shaft is
perpendicular
to the other ground marker. If someone were to count the drive shaft
rotations,
you would have the 'effective gear ratio'.
Sorry to all. My point was that there was not a 'direct' relationship
between the
effective gear ratio and one complete revolution of the tire. The
relationship
is between the circumference of the stock tire and how many revolutions the
different sized tire needs to cover that circumference.
I still don't know how this could be found with the rear end jacked up though.
Bob
Burlington, Ontario
'97 CC Sport, 5.2L, 3.55, auto., 4x2, flame red
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