Re: Wind resistance; was: High Rpm's etc.

From: L. J. Morris (ljohn@alltel.net)
Date: Tue May 13 1997 - 10:29:00 EDT


OK Sam here goes; (a little crude)

From your drive line information, (RPM's at 70 mph), your tire
circumference works out to be 7.1909 ft.

At 4600 rpm / 2.86 final drive ratio = 1608.39 rpm/tire x 7.1909 ft. =
11,565.78 ft/min.

11,565.78 ft/min x 60min/hr = 693,046.99 ft/hr / 5280 ft/mile = 131.43 mph

Obviously this does not take into account wind resistance, positive grade
or negative grade of road, etc., etc., but it'll get you in the ball park.

===================='97 Dakota SLT+=====================

                           L. John Morris<ljohn@alltel.net>
                        !!!!! ... your local *HEMI* owner ...!!!!!
           White/Driftwood/CC/4x4/5.2L/5spd/3.92SG/FIPK clone
                         soon to have "Edelbrock" cat-back
==================== Reply Separator ====================

----------
> From: Sam Parthemer <samp@cts.com>
> To: dakota@ait.fredonia.edu
> Subject: Re: Wind resistance; was: High Rpm's etc.
> Date: Tuesday, May 13, 1997 10:01 AM
>
>
> Anyone know how to figure what 4600 equals???
>
> ..Sam '95 SLT
>

 



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