In a message dated 98-12-03 18:43:54 EST, Mostrom.Paul@principal.com writes:
> If:
>
> 3 /----------
> --\ / hp x 234 = S
> \/ ----------
> (wc+wd)
>
> where
>
> hp = horse power
> wc = weight of car
> wd = weight of driver
> S = speed at end of 1/4 mile
>
> Would it then follow (for those of us who don't have dyno's) that....
>
> Speed in a 1/4 mile = 79
> Weight of car = 1600
> Weight of driver = 196 (yes, I really do fit in a Spitfire!)
>
> then:
> Calculated hp ~= 70
>
> Everything else being fairly equal (driver reaction, traction, ect. ect.).
Paul,
Yes, that is correct. As a matter of fact, the original equations were
developed to determine HP from speed and elapsed time. Your question points
out an error I made in my posts - I forgot to take into account the weight of
the driver! In my case, that adds 0.22 seconds to my time, and knocks 2.2 MPH
off my speed.
Dan Masters,
Alcoa, TN
'71 TR6---------3000mile/year driver, fully restored
'71 TR6---------undergoing full restoration and Ford 5.0 V8 insertion - see:
http://www.sky.net/~boballen/mg/Masters/index.html
'74 MGBGT---3000mile/year driver, original condition - slated for a V8 soon
'68 MGBGT---organ donor for the '74
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