Date: Wed, 17 Aug 2005 13:46:40 -0700
Reply-To: Gabriel Ross <gabeross@ORO.NET>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: Gabriel Ross <gabeross@ORO.NET>
Subject: Re: How much fuel efficiency gained by shedding 300 lbs?
In-Reply-To: <32158.161.19.64.5.1124310221.squirrel@www.dynamichosting.b iz>
Content-Type: text/plain; charset="us-ascii"; format=flowed
Boy, you're right; Wrong equation. I was/am working off of memory from a
physics course while driving CD, and the guy was very specific about the
point I was trying to make. I do remember that one side of the equation was
mass X velocity squared, so the chart is still valid. I guess the grey
matter is starting to muddle. Thanks for the correction.
Yes, I agree that wind resistance is a big factor, but a Westy going at 45
mph is consuming much less energy than a Westy going 65. That's the point I
was trying to make.
Gabriel
At 05:23 PM 8/17/2005 -0300, Brandon wrote:
>This seems like some pretty hokey math to me. For one thing the 'C' in
>E=MC2 stands for the speed of light (as in constant i belive)...
>
>It looks to me like you end up using a bit of a momentum equation but in
>any event it doesn't take into account the biggest factor which is WIND
>RESISTANCE! I'm sure there is a correct equation out there some where but
>it would have to factor in each vehicles unique wind resistance to be
>accurate.
>
>As a general rule anything above 60-65 MPH starts to become less and less
>fuel efficient the faster you go. It's different for each vesicle due to
>weight, shape, engine, and Transmition. Unless you are independently
>wealthy and have nothing better to do with your time and money just stick
>to that general rule...
>
>Brandon,
>Calgary
>87 Westy
>
> > Figure 10 lbs drag per 1,000 lbs of weight on your tires.
> > It's fairly independent of speed.
> >
> > Thusly: 300 lbs less load = 3.33 lbs less drag (rolling component)
> > Your aero drag is negligibly affected.
> >
> > Randy Bergum
> > 1990 Carat
> >
> >
> >> [Original Message]
> >> From: Gabriel Ross <gabeross@ORO.NET>
> >> To: <vanagon@GERRY.VANAGON.COM>
> >> Date: 8/17/2005 9:36:45 AM
> >> Subject: Re: How much fuel efficiency gained by shedding 300 lbs?
> >>
> >> Well, maybe it's not that germane for us vanagon owners, but ol'
> > Einstein's
> >> equation (E=MC squared) (don't know how to make my computer put that
> > little
> >> 2 above the C) is the basis of this. So that's Energy equals Mass times
> >> Velocity squared, and since the mass is a constant regardless of the
> > speed,
> >> you can easily see why it costs more to go faster. For the math
> >> impaired, here's a chart:
> >>
> >> E = M(25 X 25) 625M
> >>
> >> E = M(35 X 35) 1225M
> >>
> >> E = M(45 X 45) 2025M (that's over 3X the energy required at 25
> > mph
> >>
> >> E = M(55 X 55) 3025M
> >>
> >> E = M(65 X 65) 4225M
> >>
> >> E = M(75 X 75) 5625M (that's 9X @ 25 mph)
> >>
> >> So, going 75 takes almost double the energy as going 55. As fuel prices
> >> go
> >> up, this is more and more relevant. We're lucky that we've got a
> >> built-in
> >> governor, eh? The "eh" is for the Canadian list members, since I didn't
> >> do
> >> kph. :>)
> >>
> >> Gabriel
> >>
> >> At 10:35 AM 8/17/2005 -0400, Roger Sisler wrote:
> >> >I just heard on the radio that driving over 60 mph is like paying .15
> > cents
> >> >for each mile per hour over this figure.If gas is $2,50/gallon(dont we
> >> >wish), driving 75 mph is like paying $4.75/gallon.I havent tried this
> >> >myself,but it think it might be a bit high.75mph is a place I only
> > approach
> >> >by passing slower vehicles,and this only takes about 5 minutes.
> >
> >
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