Date:         Wed, 17 Aug 2005 17:23:41 -0300
Reply-To:     Brandon <Brandon@CONCILIO.CA>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Brandon <Brandon@CONCILIO.CA>
Subject:      Re: How much fuel efficiency gained by shedding 300 lbs?
In-Reply-To:  <410-22005831720559187@earthlink.net>
Content-Type: text/plain;charset=iso-8859-1

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.
>
>