Date: Mon, 19 Apr 2010 15:45:54 -0700
Reply-To: Al Knoll <anasasi@GMAIL.COM>
Sender: Vanagon Mailing List <vanagon@gerry.vanagon.com>
From: Al Knoll <anasasi@GMAIL.COM>
Subject: Re: Why go to big tires?
In-Reply-To: <C7F2282C.A882%mwmiller@cwnet.com>
Content-Type: text/plain; charset=ISO-8859-1
It's all in the minus sign. If the bigger wheel makes your (+) acceleration
worse, changing the (+) to a (-) means it makes your (-) acceleration (read
deceleration) equally worse.
Your van stopped WAY better because you spent more money on it. Runs better
after you wash it too.
Pensionerd.
On Mon, Apr 19, 2010 at 3:33 PM, Mike Miller <mwmiller@cwnet.com> wrote:
> Thank you! Many people whom I was sure knew what they were talking about
> said that bigger wheels mean not as good stopping but I couldn't understand
> why, because the added weight of the wheels/tires is negligible compared to
> the weight of the vehicle.
>
> This explains it so even I can understand it.
>
> Hooray.
>
> Anyway I knew my syncro/countryhomes/westy stopped WAY better with the
> G60's
> up front and Audi 5000 turbo rear disks and calipers. Now I know why.
>
> Thanks again, Sudhir [Does that mean "smart" in some language other than
> English?]
>
> Mike
>
>
> On 4/19/10 2:39 PM, "Sudhir Desai" <sudhir.desai@GMAIL.COM> wrote:
>
> > Sorry about the simplicity, I figured I would gloss it over HARDCORE.
> >
> > the 0 represents the hub, the ) the brake rotor/drum, the | the
> > outside of the tire.
> >
> >
> > STOCK
> > distance of force from hub:
> > 0--5"----//---12.5"
> >
> > direction of force:
> > 0--up-------down
> >
> > illustration:
> > 0--)-----//----|
> >
> > lets say the car is rolling along, and the torque needed to stop the
> > wheels abruptly (ignoring/subtracting any sort of rotational inertia,
> > and friction, etc) is 5000in-lb.
> >
> > that means, we would have to exert a force of 5000in-lb/12.5in = 400lb
> > (at 12.5" out from the hub) to stop the wheels abruptly.
> >
> > now, we go in more to where we can stop the wheels, the brakes. at 5"
> > out from the hub, that 400lb force needed is now (400lb*12.5in) =
> > (xlb*5in), or x = ((400lb*12.5")/5"), so
> >
> > x = (5000in-lb)/5" = 1000lb
> >
> >
> > SO... to stop the wheel abruptly, we need 1000lb of force at 5" out
> > from the hub, and that's the torque our stock brakes can put out.
> >
> >
> >
> > when we increase the tire size to a 28" tire:
> >
> > OVERSIZE
> > distance of force from hub:
> > 0--5"----//---14"
> >
> > direction of force:
> > 0--up-------down
> >
> > illustration:
> > 0--)-----//----|
> >
> >
> >
> > SO NOW WE GO BACKWARDS!!!! lol~
> > due to our original assumption of 1000in-lb,
> > we have just increased the diameter of our tires to 28".
> >
> > this means that our 1000 lb force (5000in-lb) that was enough to stop
> > the 25" tire will not be enough to stop a 28 inch tire by the
> > following equation.
> >
> > 5000in-lb/12.5in = xin-lb/14in, or xin-lb = (5000in-lb*14in)/12.5in
> >
> > x=(5000*14)/12.5 = 5600in-lb is our new stopping torque needed.
> >
> > our stock brakes can only exert 1000in-lb of torque, so we won't be
> > able to stop the wheel abruptly anymore.
> >
> > hopefully y'all are following me still...
> > we just increased the amount of force needed to stop the wheels, and
> > we see that our stock brakes are not enough to do so.
> >
> > NOW, we need to calculate the size of brakes we need to stop the tire
> > abruptly (ignoring all other factors).
> >
> > we take the known stopping torque, and our original brakes (10" brakes
> > LOL), and see how that compares with our new 5600in-lb
> > x-in/5600in-lb = 5in/5000in-lb
> >
> > x = (5*5600)/5000 = 5.6in
> >
> >
> >
> > so we'd need 11.2" diameter brakes to put out the 5600in-lb torque
> > needed to stop the wheels abruptly.
> >
> > which is another inch diameter of brake needed.
> >
> >
> > I hope my simple explanation (with calculations to show i wasn't just
> > pulling numbers out of the air) was satisfactory. :)
> >
> > Sudhir
>
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