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Date:         Mon, 19 Apr 2010 18:32:10 -0400
Reply-To:     Sudhir Desai <sudhir.desai@GMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Sudhir Desai <sudhir.desai@GMAIL.COM>
Subject:      Re: Why go to big tires?
Comments: To: Jake de Villiers <crescentbeachguitar@gmail.com>
In-Reply-To:  <l2l71d9cdf91004191507o77fe61easab3a6f00b26b8f9a@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8

yeah. my calculations assumed that even the stock tire/wheel/brakes/hub had no mass. therefore, adding a multiplier to 0, would still be 0.

not trying to be an ass, just trying to eliminate all trivial (for the sake of explanation) variables. :)

there is the weight(mass*gravity), but if we're talking real world, we need to take into account rotational inertia, linear inertia, friction, wind resistance (along the larger plane of the wheel, as well as underneath the van, as well as the disruptions in airflow caused by a greater ride height, etc), gravity, mass, etc. which I didn't think were necessary to illustrate the fact that going to a bigger (weightless)tire will not allow stock size brakes to stop the tire abruptly (as in our fake world example) ;)

(as far as tire/wheel combinations go, you can always get a lighter wheel to somewhat offset the heavier tire. the stock steelies are pretty heavy, as are the stock alloys)

I'm not going to offer to do the actual calculations, as I'm sure there is a solid modeler out there that could do ALL of them correctly. (I had to rewrite that email four times because of bad math on my part) ;) Sudhir

On Mon, Apr 19, 2010 at 18:07, Jake de Villiers <crescentbeachguitar@gmail.com> wrote: > Nice explanation Suds but... you've left out the other half of the equation. > >  As the wheel diameter grows the weight of the wheel also grows, especially > the tire mounting section which is at the far end of the 'lever' system > which is the tire's diameter. > > So that 5600in/lbs is no longer sufficient to stop the more massive rotating > 'lever', you need even more brakes. > > On Mon, Apr 19, 2010 at 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 > > > > -- > Jake > > 1984 Vanagon GL 1.9 WBX 'The Grey Van' > 1986 Westy Weekender/2.5 SOHC Suby 'Dixie' > > Crescent Beach, BC > > www.thebassspa.com > www.crescentbeachguitar.com > http://subyjake.googlepages.com/mydixiedarlin%27 > >


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