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*To*: Eliot Lim <eliot@u.washington.edu>*Subject*: Re: Al rotors*From*: Jeremy R King <kingjer@eng.auburn.edu>*Date*: Wed, 14 Feb 1996 02:50:04 -0600 (CST)*cc*: quattro mailing list <quattro@coimbra.ans.net>*In-Reply-To*: <Pine.A32.3.91j.960213233726.29691B-100000@homer02.u.washington.edu>*Sender*: quattro-owner@coimbra.ans.net

Ok. After much prodding, I've finally done "THE MATH". Please keep in mind that it' 2:00 in the morning. I'm also going to verify my calculations tomorrow (later today) in my Mechanical Engineering Design class. Here are my assumptions and calculation criteria. Since the primary purpose was to see the difference between the mass of aluminum and cast iron rotors, I've greatly simplified the model to compare two identical setups (save the rotor material of course) and see the difference. Rotor size - solid 14" diameter, 3/8" thick disc Overall wheel size (this determines wheel rpm's) - 25" Vehicle mass - 2200 lbm Vehicle speed - 100 mph These may not be the most accurate assumptions to depict the Viper/Prowler scenerio. But they will do just fine for comparison purposes. They are reasonable numbers for a vehicle, and the solid rotor assumption would tend to HELP eric's arguement more than hurt it. Looking up the formula for kinetic energy due to rotation T = 1/2 I w^2 where I is the mass moment of rotational inertia and w is the angular velocity I = m d^2 / 8 and m = (pi d^2 t roe) / 4 where d is diameter t is thickness and roe is density of the rotor Plugging all the values in, checking units for consistency, and using a density of .1 lb/in^3 for aluminum and .26 lb/in^3 for cast iron. T(al rotor) = 9,375.45 [lb ft^2 / s^2] T(ci rotor) = 25,312.17 [lb ft^2 / s^2] of course, these would be multiplied times 4 in a simple whole car model to yield: T(al) = 37,501.8 T(ci) = 101,248.68 Remember, this is for equal size rotors. This shows what eric was talking about with the greatly reduced rotational inertia of the aluminum. However, this is only for the al rotor compared to the ci one. The point Glen and myself have been trying to make is that these numbers are completely insignificant compared to the kinetic energy (linear momentum) of the mass of the vehicle. Here are those numbers: T(car) = 1/2 m v^2 where m is mass of car and v is velocity T(car) = 23,662,222 [lb ft^2 / s^2] Now, eric. Tell me exactly how significant two hundred thousand is compared to twenty three and a half million. To quantify it, it's almost exactly 1/2 of 1%. In engineering we call that insignificant. In racing it's very possible that that 1/2% may add a tenth or even more. But I guarantee you that 1/2% DOES NOT MAKE AS MUCH AS A FOOT DIFFERENCE in braking distances between the Viper and the Prowler. Probably not even an inch. And remember, I didn't even calculate the numbers for the wheels and tires, which would make that percentage even smaller. Maybe this will satisfy some of our doubting Thomases. If anybody would like to see more math, or perhaps has actual numbers that can be used on rotor size and vehicle math, I'll be happy to crunch those numbers too. My HP is always more than willing. And eric, I calculated these numbers at a high speed thinking this might be where the rotational inertia makes the biggest difference. But numbers were very similar for different speeds. I'm afraid you need to go back to math class. Jeremy '86 VW Quantum GL5 Auburn University, Alabama, USA Hometown - Reidville, South Carolina, USA On Tue, 13 Feb 1996, Eliot Lim wrote: > > ya know, those chrysler dweebs could have saved everybody here a lot > of headaches if they just put carbon fiber brakes in that car instead > of these marginally better (or much better, whatever) Al brakes... > > we should just start cc:'ing the chrysler engineers the flames and > then they'll be sorry.... > > so everybody's bragging about the math, but where is the math??? > nobody wants to do the math??? then everybody's wrong!!! > > eliot >

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