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Old 03-14-2008 | 09:36 AM
  #16  
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Great discussion. Confirming the others, a radian is any circle's circumference divided by its diameter. Units are meter/meter, in/in, mm/mm, etc. The units cancel regardless of the unit used.

You cover only the case of an instant where the acceleration is 41.8 rad/sec or averages 41.8 rad/sec. Assumptions are necessary to simplify calculations, and that is a good assumption to use. In reality the angular acceleration is variable. I drove deoxford's LWF car back-to-back with my DMF car. There is a noticeable difference in the way the revs drop during gear changes. The LWF also makes a big difference if you blip the throttle - the engine revs noticeably more quickly with the LWF. The car's overall acceleration depends on power output from the engine, and total inertia (both linear and angular) of the car. There we did not see much difference.

I want the LWF in my car - just for the more responsive feel. I am not going to drop the engine to make the switch, but I will have the LWF first time the engine is out for any reason. If you have the engine out.....
Old 03-14-2008 | 02:01 PM
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Precisely - you have to choose a specific acceleration to result in some amount of torque.

In other words, choose a beginning RPM, ending RPM, and the time it takes to get from one to the other, or the math doesnt' work.

I chose 4-6k in 5 seconds because I thought that might represent an approximation of flat out in 3rd or 4th gear. If you make the rev change larger and/or the time duration shorter between the specified revs, the acceleration is much greater and so would the resulting torque delta. Free revving is a very large RPM delta in very small time delta, so I would expect a pronounced difference between the 2 in that case. AKA more responsive for rev-matching during shifting.
Old 03-14-2008 | 03:43 PM
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Originally Posted by Bituman
Yeah I was going to say above "mental masturbation" instead of "mental exercise" but I couldn't spell **********, thanks.
You couldn't spell it with just one hand you mean..?
Old 03-14-2008 | 03:48 PM
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Originally Posted by elbeee964
your angular moment of inertia (AMI) approximation to a cylinder of 4" radius... is a killer misapproximation. Real life, it's a big ol' beefy platter. A platter's AMI is proportional to the square of the platter radius. Meaning, an oz. of material 2" away from the crank centerline is only *1/9th* the AMI contributor of his "brother"-oz., 6" out from centerline -- though both are equal distance from that 4 inch cylinder line.
Chance, how did you calculate the 4" assumption? I agree with Lonnie that this tripped my internal error checking algorithm because of the square, however there is a lot more mass further out on the wheel than there is close in. Can you elaborate on your method of collapsing the big platter to a big cylinder?
Old 03-14-2008 | 03:51 PM
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Old 03-14-2008 | 04:21 PM
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I want a math version of this:



T-shirt available here.
Old 03-14-2008 | 04:28 PM
  #22  
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First approximation is that we can eliminate the "axial" dimension, and for the purposes of inertia, treat it as a 2-dimensional object, like a disk.

I could either choose the moment of inertia of a:
disk, I= (mr^2)/2
"infinitely thin-walled hollow cylinder", I=mr^2
or
"thick walled cylinder" I= .5m(r1^2 + r2^2) r1 and r2 being inner/outer radius

Clearly, disk is the worst choice because of the empty space in the center. Casual glance indicates most of the mass is in the friction surface region (througout the axial dimesion of course), then there's the ring gear and sensor teeth outside of that. The thick-walled cylinder would probably have been a better formula to use. I chose thin-walled because the math was easier and I just kind of guessed where the "radial center" of mass would be at 4 inches. This was a total swag - but I know there is indeed some radius of thin-walled cylinder that makes the moment exactly match that of the probably more realistic thick-wall cylinder moment. Re-doing this with the thick-wall cylinder formula *might* yield better accuracy but you still have to take a swag at what the inner and outer radii are, accounting for the fact that there is an irregular mass distribution as you move along the radius of the flywheel.
Old 03-14-2008 | 05:05 PM
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So i was right in the first place
Old 03-14-2008 | 05:15 PM
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Originally Posted by chancecasey
First approximation is that we can eliminate the "axial" dimension, and for the purposes of inertia, treat it as a 2-dimensional object, like a disk.

I could either choose the moment of inertia of a:
disk, I= (mr^2)/2
"infinitely thin-walled hollow cylinder", I=mr^2
or
"thick walled cylinder" I= .5m(r1^2 + r2^2) r1 and r2 being inner/outer radius

Clearly, disk is the worst choice because of the empty space in the center. Casual glance indicates most of the mass is in the friction surface region (througout the axial dimesion of course), then there's the ring gear and sensor teeth outside of that. The thick-walled cylinder would probably have been a better formula to use. I chose thin-walled because the math was easier and I just kind of guessed where the "radial center" of mass would be at 4 inches. This was a total swag - but I know there is indeed some radius of thin-walled cylinder that makes the moment exactly match that of the probably more realistic thick-wall cylinder moment. Re-doing this with the thick-wall cylinder formula *might* yield better accuracy but you still have to take a swag at what the inner and outer radii are, accounting for the fact that there is an irregular mass distribution as you move along the radius of the flywheel.
I agree that a thin walled formula will get you dang close. I was just interested in where the 4" came from. I agree that the thick wall formula is just too dang convoluted. I like the parsimonious nature of the simpler formula because I prefer to eschew obfuscation when at all possible. *snicker snicker*
Old 03-14-2008 | 05:56 PM
  #25  
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[God, how I promised myself not to jump back into this physics heavy-lifting... ]

An old engineer I respect once cautioned me, "Mother Nature doesn't make bookkeeping errors." -- by which he meant, if your numbers don't match reality, reassess your numbers and presumptions -- don't try to argue w/ reality.

(It cautioned me that Mother Nature isn't tolerant of My bookkeeping errors -- they'll be revealed -- always. Anyway...)

I got to thinking -- LWFs really come into their own during a down-shift throttle blip, more so than a flat out acceleration run. And your engine's response isn't a linear ramp up to that blipped rpm. Instead, it's more S-shaped: a 'tip-in' relatively slow to action rise, followed by a *real* steep climb of rpms, and trailing off to a 'muted' topping out arrival to that upper rpm.
And it's that middle section's steep slope where the LWF really shows itself in real numbers. (Indeed, slope of the rpm vs. time curve is one definition of angular acceleration.)
Least, that's the story I'm going with for now ...
Old 03-14-2008 | 06:14 PM
  #26  
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Originally Posted by elbeee964
[God, how I promised myself not to jump back into this physics heavy-lifting... ]

An old engineer I respect once cautioned me, "Mother Nature doesn't make bookkeeping errors." -- by which he meant, if your numbers don't match reality, reassess your numbers and presumptions -- don't try to argue w/ reality.
I like that quote. As an economist I always advise those I work with to "become a skeptical empiricist." Interestingly enough, I think that's the same advice you were given. Believe reality... its empirical!




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