Need engineering help to beat a dead horse
#1
Need engineering help to beat a dead horse
I know this topic is SO overdone - this might actually qualify for "off topic". But I was bored and decided to try and actually do some calculations of DMF vs. LWF. Perhaps someone can see where I went wrong in my calculations...
DMF = 13.6 kg
LWF = 5.5 kg
to simplify moment of inertia calculations, I eye-balled the flywheel
and assumed it's roughly equivalent to a hollow cylinder of a 4 inch radius,
meaning roughly half the mass was outside and half inside of a 4 inch radius
moment of inertia of hollow cylinder:
I = mr^2
So for DMF, I = 13.6 kg * .0103m^2
LWF, I = 5.5 * .0103m^2
DMF I = .140 kg m^2
LWF I = .0567 kg m^2
torque:
tau = I * alpha (tau is torque, I is moment of inertia, and alpha is angular acceleration)
An example accelaration, let's take 4000-6000 RPM acceleration in say 5 seconds. 1 RPM = 2pi radians per minute.
419 radians per second to 628 radians per second.
Difference is 209 radians per second in 5 seconds,
angular acceleration is 41.8 radians per second per second
DMF tau is 5.85
LWF tau is 2.37
BUT, the units are:
kg meters meters radians
------------------------
seconds seconds
simplifying with Newton the units are:
Newtons * meters * radians
the difference is about 3.5 "newton meters radians"
3.5 newton meters is about 2.6 pound feet
So I would assume that the engine requires 2.6 "pound feet radians" less torque to achieve the same acceleration. Since torque units are pound feet, not pound feet radians, I'm not sure if my calculations are correct - but if they are and I can drop the radians unit, that's 2.6 foot pounds which is a little over 1% of the max torque rating of 228 pound feet - sounds reasonable, but I'm confused why I have an extra radians unit. What did I do wrong?
DMF = 13.6 kg
LWF = 5.5 kg
to simplify moment of inertia calculations, I eye-balled the flywheel
and assumed it's roughly equivalent to a hollow cylinder of a 4 inch radius,
meaning roughly half the mass was outside and half inside of a 4 inch radius
moment of inertia of hollow cylinder:
I = mr^2
So for DMF, I = 13.6 kg * .0103m^2
LWF, I = 5.5 * .0103m^2
DMF I = .140 kg m^2
LWF I = .0567 kg m^2
torque:
tau = I * alpha (tau is torque, I is moment of inertia, and alpha is angular acceleration)
An example accelaration, let's take 4000-6000 RPM acceleration in say 5 seconds. 1 RPM = 2pi radians per minute.
419 radians per second to 628 radians per second.
Difference is 209 radians per second in 5 seconds,
angular acceleration is 41.8 radians per second per second
DMF tau is 5.85
LWF tau is 2.37
BUT, the units are:
kg meters meters radians
------------------------
seconds seconds
simplifying with Newton the units are:
Newtons * meters * radians
the difference is about 3.5 "newton meters radians"
3.5 newton meters is about 2.6 pound feet
So I would assume that the engine requires 2.6 "pound feet radians" less torque to achieve the same acceleration. Since torque units are pound feet, not pound feet radians, I'm not sure if my calculations are correct - but if they are and I can drop the radians unit, that's 2.6 foot pounds which is a little over 1% of the max torque rating of 228 pound feet - sounds reasonable, but I'm confused why I have an extra radians unit. What did I do wrong?
#7
Oh man I like that kind of stuff, thanks for the mental exercise. Looks to me like your units are resolved. I think you may have answered your own question in that you can drop the radian, it is a dimensionless parameter. As you know (probably better than me!) radian is ratio of the length of an enclosed arc of a circle to the length of the circle's radius, and therefore dimensionless. And your number, 2.6 ft-lbs torque, does seem to be of the right order magnitude although it doesn't seem to say much for the benefits of a LWF.
But let's see if I can't help overthink this with you. Your scenario assumes the engine is turning in the range from 4000 to 6000 rpm, which is in the vicinity of max torque, as you say around 220 ft-lbs. But realistically, the real benefit of the LWF may be in the range from say, 0 to 2 or 3000 rpm, where the engine is not as torquey. If so, the torque savings would be a higher percentage of the engine torque output in that rpm range, which seems to suggest that there is some benefit to the LWF.
On the other hand, if I could just get my LWF to quit rattling (saw a thread that called it coffee grinder effect) I would happily forget about moments of inertia, angular acceleration, etc!
Thanks again.
Bob
But let's see if I can't help overthink this with you. Your scenario assumes the engine is turning in the range from 4000 to 6000 rpm, which is in the vicinity of max torque, as you say around 220 ft-lbs. But realistically, the real benefit of the LWF may be in the range from say, 0 to 2 or 3000 rpm, where the engine is not as torquey. If so, the torque savings would be a higher percentage of the engine torque output in that rpm range, which seems to suggest that there is some benefit to the LWF.
On the other hand, if I could just get my LWF to quit rattling (saw a thread that called it coffee grinder effect) I would happily forget about moments of inertia, angular acceleration, etc!
Thanks again.
Bob
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#8
Cool - I was hoping that was the answer!!!
Happy to take any LWF for LUK DMF with 8k miles
Yeah, I know, this is pathetic - my car is in a million pieces, I'm kind of waiting on some stuff to continue my cage...
Happy to take any LWF for LUK DMF with 8k miles
Yeah, I know, this is pathetic - my car is in a million pieces, I'm kind of waiting on some stuff to continue my cage...
#9
I know what you mean. My new/old Bronco is also in a million pieces. Can't seem to find the time and energy to reassemble. I only got a month or two left to do outside stuff (running, biking) before they turn the oven on here in Phoenix. Been trying to do those as much as possible before it gets hot.
Oh yeah and it's not really the LWF that is rattling on my car, somebody told me it was the LW clutch springs. I'm going to quit whining about that now because it really does run great and I seem to be managing the stalling thing.
Thanks again Chance, nice thread.
Bob
Oh yeah and it's not really the LWF that is rattling on my car, somebody told me it was the LW clutch springs. I'm going to quit whining about that now because it really does run great and I seem to be managing the stalling thing.
Thanks again Chance, nice thread.
Bob
#12
[wandering in after the parade's gone by]
...bitumen got the straight dope, there, with his input -- but there's one thang that lept out at this geek as I trudged through the physics glop: 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.
And radians? Nice thing about 'em is you can treat 'em rough -- they're dimensionless -- so you can kick the stray one to the curb if it makes your units look funny.
-Lonnie
...bitumen got the straight dope, there, with his input -- but there's one thang that lept out at this geek as I trudged through the physics glop: 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.
And radians? Nice thing about 'em is you can treat 'em rough -- they're dimensionless -- so you can kick the stray one to the curb if it makes your units look funny.
-Lonnie
#14
Well with respect to the approximation, it may not be the greatest, but maybe not too bad either. Technically, wouldn't you integrate (r^2)dm to get I? I think what Chance was doing with his 4" hollow cylinder was to define (foggy memory here) moment of inertia about a center of percussion; like the sweet spot on a bat? Mass far away from center of rotation has greater effect on rotator than mass close to rotator, or something like that. Or to put it another way, it's either Chance's hollow cylinder or Lonnie's solid disc. Either way still pretty neat.
Actually tbennett don't feel bad because I am disadvantaged in several ways. Because I'm a civil engineer, I'm not too good at that math either. My niece a chemical engineer claims we are the dumb brand of engineers. And because I'm fast approaching geezer status, a successful M-thing is starting to mean less thrills than a warm, comfortable place to take a crap, or so my doctor has been trying to warn me.
Bob
Actually tbennett don't feel bad because I am disadvantaged in several ways. Because I'm a civil engineer, I'm not too good at that math either. My niece a chemical engineer claims we are the dumb brand of engineers. And because I'm fast approaching geezer status, a successful M-thing is starting to mean less thrills than a warm, comfortable place to take a crap, or so my doctor has been trying to warn me.
Bob
#15
can we please get back to those ***-engined **** slot cars!
Hadn't heard of 'percussion' centers before, so I'll let someone else bang that drum...
Sweet spot, hmmm? Like the centroid of some liquid in rotation (--long as we're in the area of "warm places" lol -- that inertial sweet spot/liquid centroid is probably 2/3 the distance away from the center of spin.