Wart said:When I wrote about the position of the weight I was referring to the weights distance from the center of the plate. The closer to the center the more mass needed to create a simular force as a weight placed farther from the center.
Johan said:That would be clear if we bothered to use the right terminology for it. The late model flywheel is a 50oz-in, not a 50oz flywheel. 50 ounces x 1 inch from centerline = 50oz-in, or 25 x 2, or 10 x 5, or whatever it really is x wherever it really is = 50 ounce-inches.
I did a quick bit of searching, and almost everybody just says XXoz. It's kinda like saying I have 26lbs of air in my tires instead of saying 26psi.
Blue Thunder said:The centerline of the flexplate has to be the same as the centerline of the crankshaft, otherwise,(among other things) the starter would have serious issues with the ring gear.
COBRAIIW said:Centerline of the crank, the flexplate, TDC? Which centerline?
This is turning into an engineering thread! YEAH !!!!
Wart said:I wasn't sure if I wanted to bring roe (radius) into this.
Formula for inertial force is F= (mass * velocity squared)/ radius , or F=MV^2/R
If you wish to find the mass needed to create an equivalent force at a different Radius:
m(1)v^2(1)/r(1)=m(2)v^2(2)/r(2)
Velocity is lineal and figured by pi * d. I think it's not necessasary to put time into the equasion as the period is the same.
COBRAIIW said:Thanks guys, I decided to take the flexplate I have to a local machine shop to have it rebalanced to 50 oz.
He told me about 2 hrs, and 75 bucks.
COBRAIIW said:Thanks guys, I decided to take the flexplate I have to a local machine shop to have it rebalanced to 50 oz.
He told me about 2 hrs, and 75 bucks.
Dano78 said:That was probably me, Mark. (Ref: the ordeal of hte Art Carr flex plate.)
Dano78 said:Sorry if I came out a little gruff.
OTOH I didn't realize whe had quite the group of mathematicians!?!
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