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balancing fly

  • Thread starter Thread starter Marco93!
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Marco93!

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hi, I have already searched for information about the forum but I couldn't find an answer.

I have a symmetrical flywheel that turns to 2000 rpm of 20kg diameter 500mm.
they brought it to balance and told me that the machine communicated that to 2000 rpm they were added:

2 grams at 80 degrees

4 grams at 180 degrees

how do I trace to the degree of balancing g?

I hypothesize that these grams are to be put on the outer diameter, correct me if I'm wrong.
Moreover they should have chosen a plan on which to place the masses for example on one of the two internal or external faces of the flywheel, even here correct me if I am mistaken.
 
Last edited:
and ask them?
It would have been much better.

I calculated so the static moment given by the masses to add

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

the norm iso 1940 defines
the imbalance u the product of eccentricity and the center of the body for the mass m of the body itself

s(1) u(1) = e(1)*m

so and(1) = s(1)/m=1500/(20*1000)=0.075mm

always the norm defines g= angular speed*eccentricity embroidery

g = e(1)*pigreco*n/30= 0.075*3.14*2000/30=16

but I am not sure of the procedure
 
I believe that we are talking about a dynamic balancing, with two correction plans (where the balance masses are allocated) placed at a certain distance from the center of the fly (otherwise it would not make sense to add two masses to different angles on the same plane). first it would be appropriate to check the position of these two planes and if actually the mass is added on the outer diameter (often but not always so). from the knowledge of static moments on the two floors and from the position of these regarding the center of gravity, can be traced back to the resulting imbalance and therefore to the degree g
 

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