Can Vector Analysis Help Find Replacement Weights for Spin Balance?

[jdk944]

TL;DR Summary

Given the mass-radius and angular location of a single correction weight, what methods can be used to replace the correction weight with two or more correction weights with different mass-radius and angular location but still maintain static and dynamic balance?

I had a test article go through a spin balance service. The mass-radius and angular location of the correction weight, to achieve static and dynamic balance, interferes with another component on the test article (I'm unable to place the correction weight at the specified location).

Using vector analysis, I chose a desired mass and radius to find the angles for the "replacement" weights (assumed two correction weights to replace the one) but was unsuccessful in doing so.

Any input or references regarding this topic would be very helpful.

Please let me know if additional information is needed for clarity.

 

[jdk944]

Attached is my work thus far. theta1 and theta2 were drawn arbitrarily and are unknown. Their masses are unknown as well.

If I assume the two replacement weights to have the same radius and on the same plane, how can I solve for their masses and angular locations?

In my mind, I'd like to run a Matlab script with a condition that outputs a specified range of masses and their angular location. I'm not sure if this is even possible, but, for example, if I want my replacement weights to be between 4 to 7 lbs each, Matlab would output all masses in that range along with their angular location.

 

[JBA]

First, for simplicity, since rc = r1 =r2, so you should eliminate r from your equations and since you are measuring your angle from the x-axis (not the y axis) and mc is at the y-axis your y equation's left value, with r eliminated, will be 10 x Sin (90°) = 10, not 0, as you show in your y equation.
Using that equation, you have to choose both m1 and m2 and either the Θ1 or Θ2 value. Once you have done that your equation will give you the required angle for the remaining unknown Θ angle.
Alternatively, you can choose both Θ1 and Θ2 and either m1 or m2 and the equation will give the required value for the remaining unknown m.

All of the above being said, unless you make the mass of m1 and m2 and their angles from the mc weight position equal, and relatively close to the mc position you will be still be unsuccessful because you will create an imbalance relative to the locations 180° from the positions of the m1 and m2 weights.

 

[Baluncore]

Welcome to PF.

A practical alternative is to upset the initial balance.
When a balance fails due to a weight being required in a prohibited position, add the prescribed weight to one side of the prohibited position, remove all earlier balance weights, then perform a re-balance.

 

[jdk944]

First, for simplicity, since rc = r1 =r2, so you should eliminate r from your equations and since you are measuring your angle from the x-axis (not the y axis) and mc is at the y-axis your y equation's left value, with r eliminated, will be 10 x Sin (90°) = 10, not 0, as you show in your y equation.
Using that equation, you have to choose both m1 and m2 and either the Θ1 or Θ2 value. Once you have done that your equation will give you the required angle for the remaining unknown Θ angle.
Alternatively, you can choose both Θ1 and Θ2 and either m1 or m2 and the equation will give the required value for the remaining unknown m.

All of the above being said, unless you make the mass of m1 and m2 and their angles from the mc weight position equal, and relatively close to the mc position you will be still be unsuccessful because you will create an imbalance relative to the locations 180° from the positions of the m1 and m2 weights.

Thanks for pointing out the angles relative to the prescribed orientation. I confused myself with a reference I'm using...

For better accuracy, r1 = r2 but do not equal rc.

When you mentioned that both masses (m1 and m2) and their angles must be positioned equally from the mc weight to maintain balance, what type of imbalance would occur if they were unequal? If the two replacement weights share the same plane as the correction weight (mc), does that rule out dynamic imbalance? Furthermore, if the two replacement weights are placed < 180° apart and share the same plane, would the test article experience an imbalance?

Also, since I have two equations, can I just assume two of the four variables, instead of three?

Is it possible to find the least heavy pair of replacement weights to displace mc?

 

[JBA]

When you mentioned that both masses (m1 and m2) and their angles must be positioned equally from the mc weight to maintain balance, what type of imbalance would occur if they were unequal?

Any added weights placed anywhere around the cylinder will represent an imbalance relative to their 180° opposite point on the cylinder's circumference.

Furthermore, if the two replacement weights are placed < 180° apart and share the same plane, would the test article experience an imbalance?

If the two weights are placed 180° apart they will balance each other. That is the basis for placing the one weight 180°m from the original imbalance point of your cylinder; but, they will do nothing to correct for your original imbalance.

If place both weights equal apart from the Mc position such that M1 = M2, r1=r2 and (Θ1 = Θ2 as measured from Mc), then your equations reduce to:
90 = 2 M r cos(Θ) or 90/2 = M r cos (Θ)
you only need to select a value for either M or Θ.to determine the value of the other. Those equations also demonstrate that the value of M will always increase as Θ increases and the sin Θ reduces, so the sum of 2 M weights will always be greater than Mc.

If you use the original equation with unequal angles and masses, you will always need to select 3 of the 4 unknowns.
(Standard mathematics rule: One equation can only have one unknown.)