Balance Wheel inertia/torque issue.

[horologist101]

I am a watchmaker and need some help with the formula and correct units to derive the Frequency of the Balance wheel. I'm am trying to prove the formula with a known Balance Wheel/Spring and Frequency but using all known values still cannot get the formula to work. When I have a working formula then I can make changes to the wheel and spring and predict what the result will be.

Formula given:

T(Hz)= 2∏√[12.M.r.sup.2.l/E.h.e.sup.3]

Issue 1.
in the text
M is described as Mass
'r.sup.2' is described as Radius of Gyration (Rg)
however
M.r.sup.2 is the formula for 2nd Moment of Inertia (Mass x radius)...so I'm not sure if I should be finding Rg or 2nd Moment of Inertia then multiply by the Mass again (M.Rg.sup.2).

Issue 2.
E(Modulus of Elasticity) should be in the units Nmm.sup.2/rad but I can only find E in Gpa ...is there a difference?

Issue 3.
The (Hair)spring is a flat spiral spring (example: 0.30mm high x 0.12mm thick x 260mm long)but when comparing 2 formula for Elasticity 'e' and 'h' are confused so that it is not clear which dimension should be 'e' or 'h' (height and thickness)

0.30 x 0.12.sup.3 = 0.0005184
0.12 x 0.30.sup3 = 0.00324


Issue 4.
The first formula seems to ignore phi θ for some reason (because we are not exceeding the elastic limit of the spring?)
Modulus of Elasticity= 2∏√[12.l.θ/E.h.e.sup.3]

Issue 5.
The Frequency (F) in Hz is also derived from this formula:
F= 1/2∏√(K/I)
Where
K is the spring Constant (Nm/rad)
I is the Mass (Kgm.sup.2)

K is derived from the formula:

K= M/.phi.

Where M = Modulus of Elasticity (back to my problem of working out M)
Where phi θ is described as the 'unit angle of twist' (of the spring) but I've no idea what value this should be (Oscillation of the balance wheel is 270°) and in what units (degrees or radians).

Finally here are the known dimensions of my Balance Wheel and Hairspring:
F(Frequency) = 2.5Hz

Balance Wheel
Wheel Diameter = 17.4mm
R1 (Outside rim radius) = 8.6mm
r2 (Inside rim radius) = 6.8mm
Mass = 0.56gms
Rim Height = 0.75mm
Rim Thickness = 1.8mm
Density = 0.00825 (approx)

Hairspring
E (Modulus of Elasticity) = 221Gpa (this value is a guess based on a range of materials used within the industry for Hairsprings)
l (Length) = 260mm
h (height) = 0.30mm
e (thickness) = 0.12mm
θ (oscillation?) = 270°

I hope this doesn't give anyone a head ache! Any assistance would be appreciated.

 

[Navin A S]

The formula given is correct, but you need to make sure that all of the units are consistent. The Modulus of Elasticity (E) must be in Nmm2/rad, and the Radius of Gyration (r) should be in mm. To calculate the Mass Moment of Inertia (M.r2), you can use the formula M.r2=m.R1^2+m.R2^2, where m is the mass and R1 and R2 are the outer and inner radii of the balance wheel. The units for the Hairspring parameters (e, h, and l) should also be in mm. The hairspring oscillation (θ) should be in radians (since you mentioned the oscillation is 270°, this would be θ = 4.71 rad). For the spring constant (K), you can use the formula K = E.h.e^3/(2pi.l.θ), where E, h, e, l, and θ are all in the same units as above. Once you have all of these values, you can then use the formula T(Hz)= 2∏√[12.M.r.sup.2.l/E.h.e.sup.3] to calculate the frequency of the Balance wheel.