Storing Power in a Spring - Get Help Now!

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SUMMARY

This discussion focuses on the calculation of energy stored in a spiral power spring, specifically for a project involving solar-powered motors and dynamos. The potential energy (PE) in a spring is calculated using the formula PE = 1/2 k x², where k is the spring constant and x is the amount of compression or extension. For torsional springs, the energy can be calculated using the angular form of Hooke's Law: PE = 1/2 κ θ², where κ is the torsion coefficient and θ is the angular displacement in radians. Caution is advised to prevent over-winding, which can lead to energy loss due to friction.

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  • Understanding of Hooke's Law and its applications
  • Knowledge of torsional springs and their coefficients
  • Familiarity with energy storage concepts in mechanical systems
  • Basic calculus for integrating force over distance
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Hi,

I'm doing a project where I need to store energy in a spring with the use of a motor powered by solar energy. The energy must be stored and after a while released to a dynamo which converts the energy to electrical energy. I would like to use a spiral power spring, but I can't find any formulas on how to compute the amount of energy stored in such a spring. Please help, any other suggestions will also be much appreciated.

Thank you!
 
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well the potential energy stored in a spring can be calculated using PE=1/2kx^2 where PE is potential energy, k is the spring constant and x is the amount of compression or extension (length).

Power is work over time...
 
Thanx, but a spiral spring does not extend or compress, it just gets wound up more tightly, like the spring in a watch. What equation will I use in this case?
 
I would image that
k = f(d\theta)
 
The energy stored in a compression (or tension) spring with a spring constant is determined by integrating the spring's force constant times distance.

Hooke's Law:
F = k*x

So:
PE = \int k*x dx = \frac{1}{2}k*x^{2}

For a torsional spring if you know the spring's torsion coefficient, a similar calculation can be used with the angular form of hooke's law:

Angular Hooke's law:
\tau = \kappa*\theta

PE = \int \kappa*\theta d\theta = \frac{1}{2}\kappa*\theta^{2}

Where Kappa is the spring's torsion coefficient, and theta is the displacement in radians. Kappa's units would be Newton-Meters per Radian.

http://en.wikipedia.org/wiki/Torsion_spring
 
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One thing to watch in using a power spring (like a watch spring) is not to let it get wound too tight. If it is over wound, the leaf to leaf friction will hinder the unwinding action and absorb some of the stored energy, thus robbing you of some of the input effort.

The V = (1/2) K theta^2 law works just fine for a power spring.
 
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