Bandsaw Blade Tension Jig |

The jig works by measuring the amount of stretch in the bandsaw blade as force is applied by the bandsaw tension spring. Every 0.001" of blade stretch is equal to 5,800 psi of tension over 5" of blade. (See the build and demo video) When resawing, blade manufacturers recommend a blade tension of 15,000 psi. Therefore, you would need 0.0026" of blade stretch in order to reach the recommended amount of tension. The wider the blade, the more force will be required by your bandsaw tension spring to reach the amount of stretch required for 15,000 psi of tension. The amount of tension (15,000 psi) should be the same regardless of the width of the blade; the amount of force required to achieve it will be different. The 5,800 psi above is derived from the following equation using Young's modulus for steel:29,000,000 = Young's modulus for steel.
S = blade stretch in inches.
D = length of blade measured in inches.
If we plug in a blade stretch of 0.001" for S and a distance of 5" for D we get : |

About | Contact | FAQ | Privacy | Disclaimer |

Copyright © 2006-2022 Garage Woodworks |