Replies

1. 
   JohnWW | Jun 22, 2010 03:20pm | #1

   It isn't at all intuitive, but if you work out the physics and math of blade tension it turns out that any blade, no matter what the size, will stretch the same amount per inch of length when brought to a certain tension.  The figures I gave in the article and video work for any size blade, when the blade stretches the amount given, about 3 thousandths of an inch over the 5" distance between the clamps, the blade is tensioned to just above 15,000 psi whether the blade is 1/4" wide or 2" wide.

   The only question then is what tension you want, and this turns out to be related to the the capacity of the machine, not the blade type, size, or the type of work.  The blades can easily handle 30,000 psi, and this what is used on large industrial saws, but this is more stress than the frames, springs, and bearings of smaller band saws were designed to handle so the figure of 15,000 psi was more or less arbitrarily chosen.  I found in fact that most machines couldn't reach even this lower psi because the tensioning springs are undersized.  Since the article was written most manufacturers have gone to beefier springs so that most modern small shop saws can be tensioned to around 15,000 psi without collapsing the spring which isn't good for the machine since a crushed spring no longer works as a shock absorber.

1. 
   WillGeorge | Jun 23, 2010 08:57am | #2

   JohnWW

   ..... if you work out the physics and math of blade tension it turns out that any blade, no matter what the size, will stretch the same amount per inch of length when brought to a certain tension .....

   Sir, could you direct me to a topic/white paper (whatever) that shows the physics and math? I have looked but I did not find anything. At least that I understood.    :>)

   I am NOT questioning the validity of your statement! NO.. I was not a physics major. I just want to understand that property of metal. I assume that 'no matter what the size' means any width and thickness? Does this apply to any grade/type of steel blade?

   Thanks in advance.

1. 
   JohnWW | Jun 29, 2010 10:38am | #4

   I worked out the math from the strength of materials section of my "Machinery's Handbook" which is the small very thick book that is supposed to go in that square drawer in the middle of machinist's tool boxes.  The stretchability of steel doesn't vary that much between types of steel so it turns out not to be an important factor.  The formulas work no matter what size the blade is, a thicker or wider blade will require more tension than a small one, but the blades will all stretch the same amount over the length between the two attachment points for the jig when the blade reaches the proper tension.

2. 
   jg0258 | Jun 24, 2010 05:55pm | #3

   Thank you

   Makes things easier. 

2. 
   aaron_k | Aug 17, 2010 08:09am | #5

   stress-strain curves

   The reason for the simplicity of calculation is based on the fundamental relationships between elongation of materials and the applied load. This relationship is depicted in stess-strain curves:

   http://en.wikipedia.org/wiki/Stress%E2%80%93strain_curve

   Basically, if you assume the same material - and john says that they're about the same, then for a given strain (elongation) a coresponding stress (pressure) needs to be applied. The reason this relationship is so useful is that you don't need to factor in geometry. The stress part of it, which has units of pressure, normalizes the applied load (force) for any cross section.

   Of course, now if you want to figure out how FAR you need to compress the spring, you need to know a lot more exacting details.