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Bandsaw tension gauge

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  • Bandsaw tension gauge

    Hi everyone, Just finished my latest project, a band saw blade tension gauge based on a Starrett gauge. I printed a picture of one and cut it out to use as a model. Mine uses a .001" cheap dial indicator, so the scap piece of aluminum I used to mill body of the gauge needed to be modified to house the indicator. The Starrett gauge uses an indicator that reads directly in tension, and appropriatly so, list price $340.00 US dollars. A friend recently read an article, relating to a woodworking bandsaw about a home made tension gauge made from a couple sticks of wood, steel pins, some clamps, and some feeler gauges. In the article it stated that on a .025" thick blade, a .001" stretch in 5.0" length equates to 6000 psi tension. Not sure how to verify this, so I just assumed it to be correct. Using a second dial indicator, I have determined that a .001" movement on the pivoting leg on my gauge at the blade clamping thumb screw, causes a .001" deflection of the needle on the dial indicator housed in the body of the gauge, which is a one to one ratio. Next, I need to find out what the proper tension should be on say, a .025" x 3/8" wide blade, or any other blade for that matter. I'm hoping that this will turn out to be a usefull instrument for my friend. It sure was a fun project to make. Could anyone confirm any of these figures? Any comments would be welcome. Thanks Mike Green
    Mike Green

  • #2
    Thanks for sharing. Nice looking job. Have the same on my to-do list.

    Found an old article in Fine Woodworking Feb 2001 about the subject. Regardless of the cross sectional area of the blade, a tension of 6,000 psi will result in a strain of 0.001â€‌ in 5â€‌. The FORCE required will increase linearly with blade area.


    • #3
      GKman, Thanks for the reply. That must have been the article my friend was refering to.
      You may notice that the shaft of the indicator is shifted to the side of the slot milled in body of the gauge. This is due to the pivot pin position not being equidistant between the thumb screw center and center line of the slot. I should have thought about that issue before I picked a spot to locate it.
      Mike Green


      • #4
        simply fantastic! Great job. Does your friend realize how good of a friend he has? You didn't knock that out in no hour !! Let us know how it works out.

        The one thing I'm confused about is does it meaure twist or tension? I thought a tension measurement would require 3 points for the blade to come into contact with and the measurement is taken against a fixed load against the center point? Up till now my tension indicator has always been by how red my palm gets from cranking on the adjustment
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        • #5
          Blade Tension

          Check out "The Home Metal Shop Club" newsletter for an article on "Band Saw Tension Readout" designed and built by a Club member. He fabricated a hydraulic load cell and is able to get a very accurate load reading on his band saw.



          • #6
            YOD, Thanks for the kind words. We all have at least one good friend that we are willing to help out in any way we can without hesitation don't we?
            Your right, it did'nt take an hour and that's alright with me.
            Besides, while working on projects like this, I get to practice and learn along the way as I'm still new at this stuff. Got a SB heavy 10 about a year ago and a small knee mill about three months ago.
            The gauge measures tension. As GKman stated, a tension of 6,000 psi will result in a strain of 0.001â€‌ in 5â€‌. For example if a blade requires a tension of 18,000psi my gauge(my friend's gauge) should read .003"

            JRW I checked out the article you referred to, and found the gauge very clever. With his system you get an instrument built right into the machine, very convenient!
            Thanks for the suggestion, looks like I may have another project on my hands.
            Mike Green


            • #7
              Beautiful work! I'm new at this stuff too; i've had a three-in-one machine for just over a year but there's very little chance i could do something that nice!

              I would get to the final cut in the piece and then bugger something up...


              • #8
                "I thought a tension measurement would require 3 points for the blade to come into contact with and the measurement is taken against a fixed load against the center point?"

                What you're describing is a tool for measuring tension ON the blade, measured in pounds. MGREEN's very nice tool measures tension IN the blade, measured in pounds per square inch.

                The three-point tool is commonly used in aircraft rigging for adjusting the control cables. It looks like this:

                Last edited by winchman; 11-06-2006, 05:23 PM.
                Any products mentioned in my posts have been endorsed by their manufacturer.


                • #9
                  Nice work, I never knew I needed one of those, I always just use Your old dogs method. Might be something to make when the shop is slow though.


                  • #10
                    Fasttrack, Thanks for the compliment. The gauge is'nt very pretty, but seems to work just fine. I"ve had my share of buggers, and am sure I have'nt had my last.
                    The cylindrical pocket for the dial indicator was turned on the lathe in the 4 jaw chuck,most of the rest of the work was done on the mill, with the exception of the outer curved section, which was band sawn following the circle that I scribed in while still on the lathe and then smoothed on the stationary belt sander. Mike
                    Mike Green


                    • #11
                      winchman, Thanks for clarifying the way the gauge functions for YOD.
                      I knew someone could put it in better words than mine.

                      mochinist, Thanks. There's probably lots of things we see on this forum that we did'nt know we needed til we see them. lol
                      This one did'nt cost no $340.00 niether!
                      Mike Green


                      • #12
                        Question on Tension.

                        That is a great gage. It is just what I’ve been looking for. I have a couple of questions. GKman says that “The FORCE required will increase linearly with blade area.” If I have a ¾ inch blade what should the indicator read for say 6000 lbs of tension? Would you take the .75 times 5 inches and get 3.75 for the area? I think that is about two times the area of the 3/8 blade. Would .001 equal 12000 lbs of tension? My other question is about the pivot arm. What is the correct distance from the blade attachment point to the pivot, and how far is it from the pivot to the indicator feeler?
                        Thanks again for sharing this design.


                        • #13
                          Very nice gage. I only have one point of contention. If you are looking for values in the thousanths .001-.010" you should be using an indicator with 10% resolution ie .0001" for readablity and error prevention.

                          Other than that... I like it.
                          Ignorance is curable through education.


                          • #14
                            If I have a ¾ inch blade what should the indicator read for say 6000 lbs of tension?
                            The value given above was .001" per 5 inchs for 6000 psi of stress. This isn't the same as 6000 lbs of blade tension. .001 is the amount of elongation at 6000 lbs tension per sq in unit area of material so to determine how much tension that is for your blade you need to calculate the cross sectional area of the blade and divide accordingly.
                            Free software for calculating bolt circles and similar: Click Here


                            • #15
                              neat project. to be clear it is a strain guage - its' measuring the elongation of the blade caused by force, stress and strain. Hooke's law says that the amout something will stretch depends on its length so the variables are load on blade, cross sectional

                              here's the formula

                              total strain = (load*length)/(cross sectional area*modulus of elasticity)

                              for example, if you put 200 lbs force on the blade, and modulus of elasticity is say 30 000 000 for steel (most steels are very close to this) and blade dimension are .02 *.5 and length between guage points are 6", then:

                              (btw complete guesswork on load & blade dimensions)

                              strain = (150* 6)/((.02*.5)*30 000 000)
                              = .003

                              you'd expect the blade to be .003 inches longer over 6" with a 150 lb load. that's the theory anyway

                              btw that 150 lb load on blade that dimensions .01 sq inches = 15000 psi
                              Last edited by Mcgyver; 04-25-2007, 11:33 AM.
                              in Toronto Ontario - where are you?