Blade Tension Gage ?

[David Ragan]

Hi everyone,

I have a question that has been bothering me for some time; have researched it-still stumped.

Bandsaw blade tension gage.

I know about Young's modulus of elasticity coefficient (new word), and it is a lot of math.

The deal is...........you put the two arms of any size/thickness/width blade b/t the two arms of the gage, crank up the tension, and the gage measures the amount of stretch of the blade.

Of course, if you stretch a flimsy 1/4" blade 0.003" that could be, say, 10K psi
If you stretch a 1" blade 0.003", that would take a lot more tension, of course, and may represent 25 psi

The psi numbers above are intended for illustrative purposes only.


So..........if the tension gage senses the same amount of distance moved between it's two arms in each blade-how does it know that one blade is 1/4" and the other is 1", and much more tension is there?

If you had to take the reading and consult a table supplied by the manufacturer, that would make more sense, as the main variables are the cross section of the blade, then you have the material/steel that is being stretched. But there is no such table..........just the distance. What am I missing here?

As always, thank you!

[Bill Space]

I think the answer relates to the pull on the blade in pounds, versus the psi felt by the blade. In your example you are stating pull in PSI, but I think you meant to say pounds.

A larger blade requires more pounds pull to stretch the same amount as a smaller blade.

The amount of stretch is related to PSI applied to the blade. A larger blade requires more pounds applied to result in the same PSI.

Therefore, for blades made with identical materials, stretch equates to PSI. A larger blade requires more pounds pull on it to feel the same PSI, and to stretch the same amount as the smaller blade.

This is my take on it...

Bill

[David Ragan]

I was able to find a thread from SMC 2007, or thereabouts and am reading it.

Seems like Young's modules of "steel" is 29,000,000. The cross sectional area of steel being stretched, is seemingly a very minor or insignificant factor. Also, I take it, the alloys of the steel. the length being stretched is the same for any particular gauge.

So, what you wind up w is a very expensive, but super accurate vernier caliper that can measure down to the 0.001". I have read that the 0.001" is equal to 2K psi to 6K psi.

Mark Duginske initiated the article in Jan of 2007. He has, of course written two excellent bandsaw books. I have one of them. Am on the third page of 22 of that thread.

[Dennis Ford]

Yes the math is confusing. The good news is that you only have to do some math once, just measure the stretch. The math required has to do with the length over which you are measuring the stretch. Typically you would extend your calipers to ?? (something less than full extension), clamp them to the blade, zero the display and then tighten the blade. As long as you start with the same extension each time; the same amount of stretch is the same strain regardless of the blade size.
This does not require a super expensive caliper, a inexpensive digital caliper will be good enough if you are not silly about the precision needed.
An adequate job can be done by the sound the blade makes when you pluck it with a little experience.

[Bill Space]

Seems like only the two of us find this question interesting!

I think the answer to your question is pretty straightforward.

Young's modulus (E) is a number that shows the relationship between force applied in PSI and elongation of the material.

E = ( PSI) / (elongation )

If E & (elongation) are known, PSI can be easily calculated:

E x (elongation) = (PSI)

E is know for steel (and MANY other materials)

(elongation) is measured by dial indicator or other method

PSI is therefore easily calculated.

PSI can be displayed on the dial indicator face for ONE material (one Young's modulus value). Increased stretch represents increased PSI.

Because Young's modulus represents the relationship between PSI and elongation, the actual size (cross sectional area) of the blade does not matter.

The designer of the tension gage must take into consideration the length of the blade section being measured when determining how to display PSI on the dial indicator face.

If 0.001" could equal a range of 2K to 6K PSI, as you have read, I would then think a more sensitive measurement device is called for. Such as a diall indicator that measures to 0.0001 inch rather than 0.001 inch.

Edit: I see there are three of us now...

Bill

[John TenEyck]

The math is straight forward, but I won't bore you with it. The key is that at any level of spring force the stress in the blade is inversely proportional to it's cross sectional area. So a 1/4" blade will have twice the tension of a 1/2" blade at the same spring force, assuming they have the same thickness and Young's modulus.

I'm happy to review the math in detail if anyone is interested. I've done it at least a couple of times, and you can probably find it by doing a search. It sounds like the OP may already have found it. Dennis described the procedure you would follow to measure blade tension directly. It's not hard, nor expensive, and fun to do once, but pretty much unnecessary to getting your saw to cut well. Use a quality blade appropriate for what you are cutting, use a quality spring, and adjust your saw so that the blade cuts straight and parallel with the miter slot. There's not much more to it.

John

[Bradley Gray]

+1 on using pitch to set tension. A person could trial and error perfect tension and then check the blade for pitch with a chromatic tuner. i have a fairly good sense of pitch and just do it by ear.

[Chris Padilla]

I just use my finger and deflect the blade until I can't push it past a 1/4". This seems to work fine on my MM20 band saw.

[Jerry Thompson]

With my acute hearing I tighten the blade until I hear the atoms start to moan and back off one DBa.

[Alan Schaffter]

All that engineering stuff and expensive tension gauges are all good and well, but to what end? What determines the recommended tension for a specific blade, other than possibly some number provided by the blade (or bandsaw) manufacturer? And how does the blade (or bandsaw) manufacturer determine that number? Really, does each blade have a magic number where it cuts best, and if so what criteria and method was used to determine it? The only way is subjective testing- what setting allows a particular group(?) of testers or woodworkers to make good cut(s) in test wood(s), over some number of test cuts. But the setting will likely change with stock parameters- species, thickness, moisture content, grain orientation; and blade parameters- grind, hook angle, set, cleanliness, sharpness, steel alloy, stress hardening, etc., etc. So, the only thing a precision blade tension gauge gives you is a discrete number that may or may not mean anything in an imprecise application. Plus, it is only as good as the original determination, as good as the people who determined it under their test conditions.

Assuming the guage is indeed precise and repeatable (it really doesn't need to be accurate), it may give you a reasonable ability to return the same blade to the same tension setting which can be helpful if other conditions are the same. It will certainly get you in the ball park. You sure can spend a lot of money to get in the ball park, or as Duginske suggests, you can just use one of the traditional manual methods or the marks on the saw, then adjust as necessary until you are getting a cut that is "good for you."

[David Kumm]

lAbsent the math, the usefulness depends on the saw and blade to some extent. I run 1" carbide blades on 30 and 36" saws so the blades are long and expensive. Each saw and blade has its " sweet spot " and the only way I can guarantee repeating it is with a gauge. If I used the saw every day, had any musical ability, or even good hearing, sound might work. Different tensions can cause a blade to track differently on crowned tires and $300 blades are not only expensive to break, but can chew up the crowned tires which are a total pain to replace. That can cost more than the blade. In some situations, the gauge is cheap insurance. Dave

[lowell holmes]

When I'm concerned about blade tension, I use a technique I read somewhere, and it works well for me.

I over tension the blade a bit, then with the saw running, back off the tension until flutter takes place. Then I tighten the tension until the flutter stops. I then tighten the tension one quarter turn. I didn't invent this, but saw it somewhere. It works for me.

I will be behind the saw when doing this. Watch the video below.

https://www.youtube.com/watch?v=chyo9chuwJs

[David Ragan]

I'm still here, and feeling better all the time about putting my Starrett gauge up for sale.

Am also reading through this topic on SMC Feb 2007.

I love all the engineering input-can't grasp a lot of it, but take some comfort that folks who know more than me have been over this ground before.

How disappointing to finally have confirmed that all I have is a very expensive fine resolution vernier caliper.

[John TenEyck]

Cheap verniers work just as well:

26556629516_2f75d2d010_h.jpg

Used it once to determine that the tension gage on my saw was about right with an Iturra spring. But since it's just a standard set of 6" verniers, I use it for lots of other things.

John