Timberwolf BS blade technical question

[Michael Adelong]

Does anyone know the actual (psi) of a properly tensioned Timberwolf BS blade?

I find it a little odd that TW publishes a lengthy process for tensioning their blades that is basically a guide to making a "controlled guesstimate" of the proper tension.

It would be nice if they included a statement like "If you own a tension guage or dial indicator that can accurately measure blade stretch, .00Y" inches of stretch over Y inches of blade will yield ZZZZ pounds, which is the correct tension for our blades.". They must know what the actual numbers are. Right now, all of their customers are using random tension settings based on each users' interpretation of "flutter". Wouldn't they be better off if some of us were using the actual, correct tension? Maybe a lot of their users would have better results, or at least more consistent results. They might end up selling even more blades.

Does the silicon steel in the TW blades use the same 5" / .001" / 6000 psi numbers as other blades? 30K is "normal" tension. What is "low tension"? 15K? 20K? 25K?

Thanks,
Mike

[Allen Bookout]

I am sure that someone here knows the answer but not me. I have never had the need to know as the flutter method seems to work perfectly as far as I am concerned. It could be that they do not list a figure as they do not want customers overtensioning their blades and that is certainly a possibility with all of the methods of checking tension using deflection, strech, etc.

I am sure that someone here will come up with some numbers for you.

Allen

[Ken Garlock]

Michael, call up Ittura Designs, and get a copy of their catalog. Their phone number is in the Ittura ad in most WW mags.(no website.) It is 80% bandsaw education, and and 20% sales.

In the interim ask yourself, if the silicon steel was the answer to manufacturing bandsaw blades, why doesn't everyone use silicon steel? :confused:

[lou sansone]

hi guys
good question and one that points out the issues with band saw blade tension. I have yet to have anyone who believes in tension gages give me a good engineering reason for using them. Dale Thompson showed some time ago that given Young's modulus for steel at 29,000,000 psi that when you tension a blade to 15,000 to 30,000 psi what you are doing is stretching the blade in the range of 1/20 to 1/10 of 1%. What it the engineering significance of this??? Is there anyone here who thinks they know? I would be very interested in the reason.

Now the flutter method on the other hand is a little more intuitive to my mind. I guess the way I look at it is you are trying to "detune" the blade so that you definitely do not have it tensioned at its fundamental frequency. At that frequency you are definitely going to have maximum blade displacement and probably poor performance due to excessive vibration. I am not sure if you are simply detuing the blade or making sure it is tuned to a much higher frequency where the nodes are spaced so close that you think the blade is really standing still. Maybe someone else wants to get into this conversation with a little more expertise in vibrations. I will try to claim ignorance being an EE.

anyway. If the resonance standing wave phenomenon is the reason behind tension of band saw blades then I have to believe that what you really want is to look at the blade to see when that resonance disappears .

Lou

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html#c3

[lou sansone]

wow

I would have thought that someone would have something to add to my post. maybe some of you are reading it and thinking that I copied some gobbde-gook from a google search and pasted it in for fun. Well I didn't. Just in case some think I am trying to be a smart alic, it is a serious response. I really hope some other folks here here might have some further insight in to this topic. What is it about using 15 to 30 kpis that stretches the blade 1/10 of 1% that makes this the ideal tension ?

I know that some have said that they approach blade tension by adjusting the blade until it seems to cut the best and then mark the tension on the band saws Gage for the next time. I agree 100% with that approach. I honestly believe that is exactly the flutter method without actually watching the blade "stop fluttering".

The only reason I can possibly see for a band saw tension Gage is to use it only after you have found the correct tension either by trial and error and looking at the quality of the cut, or by using the flutter method first and then backing off the tension and then reapplying the same tension, this time with the Gage attached to the blade.

lou

[nic obie]

Hi lou,

I think blade tension expressed in psi is directly related to band bowing and feed rate. I have a feeling that the manufactures might think the same. However, I've been known to be wrong before.:D

It's just a easy to read measurement. Think digital vs analog.

[Ted Christiansen]

Youngs Mod (E) = Stress (sigma) / Strain (epsilon)

Youngs Modulus (E) is fairly constant for all types of steel. E is not an indicator of strength, but of stiffness - the steeper the curve the stiffer the material is. Strength on the other hand is a measure of how much stress the material can undergo before deforming permanently (yield strength), or breaking (ultimate tensile strength).
<O:p</O:p

E represents the slope of the stress-strain curve, and for steel is 29x10<SUP>6</SUP> psi. A typical stress-strain curve is shown in the attached image.


For 15,000 psi tension:

strain=stress/E = 15,000 psi/29000000 psi = 0.0005172 inch per inch. For a five inch initial length (i.e. dial caliper set to five inches and clamped to the untensioned blade), you need 0.0025862 inches on the dial caliper to get this tension.


In force terms (force in the blade causing this tension) for this same 15000 psi tension:
1" blade .05" thick: F=stress * area = 15000 psi * (1" * 0.05") = 750 pounds
.5" blade .04" thick: F=stress * area = 15000 psi * (.5" * .04") = 300 pounds

[Allen Bookout]

Well Lou, May be few responses indicate that many of us agree with you. It is a little deep for me but I will agree the best that I can. I certainly understand the last two paragraphs and unequivocally agree.

Four steps 1.Tension with the flutter 2.Test the cut 3.Mark the scale 4.Retension to the same mark the next time. Works every time. No formulas to figure out or test do dads to fiddle with. Piece of cake.

At least that is the way I see it. Allen

[Mark Singer]

wow

I would have thought that someone would have something to add to my post. maybe some of you are reading it and thinking that I copied some gobbde-gook from a google search and pasted it in for fun. Well I didn't. Just in case some think I am trying to be a smart alic, it is a serious response. I really hope some other folks here here might have some further insight in to this topic. What is it about using 15 to 30 kpis that stretches the blade 1/10 of 1% that makes this the ideal tension ?

I know that some have said that they approach blade tension by adjusting the blade until it seems to cut the best and then mark the tension on the band saws Gage for the next time. I agree 100% with that approach. I honestly believe that is exactly the flutter method without actually watching the blade "stop fluttering".

The only reason I can possibly see for a band saw tension Gage is to use it only after you have found the correct tension either by trial and error and looking at the quality of the cut, or by using the flutter method first and then backing off the tension and then reapplying the same tension, this time with the Gage attached to the blade.

lou

Lou,
You are amazing! When I studied engineering...I learned Youngs Modulus and the Modulus of elasticity.....but I don't remember much if any of it..
I think a simply supported beam stress is wl squared divided by eight...how do you remember this stuff:confused:...Moments about a point equal zero...6.02 x 10 23 is Avagadros number...wait my dryer went into a Carnot cycle:confused:

[Bernie Weishapl]

Michael I have been using the flutter method since I got both bandsaws. Have had no reason to own a gauge and don't care. Both saws cut straight and resaw is great so I see no reason to know how many psi my blade has on it. I put a new 1/2" Woodslicer blade on my Griz, used the flutter method to tension the blade two weeks ago. One of the gentleman at Highland Hardware suggested that method for their blade also. After I got the blade tensioned I cut just a hair over 1/16" walnut veneer for a friend who was visiting. He told me all he did was some light sanding and scraping after he glued it. Finish resawing the rest into 4/4 walnut lumber. All I know is it works.

[Allen Bookout]

I agree Bernie. It kind of reminds me of what I did for a living all of my working years. I flew all kinds of aircraft and never knew the chemical make up of the fan blades in the engines. But you know what------they seemed to run fine just the same.

I am not putting down the technical people on this forum as I know that for chemist, engineers, designers, etc. this type of information is vital to their various vocations but for the end user there is a point where it is not necessary to obtain excellent results.

Allen

[John Towns]

Wow!
This thread is yet another example of the excellent information and knowledge transfer that is available to us here at SMC. I can't wait for another day so I can read up on whatever topics are on everyone' minds!

John

[Allen Bookout]

Welcome John!!!!! Allen

[Lee DeRaud]

In force terms (force in the blade causing this tension) for this same 15000 psi tension:
1" blade .05" thick: F=stress * area = 15000 psi * (1" * 0.05") = 750 pounds
.5" blade .04" thick: F=stress * area = 15000 psi * (.5" * .04") = 300 pounds(scratches head) I think there's a factor of two missing: the cross-sectional area of the blade has to be doubled...it's not just a straight piece of steel, it's a loop.

[Bruce Shiverdecker]

Going through all the dynamics of tentioning is good for everything BUT a Timberwolf Blade. It is a LOW tention Blade. If you tention it to the guide on your saw, IT WILL BREAK WAY BEFORE IT'S NORMAL LIFE CYCLE. How do I know.........................The people at Timberwolf told me, through my boss.........The owner of the local Woodcraft store. (Mentioned above as "The Flutter Method")

WHY? Haven't the foggiest idea, but when a manufacturer tells my to snug up the blade, turn the saw on and ONLY tighten it till the wobble goes away, or else you will break it..................I listen. That's why I'm adding my.02.
Hope it helps someone get the full life expectancy out of a blade.

Bruce

[Kirk (KC) Constable]

Michael, call up Ittura Designs, and get a copy of their catalog. Their phone number is in the Ittura ad in most WW mags.(no website.) It is 80% bandsaw education, and and 20% sales.

In the interim ask yourself, if the silicon steel was the answer to manufacturing bandsaw blades, why doesn't everyone use silicon steel? :confused:

???

Timberwolf is the only 'good' blade I've used...but they're far and away, hands down, without any question, vastly superior to 'name brands' availble in stores. I've never paid any attention to the flutter thing...I put the blade on, tension until it 'feels right' when I wiggle it, and cut away.

[lou sansone]

Youngs Mod (E) = Stress (sigma) / Strain (epsilon)

Youngs Modulus (E) is fairly constant for all types of steel. E is not an indicator of strength, but of stiffness - the steeper the curve the stiffer the material is. Strength on the other hand is a measure of how much stress the material can undergo before deforming permanently (yield strength), or breaking (ultimate tensile strength).
<O:p</O:p

E represents the slope of the stress-strain curve, and for steel is 29x10<SUP>6</SUP> psi. A typical stress-strain curve is shown in the attached image.


For 15,000 psi tension:

strain=stress/E = 15,000 psi/29000000 psi = 0.0005172 inch per inch. For a five inch initial length (i.e. dial caliper set to five inches and clamped to the untensioned blade), you need 0.0025862 inches on the dial caliper to get this tension.


In force terms (force in the blade causing this tension) for this same 15000 psi tension:
1" blade .05" thick: F=stress * area = 15000 psi * (1" * 0.05") = 750 pounds
.5" blade .04" thick: F=stress * area = 15000 psi * (.5" * .04") = 300 pounds



as another has said, you need to double the total value for the appropiate spring ( 1" blade will really need 1500 pounds of force )

now on to the reason to strain the blade 1/20 of 1% ( 0.0005172 inch per inch. ) !!! why that value. what makes it sacred? Is it because it is right below the yield point of the steel? I don't know.

thanks lou

[tod evans]

i`m a "plucker", i pluck the blade like a guitar string and stop when it sounds right......and no i couldn`t tell you what note that is....after a couple hundred blades you know...02 tod

[lou sansone]

ok guys I finally figured it out on this band saw tension thing..

it turns out that many steels tend to yield at about 1/3 of 1% strain. so it looks like band saw blade mfgs simply play it real safe with the blade tension in the 15kpsi to 30kpsi range ( 1/20 to 1/10 of 1% strain ). this gives the blade a decent amount of tension, but pretty far away from actually yielding the steel for plastic deformation.

So there is actually a real engineering reason for the recommendation, but I believe that the recommnedation is "caveman like" because it still does not remove the very real possibility that the blade will be accidentally tensioned at the fundamental resonance of the blade. This would cause the worse possible cut.

Lou

[Mark Singer]

OK for the non engineers...and I may be wrong ....it has been a few years:confused:
Youngs Modulus is the same as the Modulus of Elasticity and is the coefficient of elasticity of a material that describes its characteristics under stress, until it reaches the Yeild Point where deformation and strain occurs...more or less:rolleyes:

[Mark Singer]

i`m a "plucker", i pluck the blade like a guitar string and stop when it sounds right......and no i couldn`t tell you what note that is....after a couple hundred blades you know...02 tod

Can you play the tune from "Deliverence":confused:

[Vaughn McMillan]

hi guys
good question and one that points out the issues with band saw blade tension. I have yet to have anyone who believes in tension gages give me a good engineering reason for using them. Dale Thompson showed some time ago that given Young's modulus for steel at 29,000,000 psi that when you tension a blade to 15,000 to 30,000 psi what you are doing is stretching the blade in the range of 1/20 to 1/10 of 1%. What it the engineering significance of this??? Is there anyone here who thinks they know? I would be very interested in the reason.

Now the flutter method on the other hand is a little more intuitive to my mind. I guess the way I look at it is you are trying to "detune" the blade so that you definitely do not have it tensioned at its fundamental frequency. At that frequency you are definitely going to have maximum blade displacement and probably poor performance due to excessive vibration. I am not sure if you are simply detuing the blade or making sure it is tuned to a much higher frequency where the nodes are spaced so close that you think the blade is really standing still. Maybe someone else wants to get into this conversation with a little more expertise in vibrations. I will try to claim ignorance being an EE.

anyway. If the resonance standing wave phenomenon is the reason behind tension of band saw blades then I have to believe that what you really want is to look at the blade to see when that resonance disappears .

Lou

http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html#c3 Lou, I think you've nailed it on the head. By "detuning" the blade to just above (or below) the natural harmonic frequency, you're putting it at a tension that's least likely to resonate. (Guitar players: Think of trying to sound a harmonic over the first fret. Or halfway between there and the nut.) It's been a long time since I slept through my acoustical physics class in high school, but practical experience and observation with guitar strings tells me your explanation is correct.


Can you play the tune from "Deliverence":confused:
Why, yes I can. With a guitar pick on my front teeth, for that matter. William Tell Overture, too. :D

- Vaughn

[tod evans]

Can you play the tune from "Deliverence":confused:

no mark i can`t carry a tune in a bucket! and yes there really are some pretty scarry folks down here in the sticks, just go to wally-world on the 1st, they`re out in droves.:).....tod

[lou sansone]

OK for the non engineers...and I may be wrong ....it has been a few years:confused:
Youngs Modulus is the same as the Modulus of Elasticity and is the coefficient of elasticity of a material that describes its characteristics under stress, until it reaches the Yield Point where deformation and strain occurs...more or less:rolleyes:

exactly right mark.. the blade manufactures must want to keep a reasonable amount of tension in the blade, but make sure that it is well away from the yield point of the steel so you don't start permanently deforming the blade.


Vaughn is also correct by his analogy in the idea of the guitar string and its natural resonance.

All of this further supports with real science, that other than telling you that you are at the ideal tension, based upon observation using the flutter method, they are useless.

So to the original point of this thread, Suffolk machinery is spot on telling folks to use the flutter method as the best way to tension a blade.

lou