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Lee DeRaud
07-08-2005, 8:47 PM
Kurt's thread on tension gauges sparked a question deep in the moldy recesses of my brain...not wanting to hijack his thread, here's a new one.

Say I've got a 1/2" blade, 0.035" thick...call it 1/32" for convenience. That means there's 1/64 square inches of blade cross-section under tension. Multiply that by 15000PSI and I get about 235lbs of "pull" required to get that tension...and that's the minimum value I should be using. Ok, so far so good.

So I go out to the garage and look at the spring on that bandsaw: if I believe the markings, it's compressing less than an inch to get to the "1/2" setting. There is simply no way that dinky little spring is in the 300 lbs/inch range.

What's wrong with this picture?

Chris Barton
07-08-2005, 8:53 PM
Hi Lee,

You have come upon one of the universal truths about bandsaws. This is why Carter sells an after market spring for that very reason. I tension my blades to where they have less than 1/2" deflection under thumb pressure. I don't really know what that comes out to be. I have tuned my fence and wheels to deliver a clean and strait cut at this setting. You can buy some fancy guages to set this but, fingers seem to work best in my case.

chris

Lee DeRaud
07-08-2005, 9:02 PM
I was off by a factor of two: there is the same tension on the "up" leg of the blade as on the "down" leg, so there is twice as much cross-section to be tensioned, and therefore twice as much spring required.

My head hurts.:eek:

lou sansone
07-08-2005, 9:12 PM
you are right about doubling the total upward force.

lou

Mike Cutler
07-08-2005, 9:47 PM
Lee. There has been a lot of discussion concerning bandsaw tension, and the "proper amout of tension". I'm not sure that your example is 100% correct. I would put a 1/2" blade at .023 and the total cross sectional area at 1/2 that minus the area of the gullets. The force is exerted equally in all directions so I wouldn't double it. If memory serves me correctly, I got real anal about thes a few years back and computed the total force required to tension a 1/2" blade on a 14" BS at ~185lbs force, and took the stock Jet spring to work and put in it a force calibrator to see if it could deliver this value. Long story short, the Stock spring could only tension the blade to about 9000 psi max. I replaced it with the upgraded coil from Ittura. I tested this one and determined that it would supply a max of about 16,000 psi to the blade ( All this testing was done on NIST traceable devices). With this spring the "tension gauge" on the Jet became somewhat usable.
I tension my blades only to the point necessary to get good performance. My theory is why stress the blade, bearings and spring unnessesarily.
Bill Crofutt, from Grizzly posted some good info on bandsaw blade tension last year, it's worth the search.

Dale Thompson
07-08-2005, 10:09 PM
Lee,
Where did the 15,000 psi come from? Even mild steel alloys have a tensile strength of 50-100 kips/sq in.. That translates into 100,000 - 200,000 psi.

I agree with your post that the "spring" force has to be divided in half to achieve the correct tensile stress in your blade. :) With that said, however, even if we reduce the strength of the steel by a ridiculous 50% due to the annealing effect of the weld, we still have a tensile strength of 50,000 to 100,000 psi. :eek:

I'm not a metallurgist. I'm much closer to being pegged as the local "village idiot". :o On the other hand, I would be much embarrassed if I did not suggest that the failure of band saw blades that are left under tension is far more likely to be caused by "strain hardening" as a result of their taking a static "set" than from simple tensile failure. :confused: DUH!? What do I REALLY know anything about - NOTHING!! :) ;)

Dale T.

John Hart
07-08-2005, 10:27 PM
Just so I understand you correctly...with your calculations, are you calculating the theoretical force required to achieve a zero-deflection laterally of the blade?So far, you've convinced me that the spring can not achieve the theoretical tension that you want....but, with that said, the Bandsaw folks must know that...So, what is it that they are trying to achieve? I'm not trying to be a smart-aleck...I seriously would like to know, because I can't seem to get my tension right and I'm ruining blades.

BTW Dale...for a village idiot, you sound like you could do calculus in your head;)

Lee DeRaud
07-08-2005, 11:03 PM
Where did the 15,000 psi come from? Even mild steel alloys have a tensile strength of 50-100 kips/sq in.. That translates into 100,000 - 200,000 psi.That was the figure Kurt quoted for minimum tensioning required for a bandsaw blade to work properly, not the tensile strength of the blade itself, which I agree (and certainly hope) is at least an order-of-magnitude higher.

Dale Thompson
07-08-2005, 11:18 PM
BTW Dale...for a village idiot, you sound like you could do calculus in your head;)

John,
When you are starting with a blank slate (or head), anything is possible. ;) Calculus, biker babes :cool: , physics, biker babes, quantum physics, biker babes, metallurgy, biker babes, electronics, biker babes, old age with no biker babes. :( ;) BUMMER!! ;)

I know NOTHING about any of the above but, as the commercial says, "I stayed at a Holliday Inn Express last night". :) :D

Dale T.

Lee DeRaud
07-08-2005, 11:19 PM
Just so I understand you correctly...with your calculations, are you calculating the theoretical force required to achieve a zero-deflection laterally of the blade? So far, you've convinced me that the spring can not achieve the theoretical tension that you want....but, with that said, the Bandsaw folks must know that...So, what is it that they are trying to achieve? I'm not trying to be a smart-aleck...I seriously would like to know, because I can't seem to get my tension right and I'm ruining blades.No, I was just calculating the required strength of the tensioning spring for the blade tension values the vendors think are necessary for proper tracking or blade life or whatever. (I'm not sure "zero-deflection laterally of the blade" is even theoretically possible.)

Concensus seems to be that, no, the stock springs can't do that. Quite frankly, I'm surprised an aftermarket spring will do a lot better, given the limited dimension it has to fit into, at least on the 14" Delta.

That said, there also seems to be some "slack" in the equation, as some blade/saw combinations will apparently do what's being asked of them with tension values the blade vendors think are inadequate.

One think that tickles my brain cells is the tensioning procedure the Timberwolf people advocate, because the end goal there seems to be to 'tune' the blade just above its installed resonant frequency, pretty much without regard to static tension. (Yes, I know they're related, but I suspect the tension value you end up with using that procedure varies depending on the saw speed.)

Cecil Arnold
07-08-2005, 11:35 PM
Like Dale, I know more about biker babes, but trust Louis Ittura's work when he states that you need at least 15,000 lbs. tension and 25-30,000 is accptable (better). At 25,000 (on a gauge) I can not get 1/2" difelction on a 1/2" blade, even with the guides all the way at the top (12 1/2") on my MM16. As I recall from Louis' catilog (everyone should order a copy if you don't already have one) no stock spring will deliver the required tension and the after market springs, while better, are still marginally accptable.

Kurt Loup
07-08-2005, 11:43 PM
That was the figure Kurt quoted for minimum tensioning required for a bandsaw blade to work properly, not the tensile strength of the blade itself, which I agree (and certainly hope) is at least an order-of-magnitude higher.


Lee,

All I know about the tensioning issue comes from the Fine Woodworking article and the Iturra Design catalog. Iturra sells Lenox blades and the recommended tension for Lenox blades is 30,000 psi. I used 15,000 psi as my goal as Iturra used that number as a compromise tension for 1/2" blades on 14" saws. Apparently in their tests, they could not tension a 1/2" blade to 30,000 psi on a 14" saw and the compromise tension of 15,000 psi resulted in good results. The FWW article states blade makers recommend 15,000 psi for consumer grade saws, but 30,000 psi on industrial quality saws used for resawing. I have a MM E16 and had no trouble tensioning up to 15,000, but I didn't try to see if it will tension up to 30,000. There is no comparison between the size of the spring in the MM vs. my 12" Jet.

Kurt