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

[ 11-04-2006, 11:40 AM: Message edited by: Mike Green ]

Mike,

When you first posted this, I don't believe the picture was showing up, and I had only a vague idea of what you were talking about. Seeing the picture now it makes sense. A very clever design.

I'm afraid I don't have any experience about correct blade tensions, but it looks like you did a very nice job making the gage. Congrats!

Paula

Paula,
Thank you for the kind remarks.
I thought perhaps nary a sole had replied because I had commited a Faux Pas in posting the gauge, it being non specific to South Bend.
I posted as well on another forum, and the data in question seems to be cofirmed.
Although the gage is not specific to this forum, I did use my South Bend 10L to bore the pocket for the body of the dial indicator I didn't think to take a pic of the set up in the 4jaw at the time, but do have a pic of the gauge without the indicator in it. Again thanks for the reply.
Mike

There is a solution to bandsaw blade tension that is being used and has been for many years. My Craftsman horizontal metal cutting bandsaw that is at least 30 years old has a coil spring with a load dimension indicator to give the proper blade tension. A lot of vertical bandsaws have a similar feature and replacement springs are available at several places.
My cheap Harber Freight had no such feature, so you had to guess at the correct tension. I fixed mime with an easily installed Die spring that is calibrated for load so you can easily adjust for proper blade tension. Mine cost about $ 5 at a local Reid suplier.
To get the load say for a 6000 psi tension I use the folowing formula.
Tension equals the load divided by the 2 blades times blade thickness times blade width.
So in my case solving for load in pounds equals 6000 X 2 X .o25" X .5" which equals 150 pounds load.
The Reid Die springs are calibrared for load in pounds per one tenth inch deflection. So I picked a spring JS- 765A. with a load of 31.2 pounds per one tenth inch and tightened the load screw to compress .5 " which can e done with a simple indicator pointer.
In my case all I had to do was lengthen the load screw 3" to take the 3" long, 3/8 ID, 3/4" Od spring.
Works great.
Walt

You mean tapping the blade with my fore-finger, like I've been doing for the last 40+ years, isn't the right way to determine blade tension?

Mike,
You're on the right track.

Your numbers sound right (but I don't have the Fine Woodworking article handy that was probably the source).

The reason your strain gauge (i.e., you're measuring blade stretch)is handy, is because one setting works for ALL BLADE SIZES. Here's the reasoning:
Blades are designed to run at 18,000 psi tension. If you multiply (divide?) this by the Young's (elastic) modulus of steel, you get a strain of 0.0006 in/in, or 0.030" over the 5" length of your gauge. If your blade is bigger, e.g., twice the cross section, you'll have to apply twice the force, but you are still getting the same strain.

I've made the simpler wood + metal peg gauges that use feeler guages to measure the 0.003" deflection.

The reason using force gauges (e.g., measuring the deflection of a spring, as installed in all bandsaws) isn't as handy, is that you have a different setting for each blade cross section (which you have to estimate) and springs take a set, i.e., lose calibration if left tensioned.

One suggestion: You probably want a dial indicator that sensitive enough to read 0.1 thou per mark vs. 1.0 thou. But they're a lot more expensive, which is why I haven't built a gauge like yours and have actually found a surplus electronic strain gauge to measure force, will add a microprocessor and display to show force up to about 1,000#. An advantage here is that it works while the saw is running

Mike W

McSquared,
I believe you meant

"strain of 0.0006 in/in, or 0.003" over 5" length" rather than 0.030"

The problem is that very small stretches are being measured and thus you really need the 0.1 thou per division type of dial indicator $$$.

Pete

GreatBot,

Yes, I meant "0.003" (as correctly stated in the next paragraph

Agreed. You really need a 0.1 thou per division dial indicator. (Anybody know of a good source for these under ~$100?)

Mike

mcsquared how about making the blade clamp to pivot ten times shorter than indicator to pivot.

here you go http://www.taunton.com/finewoodworking/pages/w00013.asp?
Reg...