Laurie Tedcastle
Member
- Messages
- 125
- Location
- NW Northants (UK)
Part 2
If you are interested in the actual force applied by the tensioning screw,
then: Tension force = tension stress x cross section area
The cross section of my blade is 1/2" wide (from the bottom of the tooth
gullet to the back of the blade) x thickness 0.025".
Thus the force = 18000 x (0.5 x 0.025) x 2 = 450lb. The blade operates on
both sides of the tension wheel - hence the multiplier 2.
To take the elongation measurement, remove all tension from the blade and then
take up the slack. Fit the gauge and set the dti scale to zero. Increase the
blade tension until the dti registers 0.002" - simples!
Lennox recommend that the blade tension should be released if the saw is not
going to be used for some time. I tend to forget!
If you use bimetal blades on these small saws, I would still stick to the lower
stress range. A particular weak point on my saw is the tension wheel slider
block. The top of this cast iron block is not very thick and has the 2 grooves
machined in the sides for the sliding location. Underneath, is a lug (threaded
for the tensioning screw) and a separate lug for the tensioning wheel spindle.
The area between the 2 lugs may fracture if excessive tension is applied.
Blade Temperature:
I've been wondering for a while just how hot a blade gets during cutting and
what the corresponding blade extension would be. So did 2 test cuts 6" (150mm)
long through 25mm thick steel plate. 1 cutting dry, 2nd with drip feed neat
cutting oil. Blade allowed to return to ambient in between each cut. I have a
damper fitted to my saw to reduce vertical vibration and help control downfeed.
I suspect the downfeed rate should have really been significantly increased for
this length of cut. The chips were rather fine.
Temperature rise about 7degC in both cases, measured with an infra red
themometer.
Now, linear expanson of steel = coefficient of expansion x temp rise x length.
So over my gauge length we get 0.000012 x 7 x 3.150 = 0.00026"
The reduced tension stress is now 18000 x 0.002 / 0.00226 = 15929psi.
So still above the minimum recommended 14500psi. But if you are doing multiple
consecutive cuts, you may get sufficient expansion (and loss of blade tension)
to cause the blade to slip on the driving wheel, or a poorly controlled cut.
For setting up 6x4 bandsaws, the guide (search the web for the pdf)
"blade tracking and adjustment for 4x6 bandsaws" by John Pitkin, is excellent.
Just be aware that the guide refers to adjusting the tension wheel tilt and
many saws do not have the feature as described. Much of the guide may also be
applicable to the new smaller saws on the market today.
If you suffer from blades sliding off the blade wheels, check the blade wheels
for roundness. I had to remove and skim mine in the lathe to remove up 0.010"
ovality - problem cured (thanks to Peter King in New Zealand for the tip in
Model Engineers Workshop magazine many years ago).
Laurie
If you are interested in the actual force applied by the tensioning screw,
then: Tension force = tension stress x cross section area
The cross section of my blade is 1/2" wide (from the bottom of the tooth
gullet to the back of the blade) x thickness 0.025".
Thus the force = 18000 x (0.5 x 0.025) x 2 = 450lb. The blade operates on
both sides of the tension wheel - hence the multiplier 2.
To take the elongation measurement, remove all tension from the blade and then
take up the slack. Fit the gauge and set the dti scale to zero. Increase the
blade tension until the dti registers 0.002" - simples!
Lennox recommend that the blade tension should be released if the saw is not
going to be used for some time. I tend to forget!
If you use bimetal blades on these small saws, I would still stick to the lower
stress range. A particular weak point on my saw is the tension wheel slider
block. The top of this cast iron block is not very thick and has the 2 grooves
machined in the sides for the sliding location. Underneath, is a lug (threaded
for the tensioning screw) and a separate lug for the tensioning wheel spindle.
The area between the 2 lugs may fracture if excessive tension is applied.
Blade Temperature:
I've been wondering for a while just how hot a blade gets during cutting and
what the corresponding blade extension would be. So did 2 test cuts 6" (150mm)
long through 25mm thick steel plate. 1 cutting dry, 2nd with drip feed neat
cutting oil. Blade allowed to return to ambient in between each cut. I have a
damper fitted to my saw to reduce vertical vibration and help control downfeed.
I suspect the downfeed rate should have really been significantly increased for
this length of cut. The chips were rather fine.
Temperature rise about 7degC in both cases, measured with an infra red
themometer.
Now, linear expanson of steel = coefficient of expansion x temp rise x length.
So over my gauge length we get 0.000012 x 7 x 3.150 = 0.00026"
The reduced tension stress is now 18000 x 0.002 / 0.00226 = 15929psi.
So still above the minimum recommended 14500psi. But if you are doing multiple
consecutive cuts, you may get sufficient expansion (and loss of blade tension)
to cause the blade to slip on the driving wheel, or a poorly controlled cut.
For setting up 6x4 bandsaws, the guide (search the web for the pdf)
"blade tracking and adjustment for 4x6 bandsaws" by John Pitkin, is excellent.
Just be aware that the guide refers to adjusting the tension wheel tilt and
many saws do not have the feature as described. Much of the guide may also be
applicable to the new smaller saws on the market today.
If you suffer from blades sliding off the blade wheels, check the blade wheels
for roundness. I had to remove and skim mine in the lathe to remove up 0.010"
ovality - problem cured (thanks to Peter King in New Zealand for the tip in
Model Engineers Workshop magazine many years ago).
Laurie
