Originally Posted by
John TenEyck
Steve, the math is simple. We start with the stiffness of steel, known as Young's modulus, which is 29,000,000 psi. Yes, 29 million. Wood, for comparison has a value of around 1.5 million psi, for comparison. We can discuss how Young's Modulus is measured if you find those numbers hard to believe but they are real values and measured every day by steel and other producers to verify product properties.
OK, Young's modulus (Y) = Stress / strain. Stress is the tension we want to apply to our bandsaw blade, the 15,000 psi you referenced, for example. Strain is the amount the blade has to be stretched to put 15,000 psi on it. Strain is measured as stretch per unit length, so inches/inch, for example.
Since you picked 15,000 psi for the stress, we can use the above equation to calculate how much we need to stretch the blade to get that stress.
Strain = Stress / Y = 15,000 psi / 29,000,000 psi = 0.0005 inches/inch. Most all the tension meters you see measure strain over some defined distance, known as gage length. My tension meter has a gage length of 12 inches, so to put 15,000 psi on a blade I need to see a stretch of 0.0005 inches/inch x 12 inches = 0.006" on the dial gage. I typically run 24,000 psi on my carbide blade, which means I'm looking for the dial gage to move 0.010".
You'll note the tension on the blade is irrelevant to the size of the blade. It doesn't matter if the blade is 1/4" wide x 0.020" gage or 1-1/4" x 0.042", if my meter shows a stretch of 0.006" over 12" it will have 15,000 psi on it. Of course, the spring force required to apply 15,000 psi to those two blades will be very different. But how much? OK, now we need to know the cross sectional area of the blade. If we forget about teeth and just say a 1/4" x 0.020" blade is a rectangle of those dimensions then the cross sectional area of it is 0.005 square inches. The 1-1/4" x 0.042" blade, again, disregarding the teeth, has a cross sectional area of 0.0525", more than 10X higher.
OK, we're nearly there. Tension = Force/Area, or Spring Force = Blade Tension x 2 x Cross Sectional Area. For your 15,000 psi on the 1/4" x 0.020" blade, 15,000 psi x 0.005 sq. inches x 2 = 150 lbs. Where'd the 2 come from you ask? Because the blade is a loop the spring has to apply 15,000 psi of tension to both sides of the loop so it takes twice the force to do that. So you need a spring capable of applying 150 lbs of force to the blade loop to get a tension of 15,000 psi. To do the same on the 1-1/4" x 0.042" blade the spring would have to be capable of applying a force of 1575 lbs. No car lifting involved, but still a lot of force.
Nothing irrational about it once you understand the math.
John