Thread: Bandsaw Blade tension - psi to lbs of force?

Hi Dean, the saw is looking the goods,
For a blade tension of 3600lbs by my rough back of the envelope working you need 2371psi, from what I can see and what you have said I have taken the cylinder to be applying the force on the extend stroke therefore the dia for the area calc is 50mm, if the force was being applied via the retract side your psi will be a lot higher, I have also taken the accumulator spring to be supplying no additional forces other than to act to oppose the force applied by the cylinder, if I havent got the mechanics of your saw setup correct then the psi reqd. will be different but I think I have and I have neglected any friction effects
The workings out
Attachment 193599



Pete

Thanks Pete,
I just found a cylinder manufactuters chart with dia/psi/force data which confirms your formula. Only thing I didnt make clear was the 3600lbs was for already both sections of the blade (1800 x 2), therefore we halve the psi to get a ball park figure of 1150psi. I'll get a 2000 rated gauge.
I also found the hand pump has an adjustable relief valve, this will be useful as the tension will easily be repeatable.

For those who have an agricultural machinery background, The cylinder came off an old Lely hay mower with the modular discs, also the spring came of the linkage safety breakaway, & the 30mm shaft on the idler band wheel was the main disc retension shaft (seriously strong bit of steel, I can rember Lely advising the nut tension was 100kg down force at 2 meters from the nut, I remember we used to do that with a pipe & crowbar attached to big stilson wrench

cheers,
Dean.

The other thing I was thinking was that if you knew the spring rate you can determine the force by how much the spring compresses. Spring rate is simply the force required to compress the spring a given amount, say 20kg/mm, so for a total deflection of 10mm you would have a mass of 200kg or a force of 2000N
A spring manufacturing place should be able to test it for you or rig something up yourself,



Pete

Pete, That becomes quite simple. The deflection of the spring can be measured by the reverse travel of the rod that the spring is carried on as I can lock the band wheel slide mechinism in place diverting all ram travel back against the spring, once I have the pressure gauge in place I'll know the oil pressure hence the lbs of force against the spring

Whilst at this stage I'm not sure whether the blade is correctly tensioned, I'm compressing the spring to about 90% of its capacity. I'm guessing this is a bit to high?
My instinct tells me I may need to use a spring that is slightly stronger but I'll know this when we test it cutting.
Or I guess I could just over-ride the springs capacity by continuing to pressurize the ram however blade tension will increase very rapidly & be much harder to keep a constant, thus defeating the role of the spring to absorb & store the hydraulic energy.

cheers,
Dean.

The advantage with knowing the spring rate beforehand is that once the saw is all together you then have two systems that should confirm what the other one says, it's just a bit of a double check on how much blade tension you have in a static situation at least cos once you start to cut something it all changes, and realistically all you need is enough tension to keep the blade on the wheels and to give you a straight cut, the least amount of tension is best for long blade life.
The spring should still be ok, change it tho if it compresses all the way before enough tension, as you say you will know....



Pete

Originally Posted by pjt

The spring should still be ok, change it tho if it compresses all the way before enough tension, as you say you will know....

Just thought of something, If it needs a bit more spring resistance I'll just mount a 2nd spring behind the rear rod to work as a helper spring.
Guess what!! My neighbour has 2 lely hay mowers, one is buggered & he only kept it for spare parts, I dont think he'll need the spring off it though

cheers
Dean

If it helps as a guide, I run my woodmizer blades at 2500 to 2700 psi.

Thanks nifty, I read somewhere on the forestry forum that the woodmizer cylinder has a 1" piston diameter, this would be why you use a higher pressure. the ram I am using has a 50mm piston.
However I would need to convert the surface area to compare the pressures. The chart I have says a 1" dia has .785" squared and a 2" dia has 3.142 squared.

cheers

Since you know you want 3600#, the cylinder has to give you that. When acting on the wheel axle, you get 1800 per blade side of the wheel.

Piston area is PI(r2) = 3.14 in2

pressure is force/area, meaning 3600#/3.14 = 1146psi or 1273 psi for 4000#

change piston area to suit. WOrking from a blade stress is just an equivalence of blade pressure*area and hydraulic pressure*area with allowance for lever arms etc.

pjt - why would the pressure be higher if you're running it from the retract side? Is there a lot of material attached to the piston on that side?

Originally Posted by bonox

pjt - why would the pressure be higher if you're running it from the retract side? Is there a lot of material attached to the piston on that side?

Yes, the retract side has the rod area which needs to be subtracted from the piston area, this area is called annulus area and as such is a smaller area so for same force output from both sides the pressure needs to be larger, alternatively you could pressure reduce the extend side if there was a requirement for equal force from extend and retract.


Pete

Originally Posted by spray-tech

Thanks nifty, I read somewhere on the forestry forum that the woodmizer cylinder has a 1" piston diameter, this would be why you use a higher pressure. the ram I am using has a 50mm piston.
However I would need to convert the surface area to compare the pressures. The chart I have says a 1" dia has .785" squared and a 2" dia has 3.142 squared.

cheers

spot on