Metal band saw blades

[chucketn]

I have an old Atlas 912 12" band saw I would like to use for cutting aluminum plate, brass, and mild steel. I understand cutting speed requirements of metal, and am installing a jack shaft arrangement to slow it down.
The saw takes an 83" or 6" 11" blade. I have not found this length blade online. What sources for custom length metal blades do you use/recommend? I do not have the ability/capability to weld my own.

Chuck in E. TN

 

[Stan]

Try searching the Yellow Pages under Machine Tools. Most medium to large cities have at least one business that makes bandsaw blades. I get blades for my 4x6 bandsaw in any conceivable pitch or style for about the same price as the catalog blades that are only available in limited pitches.

 

[JimM]

For my 4x6 I just buy longer blades of the correct width, cut them down to length and silver solder the joint. Made one of these jigs and it works a treat.

http://www.rcgroups.com/forums/attachment.php?attachmentid=2257825

If you are buying a new blade then splash out the extra cash and get a M42 / bimetal one, it makes a world of difference to the cut.

HTH

Jim

 

[gwolf]

I use a@h abrasives 1-800-831-6066
Very good quality, will make blades any length.

 

[tgss2000]

I can recommend Lenox at http://www.lenoxtools.com/Pages/Category.aspx?category=Bi-Metal , which is the link to their page on bi-metal blades. They will make them up to whatever length you want. Like Jim I would recommend bi-metal, but they do like to run with a lot more tension than carbon steel woodworking blades. Depending on your bandsaw you may need to change the spring or make other mods to get the preferred tension. There have been cases in which people have changed to bi-metal on a saw designed for woodworking only to find that the frame of the saw is not really beefy enough. But I think that's a pretty rare occurrence.

Good luck
Tom

 

[Omnimill]

As an aside, can I ask what the criteria is for blade size on this type of machine? I've seen 1/8" and 3/16" bandsaw blades (wood cutting) offered but not seen many machines that list the blade size as going that low, most are 1/4"?

Vic.

 

[tgss2000]

I can't provide a definitive answer Vic, but I've always assumed that if the saw can take wider blades, then it can also run narrower ones, providing you adjust accordingly - tracking, blade tension, whatever. I know my bandsaw's tension adjustment doesn't show anything smaller than a 1/4" blade, but the manual quite happily discusses the use of 1/8" and 3/16" blades, what they are used for, what the minimum cutting radius is etc. Doesn't tell you how to adjust the tension though :shrug:
Tom

 

[Stan]

My limited knowledge of metal cutting bandsaws is that they use a specific width of blade because the blade rides on metal wheels (no rubber tire) with the teeth extending past the edge of the wheel. That is the blade is wider than the rim of the wheel.

 

[tgss2000]

Well, you learn something every day ;D I wasn't aware of that Stan, but then my bandsaw isn't really a metal cutting saw. It's a 16 inch Delta 3 wheeler which is supposedly good for wood and metal. To switch to metal cutting mode you remove a drive pulley and install a shorter drive belt, thus reducing the cutting speed. But the wheels have tires, which are wider than any blade the saw will accommodate, and the blade rides in the middle of the wheels, which are crowned.
Tom.

Edit: O.K. I'm an idiot *club*. I just used my bandsaw with a 1/4" wide blade, and while jacking up the tension realized that the tension scale goes down to 1/8" wide blades. A senior moment on my previous post. As far as using bimetal metal cutting blades on it is concerned, it is also calibrated for 0.020" thick blades, and all the bimetals I've used are about 0.025" thick. So the calibration on the saw is doubly useless when using bimetal blades - it's calibrated for the wrong material and the wrong thickness. I have done my best to allow for this when using bimetal blades, and it seems to work OK, but to some extent I'm working in the dark.