Awl --

Proly could calc this on a program called BeamBoy, but I'd have to dig it up, and re-figger how to use it, etc.
Suppose I have 28" vise jaws (say, 1x2 alum), and two 6" Kurt vises.

Where is the best place to put the vises, in X?

28 - (2x 6) = 16, so there's 16 unsupported inches.

One could then divide by 4, and allow 4" at each end, and 8" in the middle.

But, the middle is "tethered", so I

So mebbe 10" in the middle, and 3" at each end? 12 and 2??

I could "T" the middle span for extree rigidity, with another 1x2, but I'm tryna do this as quick and dirtily as possible (as usual).

Not a critical application, as these jaws will be holding a 27" length of 5/8 x 1.25 alum, for mostly drilling, and a little interpolation (3/4" c-bores)

Proly could calc this on a program called BeamBoy, but I'd have to dig it up, and re-figger how to use it, etc.

Where is the best place to put the vises, in X?

28 - (2x 6) = 16, so there's 16 unsupported inches.

One could then divide by 4, and allow 4" at each end, and 8" in the middle.

But, the middle is "tethered", so I

***know***the middle span can be more than twice as long as the end spans, as the ends are flapping in the proverbial breeze.So mebbe 10" in the middle, and 3" at each end? 12 and 2??

I could "T" the middle span for extree rigidity, with another 1x2, but I'm tryna do this as quick and dirtily as possible (as usual).

Not a critical application, as these jaws will be holding a 27" length of 5/8 x 1.25 alum, for mostly drilling, and a little interpolation (3/4" c-bores)

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