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Hi everyone, I'm a new boy.

I am a retired fabricator/welder, used to most steel structure work & the mathematics thereof. Most things can be drawn - projections, geometry, onto a steel worktable & in 2D.

I now need to know how the helix of a banister for a spiral staircase can be drawn, say in 60 degree parts. This means one revolution of the stair would need 6 similar arcs welded together for the banister.

The length of the whole banister (per rev) is the hypotenuse of the rt angle triangle made by the horizontal of Pi x staircase diameter & the vertical rise of the (usually) 12 steps in one revolution. This hypotenuse can then simply be divided by 6 to give the length of arc of two risers, ie, 60 deg of turn. Dividing the whole hypotenuse by Pi gives the working diameter of the arc just described. This can now be laid out on the table & the steel bent to this shape. Repeat this until the whole banister length has been acheived.

But this procedure becomes increasingly inaccurate as the arc length is increased - the circle principle collapses to something like a part of an ellipse, or worse, a section of sine. Lots more metal bashing required!

Is this simpler than I make it sound? No calculus please - its not much used in the welding industry...

Many thanks for any help or comments, Mike.

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Just to confirm, you are solving by finding the length of the banister as:

where:

$d = diameter \; of \; the \; stairwell$

$c = circumference$

$h = height$

$l = length \; of \; the \; banister$

$c = \pi d$

$l = \sqrt{c^2 + h^2}$

divide the $l$ by 6 and you have the exact length of the pieces.

You are anticipating some stretching in the material and you are compensating to some extent. The stairwell diameter is being taken as the ID from the inner side of the banister and this in fact constitutes a major portion of your compensation!

If this is the case then it would seem to be more an issue with your bending technique. I hope I haven't overlooked anything or I might look foolish :/

You aren't trying to bend them into their circular form and then twist them are you? Because I don't think this will work for you . . .

 changed d to h 

Edited by Xittenn
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Just to confirm, you are solving by finding the length of the banister as:

where:

$d = diameter \; of \; the \; stairwell$

$c = circumference$

$h = height$

$l = length \; of \; the \; banister$

$c = \pi d$

$l = \sqrt{d^2 + c^2}$

divide the $l$ by 6 and you have the exact length of the pieces.

You are anticipating some stretching in the material and you are compensating to some extent. The stairwell diameter is being taken as the ID from the inner side of the banister and this in fact constitutes a major portion of your compensation!

If this is the case then it would seem to be more an issue with your bending technique. I hope I haven't overlooked anything or I might look foolish :/

You aren't trying to bend them into their circular form and then twist them are you? Because I don't think this will work for you . . .

You have the facts exactly. Yes the technique stated will work, brute force & oxy heat will shape almost anything, but its not subtle!

A better technique but more impracticable would be to roll a circular coil of 1 rev of diameter "hypotenuse", then stretch it like a spring the distance of 12 risers to get the correct spiral. Between a tree & a 4-wheel drive vehicle might do it. Health & safety would stipulate 'eye protection' of course!

This stretch would decrease the diameter back to the helix base dimension of Pi x OD & should fit exactly.

But neither is the elegant solution I seek for my work table. (I do not have a 4-wheel drive either)..

Mike.

You have the facts exactly. Yes the technique stated will work, brute force & oxy heat will shape almost anything, but its not subtle!

A better technique but more impracticable would be to roll a circular coil of 1 rev of diameter "hypotenuse", then stretch it like a spring the distance of 12 risers to get the correct spiral. Between a tree & a 4-wheel drive vehicle might do it. Health & safety would stipulate 'eye protection' of course!

This stretch would decrease the diameter back to the helix base dimension of Pi x OD & should fit exactly.

But neither is the elegant solution I seek for my work table. (I do not have a 4-wheel drive either)..

Mike.

Sorry Xittenn, the hypotenuse L = root [(Pi OD)sqd + (height)sqd]

not what you put.

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Sorry the d was supposed to be an h. . . the only way to compensate if you are trying to bend it circularly and then twist would be to shape it as an ellipse. The equations for an ellipse that will ensure proper form after twisting are pretty complicated and rely heavily on the properties of the material. This isn't really a math question this is something I would suggest you research whithin the trade itself.

MHO!

and ID was simply my thoughts on the issue with stretch, as I had posted .. . .

Edited by Xittenn
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Sorry the d was supposed to be an h. . . the only way to compensate if you are trying to bend it circularly and then twist would be to shape it as an ellipse. The equations for an ellipse that will ensure proper form after twisting are pretty complicated and rely heavily on the properties of the material. This isn't really a math question this is something I would suggest you research whithin the trade itself.

MHO!

and ID was simply my thoughts on the issue with stretch, as I had posted .. . .

Thanks very much for your replies Xittenn, most helpful. I thought ellipse in part at least the answer as any one point on the helix at any time is touching the minor axis of the oblique cutting plane ellipse of a cylinder at the helix angle. Sufficiently true for a small arc but getting progressively worse without 'recutting' the plane through the cylinder repeatedly towards infinity.

Bye, Mike. PS, I think a small elephant might be more outre to pull the rolled steel coil than a 4-wheel drive!

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