Lathe Steady Rest

The issue with skateboard wheels is how close you can get them together. Unless they are not on the same plane you can’t get them close enough to touch all sides of an 11mm shaft. That’s why I’m suggesting the design use bearings instead, with a wide rubber band around the outside to give the ‘cush’ that provides protection to the wood surface. Backlash shouldn’t be a problem since we’re not really having it move around, the iris design should help it self center while spinning…

A spring could take up any slack or backlash and an O-ring wrapped around a bearing with a little E-6000 to hold it in place.

I don’t know how to contact each other but I think I have a winner that’s based on this. I’m ecited about this project because I could also use a steady rest


Bugger ! I just did what you did Bill, clicked on Josh’s link and lost what I had wrote, Oh well.
Attached are two DXF files one of my steady rest and some layout for Bill’s.
Mine has a line drawn under it at about 9" from the center to give an indication of size if it was used on your lathe, it will support about 170mm dia max.
Bill do you have a means of opening and measuring dimensions of a DXF file.
Bill, the lathe is what is called a Round Bed, does it have a grove down the length ?.
I really don’t think rubber rings would work well.

Josh I only had a quick look at your link, it reminded me of the method Lace Bobbin turners use, one hand holds the work piece and the other holds the chisel, a very delicate touch required.

I would find it hard to put my heart into some of the complicated options mentions, I believe in KISS.


MiniL-Srest-dxf.txt (107 KB)

Steady-rest-dxf-1.txt (91.2 KB)

That’s why the top one is offset. The two on the bottom are almost touching, so 11mm will just ride lower between them. The top just keeps down pressure on the que so it doesn’t hop out.

Josh has been a busy boy! I kind of like this one, it looks like the right idea. The center wheel moves the arms in and out, using the iris design. As Barry suggests we could just stagger the wheels to allow for tighter clearances without smaller diameters than easily available. Perhaps a four wheel with NE and SW in front and NW and SE in back and a hard limit of, say, 10mm apart? If we use springs to hold the iris in we could just manually pull it open to insert the work piece and let the springs pull the wheels in to fit.

This one is another example of the same type of design.

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Hmm, a Longworth chuck is pretty similar as well, just leave the center open.

Frank posted his build.

OK, this thing has been percolating through my head for a while now, so I think I have a better handle on the design. There are six part types used to create it: The two base pieces, the two ‘ankle’ pieces, the outer stand, the inner ring, four arms and four petals. There will also be some hardware, four skate wheels and potentially some pivot dowels.
The base, ankle and outer stand are each held at right angles to the other pieces in order to securely form a stand. Each base at the bottom, laying flat, each ankle upright forming a wall front to back on each base, the outer stand then attaches to both base/ankle to be the main stand. The outer stand is somewhat an inverted U shape with the upper portion a circle 13" in diameter with a hole 7" in diameter in the middle. The circle has four holes drilled equidistant around the perimeter 1" from the edge that will form the pivot points for the four arms. The two legs of the inverted U are each 6" wide, 7" apart and at the appropriate length to place the middle of the circle even with the lathe center. The outer stand has vertical grooves cut to hold the ankles, the outer stand and ankles have toes cut to fit grooves in the bases, the bases have grooves cut to fit the toes.
Now for the iris assembly. The four arms are curves with a radius of 12", forming about 60° of an arc, with the arc extended in all directions to be 2" wide. There are holes drilled in each end to form the pivot points. One end of the arm is attached to the outer stand, the other will be placed at pivot points on each petal. The petals have an outer arc (or potentially a straight edge) that is also 60°, the outer corners each have a hole drilled 1" from the edges to form pivot points. When the petals are placed in four quadrants of a circle the clockwise pivot will attach to the arm and the counterclockwise pivot will attach to the inner ring. The inner ring is a circle 13" outside and 7" inside, just like the outer base, but with four equidistant holes drilled 1" from the inside edge. These holes form the inner pivots connected to the counterclockwise pivots on the petals. The outer base outer edge has four protrusions that are each drilled to fit a spring. The inner ring also has the four extrusions drilled for the other end of the springs, plus a larger extrusion that forms the sizing handle. The petals each form a triangle such that if the outer curve were the base the triangle would be an isosceles with the height appropriate to place wheels near the center of the steady rest. There are holes drilled at the inner point such that there is 1" of space surrounding the center of the hole. There are four skateboard wheel, two mounted to the top surface of the petals and two mounted to the bottom surface of the petals.
When fully assembled there are four layers of material surrounding the center line of the lathe. Two skate wheels touch each other at one side and two skate wheels touch each other on the other side. The springs at this point are relaxed, but not fully relaxed since we want some tension even when fully in. As you pull the sizing handle the springs are stretched and the skate wheels are pulled equally apart until you have enough room for your work piece to fit. As the piece rotates in the lathe the wheels are pushed, along with spring pressure, tighter in toward the center. I might have my clockwise-counterclockwise directions above reversed, but you can just turn the steady rest around backwards if that is the case. when done with your work piece you pull the sizing handle to release the wheels and remove your work.

What am I failing to describe? I’m thinking there’s a reasonable chance this can be cut from two 2’x2’ pieces of plywood, 1/4" or 1/2" should be plenty, but the grooves cut into the bases, ankles and outer stand will need to reflect the thickness chosen.

Bill I’m sorry I lost focus on this, weather has been raising hell with arthritis. I have just acquired Autodesk LT and am loving it for cnc. I can almost justify the $390 a year subscription. If you can provide a rough sketch I could try and get something drawn up for you.

Using rubber wheels doesn’t seem like a good idea to me, but otherwise it’s a nice design.

I warned about my drawing skills. :slight_smile:

[attachment file=84311]

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The wheels aren’t really rubber, skate wheels are much harder than that. They should have very little give to them, yet have enough that they won’t mar the wood pinched between them. I will, of course, do lots of testing with cheap pool cues before attempting anything with my good ones. :slight_smile:

Here’s a good one, from RichQ, vintage mid 70s…

[attachment file=84314]

That has a double full splice butt, giving eight points. Lots of inlays, including some more complex than squares and dots. It was their top model for the year, though they did simplify it some for the next year.

And I’ve tried drawing it myself using Inkscape, which just hammers into my head again how much trouble I have with drawing… I also found I’m going to have to tweak a bit from perfect. The center of the lathe is 9.5" from the table, the top of the pipe is 3.5", which gives us the 12" piece size defined for this model lathe. The problem is that if the outer stand is 13" outside and 7" inside we’re going to overlap the pipe by an inch. So … we need to cut away that inch at the bottom, meaning we likely want the pivot holes for the arms at 45°, 135°, 225° and 315° to make sure the arm doesn’t hit the pipe. We are also likely to have the arms curved toward the center at full retraction, again so they don’t hit the pipe (it might work to have them straight as well). We then need the inner ring to be designed so that it’s outer diameter is less than 13", likely 11" instead. Since it’s pivot points are 1" off the inner edge it won’t interfere with the pipe. Because the main piece has that bit cutaway at the bottom we can’t put a spring down there, so we’ll likely use three springs to give the return tension (they should be just off relaxed when the wheels touch in the middle, so there is always some tension). We need fingers sticking out to hold each end of each spring on both the outer stand and the inner ring and we need the spring fingers on the inner ring to miss the pipe. We may find ourselves dealing with springs rubbing on arms but at first glance I don’t think it’ll be a problem.

I have been picking at this and I think three rollers would be better mainly because it’s easier to set the points of contact. Do you want to be able to steady a 12" workpiece in this? Do you know what size the rollers will be?

With four rollers we can have each pair on a different plane, so they can get down to the minimum spacing (they can actually all touch in the middle with zero spacing, though of course they’d have issues spinning in that case). With three rollers you have them 120 degrees apart from each other, so with the three petals in their minimum position there will be a gap sized by the wheel diameter. IIRC when I did the calc for that skate wheels were too big to get the spacing down to 10mm or under, but lets see for real. :slight_smile:
The wheels Ryan offers in his shop for the Low Rider are 60mm diameter. There are smaller wheels available, but even the small wheels used on traditional skates are 50mm or larger. Soddy’s circles equation tells us for three identical circles (curvature y = 1/radius x) placed to each is tangent to the other two, the center and outer circles tangent to all three (radius s) are defined such that (3y+s)^^2 = 2(3*y^^2+s^^2). So, with x=30 we have:

(1/10+1/s)^^2 = 2*(3*1/900+1/s^^2)

Now ignoring the special case where s=0 and using the formula for quadratic equations that gives:
s=4.641016 or s=−64.641016
or the inner circle at just under 7 9mm, which meets our spec.
That looks like it either works or I can no longer do math. :slight_smile: Maybe Ryan will take three wheels, set them next to each other and do a sanity test.
I don’t see the need to do 12", that would require the petal/arm to use nearly no width around the circle, mechanically that would give very little to no leverage to the petals and I’d be afraid the vibration would be enough to overcome the spring tension. Hence my suggestion of an inside diameter of 7" with the petal pivots to the inner ring at 8" diameter. We can do close to half the full range there.

Though I suppose if we changed the design to have 16" outer diameter and 12" inner diameter we could cut a slot in the table and arrange things such that the arms and petals don’t quite hit the pipe, allowing for a full 12" part…

Not quite high on my list, I think. :slight_smile:

Oh, and your design there is missing the arms. Arms sit between the outer ring and the clockwise edge of the petal, the counterclockwise side of the petal pivots on the inner ring. Four pivot points in total for each petal/arm combo.

I’m thinking of something like this and this for the pivots.

[attachment file=86232]
[attachment file=86233]

The OD of the spacer is 10.5mm and would easily work. I have a tapered pen and it seems to catch at just a hair under 10mm. You probably need to stagger the top one a tiny bit to be safe.

Thanks Ryan! Just what the calc actually said. I shouldn’t have been using my ‘rule of thumb’ calc all this time.

I saw equations and got the chills, cold sweats, panic set in…straight to CAD and I felt better.