I have used round column mills so I am familiar with the alignment problem. While reading yet another post on this question/problem, an idea came to me. I have not seen this anywhere before, but it would not surprise me if it has been thought of before so I do not make any great claims. I do not have a round column mill now so I can not test this. Anywhere, here it is:

First the rough, 3D sketch:

Now some explanation.The four crossed circles are pulleys. The lines labeled A, B, C, and D as well as the two almost vertical lines leading from the uppermost pulleys to the weight are all small diameter steel cable. Or perhaps some heavy fishing line would do? The A and B runs are from the mechanism of Detail A, on the front or side of the mill to the first two pulleys. These two runs are approximately equal in length and are at approximately equal angles from the vertical. They are also in the same vertical plane, so those first two pulleys are equal distances to each side of Detail A and probably mounted on the ceiling.

The cable runs marked C and D are horizontal runs across the ceiling and they go to two more pulleys that are mounted as close to each other as possible, without rubbing. These runs do not have to be equal in length. But their lengths are constant so they can be neglected.

The two near vertical runs from these pulleys to the weight are, by definition of equal length. I show these pulleys separated by a small distance, but you probably want them as close as possible to each other so the runs to the weight are as close to vertical as possible.

Detail A shows the indicator mechanism which is fixed to the head of the mill. The two cables are would around the shaft on opposite sides of a central hole which prevents any slippage. A number of turns are needed to provide a sufficient amount of movement left and right. Thus, the overall length of both of the cables remains the same. Since the vertical runs to the weight are always equal and the C and D runs are always the same, if the mill is moved to one side or the other, the A and B lengths will be different and the only way for that to happen is for the dial to move. If we keep the dial reading the same, then A and B are equal and the mill head is in the same rotational position.

The Weight allows the head to be moved up and down. If the head moves down, the weight moves up. But both cables are moving the same amount so if the rotation of the head is the same, then the indicator stays the same. Vice-verse for moving the head up.

I would make the indicator arm as long as possible for the best sensitivity. A rough calculation of the important factors follows:

Assume 10" indicator length. That puts the scale at a 10" radius. This may seem a bit long, but I needed it for a reasonable shaft size below.

Assume divisions about 0.04" apart on that scale and let them represent 0.0005" of head movement.

That puts the divisions about 0.229 degrees apart. And a +/- 0.025" scale would be about 4" wide.

Assume the cables at A and B are at about 45 degrees. (It can be any angle, but 45 is a good one so I use it.)

If the head moves 0.0005" to the side, then the cable runs at A and B will each change by about

. . Change in A = change in a / sin (45)

. . Change in A = change in a / 0.707

Or . . Change in A = 1.414 X change in a

Thus, the cable lengths change by about 1.4 times as much as the left to right movement.

We want about a 0.229 degree rotation in the shaft with a 0.0005" horizontal movement (change in a).

That means a 0.229 degree rotation in the shaft with a 0.0007" change in the cable length.

The effective radius of the cable wound on the shaft is the radius of the shaft (R) plus the radius of the cable (r). This is a bit of an assumption, but it should be a fairly good one.

Assume we are using 3/64" diameter cable: that was the smallest I could find with a quick search. It's radius is 0.0234".

The effective radius we want is given by: Re = 0.0007/sin(.229)

. . Re = 0.1768"

Now for the actual radius of our shaft:

. . R = Re - r

. . R = 0.1768" - 0.0234"

. . R = 0.1533"

And our shaft diameter is 0.3067"

That is one set of numbers that would work. I did a spreadsheet to be able to fiddle with them so if you have any suggestions for changes, I can easily evaluate them. One that occurs to me is to shoot for a 0.250" shaft diameter.

The mill would probably have to be anchored to the floor to keep it in the same position, but that is probably desirable anyway.

One of the biggest advantages of this scheme is the indication of the error is right at the mill head itself so it should be easy to use while tightening the bolts that secure the head to the column. And it is all mechanical: no batteries to replace or electric cords to route.

There is the obvious disadvantage is you will have a couple of cables in front of the mill. This can be minimized by mounting the shaft of the indicating mechanism as high as possible. Perhaps the pointer arm would be point down instead of up. I would think that they would not be too objectionable. And you probably would want some kind of disconnect or adjustment in case you want to swing the mill to one side or the other.

I do not see any reason why this would not work.

Perhaps one of you who has a round column mill would like to try this. If you do, please post the results as I am highly curious.

First the rough, 3D sketch:

Now some explanation.The four crossed circles are pulleys. The lines labeled A, B, C, and D as well as the two almost vertical lines leading from the uppermost pulleys to the weight are all small diameter steel cable. Or perhaps some heavy fishing line would do? The A and B runs are from the mechanism of Detail A, on the front or side of the mill to the first two pulleys. These two runs are approximately equal in length and are at approximately equal angles from the vertical. They are also in the same vertical plane, so those first two pulleys are equal distances to each side of Detail A and probably mounted on the ceiling.

The cable runs marked C and D are horizontal runs across the ceiling and they go to two more pulleys that are mounted as close to each other as possible, without rubbing. These runs do not have to be equal in length. But their lengths are constant so they can be neglected.

The two near vertical runs from these pulleys to the weight are, by definition of equal length. I show these pulleys separated by a small distance, but you probably want them as close as possible to each other so the runs to the weight are as close to vertical as possible.

Detail A shows the indicator mechanism which is fixed to the head of the mill. The two cables are would around the shaft on opposite sides of a central hole which prevents any slippage. A number of turns are needed to provide a sufficient amount of movement left and right. Thus, the overall length of both of the cables remains the same. Since the vertical runs to the weight are always equal and the C and D runs are always the same, if the mill is moved to one side or the other, the A and B lengths will be different and the only way for that to happen is for the dial to move. If we keep the dial reading the same, then A and B are equal and the mill head is in the same rotational position.

The Weight allows the head to be moved up and down. If the head moves down, the weight moves up. But both cables are moving the same amount so if the rotation of the head is the same, then the indicator stays the same. Vice-verse for moving the head up.

I would make the indicator arm as long as possible for the best sensitivity. A rough calculation of the important factors follows:

Assume 10" indicator length. That puts the scale at a 10" radius. This may seem a bit long, but I needed it for a reasonable shaft size below.

Assume divisions about 0.04" apart on that scale and let them represent 0.0005" of head movement.

That puts the divisions about 0.229 degrees apart. And a +/- 0.025" scale would be about 4" wide.

Assume the cables at A and B are at about 45 degrees. (It can be any angle, but 45 is a good one so I use it.)

If the head moves 0.0005" to the side, then the cable runs at A and B will each change by about

. . Change in A = change in a / sin (45)

. . Change in A = change in a / 0.707

Or . . Change in A = 1.414 X change in a

Thus, the cable lengths change by about 1.4 times as much as the left to right movement.

We want about a 0.229 degree rotation in the shaft with a 0.0005" horizontal movement (change in a).

That means a 0.229 degree rotation in the shaft with a 0.0007" change in the cable length.

The effective radius of the cable wound on the shaft is the radius of the shaft (R) plus the radius of the cable (r). This is a bit of an assumption, but it should be a fairly good one.

Assume we are using 3/64" diameter cable: that was the smallest I could find with a quick search. It's radius is 0.0234".

The effective radius we want is given by: Re = 0.0007/sin(.229)

. . Re = 0.1768"

Now for the actual radius of our shaft:

. . R = Re - r

. . R = 0.1768" - 0.0234"

. . R = 0.1533"

And our shaft diameter is 0.3067"

That is one set of numbers that would work. I did a spreadsheet to be able to fiddle with them so if you have any suggestions for changes, I can easily evaluate them. One that occurs to me is to shoot for a 0.250" shaft diameter.

The mill would probably have to be anchored to the floor to keep it in the same position, but that is probably desirable anyway.

One of the biggest advantages of this scheme is the indication of the error is right at the mill head itself so it should be easy to use while tightening the bolts that secure the head to the column. And it is all mechanical: no batteries to replace or electric cords to route.

There is the obvious disadvantage is you will have a couple of cables in front of the mill. This can be minimized by mounting the shaft of the indicating mechanism as high as possible. Perhaps the pointer arm would be point down instead of up. I would think that they would not be too objectionable. And you probably would want some kind of disconnect or adjustment in case you want to swing the mill to one side or the other.

I do not see any reason why this would not work.

Perhaps one of you who has a round column mill would like to try this. If you do, please post the results as I am highly curious.

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