What's this 9.5" long Test bar telling me?

[Susquatch]

9.5" long, 1" dia. test bar.

@thestelster is correct. Did you get rid of bed twist first?

And as @Rauce said,what is your bed wear like?

 

[thestelster]

Craig, go the Learning Library in this Forum and access the "Testing Machine Tools" by Georg Schlesinger PDF. Go to the lathe section pg. 49 and test in his given sequence, don't move on to the next sequence until you have measured and adjusted as necessary. He has different criteria for different types of lathes, tool-room lathes having the highest degree of accuracy.

 

[PeterT]

And now we have come full circle on the bed twist or bed wear showing up as deflection on an indicator reading. This was the reason for the CAD math earlier. But the number is crazy in terms of how much distortion would have to occur. Example, assume:
- perfectly straight 1" OD test bar aligned to spindle axis
- we set the indicator exactly at mid point of test bar, at 9:00 position viewing bar from TS end (red arrow)
- we traverse carriage down to end of test bar & needle reads 0.003"
- so we assume some kind of bed twist condition at TS end. I'm showing 2 example sketches that could result in the same measurement condition. 1) a bowed bed where TS end is elevated relative to HS but the TS ways are in the same horizontal plane as HS ways 2) a twisted bed where TS end is rotated relative to HS end
- indicator is clamped to carriage so it goes along for the ride as carriage is traversed. Indicator ball therefore rides up the circular section test bar, extends & gives us an incremental reading
- now to the CAD diagram. Indicator starts out at red arrow & zeroed. Indicator translates to green arrow extends & reads 0.003". How much vertical distance Y would be required on a 1" diameter bar to yield horizontal displacement of 0.003". Answer = 0.055" That is a crazy amount over short bar. It would have to be larger yet on the actual TS feet. Same thing for bed wear, nobody is going to have that kind of material loss in the bed.
- (ps the lathe bed twist example isn't quite correct because it depends on where the mean rotation center occurs & the indicator stem is no longer horizontal, but we're just visualizing here)

So we talk about wear & twist, but we should be able to back this up numerically. I must be off base, but where?

 

[Susquatch]

So we talk about wear & twist, but we should be able to back this up numerically. I must be off base, but where?

I think maybe your mind got trapped by the earlier discussion.

I'm going to say that the measurement you are talking about does not matter. Nor does its relationship to the vertical displacement of the bar matter. It is sooooooo small as to be meaningless.

In my view, only wag (side to side) misalignment is measured by traversing the side of the bar from end to end. Any up and down misalignment as per your drawing is ignored.

All nod (up and down) misalignment is measured by traversing THE TOP of the bar from end to end. And coincidentally all side to side misalignment is ignored here.

I wish I had your cad skills to create a drawing for you.

I should add that this only works if there is no wobble in the bar. It must be concentric. If it is not concentric, then some math is involved to factor out the concentric error.

 

[Susquatch]

@PeterT -

Here is a markup of your drawing

After reading your note several times, and also looking at your photos, I think I see where you are going with this.

The purpose of the test bar is to adjust head nod and wag. I don't think it is the best way to align (remove twist) in the bed. But I like how you think. Maybe some combination of the wag/nod approach can be used to evaluate and correct the bed too. I will have to noodle that separately.

In any event, bed twist is typically removed using a machinists level - not using indicators. Bed twist must be done before head alignment.

First one evaluates and corrects bed alignment (twist), then one evaluates and corrects head spindle alignment (wag & nod), and then one evaluates and aligns the tailstock. Not all lathes can be adjusted in all three areas. But even if they can't, I think it's important that the operator knows what they have so they don't chase their tails trying to machine something to tolerances that are not possible on their machine.

 

[Susquatch]

@PeterT -

OK, I quickly noodled your conundrum. I believe it is possible to measure twist the way you suggest. But without actually doing it, I think it gets really complicated.

The relationship between wag/nod and bed twist is quite intimate. I'm sure one could write the equations to determine the precise relationship and solve for the alignment errors. But I don't think the method you described is the appropriate way to do it. Very few of us have measuring equipment that is accurate enough for that. I certainly don't.

However, I can very accurately measure both top and side and then use geometry to determine the center of the bar very accurately (ignoring the tiny error you described so well). If I know the precise location of the center of the bar - relative to the bed - and if I know the concentricity and sag of the bar, and if I measure the bed using a different setup between centers (both up/down and sideways), I can calculate all the unknowns.

HOWEVER - I think this is all WAY too complicated for the benefit it provides. I can't begin to describe how painful it would be to measure, calculate, adjust, measure, calculate, adjust.... ad nosia until satisfied. I think the conventional methods are much easier to do, are more direct, much faster, and quite good enough.

 

[thestelster]

Last year, my lathe was turning a taper, which it shouldn't have been doing. So I decided to re-level and check. I have a 12" Starrett #98 level @0.005" per foot, which is accurate, but not even close to the accuracy of the Starrett 199, (0.0005" per foot!). So I shimed the feet until it was as perfect as my level would allow. Using my dumbell test jig, only held by the chuck, I took very light cuts off the aluminium rings and measured. See the results in the photo. So my lathe even after leveling was cutting a taper. So more than likely my headstock is out of alignment with the bed. There is no feature to adjust head stock alignment. It would need to be scraped true. I'm not doing that. I decided to purposely impart twist in the lathe. I did a little trigonometry, and shimed up the TS end front foot. See results.

 

[Susquatch]

Last year, my lathe was turning a taper, which it shouldn't have been doing. So I decided to re-level and check. I have a 12" Starrett #98 level @0.005" per foot, which is accurate, but not even close to the accuracy of the Starrett 199, (0.0005" per foot!). So I shimed the feet until it was as perfect as my level would allow. Using my dumbell test jig, only held by the chuck, I took very light cuts off the aluminium rings and measured. See the results in the photo. So my lathe even after leveling was cutting a taper. So more than likely my headstock is out of alignment with the bed. There is no feature to adjust head stock alignment. It would need to be scraped true. I'm not doing that. I decided to purposely impart twist in the lathe. I did a little trigonometry, and shimed up the TS end front foot. See results.

Great example of accepting what you have and making the best of it! I love it!

Who else would have thought to deliberately twist their bed to compensate for spindle misalignment!

Kudos to you @thestelster!

 

[Susquatch]

I decided to purposely impart twist in the lathe. I did a little trigonometry, and shimed up the TS end front foot. See results.

Of course, Mr Einstein would have said: It's all relative......

Did you really twist the bed relative to the head or did you just shift the relative reference for the spindle to adjust the spindle? It is after all the same result! LMAO!!!

 

[Susquatch]

Using my dumbell test jig, only held by the chuck, I took very light cuts off the aluminium rings and measured.

I wish to add here that your use of a collared test bar is my preferred method too.

There seems to be a few different opinions on this out there. Some love a precision bar mounted to the spindle MT. Others just chuck a bar of any description and use geometry to extract the alignment numbers. And others use a dumbbell machined to the axis of rotation as you did.

I have a project on my to do list to compare these methods objectively. Who knows what I will find. Instinctively, I like the direct outcome of your machined dumbbell.

 

[thestelster]

Of course, Mr Einstein would have said: It's all relative......

Did you really twist the bed relative to the head or did you just shift the relative reference for the spindle to adjust the spindle? It is after all the same result! LMAO!!!

I actually twisted the bed. There are four bolts at the headstock base anchored into concrete, and they remained tight. Loosened the two bolts at the TS, and used a pry bar to lift it high enough to put the shim in place then bolted them back down.

 

[Susquatch]

I actually twisted the bed. There are four bolts at the headstock base anchored into concrete, and they remained tight. Loosened the two bolts at the TS, and used a pry bar to lift it high enough to put the shim in place then bolted them back down.

Too funny! I was kidding of course!

To continue the joke, I believe you are suggesting that you twisted the bed relative to itself. But if the tailstock is orbiting the sun at 36000 thousands further away than the headstock as measured from saggitarius A, the distance is stretched in time space relative to an independent coordinate system. So were you measuring space time or time space or bed twist or head spindle alignment?

It's all good @thestelster, I was, and still am, just having fun.

I think you did very well looking at it the way you did. Prolly better than 99% of the self annointed lathe geeks out there! Including me!

 

[PeterT]

@Susquatch Thanks for the drawings compliment. But just to be clear, the CAD diagram is trivial. I’m just using it as a glorified calculator. The sketching is just low brow annotation tools in Excel.

But I think you are missing the point I’m trying to make, so I’ll try again (with even more sketches! LOL). Don’t get hung up on test bar or cutting coupon choice right now. Let’s just define hypothetical simplifying conditions we can visualize & then invoke real world complexities into the picture one by one. We should be able to agree on cause & effect just pictorially. Then we introduce physical measurements. If these corroborate, then we have confidence in our understanding. If they don’t match, then something is amiss & needs to be explained or re-evaluated. Standard stuff.

Just for now pretend what we are calling test bar is just an extension of the spindle shaft axis pointing outward towards the TS. Let’s just call it a perfect cylinder. Why does it have to extend from HS? Because we are trying to establish how potentially variable outboard bed conditions are aligned or misaligned to this axis datum.

Picture the HS being anchored. It has this perfect cylindrical bar extending out coincident to the spindle axis. Picture the bed ways like a flexible ladder frame. The 2 ladder rails represent the front & rear lathe bed ways. The ladder rungs simply preserve the distance between rails & are proxy for bed webs. This is basically the same as Tony, Joe Pi & others mocked up with cardboard & sticks. So all we know right now is that the spindle aligned test bar is pointing out in space. The rails could be doing anything because by our definition, they are flexible.

Base case. Someone ensured us they set the rails perfectly straight & perfectly square to HS axis in 3 primary planes. How can we verify this? We can put an indicator on the carriage & traverse it down the bar. If indicator runs down side of bar it reads zero. If indicator runs on top of bar it reads zero. Alternately, if we replace the indicator assembly with a cutting tool & replaced the test bar with a cutting coupon & assumed no adverse cutting forces & no adverse material properties, the resultant left & right coupon diameters would be equal. We have proven the exact same thing because they are dimensionally equivalent to each other. If we had a magic sensor that could pick up on a laser beam aligned to spindle it would say the same thing. We could also put a precision level stepwise down both ways & across the ways at specific intervals & attempt to map the topography. All these methods have pros & cons. But we work with what we have so let’s carry on with indicator setup just for discussion simplicity.

Now we introduce bed changes to this model. Picture holding the ladder rails on the TS end, one in each hand. I can raise or lower the ends together which will curve the bed up or down respectively from the anchored HS. I can also twist left or right which will cause one rail to curve up or down relative to one another. I can displace left or right in same horizontal plane, but let’s not go there for now because the aforementioned conditions are achievable by real life jacking feet whereas this condition is not without anchoring. I can also lift and twist in combination, but again keep it simple for now.

So for each of these bed distortion conditions we should be able to at least visualize what would happen. An indicator connected the travelling carriage, pre-zero registered on the test bar should respond & display a resultant reading. So we need a geometric distortion model that yields the same reading.

Bed wear (sketch) is similar but different. I’ve sketched an erosion of the rail top surfaces close to the HS & then it returns to it virgin condition outboard towards HS. So the carriage rides this valley profile & as above, signals some reading.

At risk of repetition, I figured I would start with the 0.003” needle deflection against a known & relatively simple defined cylinder & work backwards as to what bed distortion conditions would mathematically equate to this by the indicator ball moving to a different position. And the resultant 0.055” does not go around by any stretch of imagination, which I humbly admitted. So something is wrong, but what? I invite any of you to take a crack at it. This was actually a gut check pre-calculation if where I was thinking of going – a relatively simple 3D model of lathe with spline curve distorted ways & resultant dimensions could be picked off. The numbers can’t lie, but the basic principles & geometry has to be set up correctly or else garbage in – garbage out syndrome.

 

[Former Member]

I'm going to sit back and watch this thread play out....

 

[Susquatch]

@Susquatch Thanks for the drawings compliment. But just to be clear, the CAD diagram is trivial. I’m just using it as a glorified calculator. The sketching is just low brow annotation tools in Excel.

But I think you are missing the point I’m trying to make, so I’ll try again (with even more sketches! LOL). Don’t get hung up on test bar or cutting coupon choice right now. Let’s just define hypothetical simplifying conditions we can visualize & then invoke real world complexities into the picture one by one. We should be able to agree on cause & effect just pictorially. Then we introduce physical measurements. If these corroborate, then we have confidence in our understanding. If they don’t match, then something is amiss & needs to be explained or re-evaluated. Standard stuff.

Just for now pretend what we are calling test bar is just an extension of the spindle shaft axis pointing outward towards the TS. Let’s just call it a perfect cylinder. Why does it have to extend from HS? Because we are trying to establish how potentially variable outboard bed conditions are aligned or misaligned to this axis datum.

Picture the HS being anchored. It has this perfect cylindrical bar extending out coincident to the spindle axis. Picture the bed ways like a flexible ladder frame. The 2 ladder rails represent the front & rear lathe bed ways. The ladder rungs simply preserve the distance between rails & are proxy for bed webs. This is basically the same as Tony, Joe Pi & others mocked up with cardboard & sticks. So all we know right now is that the spindle aligned test bar is pointing out in space. The rails could be doing anything because by our definition, they are flexible.

Base case. Someone ensured us they set the rails perfectly straight & perfectly square to HS axis in 3 primary planes. How can we verify this? We can put an indicator on the carriage & traverse it down the bar. If indicator runs down side of bar it reads zero. If indicator runs on top of bar it reads zero. Alternately, if we replace the indicator assembly with a cutting tool & replaced the test bar with a cutting coupon & assumed no adverse cutting forces & no adverse material properties, the resultant left & right coupon diameters would be equal. We have proven the exact same thing because they are dimensionally equivalent to each other. If we had a magic sensor that could pick up on a laser beam aligned to spindle it would say the same thing. We could also put a precision level stepwise down both ways & across the ways at specific intervals & attempt to map the topography. All these methods have pros & cons. But we work with what we have so let’s carry on with indicator setup just for discussion simplicity.

Now we introduce bed changes to this model. Picture holding the ladder rails on the TS end, one in each hand. I can raise or lower the ends together which will curve the bed up or down respectively from the anchored HS. I can also twist left or right which will cause one rail to curve up or down relative to one another. I can displace left or right in same horizontal plane, but let’s not go there for now because the aforementioned conditions are achievable by real life jacking feet whereas this condition is not without anchoring. I can also lift and twist in combination, but again keep it simple for now.

So for each of these bed distortion conditions we should be able to at least visualize what would happen. An indicator connected the travelling carriage, pre-zero registered on the test bar should respond & display a resultant reading. So we need a geometric distortion model that yields the same reading.

Bed wear (sketch) is similar but different. I’ve sketched an erosion of the rail top surfaces close to the HS & then it returns to it virgin condition outboard towards HS. So the carriage rides this valley profile & as above, signals some reading.

At risk of repetition, I figured I would start with the 0.003” needle deflection against a known & relatively simple defined cylinder & work backwards as to what bed distortion conditions would mathematically equate to this by the indicator ball moving to a different position. And the resultant 0.055” does not go around by any stretch of imagination, which I humbly admitted. So something is wrong, but what? I invite any of you to take a crack at it. This was actually a gut check pre-calculation if where I was thinking of going – a relatively simple 3D model of lathe with spline curve distorted ways & resultant dimensions could be picked off. The numbers can’t lie, but the basic principles & geometry has to be set up correctly or else garbage in – garbage out syndrome.

My head hurts...... LOL!

I know that @YYCHM started this thread with the question "what is this test bar telling me". So it's only fair that you are approaching this the same way - or at least you appear to be.

But I confess that I am struggling with trying to understand where you are going with this. I believe I followed you right up till you got to your last paragraph which begins with "at the risk of repetition". Then I struggled. I do follow your renderings though.

Please allow me to rephrase your question to see if I understand it. At the very least, it might help you see what I am having trouble following.

Are you asking what kind of external conditions would cause an indicator (that is mounted in front of a perfect concentric test bar such that it reads displacement towards the operator) to display a displacement of 3 thou?

If not, then ignore the rest of my response but try again to ask the question a different way.

If so, then the answer is simple. The spindle is misaligned such that it points toward the front of the lathe by 3 thou at whatever the distance is that it was measured at.

But perhaps it's better to describe it using your model which has a fixed head. In this case, a 3 thou reading will result from a bed that is pivoted at the head such that the bed points to the rear of the lathe.

 

[Susquatch]

@PeterT - part of the reason I think I am having trouble understanding your question is fundamental to the assumptions. Your model assumes a fixed head and spindle with a bed that is permanently (and perfectly) attached to it. As a result, you wonder how this bed has to bend or twist to produce various readings.

However, I start with a different initial condition where the head and spindle are NOT permanently or perfectly attached to the bed. Thus, my head and spindle can be rotated on the bed (or perhaps relative to it). In fact, it might be better to visualize my head as a gimbal mounted on the bed. The gimbal can be rotated up or down or left or right. Of course a gimbal is a very exaggerated way to describe it. Obviously it has very little ability to move.

If this difference in our perspectives is accurate, it might also help us zero in on your question.

 

[PeterT]

If so, then the answer is simple. The spindle is misaligned such that it points toward the front of the lathe by 3 thou at whatever the distance is that it was measured at.
But perhaps it's better to describe it using your model which has a fixed head. In this case, a 3 thou reading will result from a bed that is pivoted at the head such that the bed points to the rear of the lathe.

I am not disagreeing with this potential condition. I have mentioned this condition many times in the past which could easily satisfy or explain an indicator reading & identified many lathe styles where this could mechanically occur. So you are preaching to the choir. It may even be the predominant issue in all lathes but I digress.

In #53 I went through pains to set out initial conditions that specifically precludes HS rotation relative to the bed ways when viewed from the top. Please go back & read it again. It says assume the HS bar points straight down the mean axis of the bed, nothing about HS rotation. Now we are only going to distort the TS end of the bed like one of the cartoon examples. Train your eye on end view aspect of the cartoon, snip attached. Its hard to draw in bloody Excel but I'm trying to exaggerate it & make it obvious. Can you see how the black box (indicator reference) is now positioned above the centerline of test bar (red cross) whereas in all cases it started out at the 9:00 position = center of bar? So what does the indicator do? It extends until it contacts the tangent of the circle above the centerline. That registers a reading. And that condition constitutes an alternative explanation that has nothing to do with HS rotation.

ps - I used the phrase HS points to 'mean axis of bed' as a bit of CYB weasel words. In a perfect lathe (my first example) spindle axis is exactly coincident or parallel to bed axis in all 3 primary planes. But depending on what mode of distortion we then impart to 'the ladder' this may no longer be progressively true. By lifting end of lathe the spindle axis remains coincident to the bed vertical plane, but no longer parallel to horizontal plane because it is now at some angle relative to HS end.

 

[PeterT]

However, I start with a different initial condition where the head and spindle are NOT permanently or perfectly attached to the bed. Thus, my head and spindle can be rotated on the bed (or perhaps relative to it).

Yes exactly. But that's not a very efficient problem isolation methodology IMO. Lets say we are talking about observed elongation of a metal bar. It could be due to force and it could be due to temperature. I'm suggesting a test: assume constant temperature & observe elongation as function of force only. Or assume constant force & observe elongation as a function of temperature only. Now we have more information to observe relative impact.

 

[Susquatch]

Yes exactly. But that's not a very efficient problem isolation methodology IMO. Lets say we are talking about observed elongation of a metal bar. It could be due to force and it could be due to temperature. I'm suggesting a test: assume constant temperature & observe elongation as function of force only. Or assume constant force & observe elongation as a function of temperature only. Now we have more information to observe relative impact.

I am very familiar with isolating variables. In a perfect world that is the ideal way to find a solution to a simple problem. However, in the real world, most problems are not simple and cannot be solved that way. To use your example, if a change in temperature causes a change in force, then one cannot be held constant while the other changes. However, the problem can still be solved by substituting an equation for one in terms of the other. Simple algebra. But I digress.

Please let me ask a different question.

Why are you constraining the initial conditions to perfect alignment when perfect alignment will not reasonably produce the results you want to explain?

Or...... does that answer your question? Lol!

 

[PeterT]

I think you understand what I'm trying to say with the example. If there is some means of isolating a multivariant problem into somewhat independent chunks, it is generally beneficial because it can help direct attention to what might be dominant cause & effect. Not to go on yet another tangent, but if we verified experimentally that the material changed a tiny amount due to temperature change alone relative to observed combined elongation, then we would be in a qualified position to say force is the dominant effect. But we wouldn't know that without some prior knowledge of temperature effect which is exactly why we would bother to test temperature as best we can.

You only have the ability to 'substitute an equation' and apply 'simple algebra' when you have some correlation or basis that lays this out in equation form. I would assume someone before you had to come up with that before it appears in a text book or journal. If I say the material is pretty strong because its black, do you have enough data to describe stress-strain in equation form? If that's all it takes to explain phenomena (he says knowingly), then I will turn revert your advice back to you. Derive us some lathe equations that explains the OP's measured result. One explanation is the HS is rotated. So equation-1: taper is a function of HS angle. The End. But wait, someone else just pointed out his HS is fixed to the bed & he jacked the feet & that adjustment alone had equivalent effect. So I would expect equation-2: taper is a function of jack elevation variables & maybe bed length, and then there's bed material... But wait. I can measure 0.005" of bed wear near the HS with a feeler gauge. Wont that have some effect relative to the test bar? Equation-3...... etc

Re your question, the perfect alignment case is simply defining starting conditions to set the stage - terminology, initial assumptions, definitions, testing methodology, blah-blah. Then I introduced distinctly different modes of bed distortion effects in isolated progression to invite observation of what might be indicator reading effects. Its just a visual what-if. If you recall, I actually did the exact same thing in another post specifically for the HS rotation model & sketched a corresponding picture. I said (paraphrasing) assume the lathe & bed are initially perfect initially & now we do 1 thing: rotate the HS towards the front of lathe, what would the indicator see at the end of the bar? These are 2 distinctly different thought evaluations, but the same what-if methodology.