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NUMOBAMS 8x16 (NU-210G) Lathe Review

Using my favourite design tool for piping - Engineering Power Tools Plus Edition (check it out)


I picked 1-1/4" Sch40 as a largest size that would fit in most large hobby lathes. I also solved for 1/2" that would likely fit in a 7/8" collet.

15" length of 1-1/4" Sch40 steel pipe 1.660 O.D. at 2.268 lbs/ft, simple cantilever, supported at one end only, uniform weight distribution, solves at 0.0002" deflection at the free end, 90.633 psi stress. 21.26 in-lbs torque at the chuck end.

1/2" Sch40 deflects 0.840" O.D. 0.0009" in the same conditions.

All mostly irrelevant, because the chances of getting a straight piece of pipe is about the same as me being elected Pope.

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All mostly irrelevant, because the chances of getting a straight piece of pipe is about the same as me being elected Pope.
Don’t worry, when I’m the ruler of the world I’ll make you pope. And I’ll give orders to initiate a research program to solve this discussion.
 
Oh, I think you totally underestimate what I know about such things..... And of course, I probably overestimate it. LOL!

Frankly, I think you and I are just worried about opposite ends of extremes. In my opinion, temp effect is an easy calculation that is insignificant for what we are talking about here. I see this as chasing rainbows in terms of reading too much into it. Don't over complicate things. LOL! (Sorry, Devil made me say that - just using the same words you used earlier for the humour effect!)

Anyway, in this case the radial expansion is well below my ability to measure it and I don't believe that longitudinal expansion matters in this application. In case you are wondering, the average coefficient of thermal expansion for steel is just 0.00000645in/in/deg F.

Rather than chasing MY tail, which is not something I'd be wanting to do if I were you, I'd suggest you wait until I complete my assessment of this whole subject in designing my new collared system. I'll have actual measurements and comparison testing of other methods at that time and I will make all of it available for peer review on my thread for that. I'd like to believe that I will be totally open to any and all criticism or suggestions for improvements that anyone cares to provide at that time.
Wow, you’re even more anal about numbers than I am. I usually did my thermal expansion calcs using 0.00006”/”/*F, but I was typically only doing steam piping deflection and the temperature range was only maybe 500* F from cold to hot.
 
Does it matter whether the test bar sags? Isn’t the primary reason for using a test bar for adjusting twist? The measurement on the vertical face of the test bar is not going to see much of a change in radius from sag.
(Edit: I did a sketch in Fusion 360 to see what 0.001” sag will do for a reading of the vertical face. It's smaller than what Fusion can display...)
(I first did the sketch with a 0.01 sag to make sure I had the sketch constraints correct, then I changed the 0.01 dimension to 0.001)

View attachment 24020

What about a carbon fibre pipe test bar? That would look very nice too!

I believe the sag matters. Twist is measured and corrected before head parallelism is evaluated and corrected (if possible).

I think it's best to look at this whole deal (no matter what words you use to describe it) as:
1. Bed alignment (twist, bending, etc). Sometimes misleadingly called levelling.
2. Tailstock alignment (usually associated with cutting tapers)
3. Head alignment with the bed. Axis of rotation parallel to the axis of the bed. Not pointing up/down/right/left.

Not all of these are adjustable on all lathes.

My focus in this discussion is strictly #3.

Carbon Fiber - YES! I bet carbon fiber would be awesome for this. It is both stiff and light. But you still need machinable collars to do the actual testing. And I'm not sure where to get (or make) the appropriate carbon fiber bars.
 
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Wow, you’re even more anal about numbers than I am. I usually did my thermal expansion calcs using 0.00006”/”/*F, but I was typically only doing steam piping deflection and the temperature range was only maybe 500* F from cold to hot.

Yes, I am anal with numbers. But only for theoretical purposes. For actual applications, I prefer testing to validate the theory (which often means pitching the theory in the trash.... LOL!)
 
All mostly irrelevant, because the chances of getting a straight piece of pipe is about the same as me being elected Pope.

Get ready for papal duty...

Straighness of the pipe doesn't matter. That's what the collars are for.

Two collars are attached to the pipe. In my original thinking, they are aluminium. One is attached just in front of the chuck jaws, the other at the far free end. Extremely shallow very low pressure cuts are made on both collars with a VERY sharp tip. The idea is to avoid deflecting the bar in the process of cutting it. It's not really possible but can be minimized.
 
And I’ll give orders to initiate a research program to solve this discussion.

Woah Woah Woah! Not so fast Oh Great Ruler of the World! Thy worthy and respected command is already being made so!

Your unworthy servant of big and hairy proportions is already doing the research to resolve this matter. He doesn't need to be taking any orders from Pope @whydontu.

Of course, everyone is welcome to join on (pile on) and provide peer review (aka feedback criticism support & debate) over on that thread.

Work on the research is temporarily on hold in favour of higher priority tasks.

Edit - Here is a link to that thread 'Lathe Alignment' https://canadianhobbymetalworkers.com/threads/lathe-alignment.4723/
 
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Using my favourite design tool for piping - Engineering Power Tools Plus Edition (check it out)


I picked 1-1/4" Sch40 as a largest size that would fit in most large hobby lathes. I also solved for 1/2" that would likely fit in a 7/8" collet.

15" length of 1-1/4" Sch40 steel pipe 1.660 O.D. at 2.268 lbs/ft, simple cantilever, supported at one end only, uniform weight distribution, solves at 0.0002" deflection at the free end, 90.633 psi stress. 21.26 in-lbs torque at the chuck end.

1/2" Sch40 deflects 0.840" O.D. 0.0009" in the same conditions.

All mostly irrelevant, because the chances of getting a straight piece of pipe is about the same as me being elected Pope.

View attachment 24030

I think you understand what I am after here. My choice of a piece of black pipe was driven mostly by availability and low cost, but also by suitability for the task.

Could you take a bit of time using your tool to assist me in selecting a better candidate for this application? Ideally the lowest deflection due to weight and tip force possible while still fitting reasonably well in the chuck jaws. I agree that 1.25" diameter is about right.
 
1" Sch10 steel pipe - 1.315" nominal O.D., 0.109" wall, 1.097" I.D.. Find a fire sprinkler contractor, common size for press fit branch runs. 0.0003" deflection over 15" span.

1" Sch40 0.133" wall, 1.049" I.D., deflection will be bit more but EPT only solves to tenths. Pipe is a bit heavier.

Pipe dimensions tend to be pretty close on O.D., but usually a little heavier than nominal on wall thickness. No pipe manufacturer has ever been sued for making the pipe a touch thicker than required by spec.

Collar suggestion - turn down the end of the 1" pipe to 1.250". Leaves 0.076" wall thickness on Sch10, 0.100" wall on Sch40. Cut a shallow taper from 1.250" to 1.315" over 1" span. Cut a matching taper in your sacrificial collars. At <4 degree taper you're well into the range for a wringing taper. Lightly press collars onto bar. Taper combined with Loctite if you're a belts and suspenders type of guy.

Sketch attached. I drew the pipe as 1.3" O.D., just because I just did a snap-to-grid at 0.025", and even if you did a 0.007" cleanup cut on the pipe O.D. you'd still have lots of meat.
 

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BTW, this forum is stretching my brain. I've been using a CAD program called CADstd (http://www.cadstd.com/history/index.html) since 1987. I've forced every employer to buy me a copy. And I've bought my own copies every time my PCs switched to new OS versions. Maybe I‘m their best return customer.

And today, realizing I wanted to do a screen capture with a white background, I had to read the manual and learn I could switch to a white background. I've been using CADstd with a black background for 30+ years because I started when it was DOS and that's where my brain stuck. Sigh.
 
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1" Sch10 steel pipe - 1.315" nominal O.D., 0.109" wall, 1.097" I.D.. Find a fire sprinkler contractor, common size for press fit branch runs. 0.0003" deflection over 15" span.

1" Sch40 0.133" wall, 1.049" I.D., deflection will be bit more but EPT only solves to tenths. Pipe is a bit heavier.

Pipe dimensions tend to be pretty close on O.D., but usually a little heavier than nominal on wall thickness. No pipe manufacturer has ever been sued for making the pipe a touch thicker than required by spec.

Collar suggestion - turn down the end of the 1" pipe to 1.250". Leaves 0.076" wall thickness on Sch10, 0.100" wall on Sch40. Cut a shallow taper from 1.250" to 1.315" over 1" span. Cut a matching taper in your sacrificial collars. At <4 degree taper you're well into the range for a wringing taper. Lightly press collars onto bar. Taper combined with Loctite if you're a belts and suspenders type of guy.

Sketch attached. I drew the pipe as 1.3" O.D., just because I just did a snap-to-grid at 0.025", and even if you did a 0.007" cleanup cut on the pipe O.D. you'd still have lots of meat.

Excellent stuff.

My career brain is wired to design for optimum and reduce cost through volume.

My farm brain is wired to use the cheapest thing I can find out in the yard and force it to work.

I'll use that info (and give you credit) including the taper fit, as soon as the project matures...... unless it dies of errors, bad assumptions, a better idea, or priority rot. Lots of ground and experiments between here and there.

How expensive is fire pipe compared to regular pipe?

I'm thinking about maybe modifying your taper idea a wee bit. I'd prefer not to thin out or weaken the chuck end to accomodate a taper. So I'm thinking about a double collar at that end. One that is solidly attached to the pipe and then another collar on top of that one that is attached with a taper for replaceability.

Thank you!
 
I'm not sure I fully understand where you guys are going with this deflection thing. In post #94 @Susquatch volunteered some deflection numbers just as an example. Lets take the longest one just for discussion purposes. S=30,000 ksi, OD=1.125", L = 24". Resultant deflection = 0.005".

Now refer to crude sketch below, side view of bar, dramatically exaggerated. If this was a precision test bar, the sag would be represented by the lower curved lines only. The black line represents a DTI trace traversing down the length. If we zeroed it on on the 9:00 quadrant on left side, the chuck side, the DTI it would end up slightly higher on the circle on the right side. So one could say the needle is deflecting as though the HS was pointing towards the rear of lathe looking down from the top potentially giving us a false signal in an otherwise perfect geometry lathe. This detail is real but typically glossed over as negligible for weekend warriors. Most test bars are shorter, say 12" so will deflect less. They might may be made from higher modulus materials & deflect less on that basis, but who knows. Cantilever deflection is non-linear, most of it occurs exponentially(?) out towards the free end.

But the aforementioned is a static setup. The bar end is drooping downward in the vertical plane under its own weight at 1G. When you guys are talking about making a cutting test bar, that is a dynamic condition. The bar is now spinning at lathe RPM. Maybe I am off base but what I visualize happening is the outboard material which is off axis (arbitrary orange element) will now see increased dynamic deflection because it sees much higher G-at some distance which is increasing outboard. Maybe a loose term is 'whip'. Visualize spinning a long wire in a drill. So the view on right is showing the deflection positions that we will now be cutting through. Now things get complicated. How much are we shaving off due to this dynamic deflection? How much are we influencing the deflection by pushing the material back towards center near the outboard right end (but different than the left supported end)? We desire a long length so that the left vs right diameter difference exaggerates the axis misalignment, but now we also have these new spinning & cutting conditions to contend with.

1653273224490.png
 
I'm not sure I fully understand where you guys are going with this deflection thing. In post #94 @Susquatch volunteered some deflection numbers just as an example. Lets take the longest one just for discussion purposes. S=30,000 ksi, OD=1.125", L = 24". Resultant deflection = 0.005".

Now refer to crude sketch below, side view of bar, dramatically exaggerated. If this was a precision test bar, the sag would be represented by the lower curved lines only. The black line represents a DTI trace traversing down the length. If we zeroed it on on the 9:00 quadrant on left side, the chuck side, the DTI it would end up slightly higher on the circle on the right side. So one could say the needle is deflecting as though the HS was pointing towards the rear of lathe looking down from the top potentially giving us a false signal in an otherwise perfect geometry lathe. This detail is real but typically glossed over as negligible for weekend warriors. Most test bars are shorter, say 12" so will deflect less. They might may be made from higher modulus materials & deflect less on that basis, but who knows. Cantilever deflection is non-linear, most of it occurs exponentially(?) out towards the free end.

But the aforementioned is a static setup. The bar end is drooping downward in the vertical plane under its own weight at 1G. When you guys are talking about making a cutting test bar, that is a dynamic condition. The bar is now spinning at lathe RPM. Maybe I am off base but what I visualize happening is the outboard material which is off axis (arbitrary orange element) will now see increased dynamic deflection because it sees much higher G-at some distance which is increasing outboard. Maybe a loose term is 'whip'. Visualize spinning a long wire in a drill. So the view on right is showing the deflection positions that we will now be cutting through. Now things get complicated. How much are we shaving off due to this dynamic deflection? How much are we influencing the deflection by pushing the material back towards center near the outboard right end (but different than the left supported end)? We desire a long length so that the left vs right diameter difference exaggerates the axis misalignment, but now we also have these new spinning & cutting conditions to contend with.


View attachment 24053

I love how your mind works. And I like that you are thinking about such things. It helps to have someone asking those kinds of tough questions.

I had considered this earlier and after a lot of noodling, I put it on the back burner as not likely. The simplest reason I can give is that I don't think the dynamic situation is different from the static one in any meaningful way. In other words, the bar droops constantly due to gravity as it turns.

The bar has very high bending resistance which also means it has a very high natural frequency. I'd expect that natural frequency to be much higher than the low speed at which it would be turning during cutting.

That said, I'm not positive of that. So it's yet another calculation and experiment to do.

To be a bit clearer on my thinking, imagine a finite element analysis vs time. The bar can be moved a bit at a time during cutting. Let's just arbitrarily say that it's 15 degrees for each iteration for discussion purposes. In a real simulation we would choose a much smaller iteration interval. For each iteration we would calculate the time for the bar to droop, twist, vibrate, and otherwise move according to the force of gravity, and cutting tool forces.

In my mind, the high natural frequency of all those movements would be totally dominated by the relatively low frequency of rotation.

But you know what they say about assumptions. Therefore, I do plan to do the calculations and then do the measurements needed to validate them.

Keep thinking Peter. It is very much appreciated.
 
Now refer to crude sketch below, side view of bar, dramatically exaggerated. If this was a precision test bar, the sag would be represented by the lower curved lines only...
That was the point of my sketch, except I didn’t present it as clearly as you have.
the DTI it would end up slightly higher on the circle on the right side. So one could say the needle is deflecting as though the HS was pointing towards the rear of lathe looking down from the top potentially giving us a false signal in an otherwise perfect geometry lathe….

View attachment 24053
My sketch suggested for a 1” test bar that has deflected 0.001”, the change in DTI reading was something like 0.000001” (too small for Fusion360 to give an actual dimension)
 
That was the point of my sketch, except I didn’t present it as clearly as you have.

My sketch suggested for a 1” test bar that has deflected 0.001”, the change in DTI reading was something like 0.000001” (too small for Fusion360 to give an actual dimension)

I had only addressed @PeterT's concern about dynamic bending. But your note suggests that perhaps you guys are also talking about something else. Or perhaps I misunderstood the dynamic bending discussion too.

Sometimes the English language sucks. On reflection, I am not at all sure that I understand what either of you are talking about when you say that a 1 thou deflection results in a 0.000001 dti reading. I suspect you are thinking that the measurement of the test bar is taken on its side where a typical tool cut is made. And the very tiny measurement you refer to is the change in the dti reading that results from a slightly higher position on the radius of the shaft caused by its droop. (that was a mouthful) If so, that isn't the idea at all.

Setting dynamic deflections aside for now, it might help if I explain a little more about what I believe we are trying to do here.

Ideally, we want the axis of our spindle to be parallel to the axis of the ways. However, this isn't always the case. The axis of the spindle could be pointing down or it could be pointing up (let's call both of those nod) and it could also be pointing forward or rear ward (let's call that wag).

A rigid and solidly mounted cutting tool held in a tool holder on the cross-slide will always slide along the ways on a path that is parallel to the axis of the ways. This is a fundamental aspect of the way that a lathe works.

If the lathe spindle axis wags inward toward the operator. The lathe will end up cutting a taper that is smaller at the far end of a bar than it is at the headstock. If it is pointing away from the operator, the far end will be fatter. Because the cutting tool is located at the vertical center of the spindles axis of rotation, the angle of the resulting taper will be proportional to the wag angle of the spindle itself. This angle can be calculated by measuring the taper. The taper measurement is taken with a Micrometer not a DTI.

On the other hand, a cutting tool that travels the length of a test bar that points up or down does not cut as big a taper. That's because the cutting tool moves up or down the radial surface of the test bar as it traverses left or right. The result is a MUCH smaller taper. Perhaps this is the 0.000001 measurement you guys referred to. And of course, you are correct - this effect is relatively insignificant. Not only that, but it's insignificance IS IMPORTANT. Because the taper caused by any up or down nod is insignificant, the test bar will be more or less cylindrical. And because it is cylindrical, the nod can be measured. Not by measuring this insignificant taper, but rather by directly running a DTI (NOT a Micrometer) from left to right along the TOP of the test bar. A spindle that points up will result in a higher measurement at the end of the test bar, and one that points down will result in a lower measurement.

Because nod is measured on the top of the bar, any deflection of the bar due to its own weight can interfere with a precise measurement. However, if the deflection is known, it can simply be subtracted from the nod measurement.

Of course, there are many other ways to do this measurement and other equipment that can be used. But fundamentally, nod is measured on the top of the bar and wag is measured on the side.

Nod and wag are two separate measurements. However, any taper caused by wag must be subtracted from or added to the nod measurement.

Fans of the MT test bar will no doubt be quick to point out that the test bar does not require measuring tapers. Instead both nod and wag are measured directly with DTIs - at the top for Nod, and on the side for Wag.

At the risk of re-opening an old debate, I believe the mt5 test bar method is much easier to do but that any concentrity error in the bar or MT5 socket, or any dirt in the MT5 connection will result in errors that need to identified and cancelled out.

On the other hand, cutting a test bar requires a very sharp cutting tool, and a complicated taper measurement.

Both will work just fine as long as you know and understand their limitations.

I hope that helps.
 
I didn't have access to my CAD tools until tonight but sketch corroborates @StevSmar example
1.125" OD bar drooping 0.005" at the end equates to a very teeny indicator delta reading (0.0000222"). Assuming I had a indicator that could measure that.
 

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I didn't have access to my CAD tools until tonight but sketch corroborates @StevSmar example
1.125" OD bar drooping 0.005" at the end equates to a very teeny indicator delta reading (0.0000222"). Assuming I had a indicator that could measure that.

Gezz Peter, I thought you understood. Now I'm not so sure. I agree with your analysis of the impact of a 5 thou droop on a reading at the side of the bar.

But, when measuring spindle nod, you don't measure at the side of the bar, you measure along the top!

At the top of the bar, 5 thou droop is 5 thou! This droop due to gravity must be substrated from whatever the measured nod is.
 
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@Susquatch I'll try to explain. The CAD measurement in post 117 was just a convenient method to quantify how much needle deflection would be seen by the indicator ball riding up or down the OD of a cylinder (round section) test bar BY WHATEVER MEANS resulting in the bar not being perfectly aligned to indicator travel. Yes if I wanted to verify the beam droop under gravity calculation (0.005" over 24" for example), I would put the indicator ball at 12:00 on HS side, traverse down the bar to the end & observe reading. But not so fast, the only way this works is if the ball ends up in the exact same 12:00 position at the bar end. If the HS was pointing slightly in or out then the ball would land towards 1:00 or 11:00. So the reading would be influenced by the circular shape, not entirely droop deflection. By how much? That's what the CAD sketch quantifies. Now If I used a rectangular section test bar, the DTI ball would occur on the same plane & this effect would not be an issue.

Anyways all post 117 says is: lets just temporarily accept that a circular section bar droops 0.005" under gravity. How much influence would my indicator see DUE TO THIS DROOP EFFECT ALONE by traversing the bar along the bar side while I am trying to ascertain HS yaw misalignment, which is still the main objective. Answer 0.0000222" = negligible. So my DTI reads 0.003" of deflection, then 99% of this reading is attributable to HS rotation, less than 1% is due to droop. IOW in a static test we don't have to worry about droop effect even to this extent.

The reason this was of interest to me is more of a generalization: the same indicator ball riding up or down the circular section effect could potentially influences other modes of measurements. For example if you have a badly worn saddle, the indicator goes along for the ride since its mounted to the saddle. So as its traversing along a cylinder test bar the ball is riding high or riding low relative to initial bar center. Exact same math. Make more sense now? I've seen people discuss bed wear effect, but have not seen much reference quantifying it relative to a cylindrical datum surface.
 
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@Susquatch I'll try to explain. The CAD measurement in post 117 was just a convenient method to quantify how much needle deflection would be seen by the indicator ball riding up or down the OD of a cylinder (round section) test bar BY WHATEVER MEANS resulting in the bar not being perfectly aligned to indicator travel. Yes if I wanted to verify the beam droop under gravity calculation (0.005" over 24" for example), I would put the indicator ball at 12:00 on HS side, traverse down the bar to the end & observe reading. But not so fast, the only way this works is if the ball ends up in the exact same 12:00 position at the bar end. If the HS was pointing slightly in or out then the ball would land towards 1:00 or 11:00. So the reading would be influenced by the circular shape, not entirely droop deflection. By how much? That's what the CAD sketch quantifies. Now If I used a rectangular section test bar, the DTI ball would occur on the same plane & this effect would not be an issue.

Anyways all post 117 says is: lets just temporarily accept that a circular section bar droops 0.005" under gravity. How much influence would my indicator see DUE TO THIS DROOP EFFECT ALONE by traversing the bar along the bar side while I am trying to ascertain HS yaw misalignment, which is still the main objective. Answer 0.0000222" = negligible. So my DTI reads 0.003" of deflection, then 99% of this reading is attributable to HS rotation, less than 1% is due to droop. IOW in a static test we don't have to worry about droop effect even to this extent.

The reason this was of interest to me is more of a generalization: the same indicator ball riding up or down the circular section effect could potentially influences other modes of measurements. For example if you have a badly worn saddle, the indicator goes along for the ride since its mounted to the saddle. So as its traversing along a cylinder test bar the ball is riding high or riding low relative to initial bar center. Exact same math. Make more sense now? I've seen people discuss bed wear effect, but have not seen much reference quantifying it relative to a cylindrical datum surface.

Yes, that all makes perfect sense to me now. I erroneously assumed you didn't agree with what I was saying.

I always do my own calculations for these effects. A while back, I saw a reference to a fellow who did a spreadsheet to calculate both nod and wag from a set of measurements. I contemplated doing that too. I may even do it yet. But fundamentally, it isn't a critical measurement. It's just something we would all like to minimize and/or at least directionally understand how it might affect our work.
 
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