# Chuck Weight?



## ShagDog (Aug 2, 2020)

I have a lathe that I bought used with an 8" swing and a 1 1/2-8 spindle. The chucks that came with it are 5"(3 jaw) and 6" (4 jaw). They are older chuck that are in the range of about 7 lbs and 9 lbs apiece. I purchased a new chinese 5" with a backplate I fit to it. The chuck is very nice in quality; but, it extends out from the spindle much further than the other 2 chucks which have a thin profile. This one has a very thick profile. The new chuck with backplate weights around 12lbs or more (Can't be sure on weight as my scale stops at almost 12lbs). I did try the chuck, and I must say it works very nicely.

Question: Is the new chuck weight going to do any damage to my spindle, bearings or anything else due to the weight?


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## Asm109 (Aug 2, 2020)

No,  cutting forces are much higher than the weight of a chuck. As long as the chuck is balanced well enough that your lathe does not try to walk around the shop you are good to go.


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## Mitch Alsup (Aug 2, 2020)

ShagDog said:


> Question: Is the new chuck weight going to do any damage to my spindle, bearings or anything else due to the weight?



Cutting forces are on the order of thousands of pounds while the differential weight is on the order of 5 pounds.
You do the math.


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## benmychree (Aug 2, 2020)

Thousands of pounds cutting force  on a lathe with an 8" swing?  Where did that math come from?


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## Tozguy (Aug 2, 2020)

I would not worry about the extra weight of the chuck simply because of the weight capacity the lathe to hold and turn. Steel is heavy and the extra weight from the chuck is equivalent to a rather small work piece.


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## ShagDog (Aug 2, 2020)

Thank's for the responses so far. I did figure out how to weigh the chuck on my limited scale capacity. Should have thought of this to begin with. I took it apart, and to the extent the scale is somewhat accurate,  the backplate weighs around 4 lbs, while the chuck weighs around 9lbs., for a total of around 13 lbs.. I guess I was pretty close.


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## ShagDog (Aug 3, 2020)

Related question. What is the heaviest unsupported (no tail stock support) material that can be turned on a lathe the size of mine, 8x, with 1 1/2- 8 spindle with tapered roller bearings.

Also, what is the formula for determining cutting force?


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## macardoso (Aug 3, 2020)

ShagDog said:


> Related question. What is the heaviest unsupported (no tail stock support) material that can be turned on a lathe the size of mine, 8x, with 1 1/2- 8 spindle with tapered roller bearings.
> 
> Also, what is the formula for determining cutting force?



You'll find rigidity to be the limiting factor on unsupported workpieces rather than maximum weight of the work. I usually say if I am 3:1 or greater on the length to diameter ratio, get a tailstock involved.


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## ShagDog (Aug 3, 2020)

macardoso said:


> You'll find rigidity to be the limiting factor on unsupported workpieces rather than maximum weight of the work. I usually say if I am 3:1 or greater on the length to diameter ratio, get a tailstock involved.



I do understand what you are saying; however, I referred to "unsupported" only to be better able to allow me to understand better the weight bearing capacity of the just the spindle on this lathe.


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## Papa Charlie (Aug 3, 2020)

I wouldn't worry too much about weight bearing. There is, however, a distinct advantage to the additional weight in my mind, in that it will help absorb vibrations and should help to provide a better cut.


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## Flyinfool (Aug 3, 2020)

Most lathes have a spec listed in the manual of the maximum weight between centers. The spindle can support half of that weight.

There are so many factors that affect cutting force like how sharp is the tool at the moment, the tool is constantly getting duller as the cut progresses causing the cutting force to climb, the hardness of the material the geometry of the cutting tool the speed and feed are all major factors some of which are nearly impossible to measure. I don't know of a formula that can calculate it. 

With all that said here is one calculator that may get you in the ballpark with a bunch of assumptions built in.




__





						Cutting Forces
					

Cutting Forces, Torque, and Horsepower for Turning Applications




					www.kennametal.com


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## Tozguy (Aug 3, 2020)

ShagDog said:


> I do understand what you are saying; however, I referred to "unsupported" only to be better able to allow me to understand better the weight bearing capacity of the just the spindle on this lathe.


I have not found the weight capacity spec for my 12x36 lathe but I am curious about it. Not that I think I will ever come close to turning that big a piece.
However it is interesting to note the work piece weight capacity spec of live centers. This one for example
https://www.kbctools.ca/itemdetail/1-435-44103
fits my lathe and is rated for 500 lbs. I expect that the headstock spindle can handle much more than a live center.


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## ShagDog (Aug 3, 2020)

Interesting links, Tozguy and Flyinfool. Good analogies. I spent some time on the links. As to the kennametal one, not sure if I plugged in the right info for calculations. As to the KBC one, I also looked up MT3 dead center that fit in my lathe spindle; however, I could not find any weight capacities on the dead centers.


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## Flyinfool (Aug 3, 2020)

I have a 13x40 lathe. so not much bigger than yours. According to the manual it weighs 1100 lbs. it lists the max weight between centers as 1340 lbs. I will need to eat a lot of Wheaties before I can lift that into place. Not to mention I would be terrified to spin something that weighs more than the machine. because of that number I figure anything I am strong enough to lift into position is good.

As for the weight rating on the live center, my tail has a MT2, the live center would need to be able to hold 700 lbs, it would take this one.




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						ROYAL PRODUCTS,2MT SPINDLE-ROYAL LIVE CENTER,1-533-10102,KBC Tools & Machinery
					

ROYAL PRODUCTS,2MT SPINDLE-ROYAL LIVE CENTER,1-533-10102,KBC Tools & Machinery




					www.kbctools.ca
				




I have the light weight that can only handle 300 lbs. as I will never even get close to that.

Go to the website for your lathe and/or contact the manufacturer to see if you can find a max weight spec.


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## Toolmaker51 (Aug 3, 2020)

Ability of spindle to take weight load unsupported, and full load is not just setting a machine on a concrete floor. It is when each footpad bears its proportion and the machine is correctly level. Weight is beneficial unless off balance. Also why iron chucks are lower RPM rating than semi-steel or forged.
The 3:1 ratio of chucking or running a center is reasonable, if on center and with properly ground cutters.


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## ShagDog (Aug 3, 2020)

Flyinfool said:


> ...
> 
> Go to the website for your lathe and/or contact the manufacturer to see if you can find a max weight spec.



I wish I could. I don't have a manual, and the company that put it's name on it is no longer in business. The lathe was made in Taiwan in 1979 or 1980, approx.

Also, my lathe only has an 8" swing, and weighs over 275 lbs with a 1 1/2-8 spindle thread.


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## mikey (Aug 3, 2020)

Much depends on the quality of the spindle bearings, the size of the spindle and the overall rigidity of the lathe. You're working with an 8" lathe so I'm going to guess your max work piece weight to be somewhere near 50#. I own an Emco Super 11 CD lathe with precision spindle bearings and a 2" spindle and it can handle just over 90# unsupported at the chuck, more with a live center. 

I would go much lower on max weight given that the chuck is extended out beyond stock distances. Cutting forces are minuscule compared to the weight of the work piece and chuck on the spindle. 

Even Logan recommended using the smallest chuck that would get the job done, primarily to spare the spindle bearings.


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## ShagDog (Aug 3, 2020)

Thank's, mikey. I'll never approach 50 lbs, even with the weight of the chuck, so your response pleases me.


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## mikey (Aug 3, 2020)

To reassure you further, cutting forces are measured with a force dynomometer. Typical cutting forces (there are three) for a single point tool in common materials found in a hobby shop are usually in the single digits (kgs), and this is with brazed carbide. If you use a well ground HSS tool, cutting forces are even lower. 

With that said, it is good to keep in mind that spindle bearings do not have an easy life. They must bear the weight of the chuck and the work piece, while also keeping the work concentric while resisting the cutting forces from the tool. This is one of the key reasons to keep your chuck size and its overhang within reason - to preserve your spindle bearings. If you had a 13-14" lathe with a huge spindle, this is not that big a deal but on a little 8" lathe with a spindle close an inch in diameter, yeah, it is a big deal.


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## ShagDog (Aug 3, 2020)

Bottom line me, mikey . Is the 14lb chuck that extends about 3" from the end of the spindle fine on an 8" swing with a 1 1/2"-8 spindle (MT3) with tapered roller bearings a little over 2.75 od? Here's some photos of the spindle and the headstock from when I had it apart. Thank's.


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## mikey (Aug 3, 2020)

Yeah, I think you'll be okay provided you don't try to turn a huge/heavy work piece. Your lathe has a decent build and should be able to handle the weight easily. Just use tail stock support when you can.

Sorry I didn't answer your question clearly in the first place.


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## epanzella (Aug 3, 2020)

Spinning a 100 lb balanced part puts less force to the spindle bearings than a 25 lb unbalanced part.


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## ShagDog (Aug 3, 2020)

Mikey, thank you so much. However, it was not you that did not answer the question clearly. It was me with my jumping around on the various issues that I brought up in this thread. Plus, I guess the pictures and extra info and more precise question I added might have helped a bit .


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## Tozguy (Aug 4, 2020)

Sometimes a heavier chuck can be an advantage. For example, when making interrupted cuts, the extra flywheel effect of a heavier chuck might take some strain off the spindle drive and smooth the operation somewhat. Maybe I'm dreaming but its a nice dream!


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## Tozguy (Aug 4, 2020)

ShagDog said:


> I could not find any weight capacities on the dead centers.



Me neither. I posted a question here at HM back in 2015 on how to choose the right size of center drill. Information abounds on the dimensions of various center drills
https://www.i-logic.com/utilities/CenterDrills.htm
 but I have never seen anything (yet) about the weight carrying capacity of each size of center hole.


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## Mitch Alsup (Aug 4, 2020)

benmychree said:


> Thousands of pounds cutting force  on a lathe with an 8" swing?  Where did that math come from?



Mild steel cuts at about 30 MPSI. 
The area being cut is the product of the Depth of Cut (DoC) and the amount cut each turn (speed of cut)
So if the area being cut is 10 thou DoC × 10 thou Per Turn;
you are applying 30,000,000 * (0.010 * 0.010) = 3,000 pounds of force on the face of the tool.

Now go back and consider the forces involved when Abomb79 taking DoC of 0.400 at 0.100 cut per turn out of 9" diameter 4140 ! It is no wonder that lathe had a 25 HP motor.....


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## mikey (Aug 4, 2020)

Okay, wait now. My little Sherline lathe can take a 0.060" deep cut in a single pass in mild steel. Granted, this is with a decent tool and is near the max capacity of its little 0.08HP motor but it has done this in multiple trial cuts. My little 11" Emco lathe has taken 0.20" deep cuts in mild steel with the same HSS tool without even slowing down its 2HP motor. It can go deeper if I bother to step down one notch in speed. 

According to your calculations, my Sherline is experiencing a cutting force of 18,000# of force at the tip of the tool? Umm, I must say I am skeptical ...


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## Papa Charlie (Aug 4, 2020)

Mitch Alsup said:


> Mild steel cuts at about 30 MPSI.
> The area being cut is the product of the Depth of Cut (DoC) and the amount cut each turn (speed of cut)
> So if the area being cut is 10 thou DoC × 10 thou Per Turn;
> you are applying 30,000,000 * (0.010 * 0.010) = 3,000 pounds of force on the face of the tool.
> ...



I would love to have Adam's Monarche.


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## 682bear (Aug 4, 2020)

Mitch Alsup said:


> Mild steel cuts at about 30 MPSI.
> The area being cut is the product of the Depth of Cut (DoC) and the amount cut each turn (speed of cut)
> So if the area being cut is 10 thou DoC × 10 thou Per Turn;
> you are applying 30,000,000 * (0.010 * 0.010) = 3,000 pounds of force on the face of the tool.
> ...



I guess you had better have some awfully big arms before you pick up the gravers...

And using a deburring tool obviously takes a lot more strength than I realize...

Geez... even using a file must be more difficult than I thought it was...

Sorry... I'm kind of skeptical, also...

-Bear


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## epanzella (Aug 4, 2020)

walterwoj said:


> Made this shelf today with some 1/16" steel.  Used the mill to drill the holes with a 1 3/8" hole saw. (the arbor bit kept coming loose so I just quite using it, the mill is rigid enough I don't need it.  ) My sister recently found me a handheld band saw, so I used that to cut out the shape. Then it was a matter of some welding and a quick paint job. Added some rubber grommets so the collets won't hurt the rack and the rack wont hurt the collets. (the collets in the bags are just ones I've never used so I've never taken the bags off)
> 
> 
> 
> ...





Mitch Alsup said:


> Mild steel cuts at about 30 MPSI.
> The area being cut is the product of the Depth of Cut (DoC) and the amount cut each turn (speed of cut)
> So if the area being cut is 10 thou DoC × 10 thou Per Turn;
> you are applying 30,000,000 * (0.010 * 0.010) = 3,000 pounds of force on the face of the tool.


That's 3000 PSI, right? That would make the force the tool pressing against the spindle 3000 x contact area of the tool in square inches.


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## westerner (Aug 4, 2020)

Precision tapered roller bearings have a fit, geometry and metallurgy that gives them a strength that is hard to appreciate. 
Look at the bearings in your boat trailer, or car hauler. 
Recognize that THOSE bearings are cheap and sloppy compared to the units in the lathe. 

Look at what your trailer bearings are designed to carry, and the environment they are forced to live in. 
Personally, I have absolutely ZERO concern for my lathe spindle bearings, as long as there is oil in their gearcase......


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## Tozguy (Aug 5, 2020)

ShagDog said:


> but, it extends out from the spindle much further than the other 2 chucks which have a thin profile


How much farther? I'm just curious.
The extra extension of the new chuck is more significant to me than the extra weight.
It is generally best to do the cutting as close to the spindle bearings as possible when not turning between centers. Nevertheless this usually relates to how far the work piece extends from the chuck and not how far the chuck overhangs the bed ways.

On a used lathe there might be some advantage to the new longer chuck because it would change where the carriage sits on the ways for a given job. It depends on the turning history of the lathe but chances are with the longer chuck the carriage will be riding on a section of the ways that has less wear.


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## Flyinfool (Aug 5, 2020)

epanzella said:


> That's 3000 PSI, right? That would make the force the tool pressing against the spindle 3000 x contact area of the tool in square inches.



I think he already doing that with the X 0.010 X 0.010 part of the equasion. It is the 30,000,000 number that he is starting with that I can not find any reference to and have a hard time believing.

That would mean it takes 30,000,000 lbs of force to make a cut that is 1 inch deep and a feed of 1 inch per revolution.  Even that still sounds high. 30M lbs of force is many times more than the force required to make diamonds.


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## epanzella (Aug 5, 2020)

Flyinfool said:


> I think he already doing that with the X 0.010 X 0.010 part of the equasion. It is the 30,000,000 number that he is starting with that I can not find any reference to and have a hard time believing.
> 
> That would mean it takes 30,000,000 lbs of force to make a cut that is 1 inch deep and a feed of 1 inch per revolution.  Even that still sounds high. 30M lbs of force is many times more than the force required to make diamonds.


I don't think that formula accounts for tool area. If you make a cut with a DOC of .010 and a feed rate of .010 with a 1/4 inch pointy tool, that's gonna take a hell of a lot less force than making the same cut with a 1/2 " form tool.


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## ShagDog (Aug 5, 2020)

Tozguy said:


> How much farther? I'm just curious.
> ...
> On a used lathe there might be some advantage to the new longer chuck because it would change where the carriage sits on the ways for a given job. It depends on the turning history of the lathe but chances are with the longer chuck the carriage will be riding on a section of the ways that has less wear.



The old chuck which is very thin profile measures 1.80" from the mounting face (back) to the front, not including the jaws. The new chuck with the rather thick backplate measures 3.68" from the mounting face to the front, not including the jaws. Almost a 2" difference. 

By the way, the old chuck is a nice chuck, "Made In England"; but, no inside jaw set. I would have loved to find an inside jaw set for it; but, it seem like looking for a needle in a haystack. Still looking though. 

Possibly noteworthy is that the new chuck adds no vibration, and seems to run very nicely balanced.

As to the ways, it appears that they are in good shape, despite the awful appearance of the cross slide top surface. The carriage seems to move nice and smooth along the entire ways. I also get no taper to speak of when turning without tailstock support   .


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## Flyinfool (Aug 5, 2020)

The pointy tool and the 1/2 form tool will both be removing the same amount of material per revolution. One will be a narrow thick chip and one will be a very thin but wide chip. but both tools will remove 0.0001 square inches of material at a time.


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## epanzella (Aug 5, 2020)

Flyinfool said:


> The pointy tool and the 1/2 form tool will both be removing the same amount of material per revolution. One will be a narrow thick chip and one will be a very thin but wide chip. but both tools will remove 0.0001 square inches of material at a time.


I don't agree. If you think the forces are the same for all tooling, try running a half inch form tool on a small minilathe. What about the difference between HSS and carbide? Negative rake vs positive? There are lots of things that affect forces that are not in that formula.


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## Flyinfool (Aug 5, 2020)

I absolutely agree with the plethora of variables that can have an effect on actual tool pressure. That is why everyone is questioning the original 30 MPSI number.


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## jmkasunich (Aug 5, 2020)

epanzella said:


> I don't think that formula accounts for tool area. If you make a cut with a DOC of .010 and a feed rate of .010 with a 1/4 inch pointy tool, that's gonna take a hell of a lot less force than making the same cut with a 1/2 " form tool.





Mitch Alsup said:


> Mild steel cuts at about 30 MPSI.
> The area being cut is the product of the Depth of Cut (DoC) and the amount cut each turn (speed of cut)
> So if the area being cut is 10 thou DoC × 10 thou Per Turn;
> you are applying 30,000,000 * (0.010 * 0.010) = 3,000 pounds of force on the face of the tool.
> ...



A bit of an error here.  30,000,000 is the modulus of elasticity of steel, not the yield strength.  That's why the numbers are coming out so high.

Yield strength for steel ranges from 30,000 psi or so (crappy mild steel) to 100,000 or so (grade 8 bolts and such).  So taking a 0.01" deep x 0.010" feed per rev cut on mild steel is in the neighborhood of 3 pounds tool tip force, and the same cut in tough steel about 10.  But keep in mind that is the theoretical minimum force needed to make a 0.01" x 0.01' piece of steel yield.  The actual cutting operation surely needs more force, due to friction and other effects.

Another way to come at the tool force calculation is to look at the spindle power.  One horsepower is 33,000 foot-lbs per minutes.  Suppose you have a little lathe with a 1/2HP motor, and you are taking a maxed out cut such that the motor is almost bogging down.  You are using 18,000 ft-lbs per minute.  If you are cutting with HSS, your surface speed is probably about 100 feet per minute.  So it would take a force of 180 lbs to consume 1/2HP.


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