Chuck Weight?

[westerner]

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......

 

[Tozguy]

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.

 

[Flyinfool]

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.

 

[epanzella]

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.

 

[ShagDog]

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 .

 

[Flyinfool]

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.

 

[epanzella]

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.

 

[Flyinfool]

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.

 

[jmkasunich]

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.

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.....

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.