Romi Lathes

Take a look at the Haas Tool Room Lathes. Have a customer that has both and uses the Haas 90% of the time. Is looking to get rid of the Romi but can't find any buyers.

"Don Harrison" wrote in news:cK6dnQVpe snipped-for-privacy@rogers.com:

Might want to read through this thread first.

Depends on if you like to be able to take a cut.

Dan

This guy seems to be taking bigger cuts on his Haas than on his Dyna Myte. If hp is a concern go to the TL-3 and get 18 hp

If quality is a concern then pick up *anything* besides a f****ng piece of shit Haas lathe. 18 hp, yeah right. Horses from the glue factory, maybe. Couldn't pull a sick w**re out of bed, that pile of crap. .060" on a side in 8620 and those jokes stall, chatter, shudder, shake and blow feed drives.

Wow, nice.

No wonder I haven't posted here for a while. You are still one miserable little man. 800 Haas machines sold per month. Thats a lot of shops buying shit. Lighten up Hamei and enjoy life for a change.

If you stop posting, then only Hamei is left to post. Even when I strongly disagree, or agree with him, I don't take his "colorfullness" too seriously,

Don't you remember Yosmite Sam? Didn't he Ever make you laugh? lol

Pete

Oh I don't take him seriously or let it bother me. Yes I still watch Bugs Bunny and Yosmite Sam and laugh at them to this day even after seeing them thousands of times. It's just a shame to go through life being so miserable and negative as it seems as if Hamei is a very knowledgeable guy and could be so much help to guys starting out in this business as well as some of the long time machinists. Don

I am helping them. I'm helping them to choose not to buy a terribly crummy cnc lathe. It really doesn't matter how many are sold each month, ya know ... if a good lathe costs $87,000 while a Haas costs $75,000, which one will sell ? Most shops now are run by accountants without klews. Or maybe they do have klews - maybe there's not gonna be any work in the US in ten years, so why worry about longevity ?

But the fact is, the lathes are crappy. The "torque tube" design they used to crow about was developed by Jones and Lamson. Haas never gave a nickel's worth of credit to J&L. In fact, they claimed it was their idea.

That is theft.

The axis drives suck. You can't repair them. That's crap.

Many of them don't have gears in the headstock. For a very lightweight finishing lathe that's a good idea. For a normal- use machine, it's garbage. The much-vaunted Vector-Drive red paint on the sheetmetal couldn't pull a sick w**re outta bed. Nice for cutting stainless, real nice. In fact I wasn't joking when I said the damn things can't even take a .060" cut in

8620 without stalling. I embarrassed the hell out of myself by laughing at some people for their pansy cuts - then I tried it on their crappy Haas lathe. Yeah, it was a joke all right. A joke on me.

I could continue but I don't think you want me to. The mills are sort of okay for the price but a lathe needs to have some beef and some balls. Haas ain't got it. They stink, no matter *how* many they sell each fortnight.

Probably depends on the diameter, among other things?

Cliff wrote in news:nlg921te7aafn1t13b9ujqme6du4m971oq@

4ax.com:

Now that is a really stupid statement/question.

Effective cutting (shear) force varies with the diameter for a fixed maximum input torque.

And fixed RPM too right? Or wrong?

I'm not a physics buff, I guess all that I'm thinking is that if the sfm is kept constant and the feed per rev is the same (which BTW is how it should be), then the volume of material being removed is kept constant (or very close). So at whatever diameter you are doing your machining, if your speeds and feeds are correct, the material removal rate (theoretically) shouldn't change. Now, hp and torque curves of each machine have an effect, and you can't always run the correct speed or feed.

I would say that most people probably have a tendancy to run small diameters at a surface speed that is too slow, and large diameters at a surface speed that is too fast - which in most cases would result in higher cutting forces at larger diameters.

But if it was possible to keep the correct speeds and feeds for all diameters, would the cutting forces vary with diameter? I'm asking not making a statement, because I'm curious and have thought about this in the past.

Actually, had to stand on my head a little, but a .060" deep cut taken at 1" diameter actually removes a little bit _More_ volume of material than a .060" deep cut taken at 2" diameter - assuming the same surface speed and feed per rev are used for each cut. That's because of the greater ratio of 1.120:1 compared to 2.120:2. Just food for thought.

Bryce wrote in news: snipped-for-privacy@4ax.com:

Bryce, Just the opposite. The same SFM at a smaller diameter results in a higher RPM. If the IPR feed rate is the same then you will be removing a higher volume of material over the same period of time. For example lets take

8620 at 600 SFM and .009" IPR at 1/2", 1", and 1.5" diameters. You get:

Dia. MRR HP at 95% efficiency

1.5 4.044 4.342 1.0 4.122 4.425 0.5 4.355 4.676

What Cliff was talking about was constant torque. Which is not a condition that is used, in other words, there is no G-code for constant torque control. So he's got his hook out hoping to catch someone who is not reading carefully, or has never considered what happens to the formulas people use for metal cutting when you plug in torque as a constant rather than SFM or IPR.

Dan

Dan

Dan:

Now that's interesting, the formula for MRR I used seems to show a different result.

pi * (D squared - d squared) * f * N MRR = --------------------------------------- 4 Where: D = Original Dia. of workpiece d = Dia. after cut f = feed in inches per revolution N = RPM

Dia. RPM @ (600 SFM) MRR

1.5 1,528 3.733 1.0 2,292 3.655 0.5 4,584 3.422

Can you please list the formula you're using. Perhaps this is a result of some rounding error.

The difference in the results is because Dan was calculating starting with a diameter of 1.620" and turning it to 1.500", starting with

1.120" and turning it to 1.000", and starting with .620" and turning it to .500". This is because the RPM calculations are done for a cut taken at the nominal diameters. It looks like the results that you got were derived from starting with a diameter of 1.500" and turning it to 1.380", starting with 1.000" and turning it to .880", and starting with .500" and turning it to .380". If this is the case, the RPM calculations would have to be refigured for the diameters that the cuts are being taken at, not the original material diameters. I didn't double check, but I think that should give you resulting MRRs that are a bit higher at smaller diameters.

Bryce:

OK, let me figure it your way.

Dia. RPM @ (600 SFM) MRR

1.620 1,415 3.745 1.120 2,046 3.679 .620 3,696 3.511

The results are not all that much different, about 2 1/2% in the worst case. I still think there is a formula difference at work between Dan's figures and mine. I'd like to know which formula is more accurate.

BottleBob wrote in news: snipped-for-privacy@earthlink.net:

BB, I usually just use the following formula 12*DOC*IPR*SFM=MRR. It's close enough for approximating power requirements, plus it's easy to remember. I'm using the same formula you are for the numbers I gave. I even put it into a spreadsheet and I still get the same numbers. Maybe you have an algebraic hierarchy type of error or ?? Just to be sure we're on the same page: SFM - 600 IPR - .009" DOC - .06" (radial depth) Finish diameters of .5", 1", 1.5"

I calculated the rpm (N) off of the finished diameter as is proper. I use a shortcut formula for RPM of (sfm/dia)*3.82=N. Maybe the error is in there? Try that out and let me know.

Dan

Dan:

OK, the discrepancy between the results of your use of the MRR formula and my use of it has been resolved. The point (OD or ID of the cut) where the RPM is figured accounts for the difference. NOW, this brings to light a more basic discrepancy in the interpretation of just where the SFPM should be calculated from. When you say "I calculated the rpm (N) off of the finished diameter as is proper." Both you and Bryce believe this to be true, but the "AS IS PROPER" just doesn't ring logically valid in my mind. What is the point of upper SFPM limits? I've always thought that the SFPM figures are where the insert or tooling has a reasonable lifespan (15 minutes, 30 min. whatever), *IF* you use the ID of the cut rather than the OD, the OD may then be too fast for a reasonable lifespan. Essentially defeating the whole point of using SFPM limits to begin with. With a large dia. piece and/or a small DOC it doesn't make much difference, but in some cases like if you're turning a .5 Dia. piece to .060 in one pass then it may make a big difference. End mill SFPM is figured from the OD of the cutter, not from the DOC the end mills is making. Why should a lathe be any different. Plus IF as you say it's "proper" to overspeed the OD SFPM, by calculating from the ID of the cut's SFPM, then why not use the higher figure as your standard OD SFPM to begin with. Then you could increase it even MORE by calculating by the ID of the cut. Where would this progression stop?