Cliff wrote: > BTW, Tell BB that his HP to cubic inches per minute calcs > are dead wrong.
Cliff:
First: You need to change your date setting, it's Nov. 2 not Oct 31. You must have caught a "trick-or-treat" virus. Or your computer burped when Daylight Savings changed. <g>
Second: Where did this "HP to cu. in. per minute" thing come from? I haven't posted about HP in quite some time.
Third: "Dead Wrong" sounds so serious & FINAL. Now I'm sure you must realize that these kinds of horsepower calculations are rough estimates, sort of just ball park figures so you don't try to do a 40 HP cut on a 15 HP machine and stall the motor. Let me go over the process and sources I've used. Here are the essentials of one post I made in 2000.
W = Width of cut H = depth of cut F = Feed rate in inches per minute
HP = UHP X MRR
HP = Horsepower UHP = Unit horsepower (number taken off chart) MRR = Material Removal Rate =============================================================
Now the MasterCam book that had these formulas was a textbook I bought to learn enough MasterCam to be able to compare it to the GibbsCam program we already had. The textbook was MasterCam 7 by Dr. S.C. Jonathon Lin & Dr. F.C. Tony Shiue Copyright 1998.
The MRR is a straight-forward volume per time calculation and unambiguous. As long as you are calculating from full width cutter entry to full width cutter exit. If you try to calculate from first contact with an 8" face mill to full exit you'll of course be off a little.
The UHP (Unit Horsepower correction factor) is a number taken off a chart created from experiments done on different materials. Now that can be influenced be any number of factors; exact hardness & micro structure of material, alloying elements, sharpness & material of the cutter, whether the cutter has a positive or negative rake, various tool coatings, polished flutes or inserts, chamfer on inserts, etc. etc. The information on these kinds of charts were probably taken from information derived from sources like the Machining Data Handbook or other similar experimental data.
Go to the bottom. I'm surprised it's still an active site after all these years.
Now here is some info from my Machinery's Handbook 26th Edition. I'm not able to use subscripts, so I'll just put them side by side.
MRR = fm*w*d
Where MRR = Material Removal Rate fm = Feed Rate in inches per minute w = Width of cut d = Depth of cut
Same formula as before with slightly different symbols.
The power formula;
Pc = Kp*C*Q*W
Where Pc = HP (or kW if using SI metric units throughout) Kp = Power constant (identical to Unit HP) taken off chart, C = Feed factor for power constant (fine adjustment) Q = MRR (Material Removal Rate) W = Tool wear factor (another fine adjustment)
Essentially the same formula as the MasterCam one, again just different symbols and some extra fine tuning.
Now if these sources are "Dead Wrong" like you suggest, PLEASE, by all means let's see what YOU believe is the CORRECT way to calculate MRR and HP per cut. I'm always open & willing to learn something new.
I've added in our sister group rcm, as this sort of thing may be On-Topic & of interest over there as well.
On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> wrote: > Second: Where did this "HP to cu. in. per minute" thing come from? I >haven't posted about HP in quite some time.
"Forces in turning are a lot higher than with milling." - Dan M. This blanket claim implies that more HP is needed per unit volume of material sheared off the stock per minute. -- Cliff
Cliff wrote: > On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>> Second: Where did this "HP to cu. in. per minute" thing come from? I >> haven't posted about HP in quite some time.
Cliff:
The entire focus of my reply to you was concerning your following comment:
"BTW, Tell BB that his HP to cubic inches per minute calcs are dead wrong."
Which, as of yet, you have failed to address.
> "Forces in turning are a lot higher than with milling." - Dan M. > This blanket claim implies that more HP is needed per unit volume > of material sheared off the stock per minute.
That was Dan Murphy's comment. It has zero to do with ME, so I'll not address it.
BUT, let me illustrate a couple of other things. Small lathes can hold work using a collet closer, 10"-12"-14" chucks, faceplates, etc. Now a mill using a collet could do small lathe type work, as Dan mentioned. But imagine trying to affix a 14" lathe chuck to a 40 Taper holder and then trying to take a serious cut on a large workpiece. I don't think I'd be far off the mark in thinking that this could be a recipe for disaster. To say nothing about the spindle bearing loads and motor loads trying spin that mass up to speed AND slow it down. And don't even TRY to put that chuck in the carousel (even an empty one), it would probably fall through to the table.
> Cliff wrote: > > On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
> >> Second: Where did this "HP to cu. in. per minute" thing come from? I > >> haven't posted about HP in quite some time.
> Cliff:
> The entire focus of my reply to you was concerning your following comment:
> "BTW, Tell BB that his HP to cubic inches per minute calcs > are dead wrong."
> Which, as of yet, you have failed to address.
> > "Forces in turning are a lot higher than with milling." - Dan M. > > This blanket claim implies that more HP is needed per unit volume > > of material sheared off the stock per minute.
> That was Dan Murphy's comment. It has zero to do with ME, so I'll not > address it.
> BUT, let me illustrate a couple of other things. Small lathes can > hold work using a collet closer, 10"-12"-14" chucks, faceplates, etc. > Now a mill using a collet could do small lathe type work, as Dan > mentioned. But imagine trying to affix a 14" lathe chuck to a 40 Taper > holder and then trying to take a serious cut on a large workpiece. I > don't think I'd be far off the mark in thinking that this could be a > recipe for disaster. To say nothing about the spindle bearing loads > and motor loads trying spin that mass up to speed AND slow it down. > And don't even TRY to put that chuck in the carousel (even an empty > one), it would probably fall through to the table.
BB I think Cliffy is just missing you<g>. Its a love hate thing LOL By debating its keeps the group "on topic" Thats a good thing.
Hey how bout a 14" chuck on a 60taper? We had a 24" face mill at Cat. Damn 60 taper, You need a jib crane to load the carousel (horizontal). The smaller tools like 4" boring bars the "old timers" would hand wrestle them to the tool setter! Thus metatarsel shoes required, my feet are still not right after that stint. Any body interested in a size 11? Not even broke in yet!
--
~o~~o~~o~~o~~o~~o~~o~~o~~o~~~o~~
Some people make things happen.... Some watch things happen... While others wonder what had happened
On Tue, 03 Nov 2009 15:50:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>Cliff wrote: >> On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>> Second: Where did this "HP to cu. in. per minute" thing come from? I >>> haven't posted about HP in quite some time.
>Cliff:
> The entire focus of my reply to you was concerning your following comment:
> "BTW, Tell BB that his HP to cubic inches per minute calcs >are dead wrong."
> Which, as of yet, you have failed to address.
It is perfectly clear.
>> "Forces in turning are a lot higher than with milling." - Dan M. >> This blanket claim implies that more HP is needed per unit volume >> of material sheared off the stock per minute.
> That was Dan Murphy's comment. It has zero to do with ME, so I'll not >address it.
You don't know about HP, torque, forces (vector quantities) & etc?
> BUT, let me illustrate a couple of other things. Small lathes can >hold work using a collet closer, 10"-12"-14" chucks, faceplates, etc. > Now a mill using a collet could do small lathe type work, as Dan >mentioned. But imagine trying to affix a 14" lathe chuck to a 40 Taper >holder and then trying to take a serious cut on a large workpiece. I >don't think I'd be far off the mark in thinking that this could be a >recipe for disaster. To say nothing about the spindle bearing loads >and motor loads trying spin that mass up to speed AND slow it down. >And don't even TRY to put that chuck in the carousel (even an empty >one), it would probably fall through to the table.
Different machines are best suited for different purposes in many cases. That's why they make different machines IIRC.
None of which has to do with his claim <G>. -- Cliff
On Wed, 4 Nov 2009 04:28:14 -0800 (PST), cncmillgil <mil...@cin.net> wrote:
>Hey how bout a 14" chuck on a 60taper? >We had a 24" face mill at Cat. Damn 60 taper, You need a jib crane to >load the carousel (horizontal). The smaller tools like 4" boring bars >the "old timers" would hand wrestle them to the tool setter! Thus >metatarsel shoes required, my feet are still not right after that >stint. Any body interested in a size 11? Not even broke in yet!
Saw some large lathes in Houston once. VTLs doing nuclear valves. ~24 foot diameter tables IIRC. -- Cliff
Cliff wrote: > On Tue, 03 Nov 2009 15:50:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>> Cliff wrote: >> The entire focus of my reply to you was concerning your following comment:
>> "BTW, Tell BB that his HP to cubic inches per minute calcs >> are dead wrong."
>> Which, as of yet, you have failed to address.
> It is perfectly clear.
Cliff:
So essentially what you're saying is that you're not able to support your original accusation? Trolling were we? Why am I not surprised. LMAO!
> You don't know about HP, torque, forces (vector quantities) & etc?
OK, let's do a simple machining HP problem. We've got a 4" diameter face mill (with 90 degree inserts), the depth of cut is .125", it's cutting full width at 250 inches per minute in aluminum. The Unit HP correction factor for aluminum is .25 (taken off the MMS Online chart at bottom of this post).
What is the approximate Horsepower that it takes to make that cut?
=========================================== MRR=W X H X F
MRR = Material Removal Rate W = Width of cut H = depth of cut F = Feed rate in inches per minute
4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
HP = UHP X MRR
HP = Horsepower UHP = Unit horsepower (number taken off chart) MRR = Material Removal Rate
.25 X 125 = 31.25 HP
Seems pretty straightforward to me. Since you are claiming that it's *Dead Wrong*; please, by all means, illustrate what YOU think is the RIGHT way, AND what the correct approx. HP for that cut is.
>> You don't know about HP, torque, forces (vector quantities) & etc?
> OK, let's do a simple machining HP problem. We've got a 4" diameter >face mill (with 90 degree inserts), the depth of cut is .125", it's >cutting full width at 250 inches per minute in aluminum. > The Unit HP correction factor for aluminum is .25 (taken off the MMS >Online chart at bottom of this post).
> What is the approximate Horsepower that it takes to make that cut?
>=========================================== >MRR=W X H X F
>MRR = Material Removal Rate >W = Width of cut >H = depth of cut >F = Feed rate in inches per minute
> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>HP = UHP X MRR
>HP = Horsepower >UHP = Unit horsepower (number taken off chart) >MRR = Material Removal Rate
> .25 X 125 = 31.25 HP
> Seems pretty straightforward to me. Since you are claiming that it's >*Dead Wrong*; please, by all means, illustrate what YOU think is the >RIGHT way, AND what the correct approx. HP for that cut is.
Is that for a mill or a lathe? Per Dan M you need more force for a lathe <G>. And we know that work = force * distance .... (well, I do anyway). And that work = HP * time ... -- Cliff
Cliff wrote: > On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote:
>> Cliff wrote: >>> On Tue, 03 Nov 2009 15:50:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>>> Cliff wrote: >>>> The entire focus of my reply to you was concerning your following comment:
>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>> are dead wrong."
>>>> Which, as of yet, you have failed to address. >>> It is perfectly clear. >> So essentially what you're saying is that you're not able to support >> your original accusation? Trolling were we? Why am I not surprised. >> LMAO!
> I said it's perfectly clear & it is.
Cliff:
Yes, I suppose it is. Since you have been unable, or unwilling, to support your original accusation after being asked like 3 times now, it probably is perfectly clear that your original accusation was unfounded.
>>> You don't know about HP, torque, forces (vector quantities) & etc? >> OK, let's do a simple machining HP problem. We've got a 4" diameter >> face mill (with 90 degree inserts), the depth of cut is .125", it's >> cutting full width at 250 inches per minute in aluminum. >> The Unit HP correction factor for aluminum is .25 (taken off the MMS >> Online chart at bottom of this post).
>> What is the approximate Horsepower that it takes to make that cut?
>> =========================================== >> MRR=W X H X F
>> MRR = Material Removal Rate >> W = Width of cut >> H = depth of cut >> F = Feed rate in inches per minute
>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>> HP = UHP X MRR
>> HP = Horsepower >> UHP = Unit horsepower (number taken off chart) >> MRR = Material Removal Rate
>> .25 X 125 = 31.25 HP
>> Seems pretty straightforward to me. Since you are claiming that it's >> *Dead Wrong*; please, by all means, illustrate what YOU think is the >> RIGHT way, AND what the correct approx. HP for that cut is.
> Is that for a mill or a lathe? > Per Dan M you need more force for a lathe <G>. > And we know that work = force * distance .... (well, I do anyway). > And that work = HP * time ...
Cliff ol' buddy, was the example problem I gave too hard for you to do? Should I make it simpler? I mean you're claiming to be able to know about calculating HP, but you have so far produced nothing more than Cliffasion, misdirection, and rhetoric. Can you do the problem, plus show that the way the MasterCam Book & The Machinery's Handbook method is *Dead Wrong*? If you can't, then your original accusation is bogus & meaningless.
On Fri, 06 Nov 2009 08:56:41 -0800, BottleBob <bottl...@earthlink.net> wrote:
>Cliff wrote: >> On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>> Cliff wrote: >>>> On Tue, 03 Nov 2009 15:50:22 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>>>> Cliff wrote: >>>>> The entire focus of my reply to you was concerning your following comment:
>>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>>> are dead wrong."
>>>>> Which, as of yet, you have failed to address.
>>>> It is perfectly clear.
>>> So essentially what you're saying is that you're not able to support >>> your original accusation? Trolling were we? Why am I not surprised. >>> LMAO!
>> I said it's perfectly clear & it is.
>Cliff:
> Yes, I suppose it is. Since you have been unable, or unwilling, to >support your original accusation after being asked like 3 times now, it >probably is perfectly clear that your original accusation was unfounded.
I said it's perfectly clear & it is. How much force (a vector quantity) does it take to shear a chip & how is that related to HP (as you seem to be confused about it)? And to Dan's claim?
>>>> You don't know about HP, torque, forces (vector quantities) & etc? >>> OK, let's do a simple machining HP problem. We've got a 4" diameter >>> face mill (with 90 degree inserts), the depth of cut is .125", it's >>> cutting full width at 250 inches per minute in aluminum. >>> The Unit HP correction factor for aluminum is .25 (taken off the MMS >>> Online chart at bottom of this post).
>>> What is the approximate Horsepower that it takes to make that cut?
>>> =========================================== >>> MRR=W X H X F
>>> MRR = Material Removal Rate >>> W = Width of cut >>> H = depth of cut >>> F = Feed rate in inches per minute
>>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>>> HP = UHP X MRR
>>> HP = Horsepower >>> UHP = Unit horsepower (number taken off chart) >>> MRR = Material Removal Rate
>>> .25 X 125 = 31.25 HP
>>> Seems pretty straightforward to me. Since you are claiming that it's >>> *Dead Wrong*; please, by all means, illustrate what YOU think is the >>> RIGHT way, AND what the correct approx. HP for that cut is.
>> Is that for a mill or a lathe? >> Per Dan M you need more force for a lathe <G>. >> And we know that work = force * distance .... (well, I do anyway). >> And that work = HP * time ...
> Cliff ol' buddy, was the example problem I gave too hard for you to >do? Should I make it simpler?
Is that for a mill or a lathe? Per Dan M you need more force for a lathe <G>. And we know that work = force * distance .... (well, I do anyway). And that work = HP * time ...
>I mean you're claiming to be able to >know about calculating HP, but you have so far produced nothing more >than Cliffasion, misdirection, and rhetoric. > Can you do the problem, plus show that the way the MasterCam Book & >The Machinery's Handbook method is *Dead Wrong*? If you can't, then >your original accusation is bogus & meaningless.
I can only assume you don't understand Dan's claim or force (a vector quantity). Or work or HP-hours or .... -- Cliff
Cliff wrote: > On Fri, 06 Nov 2009 08:56:41 -0800, BottleBob <bottl...@earthlink.net> wrote:
>> Cliff wrote: >>> On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote: >>>>>> Cliff wrote: >>>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>>>> are dead wrong."
>>>>>> Which, as of yet, you have failed to address. >>>>> It is perfectly clear. >>>> So essentially what you're saying is that you're not able to support >>>> your original accusation? Trolling were we? Why am I not surprised. >>>> LMAO! >>> I said it's perfectly clear & it is. >> Yes, I suppose it is. Since you have been unable, or unwilling, to >> support your original accusation after being asked like 3 times now, it >> probably is perfectly clear that your original accusation was unfounded.
> I said it's perfectly clear & it is.
Cliff:
I agree. And it's become more clear with every reply that you make that you are unable, or unwilling, to do the simple problem/example I gave. OR to support your original accusation that the solution in the MasterCam Book & The Machinery's Handbook is *dead wrong*.
This is another one of your machining jokes, right? LOL
> How much force (a vector quantity) does it take to shear a chip & how > is that related to HP (as you seem to be confused about it)?
Now you're just being silly. The experimental work has already been done. That's what the Unit Horsepower Tables are all about, giving the HP correction factors for the material removal rate in cu. in. per minute for various materials. There's no need to reinvent the wheel here.
> And to Dan's claim?
Why do you keep trying to deflect the issue by bringing up Dan. It's YOUR comment to ME that is the issue of this sub-thread.
>>>> OK, let's do a simple machining HP problem. We've got a 4" diameter >>>> face mill (with 90 degree inserts), the depth of cut is .125", it's >>>> cutting full width at 250 inches per minute in aluminum. >>>> The Unit HP correction factor for aluminum is .25 (taken off the MMS >>>> Online chart at bottom of this post).
>>>> What is the approximate Horsepower that it takes to make that cut?
>>>> =========================================== >>>> MRR=W X H X F
>>>> MRR = Material Removal Rate >>>> W = Width of cut >>>> H = depth of cut >>>> F = Feed rate in inches per minute
>>>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>>>> HP = UHP X MRR
>>>> HP = Horsepower >>>> UHP = Unit horsepower (number taken off chart) >>>> MRR = Material Removal Rate
>>>> .25 X 125 = 31.25 HP
>>>> Seems pretty straightforward to me. Since you are claiming that it's >>>> *Dead Wrong*; please, by all means, illustrate what YOU think is the >>>> RIGHT way, AND what the correct approx. HP for that cut is. >>>> http://www.mmsonline.com/articles/019703.html >>>> ============================================= >>>> Table I >>>> Material Unit Horsepower >>>> Aluminum 0.25 >>>> Brass (Soft) 0.33 >>>> Brass (Hard) 0.50 >>>> Bronze (Hard) 0.71 >>>> Bronze (Very Hard) 1.54 >>>> C1 (200 BHN) 0.67 >>>> C1 (Over 200 BHN) 1.00 >>>> Malleable Iron 0.80 >>>> Steel 100 BHN 1.25 >>>> Steel 150 BHN 1.43 >>>> Steel 200 BHN 1.54 >>>> Steel 250 BHN 1.82 >>>> Steel 400 BHN 2.00 >>>> ============================================= >> Cliff ol' buddy, was the example problem I gave too hard for you to >> do? Should I make it simpler?
> Is that for a mill or a lathe?
Let's say it's for a mill. Can you do the problem or not?
> I can only assume you don't understand Dan's claim or force (a vector > quantity). Or work or HP-hours or ....
You can "assume" whatever you wish. I showed you how to calculate the HP of a cut. YOU haven't yet done the problem, even with spending multiple posts wasting time with evasions, innuendo, and attempted misdirections.
Just do the simple example/problem. And show that the method that the MasterCam Book & The Machinery's Handbook illustrate is *Dead Wrong*. Or just admit that you can't.
This is getting old quick, Cliff ol' buddy. You've had about 4 chances to support your accusation, and so far haven't done so. Let's see some action, instead of evasive jibber-jabber. LOL
On Fri, 06 Nov 2009 16:21:28 -0800, BottleBob <bottl...@earthlink.net> wrote:
>Cliff wrote: >> On Fri, 06 Nov 2009 08:56:41 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>> Cliff wrote: >>>> On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>>>>>> Cliff wrote:
>>>>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>>>>> are dead wrong."
>>>>>>> Which, as of yet, you have failed to address.
>>>>>> It is perfectly clear.
>>>>> So essentially what you're saying is that you're not able to support >>>>> your original accusation? Trolling were we? Why am I not surprised. >>>>> LMAO!
>>>> I said it's perfectly clear & it is.
>>> Yes, I suppose it is. Since you have been unable, or unwilling, to >>> support your original accusation after being asked like 3 times now, it >>> probably is perfectly clear that your original accusation was unfounded.
>> I said it's perfectly clear & it is.
>Cliff:
> I agree. And it's become more clear with every reply that you make >that you are unable, or unwilling, to do the simple problem/example I >gave.
CLUE: It's not much related to the subject matter. You are on the wrong track and that light up ahead ...
>OR to support your original accusation that the solution in the >MasterCam Book & The Machinery's Handbook is *dead wrong*.
Reread said post.
> This is another one of your machining jokes, right? LOL
>> How much force (a vector quantity) does it take to shear a chip & how >> is that related to HP (as you seem to be confused about it)?
> Now you're just being silly. The experimental work has already been >done. That's what the Unit Horsepower Tables are all about, giving the >HP correction factors for the material removal rate in cu. in. per >minute for various materials. > There's no need to reinvent the wheel here.
IOW You cannot answer the question? Vectors again? Or you don't understand it?
>> And to Dan's claim?
> Why do you keep trying to deflect the issue by bringing up Dan. It's >YOUR comment to ME that is the issue of this sub-thread.
>>>>> OK, let's do a simple machining HP problem. We've got a 4" diameter >>>>> face mill (with 90 degree inserts), the depth of cut is .125", it's >>>>> cutting full width at 250 inches per minute in aluminum. >>>>> The Unit HP correction factor for aluminum is .25 (taken off the MMS >>>>> Online chart at bottom of this post).
>>>>> What is the approximate Horsepower that it takes to make that cut?
>>>>> =========================================== >>>>> MRR=W X H X F
>>>>> MRR = Material Removal Rate >>>>> W = Width of cut >>>>> H = depth of cut >>>>> F = Feed rate in inches per minute
>>>>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>>>>> HP = UHP X MRR
>>>>> HP = Horsepower >>>>> UHP = Unit horsepower (number taken off chart) >>>>> MRR = Material Removal Rate
>>>>> .25 X 125 = 31.25 HP
>>>>> Seems pretty straightforward to me. Since you are claiming that it's >>>>> *Dead Wrong*; please, by all means, illustrate what YOU think is the >>>>> RIGHT way, AND what the correct approx. HP for that cut is.
>>> Cliff ol' buddy, was the example problem I gave too hard for you to >>> do? Should I make it simpler?
>> Is that for a mill or a lathe?
> Let's say it's for a mill. Can you do the problem or not?
What if it was for a lathe? Remember Dan's claim ....
>> I can only assume you don't understand Dan's claim or force (a vector >> quantity). Or work or HP-hours or ....
> You can "assume" whatever you wish. I showed you how to calculate the >HP of a cut.
For a mill or a lathe?
>YOU haven't yet done the problem, even with spending >multiple posts wasting time with evasions, innuendo, and attempted >misdirections.
You have the wrong "problem" still <G>.
> Just do the simple example/problem. And show that the method that the >MasterCam Book & The Machinery's Handbook illustrate is *Dead Wrong*. >Or just admit that you can't.
> This is getting old quick, Cliff ol' buddy. You've had about 4 >chances to support your accusation, and so far haven't done so. Let's >see some action, instead of evasive jibber-jabber. LOL
Cliff wrote: > On Fri, 06 Nov 2009 16:21:28 -0800, BottleBob <bottl...@earthlink.net> wrote:
>> Cliff wrote: >>> On Fri, 06 Nov 2009 08:56:41 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>>> Cliff wrote: >>>>> On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote: >>>>>>>> Cliff wrote: >>>>>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>>>>>> are dead wrong." >>>>>>> It is perfectly clear. >>>>>> So essentially what you're saying is that you're not able to support >>>>>> your original accusation? Trolling were we? Why am I not surprised. >>>>>> LMAO! >>>>> I said it's perfectly clear & it is. >>>> Yes, I suppose it is. Since you have been unable, or unwilling, to >>>> support your original accusation after being asked like 3 times now, it >>>> probably is perfectly clear that your original accusation was unfounded. >>> I said it's perfectly clear & it is. >> Cliff:
>> I agree. And it's become more clear with every reply that you make >> that you are unable, or unwilling, to do the simple problem/example I >> gave.
> CLUE: It's not much related to the subject matter.
Cliff:
Cliff, Cliff, Cliff... That was like the *fifth* time I've asked you to "show your work" in calculating the Horsepower of a cut. Which obviously you haven't done. What's the problem buddy?
>> OR to support your original accusation that the solution in the >> MasterCam Book & The Machinery's Handbook is *dead wrong*.
> Reread said post.
Here's your direct quote: "BTW, Tell BB that his HP to cubic inches per minute calcs are dead wrong." Now since MY calcs are the same ones that the MasterCam Book & Machinery's Handbook uses, you were essentially saying that the way THEY calculate horsepower of a cut is *dead wrong*. Unfortunately you've presented zero evidence to support your claims, even after being asked repeatedly to do so. Again, what's the problem?
> About Dan's claim & your seeming confusion.
What confusion, I showed you how to calculate the HP of a cut from cutter width, depth of cut, and a feedrate in IPM.
>>>>>> OK, let's do a simple machining HP problem. We've got a 4" diameter >>>>>> face mill (with 90 degree inserts), the depth of cut is .125", it's >>>>>> cutting full width at 250 inches per minute in aluminum. >>>>>> The Unit HP correction factor for aluminum is .25 (taken off the MMS >>>>>> Online chart at bottom of this post).
>>>>>> What is the approximate Horsepower that it takes to make that cut?
>>>>>> =========================================== >>>>>> MRR=W X H X F
>>>>>> MRR = Material Removal Rate >>>>>> W = Width of cut >>>>>> H = depth of cut >>>>>> F = Feed rate in inches per minute
>>>>>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>>>>>> HP = UHP X MRR
>>>>>> HP = Horsepower >>>>>> UHP = Unit horsepower (number taken off chart) >>>>>> MRR = Material Removal Rate
>>>>>> .25 X 125 = 31.25 HP
>>>>>> Seems pretty straightforward to me. Since you are claiming that it's >>>>>> *Dead Wrong*; please, by all means, illustrate what YOU think is the >>>>>> RIGHT way, AND what the correct approx. HP for that cut is. >>>>>> http://www.mmsonline.com/articles/019703.html >>>>>> ============================================= >>>>>> Table I >>>>>> Material Unit Horsepower >>>>>> Aluminum 0.25 >>>>>> Brass (Soft) 0.33 >>>>>> Brass (Hard) 0.50 >>>>>> Bronze (Hard) 0.71 >>>>>> Bronze (Very Hard) 1.54 >>>>>> C1 (200 BHN) 0.67 >>>>>> C1 (Over 200 BHN) 1.00 >>>>>> Malleable Iron 0.80 >>>>>> Steel 100 BHN 1.25 >>>>>> Steel 150 BHN 1.43 >>>>>> Steel 200 BHN 1.54 >>>>>> Steel 250 BHN 1.82 >>>>>> Steel 400 BHN 2.00 >>>>>> ============================================= > What if it was for a lathe?
The process for calculating HP for a lathe cut is slightly different. (Ignoring C axis live tool cuts and the like)
Pi X (D squared-d squared) X fr X N MRR = ------------------------------------ 4
Where:
MRR = Material Removal Rate in cu. in. per minute D = Original (major) diameter of workpiece d = Diameter after cut fr = Feed in inches per revolution N = Revolutions per minute
HP = UHP X MRR X C
UHP = Number taken off chart (different chart for lathes) C = Feed correction factor (High feed rates require less Unit HP)
Well, I can't speak for Dan Murphy, but I believe he mentioned something about the forces in turning being higher than in milling. Which I took to mean that because of the larger headstock bearings on a lathe, they could take more Z and X forces than the bearings on a vertical mill and so can take heavier cuts.
You seem to think what he said was that lathes use more HP to remove the same cu. in. of material per minute.
Let's do a little calculation experiment comparing the HP it takes to remove the same amount of material on a lathe vs a mill.
3" aluminum bar, 90 degree insert, .250 Depth of cut, feed of .012 per revolution, 4,850 RPM
(Pi X (9-6.25) X .012 X 4,850)/4 = A MRR of 125.7 cu. in. per min.
I adjusted the parameters so as to arrive at a similar MRR as the former example of a milling cut. (that removed 125 cu. in. of material @ 31 HP)
HP = UHP X MRR X C
Now at .012 per revolution the C correction factor is 1, so:
HP = UHP (.25) X 125 - Or approx. 31 HP.
BUT, if we change the feed to .005 per revolution then the C correction factor is about 1.2 and the HP would be:
HP = UHP (.25) X 125 X 1.2 - OR approx. 37.5 HP
BUT, if we change the feed to .020 per revolution then the C correction factor is about .8 and the HP would be:
HP = UHP (.25) X 125 X .8 - Or approx. 27.9 HP
Now I didn't go back and change the parameters to allow for the same MRR at the different feed rates, but anyone could if they wished. It shouldn't change the results.
As regards the question of which takes more HP to remove a similar amount of stock, a lathe or a mill - the conclusion that I draw is - IT DEPENDS.
It seems at lower feed rates the lathe uses more HP, and conversely at higher feed rates is uses less.
> On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> > wrote:
>> Second: Where did this "HP to cu. in. per minute" thing come >> from? I >>haven't posted about HP in quite some time.
> "Forces in turning are a lot higher than with milling." - Dan M. > This blanket claim implies that more HP is needed per unit volume > of material sheared off the stock per minute.
Nope. I was talking about cutting forces acting on the V-flange tool such as feed force, tangential force, radial force and the ever important resultant force, not horsepower.
Horepower has got nothing to do with it.
Sheesh.
Unless you'd like to explain to us how horsepower at the spindle raises or lowers feed force acting on the V-flange tool. (This ought to be good)
LOL. This is almost as funny as your C-CRAP method for programming a lathe.
I've got some books you could read if you'd like to learn about cutting metal.
>> On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> >> wrote:
>>> Second: Where did this "HP to cu. in. per minute" thing come >>> from? I >>>haven't posted about HP in quite some time.
>> "Forces in turning are a lot higher than with milling." - Dan M. >> This blanket claim implies that more HP is needed per unit volume >> of material sheared off the stock per minute.
>Nope. I was talking about cutting forces acting on the V-flange tool such >as feed force, tangential force, radial force and the ever important >resultant force, not horsepower.
<Giggle>
You wrote: "Forces in turning are a lot higher than with milling."
NOW you are calling HP a force. AND claiming you were posting about "cutting forces acting on the V-flange tool" (an entirely new subject) !!!
>Horepower has got nothing to do with it.
Did you try telling BB that one?
>Sheesh.
>Unless you'd like to explain to us how horsepower at the spindle raises or >lowers feed force acting on the V-flange tool. (This ought to be good)
>LOL. This is almost as funny as your C-CRAP method for programming a lathe.
As compared & contrasted to bad math & bad programming & silly claims ...
>I've got some books you could read if you'd like to learn about cutting >metal.
Can you loan them to BB after you read them? -- Cliff
On Sat, 07 Nov 2009 10:15:36 -0800, BottleBob <bottl...@earthlink.net> wrote:
>Cliff wrote: >> On Fri, 06 Nov 2009 16:21:28 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>> Cliff wrote: >>>> On Fri, 06 Nov 2009 08:56:41 -0800, BottleBob <bottl...@earthlink.net> wrote:
>>>>> Cliff wrote: >>>>>> On Thu, 05 Nov 2009 09:07:43 -0800, BottleBob <bottl...@earthlink.net> wrote: >>>>>>>>> Cliff wrote: >>>>>>>>> "BTW, Tell BB that his HP to cubic inches per minute calcs >>>>>>>>> are dead wrong."
>>>>>>>> It is perfectly clear. >>>>>>> So essentially what you're saying is that you're not able to support >>>>>>> your original accusation? Trolling were we? Why am I not surprised. >>>>>>> LMAO! >>>>>> I said it's perfectly clear & it is. >>>>> Yes, I suppose it is. Since you have been unable, or unwilling, to >>>>> support your original accusation after being asked like 3 times now, it >>>>> probably is perfectly clear that your original accusation was unfounded. >>>> I said it's perfectly clear & it is. >>> Cliff:
>>> I agree. And it's become more clear with every reply that you make >>> that you are unable, or unwilling, to do the simple problem/example I >>> gave.
>> CLUE: It's not much related to the subject matter.
>Cliff:
> Cliff, Cliff, Cliff... That was like the *fifth* time I've asked you >to "show your work" in calculating the Horsepower of a cut. Which >obviously you haven't done. What's the problem buddy?
It's STILL not much related to the subject matter.
>>> OR to support your original accusation that the solution in the >>> MasterCam Book & The Machinery's Handbook is *dead wrong*.
>> Reread said post.
> Here's your direct quote: "BTW, Tell BB that his HP to cubic inches >per minute calcs are dead wrong."
Did he tell you that? For a lathe or a mill?
> Now since MY calcs are the same ones that the MasterCam Book & >Machinery's Handbook uses, you were essentially saying that the way >THEY calculate horsepower of a cut is *dead wrong*.
Or is Dan M. wrong?
> Unfortunately you've presented zero evidence to support your claims, >even after being asked repeatedly to do so. Again, what's the problem?
You do need to follow along !!!
>> About Dan's claim & your seeming confusion.
> What confusion, I showed you how to calculate the HP of a cut from >cutter width, depth of cut, and a feedrate in IPM.
For a lathe or a mill? And how does it all relate to those pesky (vector) forces?
>>>>>>> OK, let's do a simple machining HP problem. We've got a 4" diameter >>>>>>> face mill (with 90 degree inserts), the depth of cut is .125", it's >>>>>>> cutting full width at 250 inches per minute in aluminum. >>>>>>> The Unit HP correction factor for aluminum is .25 (taken off the MMS >>>>>>> Online chart at bottom of this post).
>>>>>>> What is the approximate Horsepower that it takes to make that cut?
>>>>>>> =========================================== >>>>>>> MRR=W X H X F
>>>>>>> MRR = Material Removal Rate >>>>>>> W = Width of cut >>>>>>> H = depth of cut >>>>>>> F = Feed rate in inches per minute
>>>>>>> 4" X .125" X 250 IPM = A MRR of 125 cu. in. per minute
>>>>>>> HP = UHP X MRR
>>>>>>> HP = Horsepower >>>>>>> UHP = Unit horsepower (number taken off chart) >>>>>>> MRR = Material Removal Rate
>>>>>>> .25 X 125 = 31.25 HP
>>>>>>> Seems pretty straightforward to me. Since you are claiming that it's >>>>>>> *Dead Wrong*; please, by all means, illustrate what YOU think is the >>>>>>> RIGHT way, AND what the correct approx. HP for that cut is. >>>>>>> http://www.mmsonline.com/articles/019703.html >>>>>>> ============================================= >>>>>>> Table I >>>>>>> Material Unit Horsepower >>>>>>> Aluminum 0.25 >>>>>>> Brass (Soft) 0.33 >>>>>>> Brass (Hard) 0.50 >>>>>>> Bronze (Hard) 0.71 >>>>>>> Bronze (Very Hard) 1.54 >>>>>>> C1 (200 BHN) 0.67 >>>>>>> C1 (Over 200 BHN) 1.00 >>>>>>> Malleable Iron 0.80 >>>>>>> Steel 100 BHN 1.25 >>>>>>> Steel 150 BHN 1.43 >>>>>>> Steel 200 BHN 1.54 >>>>>>> Steel 250 BHN 1.82 >>>>>>> Steel 400 BHN 2.00 >>>>>>> =============================================
>> What if it was for a lathe?
> The process for calculating HP for a lathe cut is slightly different. >(Ignoring C axis live tool cuts and the like)
> Pi X (D squared-d squared) X fr X N >MRR = ------------------------------------ > 4
>Where:
>MRR = Material Removal Rate in cu. in. per minute >D = Original (major) diameter of workpiece >d = Diameter after cut >fr = Feed in inches per revolution >N = Revolutions per minute
>HP = UHP X MRR X C
>UHP = Number taken off chart (different chart for lathes) >C = Feed correction factor (High feed rates require less Unit HP)
Gee, I must have missed force in there. What variable was it?
> Well, I can't speak for Dan Murphy, but I believe he mentioned >something about the forces in turning being higher than in milling.
After all the times I've exactly quoted him? "Forces in turning are a lot higher than with milling." - Dan M.
What did you think this subthread was about?
>Which I took to mean that because of the larger headstock bearings on a >lathe, they could take more Z and X forces than the bearings on a >vertical mill and so can take heavier cuts.
You assume or you speak or Dan? "Forces in turning are a lot higher than with milling." - Dan M.
Then you went on with HP stuff ... because of claimed bearing sizes? What does that have to do with HP consumed?
> You seem to think what he said was that lathes use more HP to remove >the same cu. in. of material per minute.
"Forces in turning are a lot higher than with milling." - Dan M.
> Let's do a little calculation experiment comparing the HP it takes to >remove the same amount of material on a lathe vs a mill.
> 3" aluminum bar, 90 degree insert, .250 Depth of cut, feed of .012 per >revolution, 4,850 RPM
>(Pi X (9-6.25) X .012 X 4,850)/4 = A MRR of 125.7 cu. in. per min.
> I adjusted the parameters so as to arrive at a similar MRR as the >former example of a milling cut. (that removed 125 cu. in. of material >@ 31 HP)
> HP = UHP X MRR X C
> Now at .012 per revolution the C correction factor is 1, so:
>HP = UHP (.25) X 125 - Or approx. 31 HP.
> BUT, if we change the feed to .005 per revolution then the C >correction factor is about 1.2 and the HP would be:
>HP = UHP (.25) X 125 X 1.2 - OR approx. 37.5 HP
> BUT, if we change the feed to .020 per revolution then the C >correction factor is about .8 and the HP would be:
>HP = UHP (.25) X 125 X .8 - Or approx. 27.9 HP
> Now I didn't go back and change the parameters to allow for the same >MRR at the different feed rates, but anyone could if they wished. It >shouldn't change the results.
> As regards the question of which takes more HP to remove a similar >amount of stock, a lathe or a mill - the conclusion that I draw is - IT >DEPENDS.
You probably have some math model problems too. Or story problem ones <G>.
> It seems at lower feed rates the lathe uses more HP, and conversely at >higher feed rates is uses less.
<Sheesh> Think of the stock/tool/chip shear zone. -- Cliff
>>>>>>>> On Fri, 02 Oct 2009 22:27:30 GMT, D Murphy >>>>>>>> <spamto...@comcast.net> wrote:
>>>>>>>>>Forces in turning are a lot higher than with milling.
>>>>>>>> Have some foorp of this?
>>>>>>>Yup.
>>>>>> Nope.
>>>>>Take that math class we've talked about and I'll explain it to you.
>>>> IOW You do not.
>>>You wouldn't want to bet on that.
>>>> That's what I thought.
>>>Feh. You haven't had a lucid thought in years.
>>>Why do I need to provide proof to you?
>>>If you think you know something different, then post it. Feel free to >>>show us how smart you are and post some numbers that show the forces >>>are the same under similar conditions.
>>>It's simple math.
>> Then show it <G>.
>Nope. I'll go through a bunch of effort, then you can manipulate the >numbers to some unrealistic feed rate or DOC just to keep your ignorant >troll alive.
You can read BB's "calcuations", right?
>How about you show us a multifunction lathe that uses V-flange tooling >like CAT40. Surely if turning forces were lower than milling then a >widely available, less expensive, tooling system would be preferable, >no? You wrote: "Forces in turning are a lot higher than with milling." >> BTW, Tell BB that his HP to cubic inches per minute calcs >> are dead wrong.
>I don't know what you're talking about. But if I had to guess you >probably didn't understand what he was talking about.
How do cutting (vector) forces relate to HP consumed? -- Cliff
>>> >>>>>> On Fri, 02 Oct 2009 22:27:30 GMT, D Murphy >>> >>>>>> <spamto...@comcast.net> wrote:
>>> >>>>>>>Forces in turning are a lot higher than with milling.
>>> >>>>>> Have some foorp of this?
>>> >>>>>Yup.
>>> >>>> Nope.
>>> >>>Take that math class we've talked about and I'll explain it to >>> >>>you.
>>> >> IOW You do not.
>>> >You wouldn't want to bet on that.
>>> >> That's what I thought.
>>> >Feh. You haven't had a lucid thought in years.
>>> >Why do I need to provide proof to you?
>>> >If you think you know something different, then post it. Feel free >>> >to >> show >>> >us how smart you are and post some numbers that show the forces are >>> >the same under similar conditions.
>>> >It's simple math.
>>> Then show it <G>.
>>> BTW, Tell BB that his HP to cubic inches per minute calcs >>> are dead wrong. >>> -- >>> Cliff
>> Why should he? He's out making a living programming machines & helping >> others maximize productivity & profit, while you spend your days >> trolling usenet. Seems you should have plenty of time to "show it". >> and then you've got the gall to attempt to invoke his assistance to >> get in to your BB battle? >> bait this, troll.
>I was a featured speaker at a conference on multifunction machining this >past week.
Does it need more force?
>The published Dr. who spoke before me, as well as the reps >from Kennametal and Sandvik who spoke after I did, didn't seem to have a >problem with the differences between the forces in turning and milling. >Neither did the application engineer for Okuma. Nor did the 40+ >attendees. But k00ky Cliff does. Go figure.
Perhaps they just needed dumb looks.
>Hey Cliff, if you would like to publish an article explaining cutting >forces in turning and milling, I'll hook you up. Especially if you'd >like to explain why V-flange tooling is the best choice for turning.
>> On Mon, 02 Nov 2009 15:42:22 -0800, BottleBob <bottl...@earthlink.net> >> wrote:
>>> Second: Where did this "HP to cu. in. per minute" thing come >>> from? I >>>haven't posted about HP in quite some time.
>> "Forces in turning are a lot higher than with milling." - Dan M. >> This blanket claim implies that more HP is needed per unit volume >> of material sheared off the stock per minute.
>Just as your blanket claim seems to imply that the forces acting on the >tool and work holding have anything to do with horsepower.
Where did I say anything about any of that?
>If that's the case, perhaps you can tell us how many pounds of horsepower >it would take to pull your head out of your ass?
I see that you have unit confusion.
>I never made any claims about metal removal over time. Only about the >forces acting on the V-flange tool holder.
You very clearly wrote: "Forces in turning are a lot higher than with milling."
>Which have nothing to do with >spindle horsepower.
"Forces in turning are a lot higher than with milling." - Dan M.
> I can rip a v-flange holder right out of the spindle >with the spindle stopped.
So what? And why do you do it?
>Say by running a program written by you for >example, so what exactly does horsepower have to do with it?
Cliff wrote: > Think of the stock/tool/chip shear zone.
Cliff:
I'm sorry, my newsreader must have lost the post where you outlined your method for calculating the HP of a milling or turning cut. Could you please repost it?
On Sun, 08 Nov 2009 07:40:53 -0800, BottleBob <bottl...@earthlink.net> wrote:
>Cliff wrote:
>> Think of the stock/tool/chip shear zone.
>Cliff:
> I'm sorry, my newsreader must have lost the post where you outlined >your method for calculating the HP of a milling or turning cut. > Could you please repost it?
Think of the stock/tool/chip shear zone. And ask the right questions of Dan. -- Cliff
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