View Full Version : SFM calculating

Marty J. Roberts

04-18-2012, 11:46 PM

Fellow machinists: my students and I are doing an experiment we are going to find as many different formulas for calculating spindle speed as we can then graph the results...we want to test to see which is the closest and to see how many there are

Currently I use RPM = SFM x 4 ÷ diameter...

What do you use?

Thanks!!!

beanbag

04-19-2012, 01:10 AM

what? :confused:

Oldbrock

04-19-2012, 01:42 AM

I use that too, have for 65 years. Peter

Paul Alciatore

04-19-2012, 03:20 AM

Well, the actual formula for translating between RPM and SFM with the diameter (d) being measured in inches is:

RPM = (12 x SFM) / (3.141 x d)

This simplifies to:

RPM = 3.820 x SFM / d

The 4 in the often quoted equation as you have stated is actually an approximation of 12 / 3.141 which actually equals 3.820. The error is only 4.72% so it is good enough for most shop purposes and a lot easier to remember and use. I doubt that most motors used on our machines are that accurate.

Jaakko Fagerlund

04-19-2012, 03:54 AM

You either get one formula (the exact right one) or multiple approximations, which usually are close enough for anything.

..so, what is the point of this?

wbleeker

04-19-2012, 03:55 AM

That is what I was taught to use CS x 4/ D where CS is the cutting speed in Feet Per Minute and D is the Diameter of the Workpiece or Cutter, whichever applies.

Will

philbur

04-19-2012, 04:38 AM

Some of you grumpy old farts have forgotten what it was all about when you were a student. The point for a student is what you learn on the journey, not what you see when you get there.

It seems like a practical exercise in understanding the relationship between rpm and surface speed, or maybe even more fundamental, the relationship between diameter and circumference.

Phil:)

Blackadder

04-19-2012, 06:05 AM

RPM = sfm( M/s ) x 320 / cutter dia in mm

note this is for metric cutters

MichaelP

04-19-2012, 09:23 AM

Marty,

Did you explain to your students the meaning (or geometric meaning with some arithmetics, if you will) of the formula you gave them? May be then the absurdity of the project will become more apparent.

I'm sure you realize that "4" in your formula is really 12 (as in inches per foot) divided by pi, don't you?

Marty J. Roberts

04-19-2012, 09:26 AM

Some of you grumpy old farts have forgotten what it was all about when you were a student. The point for a student is what you learn on the journey, not what you see when you get there.

It seems like a practical exercise in understanding the relationship between rpm and surface speed, or maybe even more fundamental, the relationship between diameter and circumference.

Phil:)

You nailed Phil! Students get frustrated when there is more than one way to skin a cat!

I use my sight and sound to judge sfm.

I use my sight and sound to judge sfm.

I used to do that too until I discovered my tool wear rate was way out of line. Having no talent for that, I bought a laser tachometer :)

Marty's formula is the one I use, too, but pretty much was self-discovered after a few sessions with the calculator.

blackadder, its 318, right? (i mean, 300 is easier.)

platypus2020

04-19-2012, 01:44 PM

http://www.monstertool.com/monster_tool_calculators.html

Blackadder

04-19-2012, 01:53 PM

blackadder, its 318, right? (i mean, 300 is easier.)

its 320 as in my post I know I am dyslectic but you made me check my post

eg 320 x 24 /12

give you 640 rpm for a 12 mm cutter in steel ( en1a )

but in practice I use a machinist pro calculator ;-) expensive but they do so much more

Stuart

Mcgyver

04-19-2012, 01:56 PM

You nailed Phil! Students get frustrated when there is more than one way to skin a cat!

I don't really get this....so they don't like more than one way to skin a cat but you've got them trying to find others.....to intentionally frustrate or make them realize there's really only formula, the others are just restating it?

sounds more like algebra class, I'd just tell them the formula and say that's it, it's what you use

gundog

04-20-2012, 12:02 AM

I use this and plug in the numbers it makes it quick to play with different chip loads RPM and feeds and it is free.

Mike

http://www.whitney-tool.com/calculatorSpeedFeed.html

Marty J. Roberts

04-20-2012, 12:38 AM

[QUOTE=Mcgyver]I don't really get this....so they don't like more than one way to skin a cat but you've got them trying to find others.....to intentionally frustrate or make them realize there's really only formula, the others are just restating it?

sounds more like algebra class, I'd just tell them the formula and say that's it, it's what you use[/QUOTE

Through experiance and time you get the formulas, I get the formulas, we get the formulas, but when students are new to it all! Exposing them to the proper formula and the variations is a little exercise that in some way prepairs them for who ever they work for and it helps them with their critical thinking and reasoning skills which many of our young people lack. Also the students must understand the range properties when it comes to speeds they want to do the calculation and plug in the number, and as you know it doesn't always work that way. The students get frustrated, that makes them listen, it brings out their inner machinist!

I don't get this. There is only one way to calculate the circumference of a circle given the radius or diameter and it is only approximate. It cannot be done precisely because pi is an irrational number. Unfortunately there are infinitely more irrational numbers than there are rational numbers. Welcome to the concept of classes of infinity. :D

MichaelP

04-20-2012, 02:20 AM

If they UNDERSTAND were this formula was derived from, you may just let them rewrite it for different variables, units or combination of those. Let's say, how the formula will look if we know radius instead of diameter? How will it look if we know surface speed in SFM and the diameter in feet? How will it look if we know SFM and diameter is in millimeters? What if we know circumference instead of diameter? What if the circumference is given in feet and we know SFM? How the formula will look if we know surface speed in meters per minute, and the diameter is in inches? Etc.

The formula doesn't change regardless of the units of measurement used.

The formula doesn't change regardless of the units of measurement used.

I had a line in my previous post to that effect and yanked it because I didn't want to get into a 20-page yak fest about the certainty of it :)

The published speeds and feeds themselves are not more than good guesses taken from commercial machine operations a very long time ago and using the best analytics of the time. Time has since moved on and the home shop/hobbyist machinist, never a target of this information, really needs to rediscover the best values for the equipment they own and operate including the acceptable range of tolerance this value has. I have a non-rigid home shop bench top mill that is in no way capable of cutting according to the standard tables. I've created my own charts for speed and feeds. I used the published values to start with and adjusted according to the capacity of my machine. It works pretty well with my charts.

MichaelP

04-20-2012, 03:11 AM

I remember a very embarassing situation when I was taking my first machining classes at a local college. Our teacher, a very nice young guy and a skillful machinist, was constantly bragging about his recently received degree in ME.

One day he's lecturing us on Feeds and Speeds, and gives us the formula that looks like this:

Speed of rotation=surface speed x 12 / pi x diameter

No mentioning of units of measure (which is absolutely unacceptable, of course).

He says that we need to know the diameter and can find surface speeds for different materials in the books.

Then he's asked what units the surface speeds are given in in the books. "No units, they're just dimensionless values"

So much for understanding of the subject and the quality if his ME degree. :)But he was a really skillful practical machinist who taught us a lot. He just shouldn't really give us any theory. Reading a book would be less harmful and more productive.

MichaelP

04-20-2012, 03:23 AM

The formula doesn't change regardless of the units of measurement used.

Really? So it'll look the same ( RPM = SFM x 4 ÷ diameter) if the diameter is given in feet? :D

philbur

04-20-2012, 07:29 AM

That should slow the grumpy old farts up a bit.

Who’s going to be the first to argue with that, go on I dare you?:eek:

There's a couple here who might actually benefit from attending your class rather than criticising it.:D

Phil:)

Really? So it'll look the same ( RPM = SFM x 4 ÷ diameter) if the diameter is given in feet? :D

Mcgyver

04-20-2012, 08:23 AM

Exposing them to the proper formula and the variations is a little exercise that in some way prepairs them for who ever they work for and it helps them with their critical thinking and reasoning skills which many of our young people lack. Also the students must understand the range properties when it comes to speeds they want to do the calculation and plug in the number, and as you know it doesn't always work that way. The students get frustrated, that makes them listen, it brings out their inner machinist!

I like it. Trying to remember my own days in grade 11 machine shop my reaction was thinking it's too big a hill to climb. Then again to a large to degree you get what you expect from people and we were expected, by and large, to be louts and do things by rote. It was more an excercise on how well you followed instructions than on building an understanding...but its the understanding that matters

For my students this would involve a whole class period (or two.)

First an explanation of why there are different cutting speeds for different materials, tools, and operations.

Second an explanation of why we use SFM (or SMM) as the base number for a given material, operation, and tooling... instead of some other number.

Third a demonstration of using the formula for different materials, operations, and tooling. Making sure they understand the usefulness of it by demonstrating what happens when one gets it right and when one gets it wrong. Since we would be moving around the shop I would choose tool and/or work diameters that are easy to calculate in ones head. (4", 2", 1", 1/2", 1/4".)

Fourth as a homework assignment due in one week they would have to make a chart with diameters on the vertical axis, RPM on the horizontal, and the cells filled in with the SFM. Similar to this one:

http://fs.bookpawnshop.com/M/ad-Metal-lathe-wall-chart-myford,-unimat,-taig-and-others/

Every single time the class meets there will be at least one quiz or problem where they have to use the calculation. They will know it is expected that they know the formula and how to use it. By the end of the class, and hopefully sooner, it will become second nature to them.

-DU-

Really? So it'll look the same ( RPM = SFM x 4 ÷ diameter) if the diameter is given in feet?

Sure. Or light years or cubits. The units used are arbitrary. Changing units is simply a matter of scaling.

philbur

04-20-2012, 02:37 PM

Priceless. :rolleyes:

The units are not arbitrary. If you don't use inches for the diameter you get the wrong answer. If you want to use feet then you either convert it to inches before you insert it in the formula or you change the formula, by substituting the value 4 with 0.3333.

So to sum up, you either use the correct units or you change the formula.

Now for tonights' homework I want you to calculate what the constant value (4) would be changed to if you use the following measures for the diameter of the work-piece

a) Light-years.

b) Cubits.

Those of you that don't have micrometers that are calibrated in light-years see me.

Phil:)

Sure. Or light years or cubits. The units used are arbitrary. Changing units is simply a matter of scaling.

MichaelP

04-20-2012, 02:42 PM

Sure. Or light years or cubits. The units used are arbitrary. Changing units is simply a matter of scaling.

Evan, the 1st of April was 20 days ago. You're joking, right? :)

P.S. A hint. SFM means "Surface Feet Per Minute".

Paul Alciatore

04-20-2012, 04:00 PM

The formula doesn't change regardless of the units of measurement used.

This is both true and untrue, depending on how you are looking at the problem. In general the formula for calculating the rotational RATE given the desired surface SPEED is

RotationalRate = SurfaceSpeed / Circumference

This is a completely generic formula which is not constrained by any units or constants. It works for English, metric and Little Green Men's units; any units at all. But this is too general for day to day use so we start to make it more specific and more convenient.

First, it is a lot easier to measure a diameter instead of a circumference so we introduce the constant Pi or 3.141..... and use the diameter instead of the circumference.

RotationalRate = SurfaceSpeed / (Pi x Diameter)

Still, no mention of any units so it is still completely general.

OK so lets do it for the units he stated, Revolutions per Minute and Surface Feet per Minute. First notice that both of these use the minute for time measurement. This is good as no translation will be needed for different time units. The unit for the diameter was not explicitly stated, but implied as inches since "Feet" per Minute was stated. It would make little sense to use metric measure in this situation. Also, diameters are generally measured in inches and decimal inches in machine shops which use SFM for cutting speeds so we would want the diameter to be entered in inches, not feet or some other unit. Problem is that the units MUST be consistent in the equation so if we are going to enter the diameter in inches, we will have to do a conversion to mingle that with a SFM number. The conversion from feet to inches is a factor of 12 or if going in the other direction 1/12. So our diameter will be expressed as (inches x 1/12) or (inches / 12). I will use "diameter" with a small "d" for the diameter in inches. We substitute "Diameter" (in feet) in the above equation with (diameter / 12), "RPM" for "RotationalRate", and "SFM" for "SurfaceSpeed" to get the following:

RPM = SFM / (Pi x (diameter / 12))

This simplifies to:

RPM = SFM / ((Pi x diameter) / 12)

And again to:

RPM = (12 / Pi) x (SFM / diameter)

Now 12 / Pi = 3.820 as I said in my previous post.

So the formula, exact to three decimal places is:

RPM = 3.820 x SFM / diameter

with the units being Revolutions per Minute, Surface Feet per Minute, and Inches of diameter.

Again, as I said in my first post, this is not as convenient as it could be so the constant 3.820 has been rounded to 4 which is a lot easier to remember and a lot easier for quick calculations.

With this rounding we finally have the commonly used formula:

RPM = 4 x SFM / diameter

That is the complete derivation of the common formula. So the curious students can see where it comes from and how it was derived and why. They or you can take the process at any step and go in a different direction. For instance, you could start with my first equation and in a similar manner, derive an equation for metric units. Of if you have tables in SFM and need to use metric diameters you could derive an equation that used these. If you simply want a plug and go equation the last one is your choice. If you want a completely general one then the first. Choices are good.

philbur

04-20-2012, 04:21 PM

This would be true if it were not for the fact that THE formula being discussed is:

RPM = SFM x 4 ÷ diameter

Not:

RotationalRate = SurfaceSpeed / Circumference

Phil:)

This is both true and untrue

Black Forest

04-20-2012, 05:10 PM

"Diplomacy is a skill in handling affairs without arousing hostility" (c) :)

P.S. It reminded me an old joke.

"May I ask a question, Sir?"

"Go ahead, private, shoot!"

"Sir, do african crocodiles fly?"

Drill sergeant, after some thinking: "Sure they fly, but very low".

Diplomacy and my name have never been used in the same sentence!

MichaelP

04-20-2012, 05:15 PM

This is both true and untrue

"Diplomacy is a skill in handling affairs without arousing hostility" (c) :)

EDIT: BF, sorry, I deleted my previous message while you were typing.

The units are not arbitrary. If you don't use inches for the diameter you get the wrong answer.

They are arbitrary. SFM is merely a name for a variable in the formula. Insert any other measure you like and the variable's meaning in the formula does not change, only the scaling. Like many other formulae, the units must be consistent in the formula. That doesn't change the formula.

This would be true if it were not for the fact that THE formula being discussed is:

RPM = SFM x 4 ÷ diameter

The formula is x=y x 4 ÷ z

The units used to replace the variables may be anything as long as they are consistent. It isn't the type of units that make a formula, it is the algorithm.

From Wolfram Alpha: Formula: A group of symbols that make a mathematical statement.

http://www.wolframalpha.com/input/?i=formula&x=0&y=0

philbur

04-20-2012, 08:36 PM

The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree? A yes or no answer will suffice.

Phil:)

The units are not relevant to the formula, only each other. I can use light years or microns in the same formula as long as they are consistent with the other variables. The formula doesn't change.

MichaelP

04-20-2012, 09:42 PM

The formula is x=y x 4 ÷ z

The units used to replace the variables may be anything as long as they are consistent. It isn't the type of units that make a formula, it is the algorithm.

;

And what are we going to do with "4" in this formula? What does it represent? :)

Mike Burdick

04-20-2012, 09:49 PM

... and next evan will educate us about what happens to the hole when the cheese is gone!

.

The formula is x=y x 4 ÷ z

See what I mean?

beanbag

04-21-2012, 01:14 AM

I agree with Phil and Paul.

Rotations per time = (Surface distance per time) / (3.14 x Diameter)

is the "true" forumla that allows you to use any units you want, as long as they are consistent.

In order for the "formula" to have a 4 in it, the surface speed units must be 12x larger than the diameter units, even if you choose arbitrary units of lightyears.

This whole formula, the derivation, as well as the concept of surface speed, is explained by the instructors at TechShop to new users of the mill and lathe in about 10 minutes.

I use radians per second.

beanbag

04-21-2012, 03:34 AM

I use radians per second.

So do I, but never around normal people.

philbur

04-21-2012, 04:12 AM

Yes or no?

Phil:)

The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree? A yes or no answer will suffice.

Phil:)

You pose the False Dilemma fallacy.

Definition: False Dilemma

A limited number of options (usually two) is given, while in reality there are more options. A false dilemma is an illegitimate use of the "or" operator.

http://onegoodmove.org/fallacy/fd.htm

Time to study logic.

Any units may be used in the formula if they are consistent. You confuse the identity of the variables with the identity of the operators. It is the operators that define the formula, not the variables. The value "4" is a scaling constant and may be replaced with any other required value.

MichaelP

04-21-2012, 01:07 PM

You pose the False Dilemma fallacy.

http://onegoodmove.org/fallacy/fd.htm

Time to study logic.

Any units may be used in the formula if they are consistent. You confuse the identity of the variables with the identity of the operators. It is the operators that define the formula, not the variables. The value "4" is a scaling constant and may be replaced with any other required value.

In this case you just invented a universal formula. You just need to substitute proper units and functions instead of the variables and coefficients given in this formula, and you will be able to use it instead of any formula known to mankind.

Evan, trust me, it's much easier and classier to accept that you were wrong and didn't understand something. Just try it once.

Evan, trust me, it's much easier and classier to accept that you were wrong and didn't understand something. Just try it once.

I have been wrong many times and have said so on this forum. This isn't one of those cases. Your failure to understand is not my failure.

You just need to substitute proper units and functions...

I said nothing about changing "functions" (operators). That is your failure.

MichaelP

04-21-2012, 01:15 PM

Of course. We're all just too dumb to notice an elephant in the room. :o

P.S. First, you decided to mangle the discussed formula just to prove your point and converted a definitive term "Surface Feet per Minute" into a generic variable (just because you feel like this). Unfortunately, you overlooked the need to substitute "4" in your own, supposedly generic, formula. Not because you didn't realize where it came from, God forbid!

Then the time comes to explain what "4" represents in your generic formula that, by definition, should work for any units of measure. Well, now it's not really "4", but can be "794" or ".0000267" depending on your mood and needs. It is a variable, really. You just called it "4" because you didn't have enough Latin or Greek letters left and using 1/pi would be too sophisticated for us.

Now you have a formula that looks similar to E=829*M*C^2 where 829 is really called a "scaling constant". It sounds scientifically enough and works for all masses, but especially well for those that are measured in 1/829 of the unit that is consistent with the rest of the units used for calculation. For those who don't understand why "829" is needed there in the first place, you'll have a few links to explain what logic is, how time continuum is defined and when baking soda should be used unstead of yeast.

Finally, I don't see why you don't want to go further and substitute any of your variables with functions. It's perfectly acceptable, isn't it? It'll give you a valid opportunity to claim that you've just invented a universal formula. Do you see any defect in my logic?

Irrelevant to the matter at hand.

I suggest you study what constitutes a formula.

This is the best resource on the Web.

http://mathworld.wolfram.com/

MichaelP

04-21-2012, 01:32 PM

Evan, I managed to graduate from a middle school that gave me enough knowledge of elementary algebra to participate in this thread.

But I also studied psychology, and that's what this conversation is really about.

philbur

04-21-2012, 01:56 PM

It's a simple, straight forward question.

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree? yes or no"

There is no False Dilemma and your supposed logic does nothing to address the question but rather tries to redefine it, presumably in order to avoid answering it, while at the same time hoping that nobody had noticed.

So what's it to be - yes, or no? You know (everybody knows) what the answer is, so why not get it over with, then we can move on?

Phil:)

You pose the False Dilemma fallacy.

http://onegoodmove.org/fallacy/fd.htm

Time to study logic.

Any units may be used in the formula if they are consistent. You confuse the identity of the variables with the identity of the operators. It is the operators that define the formula, not the variables. The value "4" is a scaling constant and may be replaced with any other required value.

The full formula is:

Revolutions per minute (RPM) equal surface feet per minute (SFM) divided by the circumference (feet)

Expressed as a formula:

RPM = SFM / circumference (feet)

Solving for circumference (feet) when diameter is known:

Circumference (feet) = pi * diameter (feet)

Therefore

RPM = SFM / pi * diameter (feet)

Changing units from feet to meters:

Revolutions per minute (RPM) equal surface meters per minute (SMM) divided by the circumference (meters)

RPM = SMM / pi * diameter (meters)

Simple math is used to scale diameters between inches and feet, or meters and milimeters.

The formula when expressed in the general case (no units declared) is:

Revolutions per unit of time = surface length (units) per unit of time divided by the circumference (units)

philbur

04-21-2012, 02:39 PM

All very true dp, I don't think anybody would disagree with you, but it is not an answer to the question as asked. Maybe you would like to answer the question:

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree? yes or no"

Phil:)

All very true dp, I don't think anybody would disagree with you, but it is not an answer to the question as asked. Maybe you would like to answer the question:

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree? yes or no"

Phil:)

The answer to your question is specified in the formula you have given. Because you have specified SFM and the constant 4, the units for diameter must be inches. The constant "4" is derived from 1 / pi * 12 = 3.8197 which is appx = 4.

Change the units and constants in the formula and you can solve for any units of length and time.

philbur

04-21-2012, 02:56 PM

So that's a yes then.

Phil:)

The answer to your question is specified in the formula you have given. Because you have specified SFM and the constant 4, the units for diameter must be inches. The constant "4" is derived from 1 / pi * 12 = 3.8197 which is appx = 4.

Change the units and constants in the formula and you can solve for any units of length and time.

This would be true if it were not for the fact that THE formula being discussed is:

RPM = SFM x 4 ÷ diameter

Not:

RotationalRate = SurfaceSpeed / Circumference

Phil:)

You have truly misunderstood the difference between the formula (n = ratio of two variables) and an implementation of the formula (rpm = ratio of instances of two variables using compatible units of length and time). Evan and I are talking about the underlying formula hence the silliness regarding a specific application of it.

Edit: Having thought on it a bit more I have to recant the "misunderstood" part. I think you're just playing "got your nose" to make a point using word games.

MichaelP

04-21-2012, 04:26 PM

Evan and I are talking about the underlying formula hence the silliness regarding a specific application of it.

It's funny when you try to prove something and then your opponent flips over and uses your own arguments to prove exactly what you tried to explain to him.

The whole argument is about using a derivative formula ESTABLISHED SPECIFICALLY FOR CERTAIN UNITS as a universal one.

This derivative formula that we discussed (RPM= 4*SFM/Dia.) is good ONLY for diameters expressed in inches. Otherwise, you'll need to change how formula looks. For example, if you prefer expressing diameter in feet, the formula will look like this:

RPM=SFM/Pi*dia or, if you want a user friendly version, RPM=SFM/3*Dia. You may also write it as PRM=0.318*SFM/Dia.

The underlying generic formula (RotationalSpeed=Surface Speed/Circumference or RotationalSpeed=Surface Speed/Pi*Dia) can be used with any units. As long as you use consistent units, there is no need to change how this formula looks.

What's here hard to understand?

philbur

04-21-2012, 05:13 PM

I've misunderstood nothing. I simply asked a straightforward question to which both of you seem to have great difficulty give a straightforward answer:

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree?"

It's you guys playing the word games, you dance all around the houses seemingly in a desperate effort to avoid giving the answer that you know is correct but that you don’t want to give.

Phil:)

You have truly misunderstood the difference between the formula (n = ratio of two variables) and an implementation of the formula (rpm = ratio of instances of two variables using compatible units of length and time). Evan and I are talking about the underlying formula hence the silliness regarding a specific application of it.

Edit: Having thought on it a bit more I have to recant the "misunderstood" part. I think you're just playing "got your nose" to make a point using word games.

This derivative formula that we discussed (RPM= 4*SFM/Dia.) is good ONLY for diameters expressed in inches.

You think so? Convert all imperial units to centimetres and see what happens.

500 sfm is the same as 15240 centimetres per minute and 1 foot is 30.48 centimetres.

4*500/1 = 2000 rpm

4*15240/30.48 = 2000 rpm

philbur

04-21-2012, 08:10 PM

But Evan the correct rpm for 500 fpm on a 1 ft diameter is 167 rpm not 2,000 rpm. The constant 4 contains the conversion of feet to inches so you will only get the correct answer if you use inches for diameter with the surface speed expressed in feet per minute. Of course if you use consistent but wrong units you will get equal answers, but always the wrong answers.

How diffcult can a person make this?

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree?"

Phil:)

You think so? Convert all imperial units to centimetres and see what happens.

500 sfm is the same as 15240 centimetres per minute and 1 foot is 30.48 centimetres.

4*500/1 = 2000 rpm

4*15240/30.48 = 2000 rpm

The underlying generic formula (RotationalSpeed=Surface Speed/Circumference or RotationalSpeed=Surface Speed/Pi*Dia) can be used with any units. As long as you use consistent units, there is no need to change how this formula looks.

What's here hard to understand?

I don't get it, either, but knew it was coming. A reason for slipping in approximations is to allow one to quickly calculate the needed RPM for a job using only inches and your fingers and toes :)

e=mc^2 is a well known formula.

e=10c^2 is an application of that formula.

These are not equivalent things. One is specific, one is general. I, and Evan, have addressed only the fundamental formula and others have tried to make the claim that the application of the formula was the subject of discussion. The would appear to believe themselves the only intelligent people here on this subject.

Nothing left to see here.

David Powell

04-21-2012, 09:10 PM

Is how i would describe most replies on here. Lets try to have a civilised discussion. Theres only one speed that matters to most home shop machinists, that is the correct one for the job and the set up. The theoretical one may well be fine for a rigid set up in a strong machine, try that on a worn out mill drill or lathe and see where you get!!I am happy to pass on threads like this. Regards David Powell.

MichaelP

04-22-2012, 02:37 AM

You think so? Convert all imperial units to centimetres and see what happens.

500 sfm is the same as 15240 centimetres per minute and 1 foot is 30.48 centimetres.

4*500/1 = 2000 rpm

4*15240/30.48 = 2000 rpm

OK, Evan. Now everything has become clear. Knowing you as a pretty smart guy, I had no doubts that you realized your had made a mistake. I thought that you were just trying to hide this fact to appear to be right.

This post of yours shows that you simply do not understand how this formula works and why it cannot be used with any units of your choice but inches for the diameter. We gave you multiple hints and explanations, Paul spent considerable time typing his tutorial, but nothing worked. Maybe it's because you didn't want to listen and made no attempt to understand us. Or maybe because you simply cannot grasp it.

Anyway, contrary to our advice, you chose to use feet instead of inches in this formula. So, instead of 4*500/12 = 167, you wrote it as 4*500/1. As a result, you got RPM 12 times higher than it should be.

But that's perfectly fine. Really! Anybody can make a mistake and have difficulties understanding even a simple concept or formula.

Phil just explained to you again why it happened. I hope you'll think it over, accept the fact that you made a mistake, and we'll finish this unpleasant conversation on a good note.

But Evan the correct rpm for 500 fpm on a 1 ft diameter is 167 rpm not 2,000 rpm.

The formula as given is mathematically incorrect.

This is correct.

RPM = (Surface Feet per Minute) x 4 ÷ ( (diameter in feet)*12 )

The difference is that the so called "formula" contains an assumed operation. It must in order to produce a correct result. The correct formula is as I have posted above. To be a correct mathematical formula it must account for differences in units.

I do admit that I didn't realize the formula as given was incorrect. I don't use it.

rpm = 15240*4 / (2.54*12)

Using the correct formula in centimetres it comes out to the same as imperial, 166.666...

Incidentally, I actually do use radians per second (RPS). It's much easier. It's also how rotating tools such as milling cutters are designed. All you need to do is give the cutter a spec in RPS and the rpm required is approximately RPS x 10. If you need accuracy then RPS x 9.55.

philbur

04-22-2012, 04:47 AM

Priceless, but still the question remains unanswered:

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree?"

Perhaps you noticed that (diameter in feet)*12 ) is actually the diameter in inches.

Phil:)

The formula as given is mathematically incorrect.

This is correct.

RPM = (Surface Feet per Minute) x 4 ÷ ( (diameter in feet)*12 )

The difference is that the so called "formula" contains an assumed operation. It must in order to produce a correct result. The correct formula is as I have posted above. To be a correct mathematical formula it must account for differences in units.

I do admit that I didn't realize the formula as given was incorrect. I don't use it.

rpm = 15240*4 / (2.54*12)

Using the correct formula in centimetres it comes out to the same as imperial, 166.666...

Incidentally, I actually do use radians per second (RPS). It's much easier. It's also how rotating tools such as milling cutters are designed. All you need to do is give the cutter a spec in RPS and the rpm required is approximately RPS x 10. If you need accuracy then RPS x 9.55.

Boostinjdm

04-22-2012, 05:23 AM

I'll bite philbur.

The answer is yes, you have to use inches because the the formula uses feet (not meters or anything metric) and the conversion portion from feet to inches. You can't mix and match measuring systems. Is that right?

The real reason though is that my equipment isn't big enough to hold diameters measured in feet. Plus all my calipers measure inches.

philbur

04-22-2012, 05:54 AM

Not difficult is it.

You can actually mix and match units, providing the formula contains the necessary conversion factor, which in this case it is 12/3 .... 4. The 12 represents the conversion from feet to inches and the 3 is an approximation for Pi.

Phil:)

I'll bite philbur.

The answer is yes, you have to use inches because the the formula uses feet (not meters or anything metric) and the conversion portion from feet to inches. You can't mix and match measuring systems. Is that right?

The real reason though is that my equipment isn't big enough to hold diameters measured in feet. Plus all my calipers measure inches.

You can actually mix and match units, providing the formula contains the necessary conversion factor, which in this case it is 12/3 .... 4. The 12 represents the conversion from feet to inches and the 3 is an approximation for Pi.

As originally written it is an equality which is not correct. It is missing the unit conversion. It simply did not occur to me that it would even be called a "formula", properly called an equality if it wasn't correct. Your explanation above is also incorrect as the conversion to inches is assumed and not included in the equation which means it is not an equation.

erhaps you noticed that (diameter in feet)*12 ) is actually the diameter in inches.

Perhaps you noticed that the "12" is only a unit scaling factor and works just fine in metric too.

philbur

04-22-2012, 12:08 PM

And still the question remains unanswered:

"The only units that can be used for diameter in the formula:

RPM = SFM x 4 ÷ diameter

are inches.

Do you agree?" Yes or No.

Phil:)

MichaelP

04-22-2012, 12:11 PM

Evan, you really should take some elementary algebra classes before engaging in further discussion. It's obvious that math is not one of your strong points.

You see, in some cases you cannot find a ready answer on the Internet or in books staying on the shelf, and you have to really show what you know or understand. In those cases it's very difficult to appear as "know it all", and ignorance becomes obvious.

Please stop making a fool of yourself, because it really looks pathetic. Don't do it to yourself.

EDIT: You may want to read my post #81 below. If you read it carefully, maybe you'll, finally, understand where "4" came from, and what this long talk was about. I love your wonderful creation

RPM = (Surface Feet per Minute) x 4 ÷ ( (diameter in feet)*12 ) ,

but it really looks like this

RPM=SFM/Pi*dia

if you know how formulae are written. If you wish, you may approximate it and show like this:

RPM=SFM/3*Dia

i maintain my position: 1000/p = 318!

MichaelP

04-22-2012, 12:32 PM

I'll bite philbur.

The answer is yes, you have to use inches because the the formula uses feet (not meters or anything metric) and the conversion portion from feet to inches. You can't mix and match measuring systems. Is that right?

The real reason though is that my equipment isn't big enough to hold diameters measured in feet. Plus all my calipers measure inches.

Absolutely. This formula was custom made for diameters written in inches.

Originally, it looked like this:

RPM=SFM/ Pi * Dia (in ft).

However, since the majority of machinists use inches for diameters, the formula was customized and became

RPM=12* SFM/ Pi * Dia.(in in.)

Since 12/Pi=4 (roughly), you can write this formula as

RPM=4*SFM/Dia (in inches)

Absolutely. This formula was custom made for diameters written in inches.

Since 12/Pi=4 (roughly), you can write this formula as

RPM=4*SFM/Dia (in inches)

And this handy customization is exactly why the OP posted it here and in the hopes of seeing more similar customizations. Then there came the thread hijacking when Evan's factual and on-topic statement that the method is units agnostic.

If ever a thread deserved a reset and do-overs to get back on topic, and it is a good one, this one is a candidate.

MichaelP

04-22-2012, 01:46 PM

Then there came the thread hijacking when Evan's factual and on-topic statement that the method is units agnostic.

DP, in case you forgot how it started, please re-read posts #22, 23, 28 and 33. This will tell you exactly what Evan said and meant. And he continues demonstrating that he still doesn't understand how this formula works. (But, of course, the problem is not in his ignorance, but rather in the wrong formula, wrong terminology, general faults of mathematics, etc. :) )

This "hijacking" was essential to the OP topic since understanding of this formula makes his project meaningless. It doesn't matter how you rewrite the formula: the result will still be the same, and it makes no sense to do graphs and compare the results. That's why, instead, I suggested asking his students to rewrite the formula for different variables and units, the task Evan failed miserably. And the reason is a lack of understanding of how the formula was born and why it looks like it does.

There should be a major difference between educating people and making them trained monkeys.

DP, in case you forgot how it started, please re-read posts #22, 23, 28 and 33. This will tell you exactly what Evan said and meant.

Evan is exactly right - the formula is units agnostic. The OP's use of the formula is a purpose-built and units-fixed solution. The OP was seeking other similar solutions.

And this "hijacking" was essential to the OP topic since understanding of this formula makes his project meaningless.

Nonsense. The world is full of handy number crunching tricks: http://www.hintsandthings.co.uk/office/shortcuts.htm

Barrington

04-22-2012, 02:16 PM

Let's try for a little rigour... the formula 'rpm = 4*SFM/diam' is NOT units agnostic by any stretch of the imagination.

Considering the general formula, aside from the issue of feet/inches/metres and seconds/minutes etc., (as has been pointed out) rotation can be expressed in rotations per unit time or radians per unit time ...

So - a completly GENERAL FORMULA using mixed units might be written:-

rotational speed = K * surface speed / workpiece diameter

where: K = (k1 * k2 * k3)

and:

k1 = (time unit of surface speed) / (time unit of rotational speed)

k2 = (length unit of surface speed) / (length unit of diameter)

k3 = (1/pi) if rotational speed is required in revolutions per unit time, else k3 = 2 if in radians per unit time.

Now for the case of revs/minute, feet/minute, and inches then, k1=1, k2=12, k3=1/pi, and the UNIT SPECIFIC FORMULA becomes:-

rpm = (12 /pi) * SFM / diameter

There is then a different UNIT SPECIFIC FORMULA for each set of units, each derived from the GENERAL FORMULA.

i maintain my position: 1000/p = 318!Yes, 318 is a better approximation than 320.

This "hijacking" was essential to the OP topic since understanding of this formula makes his project meaningless. It doesn't matter how you rewrite the formula: the result will still be the same, and it makes no sense to do graphs and compare the results.+1

Cheers

.

finally somebody said im right. thank you. i will have a drink to your health (or several).

MichaelP

04-22-2012, 02:25 PM

Evan is exactly right - the formula is units agnostic.

The formula is

RPM=4*SFM/Dia

and it is NOT "units agnostic". It must be used with inches for the diameter.

Even

RPM=SFM/Pi*Dia

is NOT "unit agnostic" because you introduced variables with definitive units there: RPM and SFM.

RotationalSpeed=Surface speed/Pi*Diameter

is "unit agnostic"(as long as the units are consistent).

P.S. Barrington, thank you. I was typing this message while you posted yours.

Let's try for a little rigour... the formula 'rpm = 4*SFM/diam' is NOT units agnostic by any stretch of the imagination.

The error continues. That is the solution for specific units of revolution, length, and time where the unknown is revolutions per unit of time where time is minutes, and the known is surface feet per minute and the diameter in inches. It result in a left-hand side that is specifed in revolutions per minute. It is an application of the underlying formula. Nobody is questioning the fixed nature of that as it is a solution and can be resolved to actual numbers and units. It is immutable. And not the subject of Evan's claim.

The formula behind the solution is agnostic to units (dimensionless) and can be expressed as an equation but cannot be resolved to a number without specifying units of length and time.

Example (again): E = mc^2. This cannot be resolved to a number. It is a formula where m and c are undefined in terms of units. The constant c can be given in metric or imperial units of length, and fortnights for time so long as m is expressed in a compatible unit.

From wikipedia for E = mc^2:

where E is energy, m is mass, and c is the speed of light in a vacuum. The formula is dimensionally consistent and does not depend on any specific system of measurement units.

This shouldn't be so hard.

MichaelP

04-22-2012, 03:18 PM

Apparently, it's hard for you.:o

Apparently, it's hard for you.:o

The underlying expression used in the OP's solution is:

angular velocity = surface velocity X circumference (Edit: inadvertantly had diameter here)

Show where this is dependent upon specific units.

This is the same formula that gives us the rotational speed of an iron wheel on a track, as in a train, among its many other uses.

The formula is

RPM=4*SFM/Dia

and it is NOT "units agnostic". It must be used with inches for the diameter.

That isn't a formula to provide the indicated answer. Without external knowledge it will not give a correct answer. It might be called a "rule of thumb" but without the units conversion it is useless as a formula.

The proper notation for that is RPM≠4*SFM/Dia

The correct formula is RPM=4*SFM/(Dia*12) This formula works with any units. The other doesn't work at all.

MichaelP

04-22-2012, 03:49 PM

The correct formula is RPM=4*SFM/(Dia*12)

You still don't get it.

Evan, please, please, please, answer the following questions:

1. Where number "4" came from, and why do you use in your formula?

2. Do you agree that your formula is really RPM=SFM/Pi* Dia. ?

3. Do you understand that since the formula ALREADY introduces rotation per minute and Feet per minute, your diameter shall be given in feet in this formula?

4. Do you realize that in order to make this formula unit-independent, you have to remove all pre-introduced units, so the generic formula will look like this

RotationalSpeed=SurfaceSpeed/Pi*Diameter

Please don't rush. Think it over and come back.

MichaelP

04-22-2012, 04:05 PM

The underlying expression used in the OP's solution is:

angular velocity = surface velocity X circumference (Edit: inadvertantly had diameter here)

Show where this is dependent upon specific units.

This is the same formula that gives us the rotational speed of an iron wheel on a track, as in a train, among its many other uses.

DP,

Do you actually read our responses? You're beating a dead horse. Please re-read my response (post#87) to you. Especially, the last two lines above the P.S.. Or you may read my post above, if you wish.

Let me repeat it one more time: the underlying formula (the one that you mention) is unit-independent. The same way, as E=MC^2, etc.

The whole conversation is around the fact that the derivative formula RPM=4*SFM/Dia is NOT unit-independent. And that is what Evan didn't understand.

YOU didn't say anything wrong in your initial responses (unlike Evan), and nobody is arguing with YOU.

Barrington

04-22-2012, 04:14 PM

...'rpm = 4*SFM/diam'.....

And not the subject of Evan's claim...Ah - the root of the problem - in post #23 Evan stated "The formula doesn't change regardless of the units of measurement used.". At that point the only formulae that had explicitly been mentioned were "RPM = SFM x 4 ÷ diameter" and various minor refinements.

Would this not suggest that this was the formula to which he was referring ? ...Evan ?

Various parties then pointed out that that statement could not apply to that formula. etc.etc. and the argument morphed out of all recognition :rolleyes:

where E is energy, m is mass, and c is the speed of light in a vacuum. The formula is dimensionally consistent and does not depend on any specific system of measurement units. The term 'dimensionally consistent' merely tells us that the energy has the dimensions ML^2T^-2. I don't understand the relevance. There haven't been any dimensionally incorrect equations ay any point in this thread as far as I have seen ?

As for "does not depend on any specific system of measurement units" . Well yes, it does not contain any 'scaling' constants.

Cheers

.

I'm sorry, but you don't get it. Mathematics has certain rules and they do not include "When you see SFM divided by Dia please use inches instead of feet".

Revolutions is a dimensionless value as it doesn't depend on units of time, speed or distance, only angular velocity. Introducing time in minutes doesn't change anything that affects the use of other units of distance. SFM introduces feet as the unit of length in question. That means Dia must also be in feet. The constants 4 and 12 work just fine with other units of length. They aren't related to the size of the units, they just convert them to work with SFM expressed in units other than feet (inches for example).

I'm not using 4 in "my" formula. I'm using it in your formula, correctly written.

Do you agree that your formula is really RPM=SFM/Pi* Dia.

Your formula as properly written is a close approximation. However, SFM may be expressed in any units, not just feet. Of course then it shouldn't be labelled SFM.

The only reason the rule of thumb exists is because people using imperial aren't very accustomed to using decimal feet. How many inches is .58333 feet?

Rosco-P

04-22-2012, 05:00 PM

Would that be 6.9996 inches?

MichaelP

04-22-2012, 05:31 PM

OK, Evan,

Since I see no reason to continue, let me ask you this.

1. Will you be willing to retract your comments in your posts #23 and #33 as being incorrect?

They are directly related to the "wrong" formula that was brought by the OP (and mentioned in almost any machining book written in the US)?

2. Are you, finally, ready to give a sound "yes" to Phil's question?

The term 'dimensionally consistent' merely tells us that the energy has the dimensions ML^2T^-2. I don't understand the relevance. There haven't been any dimensionally incorrect equations ay any point in this thread as far as I have seen ?

Expand the constant "4" from the original post and identify the components and their purpose. To fully understand the original equation you also have to accept that it came with a context of usage e.g., the diameter must be in inches for the thing to make sense.

BTW, you can do this with the original equation, too.

RPM = 4 X SFM / diameter where SFM = 5 meters/second, and diameter = one light year. As an exercise for the reader, calculate the RPM for the knowns.

Barrington

04-22-2012, 07:49 PM

The term 'dimensionally consistent' merely tells us that the energy has the dimensions ML^2T^-2. I don't understand the relevance. There haven't been any dimensionally incorrect equations ay any point in this thread as far as I have seen ?Expand the constant "4" from the original post and identify the components and their purpose. To fully understand the original equation you also have to accept that it came with a context of usage e.g., the diameter must be in inches for the thing to make sense. I think you may be confusing dimensional analysis with units of dimension ? Dimensional analysis has nothing to do with units, it's about the relationships of mass, length, time etc. on two sides of an equation.

For example the (ridiculous) equation '1 foot = 1 inch' IS dimensionally correct !

Cheers

.

beanbag

04-22-2012, 08:25 PM

Revolutions is a dimensionless value as it doesn't depend on units of time, speed or distance, only angular velocity.

Yes, but do you count your revolutions in cycles or radians? ;)

1. Will you be willing to retract your comments in your posts #23 and #33 as being incorrect?

I already have.

(subject to below comment)

2. Are you, finally, ready to give a sound "yes" to Phil's question?

Not a chance. It isn't a valid mathematical formula. There is no valid reason to substitute units that are not explicitly apparent in the equation which isn't an equality at all.

I point out that the answer of 2000 that I worked using the values I supplied is correct as the formula is written.

MichaelP

04-22-2012, 10:06 PM

Evan,

I perfectly understand where you're coming from and what you're trying to say. That's why when I told a story about my teacher, I mentioned that when this formula is given, it MUST come with a clear explanation what units to use for the diameters. Unfortunately, even in many books they forget to mention it.

But despite what you or I think about it, the formula exists and is used by the vast majority of machinists. It's just plain convenient in this format for our purposes. It was written specifically for this convenience, and it works very well.

If the surface speeds were given in books in inches per minute (SIM), this conversation wouldn't exist. In real life we have to use SFM because that is what they give us (and it's easier to remember and deal with than dealing with larger SIM values). As for the diameters, it's definitely easier to use inches. That's why this formula has evolved and is being used.

Therefore the only definitive answer to Phil's question is "Yes". You don't do anything criminal or break any rules by giving the answer. It just shows that you understand where this formula came from and why it must use mixed units to give a correct answer. Deep in your mind you may call this formula any name you want, but in reality it's a very useful tool, and you have to live with this fact.

I'm glad that we, finally, understood each other. Please accept my apology for my harsh comments. Unfortunately, I'm known for being short tempered and non-diplomatic when it comes to arguments like this.

Best regards!

philbur

04-23-2012, 03:15 AM

Absolute Classic Evan.

The equation that gives the correct answer isn't valid and the equation that gives an incorrect answer is correct.

Evan you are truly one of a kind.

Phil:)

Not a chance. It isn't a valid mathematical formula. There is no valid reason to substitute units that are not explicitly apparent in the equation which isn't an equality at all.

I point out that the answer of 2000 that I worked using the values I supplied is correct as the formula is written.

John Stevenson

04-23-2012, 04:19 AM

Well read all the way thru this post and the only post that stuck me as being relevant was this one by DP below.

Over the weekend I was just checking some figures using Machinists Toolbox on speeds and feeds but it wants me to run faster than the machine will allow.

Now if I run at the maximum speed I can I will have to reduce the feed but no way the program will allow this so basically it spits out useless information which wilst correct is no use to me at all.

The published speeds and feeds themselves are not more than good guesses taken from commercial machine operations a very long time ago and using the best analytics of the time. Time has since moved on and the home shop/hobbyist machinist, never a target of this information, really needs to rediscover the best values for the equipment they own and operate including the acceptable range of tolerance this value has. I have a non-rigid home shop bench top mill that is in no way capable of cutting according to the standard tables. I've created my own charts for speed and feeds. I used the published values to start with and adjusted according to the capacity of my machine. It works pretty well with my charts.