Taylor tool wear equation

I have been looking through all of the information I can find on this subject. I understand that that real-time testing of cutters on hand is best and that there are more evolved versions of the equation including DOC and such. I am just trying to wrap my mind around the basic equation and concept.

I understand that the insert material is constant within a range of numerical data ie. (HSS=.125-.2)-(carbide=.25-.5) and that this information can be retrieved from manufacturer of inserts, I understand T is total tool life over a given amount of time 90 minutes 60 minutes 1 minute, and that V = SFM.

What I am not understanding:
Is there a constant/standard time measurement for T? in some articles I see 90 minutes some 60 minutes and some 1 minute.

I am fairly certain that my math is way off on this excel jpeg. Just looking for some guidance as to where the errors are.

WardFleck said:

I have been looking through all of the information I can find on this subject. I understand that that real-time testing of cutters on hand is best and that there are more evolved versions of the equation including DOC and such. I am just trying to wrap my mind around the basic equation and concept.

I understand that the insert material is constant within a range of numerical data ie. (HSS=.125-.2)-(carbide=.25-.5) and that this information can be retrieved from manufacturer of inserts, I understand T is total tool life over a given amount of time 90 minutes 60 minutes 1 minute, and that V = SFM.

What I am not understanding:
Is there a constant/standard time measurement for T? in some articles I see 90 minutes some 60 minutes and some 1 minute.

I am fairly certain that my math is way off on this excel jpeg. Just looking for some guidance as to where the errors are. View attachment 157793

Click to expand...

.
tool wear constant ?? often tool wear is based like 99% or more of part and tool vibration accelerating tool wear and metal not being homogeneous but full of hard spots or slag inclusions. modern metal often has few hard spots so hitting one can surprise many machinist.
.... i am just saying you can have 120 minute tool life easily and hit one big hard slag inclusion and wipe out literally over 100 inserts (many sets) and each time cutter hits hard spot all inserts are gone in less than 2 seconds.
.
or cutter can easily last 120 minutes on a rigid setup and then trying to machine a non rigid part that starts high vibration literally all inserts can be gone in a few minutes
.
sure some setups recognized as not very rigid and speeds and feeds and even cutter size, wide and depth of cut can be lowered to lower vibration and that helps tool life.
.
but hard spots or slag inclusions you can machine 100 parts and have great tool life with 98 of then but 2 parts can cause the wear equation to be off or in error 1000%.... not sure i believe in wear constant when it can be different +/- 1000% from part to part

OK, see here first:
https://en.wikipedia.org/wiki/Tool_wear
Note that this is the most important thing "n and C are constants found by experimentation or published data; they are properties of tool material, workpiece and feed rate."

You can kiss any published data goodbye unless it is on your material on your machine.
It does work in stable cutting but you need to find three points on the curve to define these constants on your own.
T is whatever you want, we use this to optimize cutting to put 70-80% life at shift changes.

Not being able to handle hard spots or inclusions is a whole nother problem addressed by geometry and edge preps.
Change these and n and C change.
Coatings change all of this also.

It would be nice if it was this simple. My life does not work that way.
A old equation, built around HSS testing that still works, it is a very useful tool in my box but you need to run tests to use it.
Knowing where you are on the curve is important as this thing takes off like a rocket on the right hand side.
We use this a lot but only after corner shape, grade, and geometry and edge prep have been optimized first.
It will not work for catastrophic failure modes. The cutting process has to be stable.
Bob

Thank you for the information and taking the time to reply. just trying to get an idea of how a shop would add cutter insert cost into a potential larger run job.

I am sure there could be defects in materials that would ruin cutters, improper feeds/speed/doc or person runs wrong program all is lost, as there could be unseen staples, nails in lumber for a framer or unseen defects in wood for a cabinet maker. Just trying to understand how a shop would estimate into a bid the cost of inserts needed as it seems that it would be a significant amount when holding tight tolerances.