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  1. #21
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    Well, I mis programmed my finishing passes, so was leaving axial stock not radial. I left .1mm, .2 and .3 as the roughing cuts went deeper. Due to my mistake I finished it all in one go, so stock left was .1, .2, .3. It did fine. 700 RPM, 12 IPM, half in the corners, 3mm step downs. Finish is not spectacular, but will pass. Next time I get a deep pocket job like this, I'll get a reduced shank dude. Thanks all for the help. This forum is and you all are awesome.

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    Quote Originally Posted by Fal Grunt View Post
    I do a lot of work with 1/2" long reach 2"LOC and 3" LOC cutters. Almost all of it is in Steel. Fixturing has a lot to do with it too. Are you holding on the bottom 1/8" of the part? That allows the part to ring like a tuning fork.
    Parts pretty deep in the vise (DX6). The rougher is singing more than the finisher. It's 90mm long with 65mm sticking out.. not too happy. I'm going to retire it after this job. Seriously like 5 hours on the thing. I actually bought a 7/16 2" LOC for this job, but then realized the big rub wall next to the pocket, so the 1/2 3" LOC is what I have on hand. Thanks for the advice.

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    Quote Originally Posted by nccodemonkey View Post
    The feedrate you program is the feedrate for centre of tool. The cutting point travels at a much faster feedrate on internal corners.
    Now that Cosmo has his issue worked out. This is the second time I have seen this statement on this here forum, could you please explain to me how a Tool is "Feeding" faster at the periphery than at the center (or centre for people who don't know how to spell).

    The Feedrate doesn't change, it is programmed in IPM. I think I understand what you may think is true. But what you are implying is the diameter changes chip load.

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    If the tool is programmed to go in two straight lines that are connected at a 90 degree angle, and the width of cut is small, when the tool gets the the corner, the flutes fill up. So, where they were only partially engaged, maybe a few degrees of the tool, this become 90 degrees of engagement. The "filling up" to the 90 degree point makes the cutting edge get effectively more feedrate breifly, while the center of the tool does not. Seems to me, possibly more useful, would be the ability to slow the RPM as well as it seems to have more impact on chatter (given my limited experience).

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    Quote Originally Posted by CosmosK View Post
    If the tool is programmed to go in two straight lines that are connected at a 90 degree angle, and the width of cut is small, when the tool gets the the corner, the flutes fill up. So, where they were only partially engaged, maybe a few degrees of the tool, this become 90 degrees of engagement. The "filling up" to the 90 degree point makes the cutting edge get effectively more feedrate breifly, while the center of the tool does not. Seems to me, possibly more useful, would be the ability to slow the RPM as well as it seems to have more impact on chatter (given my limited experience).
    Nope, sorry. Filling up the flutes is not effected by feedrate. That is engagement. A solution to the problem may be in changing the feedrate, but the feedrate does not change at the periphery of the tool.

    Imagine a 1" Endmill and a .1" Endmill with the exact same geometry, using the exact same SFPM and the exact same chip load. To maintain the chip load you will need to increase the feedrate and the RPM for a smaller Endmill. Point is, because we calculate it starting with SFM which is determined by diameter, the chip load is in fact relative to the diameter. Or should be.

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    Quote Originally Posted by litlerob1 View Post
    Nope, sorry. Filling up the flutes is not effected by feedrate. That is engagement. A solution to the problem may be in changing the feedrate, but the feedrate does not change at the periphery of the tool.

    Imagine a 1" Endmill and a .1" Endmill with the exact same geometry, using the exact same SFPM and the exact same chip load. To maintain the chip load you will need to increase the feedrate and the RPM for a smaller Endmill. Point is, because we calculate it starting with SFM which is determined by diameter, the chip load is in fact relative to the diameter. Or should be.
    When a car goes around a corner, are the outside and inside wheels turning and the same speed?

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    Quote Originally Posted by litlerob1 View Post
    Filling up the flutes is not effected by feedrate.
    When the engagement of the periphery of the tool goes from 3 degrees to 177 very quickly, I would call that a feedrate change at the edge of the tool.

    If you did a finishing pass on a 60" bore with a 59" diameter end mill (for arguments sake), the feedrate is programmed for the tool to go in a 0.5" circle. The edge sees something far faster. I suppose you could calculate it.

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    Quote Originally Posted by CosmosK View Post
    When the engagement of the periphery of the tool goes from 3 degrees to 177 very quickly, I would call that a feedrate change at the edge of the tool.

    If you did a finishing pass on a 60" bore with a 59" diameter end mill (for arguments sake), the feedrate is programmed for the tool to go in a 0.5" circle. The edge sees something far faster. I suppose you could calculate it.
    I'm not questioning that. I'm not even questioning corners, I am questioning feed at the periphery of a tool compared to the center.

    When a car goes around a corner, are the outside and inside wheels turning and the same speed?
    That is why we calculate our RPM based on SFM. You guys are both talking about SFM without realizing it. I'm talking about feedrate.

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    Quote Originally Posted by DFrench View Post
    When a car goes around a corner, are the outside and inside wheels turning and the same speed?
    Your asking the wrong question in a good example.

    If a car is turning, it is moving radially. The same way that the outside of an endmill is going faster at rotation than the center or any point in between.

    The question would be, is the outside fender going faster than the inside fender? When in a straight line? Or a curve?

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    Quote Originally Posted by Fal Grunt View Post
    The question would be, is the outside fender going faster than the inside fender? When in a straight line? Or a curve?

    Or for my sake; by my watch does the inside fender take more time to complete the corner than the outside fender? (assuming Perp. and Parallel of course)

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    Quote Originally Posted by Fal Grunt View Post
    Your asking the wrong question in a good example.

    If a car is turning, it is moving radially. The same way that the outside of an endmill is going faster at rotation than the center or any point in between.

    The question would be, is the outside fender going faster than the inside fender? When in a straight line? Or a curve?
    Yeah dude, the outside fender is moving faster than the inside when going around a corner, the wheels in the example just make that point more obvious. Check it
    capture.jpg

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    Quote Originally Posted by DFrench View Post
    Yeah dude, the outside fender is moving faster than the inside when going around a corner, the wheels in the example just make that point more obvious. Check it
    capture.jpg
    You are absolutely correct. The wheels are turning faster, but they do it in order to maintain a Constant Surface Speed, right. If the car were traveling in a straight line, and one were going faster than the other, what then?

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    60" bore, 59" end mill. 10 IPM

    center of the tools follows a 1" circle. circumference is 2*pi*0.5 = 3.14 inches. the center of the tool travels 3.14 inches and it takes 0.314 minutes to do so. v=d/t v=10 IPM

    at the cut, the distance traveled is 2*pi*29.5=185.26 v =d/t (t= constant=0.314) v =185.26/0.314 = 590 IPM. So a 59x increase from the programmed feedrate.

    You could get crazy and imagine a single flute router bit (no helix) in a scenario where as it enters the corner, the flute cuts the entire corner without breaking contact. This entity, the "effective edge feedrate" would then equal the SFM. But, considering feedrate is not defined anywhere else except for the center of the tool, you are completely correct to call BS on all of this.

    I should have mentioned earlier, I tested 2000 RPM and did get chatter.

    I have a feedrate override knob on my machine and it didn't seem to do much 50%-200% in testing. Some machines have a spindle override knob. That might be useful. But, what I'd really like is a knob that holds chipload constant and adjusts rpm and feedrate together. That would be useful.

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    This is why my race car has a spool instead of a differential. No math! LOL

    In all seriousness, though, think of the cutting tool's centerline as running instantaneously tangent to the radius of the corner at any point. Now, ask yourself if the engaged cutting edge and the edge immediately opposite it be traveling at different velocities past the object being cut?

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    I don't even care anymore.

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    Quote Originally Posted by litlerob1 View Post
    Now that Cosmo has his issue worked out. This is the second time I have seen this statement on this here forum, could you please explain to me how a Tool is "Feeding" faster at the periphery than at the center (or centre for people who don't know how to spell).

    The Feedrate doesn't change, it is programmed in IPM. I think I understand what you may think is true. But what you are implying is the diameter changes chip load.
    Take for example a .75" bore and you want to finish the ID with a .5" end mill. If you straightened out the bore into a line it will be Pi x D, or 2.356" long. The center line of the tool or the tool path is a .25" circle so the machine is actually travelling Pi x .25" or about .785" linearly. So if you program that cut at 40 IPM, the cutter is actually doing 120 IPM at the .75" Dia.. If you want a 40IPM cut you need to program it at 13.33 IPM at the tool centerline. The formula to compensate for ID tool paths is FcompID = Flinear x (Dwork - Dcutter/Dwork). The OD is the opposite, you can feed faster. Change the formula to Dwork + Dcutter/Dwork. The closer the cutter is to the size of the work the more important this becomes. If you are machining a 10" bore with a 1/4" end mill it doesn't make much difference etc... Machining in a corner is the same principle, you just use the radius of the corner and the radius of the tool. Of course if you are using a 1/2" end mill and trying to machine a 1/4" radius, you won't have much control over the size and finish.

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    Have used the Maritools reduced shank. It worked.

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    I will add in that I use Mastecam and there are instructions out there for how to mod your post to do feedrate adjustment for arcs automatically for you. You use a miscellaneous integer for off, internal, and external.

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    Quote Originally Posted by brian.pallas View Post
    I will add in that I use Mastecam and there are instructions out there for how to mod your post to do feedrate adjustment for arcs automatically for you. You use a miscellaneous integer for off, internal, and external.
    There....finally, someone has used the correct wording. ARC. Arcs require feedrate adjustment. 90 deg turns shouldn't. The analogy of the car in a curve, does not apply.
    The reason I say this is, the Cam program is simply creating a toolpath and applies a tool radius or tool diameter offset, while following the toolpath. The increase in feedrate in a 90 deg turn is simply not required. it is nothing more then a change of direction. During an arc however slowing the feedrate down along with spindle RPM, can reduce chatter in a not so rigid setup.


    Just my two cents worth.
    Last edited by Rigor; 02-08-2018 at 07:22 PM. Reason: Aaaaa f**ck it, too tired!!!

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    Quote Originally Posted by Rigor View Post
    There....finally, someone has used the correct wording. ARC. Arcs require feedrate adjustment. 90 deg turns shouldn't. The analogy of the car in a curve, does not apply.
    The reason I say this is, the Cam program is simply creating a toolpath and applies a tool radius or tool diameter offset, while following the toolpath. The increase in feedrate in a 90 deg turn is simply not required. it is nothing more then a change of direction. During an arc however slowing the feedrate down along with spindle RPM, can reduce chatter in a not so rigid setup.


    Just my two cents worth.
    Correct, I would add that I wasn't saying "should" or not,

    R


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