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  1. #11
    I believe what the graph shows is the total measured force between the tool and the table (Fx, Fy, Fz) when using a sensor between the part and the bed, so includes the all cutting forces resolved into x,y,z:
    Feed force - in direction of cut
    Feed normal force - 90 degrees from direction of cut
    Feed passive force - feed up/down due to helix)
    all of which include any moments caused by cutting

    They do talk about another sensor which goes into spindle (RCM) and so torque can be measured directly and extracted out separately.

    But in terms of machine design I would have thought you want to know the total force Fx, Fy, Fz on the tool and a target stiffness to aim for. Then use FEA, or hand calcs to work out if you meet the required stiffness when applying the force at the tool tip.

    For stiffness targets I've seen figures quoted for commercial machines in the region of 10 - 25 N/um (10,000 - 25,000 N/mm). I think one of my DIY machines was between 1000 - 2400 N/mm and was ~OK but I would aim higher for a good machine and I think it is easily possible.
    Building a CNC machine to make a better one since 2010 . . .
    MK1 (1st photo), MK2, MK3, MK4

  2. #12
    10-25 N/μm seems reasonable, and a good number to aim for. I searched around for CNC N/μm and found a good preview on work-piece precision .

    Seems like the "hardcore" machine centers have a stiffness around;
    X axis: 30-62 N/μm
    Y axis: 30-40 N/μm
    Z axis: 67-95 N/μm
    Radial: 25 N/μm (5000rpm), 20 N/μm (25000rpm)
    This is for vertical CNC mill setup. So for a router it would be a bitt less.

    According to the link above it seems like the tools are the weakest and the least stiff part of the machine. Where they use the usual deflection calculation for a circular cross section and match it to some real measurements. 0.15 N/μm for 6mm tool.
    Last edited by PotatoMill; 17-04-2016 at 04:15 PM. Reason: Grammar

  3. #13
    fnm's Avatar
    Lives in Sitges, Spain. Last Activity: 17-08-2018 Has been a member for 2-3 years. Has a total post count of 1. Received thanks 3 times, giving thanks to others 8 times.
    If anyone is interested in prediction of cutting forces in aluminum milling I would recommend to read the article:
    "A comparison of analytical cutting force models" published by WIAS
    There you can find very simple models to compute cutting forces for aluminum end milling as a function of chip thickness and axial depth of cut.
    Here I try to resume Kinzle-Victor model for predicting forces for end milling Aluminum
    The Kinzle-Victor model predicts Fi(Fr,Ft) tangential Ft and radial Fr milling cutting forces
    Fi, according Kinzle-Victor model, can be predicted through this formula:

    Fi(Fr,Ft)=Ki(Kr,Kt)*a*h^(1-mi(mr,mt)) where

    Fi is force in N,
    a is the load per tooth in (mm) ,
    h is the depht of cut in (mm) and
    Ki and mi are constants that depend of h values

    For end milling aluminum the values of constants Ki(Kr,kt) and mi(mr,mt) are:

    Kr=39 for (0.001 < h <0.01) ;
    Kr=99 for (0.01 <= h <0.1) ;
    Kr=298 for (0.1 <= h <1)

    Kt=336 for (.001 < h <.01)
    Kt=493 for (0.01 <= h <0.1) ;
    Kt=667 for (0.1 <= h <1)

    mr= 1 for (0.001 < h <0.01);
    mr=0.8 for (0.01 <= h <0.1);
    mr=0.32 for (0.1 <= h <1)

    mt=0.53 for (0.001 < h <0.01) ;
    mt=0.45 for (0.01 <= h <0.1);
    mt=0.32 for 0.1<= h <1

    For example for a cutting depht of 0.5 mm and a load per tooth of 0.1 mm the forces for aluminum are:
    Ft=Kt*a*h^(1-mt)=493*0.5*0.1^1-0.45=69 N
    F=Kr*a*h^(1-mr)=99*0.5*0.1^1-0.8=31 N

    Model computes tangential and radial forces located at xy plane but doesnt compute axial force at z axis and I dont know if are average or peak forces

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