The LM Guide is capable of receiving loads and moments in all directions that are generated due to the mounting orientation, alignment, gravity center position of a traveling object, thrust position and cutting resistance. Reverse radial load Lateral load Radial load Lateral load MA MB MC Moment in the pitching direction Moment in the yawing direction Moment in the rolling direction Fig.1 Directions of the Loads Applied on the LM Guide Calculating an Applied Load Single-Axis Use Moment Equivalence When the installation space for the LM Guide is limited, you may have to use only one LM block, or double LM blocks closely contacting with each other. In such a setting, the load distribution is not uniform and, as a result, an excessive load is applied in localized areas (i.e., both ends) as shown in Fig.. Continued use under such conditions may result in fl aking in those areas, consequently shortening the service life. In such a case, calculate the actual load by multiplying the moment value by any one of the equivalent-moment factors specifi ed in Table1 to Table6 A. Moment load Rows of balls under a load Moment load Rows of balls under a load LM rail Ball displacement line Load distribution curve Maximum applied load on a ball Maximum ball deflection Ball displacement line Load distribution curve Fig. Ball Load when a Moment is Applied An equivalent-load equation applicable when a moment acts on an LM Guide is shown below. P = K M P : Equivalent load per LM Guide (N) K : Equivalent moment factor M : Applied moment (N-mm)
Point of Selection Equivalent Factor Since the rated load is equivalent to the permissible moment, the equivalent factor to be multiplied when equalizing the M A, M B and M C moments to the applied load per block is obtained by dividing the rated loads in the corresponding directions. With those models other than -way equal load types, however, the load ratings in the directions differ from each other. Therefore, the equivalent factor values for the M A and M C moments also differ depending on whether the direction is radial or reverse radial. Equivalent Factors for the M A Moment LM Guide PR=KAR MA Equivalent in the radial direction PL=KAL MA Equivalent in the reverse-radial direction Fig.3 Equivalent Factors for the M A Moment Equivalent factors for the MA Moment Equivalent factor in the radial direction Equivalent factor in the reverse radial direction C0 C0L = =1 KAR MA KAL MA C0 KAR= MA C0L KAL= MA Equivalent Factors for the M B Moment PT=KB MB Equivalent in the lateral direction PT=KB MB Equivalent in the lateral direction Fig. Equivalent Factors for the M B Moment Equivalent factors for the MB Moment Equivalent factor in C0T KB= the lateral directions MB C0T KB MB =1
Equivalent Factors for the M C Moment PR=KCR MC Equivalent in the radial direction PL=KCL MC Equivalent in the reverse-radial direction Fig.5 Equivalent Factors for the M C Moment Equivalent factors for the MC Moment Equivalent factor in the radial direction Equivalent factor in the reverse radial direction C0 C0L = =1 KCR MC KCL MC C0 KCR= MC C0L KCL= MC C 0 : Basic static load rating (radial direction) (N) C 0L : Basic static load rating (reverse radial direction) (N) C 0T : Basic static load rating (lateral direction) (N) P R : Calculated load (radial direction) (N) P L : Calculated load (reverse radial direction) (N) P T : Calculated load (lateral direction) (N)
Point of Selection Example of calculation When one LM block is used Model No.: SSR0XV1 Gravitational acceleration g=9.8 (m/s ) Mass m=10 (kg) =00 mm l=100 mm No.3 No. No. No.1 l LM Guide m m Fig.6 When One LM Block is Used No.1 P 1 =K AR1 l 1 K CR l =980.75 98 000.19 98 100=675 (N) No. P =K AL1 l 1 K CR l =980.137 98 000.19 98 100=133 (N) No.3 P 3 =K AL1 l 1 K CL l =980.137 98 000.06 98 100=318 (N) No. P =K AR1 l 1 K CL l =980.75 98 000.06 98 100=857 (N) When two LM blocks are used in close contact with each other Model No.: SVS5R Gravitational acceleration g=9.8 (m/s ) Mass m=5 (kg) =00 mm l=150 mm No.3 No. No. No.1 l m m Fig.7 When Two LM Blocks are Used in Close Contact with Each Other No.1 = No. P= No.3 = No. P= KAR KCR KAL KCR KAL KCL KAR KCL l l l l = = = = 9 9 9 9 0.0188 9 000.081 9 150 0.0158 9 000.081 9 150 0.0158 9 000.068 9 150 0.0188 9 000.068 9 150 =507.9 (N) =168.8 (N) =381.7 (N) =.6 (N) Note1) Since an LM Guide used in vertical installation receives only a moment load, there is no need to apply a load force ().
Double-axis Use Setting Conditions Set the conditions needed to calculate the LM system s applied load and service life in hours. The conditions consist of the following items. (1) Mass: m (kg) () Direction of the working load (3) Position of the working point (e.g., center of gravity): l, l 3, h 1 (mm) () Thrust position: l, h (mm) (5) LM system arrangement: l 0, l 1 (mm) (No. of units and axes) (6) Velocity diagram Speed: V (mm/s) Time constant: t n (s) Acceleration: n (mm/s ) α (7) Duty cycle Number of reciprocations per minute: N 1 (min -1 ) (8) Stroke length: l s (mm) (9) Average speed: V m (m/s) (10) Required service life in hours: L h (h) Gravitational acceleration g=9.8 (m/s ) l Speed (mm/s) V Duty cycle l h h1 tn Fig.8 Condition t1 tn ls Velocity diagram s mm
Point of Selection Applied Load Equation The load applied to the LM Guide varies with the external force, such as the position of the gravity center of an object, thrust position, inertia generated from acceleration/deceleration during start or stop, and cutting force. In selecting an LM Guide, it is necessary to obtain the value of the applied load while taking into account these conditions. Calculate the load applied to the LM Guide in each of the examples 1 to 10 shown below. m : Mass (kg) l n : Distance (mm) F n : External force (N) P n : Applied load (radial/reverse radial direction) (N) P nt : Applied load (lateral directions) (N) g : Gravitational acceleration (m/s ) (g =9.8m/s ) V : Speed (m/s) t n : Time constant (s) n : Acceleration (m/s ) Example α Condition Applied Load Equation LM Guide 1 Horizontal mount (with the block traveling) Uniform motion or dwell l P P = P = = l l l P = l Horizontal mount, overhung (with the block traveling) Uniform motion or dwell = l P P P = = l l l P = l Note) Load is positive in the direction of the arrow.
Vertical mount Uniform motion or dwell Condition Applied Load Equation l P = P = l 3 T F l P = = T = PT = PT P PT = T = E.g.: Vertical axis of industrial robot, automatic coating machine, lifter Wall mount Uniform motion or dwell PT l T P = P = = P = T = PT = l P T PT = T = l PT E.g.: Travel axis of cross-rail loader Note) Load is positive in the direction of the arrow.
With the LM rails movable Horizontal mount Condition Applied Load Equation Point of Selection P to P (max) = LM Guide 5 P to P (min) = l E.g.: XY table sliding fork 6 Laterally tilt mount θ h1 P PT l E.g.: NC lathe Carriage T = T= P = PT= = T= P = PT= cosθ cosθ sinθ cosθ sinθ cosθ sinθ cosθ cosθ sinθ cosθ cosθ cosθ l sinθ h1 sinθ l cosθ l sinθ h1 sinθ l cosθ l sinθ h1 sinθ l cosθ l sinθ h1 sinθ l Note) Load is positive in the direction of the arrow.
Condition Applied Load Equation Longitudinally tilt mount = cosθ cosθ l cosθ sinθ h1 T = sinθ h1 P P = cosθ cosθ l 7 P T θ l PT PT = = cosθ sinθ cosθ cosθ sinθ h1 cosθ l sinθ h1 T = sinθ E.g.: NC lathe Tool rest P = cosθ cosθ cosθ l sinθ h1 PT = sinθ Horizontal mount with inertia During acceleration = P = m α 1 l P = = m α 1 l T = PT = m α 1 PT = T = m α 1 8 Speed V (m/s) F l P T PT α t1 t t3 Time (s) Velocity diagram E.g.: Conveyance truck During uniform motion to P = During deceleration = P = P = = T = PT = PT = T = m α 3 m α 3 m α 3 l m α 3 l Note) Load is positive in the direction of the arrow.
Condition Applied Load Equation Point of Selection Vertical mount with inertia T F P l During acceleration m (gα 1) l = P = m (gα 1) l P = = m (gα 1) T = PT = m (gα 1) PT = T = During uniform motion LM Guide 9 PT Speed V (m/s) P α t1 t t3 Time (s) Velocity diagram E.g.: Conveyance lift = P = P = = T = PT = PT = T = l l During deceleration m (g α 3) l = P = m (g α 3) l P = = m (g α 3) T = PT = m (g α 3) PT = T = Horizontal mount with external force l l Under force F1 = P = P = = T = PT = F1 l5 F1 l5 F1 l 10 l5 F F1 F F3 P PT T PT = T = F1 l Under force F = P = P = = F F F l F l E.g.: Drill unit, Milling machine, Lathe, Machining center and other cutting machine Under force F3 = P = = P = T = PT = PT = T = F3 F3 F3 F3 F3 l F3 l Note) Load is positive in the direction of the arrow.