How to compare linear actuators with servo actuators, concerning a humanoid application

  • Most humanoids or bio-similar robots today use servos or other drives which are inherently rotational. If I were to try and replicate the properties of such an actuator with a linear actuator operating off-set from the joint, what would be the reasonable comparison metric?


    E.g. a servo operating on a 1 m long arm, where the servo is rated to 10 Nm, what would be the experience torque at the end of the arm?


    A linear actuator, off-set from the joint by 10 cm, if it were rated to 100 N, should provide 10 Nm at the end of the arm (right?).


    I'm sorry for asking these basic questions, I am not a roboticist :grinning_face_with_smiling_eyes:

  • Place your Ad here!
  • Nation

    Approved the thread.
  • a servo operating on a 1 m long arm, where the servo is rated to 10 Nm, what would be the experience torque at the end of the arm?

    where on the arm is servo acting and and what angle to the arm?




    consider above example where blue diagonal is linear servo,

    L1 is full arm length (1m)

    L2 is short arm length (point at which linear servo acts)

    black arc under shoulder shows angle between vertical body and linear actuator.

    M is weight of the arm (for simplicity we may opt to think of it as point mass at the middle of the arm.

    g is acceleration due to gravity (9.8m/sec^2)


    if there is no linear actuator, arm will fall due to gravity.

    torque arm weight produces is L*M*g


    suppose arm weighs 10kg and L is 0.5m (midpoint of the arm) then

    force due to gravity is F=M*g=10kg*10m/sec^2=100N.

    and torque that gravity exerts depends on arm position. above example is the worst case scenario so

    T=0.5m * 100N = 50Nm

    in other words you would need five of those actuators just to support arm from falling down.


    If you attach linear actuator close to the shoulder (say 10cm or L1=0.1m) and lets say actuator angle is 20deg. then sin(20deg) is 0.342


    and just to support the arm, you would need to counter 50Nm torque using same torque but supplied by the linear actuator.


    T=F*L1*sin(angle).

    F=T/(L1*sin(angle))

    F=50Nm/(0.1m*0.342)=1462N


    so how do we minimize this force?


    well equation is simple

    F=T/(L1*sin)


    F gets bigger when T is bigger, L1 is smaller and sin(angle) is smaller


    so how do we fight this?

    reduce T by using lighter arm

    if arm is heavy, do not stretch it fully out like that (horizontal position) because the gravity is the enemy.

    increase L1 (but there are practical limits)

    increase sin(angle) but there are also practical limits

    reduce acceleration due to gravity by letting your robot work on the moon or better yet in space.


    :smiling_face:



    ultimately you may want to add counterweight that is inside the torso and position linear actuator in such a way that optimizes angle for positions you intend to use.


    in this case weight of the arm is doubled (part inside torso is as heavy as external arm... or at least close). this way shorter but heavier part that is inside torso, works like see-saw and counterbalances the visible arm that is outside. this makes work for linear actuator MUCH easier. but the LEGS will suffer twice as much....




    pasted-from-clipboard.png

    1) read pinned topic: READ FIRST...

    2) if you have an issue with robot, post question in the correct forum section... do NOT contact me directly

    3) read 1 and 2

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account
Sign up for a new account in our community. It's easy!
Register a new account
Sign in
Already have an account? Sign in here.
Sign in Now

Advertising from our partners