### Robot Arm Torque

```ROBOT DYNAMICS
MOTORS supply the FORCE that the
robot needs to move
Rotational Force is called TORQUE
The motor needs to supply force to
• wheels
• arms
The Rolling of WHEELS
without slipping or spinning
Circum ference  2r
Everytime a wheel rotates an entire revolution, the robot travels a
distance equal to the circumference of the wheel.
Multiply that distance by the number of rotations per minute (rpm) and
you get the distance your robot travels in a minute (its speed)
v  (2r )(rpm)
For example, if your motor has a rotation speed
(under load) of 100rpm (determined by looking up
the motor part number online) and you want your
robot to travel at 3 feet per second, calculate the
wheel diameter you would need:
v  (2r )(rpm)
3  d (1.67rps)
d  0.57 ft or 6.89inches
Wheel diameter and the motor rpm are not the only
factors that determine robot velocity:
• motor torque
• robot weight
• robot acceleration
To achieve proper velocity/movement, you
must balance
• motor torque
• robot acceleration
• wheel diameter
Motor Torque and Force / Acceleration
High force is required to push other robots around,
or to go up hills, or have high acceleration.
Motor datasheet
motor torque
•
•
  F r
Acceleration
F  ma
motor speed
rpm or rps
Robot mass
Robot Motor Factor, RMF
Something to make life simpler, Can do quick
calculation to optimize your robot or select the
  F r
  rps  m a(r  rps) Wheel speed
v
v  (2r )(rpm)
  rps  m a( 2 )
RMF
(depends on motor specs)
Robot characteristics
or requirements
Robot Motor Factor, RMF
Example: You found the following 3 motors
Motor A: 2 ft lb, 1 rps
Motor B: 2.5 ft lb, 2 rps
Motor C: 2 ft lb, 4 rps
RMFA= 2 ft lb rps
RMFB= 5 ft lb rps
RMFC= 8 ft lb rps
RMF    rps
Suppose you want a velocity of 3 ft/s, an acceleration
of 2 ft/s2, and you estimate your robot to weigh 5 lbs
RMF  ma ( v 2 )  5  2  3 /(2 )
RMF  4.77 ft lb rps
Motor B & C will both work. Motor C is overkill, waste of \$
Wheel diameter
to use?
v
3
d

 0.48 ft  5.73in
  rps  (2)
Robot Efficiency
RMF is for 100% efficient systems. Gearing and friction
and many other factors cause inefficiency. General rules
for estimating inefficiency – If your robot
• has external gearing, reduce efficiency 15%
• uses treads, reduce efficiency 30%
• operates on high friction terrain, reduce efficiency 10%
Example: Tank robot on rough terrain would have
what efficiency?
Efficiency (1  0.30)(1  0.10)  0.63 (63%)
Robot Motor Factor, RMF
incorporating efficiency
Something to make life simpler, Can do quick
calculation to optimize your robot or select the
  rps  ma( v 2 )( 1efficiency)
RMF
(depends on motor specs)
Robot characteristics
or requirements
(efficiency is a decimal # ie 80% is 0.8)
Robot Arm Torque
determine the torque required at any given lifting
joint (raising the arm vertically) in a robotic arm
  FL
  mg  L
Weight
Torque
needed to hold a mass a given
distance from a pivot
L is the PERPENDICULAR
length from pivot to force
Robot Arm Torque
To estimate the torque required at each joint, we must choose
the worst case scenario
Greatest
torque
As arm is rotated clockwise, L, the perpendicular distance
decreases from L3 to L1 (L1=0). Therefore the greatest torque is
at L3 (F does not change) and torque is zero at L1.
Motors are subjected to the highest torque when the arm is
stretched out horizontally
Robot Arm Torque
You must also add the torque imposed by the arm itself
L
WL=mg
L/2
W1
Arm weight
  (m g L)  (W1  L / 2)
 L(m g  W1 / 2)
RMF (motor specs)
Robot arm torque
  rps  L(mg  W1 / 2)(rps)(1 / efficiency)
Robot Arm Torque
If your arm has multiple points, you must determine the torque
around each joint to select the appropriate motor
L3
L2
L2/2
L3/2
WL=mg
L1
W3
L1/2
Wm2
W2
Wm3
W1
Wm1
 3  (mg L3 )  (W3  L3 / 2)
 2  mg (L3  L2 ) W3  (L2  L 2 ) (Wm3  L2 )  (W2  L 2 )
3
2
 1  m g ( L3  L2  L1 )  W3  ( L1  L2  L 2 ) Wm3  ( L1  L2 )
3
 W2  ( L1  L2 2 ) (Wm 2  L1 )  (W1  L1 2 )
Robot Arm Torque
L3
L2
L2/2
L3/2
WL=mg
L1
W3
W2
L1/2
Wm2
W1
Wm3
Wm1
Gears
No good robot can be built without gears.
Gears work on the principle of mechanical advantage
With gears, you will exchange the high velocity of
motors with a better torque. This exchange happens
with a very simple equation that you can calculate:
 old  vold   new  vnew
Motor specs
Example:
Suppose your motor outputs, according to spec are
3 lb-in torque at 2000rps ,

v


v
old
old
new
new
but you only want 300rps.

3 lb-in * 2000rps = Torque_New * 300rps
3

new torque will be 20 lb-in.
Now suppose, with the same motor, you need 5 lb-in of
torque. But suppose you also need 1500rps minimum
velocity. How do you know if the motor is up to spec and
can do this? Easy . . .
3 lb-in * 2000rps = 5 lb-in * Velocitynew_
New Velocity = 1200rps
You now have just determined that at 1200 rps the selected
motor is not up to spec. Using the simple equation, you have
just saved yourself tons of money on a motor that would
have never worked. Designing your robot, and doing all the
Gear Ratios
HOW do you mechanically swap torque and velocity
with gears?
Moves slower
More torque
Moves faster
Less torque
The gearing ratio is the value at which you change your
velocity and torque. It has a very simple equation. The
gearing ratio is just a fraction which you multiple your
velocity and torque by. Suppose your gearing ratio is 3/1.
This would mean you would multiple your torque by 3
and your velocity by the inverse, or 1/3.
Gear Ratios
Example: Suppose you have a motor with output of
10 lb in and 100 rps (old=10 lb in, vold=100rps) and you
have a gear ratio of 2/3
Gearing ratio = 2/3
new=10 lb in x 2/3 = 6.7 lb in
vnew=100rps x 3/2 = 150 rps