Intro to Robot Mechanics
For Dummies!
By: Ricky Marcus
Mechanical Engineering Mentor
Team 1745 P-51 Mustangs
September 2016
Newton’s Laws of Motion
First Law of Motion:
An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
Second Law of Motion:
Force = Mass * Acceleration
Third Law of Motion:
For every action, there is an equal and opposite reaction.
What is Force?
A force can be thought of as a push or a pull and it has a magnitude and a direction.
From Newton’s Second Law, F = ma, we know that a force causes an object of a certain mass to accelerate. If all the forces on an object sum to zero, then the object is not accelerating. It could be motionless (v=0) or have a non-zero constant velocity.
Forces can come in many different forms:
A force distributed over a certain area is called pressure.
Normal Force, FN
The force that a surface exerts on an object resting on
that surface. This force prevents the object from falling through the surface.
Assuming the surface is immovable, the Normal Force is equal to the sum of all other forces acting perpendicular to the surface. Usually, this is the object’s weight.
For multi-wheeled robots resting or moving on a level floor, the weight of the robot will be split up between all the wheels. The distribution can be calculated if you know the location of the robot’s Center of Gravity. The Normal
Force at each wheel will equal the weight on that wheel,
assuming there are no other forces acting on the robot. If
there are moving parts on the robot, the CG will move too.
Friction Force, Ff
A force between two surfaces sliding relative to one another.
where 𝜇 is the coefficient of friction and FN is the normal force.
There are two types of friction:
These are both calculated the same way. They just have a different coefficients. The coefficients can sometimes be looked up in a table or found experimentally. They are entirely dependent on the two materials sliding against one another. Static friction will always be greater than kinetic friction. It will take more force to get an object to start moving than it will to keep it moving.
What is Torque?
When a force tries to rotate an object about an axis, we call it Torque.
You can think of Torque as the ‘umph’ behind a twisting motion. For example, if one wrench tightening a bolt is harder to turn than another, it requires more Torque.
Torque is defined by the following equation:
𝛕 = r x F
F is the force acting on the object and r is the distance
from the force to the axis of rotation.
The Greek letter Tau, 𝛕, is the universal symbol for torque.
r
F
𝛕
Axis of Rotation
Torque continued...
Here we see two cases:
Since, 𝛕 = r x F, the torque in the second case is half the torque in the first case.
Bonus Material:
When we have two (or more) forces going in different directions, we choose one direction (clockwise or counterclockwise) to be positive.
If the force is not perpendicular to the radius, there is some trigonometry involved as well. Only the normal portion of the force creates torque.
r
½r
F
F
CASE 1
CASE 2
Torque continued...
Torque matters a lot in robotics because many of the actuators (fancy word that means a device that causes
motion) on a robot are motors and servos.
Motors and servos turn electricity into rotational motion. That motion has an associated speed (technically called angular velocity) and torque. The electromechanical design of the motor generates this torque.
We can describe a motor/servo as ‘strong’ or ‘weak’ if it generates a lot of torque or very little, respectively.
Motors will be discussed in depth later in this presentation.
Servo
Motor
Torque continued...
Inversely compared to a wrench, a torque of a certain magnitude will apply less force the further out you want to apply it.
In this example, we have two wheels of different radii
attached to identical motors. The force at the
ground of the small wheel is twice the force of
the large one.
Keep in mind, that the larger
wheel will be able to go
faster even though getting
up to speed is harder.
r
F
𝛕
2r
½ F
𝛕
Free Body Diagram
A Free Body Diagram is when we draw ALL the forces and torques that are acting on an object. This is always the first step when solving a mechanics problem.
Here, we have two boxes. The one on a ramp has a weight, normal force, friction force, and the tension of the rope. The hanging box only has weight and tension.
Easy Example
F = m * a
F = 1000 kg * 0.05 m/s2
F = 50 N (N= kg*m/s2)
But what does this mean??
If you can push this car with a constant force of 50 N (~11 lbf), it will accelerate at 0.05 m/s2.
Assuming no friction or drag (ideal case), after one minute of pushing, it will be moving at (60 seconds * 0.05 m/s2) = 3 meters/second = 6.7 miles per hour.
Easy Example continued...
A constant acceleration causes a linear increase in velocity over time.
For our robots, the acceleration will not be constant. Our motors have a top speed and the torque (which causes the acceleration) will vary with the speed.
Motors
Instead of going onto the playing field and pushing our wheeled mechanisms around, we use electricity to power motors which move our robots for us.
Motors are one of the most important components on our robots. We use them not just to turn the wheels but they can also be used to accomplish whatever tasks are required in a particular competition.
There are many of electronic components and hundreds of lines of code behind making the motors move in a controlled way, but that is beyond the scope of this training presentation.
Motors continued...
Motor mechanics can be defined by two primary variables: torque and speed. These values vary based on the amount of current provided to the motor and the physical load attached to the motor, so we will focus on the maximum torque and maximum speed a given motor can provide.
The maximum current and motor efficiency are also important to consider, but their effect can be lessened by focusing on using motors at speeds and torques around half of their maximum values.
The following slides look at a commonly used FIRST motor called the CIM motor.
Motors continued...
Here are the specs to a CIM Motor (standard issue part for FIRST)
Free Speed: 5,330 rpm (+/- 10%)
Free Current: 2.7A
Maximum Power: 337 W (at 2,655 rpm, 172 oz-in, and 68A)
Stall Torque: 2.41 N-m (21.33 in-lb)
Stall Current: 131A
Motors continued...
Free Speed: 5,330 rpm (+/- 10%)
Free Current: 2.7A
The Free Speed is the maximum speed of the motor. The motor only reaches its maximum speed when it is free from any load (i.e. there is nothing attached to the motor. The Free Current is the amount of current the motor draws while running at the Free Speed.
Motor speeds are generally given in revolutions per minute (rpm), but when used in calculations we need the units to be in radians per second (rad/s). In this form, we call the speed “angular velocity.” The conversion is to multiply rpm by 2𝜋/60.
Motors continued...
Stall Torque: 2.41 N-m (21.33 in-lb)
Stall Current: 131A
The Stall Torque is the max torque that the motor can output. The motor only reaches its max torque when the load is so great that the motor can no longer spin. The Stall Current is the current that the motor draws in the stalled state. You can see that it is very high relative to the Free Current of 2.7A.
Generally, motors will have very high speeds and low torques. For this reason, we typically will “gear down” the motors to decrease the speed and increase the torque. Gearing will be explained in depth later in the presentation.
Motors continued...
Maximum Power: 337 W (at 2,655 rpm, 172 oz-in, and 68A)
This is where things get interesting. We’ve discussed both the
free state and the stalled state of the motor, but what happens in between?
Motors continued...
Maximum Power: 337 W (at 2,655 rpm, 172 oz-in, and 68A)
We can see clearly that as the torque goes up, the current goes up and the speed goes down. As a bonus, the relationships look fairly linear!
Power is defined as speed times torque:
P = ⍵ * 𝛕
Pay attention to units when using this equation!
Need angular velocity in radians/sec (NOT RPM)
Need torque in Newton-meters (NOT OZ-IN)
Power will be in Watts
Torque
Speed
Current
MAX POWER
Motors continued...
Maximum Power: 337 W (at 2,655 rpm, 172 oz-in, and 68A)
Remember how I said before that we want to use our motors
in the middle of their range? This is because the middle of the range is where the power is greatest.
It also give us a lot of leeway. It is hard to burn out a motor when it is running well within its limits.
It is always good to know what a motor is capable of, but we can’t expect the motor to stall for an extended period of time without damage.
Harder Example (admit it, you knew this was coming)
What is the largest weight a CIM motor can hold if the weight is hanging from a 6” diameter pulley attached to the output shaft of the motor?
Harder Example continued...
Here are the specs to a CIM Motor from before:
Free Speed: 5,330 rpm (+/- 10%)
Free Current: 2.7A
Maximum Power: 337 W (at 2,655 rpm, 172 oz-in, and 68A)
Stall Torque: 2.41 N-m (21.33 in-lb)
Stall Current: 131A
MAXIMUM TORQUE
Harder Example continued...
What is the largest weight a CIM motor can hold if the weight is hanging from a 6” diameter pulley attached to the output shaft of the motor?
Harder Example continued...
What is the largest weight a CIM motor can hold if the weight is hanging from a 6” diameter pulley attached to the output shaft of the motor?
F (or Weight) = 𝛕 / r
W = 21.33 in-lb / 3 in
W = 7.11 lbf
Harder Example continued...
But what does this mean???
If you want to lift a 7.12 lbf object and all you have is a CIM motor with a 3” pulley,
You’re gonna have a bad day!
So what are our options if we HAVE to lift this 7.12 lbf object?
Harder Example continued...
So what are our options if we HAVE to lift this 7.12 lbf object?
A pulley with a 2” radius will lift up to a 10.67 lbf object.
Add a VersaPlanetary 3:1 Gear Kit on our CIM motor and we can now lift a 21.33 in-lbf * 3 (gear ratio) / 3 in (pulley) = 21.33 lbf object!
Gear Ratios
Here our motor is attached to the big gear. If we attach a shaft to the small gear, what are the maximum torque and free speed of the driven (second) shaft?
Big gear: 60 teeth
Small gear: 10 teeth
By intuition, we can see that every time the big gear spins one time, the little gear will spin more than one time.
In fact, the number of times can be calculated by the ratio of the number of teeth on the two gears.
Gear Ratios continued...
Gear Ratio: Driven Gear / Drive Gear = 10/60 = 1:6
This means that every time the drive gear turns 1 time, the driven gear will spin 6 times.
Usually you will see this in a different format so the second number in the ratio is one:
⅙:1 = 0.167:1
So the drive gear must turn 0.167 times to turn the small driven gear one time!
Gear Ratios continued...
From our specs, we know that the free speed velocity (speed when there is nothing attached to the motor) of the CIM motor is 5,330 rpm
That is 5,330 revolutions per minute!
Or 88.83 revolutions per second! WOW!
Anyways, this will be the speed of the motor and since the big gear is attached to the shaft of the motor, it is also the speed of our big gear. (It would be physically impossible for these to spin at different speeds)
Gear Ratios continued...
CIM Free Speed: 5,330 rpm
Gear Ratio: 1:6
Free Speed of driven shaft = Free Speed of drive shaft = 5330 * 6 = 31,980 rpm
Gear Ratio
Gear Ratios continued...
But what about the max torque for the driven shaft?
Remember our Torque = Force x radius equation?
The force at the edge of the big gear must equal the force at the edge of the small gear, because forces must always add to zero*.
Drive Gear Driven Gear
Motor Torque = F = Driven Torque?
60 10
* I am swapping the number of teeth in for the radius of the gears. I can do this because the gears are designed so that the ratio of the radii and the ratio of the teeth are equal.
Gear Ratios continued...
But what about the max torque for the driven shaft?
CIM motor Stall Torque: 21.33 in-lb
21.33 = F = Driven Torque?
60 10
Driven Torque = 21.33 in-lb * (⅙) = 3.56 in-lb
Gear Ratios continued...
Drive Speed: 5,330 rpm
Drive Torque: 21.33 in-lb
Driven Speed: 31,980 rpm
Driven Torque: 3.56 in-lb
It is very important to remember that as speed goes up, torque goes down.
The opposite is true as well. If the gears are swapped:
Driven Speed: 888.33 rpm
Driven Torque: 127.98 in-lb
Gear Ratios continued...
So remember this example?
Add a VersaPlanetary 3:1 Gear Kit on our CIM motor and we can now lift a 21.33 in-lb * 3 (gear ratio) / 3 in (pulley) = 21.33 lbf object!
Now we know that even though we increased the torque by 3 times to lift a heavier object, the speed at which we can lift it is 3 times SLOWER.
Design is all about TRADEOFFS.
Belt/Chain Ratios
So now the question is how is using two pulleys with a belt between them (or two sprockets with a chain) different from using two gears?
The equations we used for the gears to calculate torques and angular velocities will be the same for the belts and pulleys.
Belts and chains can be useful when you need to mount the motor further from the mechanism you want to move. They also allow for more slop in your manufacturing. Gears must precisely mesh. Belts and Chains might have more features to set the tension.
Great Resource for Gears and Chains: http://revrobotics.com/resources/
Efficiency
Not really going to get into the concept of efficiency here since it is an advanced topic, but note that there is friction between the various components of the robot drivetrain (gears, bearings, motor innards). We generally do not directly calculate the value of the friction anywhere, but instead assign an efficiency to the entire drivetrain. Typically, the efficiency will be between 60 and 95%. This means that 5 - 40% of the input power (electricity) will be converted to heat instead of motion/torque.
Because we will design to operate in the middle of our selected motor’s operating range, we should not be generating too much heat and always have a little bit more power available when we really need it.
From the CIM motor example, we can see efficiency graph for just the motor.
Linear Velocity
So we’ve talked about the angular velocity of a motor and how torque acting over a distance produces a force. But most of the action a robot will perform will be moving linearly in space.
r
v
ω
Here we have a wheel of radius, r. In this ‘direct drive’ setup, the wheel is directly mounted to the shaft of the motor.
The motor spins at an angular velocity, ω.
The velocity of the outside of the wheel, v, is:
v = ω * r
Linear Velocity continued...
Let’s say the motor is spinning at 1000 rpm and the wheel radius is 4 in.
First we must convert rpm to radians/second.
ω = 1000 rev/min * 2𝝅 rad/rev * 1 min/60 sec = 104.72 rad/s
Then,
v = 104.72 rad/s * 4 in = 418.88 in/s = 34.91 ft/s
For a wheel with a 2 in. radius:
v = 104.72 rad/s * 2 in = 209.44 in/s = 17.45 ft/s
r
v
ω
Rolling and Friction
When the robot's wheels are rolling on a surface, the wheels are not sliding relative to the ground. In fact, we have to consider friction completely differently. This concept is known as traction and is very important in robot drivetrain design.
Because traction is related to friction, we know that it is completely dependent on the wheel and floor materials. A robot will behave very differently on ice vs. carpet.
There can be situations where the wheels are rolling and sliding at the same time, but that is outside the scope of this presentation. However, I will show how you determine the torque transition point between ‘pure rolling’ and ‘sliding and rolling.’
Traction
In a very basic sense, there is no force you
have to overcome to cause a wheel to turn.
This is why we use wheels and why civilization has been using them for thousands of years! Contrast this with a sliding block that must overcome the static friction before it can begin to accelerate.
At each infinitesimal moment in time, the wheel is rotating around a static contact point, P, between the wheel and the ground. Once the robot rotates a fraction of a degree, there is a new contact point slightly ahead of the old one and the process begins anew. This is how the robot moves forward.
Traction continued...
In order for this to occur, Point P must not move
while the wheel is pivoting around it (remember
we don’t want to roll and slide). This means that
Point P is static. To confirm that this is the case,
see this picture. At the bottom of the wheel, the
spokes are clear while at the top they are blurry!
The velocity of the center of the wheel, where the axle is attached, as it rotates around P is the linear velocity of the robot.
Question: What is keeping Point P from moving?
Traction continued...
Answer: Static Friction!
Let’s take a look at the forces here by drawing a free body diagram.
Traction continued...
The motor torque applies a force at the ground equal to the torque divided by the radius of the wheel. The ground applies a force back to the wheel ONLY if there is friction.
This force applied to the wheel is called the Tractive Force. It is maximum when the applied force equals the maximum static friction. This friction force is equal to the coefficient of friction between the wheel and ground times the normal force on that wheel. Once we exceed the maximum tractive force, the wheel will start to slip in addition to roll. Ideally, this will not happen under normal operating conditions. However, we do want the wheels to slip before the motors stall or the breakers trip.
The tractive forces on the drive wheels are only the
forces that cause the robot to move forward!
Questions?
Obviously, there is a lot more to designing a robot than just the material presented here. However, if you understand these concepts, you are well on your way toward mastering these advanced topics:
Additional Problem - Staged Gearing
The first small gear drives a large gear, which is connected to the same shaft as a second small gear. The second gear drives another large gear.
If a CIM motor is attached to the first small gear, what is the maximum torque of the second large gear?
What is the speed of the second large gear (in rpm) when the torque is equal to half of its maximum value?
Additional Resources