# A General Method for Choosing Motors and Gearings

With the plethora of requirements, motors, gearboxes, pulley systems, etc., how do you choose which motor and gearbox(es) to use? How do you know they will meet your needs? This page provides a very basic guide to what to consider and the process to do so to choose an appropriate motor and gear ratio.

The example/case study below (in the green boxes) are probably more intuitive and easier to follow than the theory presented first in each section, but the theory discusses the why and how.

## 1. Define the problem.

Typically, you have the following overarching constraints and requirements in FRC:

Constraints:

• 40A (max) breaker per motor output
• 120A breaker total (including other mechanisms)
• mass budget
• space
• specific allowable motors

Requirements

• mechanical power - combination of factors, including
• speed - how fast would you like to be able to accomplish the task?
• distance - what is the range of the mechanism?

On this page, we'll ignore mass and space constraints for the most part.

First, let's consider power and efficiency.

From physics, power is equivalently defined by the following:

P = W/t = Fd/t = Fv

where W = work, F = force, d = distance, t = time, and v = (average) velocity.

1. Figure out the force required by your system.
2. From your design, determine how fast you want the mechanism to actuate.
3. As an elimination tool, any motor combination with a total output power less than your requirement will not do the job (depending on efficiency), those motors. You can either
• reduce the speed
• increase the number of motors
4. Make sure there is some margin for the inefficiency of the system, the gearing, and other unaccounted factors.
• Efficiency is higher for directly directly systems, and for simply gearboxes.
• Efficiency lowers for gearboxes with a lot of contact points, such as planetary gearboxes.
• Efficiency can be as low as 40-50% for poorly designed worm gears.

From here on, your `effective power` is your `required power` * `efficiency`. (Likewise, `required power` = `effective power` / `efficiency`.)

Case Study: Lifting a Robot

Throughout this page, the situation of lifting a 134.885 lb (600 N) robot up 1 ft by using a grappling hook mechanism will be used as a case study to show how the process works. All other constraints apply here.

The reaction force is gravity, so our power is dependent on how fast we want to lift the robot. Lifting a 600 N robot at various speeds:

Speed (m/s)

Speed (ft/s)

Required Power (W)

0.25

0.82

150

0.50

1.64

300

1.00

3.28

600

Suppose we wanted to lift at 1 m/s (3.28 ft/s) so that the robot would climb up in about 1/3 of a second. We need a minimum of 600 W, not counting inefficiency, in order to do this. Looking at the max power ratings on several common FRC motors:

Motor

Maximum Power
(at different RPM)

CIM

337 W

Mini-CIM

230 W

775pro

347 W

We see that there is no one motor that gives enough power to do this. (It is impossible to get that linear speed with any FRC motor, regardless of gearing, against this force.)

We could

1. gear 2 motors together, or
2. decrease the desired speed.

1. More Motors
2 CIM's, 3 mini-CIM's, or 2 775pro's would each have enough power to achieve that speed.

Suppose we gear 2 CIM's together with a gearbox. Their theoretical combined power is 674 W, which is theoretically enough to achieve that speed, since it's more than the required 600 W.

Considering potential inefficiency, this 2-CIM combination can only accommodate at least a 90% efficiency at the max power. Any lower, and the motor may not be sufficient.

Suppose that we've decided irrevocably to use a worm gearing. Since for this situation we will need multiple motors, let's assume an efficiency of 40% (worm gears can be inefficient, plus motors' powers don't quite add efficiently either). From above, our `required power` is 600 W / 40% = 1500 W from motors. (By the way, this is a huge number.)

In order for this (admittedly, extreme in many regards) case to work, we'd need to combine the outputs of at least 5 CIM motors together.

2. Speed Decrease
Instead, we could rationalize that 3.28 ft/s is a really fast speed to climb up a foot.

Let's assume we want to use a 775pro due to other reasons (perhaps mass budget issues). What is the maximum speed that we can achieve with the 775pro?

v = P/F = (347 W) / (600 N) = 0.58 m/s or 1.9 ft/s

Is 1.9 ft/s fast enough? Definitely. Even with an efficiency of 50%, is 0.95 ft/s fast enough? Probably.

## 2. Understand the requirements of the system.

In section (1), we considered power and efficiency as the basis for understanding the problem. Next, we consider mechanical systems that can accomplish our needs.

Since we don't exactly know the efficiency of the final motor+gearing combination, we still need to consider our `effective power` = `required power` * `efficiency`.

We also need to consider

• torque requirements
• speed requirements
• current limitations

for the

• motor
• gearing, etc.

We'll need to analyze:

1. required torque and available torque

--> torque determines motor speed, which determines your actual output/linear speeds

1. current draw

based on a choice of

• motor
• gear ratio

Torque Analysis
The torque/force induced by your load is the driving force for the kinematics of the mechanism. From the system, you can figure out the particular force

Back to lifting the 600 N robot.

Let's assume we're going to go with a 775pro. Useful information about the problem and the motor from the above box:

 Max Power of the 775pro 347 W 1 ft/s = 0.305 m/s ~ 180 W

Rationale for Desired Linear Speed: Climbing in about 1 s is decently fast, and even if the efficiency is about 50%, the motor+gearing will be able to accomplish that.

This next part will require some trial and error, trying different combinations of gearing and output radius:

• Suppose our design involves a winch with a 4 in. (.1 m) diameter drum --> radius of drum is 2 in. (.05 m).

The force available to lift the 600 N robot at the output will be divided by the radius (read: smaller radii enable more force with less deflection).

• The following gear ratios are available in stock (either in the shop or from a manufacturer) in the form of a "standard" gearbox:
• 1:100
• 1:80
• 1:60
• 1:20

Let's take a loop at the motor curve for the 775pro:

Torque Analysis
From the design of a 2 in. (.05 m) radius drum at the output, we need an output torque of at least 30 Nm. Note that `output torque` is related to `motor torque`:

`output_torque` = `motor torque` / `gear ratio`

where in this case gear ratio is expressed as 1:n (sometimes it is represented as the opposite; sometimes it is the opposite)

@ Stall Torque of 0.71 Nm: A 1:60 gearbox would get us there.

`motor torque` = 30 Nm * (1/60) = 0.50 Nm needed

But stall torque generally serves only as an upper limit (if stall torque + gear ratio isn't enough, it can't be done; but if it is enough, you still need to crunch numbers a bit).

For the hypothetical gear ratios available for this motor:

Torque at Output

Gear Ratio

Torque at Motor

Speed at Motor
(from Motor Curve)

Speed at Output

Linear Speed

30 Nm

1:100

0.3 Nm

10,800 RPM

108 RPM

0.57 m/s = 1.87 ft/s

30 Nm

1:80

0.375 Nm

8500 RPM

106.25 RPM

0.57 m/s = 1.87 ft/s

30 Nm

1:60

0.5 Nm

5500 RPM

91.67 RPM

0.48 m/s = 1.57 ft/s

30 Nm

1:20

1.5 Nm

0 RPM

0 RPM

0 m/s = 0 ft/s

From this table, we see that we can't use the 1:20 gearbox, as it doesn't provided enough output torque given our load and design. But we can use any other others: 1:100, 1:80, and 1:60. These give us 2 distinct speeds in this to choose from.

Current Analysis
The other factor we need to take into account is the current draw of the motor. Circuit breakers will trip if both

1. a current threshold is reached, and
2. current is above for a certain amount of time.

That is, breakers won't trip for short peaks.

These values are represented as datapoints or as a graph in fuse datasheets, like this one for the 40A FRC Circuit Breaker.

For each of the motor speeds above, there is an associated current; our task now is to check the current draw and make sure the breaker won't trip:

Speed at Motor

Current Draw
(from Motor Curve)

Percent of Current Rating

Trip Time for 40A Breaker
(from Spec Sheet)

10,800 RPM

60 A

150%

3.9-47 s

8500 RPM

70 A

175%

2.2-9.2 s

5500 RPM

90 A

225%

1-3 s
(interpolated)

In this particular situation, we are not going to be pulling that current for more than a second, at most 2 seconds, so we could choose any of the three gearings, but it'd be best to not choose the 1:60 speed. 1:80 gear ratios and above are your best bet in this situation.

• mass