# Understanding Motor and Gearbox Design

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## Introduction: Understanding Motor and Gearbox Design

Why Spend Time Choosing the Right Motor and Gearbox?
Choosing the correct combination of a motor and a gearbox for a given application is very important, both in the FIRST Robotics Competition (FRC) and in actual engineering projects.  Without appropriate motor-gearbox combos, your team will find that your robot does not function as quickly and effectively as intended, and may have a tendency to burn out motors.

This tutorial will teach you the fundamentals of gearbox design and implementation.  First, I will teach you about motor characteristics.  Next, I will discuss how to choose a motor and gear ratio given application requirements.  I will then provide information about choosing a gearbox, followed by an overview of the motors and gearboxes available in FRC.  Finally, I will demonstrate how to use what you learn in this tutorial in an example problem and point out extra tools and resources if you want to learn more.

This tutorial was made through the Autodesk FIRST High School Intern program.

Prerequisites
A basic understanding of physics – e.g. force, torque, power, and gear systems
A willingness to learn

## Step 1: Motor Characteristics

There are several important characteristics of motors that provide information about a motor and its capabilities.  They are the motor’s output torque, its current draw, its output speed, its power, and its efficiency, each of which I will discuss in turn.  These characteristics are interdependent and can all be derived from four values: the motor’s stall torque, stall current, free current, and free speed.

Torque
A motor’s output torque is the amount of force with which its output shaft can rotate.  If too much torque is applied to a motor, its output shaft will stall, or stop turning.  Other motor characteristics are commonly written as a function of torque.  It is usually measured in N-m when metric units are required and oz-in when English units are required.

Current Draw
The motor’s current draw is the amount of electrical current the motor draws at any given load.  As the load on the motor (the torque) increases, the amount of current that the motor draws increases linearly.  This relationship can be written as
 (1)

 Symbol Name Units Description I Current Amps (A) The amount of current drawn by the motor Istall Stall current Amps (A) The amount of current drawn when the motor is stalled Ifree Free Current Amps (A) The amount of current drawn when the motor has no load placed upon it τstall Stall Torque Newton Meters (N-m) The amount of torque required to stall the motor τ Torque Newton Meters (N-m) The amount of torque applied to the motor output shaft

Speed
The motor’s output speed is the rotational velocity at which the output shaft spins.  As the load on the motor increases, the output speed decreases linearly.  This relationship can be written as
 (2)

 Symbol Name Units Description ω Speed Rounds per Minute (rpm) The rotational velocity of the motor’s output shaft ωfree Free Speed Rounds per Minute (rpm) The speed at which the motor spins when it has no load place upon it τstall Stall Torque Newton Meters (N-m) The amount of torque required to stall the motor, or prevent its output shaft from rotating τ Torque Newton Meters (N-m) The amount of torque applied to the motor output shaft

Photo Credit: http://www.engin.umich.edu/group/ctm/examples/motor/motor.html

## Step 2: Motor Characteristics (continued)

Power
A motor’s power is the rate at which the motor can do work.  It is essentially a measurement of how fast a motor can get a job done.  Its value in watts is given by the equation
 (3)

 Symbol Name Units Description P Power Watts (W) The amount of power supplied by the motor τ Torque Newton Meters (N-m) The amount of torque applied to the motor output shaft ω Speed Rounds per Minute (rpm) The rotational velocity of the motor’s output shaft

Efficiency
Motor efficiency is a measurement of how much of the electrical energy put into a motor is converted to mechanical energy.  Much of the remaining energy is converted into heat, which can cause a motor to burn out if it is operated at a torque/rpm where its efficiency is very low.  Efficiency is given by the equation
 (4)

 Symbol Name Units Description η Efficiency Percentage (%) The percentage of electrical energy input into the motor that is converted to useful mechanical energy Pout Power Output Watts (W) The motor’s output power at a given torque and speed Pin Power Input Watts (W) The amount of electrical power supplied to the motor I Current Amps (A) The amount of current drawn by the motor V Voltage Volts (V) The voltage at which the motor operates

Photo Credit: http://www.engin.umich.edu/group/ctm/examples/motor/motor.html

## Step 3: Motor Curves

A motor’s speed, current draw, power, and efficiency are often plotted against the output torque to make their values easier to visualize.  The equations for these curves are all derived from the four specifications discussed above using equations 1 through 4 of the previous few pages.
The graph on this page shows the motor curves for a CIM motor, one that is very common in FRC.

## Step 4: Choosing a Motor and Gear Ratio

Now that you understand the specifications that distinguish motors, you can work on choosing a motor and gear ratio for your application.  Which motor is most appropriate for a given job is entirely dependent on the application’s requirements.  This means that you must determine end results such as how big of a load are you moving and how fast do you want it to move, and then translate these into requirements such as output torque and speed.

Start by looking at the specifications of the available motors.  The Motor Spec Sheet for the 2012 FRC season is included on this page.  There are many factors to consider when choosing a motor and gear ratio, including:
• How gearing will affect the motor’s output torque and speed.  Usually, gears will be used to decrease speed and increase torque.
• Inefficiency in power transmission – each stage of gearing or chain run is approximately 90% efficient.
• Differences between theoretical and actual performance.  Because theoretical performance is usually better than actual performance, even after accounting for inefficiency, it is important to choose motors and gear ratios with a healthy safety factor.  That is, make sure that they will be able to handle more than the expected load at a faster than required speed.
• The amount of current that a single motor can draw is limited by the circuit breakers on the power distribution board.  When using 40 amp breaker, your current draw is limited to a maximum of 40 amps, meaning you should design the motors to draw less than 40 amps under the expected load.  In addition, the robot can draw a maximum of 120 amps at a time, as limited by the main circuit breaker.
• Running motors at or near stall load, the maximum amount of torque they can output, will cause them to burn up because much of the energy supplied to the motor will be turned into heat.  The amount of heat that a motor can handle is directly related to its total mass.  For this reason, heavy motors like CIMs are much less likely to burn out than smaller ones like the Fisher Price motors.
• If no single motor will fulfill your requirements, consider pairing motors.  When combining two motors, the output torque and current draw are additive, while the output speed does not change.  If two different motors are matched together, their free speeds must be matched through a gear reduction.  For example, combining a Fisher Price and CIM motor would require an extra 3:1 gear reduction for the Fisher Price motor because its output speed is approximately 3 times faster than a CIM’s.  If the output speeds are not matched, it will cause added resistance in the gearbox and negate any benefits of having multiple motors.

By accounting for all of these factors in your calculations when choosing a motor and gear ratio, you will ensure that your robot works as you intend the first time around.  The example problem at the end of this tutorial will demonstrate how to go through the process of making these calculations.

## Step 5: Available Motors

This section of the tutorial will describe some common use scenarios for the different motors allowed in the FIRST Robotics Competition.

 Motor Name Picture Notes RS-500 Series: AndyMark 9015 Fisher Price BaneBots RS-550 These three motors are all very similar – the only distinguishing factors are their performance specifications.  Generally, they are used in manipulators such as powering a pitching wheel, elevator, or conveyor/collector system. BaneBots RS-775 The RS-775 is a larger and more powerful version of the RS-500 series of motors.  It is also commonly used for manipulators.  However, the RS-775 motors have a history of developing case shorts, causing some teams to avoid using them. CIM The CIM is the largest, most powerful, and most reliable motor supplied to FRC teams.  Because teams are only allowed to use 4 of them, they should be allocated to the drivetrain, where their power and reliability are most needed. Denso (Window Motor) The Window Motor is a motor that comes with a worm gearbox attached.  Its high torque and low speed are often used in manipulators.  Due to the worm drive, they cannot be backdriven, a capability that is desirable for some applications. Vex 393 The Vex 393 was a new motor for the 2012 season.  For this reason, it has not seen much use.  However, its small size and relatively high torque make it well suited for secondary functions in manipulators.  As teams become more familiar with it, this motor will likely see wider use in more applications.

Photo Credits:
http://www.andymark.com/product-p/am-0316.htm
http://www.andymark.com/product-p/am-0912.htm
http://www.andymark.com/CIM-motor-FIRST-p/am-0255.htm
http://www.vexrobotics.com/products/accessories/motion/276-2177.html

## Step 6: Choosing a Gearbox

Now that you have a motor and gear ratio chosen, you need to choose a gearbox.  The first requirement for choosing a gearbox is that the chosen motor must fit on the gearbox.  Though most motors have unique bolt patterns, the BaneBots RS-550 motor, Fisher Price motors, and AndyMark 9015 motor all belong to the RS-500 series of motors and therefore have the same mounting pattern.

Next, the gearbox must have the gear ratio you have chosen.  However, there is more leeway in this requirement.  Some gearboxes can be “stacked” together, creating greater reductions.  In addition, not all reduction needs to happen in the gearbox and can instead be achieved through power transmission systems such as sprockets and chain.  It is also possible that the exact gear reduction that you want is not available, in which case close enough is usually good enough.

Finally, the gearbox must have an output shaft that you can use.  Though various sizes of keyed shafts are most common, hexagonal shafts are becoming more and more popular in FRC.  There are also many different hubs to accommodate the various styles of output shafts.  Ultimately, this is the least restrictive requirement when choosing a gearbox.

## Step 7: Available Gearboxes

This next section provides a list of commonly used gearboxes that are compatible with common motors.

 Motor Name Compatible Gearboxes RS-500 Series: AndyMark 9015 Fisher Price BaneBots RS-550 Banebots RS-500 PlanetaryAndyMark CIM-SimAndyMark 3 Stage ToughboxAndyMark PlanetaryAndyMark Double Doozy Planetary BaneBots RS-775 BaneBots RS-700 PlanetaryAndyMark PG71 Planetary CIM AndyMark ToughboxAndyMark Toughbox MiniAndyMark Toughbox NanoAndyMark 3 Stage ToughboxAndyMark CIMple BoxAndyMark ShifterAndyMark Super Shifter

Photo Credit: http://www.andymark.com/product-p/am-0114.htm

## Step 8: Using What You Learned

Now I will work through an example problem to demonstrate how to go through the process of designing a gearbox.  The drawing above shows a picture of a two stage elevator, an element of a manipulator commonly found in FRC.  The challenge is to design a gearbox that is capable of driving the 3 inch diameter winch and lifting the elevator to its maximum height of 84 inches high in a time of 1.5 seconds.  For the purpose of the problem, we will make two major simplifications: first, we will assume that the 18 pound load is applied for the entirety of the elevator’s travel, when in reality the winch must lift the weight of the first stage for only half of the distance.  Second, we will ignore acceleration and deceleration time, as these calculations are beyond the scope of this tutorial.

First we will convert all units to metric because metric units are much easier to work with.

Next we must turn our end goals into requirements that can be used to choose a motor and gear ratio.
Calculating the required rotational velocity of the winch:

Number of rotations to raise elevator:

Calculating the load on the winch:

## Step 9: Using What You Learned (continued)

Now we must choose a motor and gear ratio.  We’ll start by looking at the specifications of the available motors and make a guess about which motor may work well for the job.  We’ll try using a single BaneBots RS-550 as our starting point because of its high power, meaning it will be able to get the job done faster.  In addition, it is commonly used in applications such as this, meaning that it is probably a good fit for the job in general.  To make estimations easier, I made a motor curve graph for the RS-550.

First, we want to make sure that the motor won’t draw more than 40 A and blow a circuit breaker.  Looking at the graph, we can visually see that it takes a load of .23 Nm for the RS-550 to draw 40 A.  To ensure that the motor won’t reach this, even under heavy load, we will try designing for a current draw of 20 A.  Looking at the graph again, we see that this corresponds to a torque of .115 Nm.  Now, we can calculate the reduction we would need to achieve the necessary torque of 3.05 Nm.
Gear Reduction:
We have now chosen a gear reduction of 26:1, which means we can calculate the exact load our elevator motor should encounter.
Now, we can use equation (1) from “Motor Characteristics” to calculate the current we would expect the RS-550 to draw at this load:
Current Draw:
Our estimated current draw, 21.0 A, is well within our acceptable bound of 40 A.  Next, we will determine the rotational velocity of the gearbox output shaft using equation (2).  We will account for the 75% gearbox efficiency at this stage in the calculations.
Motor Speed:
Now we can check to see if our chosen gear ratio will allow us to achieve our desired output speed, 357rpm.
Gearbox Speed:
Finally, now that we have verified that our gear ratio satisfies our requirement, we can calculate how long it should take for the motor to raise the elevator.
Lift Time:
We have now completely verified that our RS-550 motor and 26:1 gearbox will achieve or exceed our original goals.  Because real world performance is often worse than the theoretical performance, it is wise to “overdesign” these systems.  Doing so also ensures that our simplifications do not cause our system to perform much worse than expected.

When you first go through this process, you may have to go through the calculations multiple times as you try different motors and gear ratios.  As you gain experience, you will gain an intuition of which motors and ratios will work well for a job.

The final step in this process is to choose a gearbox.  In this example, choosing the RS-550 version of Banebot’s P60 gearbox with a 26:1 reduction makes a lot of sense.  Not only is it compatible with our motor, but it also has the right gear reduction and a common .5 inch keyed output shaft.

Hopefully this example problem has helped you understand the process of choosing a motor and gearbox.  In addition, I hope that it has shown you how to properly apply the theory you learned earlier in this tutorial.

## Step 10: Reference Information

This section of the tutorial is meant to provide some additional resources for learning about motors and gearboxes, as well as some tools that can expedite the design process.  However, DO NOT use the tools in place of understanding the theory.  Instead, use them because you have verified them against your own calculations and because you understand how they work.

John V-Neun’s Design Calculator: This spreadsheet can significantly expedite the process of choosing a motor and gear ratio.  However, only use it once you understand the theory behind the calculations.

A FIRST Encounter with Physics: This lesson teaches some of the fundamental physics concepts encountered in FRC.  I looked at its section on motor and gearbox theory to ensure that I had all of my information right for this tutorial.  However, this tutorial goes into a bit more detail than its chapter on motors and gearboxes.

Photo Credit: http://aprettybook.com/2011/09/18/future-engineers/

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## 23 Discussions

what kind of gearbox will be suitable for motor allowable RPM range: 500-1500RPM. The output shaft having • RPM: 1 RPM +/- 5% • Torque: 20Nm +/- 1Nm (at 1 RPM) • Radial rotating force: Maximum 300N at tip • Movement: Bi-directional (open/close of valve) – 360O in both directions (no fixed stop) • Hysteresis:

Your data ( graph ) for the motor to lift your elevator makes no sense. The numbers quoted are too high: for example the stall current is 85A, but the stall current for a 550 motor is 8A and would burn out very quickly at that. I'm baffled.

How the Gear shifts without a Clutch Assembly.

Same way I can spin up a shift car..

this is sound crazy but i want to know if is possible chose a motor of 1 horse power and chose a gearbox to move or generate on the other side 27 horse power at 3600 rpm?

YES IT IS CRAZY.. You cannot create horse power.. One horse power cannot make even another horse power because of gear friction you will only be able to get 0.9HP !!

Let alone 27 times more..

am working over a lift and rotate equipment and i want to replace the existing manual hand wheel with a motor and gearbox for making ease and less effort to the operator.

The weight to be rotated is equal to 200 kg. so what is the HP of motor and the equivalent gearbox to be efficiently running in a smooth way.

How can i maje a solar powered tilling machine?

I am working on project of making small hydro generation unit. In which I have made hydro turbine of 'Pelton Wheel' type. It rotates at 300 rpm but as rated speed of synchronous generator which is to be connected is 1500 rpm, so to generate electricity I have to step up the speed from 300 rpm to 1500 rpm. For that purpose is it feasible to use gear type arrangement?? Which type of gear to be used?? How will it work?? Explain me in detail......

It depends on what type of shaft you're putting it on. Let's be practical.

You're probably going to use spur gears. We need a ratio of 5:1 (1500/300=5)

You can do 2x 2.236:1 or 1 5:1. What is the shaft size of your pelton wheel and what material is it? If it's metal or a shaped (like a square or hexagon beam) you can buy a spur gear (or make one if your pelton wheel is large) and as the main shaft spins you will spin your motor (what sized shaft?) 5 times as fast.

Things to take into consideration:
Shaft size helps determine what gear PITCH you use. (The size of the teeth). If you have a really tiny shaft you can't use a large pitch, for power transfer you want a large shaft and large teeth so your gears don't strip themselves.

Gear material is also determined, wood has the potential to expand and contract, plastic probably won't hold up to how much stress you are putting on your wheel. I would go with metal for you. The gears will be really expensive so honestly, I would go old school and use pressure treated wood to make huge gear teeth (like an old fashion water wheel) which is what I think you're using.

So all in all, you need a gear ratio of 5:1 (or close to it, will your wheel always be moving at 300 rpm? if the stream wanes you still want 1500 rpm! If the motor can handle more, think about changing the ratio!)

When I'm saying "huge gear" I'm imagining your wheel to be a foot or two (or more) diameter and you can cut the teeth using a pattern, then another gear. The reason larger is better is because there is less wear, but more importantly the larger the gear the less your human error impacts the gear. (No way you're going to machine a super tiny 188 tooth gear!, but make a 24 tooth gear that's big, you got that!)

If you pelton wheel is rinky dinky and the shaft is only 1 inch diameter or so you can just buy a gear or two from servocity or another gear provider. Make sure the gear pitches match (mod and pitch are the same thing so match 1 mod with 1 mod or 24dp with 24dp, etc.) Then make sure you can attach the pinion gear to your motor, (press fit, glue, welding).

Use search words like "spur gear" "bore diameter xx" "pitch xx, or xxP" "pinion gear"

i have a 44 ft long wooden boat that i made it by myself . the boat beam is 15 ft and hull depth 9 ft .Boat displacement is around 22000 lb . the boat is powered by two volvo penta AQAD40 marine diesel engines that produces each 165 hp and a max of 3000 rpm .. i need to install 2 gearboxes could you pls advise about the best ratio suitable for this boat

Hello, I'm having trouble viewing the images (formulas) when downloading this as a PDF. Is anyone else having trouble and if so, found a workaround? Thanks!

This is good. Would love to see a follow-up about actually building geartrains - best practices for making a "breadboard" with parallel plates, sources for gears, axles, bearings, etc.

I posted an Instructable about how I made a gearbox here - it's kind of the same topic, but doesn't go into quite the detail it sounds like you want.
It's definitely a good idea though - I'll put it on my list of ideas for future tutorials.

Great tutorial! Oh yes, I definitely would like to learn more about how to connect wheels to a motor and bearings as well :-) Good luck with your study!

What is the difference between torque and stall torque when dealing with an electric motor? As I recall, the highest torque rating of an electric motor is it's stall torque. I thought separating them out was more a function of usable power in an IC engine rather than an electric motor. Or is that formula generic, it's been a dozen plus years since I've even had to look at that stuff.

Also, are you going to get into the differences between DC and AC motors? I seem to recall there being a couple of differences when you start talking about what you want the motor to actually do. BTW, not to sound like a jerk, but I do mean differences beyond what source of electricity is handy.

To answer your second question, the reason why I did not discuss AC motors is twofold. First, this tutorial was meant for students who are part of the FIRST Robotics Competition, which only allows brushed DC motors, though the theory is definitely applicable to any project that uses DC motors. Second, I would have no idea what I was talking about if I tried to cover AC motors. I have never used them, so I might be giving unreliable information if I tried to teach about them.

If I understand your first question correctly, a brushed DC motor's output torque is not equal to its stall torque. Instead, it applies as much torque as is necessary to rotate the motor's output shaft. In other words, stall torque is a constant that is a characteristic of the motor, while torque is the amount of torque the motor is outputting at a given speed/current.

I'd been waiting for an instructable like this... For a long time I've wanted to attempt building a solar powered plant stand/turntable which would slowly rotate (about 180Â° in 24 hours) but I never knew where to start. Though it is still daunting, at least now I know where to begin!

I'm no expert but I think a servo might be better for something like that.