As stated, the purpose of the power transmission is to reduce the speed of the mo-tor to some usable speed for the robot and to transmit the power to the wheels.
The speed of a robot is a function of the rotational speed of the wheels and the di-ameter of the wheels. Equation 1 shows this relationship, wherevis the velocity of the robot,Dis the diameter of the driven wheels, andNis the rotational speed of the wheel. So, to determine the required rotational speed of the wheel, Equation 1 is solved forN, which is shown in Equation 2.
If your robot has 10-inch-diameter wheels and the rotational speed of the robot is 300 RPM, the speed of the robot will be 9,425 inches per minute, or about 8.9 miles per hour (MPH). If you want your robot to move 20 MPH, this same wheel will have to spin at 673 RPM. This is one fast robot.
After you have an idea of the wheel speed you want, you need to determine how much of a speed reduction in the power transmission you will need to convert the motor speed to the wheel speed. This is done by using a combination of different sprocket diameters, pulley diameters, or gear diameters. The speed ratios of a gear train are just a ratio of the gear diameters.
Figure 6-1 shows a sketch of the same type of speed reduction. The leftmost sketch shows two gears in mesh, and the sketch on the right shows a belt/chain gear reduction. One thing to note here is that with the gear reduction, the direction of the driven gear is opposite of that of the driving gear. With the belt/chain sys-tem, the directions of both pulleys/sprockets are the same.
Grant Imahara and Deadblow (continued)
Fortunately, the Washburn family was nearby in the contestant stands. Shane Washburn, Grant explains, was a co-worker at ILM and he had fought against Grant with his botRed Scorpionin previous years. Moreover, Shane’s father, Ray, was a welder and hydraulics expert, and his brother, Jon, was an emergency medical technician. “They heard the air line rupture and were immediately at my side. While I was desperately trying to turn off the SCUBA tank, the Washburns and my crew were taking the screws out of the top cover. We wheeled my robot out of the way and theBattleBotspeople and Team Poor College Kids graciously allowed another match to go before us. This bought a little time, but not much. Ray ran all the way back to the pits and grabbed all the air fittings from my toolbox. We fixed it there on the spot in about five minutes—I couldn’t have done it without their help.”
Despite the catastrophic failure, Grant adds that he went on to beatKeggerwith just a single onboard air tank. But there’s a lesson in the story: “Always inspect all of your equipment for wear and damage, even if you don’t think you had any.”
6.1
6.2
Equation 3 shows how the speed of the output gear relates to the speed of the input gear.
In Equation 3,D1andN1are the diameter and rotational speed of the driving gear, andD2andN2are the diameter and rotational speed of the driven (output) gear.
WhenD1is greater thanD2, the output gear will spin faster than the driving gear;
whenD1is less thanD2, the output gear will spin slower (gear reduction) than the driving gear. When driving two shafts together, such as a front and rear axle being driven with only one motor, the gear/sprocket diameters between the two axles must be the same or the wheels will spin at different speeds.
If you have a 3000 RPM motor and you want a wheel speed of 300 RPM, you will have to reduce the speed of the motor by a factor of 10. By looking at equa-tion 3, you can see that the output gear,D2, will have to be 10 times bigger than the input gear,D1. This is a pretty big gear reduction with only two gears. If you were using a 1.5-inch-diameter gear on the motor shaft, you would have to use a 15-inch-diameter gear on the wheel. If the wheel is only 10 inches in diameter, the gear’s diameter will cause the gear to strike the ground, since it is larger than the wheel. When this type of situation occurs, three or more gears/pulleys/sprockets must be used together.
Figure 6-2 shows a more complex speed reduction.
Though the configuration shown in Figure 6-2 seems complicated, it can be simplified by looking at it as two separate two-gear systems. In this example, the speed of gear number 2 is the same as what is shown in Equation 3. The speed of gear number 4,N4, is first shown in Equation 4 that follows. It has the same exact form as what is seen in Equation 3. Since gears numbers 2 and 3 are physically at-tached to the same shaft, they will both spin at the same speed, which is shown in Equation 5. Because of this, you can substitute Equation 3 into Equation 4 to de termine the final speed of the output shaft. Equation 6 shows the speed reduction for
6.3 FIGURE 6-1
Simple speed reduction schematic
the gear reduction shown in Figure 6-2. TheD1/D2is the first gear reduction ratio, andD3/D4is the second gear reduction ratio.
In the previous example, you looked at a speed reduction of 10. With the dou-ble-speed reduction system, you have a lot of options for choosing gear diameters.
The product of the first and second stages in the speed reducer must be 10. For ex-ample, you can choose the first gear reduction to be 4 and the second gear reduction to be 2.5. In this case, you can use the same 1.5-inch-diameter gear on the motor shaft, and then the second gear should be 6 inches in diameter. This is smaller than the 10-inch-diameter wheels used in this example. The third gear could be a 2-inch-diameter gear, which would mean that the last gear should be 2.5 times larger or 5 inches in diameter. These gear sizes are much more manageable than trying to do this entire gear reduction in one step.
FIGURE 6-2 Schematic of a
double speed reduction system.
6.4
6.5
6.6
As a general rule, the greater the gear reduction, the more gears you will have to use to achieve the gear reduction. In the real world, you may not find the exact gear and sprocket diameters you want. This may be because the actual sizes do not exist. For example, if you are using sprockets instead of gears, it is rare to be able to find a sprocket that has a diameter 10 times greater than the driving sprocket.
You will usually have to choose components that are close to the values you want.
Thus, the speed reduction will be a little lower or higher than what you want.
Torque
The output torque is also a function of the gear ratios, but the torque and gear ratios have an inverse relationship. When the speed is reduced, the torque on the output shaft is increased. Conversely, when the speed is increased, the output torque is re-duced. Equation 7 shows the torque relationships from Figure 6-1. The direction in which the torque is being applied is identical to the rotational directions.
T1andD1are the torque and the diameter of gear 1, andT2andD2are the torque and diameter of gear 2. IfD2is greater thanD1, the output torque is increased.
From Figure 6-2, the output torque is shown in equation 8.
In the previous example, where we were looking for a 10-to-1 speed reduction, this will increase the output torque by a factor of 10.
During the robot design process, the power transmission must be considered at the same time while you’re selecting the motors. The number of gears, sprockets, and pulleys and their sizes can have a significant impact on the overall structural design of the robot. To simplify the overall power transmission design, you should choose a motor that has the lowest RPM so that the number of components in the power transmission (or speed reducer) can be minimized.
Force
The robot’s pushing force is a function of the robot’s wheel diameters and the out-put torque on the wheel, and the coefficient of friction between wheels and floor.
By definition, torque is equal to the force applied to some object multiplied by the distance between where the force is applied and the center of rotation. In the case of a gear, the torque is equal to the force being applied to the gear teeth multiplied by the radius of the gear. Equation 9 shows this relationship, where T is the torque,Fis the applied force, andris the distance from the center of rotation and
6.8 6.7
where the force is being exerted. Equation 10 shows how the force is related to the applied torque.
Using this relationship, you might think that your 500 in.-lb. torque motors and your 10-inch-diameter wheeled robot would have a pushing force of 100 pounds (100 pounds = 500 in.-ibs. / 5-inch radius). But this isn’t the case. Wheel friction becomes part of the equation. Without friction, powered wheels will never move a vehicle, and turning the vehicle would be virtually impossible. In most mechanical devices, friction is undesirable; but for wheels, friction is good. For combat ro-bots, the more friction you can get the better your robot can push. The frictional force to move an object across a horizontal floor is equal to the product of the co-efficient of friction between the floor and the object’s surface and the weight of the object. Equation 11 shows you how it works:
whereFfis the frictional force,µis the coefficient of friction, andFwis the weight of the object.
Figure 6-3 shows a schematic of the various forces acting on a wheel.Fwis the weight force acting on this wheel. For a really rough approximation, this value could be estimated by dividing the robot’s total weight by the number of its wheels. This applies only a rough estimate to the weight of a wheel, and it is true only if the robot’s center of gravity is at the geometrical center between the wheels.
Computer-aided design (CAD) software can help provide the actual values for the wheels, or they can be directly measured by putting a scale under each wheel.
6.11
FIGURE 6-3
Schematic showing reaction forces on
a wheel.
6.10 6.9
So how much can a robot push? The maximum pushing force will be equal to the sum of all the frictional forces,Ff, for all of the wheels. When the reaction forces of an immovable object, such as a wall or a bigger robot, exceeds the total frictional forces, your robot will stop moving—and, in this case, your robot could actually be pushed backward! By combining Equations 9 and 11, the torque re-quired to produce the maximum pushing force will be as shown in Equation 12.
For a robot with all identical wheels and motors that can deliver all the torque it could need, the total maximum pushing force,Fmax, will become the product of the weight of the robot and the coefficient of friction. Equation 13 shows this.
If the motor torque can produce a force greater than the frictional force, the wheels will spin. If the maximum torque of the motors cannot produce forces greater than the frictional forces, your robot’s motors will stall when you run up against another robot or a wall. In Chapter 4, you learned that stalling a motor is not a good idea, so it is a better idea to have the wheels spin rather than being stalled. Equation 14 shows the stall torque relationship for each wheel. This infor-mation can be used to help you determine the speed reduction in the power trans-mission and help you pick the right-sized motors. Equation 13 is a rather interesting equation. This maximum force is the maximum force your robot can exert, or it is the force another robot needs to exert on your robot to push it around. This force is a function of two things: weight of the robot and the coeffi-cient of friction between the robot’s wheels and the ground. So, this tells you that increasing your robot’s weight can give you a competitive advantage.
One of the difficult tasks in determining the pushing force is determining the coefficient of friction. The coefficient of friction between rubber and dry metal surfaces can range from 0.5 to 3.0. In your high school science classes, you probably learned that the coefficient of friction cannot be greater than 1.0. This is true for hard, solid objects; but with soft rubber materials, other physics are involved. It is not uncommon to find soft, gummy rubber that has coefficients of friction greater the 1.0, and some materials have a coefficient of friction as high as 3.0. For all practical purposes, the coefficient of friction for common rubber tires and steel surfaces is between 0.5 and 1.0.
The other factor that affects the coefficient of friction is how much dirt is on the surface. A dirty surface will reduce the overall coefficient of friction. This is why off-road tires have knobby treads to help improve the friction, or traction.
As a worse-case situation, assume that the coefficient of friction is equal to 1 and size all your components so you will not stall the motors in these conditions.
This will give most robots a small safety margin. If you want to be more conserva-tive, use a coefficient of friction greater than 1.
6.14 6.12
6.13
Location of the Locomotion Components
Most combat robots are fairly simple, internally. They consist of a power source, a set of batteries; motors for the wheels; a radio-controlled (R/C) system receiver;
controllers to take an R/C signal from the receiver and send power to the motors;
and a weapon system’s actuators, if they’re required in your design. Other compo-nents appear in various robot designs, such as microcontrollers to process incom-ing data to pulse width modulation signals, DC to DC converters, fans to cool controllers, and so on, but these are generally smaller and can be placed in tight-fitting places.
The location of the main drive motors is the most critical concern in the placement of large robot subsystems. Usually, these motors are quite large. The other large subsystems, such as batteries and controllers, can be located “wherever possible.”
Motors have to be close to the wheels, and their position and orientation is critical.
Quite often they are mounted in the lowest part of the robot. The motors must be positioned accurately, especially if a series of gears are used to transmit the power to the wheels, and the chains or gears need to be aligned in the same plane as the wheel system.
Mounting the Motors
Mounting of the motors in any application is important, but combat robots pres-ent another magnitude of problems for their motors. The motors are trying to wrench themselves out of their mounts from extreme torque conditions. At the same time, their mounts are being shaken so intensely that the mounting screws can be sheared in half. So you must design your robot to handle such extremes.
Quite often, a DC motor you might find in a surplus catalog has several threaded holes in the front face where the output shaft is located. Using these mounting holes for screwing the motor to a plate is okay for the types of applications for which the motor was originally designed, but using these holes may not suit an extreme situa-tion in combat robots. To determine whether these holes are suitable, you may need to subject the motor / mounting brackets to a shock test. The large inertial mass of the motor may just shear off the screws as you slam the assembly into your garage floor. Unfortunately, you might have to use an “easy-out” to remove the remaining portion of the screws. Use your judgment here.
You’re in a far better situation if your motors have a flange mount around the front face of the motor. If you need more strength, you can drill out the threaded holes and make larger holes for through-hole, high-strength bolts. A flanged base mount can be found on some older motors. Flange-based motors offer a higher strength method of mounting compared with the threaded face hole method.
Another method to use for mounting motors is to secure the face with the exist-ing mountexist-ing holes to a motor bracket you’ve fabricated, and then secure the back part of the motor with several high-strength clamps and a machined block in which to rest the motor. Use high-strength hose clamps that have a machined screw—not the “pot metal” types found in some hose clamps. This back clamping will prevent the heavy motor from moving. See Figure 6-4.
Thermal Considerations for the Motor
One of the drawbacks of using a higher-than-recommended voltage on a DC motor is the possibility of overheating. Even though combat matches generally last only a few minutes, intense heat built up in a motor can destroy it. This is not a power transmission issue, but it certainly is a mounting consideration. Some motors use a fan at one end to draw in air for cooling, but the intermittent action of the motor may mean that the motor is cooking in its built-up heat while it is off. You must also remember that the windings that heat up are in the armature, which is the ro-tating component that is isolated from the case, so heat sinks are not as effective as one might think. If the armature heats up too much, it can begin to disintegrate, slinging wire pieces all over the inside of the motor. If that happens, you’re in for a bad day.
How can you keep these motors cool? If you’ve run the motor on your bench while under load and you’ve noticed that the case gets extremely hot, you may want to mount it in a machined aluminum block to absorb and conduct the excess heat away from the motor. Some competitors have also used a small blower to force air through the motor to augment the fan. Have the fan run even when the motor is off to continue the cooling process as much as possible.
FIGURE 6-4 Clamping method to produce a secure motor mount.
Methods of Power Transmission
In previous chapters, several methods of interconnecting the motors with the wheels have been discussed. In direct-drive methods, the motor or gearmotor’s output shaft is connected directly to the wheels (see Figure 6-5).
Indirect-drive methods include a chain, belt, and even a series of flexible cou-plings. The following sections will discuss various chain and belt drive systems.
Indirect-drive methods include a chain, belt, and even a series of flexible cou-plings. The following sections will discuss various chain and belt drive systems.