## Archive for February, 2014

### Gear Ratio Formulas

Sunday, February 23rd, 2014
 Last time we introduced a way to convert individual gear speeds in relation to one another within a gear train by employing a conversion tool known as the gear ratio.    Today we’ll introduce the gear ratio formulas, of which there are two types.       The first formula for determining gear ratio is based on knowing the driving gear revolutions per minute (RPM), notated as nDriving, and the driven gear RPM, nDriven.     Given that knowledge we can calculate the gear ratio, R, that exists between them by the formula: R = nDriving ÷ nDriven             (1)       The other way to determine gear ratio, R, is by knowing the number of teeth on both the driving gear, NDriving, and the driven gear, NDriven.    That’s right, it all boils down to simply counting the number of teeth on each gear.  In this instance the gear ratio is calculated by the following formula: R = NDriven ÷ NDriving             (2)       Equations (1) and (2) may look virtually identical, but they’re not.    In mechanical engineering calculations, lower case n is typically used to denote the RPM of rotating objects such as shafts, wheels, pulleys, and gears.    Upper case N is typically used to denote the number of teeth on a gear.       Next time we’ll see how to manipulate these two equations so as to arrive at a particular gear ratio. _______________________________________

### When Do You Need To Modify Gear Ratio?

Wednesday, February 19th, 2014
 Last time we saw how the involute profile of spur gear teeth ensures smooth contact between gears when they rotate.   Today we’ll see why it’s important to be able to change the rotational speed of the driven gear in relation to that of the driving gear by modifying their gear ratio, the speeds at which gears move relative to one another.       Why would we want to modify the rotational speeds of gears relative to one another?   One reason is to compensate for the fact that alternating electric current (AC) motors drive most modern machinery, and these motors operate at a fixed speed determined by the 60 cycles per second frequency of electricity provided by the utility power grids of North America.   By fixed speed I mean that the motor’s shaft revolves at a single, fixed rate.  It can’t run any faster or slower.   This is fine for some motorized applications, but not others.       Basic machinery such as wood cutting saws, grinders, and blowers function well within the parameters of the AC motor’s fixed speed, because their working parts are intended to rotate at the same rate as the motor’s shaft.   As a matter of fact, in this instance there’s often no need for a gear train, because the working parts can be connected directly to the motor’s shaft, and the machinery will be powered and function correctly.   There are many instances however in which a fixed speed does not match the speed required for more complex machinery to correctly perform precise, specialized tasks.       Take a machine tool meant to cut steel bars, for example.   It has a rotating part meant to cut through the steel during machining, and to properly do so its cutting tool bit must turn at 400 revolutions per minute (RPM).   If it turns any faster, the cut won’t be smooth and the tool bit will overheat and break due to increased friction.   If the AC motor driving the machine tool turns at 1750 RPM, a common speed for such motors, then the tool bit will be turning at a much faster rate than the desired 400 RPM, and this presents a problem.       To solve the problem we need only add a gear train between the motor and the part containing the tool bit, meaning, we must connect the gear train’s driving gear to the motor’s shaft and a driven gear to the part’s shaft.   But in order for this arrangement to work a conversion must take place, that is, we must design the gear train to operate at a specific gear ratio.   By gear ratio, I mean the speeds at which the two gears will rotate relative to one another.       Next time we’ll introduce the gear ratio formulas that make it all work. _______________________________________

### Spur Gears In Motion

Wednesday, February 12th, 2014
 Last time we learned about forces generated when spur gear teeth mesh and move along a specific line of action.   Today we’ll see them in movement.       Looking at the illustration below it might appear that there are three teeth in contact, but this isn’t the case.   As the gears rotate, only two teeth make contact at any given time, although the third tooth comes very close.   The actual point of contact between the teeth is represented by a black dot on the illustration below.   This is where two opposing forces, F1 and F2, meet.       Now let’s animate the illustration to see how the line of action remains constant the entire time the gear teeth are in motion.   By constant I mean that this imaginary line’s position and angle does not change relative to the gears throughout the course of their movement.       In the animation, the point of contact moves along the line of action as the gears turn.   Each tooth’s involute profile ensures that smooth contact is maintained along the faces and flanks of the gear teeth.   The involute profile’s unique shape facilitates opposing teeth remaining in constant contact along the line of action for the duration of their movement together.       If the gear tooth profile wasn’t involute in its shape, say for example it was square or triangular, the forces acting upon the meshed teeth during movement would vary in direction and intensity as a result of uneven contact between the teeth.   For example, consider the square shaped tooth profile in the gear train below.       As the gears rotate, the pointed tip of one tooth strikes the flat face of another.   As they  continue to turn, the two flat faces of the teeth slap together, then the pointed tip of one tooth will strike the flat face of the other tooth, and so forth.   The result is movement that is jerky and destructive.   There would be excessive vibration and wear and tear on the teeth, resulting in rapid gear tooth erosion and decreased efficiency overall.       Next time we’ll introduce the gear ratio, a formula which allows us to alter the rotational speed of the driven gear in relation to that of the driving gear, something which comes in handy when designing things that require this differential. _______________________________________

### Overcoming Inertia

Monday, February 3rd, 2014
 Inertia.   It’s the force that keeps us in bed after the alarm has rung.   It seems to have a life of its own, and today we’ll see how it comes into play in keeping other stationary objects at rest.       Last time we identified a specific point of contact between spur gear teeth in a gear train and introduced the opposing forces, F1 and F 2, generated there.   Today we’ll see what these forces represent, identifying one of them as inertia.       So where do these forces come from?   They’re forces generated by different means that converge at the same point of contact, the point at which gear teeth mesh.   They follow a very specific geometric path to meet there, an imaginary straight line referred to as the line of action.       F1 is always generated by a source of mechanical energy.   In our locomotive example introduced earlier in this blog series that source is an electric traction motor, upon which a driving gear is mounted.   When the motor is energized, a driving force F1 is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear.       Force F2 is not as straightforward to understand, because it’s not generated by a motor.   Instead, it’s the resisting force that the weight of a stationary object poses against its being moved from an at-rest position, known as inertia.   The heavier the object, the more inertia it presents with.   Trains, of course, are extremely heavy, and to get them to move a great deal of inertia must be overcome.   Inertia is also a factor in attempting to stop objects already in motion.       To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle.    At their point of contact, the driving force of the motor, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2.   If F1 is less than or equal to F2, then the train won’t leave the station.       Next week we’ll animate our static image and watch the interplay between gear teeth, taking note of the line of action during their movement. _______________________________________