Last time we developed torque equations for the driving and driven gears within a simple gear train. They are,
× F
Fwhere, T are the driving and driven gear torques, _{2}D and _{1}D are the driving and driven gear pitch radii, and _{2}F is the resultant Force vector, the common factor between the two equations. Now we’ll combine these two equations relative to As a first step we’ll use algebra to rearrange terms and place the two equations equal to
In a similar fashion, we’ll do it for the driven gear by dividing both sides of the equation by the pitch circle radius,
Since
which means that,
Next time we’ll see how to use this equation to manipulate our gear train so that it acts as a torque converter by increasing T and the ratio of _{1}D to _{1}D, thus providing a mechanical advantage to the electric motor the gear train is attached to._{2}_______________________________________ |

## Posts Tagged ‘gear train’

### Equating Torques and Pitch Circle Radii Within a Gear Train

Thursday, May 29th, 2014### Gear Train Torque Equations

Thursday, May 22nd, 2014
In our last blog we mathematically linked the driving and driven gear Force vectors to arrive at a single common vector D. We can then use this commonality to develop individual torque equations for both gears in the train._{2} In this illustration we clearly see that the Force vector, D. Let’s see why this angular relationship between them is crucial to the development of torque calculations._{2}First a review of the basic torque formula, presented in a previous blog,
By inserting F, and ϴ = 90º into this formula we arrive at the torque calculation, T, for the driving gear in our gear train:_{1 }
From a previous blog in this series we know that
By inserting F, and ϴ = 90º into the torque formula, we arrive at the torque calculation, T, for the driven gear:_{2 }
Next week we’ll combine these two equations relative to F, the common link between them, and obtain a single equation equating the torques and pitch circle radii of the driving and driven gears in the gear train. _______________________________________ |

### The Mathematical Link Between Gears in a Gear Train

Wednesday, May 14th, 2014
Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them. From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F, then use that linking equation to develop a separate torque formula for each of the gears in the train._{2} We learned in the previous blog in this series that F travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value _{2}ϴ with respect to their Distance vectors D and _{1}D This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation:_{2. }
where Fcancels out some of the positive force of _{2 }F._{1}Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train. _______________________________________ |

### Distance and Force Vectors of a Simple Gear Train

Monday, May 5th, 2014
Last time we examined how torque and force are created upon the driving gear within a simple gear train. Today we’ll see how they affect the driven gear. Looking at the gear train illustration above, we see that each gear has both distance and force vectors. We’ll call the driving gear Distance vector, D. Each of these Distance vectors extend from pivot points located at the centers of their respective gear shafts. From there they extend in opposite directions until they meet at the line of action, the imaginary line which represents the geometric path along which Force vectors _{2}F and _{1}F are aligned._{2} As we learned last time, the Force vector, F follows a path along the line of action until it meets with the driven gear teeth, where it then exerts its pushing force upon them. It’s met by Force vector _{1}F, a resisting force, which extends along the same line of action, but in a direction opposite to that of _{2}F. These two Force vectors butt heads, pushing back against one another._{1} F must be greater than _{1}F, in other words, it must be great enough to overcome the resistance presented by _{2}F._{2} With the two Force vectors pushing against each other along the line of action, the angle D, is the same as the angle _{2}ϴ between F. Next time we’ll use the angular relationship between these four vectors to develop torque calculations for both gears in the gear train._{1} and D_{1}_______________________________________ |

### Torque and Force

Tuesday, April 29th, 2014
We’ve been discussing torque and how it enables more power to be available to applications such as loosening tight nuts with a wrench. Now we’ll see how those same principles apply to another application, a simple gear train. To review, the torque formula is,
where, Referring to the gear train illustration above, we see that Force and Distance vectors are present, just as they had been in our previous wrench/nut example. But instead of torque being created by way of force that’s applied to a wrench, things are reversed, and it’s the torque that creates the force. You see, in the wrench/nut example, the force applied to the wrench handle created torque on the nut. In our present gear train example, the torque applied to the motor shaft is created by an electric motor exerting pressure upon the motor shaft, which in turn exerts a force upon the driving gear teeth. The driving gear is also attached to this shaft, so torque causes the driving gear to rotate along with the motor. This rotation results in a force being exerted at the point where the teeth of the driving gear mesh with the teeth of the driven gear. In other words, in the wrench/nut example force created torque, while in the present example torque creates a force. The gear train has a pivot point, as there was in our wrench/nut example, but this time it’s located at the center of the motor shaft rather than at the center of a nut. The pivot point in both examples is where the action takes place. The motor’s shaft and driving gear rotate around it, just as the wrench jaws and handle rotated around the nut’s pivot point. In both examples, the Distance vectors extend out from the pivot points to meet up with the Force vector’s path. In the gear train example, this Force vector path is called a You will note that there is an angle Next time we’ll examine the distance and force vectors in a simple gear train. _______________________________________ |

### Vectors, Sin(ϴ), and the Torque Formula

Wednesday, March 26th, 2014
Last time we introduced a physics concept known as and vectors,sin(ϴ).
In this formula, ϴ, or thay-tah, is a Greek symbol used to represent the angle present between the Force and Distance vectors as they interact to create torque. The value of sin(ϴ) depends upon the number of degrees in the angle ϴ. Sin(ϴ) can be found by measuring the angle ϴ, entering its value into a scientific calculator, and pressing the Sin button.We’ll dive into the math behind the vectors next time, when we return to our wrench and nut example and apply vector force quantities.
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### Achieving Mechanical Advantage Through Torque

Wednesday, March 19th, 2014
Last time we saw how gear train ratios allow us to change the speed of the driven gear relative to the driving gear. Today we’ll extend this concept further and see how gear trains are used to amplify the mechanical power output of small motors and in so doing create a torque.Below is an ordinary electric drill. Let’s see what’s inside its shell. There’s a whole lot of mechanical advantage at work here, giving the drill’s small motor the ability to perform big jobs. A motor and gear train are housed within the drill itself. The motor shaft is coupled to the chuck shaft via the gear train, and by extension, the drill bit. A It’s the drill’s gear train that provides the small motor with the mechanical advantage necessary for this hand-held power tool to perform the big job of cutting through a thick steel plate. If the gear train and its properly engineered gear ratio weren’t in place and the chuck’s shaft was connected directly to the motor shaft, the motor would be overwhelmed and would stall or become damaged. Either way, the work won’t get done. To understand how operations like these can be performed, we must first familiarize ourselves with the physics concept of The nut is fastened to the bolt with When you use your arm to manipulate the wrench you apply force, a force which is transmitted at a distance from the pivot point. This in turn creates a torque on the nut. In other words, torque is a function of the force acting upon the handle relative to its distance from the pivot point at the center of the nut. Torque can be increased by changing one or both of its acting factors, force and distance. We’ll see how next time when we examine the formula for torque and manipulate it so that a weak arm can loosen even the tightest nut. _______________________________________ |

### Gear Reduction Worked Backwards

Sunday, March 9th, 2014
Last time we saw how a gear reduction does as its name implies, reduces the speed of the driven gear with respect to the driving gear within a gear train. Today we’ll see how to work the problem in reverse, so to speak, by determining how many teeth a driven gear must have within a given gear train to operate at a particular desired revolutions per minute (RPM). For our example we’ll use a gear train whose driving gear has 18 teeth. It’s mounted on an alternating current (AC) motor turning at 3600 (RPM). The equipment it’s attached to requires a speed of 1800 RPM to operate correctly. What number of teeth must the driven gear have in order to pull this off? If you’ve identified this to be a word problem, you’re correct. Let’s first review the gear ratio formulas introduced in my previous two articles:
n (1)_{Driven }
N (2)_{Driving }Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required:
n3600 RPM ÷ 1800 RPM = 2_{Driven} = This equation tells us that to reduce the speed of the 3600 RPM motor to the required 1800 RPM, we need a gear train with a gear ratio of 2:1. Stated another way, for every two revolutions of the driving gear, we must have one revolution of the driven gear. Now that we know the required gear ratio,
N_{Driving}2 =
The driven gear requires 36 teeth to allow the gear train to operate equipment properly, that is to say, enable the gear train it’s attached to provide a speed reduction of 1800 RPM, down from the 3600 RPM that is being put out from the driving gear. But gear ratio isn’t just about changing speeds of the driven gear relative to the driving gear. Next time we’ll see how it works together with the concept of
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### Gear Reduction

Wednesday, March 5th, 2014
Last time we learned there are two formulas used to calculate gear ratio, If you’ll recall from last time, our formulas to determine gear ratio are:
N (1)_{Driving }
n (2)_{Driven }Now let’s apply them to this example gear train to see how a gear reduction works. Here we have a driven gear with 23 teeth, while the driving gear has 18. For our example the electric motor connected to the driving gear causes it to turn at a speed, n._{Driven}First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other:
N_{Driving}
In gear design nomenclature, the gear train is said to have a Interestingly, the Since we have already determined that the
n_{Driven} Now all we need is one more numerical value to solve Formula (2)’s equation. We know that the speed at which the driving gear is rotating, n:_{Driven }1.27 = 3600 RPM ÷
Based on our calculations, the driven gear is turning at a speed that is slower than the driving gear. To determine exactly how much slower we’ll calculate the difference between their speeds:
– n= 3600 RPM – 2834.65 RPM ≈ 765 RPM_{Driven} So in this gear reduction the driven gear turns approximately 765 RPM slower than the driving gear. Next time we’ll apply a gear reduction to a gear train and see how to arrive at a particular desired output speed. _______________________________________ |

### 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 The first formula for determining gear ratio is based on knowing the driving gear revolutions per minute (RPM), notated as n. Given that knowledge we can calculate the gear ratio, _{Driven}R, that exists between them by the formula:
÷ n_{Driven } (1) The other way to determine gear ratio, N. 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:_{Driven}
(2)N_{Driving } Equations (1) and (2) may look virtually identical, but they’re not. In mechanical engineering calculations, lower case Next time we’ll see how to manipulate these two equations so as to arrive at a particular gear ratio. _______________________________________ |