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Last time we developed torque equations for the driving and driven gears within a simple gear train. They are, T1 = D1 × F T2 = D2 × F where, T1 and T2 are the driving and driven gear torques, D1 and D2 are the driving and driven gear pitch radii, and F is the resultant Force vector, the common factor between the two equations.
Now we’ll combine these two equations relative to F to arrive at a single equation which equates the torques and pitch circle radii of the driving and driven gears in the gear train. This type of computation is commonly used to design gear trains to ensure they perform at a given level. As a first step we’ll use algebra to rearrange terms and place the two equations equal to F. First we’ll do it for the driving gear, dividing both sides of the equation by the pitch circle radius, D1. T1 ÷ D1 = D1 ÷ D1 × F T1 ÷ D1= 1 × F F = T1 ÷ D1 In a similar fashion, we’ll do it for the driven gear by dividing both sides of the equation by the pitch circle radius, D2. T2 = D2 × F → F = T2 ÷ D2 Since F is the common term between the two equations, we can set them up as equal to each other, F = T1 ÷ D1 = T2 ÷ D2 which means that, T1 ÷ D1 = T2 ÷ D2 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 T2 with respect to T1 and the ratio of D1 to D2, thus providing a mechanical advantage to the electric motor the gear train is attached to. _______________________________________ |
Posts Tagged ‘machinery design expert’
Equating Torques and Pitch Circle Radii Within a Gear Train
Thursday, May 29th, 2014
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, D1, and the driven gear Distance vector, D2. 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 F1 and F2 are aligned. As we learned last time, the Force vector, F1, results from the torque that’s created at the pivot point located at the center of the driving gear. This driving gear is mounted on a shaft that’s attached to an electric motor, the ultimate powering source behind the torque. F1 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 F2, a resisting force, which extends along the same line of action, but in a direction opposite to that of F1. These two Force vectors butt heads, pushing back against one another. F2 is essentially a negative force manifested by the dead weight of the mechanical load of the machinery components resting upon the shaft of the driving gear. Its unmoving inertia resists being put into motion. In order for the gears in the gear train to turn, F1 must be greater than F2, in other words, it must be great enough to overcome the resistance presented by F2. With the two Force vectors pushing against each other along the line of action, the angle ϴ between vectors F2 and D2, is the same as the angle ϴ between F1 and D1. Next time we’ll use the angular relationship between these four vectors to develop torque calculations for both gears in the gear train. _______________________________________ |
