Posts Tagged ‘gear reduction’

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: R = nDriving ÷ nDriven             (1) R = NDriven ÷ NDriving             (2)       Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required: R = nDriving ÷ nDriven = 3600 RPM ÷ 1800 RPM = 2       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, R, we can use Formula (2) to determine how many teeth the driven gear must have to turn at the required 1800 RPM: R = 2 = NDriven ÷ NDriving 2 = NDriven ÷ 18 Teeth NDriven = 2 × 18 Teeth = 36 Teeth       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 torque, thus enabling small motors to do big jobs. _______________________________________

Gear Reduction

Wednesday, March 5th, 2014
 Last time we learned there are two formulas used to calculate gear ratio, R.   Today we’ll see how to use them to calculate a gear reduction between gears in a gear train, a strategy which enables us to reduce the speed of the driven gear in relation to the driving gear.       If you’ll recall from last time, our formulas to determine gear ratio are: R = NDriven ÷  NDriving            (1) R = nDriving ÷  nDriven            (2)       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, nDriving, of 3600 revolutions per minute (RPM).  Knowing these numerical values we are able to determine the driven gear speed, nDriven.       First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other: R = NDriven ÷ NDriving R = 23 Teeth ÷ 18 Teeth R = 1.27       In gear design nomenclature, the gear train is said to have a 1.27 to 1 ratio, commonly denoted as 1.27:1.   This means that for every tooth on the driving gear, there are 1.27 teeth on the driven gear.       Interestingly, the R’s in both equations (1) and (2) are identical, and in our situation is equal to 1.27, although it is arrived at by different means.   In Formula (1) R is derived from calculations involving the number of teeth present on each gear, while Formula (2)’s R is derived by knowing the rotational speeds of the gears.   Since R is the common link between the two formulas, we can use this commonality to create a link between them and insert the R value determined in one formula into the other.       Since we have already determined that the R value is 1.27 using Formula (1), we can replace the R in Formula (2) with this numerical value.   As an equation this looks like: R = 1.27 = nDriving ÷  nDriven       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, nDriving , is 3600 RPM.   We use basic algebra to calculate the driven gear speed, nDriven : 1.27 = 3600 RPM ÷ nDriven nDriven = 3600 RPM ÷  1.27 = 2834.65 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: nDriving – nDriven = 3600 RPM – 2834.65 RPM ≈ 765 RPM       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. _______________________________________