This article will be about gears, one of the important types of machine elements. We’ll give the formulas needed for gear drawing and gear manufacturing. You can review the formulas for gear drawing and example gear drawings below.
For our article explaining gear terminology such as module, pitch circle diameter, and pitch given below, and the elements of gears, see: Gear Terminology
Formulas Needed for Gear Drawing
Spur gear drawing formulas
Formulas needed for bevel gear drawing
Bevel Gear Pair Drawing
Formulas Needed for Helical Gear Drawing
Helical gear calculation formulas and calculations
Bevel helical gear drawing formulas and calculations
Worm and Worm Gear Drawing Formulas
Gear Calculation Examples
Spur Gear Calculation Example
Example Problem: In a spur gear pair to be used in a steel structure, the module is 2, the small gear has 18 teeth, and the large gear has 54 teeth. The small gear’s speed is 1410 rpm. Find the dimensions of these gears and how many revolutions the small gear will turn.
Solution:
| Given Module Number of Teeth Speed | Gear 1 m= 2 Z1= 18 teeth n1 = 1410 rpm | Gear 2 m=2 Z2= 54 teeth |
Gear 1 Calculation:
Z0= 54 teeth
t = π . m = π . 2 = 6.28 mm.
D01 =m.z1 =2.18=36mm.
Da1 = D01 + 2m = 36+2.2 = 40 mm.
Df1 = D01-2.332 m = 36-2.332.2 = 31.336 mm. hb = m = 2 mm
ht = 1.166 m = 2.332 mm.
h = hb+ht – 2+2.332 = 4.332 mm
b = 10 m = 10.2 = 20 mm.
Gear 2 Calculation:
t = same as gear 1 (6.28 mm)
D02 = m.Z2 = 2.54 = 108 mm.
Da2 = D02 + 2m = 108 + 2.2 =112 mm.
Df2 = D02 – 2.332 m = 108 – 2.332.2 = 103.336 mm. The hb, ht, h, and b dimensions are the same as gear 1.
Helical Gear Calculation Example
In a helical gear pair mounted in a cast-iron gearbox, the normal module is 2.5, the small gear has 22 teeth, and the speed is 1200 rpm. Given that the gear pair’s ratio is 1/4 and the center distance is 140 mm, find the dimensions of the gears, the speed of the large gear, and the virtual number of teeth of the gears.
Solution:
| Given Normal Module Number of Teeth Speed | Gear 1 mn= 2.5 Z1= 22 teeth n1 = 1200 rpm i = 1/4, a = 142 mm | Gear 2 mn=2.5 |
Gear Calculation
D0 = ms. Z1 = 2.5818.22 = 56.8 mm
Da1 = D0 + 2mn= 56.8 + 2.2.5 = 61.8 mm
Df1 = D0 – 2.332 mn= 56.8 – 2.332.2.5 = 50.97 mm
tn= mn . π = 2.5 . 3.14 = 7.85 mm
ts=ms. π = 2.5818 . π = 8.1 mm
hb= mn= 2.5 mm
ht= 1.166 mn = 1.166.2.5 = 2.915 mm
h = hb+ht = 2.5 + 2.915 = 5.415 mm
b = 15. mn = 15 . 2.5 = 37.5 mm
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