Question

5) A pair of standard spur involute gears has a module of 5 mm, pressure angle α=20。, center distance a=350 mm, transmission ration i12=9/5. Calculate the numbers of teeth (Z1 and Z2), reference diameters (d1 and d2), addendum diameters (da1 and da2), base diameters (db1 and db2), tooth thickness s , space width e, pressure angles on the addendum circles (αa1 and αa2) the radio of curvatures of tooth profile on the reference circles

 

(ρ1 and ρ2), and the radio of curvatures of tooth profile on the addendum circles (ρa1 and ρa2 ).

Answer 1

$a= \frac{d_1+d_2}{2}$

$a=m( \frac{Z_1+Z_2}{2})$

$\frac{Z_2}{Z_1}=\frac{9}{5} \implies Z_2 = 1.8 Z_1$

$350= 5 \frac{2.8 Z_1}{2} \implies Z_1 = 50; Z_2 = 90$

$d_1 = mZ_1 \implies d_1 = 250 mm$

$d_2 = mZ_2 \implies d_2 = 450 mm$

For 20⁰

Addendum diameter = 1 module = 5 mm

Radius of the addendum circle = r + addendum = $\frac{mZ_1}{2}+5=\frac{5*50}{2}+5=175 mm$

$d_4 = 350 \implies d_2 = [\frac{5*90}{2}+5]*2=230*2=460 mm$