Question

The following particulars relate to a symmetrical tangent cam operating a roller follower :-

Least radius = 36 mm, nose radius = 30 mm, roller radius = 21 mm, distance between cam 

shaft and nose centre = 28 mm, angle of action of cam = 150°, cam shaft speed = 750 r.p.m. 

Assuming that there is no dwell between ascent and descent, determine the lift of the valve 

and the acceleration of the follower at a point where straight flank merges into the circular 

nose.

Answer 1

$BT=OT-OB\\ BT=OP+PT-OT\\ =d+r_2-r_1\\ =25+3.2-16=12.2$

Flank radius, R

$R= \frac{r^2_1-r_2^2+d^2-2r_1 d cos \alpha}{2(r_1-r_2-d cos \alpha)}\\ R= \frac{16^2-3.2^2+25^2-2*1.6*25 d cos75}{2(16-3.2-25 cos 75)}\\ R=52.82 mm$

Flank angle $\phi$

$\frac{PO}{sin \phi}= \frac{PQ}{sin(180- \phi)}\\ sin \phi = \frac{sin(180- 75)}{52.82-3.2} \implies \phi = 29.6 ^0$

Acceleration at the end of the contact with the flank when $\theta = \phi=29.6^0$

$a=\omega ^2(R-r_1) cos \phi \\ a=(\frac{2 \pi *600}{60}) ^2 (52.82-16)*10^{-3} cos 29.6\\ a=129.39 m/s^2$

Reterdation at the beginning of the contact with the nose

$a=-\omega ^2(R-r_1) cos \phi \\ a=-(\frac{2 \pi *600}{60}) ^2 (25)*10^{-3} cos 29.6\\ a=-25.21 m/s^2$