Question #301067

A reciprocating roller follower has cycloidal motion and its



stroke of 30 mm is completed in 90°of the cam rotation.



The follower is offset against the direction of rotation by



6.25 mm and radius of the roller is 12.5 mm. determine the



base circle radius which would limit the pressure angle to



30°.

1
Expert's answer
2022-02-23T04:38:02-0500

h=30mm,β=90,e=6.25mm,Rr=12.5mm,αmax=30,r1=?.h=30mm,β=90^∘ ,e=6.25mm,R r ​ =12.5mm,α max ​ =30^∘ ,r 1 ​ =?.

For cycloidal motion,


y=f(θ)=h[θ/β(1/2)sin(2πθ/β)].y=f(θ)=h[θ/β−(1/2)sin(2πθ/β)].


=30[2θ/(1/2)sin4θ].=30[2θ/−(1/2)sin4θ].


=(15/π)[4θsin4θ].=(15/π)[4θ−sin4θ].


df/dθ=(60/π)(1cos4θ).df/dθ=(60/π)(1cos4θ).


d2f/dθ2=(240/π)sin4θ.d^ 2 f/dθ^ 2 =(240/π)sin4θ.


tanαmax=d2f/dθ2/df/dθ=(240/π)sin4θ/(60/π)(1cos4θ).tan_{α max} ​ =d^ 2 f/dθ^ 2 /df/dθ=(240/π)sin4θ/(60/π)(1−cos4θ).


tan30=4cot2θ.tan30^ ∘ =4cot2θ.


θ=40.89.θ=40.89^ ∘ .


tanα=[df/dθe]/[f(θ)+(r2e2)0.5].tanα=[df/dθ−e]/[f(θ)+(r 2 −e 2 ) ^ {0.5} ].


=[(60/π)(1cos4θ)e]/[(15/π)(4θsin4θ)+(r21e2)0.5].=[(60/π)(1−cos4θ)−e]/[(15/π)(4θ−sin4θ)+(r 2 1 ​ −e 2 )^ {0.5} ].


r1=42.22mm.r_ 1 ​ =42.22mm.


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