Developer Anchubby

   When I develop the machine, I used to use some formulars in textbook which I studied at my university. I used the Castigliano's theorems so often and several cases were already solved so I just needed to apply it to my real case. But below case was not in my textbook, which I needed this case for my idle gear development so I made it by myself. M A B = − M ′ − F a c l M B C = − M ′ + F ( c − l − a ) M_{AB} = -M' - \frac{F a c}{l} \qquad M_{BC} = -M' + F(c - l - a) ∂ M A B ∂ M ′ = − 1 ∂ M B C ∂ M ′ = − 1 \frac{\partial M_{AB}}{\partial M'} = -1 \qquad \frac{\partial M_{BC}}{\partial M'} = -1 θ A = ∫ 0 l + a M ( ∂ M ∂ M ′ ) d c E I , (where M ′ = 0) \theta_A = \int_0^{l+a} M \left( \frac{\partial M}{\partial M'} \right) \frac{dc}{EI} \quad,\quad 2M' = 0 = ∫ 0 l − F a c l ( − 1 ) d c E I    +    ∫ l l + a F ( x − l − a ) ( − 1 ) d c E I = \int_0^l -\frac{F a c}{l} (-1) \frac{dc}{EI} \;+\; \int_l^{l+a} F(x - l - a)(-1)\frac{dc}{EI} ​ = F a E I ∫ 0 l x l d ...

 I worked in R&D for transmission development. I used to study by myself to prove any formular for better understanding. Sometimes I look back my documents and surprised to myself how I did this all. Below is to find the reason why profile shifted gear addendum and tip diameter is decided by center distance correction coefficient and profile shift coefficient of partner gear. I was wondering why some gear specs is decided by partner gear's specification so I calculated to see it. Addendum and Dedendum based on gear 1's reference circle is as below. Addendum = ( 1 + y − x 2 )   m (1 + y - x_2)\,m Dedendum = ( 1.25 − x 2 )   m Total depth = Addendum + Dedendum = ( 2.25 + y − ( x 1 + x 2 ) )   m The clearance between 2 gears are same as length of AB. A  ⁣ B = ( 1.25 − x 2 ) m − { ( 1 + y − x 2 ) m − y m } A\!B = (1.25 - x_2)m - \{(1 + y - x_2)m - y m\} = 0.25   m = \boxed{0.25\,m} ​ ​ This proves that even after profile shift is applied, to meet the clearance (0.25m) between...

For beginner member in R&D, I always used to recommend to prove some formulars by themselves. By doing so, they can remember it better and apply it to another area better.  Regarding gear in machine, pressure angle and correction coefficient of center distance can be obtained from the following calculation. Note that even when gear has profile shift, base diameter doesn't change. d ′  cos α ′  =  d b (base diameter) ​ d  cos α  =  d b (base diameter) ​ ​ \Rightarrow\ d' \cos \alpha' = d \cos \alpha ⇒    d ′  cos α ′  =  d  cos α cos ⁡ α ′ = d d ′  cos ⁡ α = cos ⁡ α ⋅ a 0 a a : actual center distance a0 : standard center distance where a 0 = r 1 + r 2 2  m a_0 = \frac{r_1 + r_2}{2} \text{ ?} Finally then I can get, ∴   α ′ = cos ⁡ − 1 ( cos ⁡ α 2 λ z 1 + z 2 + 1 ) \therefore\ \alpha' = \cos^{-1} \left( \frac{ \cos \alpha }{ \frac{2\lambda}{z_1 + z_2} + 1 } \right) This working gear pressure angle is us...