biella handle on different axes

[aleset]

Good morning.
kindly, someone has a demonstration of the calculation of the race of a biella crank system where the guide is not located on the same axis of the center of rotation of the biella.
greetings

 

[antonio_sc]

Hi.
enough a little trigonometry: if you know the eccentricity "e", the length of the biella "b" and the length of the crank "m". the same way that, in the classic mechanism you have a race equal to (b+m)-(b-m)=2m, the race in the eccentric manovellism will be given by:

(m+b*cos)α)-(b)a-m)these two factors represent the position of the point d (beautiful foot) compared to (hand crank center) to the upper and lower dead point.
the corner au you can find it from the rectangle triangle (assigning the catetus "e" and the hypotenuse "b").
in any way I know that in literature we talk about eccentricity compared to the center of rotation of the crank (and not of the biella).

 

[meccanicamg]

It's called ordinary decentralized maneuver. is found enough on the books of rational mechanics and applied mechanics....googolando is located.

 

[aleset]

Thank you.

 

[aleset]

I would have another question, I'm thinking about it. You can relate the race of "d" with the angle that "m" forms with a horizontal line passing by "a". without using the alpha corner?

 

[antonio_sc]

being a cinematism to a degree of freedom, only one parameter is sufficient to identify its position. . .
once again he thinks about triangles. at the moment I can't make a pattern but, calling θ the crank angle, just note that:

e+m*sen(θ)=b*sen(au )

therefore au == sync[(e+m*sen(θ))/b]and consequently the position "s" of the biella foot will be given by:

s= m*cos(θ)+b*cos(au)..replace and get the relationship you are looking for (the race according to θ)

 

[LaboratorioDieci]

As for the position of the upper and lower dead points, it could also be calculated with the Pythagorean theorem, or am I mistaken?