Trig question

[Bogster]

Guys,

I am involved in a project were it is required to place a cylinder of 34inches in diameter through another cylinder which is 38inches in diameter. The 38" diameter cylinder is 2ft in length and sitting on its end.

Ok, sounds a stupid question but the problem is if the 38" cylinder is not sitting flat in the horizontal plane but is inclined then the effective ID that the 34" cylinder will see as it is lowered vertically into the 38" cylinder will be reduced. In effect the 38" cylinder opening becomes elipsoid in shape from a plan view.


This is pretty easy to solve if the 38" cylinder was a thin ring however the problem is compounded by the fact that the cylinder obviously has to openings at a distance (X) apart.

Can someone help by prescribing an equation which will tell me what reduction will occur in the effective diameter of a cylinder of (x) length for a given angle of rotation.

 

[astroclone]

Well, if you are lowering something vertically into the inclined cylinder wouldn't the effective diameter be
A = D X cos (angle)?
The other dimension (B) should stay the same, correct? With the B dimension axis moving by 2 ft X sin (angle)?
Maybe I'm misunderstanding, but that's my guess.
T

 

[LPPE]

Not to circumnavigate the question here, but when I have situations like that, it's easy to check your thoughts by drawing it in ACAD.

 

[desertfox]

Hi Bogster

if you multiply the Cosine of the angle of rotation of the
38" cylinder by 38" you will get the effective diameter that
the 34" cylinder will see.

ie:- 38" x Cos(x)=effective diameter.

hope this helps

regards

desertfox

 

[Ron]

Considering the length of the 34" cylinder, you need only to resolve that the RATE of change of the effective diameter of the 38 inch ID cylinder is compatible with the length of the 34" diameter cylinder. The rate of change should be constant, assuming there is no bending in the 38" diameter tube.

 

[Bogster]

Guys,

Thanks for the help. I was thinking along the same lines as Astroclone. I understood that if I multiplied the cosine of the angle by the diameter (38) I would get the change in diameter of one end of the cylinder. It was the opposite end that was causing my humble brain spin. I can see now that the reduction in diameter will be = 38-(38xCos (angle of rotation)+24XSin(angle of rotation)). So I think the greater the length of the 38" cylinder, the greater the reduction in the effective diameter. quickerThanks for the replies