gas pipe & linear bearing dimensions relationship

[zoltan]

Could somebody help me with the relationship between the gas pipe diameter and the bearing size/mounting dimensions for linear bearing. I mean which the correlation between the diameter of pipe, bearing and size of angle profile. I would like to use other diameter of pipe that JGRO machine has and I do not know how to correlate in assembly. Please, help.

[spalm]

The length of the spacer should be:
((Diameter_of_Pipe - Thickness_of_Bearing) / 2)

This assumes that the bearings are like in JGRO's where a bearing surface is in the same plane as the other side of the aluminum angle.

I can draw a picture if you want more info.

Steve

[spalm]

Sorry, I just edited the last post to subract and not multiply. Ooops.

Example:
((1.315" Diameter_Pipe - .275" Bearing_Thickness) / 2) = .52"

Steve

[zoltan]

Steve,

Thank you. I would like to ask you to make a drawing if it is possible in order to avoid confusion. If I understood correctly this is the distance from roll bearing edge to the surface of AL angle, corresponding mounting face, isn't it? I have another question, too, what is the relation for length of the AL angle or the distance between the bearings and the diameter of pipe, if there is any. Thank you. I really appreciate it.

Zoltan

[spalm]

Zoltan, once you look at this as just triangles I think you will understand.

First the relationship of the sides of a 45-45-90 triangle is the square root of two (or 1.414). Look at the first drawing. The height and width can be obtained by multiplying or dividing by 1.414. 1.25 inches is the size of JGRO’s aluminum.

Now look at the second drawing with the bearings and a pipe. Notice that the bearings are placed on the same plane as the opposite side of the aluminum. We also want the center of the bearing to be the contact to the pipe, which is why it is placed half way up the diameter of the pipe (also called the radius). This relationship creates and upside-down triangle. The same 1.414 rules apply. (I just used a 1.588” circle because it was easy to draw.) Notice the height of the center point of the pipe is the diameter of the pipe divided by 1.414.

This is like a word problem in school. First thing you do is draw it out and simplify it. Don’t get confused with formulas.

Hope this helps,
Steve

[zoltan]

Thank you. I really appreciate your effort and time spent to help me. I believe that I was blind. I could not figure out how it works. Now, I see. I missed the detail of having the roll surface in the same plane with the opposite side of AL angle. Thank you again.

[spalm]

Cu multa placere. A fost placerea mea sa te ajut. Sper ca ti-a fost de folos. Citeodata nu poti vedea padurea din cauza copacilor. Mult noroc !