Simple math question on a port location problem

[Metal Butcher]

Rather than using my best guess I need help from a knowledgeable designer.

I'm building a 50 percent larger version of Elmer's #2 Twin Vertical Wobbler. This may not really be relevant to my question other than it made some math checking necessary to upscale the jig I'll be using.

The question is about the original dimensions used on the drill jig to locate the single port holes on the cylinders and the two holes on the adjacent side of the Column.

The top end of the jig is pushed against a (1/8) .125 pin used in the crankshaft hole, and the center line distance on the jig to the outer contact edge is (13/64) .2031. This adds up to a crank center line distance of .2656.

On Elmer's drawing of the crank disc the center line distance between the crank shaft and the crank pin is shown as (1/4) .250.

Is this .0656 discrepancy just a minor glitch caused by the use of fractions,? or is this over lapping scenario a necessary part of normal port design on oscillating cylinder engines,? Or is there a flaw in my math?

Sorry, that was three questions! ;D

-MB

 

[mklotz]

First off, the discrepancy is 0.2656 - 0.25 = 0.0156 (1/64), not the 0.0656 you indicate in your post.

A straight line drawn from the center of the 1/8" crankshaft hole to the center of the 1/16" hole will not be *exactly* perpendicular to the edge of the jig so its length won't be *exactly* 13/64". Draw an enlarged diagram and you'll see what I mean.

However, if we assume it is a straight line then that would mean subtracting half a sixty-fourth (1/128") from the 13/64" jig dimension, making it 25/128". I think Elmer didn't want to work with fractions that small and simply kept everything in integral sixty-fourths. Given the nature of the wobbler design, such a small discrepancy if almost certainly moot.

Of course, Elmer could have used decimal inches but, for reasons unclear to me, he seemed to do most of his drawings in fractions. Perhaps he felt that practice would make the drawings less intimidating to the amateur.

 

[90LX_Notch]

.2656-.250=.0156 or rounding up .016. Not .0656 as you calculated.

 

[Metal Butcher]

Sorry guys, I did an incorrect edit after the posting and added the word "discrepancy".

Wow you guys are sharp and did not assume what I meant!

I understand the part about the line up of the jig to be below center and not absolutely accurate.

The answer were needed as a design parameter for my changes.

Thanks. -MB

Edit, I did it again! The scrap paper I did the math on shows A discrepancy of .0156 or about 15 thousand. This means I also made a typing error on my last sentence in the original post, it should have been .2656 with out the added word "discrepancy", not .0656, You see the similarity, whats a couple of hundred thousands.
I think my confusion is quite clear now. :big:

Maybe I should call it a day!