Because of this I used the 1/4" extension to turn a 3/8" diameter spigot at each
end. This could then be held in a collet chuck being a very much safer method. I
did though remove the bulk of the metal, converting the rectangular shape into a
circle, whilst holding one end in the four jaw and supporting the other with the
tailstock centre, also at this stage producing the spigot at each end.
Photograph 1 shows the first end having been machined and from this you should be
able to see evidence of the tool being fed in at regular intervals along its length,
also the spigot at the tailstock end. I did though initially, partially reduce the
diameter in a few stages at the smaller end to avoid too deep a cut being required
to produce the final result. Having completed the first end the con rod was turned
end on end and the process repeated. The finishing stage, filing and emery paper,
took about 30 minutes but would have been substantially reduced had I had a much
finer file. Taking out the filling marks took about 25 minutes, in all though it
was an easy process. The overall system worked well as should be evident by the photographs
of the finished barrel, Photograph 2 and 3.
Machining the barrel is though only part of the task as calculating the changes in
diameter along its length can be a lengthy task if done purely manually, with a basic
calculator of course. A programmable calculator will make the process much quicker.
Considering first the manual method, the first calculation is to determine the radius
that produces the barrel form, Sk 2 showing the basis of the calculation. In this
R = the radius required with VD the distance between the smallest and largest diameters
along the barrel and HV the difference between the smaller radius and larger radius.
VD stands for vertical distance and HV horizontal variation.
Using Pythagoras's theorem we get R² = VD² + (R - HV)²
Requiring R this can be rearranged to give the formula indicated in the sketch.
Next stage is to calculate the tool infeed at each position, this shown for a single
position in Sk. 3. Again we use Pythagoras's theorem which gives
R² = IVD² + (R - IHV)²
See the sketch to see how this can be rearranged to calculate the required value,