Gallery, Projects and General > Neat Stuff

Nut insert for repair

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RobWilson:
 :smart:


--- Quote from: vtsteam on April 13, 2015, 10:50:12 PM ---
 :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:


And then of course apply the technical term "slightly smaller than" to arrive at an exact value.

--- End quote ---


 :lol:  PRICELESS !   :lol:


Rob  :coffee:

RobWilson:

--- Quote from: mklotz on April 13, 2015, 11:15:54 AM ---To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

--- End quote ---





Just checked above with  my book " Math to build on a book for those who build "  close enough  ,but  the book did not cover the "slight smaller factor"  :coffee:


Rob

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