Gallery, Projects and General > Neat Stuff

Nut insert for repair

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bertie_bassett:

--- 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 ---


Could you not just measure the nut??

sparky961:
:poke: I'm sure I heard John go through that formula on the video.  I just couldn't understand him. :poke:

krv3000:
I think he added ten to it to get the factor of 7

doubleboost:
Sorry Lads
No magic formula I measured across the flats and just went with that
The flat bottomed hole it the important bit

I will have a play with different size nuts & post a bit video

Any further DRO installs will be made from steel
 John

vtsteam:

--- Quote from: bertie_bassett on April 13, 2015, 04:41:41 PM ---
--- 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 ---


Could you not just measure the nut??

--- End quote ---

 :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.

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