Announcement

Collapse
No announcement yet.

Ball bearing socket

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Ball bearing socket

    A friend of mine called me this morning with a machining question and I don't know how to answer his question.

    He needs to make a socket (a round bottom hole) in a steel plate that a 1" dia steel ball bearing will set into. The circumference of the top of this socket has to be .842 (+/- .005).

    Does anyone know how to figure out how deep to make this socket so that the dia at the top of the socket is at the required circumference? I don't have a clue how to figure this out, math never having been my strong point...
    The only way I would approach something like this is by trial and error, and even then I'm not sure how I would measure the circumference of the top of the hole accurately enough for what he needs.

    Any help would be appreciated. Thanks in advance
    Corm

  • #2
    0.01829" [ Approx ]
    .

    Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.



    Comment


    • #3
      Drill the 1" dia hole to a depth of .770" with a flat bottom--or a 0.5" spherical radius. That will give you exactly 0.842 at the point where the ball sets flush with the top of the hole. If the hole is going to have a 118 degree included angle in the bottom, let me know and I'll do a calc for that.
      Last edited by brian Rupnow; 12-15-2010, 03:38 PM.
      Brian Rupnow

      Comment


      • #4
        I say .871 for a round bottom.

        I calculated based on chord length and segment height.
        Cheers,
        Gary

        Comment


        • #5
          He's asking that the CIRCUMFERENCE is equal to 0.842" which makes the diameter equal to 0.2680".

          He may want what has been posted but that's NOT what he asked for.

          Edit.
          there are two possible depths, one below the centre line and one above, in Brians case he's quoted 0.780" but the same applies if the socket is shallow and below centre in which case it's 0.230 "

          I went for the shallow answer given that in Brians picture it's not possible to fit a ball.

          One of those not enough info to answer correctly.
          Last edited by John Stevenson; 12-15-2010, 03:52 PM.
          .

          Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.



          Comment


          • #6
            My revised answer is 0.129

            This is more of a dimple than a socket.
            Cheers,
            Gary

            Comment


            • #7
              Limitations

              Some seem to assume that the bottom of the hole will be a pretty accurate hemisphere.

              Calculations need to reflect the physical and practical limitations of the tools and machines.

              Comment


              • #8
                Sir John, you are correct, but I misstated what I needed. I need the 'dia' at the top of the socket to be .842, not the circumference as I mistakenly stated. My fault, please accept my apologies.

                gcude gave me the kind of answer I was looking for. Can you share the math formula you used to figure that out?

                Brian, I don't know how to interpret your response. You obviously put a lot of thought and effort into the response, and I am grateful for that, I just don't know how to read it. You start out by stating 'drill a 1" dia hole". The hole, or socket, that my friend needs has to be .842 in dia at the top of the socket, so how can he start by drilling a 1" dia hole? Please forgive me if I am missing something obvious.

                Thanks to everyone for your assistance.
                Corm

                Comment


                • #9
                  Pics please

                  I think a rough sketch scanned and posted to Photo-Bucket with the link to PB posted here would help considerably.

                  Comment


                  • #10
                    Originally posted by Corm
                    Sir John, you are correct, but I misstated what I needed. I need the 'dia' at the top of the socket to be .842, not the circumference as I mistakenly stated. My fault, please accept my apologies.

                    gcude gave me the kind of answer I was looking for. Can you share the math formula you used to figure that out?

                    Brian, I don't know how to interpret your response. You obviously put a lot of thought and effort into the response, and I am grateful for that, I just don't know how to read it. You start out by stating 'drill a 1" dia hole". The hole, or socket, that my friend needs has to be .842 in dia at the top of the socket, so how can he start by drilling a 1" dia hole? Please forgive me if I am missing something obvious.

                    Thanks to everyone for your assistance.
                    I used the formula for measuring the chords of a circle as shown in Tom Lipton's book "Metalworking Sink or Swim", Chapter 2, page 25:

                    D = (AB / C) + C

                    Where D is your 1" ball diameter, A and B are two equal halves of your hole diameter (I had used circumference as you first stated), and C is the segment height. I plugged in a few numbers in Excel and that's how I solved for the numbers I gave you.

                    I first took the .842 and divided by PI and got .26 for diameter (that we now know was incorrect), then I split that into equal halves for A and B (.13 * .13) to get .0169. Then solving for C, I arrived at .129 for my depth (second attempt).
                    Cheers,
                    Gary

                    Comment


                    • #11


                      .
                      .

                      Sir John , Earl of Bligeport & Sudspumpwater. MBE [ Motor Bike Engineer ] Nottingham England.



                      Comment


                      • #12
                        As one with little experience, this had me scratching my head too but if I understand correctly you want to be able to drop this 1" ball into a hole and have that given diameter protruding from the end (to me, like a roll on deodorant)...so I think the numbers Brian gave will remove as much material as possible before doing the final plunge with a ball nose mill...?
                        Then by controlling how deep that last plunge is, you may have to peck, you end up with what, I think, you are asking for...

                        All of this assumes that you don't want any of the 1" ball showing out the back i.e. the material dimensions allow for the ball to be completely concealed except for the .8xx showing out the "top". The part that had me scratching my head is having to insert the ball so the hole has to be that diameter plus ? to allow the ball to be inserted (that is the number I believe Brian was referring to when stating drill 1")

                        Comment


                        • #13
                          How about this?
                          Brian Rupnow

                          Comment


                          • #14
                            As Sir John shows, a picture is much better. Hopefully Tom appreciates the free ad for his book and allows the liberty of showing what value you get from it. Highly recommended.

                            Last edited by gcude; 12-15-2010, 04:33 PM.
                            Cheers,
                            Gary

                            Comment


                            • #15
                              It's this high reliance on formulae that keep people back from working it out for themselves. It's junior school trig. In fact you don't need trig at all. All you need is that old Greek chap, Pythagorus. He'll sort it all out for you.

                              What he won't do is shove a 1" ball through a .842 hole ! If that's the required arrangement, the ball will "set up" into the hole, not "set into" as per the OP's first post.

                              Anyway, John was right both times.
                              Richard

                              Comment

                              Working...
                              X