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Corm
12-15-2010, 04:11 PM
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

John Stevenson
12-15-2010, 04:30 PM
0.01829" [ Approx ]

brian Rupnow
12-15-2010, 04:32 PM
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.
http://i307.photobucket.com/albums/nn294/BrianRupnow/junk.jpg

gcude
12-15-2010, 04:36 PM
I say .871 for a round bottom.

I calculated based on chord length and segment height.

John Stevenson
12-15-2010, 04:45 PM
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.

gcude
12-15-2010, 04:53 PM
My revised answer is 0.129

This is more of a dimple than a socket.

oldtiffie
12-15-2010, 04:56 PM
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.

Corm
12-15-2010, 04:57 PM
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.

oldtiffie
12-15-2010, 05:02 PM
I think a rough sketch scanned and posted to Photo-Bucket with the link to PB posted here would help considerably.

gcude
12-15-2010, 05:16 PM
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).

John Stevenson
12-15-2010, 05:19 PM
http://www.stevenson-engineers.co.uk/files/socket.jpg

.

RussZHC
12-15-2010, 05:23 PM
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")

brian Rupnow
12-15-2010, 05:26 PM
How about this?
http://i307.photobucket.com/albums/nn294/BrianRupnow/junk-2.jpg

gcude
12-15-2010, 05:27 PM
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.

http://i330.photobucket.com/albums/l416/garycude/MeasureChords.jpg

rohart
12-15-2010, 05:39 PM
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.

Corm
12-15-2010, 06:42 PM
Both John and Brian have exactly what I was trying to ask for, and wasn't asking well. Brian's picture is just what my friend wants to end up with. Gcude, thanks for your help also.

I never had math training past basic geometry in school. I guessed that Trig would be needed for this.

Thanks to all of you for your help. If I can, I'll post a picture of his finished project.

Merry Christmas, and Happy New Year!

brian Rupnow
12-15-2010, 07:35 PM
I could have used trig to come up with youe answer, but 10 years of working with 3d cad which does it all for me, has made me lazy.

Corm
12-23-2010, 09:22 PM
I worked with my friend today to help him do some of the machining on those parts that needed the round bottom holes which you folks helped us figure out how to do. They came out exactly as needed. I took a couple of pictures to show you, but I can't figure out how to attach the pictures. The text belows comes from the FAQ -

"To attach a file to a new post, simply click the [Manage Attachments] button at the bottom of the post composition page, and locate the file that you want to attach from your local hard drive."

I can't find a [Manage Attachments] button on the page anywhere. I can find buttons that enable one to insert an image link or email link, but they need a URL link. My pictures are on my hard drive, so no URL.

What am I missing? I'm using a Laptop with Windows XP and IE Rel 8 as my browser. I've looked at my USER profile to see if there is an option there that I haven't turned on, but no luck there either.

Anyone have any suggestions?

rohart
12-23-2010, 09:40 PM
There's a sticky at the top of the list of threads that explains it.

But basically this site doesn't host pictures - it doesn't store them for you. You have to store them somewhere, and link to them in your text.

You store the pictures anywhere - Photobucket, in your storage area on your ISP's server, wherever you like.

Glad it worked out.

Corm
12-29-2010, 10:23 PM
Here's a pictures as promised. I hope it works OK, I'm new to Photo Bucket and posting pictures on this forum -

http://i1103.photobucket.com/albums/g476/cc7260/Dec2010pics049.jpg

Corm
12-29-2010, 10:27 PM
And one more -

http://i1103.photobucket.com/albums/g476/cc7260/Dec2010pics050.jpg

Thanks again to everyone for your help.

Paul Alciatore
12-30-2010, 12:54 AM
No trig is required for this calculation. Only the Pythagorean theorem and a bit of elementary algebra.

http://img.photobucket.com/albums/v55/EPAIII/BBHole.jpg