Author Topic: How to make a round ball mold (Read 4739 times)

hatchet · « **Reply #30 on:** January 02, 2009, 05:59:17 PM »

i made the above, a one inch ball mold to go with my dad's Christmas present which i also cast for him. i designed the trunnion caps in Solidworks, exported to MasterCAM, produced a pattern with the 3-Axis CNC, cast them and finished machined them to mount the cannon to the oak carriage. the wheels were made from a pattern, drilled tapped and mounted on 3/8" steel rod for axils. the trunnion caps and wheels were made from silica bronze at a later time than the barrel.

The barrel was cast and prior to casting was simulated in a finite element program designed for casting cooling analysis to avoid internal shrinks/voids.

The finished product one bronze gunmetal naval signal gun.

Victor3 · « **Reply #31 on:** January 02, 2009, 10:23:35 PM »

Quote from: hatchet on January 02, 2009, 05:52:03 PM

next take a full radius ball end mill bit and sink it to the radius of the ball you wish to create.

Yes, that is indeed the easiest way to do it on a conventional mill. Problem is, you need a fairly expensive ball endmill if you're going to do a larger pocket. The beauty of this method is the ability to use a simple flycutter or boring head to get any size cavity you want.

Again, using ratpatrol's method/formula of setting either the mill head or workpiece at 45 degrees, you are actually cutting an eliptical shape (observed at a plane cut through the center of the pocket), not a 'true' radius. This is why you have to adjust the diameter of the flycutter smaller, per his calculation. You are only using one section of the elipse formed by the cutter's swath to make your pocket, and at 45 degrees, the difference between the part of the elipse you're using to make the pocket and a true radius is negligible (for cannon ball purposes anyway).

Visualise it this way - Imagine you have a hemishere cut per ratpatrol's method - the diameter of a quarter let's say - milled into a flat-faced block of aluminum. Stick a quarter full depth into the exact C/L of the cavity so that the quarter is dividing it right in half. Fits pretty good, right?

Now - Pull your quarter out, rotate your block at 45 degrees and try to stick the quarter in on C/L of the cavity diameter at 45 degrees to the face of the block so that the OD of the coin matches the surface of the cavity. You can't do it, because the eliptical shape of the cavity will not match the true radius of the quarter.

This may sound counter-intuitive to some, but believe me it's true. The error at 45 degrees is slight, but the quarter will not fit perfectly.

However, a nickle should go in very closely at 45 degrees on C/L, and if you were to use a flycutter the diameter of a nickle using this method of machining the cavity (cutting 45 degrees to the face of the block), the resulting pocket would almost fit a quarter perfectly at the C/L. Why? Because the nickle-diameter cutter is cutting more material below the C/L of the pocket at the outside diameter at 45 degrees!

Note - edited to remove incorrect content. I was mixing apples & oranges (ball endmill vs rotary-table milled) when I originally wrote it.

Is it all clear as mud now?

kappullen · « **Reply #32 on:** January 06, 2009, 02:08:16 AM »

Quote from: GGaskill on January 02, 2009, 05:50:43 PM

Then repeat the process until you are to finish depth.

How much do you raise the knee if you are making a 2" ball?

One inch of course!

Kap

Victor3 · « **Reply #33 on:** January 06, 2009, 03:32:11 AM »

Damn...

I wish I could explain this thing better in text than I can see it in my head, but I can't. If my AutoCAD was working, it might help.

Let me muddle things up with some more words...

"The feature is formed by a radius (the cutter swath) tangent to the inside of a rotating cone set at 45 degrees to the plane formed by the radius, resulting in a hyperparabolic shaped thingamabob similar to a hemisphere that is good enough to cast cannon balls with."

I know, not much help. Where can I get some of them foam balls again?

Double D · « **Reply #34 on:** January 06, 2009, 04:25:35 AM »

Victor,

You explained how to do it and thats good enough. Let those who don't under stand go try it. It took me a day or two to wrap my brain around it.

The only other thing you could do if you feel the need to further explain is some step by step progressive pictures.

GGaskill · « **Reply #35 on:** January 06, 2009, 01:02:50 PM »

I think I understand the geometry of this finally and it is my opinion that it actually does make a hemispherical cavity, assuming accurate machining is done. The feed can be either along the spindle axis or along a perpendicular to the work piece. The reason I believe the cavity is actually hemispherical is that the finish position of the cutter is the same as a plane cutting the hemisphere, the intersection of which will be a circle, which is what the cutter cuts. So all cuts but the last are roughing cuts.

It looks like the cutter can be a center cutting end mill or a boring bar as well as a fly cutter. End mills will give a limited choice of diameters.

Now to clean off the Bridgeport.

Consider this as a cross section through the workpiece in a plane through the machine spindle that is perpendicular to the workpiece. The cutter is represented as the line D-A-B-E (if using an end mill.) The line 3 through the center of the cavity is the spindle center line. The line 1 (the one from the end of the line 3) is the center line of the end cutting ellipse. But the end is only an ellipse if viewed from off its axis; it is really circular. As the circle moves around the cavity, it has to produce a hemispherical cavity if located properly.

The 45° is critical; it probably should be set up by sweeping a sine bar. Also the diameter of the cutter is important; even endmills should be measured instead of assuming they are of nominal diameter. You can see where the .7071.. (sine of 45°) comes from. The triangle A-B-C is the one of interest. The cavity diameter (A-C) is the hypotenuse and either side (A-B or B-C) is the cutter diameter.

You can see from the drawing that the infeed distance is the diameter of the cutter (B-C) if feeding the spindle and the radius of the cavity (B-F) if feeding the workpiece.

Formulae: dia of cavity = dia of cutter / .707107 (.707107 is sin 45°)
dia of cutter = dia of cavity X .707107

dan610324 · « **Reply #36 on:** January 06, 2009, 01:43:05 PM »

please show pictures , I cant for my life understand it .
Ive read this post over and over again but still understand zero .
I must be extremely stupid

Victor3 · « **Reply #37 on:** January 07, 2009, 02:30:32 AM »

Quote from: GGaskill on January 06, 2009, 01:02:50 PM

The line 1 (the one from the end of the line 3) is the center line of the end cutting ellipse. But the end is only an ellipse if viewed from off its axis; it is really circular. As the circle moves around the cavity, it has to produce a hemispherical cavity if located properly.

Ahhhh.... But remember that to your mould block, the shape being introduced to it still an elipse because it is set at 45 degrees to the cutter's axis. Doesn't matter at what angle you happen to view the end of the cutter from; the material is still rotating about the elipse formed by the end of the tool at 45 degrees. The elipse is the only shape actually doing the finish machining.

When you do your test machining, you may look at the cavity and inspect it with some kind of a radius gauge and find that it looks dang near a true hemisphere. Put it on a CMM though, and the error (not much at 45 degrees) becomes clear.

About 20 years ago, this is how I found out that the shape is imperfect, and worked out why it is what it is.

The trig calculation only lets you "fake" a hemishere by creating an eliptical cutter swath close to the radius you want. You couldn't use a cutter smaller than the actual hemishpere to make it if this were not the case.

Victor3 · « **Reply #38 on:** January 07, 2009, 04:00:15 AM »

Okay. Here's one of those providential things that sometimes happens...

I went over to my old rollaway I used when I was a machinist, wondering if there was some helpful visual aid in there to photograph. Lo and behold, I see this formula that I posted there (on the orange card, never mind my wife feeding the horse) many years ago...

"R of cutter / R to be cut = Sine of tilt angle"

With this formula, you can use any diameter of cutter you have on hand to fake any radius you want by creating an eliptical cutter swath. I used it many a time to cut an elipse on a bar that had to fit onto the OD of a tube/pipe, or to form a socket to roughly fit a ball.

and7barton · « **Reply #39 on:** January 07, 2009, 04:17:09 AM »

What was the old method they used to make ball moulds - pre styrofoam, and NOT using any kind of machining ?
Is there some kind of low-tech method ?

GGaskill · « **Reply #40 on:** January 07, 2009, 07:22:00 AM »

I understand the "tilting the cutter to produce a larger radius" technique and, in fact, at one point did a derivation that calculated the deviation from a true circle at the edges or the center of the cut. But in this case, the cutter is actually perpendicular to the surface at the final cut (this is most obvious if you feed the spindle as it is always perpendicular to the finished surface but it is still true if you feed the workpiece, just only at the final position.)

If anyone has a CMM available, I would be interested to see a multi-point assessment of the hemispherical surface generated by this procedure and confirm one or the other theory.

seacoastartillery · « **Reply #41 on:** January 07, 2009, 08:18:37 AM »

George, After doing First Articles for Aircraft and Aerospace companies for 36 years, I suppose Mike and I could do one more. Send us your mold hemispheres and we will give them a thorough inspection, no charge. We have a CMM on the premises. We must know, however, what feature you wish us to establish as Datum -A- and also Datum -B-. Nice to know info would be the diameter of the sphere you expect. This is NOT necessary to proceed, however. It would be nice to find out if you have a sphere or an egg. AND, if an egg, how much of an egg? Some photos will be posted to explain the set-up and inspection process. Sorry, but George asked first, so his will be the ONLY freebe this year as we are very, very busy.

PM sent.

Regards,

Mike and Tracy

kappullen · « **Reply #42 on:** January 07, 2009, 09:48:26 AM »

For an illustration of this machining effect,

Take a truely round disc (for example a nickel) and drop it in a truely sherical

hole.

If this disc is .7071 percent of the hole's diameter, or smaller,

the disc will fit in the hole and touch all around its od.

At the .7071 % of the hole diameter, the disc will fill a 90 degree quadrent

of the hole.

Look at GGaskill's sketch and rotate it around the axis of line 2,

and you have generated a spherical form in your mind.

This hole is only true depending on the skill of the operator

in making the initial setup of cutter, angles, and arc of the cutter.

If the free cmm says you did it wrong, you did it wrong. The geometry works out right.

If you are off center with the tool, you will end up with a doughnut shaped form,

or tit in the center of the hole.

A boring head can be setup to a diameter by;

1) pick up the edge of the mold block with your edge finder.

2) adjust .100, or .25 (allow for half the edge finder) to center the spindle over the edge of the block.

3) move half the required diameter (.7071 X desired hole diameter /2) away from the block.

4) mount boring head, or flycutter, and adjust tool to just touch the block when
rotating the spindle by hand. The tool needs a sharp corner.

5) now set the head over 45 degrees, center the cutting tool, and spindle, over the rotary table and machine your mold.

Sounds like a hard job, but do it once, and the block will make you lots of balls.

You don't need a cmm to check your balls. Just a bore sized tube, three times the diameter of the ball.

If the ball does not roll thru, it's no good (NSSA Rule, not mine).

Kap

GGaskill · « **Reply #43 on:** January 07, 2009, 11:53:58 AM »

Send us your mold hemispheres and we will give them a thorough inspection, no charge.

I guess that means I have to clean off the mill table and setup the rotary.

seacoastartillery · « **Reply #44 on:** January 07, 2009, 02:29:18 PM »

Yep.

Victor3 · « **Reply #45 on:** January 08, 2009, 03:54:19 AM »

Quote from: GGaskill on January 07, 2009, 07:22:00 AM

...in this case, the cutter is actually perpendicular to the surface at the final cut (this is most obvious if you feed the spindle as it is always perpendicular to the finished surface but it is still true if you feed the workpiece, just only at the final position.)

Mr. Gaskill, I think you might be right. You're making a lot of sense here. Maybe I've been wrong on this one for two decades

It's quite possible that when a friend of mine (best jig bore man I've ever met) and I made a test piece years ago, the setup may have been flawed in some way, causing the error we found at inspection. We had measured error in "run-o-the-mill" parts before, and wanted to prove to ourselves one way or the other what was going on with this operation. We worked together to try to eliminate variables in the machine/setup/tool that might cause errors, but who knows if we were really able to do so? We may have duped ourselves.

Some setup variable that should be taken into account on a Bridgeport are:

1. The angle the head (if the head, not the rotary is to be set at 45) must be set exactly at 45 degrees in the X axis. Also, the head must remain at zero in the Y axis, so it's best to tram the head in on a sine plate (not just a bar set // to X) that has been set carefully at X, 45, Y, 0, so as to eliminate any machine error from 0 - 45 with the rotation of the head.

2. If you're going to be moving your quill during the setup and/or operation, the ram has to be set (via the turret) // to the Y axis..

3. Table sag, (caused by any offset weight) in the table, saddle and knee slides has to be controlled. If the slides are locked (clamped) when you do your tramming, they must be locked the same way when you take your finish cut on the workpiece.

4. The C/L of the spindle must be exactly on C/L of the rotary. This is tricky; We installed a tooling ball on C/L of the workpiece/rotary for this and indicated it in when we did our test piece.

5. Any radius on the tool nose (if using a flycutter) must be corrected for.

6. Depth of the finish cut from the face of the block must be precise.

Mr. Gaskill, I'm still not 100% convinced that you're correct. If you are, I'll just quote Yogi Berra and say...

"I never said most of the things I said."

Terry C. · « **Reply #46 on:** January 08, 2009, 07:54:10 AM »

Quote from: dan610324 on January 06, 2009, 01:43:05 PM

please show pictures , I cant for my life understand it .
Ive read this post over and over again but still understand zero .
I must be extremely stupid

Here are a couple of photos that might make things easier to visualize.

The workpiece, in this case represented by half of my 1.125" ball mold, is set at a 45° angle.

The outer edge of the penny represents the path of the flycutter. The penny is slightly undersized for this mold but it's close enough for this illustration. In practice, the cutter should reach from bottom center to outer edge.

The cutter is spinning as the workpiece is rotated. When the work has made a complete revolution, you have the hemispherical cavity.

Hope this helps.

dan610324 · « **Reply #47 on:** January 08, 2009, 09:00:02 AM »

HO HO HO HO HO Santa is here

no I'm not crazy ,
but this a little late Christmas photo from Santa Terry was enough for me to understand

dan610324 · « **Reply #48 on:** January 08, 2009, 09:20:36 AM »

Ive been looking and thinking a bit further ,
you must be able to use a boring endmill for this setup also .
if you find one with the right dimension for your decided cavity
or if you are a little fleible about the ball size .
a 1/2" endmill would give approximately an .75" ball mould
far from correct probably but I dont have any calculator here

Terry C. · « **Reply #49 on:** January 08, 2009, 09:34:00 AM »

You are correct. You could use an endmill, if you could find one that would produce the size cavity you need.

A .5" endmill would produce a .707" ball mold.

The method they're describing uses a flycutter, which is adjustable, so you can set the tool to trace the correct diameter that you need for your desired mold size.

Works just like the endmill, except there is only one 'flute' doing the cutting.

dan610324 · « **Reply #50 on:** January 08, 2009, 10:00:20 AM »

does anyone know what the shrinkage percent is on lead , zink and aluminum when cast ?
bronze has approximately 1,6 % in shrink if poured at 1050 degrees celsius .
could be interesting to know the others also .

GGaskill · « **Reply #51 on:** January 08, 2009, 11:04:08 AM »

I don't remember the exact shrinkage/expansion numbers for lead and aluminum but the cavity gets bigger as the mould heats up and the ball gets smaller as it cools down and they are roughly comparable.

I should have been more thorough in researching this. See this Wikipedia page for the numbers. The coefficient of thermal expansion for lead is 29 parts per million per °C, for aluminum is 23 parts per million per °C and for Zamak-3 is 27.4 parts per million per °C. Assuming the lead alloy moulding is the exact size of the cavity when the lead changes from liquid to solid (a true leap of faith since most lead alloys are not eutectic, meaning "the melting point is as low as possible, and that furthermore all the constituents crystallize simultaneously at this temperature from molten liquid solution"*), the moulding will be at the melting/solidification temperature of 327.46 °C* (if pure lead), and it will cool to room temperature (21°C) giving a temperature change of -306°C. Multiplied by .000029 gives .0088"/". The mould will not be as hot as the casting, so we have to guess the temperature for that or measure it with something, maybe an infrared thermometer. Just as an example, let us suppose that the mould will be 221°C; then the cavity will expand by 200 times 23 parts per million, .0046"/".

So the ball will be .46% oversize when hot but .88% smaller when cold giving a net of approximately .42% smaller than the cavity. Anyway this illustrates the math; actual melting temperature and mould temperature will give closer results.

*--Wikipedia

dan610324 · « **Reply #52 on:** January 08, 2009, 12:36:02 PM »

it could be good to know , because if its to much maybe it should been calculated when deciding what ballsize to use in a specific gun . not just make the mould to the correct size .
the windage could be unneccesary large maybe .