How do concave disk--e.g. very shallow spherical bowl?

[NortonDommi]

Take a piece of rod of the length radius you want and make the ends into points. Make a center punch mark on your headstock and another on your cross-slide. Place the rod between the two and put some pressure on it. As you wind the cross-slide while maintaining pressure on the rod the tool will describe an arc the radius of the length of rod.
Very quick and easy for a concave surface. If you want a convex surface work off the backside of the cross-slide and a stop like the tailstock. I'll see if I can dreg up an old article showing this.

 

[NortonDommi]

I think all the photo's I took of doing this are on an old computer that is not hooked up Anyhoo on page 27 of the attached book,(Pg 36 in the PDF),you will see what I mean.
Sorry I mixed up concave and convex in the above. Anyway quick and easy. I forgot to mention center your tool before placing the radius rod. I make up rods as needed and keep them in a box. I've spot drilled mounting points on my headstock, both side of my cross-slide and my tailstock so no mucking around.

 

[Bill Kahn]

I think all the photo's I took of doing this are on an old computer that is not hooked up Anyhoo on page 27 of the attached book,(Pg 36 in the PDF),you will see what I mean.
Sorry I mixed up concave and convex in the above. Anyway quick and easy. I forgot to mention center your tool before placing the radius rod. I make up rods as needed and keep them in a box. I've spot drilled mounting points on my headstock, both side of my cross-slide and my tailstock so no mucking around.

Hi, OP here. Wow. What an amazing book. Looks like the advanced Object Oriented Programming book analog of the era--material for the genius experts. Well well beyond anything in my league. It continues to fascinate me that things that I have on my mind to make can at times be so very hard to do. (And other times, with a lucky piece of ingenuity, become doabale (well, on the 3rd or try at least)). -Bill

 

[Reddinr]

This doesn't answer your original question but how about looking online at one of the 7 or 8" magnifying make-up mirrors, take it apart, make a nice wood or metal base and glue it in? Not sure what concave radius those things have but it might do the trick. BTW, thanks for the Christmas gift idea! I just bought one of the Euler's disks for a relative.

 

[RJSakowski]

I have a commercial Euler's Disk.

is one of a zillion youtube videos if you have not seen it. Yes, have made alternative disks to the 3"X.5" steel one it comes with. (Same general effect, but slightly different sounds, timing, visuals, and ending accelerations) But now I want to make my own base--the one it comes with is not as strong/stable as I'd like to try (it is not rigid steel).

(If you spin a disk on a flat metal plate the disk tends to wander a-ways, and as these are pretty heavy and energized pieces of spinning metal when they come off the steel plate they knock stuff around and dent normal household items (don't ask))

So, I want to put a gentle concave curve (think a telescope mirror shape, but I don't need anything near that sort of precision, a section of a sphere, a very shallow bowl) into an 8" diameter (by .5") steel base disk. I can make the disk (I use a 2" diameter by 1" long stub screwed in from the back to grab with the three-jaw.) This concave bowl shape keeps the heavy spinning Euler's disk safely spinning freely in a constrained space.

I have measured the sagitta from a 6" ruler, and it being .082" indicates I need a radius of like 55". I have a PM1040V lathe (and PM25MV mill).

My question--any ideas how I make a steel part like this? Geometrically I do want the shape to be a section of a sphere.

Thanks for any pointers as to where this may have been discussed before too.

-Bill

It is well known that when the head of vertical mill is out of tram, the path cut will be concave rather than flat. The radius of the concavity depends on the deviation from vertical of the head and approaches infinite as the head approaches vertical.

To cut a spherical concave surface, the work would be mounted on a rotary table with the lowest point of the cutter arc intersecting the RT spindle axis. A fly cutter set for a diameter bigger than the disk diameter would be needed. Because the path of the cutter, the cut path will deviate from the spherical at the outside of the disk but will be a good approximation at the center of the disk.

For a 6" disk, it would take a fairly substantial fly cutter and I would suggest making a custom cutter using a disk as the basis rather than a typical fly cutter. This will provide for better balance of the tool. If you envision increasing the diameter of the typical fly cutter, you are fairly well there.

Here is an article on the process. https://authors.library.caltech.edu/46457/1/1.1752357.pdf

 

[Bill Kahn]

It is well known that when the head of vertical mill is out of tram, the path cut will be concave rather than flat. The radius of the concavity depends on the deviation from vertical of the head and approaches infinite as the head approaches vertical.

To cut a spherical concave surface, the work would be mounted on a rotary table with the lowest point of the cutter arc intersecting the RT spindle axis. A fly cutter set for a diameter bigger than the disk diameter would be needed. Because the path of the cutter, the cut path will deviate from the spherical at the outside of the disk but will be a good approximation at the center of the disk.

For a 6" disk, it would take a fairly substantial fly cutter and I would suggest making a custom cutter using a disk as the basis rather than a typical fly cutter. This will provide for better balance of the tool. If you envision increasing the diameter of the typical fly cutter, you are fairly well there.

Here is an article on the process. https://authors.library.caltech.edu/46457/1/1.1752357.pdf

I may give this a try--make up a 10" steel disk on the mill with the rotary table (maybe could fit on my 1040 lathe?). Bolt in (need to puzzle just how) a carbide insert, tilt the milling head (a pain to retram) or the rotary table a bit (doing the math provided, which sounds like fun). Mount (well) the 8" steel blank disk. Find some RPM that doesn't scare the bejeezus out of me. How hard could this be? I can just imagine that 10" disk spinning around right at sternum height. (I can generally stay out of the lathe work piece plane of rotation for stuff like this). Does sound like fun. Ahh, so many projects.

 

[P. Waller]

It is well known that when the head of vertical mill is out of tram, the path cut will be concave rather than flat. The radius of the concavity depends on the deviation from vertical of the head and approaches infinite as the head approaches vertical.

This creates a radius in one axis only, it will not produce a concave spherical shape.
This work may be done in a mill but will require either a tracer or 3 Axis control. It is far more easily accomplished on a lathe since it is a round part to begin with.

A lathe would require a simple attachment as noted above. This is a quick and dirty sketch to show the idea.

If your machine is not long enough this will require building an attachment point beyond the end of the bed which is easily done, this will allow turning any size radius within its range by changing the linkage length.
Short of CNC control the other alternative is to gear the X and Z axes together much like the set up used for the helical milling of gear teeth.

 

[RJSakowski]

This creates a radius in one axis only, it will not produce a concave spherical shape.
This work may be done in a mill but will require either a tracer or 3 Axis control. It is far more easily accomplished on a lathe since it is a round part to begin with.

A lathe would require a simple attachment as noted above. This is a quick and dirty sketch to show the idea.

If your machine is not long enough this will require building an attachment point beyond the end of the bed which is easily done, this will allow turning any size radius within its range by changing the linkage length.

That is why you would use the rotary table to rotate the part. The result will deviate from the desired spherical profile at the edges but the the center section would be fairly true to the spherical profile and that is where all of the disk spinning takes place.

As to using the lathe, you will see in post #2 above, what you are suggesting is what I had proposed. The big problem, here is that the desired radius is 55" which extends well past the OP's lathe bed. It makes for a very long cutting arm and will require additional support near the cutting bit. It isn't impossible; I have done similar things.

edit: looking closer at your drawing, I see that you are attaching the near pivot point to the compound. This would provide a more stable cutter presentation. The one comment that I would add is that the far side pivot would need to be offset from the spindle axis by the same amount that the near side pivot was offset from the cutter in the x direction.

Taking your concept one step further, it would be relatively easy to make a tool which would accomplish this task. A threaded hole for a shoulder bolt for the near side pivot, some means of feeding into the work in the z direction, adjustability in the x direction for the far side pivot come to mind.

 

[P. Waller]

The result will deviate from the It makes for a very long cutting arm and will require additional support near the cutting bit. It isn't impossible; I have done similar things.

It moves the carriage just as a taper attachment moves the cross slide so there is no need for any additional tool support.
This may also be done with a template that forces the carriage to move in relation to the cross slide, this is far easier assuming that you have a way to generate a 55" radius template of the desired accuracy.

 

[RJSakowski]

It moves the carriage just as a taper attachment moves the cross slide so there is no need for any additional tool support.

I got that and it is an obvious improvement over what I had originally suggested.