Problem #1

One minor oversight, probably just an assumption, but timing of the pattern is determined by one of the square lines used to find center. That's how you align the pattern to the block.

Too easy. Wait for the next one.
 
Bill Gruby said:
This is a fairly simply problem. You have a workpiece that measures 2"X4"X4". You need to drill 6 equally spaced holes on a 3" diameter bolt circle from the center of the workpiece. Sounds pretty simple so let's complicate it some.

Your machines, only one a Drill Press.

Your Tooling, only two, a Drill Bit and an XY Table that can be bolted to the Drill Press. Nothing else but you and your brain, and the print below. (Click on the pix of the print and it will enlarge for you.)

What I want to know is how far you have to move on the X Axis and the Y Axis to get to Hole location "C" from "0". "0" being the center of the workpiece. Your workpiece is aligned so Holes A&D will be on the Y Axis. I have givin you just enough information to start, you must find the rest.

Please tell us how you arrived at the numbers (formulas etc.) This should not give the Pros any trouble so give the Newbies a chance to figure it out please. Those of you who wish to try and solve this feel free to look anywhere you wish to come up with an answer. Most of all have fun with this. Back away if you get stuck and ask questions. Good Luck.

"Bill Gruby"
Click to expand...

Easy cheesy. I won't give it away, but a good hint is to remember your "unit circle".

6 holes also means 60 degree spacing. Holes A and D are on the Y axis (90 degrees from X). FBE and C are all 30 degrees from the X axis. So, their Y coordinates are all the same magnitude and differ only in sign....

SOH CAH TOA...

John
 
The least mind invasive way would be to find the center of the block and mark that with a prick punch. Set a pair
of dividers to 3 inches and scribe a circle. Step off your estimation of 1/6 the circumference (C=2Pi x R). Set your dividers to that setting (1.57) and step off and prick punch your marks.

Any amount you are over or under in your estimation of 1/6 the circumference, divide that error by 6 and add..or subtract...it from your previous divider setting.
 
Well, flutedchamber....not quite. Since we divide the circle into 6 triangles, we form 6 equilateral triangles, meaning the radius of the circle is also the side length, and being equilateral, the "base" of each triangle is also the radius. So, instead of 1/6 of the circumference, it is the same as the radius. So punch the center, draw your circle, punch one of the quadrant axis intersection with the circle points, and using the radius, scribe the intersections along the circle.

Probably most people use just use Pi(d) rather than 2Pi*r...same thing though.
 
By prick punch do you mean the drill bit remember we only have a drill press, a drill bit and an x/y table, oh and the print.
 
Tony Wells said:
Well, flutedchamber....not quite. Since we divide the circle into 6 triangles, we form 6 equilateral triangles, meaning the radius of the circle is also the side length, and being equilateral, the "base" of each triangle is also the radius. So, instead of 1/6 of the circumference, it is the same as the radius. So punch the center, draw your circle, punch one of the quadrant axis intersection with the circle points, and using the radius, scribe the intersections along the circle.

Probably most people use just use Pi(d) rather than 2Pi*r...same thing though.
Click to expand...

But the radius is the same as 1/6 the circumference, no?
 
No, it's 6x the radius. Remember, we are now working with 6 equilateral triangles. It's a hexagon, and the sum of the sides of an inscribed hexagon (the perimeter) must be less that the circumference of the inscribing circle. Give me a few minutes and I'll post a CAD screen shot. I have a cat in my lap, and typing this is bad enough! D:
 
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