## RE: [mill_drill] new article: Drilling a Hexagonal Hole

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• Rick, your eye seems to be pretty accurate as the 1/16 hole for the 1/4 hex looks to be about right based on the theory as well.  You also make a good point
Message 1 of 24 , Mar 7, 2013
• John, You give me too much credit. I chose 1/16” because that was the smallest drill in my fractional drill set and I had spares. I tried to use my CAD
Message 2 of 24 , Mar 7, 2013

John,

You give me too much credit. I chose 1/16” because that was the smallest drill in my fractional drill set and I had spares.

I tried to use my CAD program to help me see the general case for any size polygon but it quickly got very complicated. Clearly a 1/16” hole at each point is a good start. But beyond that, it gets very tricky.

Rick

From: mill_drill@yahoogroups.com [mailto:mill_drill@yahoogroups.com] On Behalf Of John
Sent: Thursday, March 07, 2013 3:46 PM
To: mill_drill@yahoogroups.com
Subject: RE: [mill_drill] new article: Drilling a Hexagonal Hole

 Rick, your eye seems to be pretty accurate as the 1/16" hole for the 1/4" hex looks to be about right based on the theory as well.  You also make a good point about keeping a wall between the smaller holes.  That should further help to complicate the math  ;-)  -John --- On Thu, 3/7/13, Rick Sparber wrote:From: Rick Sparber Subject: RE: [mill_drill] new article: Drilling a Hexagonal HoleTo: mill_drill@yahoogroups.comDate: Thursday, March 7, 2013, 5:04 AM John, I found the math daunting. For example, set the flat to flat distance at ½” and drill 1/16” holes at each point. Then drill the center ½” hole. You will see that the excess metal to be removed is large. By eye I can pick drill sizes and hole placement to get this excess as small as I like. But doing with a program is tricky. Of course, if you drill multiple small holes, you want to keep a web of around 0.01” between holes so the drill does not break through. The large central hole doesn’t care about the smaller holes both because it is so much larger and because the smaller holes will be equally distributed around the center so no net lateral force is present. Best of luck! Rick

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