Let’s say I have a length of steel rod. It is hexagonal- the type of stuff made into Allen wrenches.
Let’s say I want to take that rod stock and cut it into , oh say, 2 inch lengths and then pin or set-screw it into place in a nice metal knob.
The knob has a round hole on the underside, NOT a hexagonal or square hole.
Let us say, that I have a 1/4" hex piece. I need to find the right sized hole to put that 1/4" hex stock into. The piece has six sides- the 1/4" measure refers to the distance from the top horizontal to the bottom horizontal face.
How do I calculate this? ( I dunno if I need 1/4" stock yet, but it’s a convenient measure). What size hole in the knob do I order, if the hex is a 1/4 measure?
The distance between two horizontal flat sides is 0.25". Thus, the distance between two opposite corners in the hexagonal cross-section is 0.25" divided by the cosine of 30 degrees which equals 0.289". The diameter of your hole needs to be about 0.289" if you want a tight fit and if I understand the question correctly.
If I’m reading the problem right, the 1/4" is twice the length of the apothem. According to this page, the radius of a regular hexagon is equal to 2/sqrt(3) times the length of the apothem, so you’re looking at about 1/7" radius.
The diameter of the hole will be the distance across the flats divided by the cosine of 30 degrees (half the angle subtended by each face of the hexagon).
Cos(30) is pretty close to 0.866. So your hole needs to be 0.25/0.866 or around 0.2887" in diameter.
A letter ‘L’ drill is 0.29" - a trifle big. A 9/32 drill is 0.281" - a tad small. I’d say use a 9/32 drill and be prepared to tap the hex stock into place with a hammer.
I’m guessing this is meant for GQ (I’ve allerted a mod)
You want the right sized hole that a 1/4 in hex will fit into?
from the center of the hex to a side is 1/8 in
draw a right triangle, where the sides are (half of) one ede, and the two segments from teh center (to the edge and to the corner)
This is a 30-60-90 triangle
Using pencil/pen and paper, draw a hexagon. Draw a rectangle in the hex using the two hoizontal hex sides as two of the sides of the rectangle. Draw a diagonal line connecting opposite corners of the rectangle. The length of that diagonal line is the unknown you want to solve correct?
Anyway, the diagonal line divides the rectangle into two triangles of 30°, 60°, and 90° angles. The diagonal line is the hypotenuse of each of those triangles. The adjacent side (the one 30° from the hypotenuse) of each triangle is ¼" in length (0.25").
H = A / COS(θ) = 0.289"
where A = adjacent side = 0.25" and θ = 30°.
On preview, I concur with the others as well.
This reminds me of those tough days of kindergarden, when they’d make you figure thse problems out without paper. I sat around for hours trying to hammer square pegs into round holes, and the teachers never congradulated me for my successes.
Good lord, how’d that happen? Thank you so much, N9Iwp.
And, my thanks to those of you providing the formula. I’ll do something I so rarely do- print a hard copy of a Straight Dope thread so I can keep this handy. When I get the hex stock in, ( which is predicated by the wrench needed to tighten down a machine screw in an existing locking collar ), I’ll be able to figure out which knob to buy.
Youse guys are the best. I got a D in Geometry, and failed Algebra I twice. ( true story. I got a sympathy D on the third try to pass and graduate. ).
Mods, you can lock this one up since many of our math-gifted Dopers seem to be agreeing on this formula as the correct one to make use of.
If the radius is .144, how can the diameter be .289? Isn’t diameter exactly twice the radius ? Shouldn’t it have come out to .288? ( remember what I just said- I’m profoundly math challenged, but even I know that 144 times 2 is 288. )
I should also note this could be dteremined empirically.
Take your hex stoxk and put in ina drill.
“drill” into something soft like wax or soap.
You now have a correct sized hole.
If you want, you could use plaster to make a copy.