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And, THIS thread seems to be about the square mileage of a hex, not unit density, or artillery effectiveness.  Or, is there a missing first poster before Anthropoid?  Perhaps his request to know what the square mileage of hex was might have prompted some of these responses.  [8|]

From my perspective this thread should be about anything that is silly and geometrical.

OK 'poid.  It's time you an' me SQUARE off.  I know you played the TRIANGLE in the school band, but I was in a CIRCLE of players on the football field.  And what's your ANGLE anyway?  It seems like you're living in a PARALLEL universe in some unknown AREA.  Well, PLANE [sic] and simple, I'm going to lay you out PERPENDICULAR.

Geometry geeks.  [:D]

This is a discussion of the highest caliber

Actually, I think the discussion was on artillery of all shapes and sizes. Not just the big guns [:D]

ADDIT: @ Brigs: I _think_ you might have made a slight miscalculation. I did not get my skinfold's calipers out to check
for certain, but I think that the angle that you've depicted represented at MOB' (by which I am guessing you are showing
"maximal organizational banzai?") should in fact be more obtuse.

Your guess about the named angle is absolutely correct, Anthropoid. However, I do believe my calculations are precisely
accurate. What you probably missed is the fact that the obtusion factor is diminished somewhat due to the earth's
rotation. Don't worry, though, because that's a common mistake. Look at my chart again and you'll see exactly what I mean.

I'm no expert, but I find a skinfold caliper to be useful only for computing spread velocities in trajectory tables. Others have
questioned my advanced techniques, though, so I could be wrong about that.

ORIGINAL: jwilkerson

The distance between opposing SIDES being 20 nm not the distance between opposing vertices.

So the 12 embedded triangles have sides of 20 nm and ~11.5 nm with the hypotenus of ~23 nm ... so the length of the "sides" of the hexagon in question are ~23nm and the area is ~1380sqnm (6 equal sided triangles with areas of ~230sqnm).

I agree with this here for area. Something around 1385sqnm give or take. My error was figuring it point to point on the individual hexagon rather than center to center across two hexagons. Silly error on my part.

ORIGINAL: Braedonnal

ORIGINAL: jwilkerson

The distance between opposing SIDES being 20 nm not the distance between opposing vertices.

So the 12 embedded triangles have sides of 20 nm and ~11.5 nm with the hypotenus of ~23 nm ... so the length of the "sides" of the hexagon in question are ~23nm and the area is ~1380sqnm (6 equal sided triangles with areas of ~230sqnm).

I agree with this here for area. Something around 1385sqnm give or take. My error was figuring it point to point on the individual hexagon rather than center to center across two hexagons. Silly error on my part.

To Err is aviation, but to see is naval gazing.

ORIGINAL: 88l71

ORIGINAL: khyberbill

---And there is a nice green ball for the first to ask about knots per hour

As an ex-submariner, I prefer to think in furlongs per fortnight.

"My car gets 40 rods to the hogshead, and that's the way I like it!"

-Grandpa Simpson

DARNIT! I was hoping to get this quote in first. Curse you 88I71! [:'(]

So no real hope of this one taking over from "The THREAD!" as the longest running silliest thread on Matrix, eh?
 
Ah well, you can't blame a guy for trying.

ORIGINAL: Anthropoid

So no real hope of this one taking over from "The THREAD!" as the longest running silliest thread on Matrix, eh?

Ah well, you can't blame a guy for trying.

You give up too easily. What if the hexes were in parsecs? In that case how big would be the planet the map was representing a part of?

PS: Provide your answer in millimeters.

@witpqs: the answer is negative-infinity millimeters, for the simple reason that, a planet that large would be too massive and would soon implode on itself, causing a massive tidal wave and forming a black hole that would make Calcutta look like a spacious well-lighted sun room.
 
Honestly I miss the "Disregard" thread; that one just had a certain "je ne sais quoi"

Sorry Antropoid but the hypothetical planet is made of ultralightium and therefore will not shrink to size of a politician's integrity. The only thing massive is the calculation required.

ORIGINAL: USS America

ORIGINAL: Mike Scholl

ORIGINAL: bjmorgan

I just did the math and found that jwilkerson is correct. I got 1385.64 NM. I haven't used trig in 40 years. Since then sin and cosin were things you did while getting a loan.

AGAIN I ask..., who the bloody devil measures LAND in nautical miles? This thread began with a question about troop density per square mile (not knot) in China. So how about some math whiz crank up his brain box and give us some useful information in statute miles?

2000 yards in 1 nautical mile.

1760 yards in 1 statute mile.

You do the math.

Hey, an entire thread on nautical miles, and only one misuse of Knot! (And that marginal depending on what exactly Mike Scholl meant . And correct usage of parsec as well:). Funny how a until of velocity (knot) gets used as a length, and a until of length gets used as a time ... knots per parsec anyone :0

ORIGINAL: Kwik E Mart

are we talking African or European swallows here?

LMAO now that is my kind of math

ORIGINAL: HMSWarspite
Hey, an entire thread on nautical miles, and only one misuse of Knot! (And that marginal depending on what exactly Mike Scholl meant . And correct usage of parsec as well:). Funny how a until of velocity (knot) gets used as a length, and a until of length gets used as a time ... knots per parsec anyone :0

Oh no! _NOT_ knots! Please do not starting inserting knots into this discussion . . .



The subject is already thorny enough without you making it knotty on top of that. I just knew that Knot Insertion would get brought up.

Knot insertion is a fundamental algorithm in curve and surface design. Its goal is to add one or more knots to the knot vector without changing the shape of the curve.