American Civil War Game Club (ACWGC)

Basic math problem.

Line 125 foot long. Pivot on right end of line until a new line is facing 60 degrees from the orginal line's facing.

How far does the left end of the line have to move?

(I slept during math class. [:D])
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Finally found a protractor.

I drew a 6" line, then drew a 6" line at a 60 degree angle to one end, then measured from the two outside ends: it was 6 5/8". This distance is @ 1.1 times longer than the line.

I drew a 1" line, then drew a 1" line at a 60 degree angle to one end, then measured from the two outside ends: it was almost 1 1/8". This is @ 1.1 times longer than the line.

Would it be safe to assume that the 1.1 times ratio is a constant then, and that with a line 125' long its outside end will have to move 137.5' (125' x 1.1) to face the line 60 degrees from its original facing?

MajGen Al 'Ambushed' Amos
3rd "Amos' Ambushers" Bde, Cavalry Division, XX Corps, AoC
The Union Forever! Huzzah!

Diameter of a circle is 3.1415....*D if I remember correctly.

Hence, it travels 60 devided by 360 (degrees in a circle). Equals 1/6 th. So I beleive we would be moving 1/6 * 125 ft* Pi., or about 21*3.1415... or if you need to be more precise about 1/6*125*Pi.

Gotta go to sons soccer practice, but I think this is right. I will check a little more closely after soccer and teacher open house conference.

MG Michael Laabs
3/III A of M

Al,

Assuming I have understood your problem correctly you can just use the arc length formula. Namely, s = r@, where s is the the length of the arc made by the pivot, r is the distance from the center (125 ft) and @ is the angle made with the axis (converted to radians).

Therefore we have, s = 125 ft x (pi/3) = about 130.9 ft.

Your experiment was an empirical proof of this formula, however your instrumentation is not accurate enough to get a reliable result. (pi/3) is approximately equal to 1.0472..., a little bit smaller than 1.1, but a constant nonetheless.

Hope that helped you.

Lt. Col. Brad Slepetz
III Corps
AoG

Am not a whiz at math either but I believe Gen. Laabs is basically correct in his thinking but may have errored in his numbers. The 125 ft radius would actually be a 250 ft diameter for the circle giving you an arc distance of 130.8997 ft approximately but the straight line chord distance from start point to end point if you drew a straight line and not an arc would be 125 ft as with a 60 degree arc an equilateral triangle would be formed. I believe this is correct and hope this helps.
Your Most Humble Servant,
Captain William Shrum
3/1/1 AoM
Magnolia brigade
CSA

Captain Shrum,

You are right, General Laabs would have had it if he remembered the diameter is twice the radius.

Lt. Col. Brad Slepetz
III Corps
AoG

Thank you gentlemen. 1.04 is a constant and I will run with that. [;)]

Thank you all for your grey cell usage.

MajGen Al 'Ambushed' Amos
3rd "Amos' Ambushers" Bde, Cavalry Division, XX Corps, AoC
The Union Forever! Huzzah!

Ouch!!! Gotta get back to doing something besides hanging out with the almost one year old and his older siblings.....ouch!!! And with a good amount of training in Physics.....OUCH!!!! Shows that if you don't use it you tend to lose it---I assume this is true with things besides math, but I don't think I want to go there. Thank you, Lt. Col. Brad Slepetz, and Captain William Shrum--130 ft then--very good.
MG Michael Laabs
3/III A of M

<blockquote id="quote"><font size="3" face="book antiqua" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by slepsta</i>
<br />Captain Shrum,

You are right, General Laabs would have had it if he remembered the diameter is twice the radius.

Lt. Col. Brad Slepetz
III Corps
AoG

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

.

Don,

Thanks, but my project needs more precision than a mere 29 places out. [:p] hehehe....

MajGen Al 'Ambushed' Amos
3rd "Amos' Ambushers" Bde, Cavalry Division, XX Corps, AoC
The Union Forever! Huzzah!

<blockquote id="quote"><font size="3" face="book antiqua" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by dradams2</i>
<br />Hmm...well I'm a little late for this one, but as said, the approximate answer is 130.89969389957471826927680763665.

[8D]

Gen. Don Adams
5th Texas "Lone Star" Cavalry Brigade
I/III ANV
Cabinet Member
http://www.rootsandsaddles.com/index.htm

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Don,

If we want to get real technical we must include a discussion of significant digits. Because the length of the line is given as 125 ft, our answer can only honestly hold three digits (assuming Al meant for 60 degrees to be an exact number). Therefore 131 ft is just as good as the long irrational mess the calculator gives you.

Maybe Phil is right. I do need to get out more. [:D][8D]

Lt. Col. Brad Slepetz
III Corps
AoG

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