A simple math problem

[Orrex]

A length of wire 1/16" in diameter is wrapped exactly once around a rod 7/32" in diameter.

How long is the 1/16" wire?

[Dangerously Amused]

1. There is no such thing as a "simple" math problem, friend.

[hunter]

2. You silly person. In mathland wire doesn't bend.

You could poke your eye out or something.

[Ptah]

3. it is

.785" long.


oops, it is .8836" as InternalDialogue says below.

:blush:

[Tuesday Afternoon]

4. uhm....

7/32"

am I missing something?

[Orsino]

6. Perhaps the thickness of the wire.

Assuming that the single loop of wire is perpendicular to the length of the rod, we might use the radius of the wire's center, rather than the inside where it touches the rod.

pi/4 inches.

[Tuesday Afternoon]

8. you know

Higher Math is like God to me. I have to take it on faith. :D

[leeroysphitz]

7. Once the wire is wrapped around the rod it becomes a hoop with an inner diameter, and outer diamter.

If you wanted to cut the wire to wrap around the rod one time with no gaps and without beveling the ends would you use the inner diameter (.2187" or 7/32") or the Outer diamter (.3438") or split the difference at .2813"?

[leeroysphitz]

68. Oops. Converted everything to Metric.

My Bad.

[Rabrrrrrr]

29. Oh my God, yes, you are missing something.

[Tuesday Afternoon]

55. sounds like I am missing just about

a cock hair worth of wire depending on the ethnicity of said cock.

[InternalDialogue]

5. .8836"

[Ptah]

13. After more thought, you are correct.

:thumbsup:

[InternalDialogue]

15. When Ptah says I'm correct, I know I did something right.

It'll probably be the only time I'm slightly quicker than you on something like that. It's a badge of honor.

[Orrex]

18. That was the answer that I got, but I wasn't sure about my method

Thanks for confirming!

[Tuesday Afternoon]

59. won't points be deducted for not giving the answer in fraction

form? :hide:

[Ptah]

60. That is expressed in fractions

.8836= 8836/10,000








:hide:
















still :hide: ing

[Tuesday Afternoon]

61. won't that reduce?

:yoiks:

[Ptah]

63. No advantage to be gained by reducing.

The simplicity of metric, yet still inches.

[Tuesday Afternoon]

64. good to know...

not sure why :D still knowledge is its own reward :fistbump:

[Rabrrrrrr]

69. No, but we will take off points for too many significant digits.

[Tuesday Afternoon]

70. Is that why Sam Neill did that to Holly Hunter in The Piano?

He thought she had too many significant digits.

[PRETZEL]

9. just a little over 3/4th of an inch?

that's my guess anyway

[Shell Beau]

10. Would it not still be 1/16"?

;)

[InternalDialogue]

11. Is it wrapped exactly perpendicular to the length of the rod?

Cause if not, I need to change my answer but would have no idea to what.

[Orrex]

23. Let's say yes

[KitchenWitch]

12. Approximately 1.2 inches

This shit would be so much easier to calculate if you used metric!

[PRETZEL]

14. Is the formula

2(pi)(radius)squared?

If so, then I'll have to change my answer to yours.

[Bertha Venation]

24. Oh, this country showed how obtuse it was when it gave up

on the metric system when we were kids! "We don't need to know anyone else's methods; ours work just fine for us and screw the rest of the world."

:hi:

[Ptah]

26. hm

A length of wire 1.5875mm in diameter is wrapped exactly once around a rod 5.55625mm in diameter.

How long is the 1.5875mm wire?

[hunter]

34. Since it's expressed in fractions, calculate in fractions...

...using 22/7 for pi, it's close enough. Slide rule stuff.

When I was in high school they still taught slide rules and tables. Calculators were expensive and rare.

[Ptah]

47. Yes, three significant digits.

A length of wire 1.59mm in diameter is wrapped exactly once around a rod 5.57mm in diameter.

How long is the 1.59mm wire?

Length of wire = 3.14 * (1.59 + 5.57)

Length of wire = 3.14 * 7.17

Length of wire = 22.5 mm

[madinmaryland]

16. Try this:

To find the circumference=2(pi)*r

The radius is 7/64 (rod) + 2/64 (wire) = 9/64"

C=pi*9/64 = 0.884"

[Orrex]

19. That's kind of how I did it

Thanks for proving another confirmation!

[madinmaryland]

22. My daughter is starting equations (5th grade) and I am relearning all that stuff.

Funny, but as an engineer, I took two years of advanced calculus and differential equations in college, but remember very little of it!

[Jamastiene]

66. same thing happened to me.

Engineering...Calculus...do what? I can't remember anything now except the blood, sweat, and tears(when I lost my 4.0 GPA). Funny, the math wasn't the reason for that either.

[Ptah]

20. pi * 9/64 = 0.4418

[madinmaryland]

21. 0.44188 * 2 = .8836

Radius = 2(pi)r

[Generic Brad]

52. Mmmmmm. Pi.

[billyskank]

17. Add the diameter of the wire and the rod, and multiply by pi.

That's your answer.

[Ptah]

27. Correct.

:thumbsup:

[Orrex]

37. I thought it was the diameter of the rod and half the diameter of the wire?

:shrug:

[Ptah]

40. billyskank is correct.

If you add half of the wire to the diameter of the rod, you are adding radius (apples) and diameter (oranges), I think.

[Orrex]

42. Ah--now I see! The apples/oranges thing makes sense to me, for once

I got as far as seeing that the length is the circumference of the circle running through the center of the wire. But I dropped the ball when I tried to derive that figure from the rod's diameter plus the distance from the rod's surface to the wire's center.

As you say, that's apples/oranges; the actual diameter of the circle running through the center of the wire is half of the thickness of the wire plus the diameter of the rod plus half of the thickness of the wire again. Or the diameter of the wire, of course...

[Ptah]

44. As you can see, that was my mistake re: #8

http://www.democraticunderground.com/discuss/duboard.php?az=show_mesg&forum=105&topic_id=8331494&mesg_id=8331566

:blush:

[SidneyCarton]

39. oops! that was my mistake.

[hunter]

71. (2/32 + 7/32) * 101/32

Fractions are our friends.

101/32 is close enough to pi.

9/32 * 101/32 = 909/1024 which is a little bit over 28/32 and reduces to 7/8

In binary math it's really fast and pretty, just shifts and additions.

binary Pi approximations:

101/32, 201/64

3217/1024 is a sweet one used in all sorts of 16 bit applications.

[pokerfan]

25. Unknown

There could be wire left over...

[Orrex]

35. Ahem. Looks like somebody didn't read the OP

:spank:

[pokerfan]

48. I read the OP

A length of wire 1/16" in diameter is wrapped exactly once around a rod 7/32" in diameter.

How long is the 1/16" wire?

There. I just read it again.

The wire could be any length.

[Orrex]

49. :eyes:

Look, it's not a trick question. Everyone else here understood that "wrapped exactly once" meant that the length of the wire is sufficent to wrap exactly once.

I suppose that my wording lacks the necessary precision if we're trying to split the atom or figure out a bar tab, but for purposes of a fun little diversion, I believe that it was adequate.

[pokerfan]

51. Then that's different

The ID and the OD of the resulting torus:
ID = pi(7/32) ~ 0.687223393"
OD = pi(11/32) ~ 1.07992247"

The length of wire required is probably going to be close to the OD.

[Tuesday Afternoon]

54. pokerfan -- That was my first thought, too

damn wire could be 12 feet long for all we know...I think we want to wrap the wire around the circumference of the rod without having any scrap left over. :yoiks:

[pokerfan]

56. vindication!

thank you! :bounce:

[Orrex]

58. Pfft. She's obviously insane.

:crazy: I mean, come on! :crazy:

[Tuesday Afternoon]

65. so tempted am I ...

to make that quote my new sigline :D

[Bake]

28. 3.214 feet.

Oh wait. Did you mean it's supposed to be TOUCHING the rod?

Bake

[Rabrrrrrr]

30. Length depends on where you measure the outside wire -

since it has to be cut at an angle in order for the planes of the two ends to match up, the length of the wire varies from 0.69 to 0.88 inches.

However, this is also assuming that a) the wire is wrapped perfectly into a circle, and b) the inside circumference of the wire is in contact perfectly with the outside circumference of the rod.

[LynzM]

72. Indeed.

Math-geek crush, yep. :D

[Writer]

31. Never mind.

Sigh.

[billyskank]

36. Can't help but mind.

[hunter]

32. So, how does a nice straight wire become a donut?

Does the part of the wire touching the rod get squished? Does the part of the wire away from the rod get stretched?

Will the length and diameter of the wire be the same if you unwrap it from the rod as it was before you wrapped it around the rod? Will the unwrapped wire still have a round cross-section?

It's not a geometry problem, it's an engineering problem. To answer the question you have to know the nature of the material the wire is made of.

In the world of simple geometry wires don't bend.

[Ptah]

33. That's exactly what happens, hunter.

The inside gets squished (compression) and the wire away from the rod gets stretched (tension).

The common rule of thumb is to consider the neutral axis (the center of the wire) is used for the length computation.

Your sentence: "To answer the question you have to know the nature of the material the wire is made of." is spot on.

[SidneyCarton]

38. 0.691 inches

Circumference = Pi x Diameter

[Ptah]

41. What diameter are you using?

:shrug:

[SidneyCarton]

45. The diameter of the dowel.

I looked up thread and saw my mistake... hence I got my degree in history. ;)

[Orrex]

43. Thanks for chiming in, everyone!

And, considering the difficulty it had given me, I'm glad that it wasn't quite as simple as I feared it might be!

[peruban]

46. Since we're dealing with such small quantities

This problem can be reduced to a simple geometric exercise. The diameters of the wire and rod are nearly insignificant. One simply has to deduce the circumference of the rod. That simple.

Now, if you want to get engineering in on the matter you can take the center axis of the wire or internal and external lengths but what's the use? The measurements are insignificant and the entire problem can be deduced using middle school geometry.

The only reason the measurements would matter is if we are dealing with a quality control issue and so the materials are optimized for economic purpurses. Otherwise it's just a futility in calculation, which I leave to calculators.

[Orrex]

50. It's to get an estimate on the linear footage of wire in a piece of chainmail

When you're talking about, say, 32,000 links of a certain diameter, the difference starts to add up.

Of course, the links aren't exactly circular, nor are they exactly uniform in size, but those differences are smaller even than the difference between rod_diameter and rod_plus_wire diameter.

[peruban]

53. If it's chain/mail we're talking about

There's a very simple process I learned when I was ren-fair geek in high school. You use a dowel to wrap the wire around in a cranking fashion throughout the length of the rod to create a coil. Then you slip the coil off the dowel and cut the links seperately, keeping in mind to make them uniform in circumference, and then link them together in a knitting fashion. You can get fancy and rivet each link shut but simple bending and meshing are enough for modern ren-fair needs.

Is your question regarding making a chain helmet or armor? It still doesn't really matter since you're not under any coporate quality control standards. Just buy a moderate gauge of wire, make a wooden or metal dowel with a cranking mechanism which is removable and create coils by wrapping the wire around the dowel in tight formation. You then cut each coil at around the same point and, presto!, uniform links for fashioning mail suits, bracelets or whatever. No mathematics really involved in this problem.

Remember, blacksmiths of old did not have the mathematical skill we have today and so most measurement was done with lengths of string or other metrics. Any spare coil can be reused later on.

It seems to me that the most important aspect of the job is to find your prefered diameter of dowel for wrapping the wire around. Consult with a local medieval historian to get an idea of what size to choose or research it yourself online. And brass or iron wire is cheap enough that you don't need to worry about QC economic concerns. Some wire leftover can always serve to mend any dislodgements or your initial patterns. Also, if you gain weight, you might want to have some loose wire handy to make alterations as needed. All you need is a pair of pliers, a wire cutter, and a crank fashioned dowel to form loops.

Hope this helps.:hi:

[Orrex]

57. Thanks for the tips, but...

If I had to make a guess, I'd say that I've joined somewhere in the neighborhood of 200,000 links in various hauberks, coifs, byrnies, gloves, and the like! I don't rivet because I haven't found any two sources that agree on a good method for use in a home workshop, so the links are butted.

The real reason that I'd like to estimate the length of the wire is so that I can answer when someone asks "how much wire is that?"


Incidentally, I've been using a Black & Decker VSR drill with a 3/8" chuck since about '94. Hand-cranking is for loonies!

[peruban]

62. Hand cranking is tedious, I completely agree.

My history teacher in high school was just very traditional in his approach. He taught us how to do it without measuring.

But if it's just a matter of amount of wire used, keep track of how many links make up the mesh and just multiply the circumferential lengths to deduce the amount of wire used. Measue your head, jawline, torso, belly, and waist line to get an idea on what size fits you best. Always keep about a yard more than you need in case you have to make alterations to the vestiment. The gauges used for chain/mail are usually insignificant and you can always calculate the amount used versus the amount you purchased since the diameter of the wire is negligible. A thinner gauge would obviously be lighter to handle and easier to mend. That's only it you plan the meshing for exhibition purposes at fairs and such. If, however, you intend to use it as an actual mock battles then use tighter gauges, tighter coils, and stronger metals. Also, riveting (as hard as it is to do) will ensure that you dont run into impalings or ruin the arnir itself. There are plenty examples online for this instruction but a degree of mastery is required to bolt the loops together - it is not easy.

Remember, chain/mail armor is very personal, so don't expect to give this as a gift, no matter how enticing.

If you have other questions along the line of thought I gave please feel free to ask. I promise to try to give as much information as I can.

[Ptah]

67. 2356 feet, approximately

[Bucky]

73. I'm going to bubble in "C" for this problem

Always go with "C"