Any Mensa DUers? 3 math/physics problems for your consideration:

[karlrschneider]

1. You lay a wire around the world at ground level on the equator, pull it tight and tie the ends
together. (Assume 25,000 miles circumference, ignore mountains etc. - it's a smooth surface)
Now you 'splice' a ten foot piece into the wire. How high above the surface will it now be assuming
it's evenly supported all the way 'round?

2. You are in a small boat in a swimming pool. In the boat you have several hundred pounds of scrap
iron which you toss overboard. Does the level of water on the side of the pool go up, down, or stay the same?

3. A cow is fitted with a collar and a 100 foot rope attached to the side of a silo 50 feet in diameter in the middle of a large field. Neglecting the length of the cow's neck, how many square feet can it
graze?

I think #2 is the easiest and #3 is the most difficult. Anybody wanna try? :D

[Richard Steele]

1. You're posting during a geometry exam, aren't you?

Is the cow's rope attached to a fixed point on the silo?

[lapfog_1]

7. No. in the classic case... the attachment point at the silo

is actually a ring, allowing the cow to graze up to a radius of 125 feet from the center of the silo in any direction.

So you calc the area of two circles, one of radius 125 and one of radius 25 and subtract the area of the smaller from the larger.

Number one is similar... you have two circles (ignore all the stuff about the earth, one is 25,000 miles in circumference and the other is 25000 miles plus 10 feet in circumference, calculate the radius of both circles and subtract the smaller from the larger.

I see number 2 has already been answered.

[canetoad]

20. As the problem was posed

I was thinking the cow grazed in a parabolic area (with a cut-out in the middle, representing the silo). 100ft north, south and east/west of tethering point. Still working on the sums

[karlrschneider]

21. Yes, I neglected to mention that, sorry!

:D

[FARAFIELD]

24. Im in MENSA

But it doesn't necessarily mean that we can ace geometry (not math and physics as you stated) rather it means that we might be a little bit better at picking up patterns, and have slightly better abstract thinking. But even that is an illusion, because not everyone that COULD be in MENSA is in MENSA. SO even though I was an astronomy major it doesnt mean my qualifications are any better than someone that didnt get their IQ (generally Stanford-Binet) taken when they were five!!! The qualifications are simply that you are in the top ten % of the top 1% of the population in IQ (right now that would be 144). But if you happen to be a very aware child (having extremely smart older siblings helps) you can pull off a gigantic IQ when you are tested when you are young. And for MENSA they dont care what age you were when you attained the score. I bet I would be middling today.

[karlrschneider]

31. No offense, I assure you, I just find these to be fun games for some who

appreciate stuff like this. :-)

[whistle]

39. Sounds like the "dumbing down" of MENSA, still an IQ of 144 is not

...all that bad :yoiks:

:sarcasm: just kidding :rofl:

[Nobody]

101. I could be in MENSA but I never needed to

My IQ was tested at age ten and I was not even trying. I demanded to know the number and was told that they shouldn't tell me. I insisted on knowing, it was my score, and I wanted to know what it was. I was bracing myself for a George Bush-esque low number the way they were going on about whether I could handle knowing.

I have not gone to college, and I skipped two grades ,not chronologically, I went to school for 12 years but only 10 of those years resulted in education. LONG story. That meant my math skills leave much to be desired, but I'm damn good at problem solving, filling in the blanks, and reinventing wheels. My biggest math accomplishment is figuring out on my own how to do functions, since I came into math class after everyone else learned how to do them and I was there just in time to get the homework assignment. If I didn't have a MENSA qualifying IQ, I'd have been hopelessly behind.

However, most of my friends can likewise boast MENSA-level intelligence as well.

I think you've got something there about smart older siblings. They were well aware that I have a gazillion hobbies. Being the youngest meant I had everyone teaching me everything under the sun. What they learned in school, they tried to teach me. By the time I was five I could read road signs as they were whizzing by, I could count to 100 in Spanish, by watching Villa Allegre which came on after Sesame Street I was picking up Spanish vocabulary and had a huge head start once I got to Spanish class in HS.

[Viva_La_Revolution]

2. #2 - stays the same

but your boat rides higher.

#1 and #3 - sorry - I've started drinking early. ask me again tomorrow.

[unblock]

18. i don't think this is correct

when in the boat, the weight but not the volume of the scrap is relevant.
when in the water, the volume but not the weight is relevant.

since scrap metal is more dense than water, i would presume the water level would drop when you throw the scrap overboard.

[karlrschneider]

Yes, you're right! Think Archimedes!

The iron being more dense than water displaces more when it's floating than when it's submerged. ;-)

[Tesha]

132. For #2, there isn't quite enough information in the problem.

For problem #2, there isn't quite enough information provided
in the problem statement.

We don't know whether the scrap iron is in a configuartion
that could/would float.

If the scrap iron were solid (so that sinking was assured),
the level of water in the pool would go down when the ballast
is tossed overboard.

But if the scrap iron were in the form of hollow spheres or
small boats or any other shape that would float (and it was
permitted to float when jettisoned), then the level of water
in the pool would be unchanged because the floating scrap
would displace as much water itself as it had caused the
boat to displace when it was on board the boat.

Tesha

[Viva_La_Revolution]

25. ack! I didn't take into account the weight of the metal vs. the wt of the water

my bad.

here, share with me :beer:

[Toots]

87. No it is about displacement

they displace the same amount of water whether riding in a boat or actually in the water itself..

[Greyhound]

94. Sorry Toots, I didn't read your reply before making mine. n/t

[Greyhound]

93. The mass of the metal while in the boat causes the boat to displace

that much more water, so there is no difference whether the metal is in the boat or in the pool. Net result; no change

[karlrschneider]

96. Nope. Something that floats displaces a volume of water equal to

it's weight, if it is sunken it displaces a volume equal to it's own volume.

[Greyhound]

104. But only if you assume that the scrap metal does not itself float, as someone

else pointed out up-thread. If the metal were a hemisphere or a sealed sphere the water displacement becomes a variable based on the density/displacement of the metal. No?

[karlrschneider]

108. Yes that's true but one wouldn't ordinarily find water-tight hollow scrap

metal. Next time I'll specify solid steel ball bearings. :D

[Greyhound]

110. Thanks, partial credit got me summa cum laude...

:rofl:

[Bucky]

112. Wouldn't the scrap in the boat also push some air in the boat below the water's surface level?

The amount of air that displaces the water would disappear when you throw the metal out of the boat, altho the mount of water that is displaced by the metal would be the same. So I think the actual amount of water displaced would go down. The water level would lower a bit. Especially if you made a big splash when you threw the iron out of the boat.

[SidDithers]

3. #1 is a bit more than 3 feet high...nt

Sid

[Viva_La_Revolution]

5. nah... that can't be right

ten extra feet? .03 inches maybe.

but like I said... I been drinkin. :cheers:

[Boojatta]

9. Isn't it about 1.59 feet high?

It sounds as though you're giving the total of the heights on both sides of the Earth rather than the height above one point on the Earth.

[SidDithers]

19. Yeah, you're right, I buggered it up...

d is 10/pi bigger, but of course we're talking about the radius, so it would be half that.

Thanks for the pick-up :)

Sid

[karlrschneider]

27. Yes, it turns out to be 5/pi feet

:D
It doesn't matter how big the earth is, the answer is the same even if it was a point of zero radius!
:D

[Greyhound]

95. Don't understand this question. If you have 132,000,000 (25,000 * 5280) feet of wire

and add 10 feet to it you end up with 132000010 feet, right? Or does the problem say that we raise the wire ten feet off the ground all around the circumference?

Either way, I can't remember the formulas that will give the answer.

[SidDithers]

100. You got it...

We've now got two equations:

132,000,000 = pi*2r
and
132,000,010 = pi*2(r + x)

where x is the increase in radius because of the bigger circumference.

solves to 5 = pi*x
x = 1.59 or so

We've now got an additional 10 feet of wire in the circumference, and that hypothetically raises the rope an additional 1.59 feet off the ground, all the way around.

Sid

[nam78_two]

4. heh-what am I missing on #3?

Wouldn't that just be pi*<(100+25)^2-(25)^2>

My instinct would be to think thats the easiest :)? What am I doing wrong?
And 2 would just stay the same, but I don't feel like attempting 1 :P.

[A Simple Game]

6. Number three is just the area of a 200 ft. diameter circle

minus the area of a 50 ft. circle.

Number 1 is the difference in diameter of a 25000 mile circumference circle subtracted from the diameter of a 25000 mile plus 10 ft. circumference circle.

Number 2 is no difference.

[nam78_two]

12. Yeah agreed on 2 and 3

But I don't quite see that as the solution for 1. If you have just 1 wire sticking out at one part of the circ, that wouldn't give you another circle-it seems like a misshapen geometrical figure then no?

[StraightDope]

13. You fogot that the point at which the rope is thethered varies by 50 feet...

Depending which side of the silo the cow is on.

[karlrschneider]

34. Yes, the rope starts wrapping around the silo once the cow gets

past 90 degrees either side of where the rope is attached!
Eventually it gets 'pulled' right up to the side of it.

[whistle]

43. It would be sort of like the affect of gravity on an orbiting object around a

...black hole

[StraightDope]

90. That changes things considerably...

I had assumed that the rope was attached to a ring the same size as the circumference of the silo. That being the case, I'm not even going to try to figure that one out. I never took calculus, or trigonometry. Hell, I haven't had geometry since the 6th grade! Turns out I was dead-on right when I asked my teachers, "When am I EVER going to need to know this stuff?" Aside from playing pool and showing off on DU of course. :)

[tridim]

17. The radius of the rope changes as the cow wraps around the silo

There's the rub.

[karlrschneider]

35. Exactly. That's why it's a hard problem, it's a doulbe involute

that overlaps itself.

[Mr_Spock]

84. It also has to be assumed the rope is attached at graze ht.

onto the silo - otherwise there will be another small error introduced.

[karlrschneider]

28. Actually, no. On #3 I forgot to mention the rope is firmly attached

to the silo, it doesn't 'slide around' it. You're close on #1 but mistaken on 2.
:-D

[BlooInBloo]

11. It's the difference of circles. 1) is an easy version of finding dD/dC

[nam78_two]

15. yeah

Thats my point :). Its just pi (r1^2-r2^2).
I thought that was the easiest one. I think # 1 is trickier.

[StraightDope]

8. #3 isn't that bad...

Total diameter of the rope+silo is 250 feet, considering that at full extension, a point A opposed by point B will be 250 feet apart. You first find the area of the total, (250/2)^2*pi, or 15625*pi. You then have to subtract the area of the silo from this figure,
(50/2)^2*pi, or 625*pi.

Thus, the cow can graze 15000*pi square feet.

[Little Wing]

89. ahh but

there's nothing in the question to suggest that the rope's connection is able to travel around the silo, so you have to subtract not only the area of the silo, but the area the cow can reach isn't a perfect circle

[stonecoldsober]

10. I win!

1) Less than a foot
2) Very close to even
3) More than she can eat

[uppityperson]

69. good answers and here is more information I need.

#1) Does "ground level" change with the height of the ground (figuring that you would need extra to go down into ocean trenches and up hills)?
#2) Is there enough water to float the boat and cover the iron that you throw into the water or is there just a think layer on the bottom of the pool?
#3) Is the cow alive?

[TheCowsCameHome]

14. I'm busy trying to find a wire that long. Be back to you later.....

after I get tanked...........

[jayctravis]

16. Uhhh.

1. I don't understand what you're asking...the wire is at ground level, so you've tied a string around the world... but splicing in ten feet somehow makes it hover above the surface? It wouldn't be above the surface...it would still lie there only be ever so minutely less tight.

2. Stay the same.

3. Assuming the rope length remains constant as the cow walks around the silo (the rope being tied in a big loop around the circumfrence) you have a silo diameter of 50', so a radius of 25'; the silo cuts out an area of 1962.5 square feet. The radius of the entire area of the silo plus the yard the cow can graze is figured from a radius of 125 so 49,062.5 square feet minus 1962.5 = 47,100 square feet of grazing area.

[Boojatta]

29. Beware relying on your intuition unless it is well-developed in the subject area.

It wouldn't be above the surface...it would still lie there only be ever so minutely less tight.

Give us the ever so minute number.

[jayctravis]

42. The question makes no sense...

Now that I'm reading the other posts I see it's another "circle minus another circle" puzzle, but the wording made absolutely no sense. If you tie a wire around the earth on the ground, then add ten feet to the wire, it's not going to magically hover around the equator. Essentially it should read "If the earth is (whatever) in diameter and a telephone wire is strung directly around the equator on ten-foot poles (assume no sag in the wire) how much taller does every pole need to be if ten more feet of wire is spliced into the existing length?"

If someone writes word-problems then there needs to be no doubt as to what information is being sought, especially if there is a "right" answer.

[karlrschneider]

44. Well, I did say "if it was evely supported above the ground..."

You correctly re-stated the problem in a slightly different way...
Sorry if it wasn't clear.

[toshirajo]

33. I agree on all points

1 would be a slightly slack wire that lies on the surface for all practical purposes.

edit to add - Jeez. I just did the math and was very surprised to find an additional radius of 1.59ft. I would have never guessed that based on the magnitude of the circumference.

2 no difference / remains the same

3 the silo is a diversion. The grazing area is pi r squared (3.14 x 100 x 100) = 31400 sf

[karlrschneider]

37. Welcome to DU! Unfortunately you missed all 3.

:D

[toshirajo]

52. Thanx Karl

Yeah. I woulda never guessed adding 10 ft to a 132,000,000 ft circumference circle would have changed the radius that much.

I'll chew on the silo problem.

And apparently the boat problem too.

Apparently. First msg here, and I'm already dazzlen 'em my sheer brilliance. not. :)

Toshi
who slinks away with his modesty intact but his intellect slightly worn

:)

[unblock]

22. #2 - water level drops

when the scrap is in the boat, it displaces water as a function of its weight.
when the scrap is in the water, it displaces water as a function of its volume.

on edit - silly brain fart in subject line

[BlooInBloo]

32. Does it matter if the scrap metal happens to be in the shape of a boat's hull? :)

[karlrschneider]

38. It might.

:evilgrin:

(I know what you're up to, smart alec) :D

[BlooInBloo]

41. who? moi? lolol!

[karlrschneider]

36. Exactly!

10 points! :D
:toast:

[BrotherBuzz]

80. Wait, wait, wait, by your logic...

if oneighty had had a really, really good day and hauled a huge load of clams on board his boat, the body of water he was fishing would actually have risen? Whoa! :scratches head and pondering the impossible: ;)

[karlrschneider]

106. It sure would! Not much though.

:D

[Viva_La_Revolution]

23. Ok karl... quit teasing

give us the answers already!

Pretty please?

[Blue_In_AK]

26. I used to be a Mensa

but being brilliant and figuring out problems like these is too much trouble. Now, I enjoy just being average.

[joe_sixpack]

30. For #1

I think the wire should be about a foot and a half above the surface. Is that correct?

[karlrschneider]

40. Very close! 10 extra points to 'show your work'

:D

[selfdestructive]

98. 1 & 2

#1 is the answer - 4.8" inches ?
#2 is down by water displacement principles

[karlrschneider]

105. Right on #2, 1 is actually about 19 inches...

welcome to DU! :toast:

[aggiesal]

130. Hope this helps with #1.

C = Circumference
D = Diameter
pi = 3.14159

C/D = pi
so
C/pi = D
r = D/2
r is what we are looking for.

(25000 miles / D) = 3.14159

(25000 miles / 3.14158) = D

1 Mile = 1760 Yards = 5280 ft
So 25000 Miles = 132,000,000 ft.

(132,000,000 / 3.14159) = D(lo)
D(lo) = 42,016,904.9763 ft.
r(lo) = 21,008,452.4881 ft.

(132,000,010 / 3.14159) = D(hi)
D(hi) = 42,016,908.1594 ft.
r(hi) = 21,008,454.0797 ft.

r(hi) - r(lo) = Delta circumference
21,008,454.0797 ft. - 21,008,452.4881 ft. = 1.5915 ft.
1.5915 ft. * 12 in. = 19.0986 in.

[bleever]

45. #3 results in an ovoid, assuming the cow can walk both

clockwise and counterclockwise.

Still working on the math for the area...

[karlrschneider]

48. I probably should have left that one out, it actually is a very difficult

geometry problem. I first heard it years ago and finally had to resort to writing a C program to solve it. Then having found the solution, I went back and did a lot of trial and error using calculus to get the answer.

[boolean]

46. Answers

#1 is 1.59 feet
#2 is inconclusive because you don't tell us the shape of the scrap iron
#3 requires advanced calculus

[BlooInBloo]

47. Are you thinking of it as a constrained maximization problem? lol

[boolean]

51. No

I'm thinking it requires taking the integral of a function that approximates the shape formed when the cow crosses thee 90 degree mark on both sides of a tight rope in both the clockwise and counterclockwise direction, then adding the value of that integral to the area of the other side where the cow does NOT cross the 90 degree mark.

Either way, I'm not doing it!

[BlooInBloo]

56. Yup - just said the same in a more jackass way...

I'm sure the greeks could systematically answer it tho - which is not to say that something calculus-y wouldn't be infinitely easier...

[karlrschneider]

50. You got #1 exactly. The shape on #2 doesn't matter as long as it sinks

and you're right on 3, I probably should have not posed that one.
:-)

[BlooInBloo]

53. ah - I see the problem with 3 - the poverty of available paths on the "silo side"...

Glancing at problems was always my biggest weakness :(

[boolean]

55. What if....

What if the volume of the scrap metal is greater than the volume displaced by the boat? I'm just sayin' :evilgrin:

[karlrschneider]

63. The boat wouldn't float in the first place!

You tryina be a smartass? :D :D

[heidler1]

81. I believe that if the scrap iron was in the shape of sphere or cylinder

and air tight plus hollow it could be, say one once heavier than the displaced water at some given temperature. Then the water level in the pool would still go down very slightly and be hard to measure.

The first time I heard this a ship was hung up on the bottom of a canal lock so what to do? Throw the ballast overboard. This lowered the water level, but this float ed the boat more and freed up the boat. I always wondered what would prevent the ballast from getting under the ship and causing more trouble.

[porkrind]

62. excellent point on #2

I assumed the displaced volume would be taken up by solid iron, which may not be true. If the iron enclosed a space which could not be filled with water, the displacement is unknown.

If the scrap iron assemblage sinks, then the water level will be lower.

If the scrap iron assemblage floats, then the water level will remain the same.

[karlrschneider]

66. Okay you got me. If I ever pose this again I will carefully specify

the 'scrap iron' as steel canonballs! :D

[ChairmanAgnostic]

49. and now for the REAL answers:

1) it's god's will.
2) It's god's will, but only if you pray hard enough.
3) It would have been god's will, but for the fact that you did not pray hard enough, leading a very angry god to let those heathen, sinning, bastard, baby-killing, peace-loving, terra-ist coddling, anti-american, muslim loving, anti-jesus, gay, transvestite, pot-smoking, devil-worshiping, democrat party members win two weeks ago.

[toshirajo]

57. Well, there you go

Faith based math.

God revealed himself when the inspired the word MENsa too. Clear as day. Thank you Jesus. Pass the alphabet please.

[ChairmanAgnostic]

58. we already have faith based physicks, geology, and genetics.

Remember that Pi is actually 3; the universe is 6,016 yrs old; Adam and Eve walked hand in hand with their pet T-rex Dino; contraception and family planning is the devil's work; we are not members of the tadpole family; except for my ex-brother in law, we are not related to apes. Oh, and the founding fathers were all fundie christians, who like their god forgetting to make Adam his Eve initially, they just forgot to mention it in the constitution.

[spacelady]

54. Small boat in a swimming pool with all that scrap would sink before you

could throw it overboard. I would not be in that boat in the first place.

[bleever]

59. In #3, consider the point where the rope is attached

as the center of another circle, because from that point the cow can reach 100 feet straight out, to the left, and to the right. It's a half circle, easily calculated.

Wrapping around, how far does to cow reach from the point of the first circle? Basic geometry gives me (and my glass of bourbon)a distance of something like 76.5 feet. So the back half of the cow's range is half an oval, with a radius of 100 feet on the sides shrinking to 76.5 at the apex.

So, it's 100 squared times pi, minus 25 squared time pi, minus the crescent on the back side between 76.5 feet from the point of attachment and 100 feet out.

What that is exactly, I can't do the calculus for right now.

It's an open bar at Mensa functions, isn't it?

[TahitiNut]

82. The shape of the grazing area looks a bit like round-bottom heart with a big hole in it..

(For the purposes of description, consider the rope attached at the 180 degree point of the silo. South.)

The 100' rope is more than long enough to reach beyond the opposite side of the 50' silo. Thus, the rope is tangent to the surface of the silo when the cow crosses the "meridian" north of the silo. The point at which it 'last' touches the silo is then somewhere (I'm guessing) around the 280-290 degree point (or 70-80 degree point if going counter-clockwise), and the distance from that point to the cow (at the N-S meridian line) becomes the smallest radius; but the center of that (constantly decreasing) 'arc' shifts along the side of the silo where the rope touches.

[porkrind]

60. Tricky!

Problem #1: 10/(2*PI) ft, or ~1.6 ft.

Problem #2: Water level in pool goes down.

Problem #3: ~21,230 ft^2.

:spank:

[porkrind]

61. oops, I assumed the rope was fixed at one point on the silo

which makes it a tricky semi-involute that overlaps. Very naughty, but I solved it. I like it better that way, it's more fun.

[karlrschneider]

65. Nice! That #3 took me hours when I first saw it years ago!

:D

[BlooInBloo]

67. Wouldn't be the the worst Ponder This question I've ever seen....

I'm still trying to figure out how the greeks would've gone about it... Damn I wish I hadn't lost my Heath translation... sigh.

[porkrind]

68. Thanks for the fun thread!

It made my evening!

[Gregorian]

64. Without actually pulling out a calculator...

1 No matter what you do, circumference is just the ratio of 2(pi)radius. When you do the math, it's surprising how big a change in radius that small addition makes.



Argh! I screwed up the last two.

Number two shows that I didn't understand Achimede's principal. Since the steel is heavier than water, it'll sink. But in the boat, since it's heavier than water, it would be able to displace more than it's own volume, due to the geometry of the boat. That is because the boat floats due to the fact that it incorporates enough air to make the entire structure lighter than the water that it displaces. Sheesh. It's been way too long since I've had to actually think about something other than George!

There's no way the third problem is a fixed leash. That's much too complex for just doing in one's head.


2 You haven't changed the mass in the swimming pool, whether the scrap iron is in the boat or in the pool. That's not really a good answer. It has more to do with the fact that it displaces as much water in the boat as out of the boat because it's more dense than water.

3 This is a good one. A 100 foot rope give the cow a 50 foot radius of grazing. Since the silo is also a 50 foot radius, it effectively cuts the cow's grazing area in half. Wrong. I just realized I didn't see this correctly.

[alittlelark]

70. #3 - 49,090 +/-

[ThomWV]

71. I Would Expect Any Teenage Schoolchild To Be AbleTo Answer Each Of Those

Questions without any problem. It certainly doesn't take a Mensa member to tackle them, just some very simple arithmatic for two of them and basic knowledge for the other.

[alittlelark]

72. #2 - 19" (w/in 1% ) above the surace.

[alittlelark]

73. #2 goes down

[alittlelark]

74. Do you have any more problems?

we're having fun here!

[BlooInBloo]

75. Along the lines of (3). Citation after solution/give up - fair warning - it's a bit harder...

But I enjoyed it. I used an elephant gun on it (calculus of variations), but an "elementary" solution is possible, and is what I'll post, if there's interest.

And yah, it's easy to Google - but is that really the point?

Problem:

We have 100 meters of fencing, which we use to enclose a pasture. Our fence must begin and end at the oak tree. Ground south of the oak tree, or less than 20 meters north of the oak tree, is worth $100 per square meter; but ground more than 20 meters north of the oak tree is worth $200 per square meter.

What shape maximizes the total value of the enclosed pasture, and what is this value?

Clarifications:
The "oak tree" is a single point. "20 meters north" refers to an east-west line passing 20 meters north of the tree. "Begin and end at the oak tree" means that both ends of the fence touch this point. All of this takes place on a flat Earth.

[Gregorian]

77. I'd like to see the elementary solution.

My ego is highly bruised right now. I used to be so good at this stuff.

I see a half circle that the cow can graze. The silo side of things just isn't giving me any geometric solutions. I see complex ones.

[BlooInBloo]

78. elementary != easy. just means doesn't use concepts beyond typical high school...

... nevertheless, applied problems can still be quite challenging.

EDIT: Will pm ya, if you like - lemme know!

[Gregorian]

79. Yes. Please.

I hate being stumped.

My poor unexercised brain needs any stimulation it can get. :)

[alittlelark]

83. The tether attached to the silo does not say it is fixed

into position..... makes it easy once you see it that way.

[muriel_volestrangler]

86. Do you mean the 'elementary' solution is an approximate one

eg using 2 circular arcs, which gets close to the maximum, while the 'true' solution involves something very complicated, but which gets a slightly larger value? Or does the 'elementary' solution get the correct answer? Because I can't yet convince myself it shouldn't involve ellipses, or maybe something even harder to work with.

[BlooInBloo]

88. A solution gets the right answer by definition...

... What I mean by "elementary" - here, at least - is that there is a solution that doesn't make use of any techinical concepts beyond what the ancient greeks had available to them. In particular, no calculus stuff. (Please, no pedantic debates about Archimedean approximations & what-not.)

[muriel_volestrangler]

91. In that case, I'll give up

because I can't even contemplate solving it for 2 circular arcs unless I use calculus; and I can't prove that circular arcs are the maximising solution anyway.

[Gregorian]

92. :)

After two days of thinking about this, I see no way to do this without calculus.

I had years of math, and the only way I see to do this is with the equations for the silo circle and the equation for the cow leash, where the relationship is such that for any given distance of circumference along the silo, the same amount of distance is removed from the radius of the leash. From zero to pi radians.

It's a mistaken post. As I am a lot, I could be wrong. But this really doesn't appear to be an elementary problem on the order of the first two questions.

I'm actually really bummed out by all of this. It's bringing back the fact that I didn't learn nearly what I went to college to learn. Twenty years later, and I'm still really upset by it. I should have graduated, after FIVE years of calculus, with the ability to use it. Nope. Oh well, I still made a living as a design engineer.

[BrewerJohn]

103. I found a fairly simple integral that should calculate

the area within the region where the leash becomes restricted. The big problem I found
is figuring out how to subtract off the wrapped-around part that goes past the point
opposite where the leash is tied, to prevent it from being counted twice. That does not
seem to be simple at all, because I have to use a different integration variable and
things suddenly get tangled up. I gave up on it about three times yesterday, but I see
I'm not the only one it's still haunting!

[Gregorian]

115. Haha.

That's why I was saying zero to pi. If the relationship betbween the two area functions can be represented using those limits, then subtracting off that double counted area wouldn't be needed.

The way I see this is that the radius of the leash at any given instant, is like the radius of a circle. But an instant later, it's shortening as it moves along the silo wall. I'm thinking out loud here. I think the involute can be represented as a moving circle. Maybe it's three equations and not just two. Yuck. I'm brain dead after all of these years of not using my math. I really retained a lot for the first ten years.

I may go run out to my books and...no wait, there's this thing that has been invented since I was in school. The internet. Maybe I can find some equations relating to "involutes". That is if I really get bored. :)




Edit- And involute of a circle is a spiral. (See Wiki for a pretty good discussion.) So this is a spiral with a radius which decreases as a function of the distance covered of the circumference of the silo.

x = a(cos(t) + tsin(t))
y = a(sin(t) - cos(t))
Where: t is the angle and a the radius

Yuck.

Edit number two- It's called an Involute on a Base Circle.

[ChaoticSilly]

76. #3 is the area of the silo

assuming of course the leash is attached to the inside wall of the silo, there are no doors or the doors are all closed and locked, the cow doesn't have a chain saw or lock pick and the farmer has installed grow lights and a sprinkler system inside the silo to help the grass grow.

Wait, what kind of "grass" are we talking about here and is it legal in the farmer's state?

:evilgrin:

[Mr_Spock]

85. 1,2,3

#1 19.05"
#2 down
#3 1/2 of the area of a 100' dia. circle, + some shit & minus some shit - assuming a fixed grazing height attachment to the side of the silo and assuming the silo is indeed a closed building.

[stepnw1f]

97. #2 is Down

Reason: the more surface in the bought pressing down on the surface of the water.

#1 - 0.474 in

#2 is Down

#3 - 5000 sq ft?

[selfdestructive]

99. 1 & 2

#1 is the answer - .0004 of a foot OR 1/2,500 of a foot OR .0048" inches?
#2 is down by water displacement principles


#3 is misleading because are we to assume that the cow can move all the way around the outside perimeter of the silo with the full 100 feet?

[karlrschneider]

102. Geometry & simple algebra solution to #1:

String around the world, the basic solution:


Okay we start with the wire tight around the smooth earth so

c1 = circumference
r1 = radius from earth center to wire
h = height of wire = 0 (it's on the surface)

After adding in the 10 feet

c2 = new circumference
r2 = new radius from earth center to wire
h = new height (this is what we need)

r1 = c1 / 2*pi (half of diameter of any circle)

and

r2 = c2 / 2*pi

but c2 = c1 + 10

so r2 is (c1 + 10) / 2*pi
and
h = r2 - r1

then h = (c1 +10) / 2*pi - c1 / 2*pi

multiply all terms by 2*pi

now 2*pi*h =10

and h = 10/2*pi or 5/pi, about 1.6 feet

The size of the circle doesn't matter at all!

:D

[selfdestructive]

109. *shakes fist*

woops on #1 lol

[immoderate]

114. Right, another way...

The radius is in direct ratio to the circumference by a factor of 2 x pi. If you increase the circumference by 10" the radius is increased by 10" / 2 x pi. 1.6" sounds about right.

On the boat problem, let's say that it's 80 pounds of iron, and that iron is 8 times the density of water. The iron on the boat displaces 80 pounds of water. The iron off the boat displaces 10 pounds of water. The water level goes down.

If the cow is tethered so that the rope slips around the silo, the area is the 150 foot circle minus the 50 foot circle. If the cow is tethered at a fixed point, the limit of the cow involves calculating the area under a function that calculates the involute (wrapping) of the circle. I would have to hit the calculus books for a couple of days to come up with that integral.

--IMM

[karlrschneider]

118. Very good!

Here's a bit on the involute...the total area is one half that of a circle with a 50 foot radius
plus 2 times the integral of the involute from 0 to pi/2. :D


http://www.answers.com/topic/involute

[immoderate]

119. I think you mean...

The area of a of a half circle of 100 foot radius, plus twice the involute. I'm assuming that the involute starts when the rope starts wrapping.

--IMM

[karlrschneider]

121. Well, there are actually 2 involutes

CW and CCW and they intersect at 180 degrees from the attachment point. The plan view almost looks
like a cardiod at that point. Yes you're right the starting point is at the 90 and 270 degree points
(or plus/minus 90 if you prefer)
:-)

[Greyhound]

107. #3 is the rope fixed to a single point on the silo? If so, how far above the ground is

it fixed?

[Greyhound]

111. I think it is really indicative that this thread has generated so many

responses. Shows how un-amerikan DUers are. Right on! :hi:

[Bucky]

113. I wish I could join Mensa, but I am very stopid

I mean stupid.

[StraightDope]

116. In other words, you're a "moran".

:rofl:

[selfdestructive]

117. let's just say

you have a case of beer at 11am on the 1st day of the week.
you need to figure out how much you've drank after 4 hours.
if you fail to do this you will be indefinitely tethered in a silo with a cow.
afterwards, if you are still alive, you will be made to walk 25,000 miles, which at the end of, if you make it of course, you'll get some pie...

p.s.
there are no empties to count or they have blended in with empties from the previous two days from which you have no idea of which how many you have actually consumed

p.s.s
the A-team is not available so please don't go there

[Cookie wookie]

120. Question 1: 23.097744 nanometers

That is if the wire is raised an equal amount around the earth.

[Cookie wookie]

122. Question 2. Stays the same

[Bucky]

126. Ha! You forgot the water displaced by the air in the boat.

Now all the Mensa people will laugh at you and call you mean names...

which they learned from kids who taunted them on the bus back when they were in school.

[Cookie wookie]

134. well, I'm not a Mensa, but what the hey

used to be married to one...does that count?

Still can't resist a puzzle.

[porphyrian]

123. I've never joined Mensa, but I'll try.

1. No more than it was. Adding that little rope to that much would simply make it a little less tight, but probably not noticeably so.

2. My instinct is to say it would stay the same, but I think it may actually go down, since, while in the boat, the scrap iron has a bunch of space around it filled with air (the hull of the boat is displacing the water, not the metal alone). However, once it's tossed into the pool, the iron displaces only each piece's exact area, and the boat rises in the water.

3. I don't really do math on my day off, but the maximum possible area would be an ellipse with the point straight out from where the rope is attached being the farthest away (at 100 feet) and the point on the far side of the silo from where the rope is attached being closest to the silo. Cows are fairly stupid, too, so they may not realize that they can turn around and get more slack on the rope, so they may only go one way around.

[BrewerJohn]

124. Here's what I settled on as a solution to (3)

The first figure shows how I divided up
half the total area into three regions: region 1 is just a quarter-circle of radius 100 ft,
region 2 is the area over which I integrate the involute, and region 3 is the little sliver
left over when I stop the integration.


The second figure shows how to do the integral, labeling the radius of the silo r and the
length of the rope L. I use the formula that the area of a wedge
of a circle is (1/2)r^2 times the included angle. The infinitesimal area is not quite a
circular wedge, but it only differs from one by a factor of d(theta). Since the area is already
proportional to d(theta), and second-order infinitesimals always vanish in the limit, this
should be OK. I integrate this dA from theta = 0 to theta = theta_m, where theta_m is such
that the rope left over just touches the dividing line between north and south.


In the third figure I show a closeup of region 3. This area can be calculated by taking
the difference between the area of the triangle (known from simple geometry) and the
area of the circular wedge.

The only really sticky part is that theta_m can't be solved for analytically. In equation
form, (L/r) - theta_m = tan(theta_m). This has to be solved either numerically or graphically.

[karlrschneider]

128. You have the right approach but your region 3 is actually part of region 2!

(meaning, the cow can 'go' there) You don't need to calculate "region 3"!

The reason for stopping at pi radians from the attachment point is to make it simpler, just
integrate to that point and double it then add the area of a semicircle of 100 foot radius! :D

[BrewerJohn]

129. Uh, if the cow can go there, I need to calculate it, no?

The thing is, I'm not stopping the integral at pi radians; I have to stop before I get there, or else
part of the area element as I've defined it will stray into the upper half-plane, and then that part
will get counted twice if I just double. Stopping it there leaves the little sliver the cow can get to, but hasn't
been covered by the integral.

What might be confusing is that I'm not using an area element that just goes out radially from the
circle, but one that shoots out tangent to it.

[karlrschneider]

131. Okay I see what you're saying but my solution considered all area alements

as radial-rho slices perpendicular to the tangent at each point on the side of the 'silo'. I left my notes over at the shop and don't have them handy here in the house...if you consider the 100 foot rope will 'wrap' around ~229 degrees before it jams up against the wall it would seem to be simpler to divide the whole 'back side' into 2 equal areas rather than integrate to some odd figure like 1.63pi then repeat the whole thing with another smaller endpoint and add them up. But now you're making me wonder if I actually managed to calculate it properly from the beginning! Maybe the best way to do this is with a tin can, a string and some graph paper! :D

[Gregorian]

125. Here's the approach to answering #3

http://mathworld.wolfram.com/GoatProblem.html

I ended up reviewing gear design, after all of this thinking about involutes. Then it happened- I never thought it would, but I found myself Googling- "formula area involute goat".

I had already come up with the basic solution. It's almost the same as this website states. But they eliminated the overlap of grazing area through the problem statement constraint that the goat only have a leash long enough to end at the opposite side of the silo. So if you skip the first paragraph, which is a separate problem, and account for the limit on the formula to only the opposite side of the silo, then this is how you would see the math worked out.

The involute curve is a function of the location on the silo circumference. The involute defining line length is a direct relationship to it's location on the silo circumference. That's a crappy way to state it. Oh well. I'm not getting graded. Just humiliated in public.

Phew. I'm outta here! Enough of this cow thing. :)



Edit- Damn you karlrschneider!!! (:

[BrewerJohn]

127. I think my solution reduces to Wolfram's

when the rope becomes <= pi*r so that I can integrate right up to L/r and not have to bother with any region 3.

I like mine better though, because I didn't need to use the equation of the involute!

[Ganja Ninja]

133. I'm not going to try to answer the first one.

The answer to the second one is the water level goes down.

The answer to the third one is 47,100 sq ft assuming the cow can graze the full 100 ft for 360 degrees around the silo.