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2 votes (10.52%)
5 votes (26.31%)
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19 members have voted

Wizard
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Wizard
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January 8th, 2019 at 6:33:21 PM permalink
Quote: unJon

While I too have heard that girth matters, it appears that length trumps all, when it comes to the catenary position.



I have heard girth is just as important, if not more.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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January 9th, 2019 at 3:56:14 PM permalink
I tried to figure this one out, but something just doesn't make sense.


If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:
The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.
f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)
f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh-1 5
Excel says this is 22.9243, so the distance between the poles would be 45.8486
However, I am not sure if that value is in degrees or radians.

Ace2
Ace2
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January 9th, 2019 at 4:23:19 PM permalink
Quote: ThatDonGuy

I tried to figure this one out, but something just doesn't make sense.


If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:
The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.
f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)
f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh-1 5
Excel says this is 22.9243, so the distance between the poles would be 45.8486
However, I am not sure if that value is in degrees or radians.

Thatís the answer I got as well.
ChesterDog
ChesterDog
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January 9th, 2019 at 4:58:12 PM permalink
Quote: ThatDonGuy

I tried to figure this one out, but something just doesn't make sense.


If I am reading the problem correctly, the length of the rope isn't going to be 100.

Here's what I get:
The curve is f(x) = a cosh (x/a), with f(0) = 10 and f(d) = 50, where 2d is the distance between the poles.
f(0) = 10 = a cosh 0 = a -> a = 10 -> f(x) = 10 cosh (x/10)
f(d) = 50 = 10 cosh (d/10) -> 5 = cosh (d/10) -> d = 10 cosh-1 5
Excel says this is 22.9243, so the distance between the poles would be 45.8486
However, I am not sure if that value is in degrees or radians.




Here's the Wizard original question:

Quote: Wizard

A 100 meter rope is suspended from the top of two 50-meter poles. The lowest point of the rope is 10 meters from the ground. How far apart are the poles?...



Here's a hint on how to check your solution: If you change the 50-meter poles to 40-meter poles, the lowest point of the rope would then be tangent to the ground. But the distance between the poles would remain the same. When we solve this problem, the answer should be the same as for the 50-meter poles.
ThatDonGuy
ThatDonGuy
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January 9th, 2019 at 5:38:40 PM permalink
Quote: ChesterDog

Here's a hint on how to check your solution: If you change the 50-meter poles to 40-meter poles, the lowest point of the rope would then be tangent to the ground. But the distance between the poles would remain the same. When we solve this problem, the answer should be the same as for the 50-meter poles.


That's another reason I'm pretty sure I'm going about it the wrong way; if I use f(x) = a cosh(x/a), but f(0) = 0, then a = 0 - but a can't be zero because then x/a is undefined.
ChesterDog
ChesterDog
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January 9th, 2019 at 5:57:42 PM permalink
Quote: ThatDonGuy

That's another reason I'm pretty sure I'm going about it the wrong way; if I use f(x) = a cosh(x/a), but f(0) = 0, then a = 0 - but a can't be zero because then x/a is undefined.




The Wizard gave us the general equation for a catenary as y = a cosh(x/a) - a. You can see that this is the equation for the catenary when it is tangent to the ground since when x = 0, then y = 0.


I worked on the problem many hours before giving up and Googling. The solution is so obvious to me in hindsight.
ThatDonGuy
ThatDonGuy
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January 9th, 2019 at 6:29:35 PM permalink
In that case...

f(x) = a cosh(x/a) - a
f'(x) = sinh(x/a)

Let 2d be the distance between the poles; the rope goes through (0,0) and (d,40)

f(0) = 0
a (cosh (d/a) - 1) = 40
a cosh (d/a) = 40 + a
a2 cosh2 (d/a) = a2 + 80a + 1600
a2 (cosh2 (d/a) - 1) = 80a + 1600
a2 sinh2 (d/a) = 80a + 1600

50 = INTEGRAL(0,d) {sqrt(1 + sinh2(x/a)) dx}

50 = a INTEGRAL(0,d) {cosh(x/a)) d(x/a)}
50 = a INTEGRAL(0,d/a) {cosh t dt}
50 = a sinh (d/a)
2500 = a2 sinh2 (d/a)

Thus 80a + 1600 = 2500 -> a = 45/4

45/4 cosh (4d/45) = 205/4
cosh (4/45 d) = 205/45 = 41/9
d = 45/4 cosh-1 41/9
Distance between poles = 2d = 49.4376

ChesterDog
ChesterDog
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January 9th, 2019 at 6:50:32 PM permalink
Quote: ThatDonGuy

In that case...


f(x) = a cosh(x/a) - a
f'(x) = sinh(x/a)

Let 2d be the distance between the poles; the rope goes through (0,0) and (d,40)

f(0) = 0
a (cosh (d/a) - 1) = 40
a cosh (d/a) = 40 + a
a2 cosh2 (d/a) = a2 + 80a + 1600
a2 (cosh2 (d/a) - 1) = 80a + 1600
a2 sinh2 (d/a) = 80a + 1600

50 = INTEGRAL(0,d) {sqrt(1 + sinh2(x/a)) dx}

50 = a INTEGRAL(0,d) {cosh(x/a)) d(x/a)}
50 = a INTEGRAL(0,d/a) {cosh t dt}
50 = a sinh (d/a)
2500 = a2 sinh2 (d/a)

Thus 80a + 1600 = 2500 -> a = 45/4

45/4 cosh (4d/45) = 205/4
cosh (4/45 d) = 205/45 = 41/9
d = 45/4 cosh-1 41/9
Distance between poles = 2d = 49.4376



Nice! That agrees with CrystalMath's answer.

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