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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,963

A string is wrapper around the Earth's equator(i.e, the circumference) and the two ends of the string just touch. Now suppose that another string is tied to the original string so that it is 100 ft longer. If this new string is placed around the equator and pulled tight so it is suspended over the earth, how high will the string be above the ground?

a. An atom high

b. A bowling ball high

c. A Mack truck high

d. Exactly 100 ft.

I believe it is C.

Explanation:

Let CE = Earth's circumference and let CS = String's new circumference

Let RE= Radius of the Earth and let RS = Radius of the string

CE = 2πRE

and CR+100 = 2πRS.

We want to find the difference between the two radiuses(aka height.)

C/2π = RE

C+100/2π= RS

thus...

RS - RE = (C+100-C)/2π.

RS - RE = 100/2π = 50/π

50/pi ≈ 17.

Therefore Mack truck, aka C, is the solution.

I would LaTeX the aforementioned equations but I'm in a hurry.

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,963

Correct? Incorrect?

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Hi Mathegocart,

your working is correct.But I don't know about Mack truck.

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

hi Mathegocart and thickhead

I've searched their site but cannot find any dimensions for their trucks. Strange! Wouild you buy a vehicle without knowing how big it is? Pictures with a trucker in shot suggest they are only about 12 ft tall. If a truck really was 17 ft tall it wouldn't fit under UK motorway bridges.

Bob

ps. LATER EDIT: This is how to cut off the main route from the rest of England to the Dover continental ferry port and the Channel Tunnel:

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**Mathegocart****Member**- Registered: 2012-04-29
- Posts: 1,963

bob bundy wrote:

hi Mathegocart and thickhead

I've searched their site but cannot find any dimensions for their trucks. Strange! Wouild you buy a vehicle without knowing how big it is? Pictures with a trucker in shot suggest they are only about 12 ft tall. If a truck really was 17 ft tall it wouldn't fit under UK motorway bridges.

Bob

ps. LATER EDIT: This is how to cut off the main route from the rest of England to the Dover continental ferry port and the Channel Tunnel:

What a calamity! They look as if to fit under most US bridges here..

The integral of hope is reality.

May bobbym have a wonderful time in the pearly gates of heaven.

He will be sorely missed.

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,606

Hi all;

I understand (maybe wrongly) from Mathegocart's explanation that the shape for the longer string version is circular (like the original string), but in an experiment in which I used my basketball the shape is quite different (see image).

The string curves around the ball tightly from the base to the tangent points (the black dots), and then both sides continue as straight lines up to the suspension point...the apex.

Transferring that shape to Mathegocart's problem I've tried to work out the distance from the top of the earth to the string's apex, using the popular integer measurement of 40075km (that I converted to imperial) for the earth's equatorial circumference and basing the diameter on that.

But...I'm getting an answer that doesn't make sense to me.

What answer do you guys get?

Thanks!

*Last edited by phrontister (2017-02-23 19:03:36)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

That is very clever. You used EM to help you understand the problem. Were you able to get it all into Gebra or M? If so, may I see the work?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Actually I did not read the problem properly. I was carried away by Mathegocart's working which looked o.k. Now I get the height as 3888 ft as per the model given by phronister. None of the answers to be chosen fits it.

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,606

That's nearly exactly what I got, thickhead.

My answer in Geogebra was 3890 feet, rounded to the nearest integer.

I was going to try mathematically, but haven't worked out a solution method yet.

*Last edited by phrontister (2017-02-27 11:20:15)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You might want to be careful because this problem is similar to a well known problem that Forman S. Acton poses in one of his books. It is extremely ill conditioned.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,606

Maybe that's why my answer doesn't make sense to me, as I said earlier. I'd expected the correct answer to be closer to the height of an atom (Mathgocart's answer *a*).

According to an online calculator into which I entered the earth's diameter (or radius?) and some other figure I don't recall, 3890 feet also happens to be the perpendicular distance from the top of the earth down to the chord between the two tangent points. I thought that was rather coincidental, and so I started to doubt my understanding of my version of the problem...but the huge scale isn't helping me to get my head around it.

Sorry...have to go out for 5 hours or so.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

This is where subtractive cancellation will occur. You were wise to seek help with a Mclaurin series but you truncated too early. Taking another term gets 3889 which is much closer to the answer you seek.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

I can get exact value

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

That is not what I am getting for the exact value.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

That depends on what value you take for R.I took it as 3959 miles.

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I took R = 3959 * 5280 and am getting 3889.000315217549 which agrees with the two term Mclaurin series.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

The source of error may be in the calculation of theta.For h the formula is perfect.

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,442

The Earth is not exactly spherical; it is an oblate spheroid (bulges at the equator). So the diameter varies depending on where you measure it. So I wouldn't worry about small discrepancies in answer here.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Bob;

In the real world you are correct. The measurements we are using are not and will never be exact. But in computation, once we fix them as constant and compute from there it is important to get answers that do not suffer from numerical error.

Hi thickhead;

The correct answer to this can be found at Wolfram Alpha. Just enter this:

R = 3959*5280;t = (150/R)^(1/3);N[(2 R*Sin[t/2]^2)/Cos[t], 100]

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,606

Hi Bobby;

I haven't been able to follow your workings or thickhead's as the maths is too advanced for me, but I thought that as I initiated this version of Mathegocart's problem I'd better try to find a solution to it.

I found these formulas on a surveying website:

Where:

r = earth's radius

c = central angle;

h = height of string apex above top of earth

t = length of tangent from earth to string apex

In M, I tried to 'Solve' for c, but couldn't get that to work. However, 'FindRoot' did, with the following result in M:

```
Input:
r=3959*5280;
c=c/.FindRoot[2r Tan[c \[Degree]/2]-100-r*c \[Degree]==0,{c,2}];
h=r/Cos[c \[Degree]/2]-r
Output:
3888.614460591227
```

I'm pleased that this is pretty close to my G's metric answer (for which I used r=40075km), which meant I was on the right track back then...

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

There are differences between the formula I used

R = 3959*5280;t = (150/R)^(1/3);N[(2 R*Sin[t/2]^2)/Cos[t], 100] post #14 which I computed accurately to 100 digits and the one your surveyor site gave. The one I used comes from the two term Mclaurin series, it will just be an approximation.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

Bobbym is right.After getting theta I had truncated it. Using it directly (copying) I get 3889.000315 in excel.

(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha

{Gods rejoice at those places where ladies are respected.}

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,606

When I plug Bobby's W|A result (from post #20's formula) into G - which I hadn't tried before - I get 3889.00031521 (correct to 8 decimal places), so my G method seems to be good.

The error looks like being the central angle, for which I get different answers depending on whether I use Bobby's formula or the surveyor site's.

I've probably messed up the surveyor formula somehow...

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

What is the name of the site?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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