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blackpearl

The Devil
I saw this on Discovery channel and want to share it with you.

Suppose we take a long ribbon and wrap it around earth, around the equator, so tight that not even a piece of paper can go between the ribbon and the earth. Now we increase the length of the ribbon by just 1 metre so that it becomes slack. Now its possible to raise the ribbon from the surface of the earth. You have to tell me by how much the ribbon can be raised from the earth. Remember, the ribbon is raised not at one point on the earth but all around the earth, equally. I have made a small diagram to make it clear.

681gpki.gif


Do not take out your calculator or head to google. Just make a guess. By how many millimeter/cm/meter the ribbon can be raised.

Scroll for the answer

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The answer is 16 cm. Sounds unbelievable, isn't it? The circumference of the earth is approximately 40,07516 metre, and so is the length of the ribbon. How can by increasing the ribbon length by just a tiny 1m enable it to be raised by 16cm all around the earth?

Here is the math:

Radius of earth = r (meters)
Circumference of earth = 2πr
Initial Length of ribbon = 2πr
New length of ribbon = 2πr + 1

Height by which the ribbon can be raised = (Radius of the circle of ribbon) - (Radius of earth)

H = (2πr + 1)/2π - r
= r + 1/2π - r
= 1/2π
= 16 cm

Notice that H doesn't depend on the radius of the earth, which means that whether you wrap the ribbon around a football, or around earth, or around the sun, if you increase the lenght by 1m it can always be raised by 16 cm!!

Amazing, isn't it?
 
Last edited:

cyberscriber

-- .- -. ..
blackpearl said:
Here is the math:

Radius of earth = r (meters)
Circumference of earth = 2πr
Initial Length of ribbon = 2πr
New length of ribbon = 2πr + 1

Height by which the ribbon can be raised = (Radius of the circle of ribbon) - (Radius of earth)

H = (2πr + 1)/2π - r
= r + 1/2π - r
= 1/2π
= 16 cm

Notice that H doesn't depend on the radius of the earth, which means that whether you wrap the ribbon around a football, or around earth, or around the sun, if you increase the lenght by 1m it can always be raised by 16 cm!!

Amazing, isn't it?

With what unit you measure 1/2π? It is a constant number. not a measurable unit.
And again, use your logic, in the case of big spheres like sun, 1 meter is negligible. You definitely can't increase 16 cm.
And finally, you can't have seen these type silly things in discovery channel.
 

fun2sh

Pawned!... Beyond GODLIKE
@cyberscriber
i think u need to study maths a bit more coz u cant understand puch a simple solution. THIS THING IS A TRUE FACT AND NOT A SILLY THING
 

rocket357

Security freak
Notice one thing about this:

The radius increases a set amount with a set increase in circumference. In other words, the radius of a circle with circumference = 0 m would be 0 m, right?

What then, is the radius of a circle with circumference = 1 m?

C = 1 m
r = 1 m/(2(pi)) = 15.9 cm

What if the circle has a circumference of 2 m?

C = 2 m
r = 2 m/(2(pi)) = 31.8 cm = 2(15.9cm)

3 m?

C = 3 m
r = 3 m/(2(pi)) = 47.7 cm = 3(15.9cm)

See the pattern? It doesn't sound logical, but blackpearl's right.
 

Rollercoaster

-The BlacKCoaT Operative-
according to math we get a equation as follows
(forget earth, take a arbitrary circle with a concentric circle inside with radius R, D is the difference in the Circumference of the circles and H is the difference in the radius of the two)

i.e

Big circle Circumference = Small circle Circumference + Increase in Circumference

2 x Pi (R + H) = (2 x Pi x R ) + D
2PiR + 2PiH = 2PiR + D
2PiH = D (2PiR canceled on both sides)
H = D / 2Pi
H = D / (2x22/7)
H = D / (44/7)
H = (7/44 ) x D
H = 0.159 x D

note that H will have the same unit as D

H (for D = one meter) is = 0.159 x 100cm = 15.9 cm

It doesnt seem likely to be true for all circles. lets check

lets reverse (all units in cm)
Small circle:
radius = 1
so Circumference = 2PiR = 2Pi

Big circle:
Circumference = Circumference of small circle + 100cm = 2pi + 100 (as given increase by 1 meter)
so radius = (2pi+100)/2pi = 1 + 100/2Pi = 1 + 100/(44/7) = 1 + 15.9 = 16.9

so difference between radius i.e H = big radius - small radius = 16.9 - 1 = 15.9

another one:
Big circle:
radius = 1,000,000,000
Circumference = 2,000,000,000 x Pi

Small circle:
Circumference = (2,000,000,000 x Pi ) - 100 (... 1 meter decrease )
radius = ((2,000,000,000 x Pi ) - 100 )/2Pi = 1,000,000,000 - 100/2pi = 1,000,000,000 - 15.9 = 999999984.1

difference in radius = big radius - small radius = 1,000,000,000 - 999999984.1 = 15.9 !!!!

:) hence proved

* sorry guys i typoed the Circumference as diameter.. i have corrected it now.
 
Last edited:

Pathik

Google Bot
Rollercoaster said:
according to math
we get a equation as follows
(forget earth, take a arbitrary circle with a concentric circle inside with radius R, D is the difference in the diameter of the circles and H is the difference in the radius of the two)

i.e

Big circle diameter = Small circle diameter + Increase in diameter

2 x Pi (R + H) = (2 x Pi x R ) + D
2PiR + 2PiH = 2PiR + D
2PiH = D (2PiR canceled on both sides)
H = D / 2Pi
H = D / (2x22/7)
H = D / (44/7)
H = (7/44 ) x D
H = 0.159 x D
hey but if we take H and D as u say.... then H shd be 2*D as H is the Diff in the radiuses and D is the diff in diameters!! or did u confuse Diameter wit circumference???
 

abhi_10_20

Cool and Calm
yeah...
shouldnt it be............

Big circle diameter = Small circle diameter + 2*(Increase in diameter) ??
 
OP
blackpearl

blackpearl

The Devil
If you are finding it difficult to believe, do your own calculation. Lets take some arbitrary value.

R= 100m = 10000cm
Circumference = 2x3.14x10000 = 62831.85 cm.

Increase this circumference by 1m i.e. 100cm
New circumference = 62831.85 + 100 = 62931.85 cm

New R = 62931.85/(2x3.14) = 10015.9 cm

Difference in radius = 15.9cm!!

Lets do again in reverse. This time R = 5869cm (really arbitrary)

Circumference = 2x3.14x5869 = 36876.01 cm

Now lets increase R by 15.9cm.

New R = 5869 + 15.9 = 5884.9 cm
New circumference = 2x3.14x5884.9 = 36975.91 cm

Difference in circumference = 36975.91 - 36876.01 = 99.9cm = 1m!!

Do you need any more proof? :)
 

infra_red_dude

Wire muncher!
blackpearl said:
H = (2πr + 1)/2π - r
= r + 1/2π - r
= 1/2π
= 16 cm
the math is absolutely correct but the error is in the last line. as somebody pointed out 1/2Pi is a constant and has no units! what if the circle radius has units of 'km'? then H = 1km/2Pi = 159.15m.. hence the answer should be:

H = 0.15915 units. it is constant for all radii.
 
OP
blackpearl

blackpearl

The Devil
^^ No. I said initially R is in cm. So the final answer is also in cm.
Its actually 1/2π cm. I didn't write "cm" in every line, jujst in the final answer.

If you take R in km, then the equation will be:

H = (2πr + .001)/2π - r Km
= r + .001/2π - r Km
= .001/2π Km
= 16 cm
 
F

fannedman

Guest
In general, increase in height is directly proportional to increase in length(obviously) with the proportionality constant being 1/(2pi)
i.e
H=X/(2pi)
where X is increase in length

needless to say whatever units X is in, the same are the units of H
 

infra_red_dude

Wire muncher!
blackpearl said:
^^ No. I said initially R is in cm. So the final answer is also in cm.
Its actually 1/2π cm. I didn't write "cm" in every line, jujst in the final answer.

If you take R in km, then the equation will be:

H = (2πr + .001)/2π - r Km
= r + .001/2π - r Km
= .001/2π Km
= 16 cm
you are correct!!! my mistake! i read the whole thing again..... you said that the cirumference was increased by 1 m. thats correct. i wrongly read it as increase in 1 unit. if its 1m then the answer is 0.15915m = 15.915cm. if the increase was 1 km then the answer wud be H = 0.15915km = 159.15m. that 1 m is the key there :) the answer is independent of the dimensions but dependent on the unit of increase of circumference.
 

subratabera

Just another linux lover.
Well...if you think that's strange...try this...

here's the problem:

A railway track is a mile long. Well, let's say 1000 metres, like so
It is one continous ribbon of steel, pinned down at both ends, but free to move in between.

The temperature rises and the track expands to 1002 metres.
If the track buckles (as shown), how high would you think the centre rises?

curious-1.gif


Scroll for the answer
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Okay, let's estimate that height.
Assuming a right triangle, we have that h2 = 5012 - 5002 = 1001.
So our estimate is h = √1001 = 31.6 metres.

curious-2.gif


Well, it's just a simple estimate ... but, as I recall from the actual solution to this problem, it's within 10%.

Roughly. That's' close to 100 feet.

Interesting, eh? Most people would guess as you have. Something small. when it's actually HUGE!


SOURCE(worth a visit): http://gummy-stuff.org/curious.htm
 

infra_red_dude

Wire muncher!
^^^ whatever but its jus a approximation. as its not an actual hypotenuse but an arc. of corz the answer won't change much tho if its considered an arc.
 

anantkhaitan

Burning Bright
blackpearl said:
Notice that H doesn't depend on the radius of the earth, which means that whether you wrap the ribbon around a football, or around earth, or around the sun, if you increase the lenght by 1m it can always be raised by 16 cm!!
Absolutely, i don't know why some members disagree with the fact
do it the variable way

Assume radius be : r (meters)
circumference = 2πr (meters)
new circumference= 2πr + 1 (meters)
new radius = (2πr + 1)/2π (meters)
Height (H) = new radius - old radius (meters)
Code:
= (2πr + 1 )/2π - r (meters)
= r + 1/2π - r (meters)
= 1/2π (meters)
= 15.923566879 (cm)

So conclusion whatever may be 'r' , H remains same i.e. as mention above
 
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