Straightforward Easy Maths Problem

[renormalised (Carl)]

Got a simple maths problem for you all

If one day can be represented by a sheet of A4 paper (in thickness = duration), how high would the pile of paper be representing the age of our planet (4.56 billion years)?

Should be easy

[mithrandir (Andrew)]

What weight paper?

500 sheets of the usual grade of A4 paper is about 5cm high. That's 0.1mm per sheet.

0.1 mm * 365.2425 days * 4.56e9 years is ~1,665,000Km

Now my copy of the Lord of the Rings is on rice paper. 21mm for 1,200 pages. That makes each sheet 0.0175mm

0.0175 mm * 365.2425 days * 4.56e9 years is ~291,000Km

[RickS (Rick)]

Originally Posted by mithrandir

0.1 mm * 365.2425 days * 4.56e9 years is ~1,665,000Km

Out by a factor of 10 somewhere? I got 183,205km assuming a thickness of 110um (80gsm paper).

[sjastro]

1 day = 0.0027 year.

Total number of sheets = (4.56/0.0027) X 10^9 (assuming English billion). = 1.69 X 10^12 sheets.

Assuming a sheet has a thickness of 0.1mm.

Total thickness = (1.69 X 10^12) X 0.1mm
= 1.69 X 10^11 mm
= 1.69 X 10^ 8 m or 169,000,000 m.

Regards

Steven

[Nesti (Mark)]

ummmmm...42?!

[mithrandir (Andrew)]

Originally Posted by RickS

Out by a factor of 10 somewhere? I got 183,205km assuming a thickness of 110um (80gsm paper).

Doh.

1mm=1e-3m. cm vs. mm confusion.

So yes, 166,000Km for 80gsm and 29,100Km for the rice paper.

[Robh (Rob)]

One sheet ~ 0.1mm.
4.56 billion years ~ 4.56 x 10^9 x 365.256 = 1.666 x 10^12 days
1 mm = 10^-6 kms
Thickness = 1.666 x 10^12 x 0.1 x 10^-6 = 166600 km
which pretty much agrees with Andrew.

If thickness of sheet ~ 0.11mm then answer would be
166600 x 0.11/0.1 = 183260 km
which pretty much agrees with RickS.

[sjastro]

The answer has a dimension of length L, the height of the pile of paper.

Multiplying the time in days T, by thickness L does not give a length measurement as the dimension is TL.

Regards

Steven

[Coen]

The question as phrased indicated represent one day by a fixed quantity which has dimension length L - there is an implied per day in that construct. When multiply by a certain number of days the length units of L maintained as days cancel out.

4,560,000,000 * 365.24219 = 1665504386400 days (using 365.24219 days in a year and assuming the Earth has maintained the same rotation rate on its axis and maintained its orbital size - bold assumptions)

Thickness of the paper is - take your pick but scale it by 1670000000000 (1665504386400 round to two sig figs) and you'll get your answer with all the assumptions made.

Given the boldness of the assumptions the answer is pretty much meaningless.

[Steffen]

Originally Posted by Coen

Given the boldness of the assumptions the answer is pretty much meaningless.

Not to mention its practical utility

Cheers
Steffen.

[renormalised (Carl)]

Originally Posted by Coen

The question as phrased indicated represent one day by a fixed quantity which has dimension length L - there is an implied per day in that construct. When multiply by a certain number of days the length units of L maintained as days cancel out.

4,560,000,000 * 365.24219 = 1665504386400 days (using 365.24219 days in a year and assuming the Earth has maintained the same rotation rate on its axis and maintained its orbital size - bold assumptions)

Thickness of the paper is - take your pick but scale it by 1670000000000 (1665504386400 round to two sig figs) and you'll get your answer with all the assumptions made.

Given the boldness of the assumptions the answer is pretty much meaningless.

There's no point in becoming too cerebral about the question. If you take things too literally, you would spend forever trying to work it out as the earth's rotational rate has varied constantly through time. Plus, you would have to take into account the changes in orbital eccentricity over time etc etc etc. That's why I made the assumption of a "standard" day length for the whole period of time. Makes the calculations so much easier and prevents an aneurysm trying to be too literal about the question

[renormalised (Carl)]

In any case, the answer is this...

Given 1 sheet of paper is 0.1mm (10^-7km) thick:

(4.56 x 10^9 x 365) x 10^-7

= 166,440km

[Coen]

Or not quite halfway to the Moon.

[renormalised (Carl)]

Ok, what is the total amount of energy stored in that pile of paper??

And what is it equivalent to??

[sjastro]

Originally Posted by renormalised

In any case, the answer is this...

Given 1 sheet of paper is 0.1mm (10^-7km) thick:

(4.56 x 10^9 x 365) x 10^-7

= 166,440km

Sorry to be nitpicky Carl.

While the magnitude of the answer is correct it is expressed in km.day. not km.

To cancel out the day unit, work out the number of sheets "n" in the stack instead.
This is simply (4.56 X 10^9 X 365)/1 = 1.664 X 10^12

While dividing by one (one day) seems trivial, the number "n" is a dimensionless quantity. Multiplying "n" by thickness gives the correct unit.

Regards

Steven

[renormalised (Carl)]

Originally Posted by sjastro

Sorry to be nitpicky Carl.

While the magnitude of the answer is correct it is expressed in km.day. not km.

To cancel out the day unit, work out the number of sheets "n" in the stack instead.
This is simply (4.56 X 10^9 X 365)/1 = 1.664 X 10^12

While dividing by one (one day) seems trivial, the number "n" is a dimensionless quantity. Multiplying "n" by thickness gives the correct unit.

Regards

Steven

That's Ok...I assumed the number of sheets from the workings and didn't bother with defining "n"...though that's not technically correct

[multiweb (Marc)]

... I'll never look a Reflex A4 box the same way... I'm scared for life...

[Coen]

As long as the paper is straight, we can go forward. (yes I am a Dad...)

Regarding amount of energy stored in the pile paper: what conversion process? Talking potential energy? Chemical energy? Nuclear Energy? Or perhaps how much energy can be expended filling up every page with cerebral thoughts?

[mithrandir (Andrew)]

Originally Posted by sjastro

Sorry to be nitpicky Carl.

While the magnitude of the answer is correct it is expressed in km.day. not km.

To cancel out the day unit, work out the number of sheets "n" in the stack instead.
This is simply (4.56 X 10^9 X 365)/1 = 1.664 X 10^12

While dividing by one (one day) seems trivial, the number "n" is a dimensionless quantity. Multiplying "n" by thickness gives the correct unit.

Regards

Steven

You've been missing the implied unit in the paper. 1 sheet per day.

thickness of 1 sheet * sheets/day * days/year * years all cancels leaving whatever linear unit the sheet thickness is in.

The assumption for days/year (365, 365.25 Julian calendar, 365.2425 Gregorian calendar, 365.242189 mean tropical year for 2000 according to Laskar) mean whatever value you come up with can only be an estimate.

And there's no allowance for the changes caused by the impact that created the moon (which would have made a large change), or meteor/asteroid/comet impacts, or tidal slowing of the rotation.

But it's an amusing interlude.

[renormalised (Carl)]

Guys....I think you need to remember something which is very important in science...sometimes thinking too much about a simple problem can get you into trouble