thom
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2 Strange mathematical facts .
- Thread starter Dave7
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Maz
Guru
thom
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As people seemed to be mostly interested in numbers here (not really my primary interest) I thought I would post an interesting fact that I don't think gets mentioned enough.
There are four lots of what many people might regard as 'numbers' and their dimensions are 1,2,4 and 8. Most people only know about the first two. They are real numbers, complex numbers, quaternions and octonions of respective dimensions 1,2,4 and 8. The reason for this among other things is a theorem to do with Normed Division Algebras sometimes called (confusingly) Hurwitz's Theorem.
There are four lots of what many people might regard as 'numbers' and their dimensions are 1,2,4 and 8. Most people only know about the first two. They are real numbers, complex numbers, quaternions and octonions of respective dimensions 1,2,4 and 8. The reason for this among other things is a theorem to do with Normed Division Algebras sometimes called (confusingly) Hurwitz's Theorem.
Fnaar
Smutmaster General
- Location
- Thumberland
Just got it
threebikesmcginty
Corn Fed Hick...
- Location
- ...on the slake
Shouldn't that be 58008?
Is there any significant compression when you coil it around? Also it would help if you knew the width of the foil or even the total number of times it is wound round.
The outer circumference is 2πr = 2 x π x 120mm which is about 754mm.
The inner circumference is rather similar and 2 x π x 25mm which is about 157mm.
You could get a crude approximation by using an arithmetic progression.
We know the total length is 250,000mm. We know the first term is 157mm and the last term is 754mm
S_n = (n/2) (a_1 + a_n)
250,000 = (n/2) (157+754)
500,000 = 911n
n=549 coils. And this would mean each rotation added about a third of a mm to the thickness.
It is obviously somewhere between 330 coils and 1600 coils.
Again you'd then feed it back into the arithmetic progression.
a_n = a_1 + (n-1)d and so on.
a_2 = 157 + (2-1)d
Otherwise you could use an Archimedean spiral with the above information I guess
r = a + bθ and we have b from earlier
Anywho, sorry quite busy. Sure someone else will think of something better.
The outer circumference is 2πr = 2 x π x 120mm which is about 754mm.
The inner circumference is rather similar and 2 x π x 25mm which is about 157mm.
You could get a crude approximation by using an arithmetic progression.
We know the total length is 250,000mm. We know the first term is 157mm and the last term is 754mm
S_n = (n/2) (a_1 + a_n)
250,000 = (n/2) (157+754)
500,000 = 911n
n=549 coils. And this would mean each rotation added about a third of a mm to the thickness.
It is obviously somewhere between 330 coils and 1600 coils.
Again you'd then feed it back into the arithmetic progression.
a_n = a_1 + (n-1)d and so on.
a_2 = 157 + (2-1)d
Otherwise you could use an Archimedean spiral with the above information I guess
r = a + bθ and we have b from earlier
Anywho, sorry quite busy. Sure someone else will think of something better.
threebikesmcginty
Corn Fed Hick...
- Location
- ...on the slake
srw
It's a bit more complicated than that...
In which case it's not foil.
[edit]
The length is 250,000mm, not 25,000 - so your figures are out by a factor of 10.
Ta. Suppose you shouldn't do things in a hurry. Makes the numbers a lot better at about 549 rotations which makes me a lot happier guessing and the thickness is about a third of a mm.
edit: I have seen thicker foil in a workshop, albeit it only about 1mm thick.
edit: I have seen thicker foil in a workshop, albeit it only about 1mm thick.
thom
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[QUOTE 2091017, member: 9609"]As there are a few mathematically minded people on here can anyone help me out with an equation for the following problem
We have a roll of foil at work and I would like to estimate how much we have left from either the circumference or the diameter. When it s new it is 250m long, the inner diameter is 50mm and the outer diameter is 240mm. I would like some formula that would tell me how much is left from the diameter.[/quote]
Yeah it's quite easy, at least for a rule of thumb.
Your remaining length will be (in meters) near enough
250 x ( d^2 - 5^2) / ( 24^2 - 5^2) = 250 x ( d^2 - 25 ) / 551
where d is the current diameter measure in centimetres.
(Just for the sake of pedantic clarity, d^2 is d squared, i.e.. d times d. I only say that to confirm notation.)
First person to proove this right gets secondary bragging rights ;-)
We have a roll of foil at work and I would like to estimate how much we have left from either the circumference or the diameter. When it s new it is 250m long, the inner diameter is 50mm and the outer diameter is 240mm. I would like some formula that would tell me how much is left from the diameter.[/quote]
Yeah it's quite easy, at least for a rule of thumb.
Your remaining length will be (in meters) near enough
250 x ( d^2 - 5^2) / ( 24^2 - 5^2) = 250 x ( d^2 - 25 ) / 551
where d is the current diameter measure in centimetres.
(Just for the sake of pedantic clarity, d^2 is d squared, i.e.. d times d. I only say that to confirm notation.)
First person to proove this right gets secondary bragging rights ;-)
Fnaar
Smutmaster General
- Location
- Thumberland
The important thing User9609 wants to know (possibly) is will there be enough to go over the turkey at Christmas 
