Assume the world is perfectly round with no mountains and valleys and you tied a piece of string around it with the ends just coming together.
If you added another 3 foot to it. How far off the surface of the world would it be all the way around.??
Answer= nearly 6" off the surface all the way around.
Now that's surprising, unless you know different?
Now if you tied a piece of string around one of your E type wheels and added 3 ft to it. How far would it be off the surface of the wheel?
Answer . Exactly the same, nearly 6"
That's assuming you do not have square wheels.
The boffins amongst you can now work it out exactly for us. I will accept to two decimal places
Brain Teaser for Christmas
#1 Brain Teaser for Christmas
Tony (E typed)
1962 E Type Series 1 Roadster (OTS)
Tony
1962 E Type Series 1 Roadster (OTS)
Tony
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#2 Re: Brain Teaser for Christmas
Tony wrote:Assume the world is perfectly round with no mountains and valleys and you tied a piece of string around it with the ends just coming together.
If you added another 3 foot to it. How far off the surface of the world would it be all the way around.??
Answer= nearly 6" off the surface all the way around.
Now that's surprising, unless you know different?
Now if you tied a piece of string around one of your E type wheels and added 3 ft to it. How far would it be off the surface of the wheel?
Answer . Exactly the same, nearly 6"
That's assuming you do not have square wheels.
The boffins amongst you can now work it out exactly for us. I will accept to two decimal places
X=distance of string from surface
R=radius of earth or wheel or any other circular thingy.
C=circumference of above thingy.
1) 2piR=C
2) 2pi(R+X)=C+3
1) R=C/2pi
Substituting for R in 2):
2pi(C/2pi+X)=C+3
2piC/2pi+2piX=C+3
2piX=(C+3)-2piC/2pi
X=(C+3)/2pi-(2piC/2pi)/2pi
X=(C+3)/2pi-C/2pi
X=C/2pi+3/2pi-C/2pi
X=3/2pi
X=3/6.284
X=5.73ins (say 6)
Pete
'71 S3 2+2
'71 S3 2+2
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christopher storey
- Posts: 5698
- Joined: Sun Mar 09, 2008 3:07 pm
- Location: cheshire , england

#3
Or, put more succinctly,( I'm only a 6 cylinder man!) since radius = circumference divided by 2pi , the radius increases by 36/6.2832 inches = 5.73 inches
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#4
But why use one line when a dozen will do?christopher storey wrote:Or, put more succinctly,( I'm only a 6 cylinder man!) since radius = circumference divided by 2pi , the radius increases by 36/6.2832 inches = 5.73 inches
As an aside, where have all the "cartoons" gone from threads recently? I don't need sunglasses to read with anymore.
Pete
'71 S3 2+2
'71 S3 2+2
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#5
Heuer wrote::bigangry: :laola: :bananadance: :bang: :bounce2: :hatepc: :celebrate: :santa:
Oh no, Disney's back!
My brain hurts!
Pete
'71 S3 2+2
'71 S3 2+2
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#6
Hey, it ok for you clever blokes, I did it the hard way. Anybody want to buy a very, very large ball of string?
Tony (E typed)
1962 E Type Series 1 Roadster (OTS)
Tony
1962 E Type Series 1 Roadster (OTS)
Tony
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#7 Re: Brain Teaser for Christmas
I have long since thought 'boffins' can prove anything they want due to the fact that no one else can understand algebra (of the Einstien type).
I once heard that in the 'great theory of everything' a boffin had a fantastic calculation that revolutionised the thinking on the creation of the universe. But for it to work, there must be 14 (or whatever it was) different dimensions in the universe and therefore they concluded that there are 14 dimensions in the universe. It didn't seem to occur to him/them that there may only be 3 dimensions in existence and the calculation was just wrong.
I once heard that in the 'great theory of everything' a boffin had a fantastic calculation that revolutionised the thinking on the creation of the universe. But for it to work, there must be 14 (or whatever it was) different dimensions in the universe and therefore they concluded that there are 14 dimensions in the universe. It didn't seem to occur to him/them that there may only be 3 dimensions in existence and the calculation was just wrong.
Andrew Day. Former owner of S1A 4.2 2+2. Current cars; Aston Martin DBS 1968, Ferrari F355 & Fiat Coupe 20vt. Flag of choice; EU
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