How long is the piece of wire?

Thread Starter

Hymie

Joined Mar 30, 2018
762
In determining the answer to the question below, you do not know the value of pi and cannot use the formula for the volume of a sphere, or the formula for the area of a circle.

I have a solid steel sphere, 600mm in diameter and a machine that can convert the sphere into a single length of wire, with no wastage of material.
The produced wire has the same density as the original sphere.

I have set the machine to produce a wire of 1mm in diameter - exactly how long is the 1mm diameter wire, produced from the 600mm diameter steel sphere?

In giving your answer show your working.
 

wayneh

Joined Sep 9, 2010
16,271
The volume of a sphere is 4/3 π r^3. The volume of a column is π r^2 L. You've given both r values. Solving for L is just algebra.
 

Thread Starter

Hymie

Joined Mar 30, 2018
762
The volume of a sphere is 4/3 π r^3. The volume of a column is π r^2 L. You've given both r values. Solving for L is just algebra.
Perhaps I should add that the volume for a column is not permitted either – it being the area of a circle multiplied by the height.
 

crutschow

Joined Mar 14, 2008
24,397
WTF? The rules weren’t there when I first answered. Do I also have to stand on my head and use only my toes to produce an answer?
It's stated in the TS's first paragraph, and the post doesn't show it having been edited, so I would think it's been there from the beginning. :rolleyes:
 

Thread Starter

Hymie

Joined Mar 30, 2018
762
My answer is 480m.
pi cancels out. Hence you don't need to use pi.
I cannot see how you can solve this without knowing the volumes of the sphere and cylinder.
Yes, it’s relatively easy knowing the formula for the volume of a sphere and cylinder – all will be revealed within a few days if no one posts the correct answer.
 

wayneh

Joined Sep 9, 2010
16,271
It's stated in the TS's first paragraph, and the post doesn't show it having been edited, so I would think it's been there from the beginning. :rolleyes:
I’ll post a screenshot when I get back to my laptop, if the TS doesn’t confirm the change first.
 
Assume your machine works by forcing the sphere through a series of ever smaller holes. This wouldn't really work, but this is a puzzle, not an engineering design.
The first hole is 300 mm in diameter; the resulting blob is 4 * 600 mm.
Next hole is 30 mm in diameter leaving a blob of 100 * 4 * 600 mm
Next is 3 leaving a blob of 100 * 100 *4* 600 mm.
Finally a 1 leaving a wire of 9 * 100 * 100 * 4 * 600 mm.

I don't know if I'm using an outlawed piece of knowledge by knowing the hole to volume increase is ratio of the holes squared.
 

jpanhalt

Joined Jan 18, 2008
8,766
Just to clarify, when you say "can't use the formulas for area or volume" do you mean you cannot use any similar formula (e.g., the volume is proportional to the diameter^3) or that you cannot use the (4/3)( pi) *(r^3) specifically, but you can assume the volume is proportional to r^3? In other words, getting rid of pi in the calculation is extremely simple.
 

Thread Starter

Hymie

Joined Mar 30, 2018
762
Assume your machine works by forcing the sphere through a series of ever smaller holes. This wouldn't really work, but this is a puzzle, not an engineering design.
The first hole is 300 mm in diameter; the resulting blob is 4 * 600 mm.
Next hole is 30 mm in diameter leaving a blob of 100 * 4 * 600 mm
Next is 3 leaving a blob of 100 * 100 *4* 600 mm.
Finally a 1 leaving a wire of 9 * 100 * 100 * 4 * 600 mm.

I don't know if I'm using an outlawed piece of knowledge by knowing the hole to volume increase is ratio of the holes squared.
I like your thinking – it is not an outlawed piece of knowledge that you are using (only the formula for the volume of a sphere/column/area of a circle are not allowed).

But your solution gives a result of 216,000 metres for the wire length – does this agree using the disallowed formula?
 

Thread Starter

Hymie

Joined Mar 30, 2018
762
Using proportions and neither of the "outlawed" formulas per se or pi, I get 144,000 meters. Is that legal (see post #12)?.
That’s hardly showing your working – claiming that you have not used the outlawed formula and giving an answer.
 

Thread Starter

Hymie

Joined Mar 30, 2018
762
Likewise you have not responded to post #12. Rest assured, I have the 6th grade algebra worked out.
It has already been explained above that using the disallowed formula does not require knowing the value of pi – show your working; I and others will then no doubt point out if you have used the formula for the volume for a sphere/column.
 

jpanhalt

Joined Jan 18, 2008
8,766
If you use algebra and can assume the volume of the sphere is proportional to diameter^3 (D^3), area is proportional to diameter ^2 (D^2), and length = L in mm), you get:
kD1^3 = mD2^2 x L , so the length is (k/m)D1^3 mm
I assume someone can derive from first principles the proportionality constants. It is clear you don't need to know pi. Now, (4/3)/2^3 = 4/24 = 1/6 for k; and following similar reasoning, m= 1/4. So, L = 4/6x600^3 = 144X10^3 m

That solution, if allowed, makes the problem very similar to the age-old problem of how much string would you need to add to a string that was wrapped around the Earth to raise it 1 foot. Earth radius = 3,959 miles. Anyone who didn't have an umpteen digit calculator (i.e., used a slide rule) and solved it knew how to solve the simple algebra.
 
Last edited:

Thread Starter

Hymie

Joined Mar 30, 2018
762
If you use algebra and can assume the volume of the sphere is proportional to diameter^3 (D^3), area is proportional to diameter ^2 (D^2), and length = L in mm), you get:
kD1^3 = mD2^2 x L , so the length is (k/m)D1^3 mm
I assume someone can derive from first principles the proportionality constants. It is clear you don't need to know pi. Now, (4/3)/2^3 = 4/24 = 1/6 for k; and following similar reasoning, m= 1/4. So, L = 4/6x600^3 = 144X10^3 m

That solution, if allowed, makes the problem very similar to the age-old problem of how much string would you need to add to a string that was wrapped around the Earth to raise it 1 foot. Earth radius = 3,959 miles. Anyone who didn't have an umpteen digit calculator (i.e., used a slide rule) and solved it knew how to solve the simple algebra.
Very good – but you have used the formula for the volume of a sphere to determine the value of k in your answer – which is not allowed.
 

jpanhalt

Joined Jan 18, 2008
8,766
Very good – but you have used the formula for the volume of a sphere to determine the value of k in your answer – which is not allowed.
Why didn't you answer that at post #12? Let's see whether anyone can do it without using a simple derivation of those formulas (i.e., the ratio of volume to bisecting area).
 
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