How long is the piece of wire?

WBahn

Joined Mar 31, 2012
29,976
There is a known ratio of sphere volume to cylinder volume.

Edit: just like cone volume.
First off, there is no "known ratio" since the volume of a sphere is a function of one variable and the volume of a (right-circular) cylinder is a function of two.

But even if you constrain things so as to get your "known ratio" of 4/3, you can't use it for this problem because it is derived from the formulas for the volume of a sphere and the volume of a cylinder, which you are not allowed to use either directly or indirectly. You aren't even allowed to derive those formulas from first principles. So unless you can establish how to get that 4/3 ratio via some means that is independent of the volume formulas, you can't use it.
 

BR-549

Joined Sep 22, 2013
4,928
These fundamental ratios have been known for a very long time. You make check it yourself.

Calculate the volume of ANY sphere. Encapsulate sphere in cylinder. Calculate cylinder volume.

Sphere volume = .66 cyl volume.

Encapsulate cone in cyl. Cone volume = .33 cyl volume.
 

jpanhalt

Joined Jan 18, 2008
11,087
I think these volume ratios were found with water displacement.......not mathematically derived.
BR-549 has all but solved the problem, but not finished the calculations with this knowledge.
If such experimentally derived ratios are allowed, then the answer cannot be "exact" and the original question asked for an exact result,
Hymie said:
...exactly how long is the 1mm diameter wire
John
 

BR-549

Joined Sep 22, 2013
4,928
What is the volume of a 300Rmm, 400Hmm cylinder?

150Rmm, 1600Hmm?

75Rmm, 6400Hmm?

Why can't we get exact? We only have an additional proportionality, on the last step........to get a 1 mm dia. Should be able to get pretty close.

Check it with your math......my desktop is too short for my dough.
 

BR-549

Joined Sep 22, 2013
4,928
New question. What is the spherical volume of a 1mm dia wire 2000 meters long?

Show your work. You may use calculus......or symmetry.

Hint.....convert length to cyl......convert cyl to sphere.
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
Ok. Could you finally give the full answer to the excercise you proposed?
Since the volume of a sphere is equal to a cylinder of the sphere’s diameter of length 2/3 its diameter – the 600mm diameter sphere is equivalent in volume to a 600mm diameter cylinder of 400mm in length.

Given that areas of circles are in the same proportion to each other as the square of their diameters – so the ratio of 0.6m to 0.001m (by their squares) becomes 0.36 to 0.000001, or 1:360,000. So the length of the wire is 400mm x 360,000 = 144,000 metres.

Obtained without using pi or the formula for the volume of a sphere/column/circle.

Confirming the answer to be correct:-

The volume of a sphere is given by 4/3 x pi x r˄3
The volume of a column is given by H x pi x r˄2

So we get 4/3 x pi x 0.3˄3 = H x pi x 0.0005˄2 - (solve for H)

4/3 x 0.027 = H x 0.00000025

0.036/0.00000025 = H

H = 144,000 metres
 

jpanhalt

Joined Jan 18, 2008
11,087
Since the volume of a sphere is equal to a cylinder of the sphere’s diameter of length 2/3 its diameter – the 600mm diameter sphere is equivalent in volume to a 600mm diameter cylinder of 400mm in length.

Given that areas of circles are in the same proportion to each other as the square of their diameters – so the ratio of 0.6m to 0.001m (by their squares) becomes 0.36 to 0.000001, or 1:360,000. So the length of the wire is 400mm x 360,000 = 144,000 metres.

Obtained without using pi or the formula for the volume of a sphere/column/circle.

Confirming the answer to be correct:-

The volume of a sphere is given by 4/3 x pi x r˄3
The volume of a column is given by H x pi x r˄2

So we get 4/3 x pi x 0.3˄3 = H x pi x 0.0005˄2 - (solve for H)

4/3 x 0.027 = H x 0.00000025

0.036/0.00000025 = H

H = 144,000 metres
Show us your work, particularly how you arrived at the ratio of 2/3, which is exactly the ratio that I used. Should I have just said, "It's common knowledge?"
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
Show us your work, particularly how you arrived at the ratio of 2/3, which is exactly the ratio that I used. Should I have just said, "It's common knowledge?"
Archimedes discovered this, so it has been known for some time.
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
Please give a link. Archimedes did not use calculus. You claimed the relationship was not empirical.
Given that calculus was developed in 17th-century by Newton and Leibniz – it is unlikely that Archimedes used it (in the modern sense).
 

jpanhalt

Joined Jan 18, 2008
11,087
That (post #55) does not answer the question of showing your source/work. How did Archimedes solve that relationship by accepted geometric analysis and without using the proportionality constant we call pi.

Your link, "including the area of a circle, the surface area and volume of a sphere, and the area under a parabola." would not satisfy even the most benevolent reviewer. Of course, everything in Wikipedia goes back to philosophy (https://en.wikipedia.org/wiki/Wikipedia:Getting_to_Philosophy ). That is not a do all reference for academic work.

In sum, waste of time.

John
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
That (post #55) does not answer the question of showing your source/work. How did Archimedes solve that relationship by accepted geometric analysis and without using the proportionality constant we call pi.

Your link, "including the area of a circle, the surface area and volume of a sphere, and the area under a parabola." would not satisfy even the most benevolent reviewer. Of course, everything in Wikipedia goes back to philosophy (https://en.wikipedia.org/wiki/Wikipedia:Getting_to_Philosophy ). That is not a do all reference for academic work.

In sum, waste of time.

John
Then perhaps I will claim to have discovered that the volume of a sphere is equal to a cylinder of the sphere’s diameter of length 2/3 its diameter and not Archimedes.
 

WBahn

Joined Mar 31, 2012
29,976
Archimedes discovered this, so it has been known for some time.
I would argue that in determining the formula for the excluded volume of a sphere, you have effectively found a formula for the volume of a sphere – and therefore this is not allowed.
Why wouldn't you argue that by using a "well-known" ratio for the volume of a sphere to the circumscribing cylinder that you have effectively found a formula for the volume of a sphere -- and therefore this is not allowed.

I would think that the use of the formula for the perimeter of a circle or the surface area of a sphere are allowed – with the proviso that they are not a surrogate for the volume of a sphere or the area of a circle (and of course you cannot use the value of pi).
How is using a "well-known" ratio for the volume of a sphere to the circumscribing cylinder not a surrogate for the volume of a sphere?
 

jpanhalt

Joined Jan 18, 2008
11,087
Then perhaps I will claim to have discovered that the volume of a sphere is equal to a cylinder of the sphere’s diameter of length 2/3 its diameter and not Archimedes.
Well, if you are comfortable with that guess as a justification, then I am even more comfortable calling this a waste of time.
 
Top