I Need Some Help, Please

[AV1611VET]

I'm trying to ascertain the radius of a five-mile thickness of water, if it was sheared off at the earth's surface and ballooned out until it had a thickness of 1cm.

Can anyone help me with this?

In other words, with a starting point of 3955 miles inner radius, and a 3960 miles outer radius, how far out would this torus (?) of water have to be ballooned, in order for it to have a thickness (depth) of 1 centimeter?

 

[NailsII]

I'd love to help but maths isn't my strongest point AV, that's why I became a biologist ;-)

If I get time later tonight I'll have a stab with my pencil & paper if no-one else has a reasonable answer by then.

PS - do you want me to use pi=3 or pi=3.14 etc. (only kidding)

 

[AV1611VET]

I'd love to help but maths isn't my strongest point AV, that's why I became a biologist ;-)

If I get time later tonight I'll have a stab with my pencil & paper if no-one else has a reasonable answer by then.

Thanks, Nails --- I'd appreciate it!

PS - do you want me to use pi=3 or pi=3.14 etc. (only kidding)

 

[Washington]

I got as far as the volume of the shell you're talking about: 52[sup]11[/sup] Cubic miles (5,200,000,000,000, (5.2 trillion) cubic miles) at which point the numbers got to large to deal with.

 

[plindboe]

4/3 * pi * (3960*1.6*100000)^3 - 4/3 * pi * (3955*1.6*100000)^3 = 4/3 * pi * (r+1)^3 - 4/3 * pi * r^3 <=>

(3960*1.6*100000)^3 - (3955*1.6*100000)^3 = (r+1)^3 - r^3 = 3r^2 + 3r +1 <=>

3r^2 + 3r +1 + (3955*1.6*100000)^3 - (3960*1.6*100000)^3 = 0

Solution is r=566351335008.6678 cm or 3539695,8438 miles.

Perhaps someone else should try to calculate it to make sure I did it right. It's confusing to play with such big numbers, so I'm sure I could have slipped up at some point.

Peter

 

[Tinker Grey]

The volume of water of the flood to 5 miles over the average radius of 6380 kilometers is computed thusly:

1) Convert 5 miles to kilometers &#8211; we&#8217;ll say 8 kilometers for convenience.
2) The volume of a sphere is 4/3 x pi x r[sup]3[/sup].
3) So the volume of water is 4/3 x pi x 6388[sup]3[/sup] less the volume of the Earth or 4/3 x pi x 6380[sup]3[/sup]. The volume of the water is 4/3 x pi x (6388[sup]3[/sup] &#8211; 6380[sup]3[/sup]) = 233,511,593 cubic kilometers.
4) What we now want is 4/3 x pi x (r[sup]3[/sup] &#8211; (r-0.00001)[sup]3[/sup]) = 233,511,593 cu.km.

• So, (r[sup]3[/sup] &#8211; (r-0.00001)[sup]3[/sup]) = 978,131,072
• For convenience, let y = 0.00001
• (r[sup]3[/sup] &#8211; (r-y)[sup]3[/sup]) = r[sup]3[/sup] &#8211;(r³-2r²y+y²r-r²y+2ry²-y³) =
• 2r²y - y²r + r²y - 2ry² + y³ = 978,131,072

5) Let&#8217;s agree that this is messy. Further let us agree that subtracting y[sup]3[/sup] (0.00001[sup]3[/sup] = 10[sup]-15[/sup] cu. Km.) isn&#8217;t worth the effort. After all, that&#8217;s one cubic centimeter.
6) So, 2r²y - y²r + r²y - 2ry² = 978,131,072
7) Or, 2r² - yr + r² - 2ry = 978,131,072/y = 97,813,107,200,000
8) Or, 3r[sup]2[/sup] - ry &#8211; 2ry = 97,813,107,200,000
9) Or, 3r[sup]2[/sup] &#8211; 3ry = 97,813,107,200,000
10) Or, r[sup]2[/sup] &#8211; ry =&#61472;32,604,369,067
11) Using an online &#8220;complete the square calculator&#8221;, r would be about 180,567 km.

So, the distance from the surface of the earth to the shell would be 174,187 km or 108,258 miles or a little less than ½ the distance to the moon.

I used a slightly different set of radii that you did, AV1611VET. If you wish to recompute, here is the complete the square computer I found: http://www.algebrahelp.com/calculators/equation/completingthesquare/

Figuring backward, I see that I am off by a factor of 1000. I don't know if that is my fault or that of the above website. However, 18056700[sup]3[/sup] - 18056699.99999[sup]3[/sup] yields the desired 978,131,072 from 4(a) above. So, the real answer is 18,056,700 km or 11,222,312 miles which is 50 times the distance to the moon or 1/10 the distance to Mars or 1/4 the distance to Venus.

Sorry for the inconvenience.

 

[plindboe]

Hmm, wonder who made the right calculation.

Peter

 

[AV1611VET]

Mamma mia!

Thank you very much, everyone, for the answers!

(For the record, I hope Tinker is right.)

 

[plindboe]

Mamma mia!

Thank you very much, everyone, for the answers!

(For the record, I hope Tinker is right.)

I think I'm right though as it appears Tinker didn't add the right numbers to the online calculator. When one uses the numbers from his equation 10), then one gets 5710023 kilometers which is the result I got. That means that the bubble would have a radius of 3,5 million miles, which is almost 15 times the distance to the Moon.

You can do the calculation online yourself:

http://www.math.com/students/calculators/source/quadratic.htm

A=1
B=-0.00001
C=-32604369067000

Peter

 

[MaxP]

AV and evolutionists working together?

 

[Tinker Grey]

*deleted*

ETA: plindboe is correct ... my line 10 dropped an extra ",666" from C=-32604369066666 or C=-32604369067000 (as plindboe had it).

 

[Tinker Grey]

Oh, and thanks for finding my error, pindboe! It was bugging me.

 

[plindboe]

Oh, and thanks for finding my error, pindboe! It was bugging me.

Me too. Had to go over it a few times to make sure. It's so easy to make errors in math, and it's such a pain when it doesn't turn out right.

But at least I've proved that contrary to NailsII's statement earlier, some biologists do know math.

Peter

 

[Tinker Grey]

Oh the shame of it all. An engineer being shown-up by a biologist!!!!!

 

[plindboe]

Hehe.

 

[Chalnoth]

By the way, for quick-and-dirty calculations, I recommend using Google. It even does units!

(e.g. "1/(72(km/sec/Mpc)) in years" put into a Google search gives the current Hubble time)

 

[AV1611VET]

AV and evolutionists working together?

42

 

[NailsII]

Hmmmmm, maybe this biologist is pretty crap at maths.

I got 41,017,122,522,352cm3 of water which is 6,601,049,596miles.....

That's like over half the distance to pluto at its maximum distance from the sun....

I just punched the numbers into excel and hoped for the best though!!

 

[Chalnoth]

Edit: Nevermind, my mistake.

 

[Chalnoth]

The volume of water of the flood to 5 miles over the average radius of 6380 kilometers is computed thusly:

1) Convert 5 miles to kilometers &#8211; we&#8217;ll say 8 kilometers for convenience.
2) The volume of a sphere is 4/3 x pi x r[sup]3[/sup].
3) So the volume of water is 4/3 x pi x 6388[sup]3[/sup] less the volume of the Earth or 4/3 x pi x 6380[sup]3[/sup]. The volume of the water is 4/3 x pi x (6388[sup]3[/sup] &#8211; 6380[sup]3[/sup]) = 233,511,593 cubic kilometers.
4) What we now want is 4/3 x pi x (r[sup]3[/sup] &#8211; (r-0.00001)[sup]3[/sup]) = 233,511,593 cu.km.

• So, (r[sup]3[/sup] &#8211; (r-0.00001)[sup]3[/sup]) = 978,131,072
• For convenience, let y = 0.00001
• (r[sup]3[/sup] &#8211; (r-y)[sup]3[/sup]) = r[sup]3[/sup] &#8211;(r³-2r²y+y²r-r²y+2ry²-y³) =
• 2r²y - y²r + r²y - 2ry² + y³ = 978,131,072

For a little lesson in both approximation and using the Google calculator, here:

First, since the Google calculator is going to take care of our units, let's leave those in and just do the following. I'll rewrite the last line in the above train of thought, combining terms on the left, and writing the expression the number on the right came from:

3r[sup]2[/sup]y - 3ry[sup]2[/sup] + y[sup]3[/sup] = (3960[sup]3[/sup] &#8211; 3955[sup]3[/sup])miles[sup]3[/sup]

with:
y = 1cm

Now, first of all, we know that the answer to r is going to be pretty big: it's got to be significantly larger than the nearly 4,000 miles we already have, which is vastly larger than the number y (1cm). Since r is going to be something like a million times larger than y, we are only affecting things around the sixth decimal place (or further out) by just dropping all but the term with the greatest power in r. So we can simply rewrite the above as:

3r[sup]2[/sup]y = (3960[sup]3[/sup] &#8211; 3955[sup]3[/sup])miles[sup]3[/sup]

Rewrite in terms of r:

r = sqrt((3960[sup]3[/sup] &#8211; 3955[sup]3[/sup])miles[sup]3[/sup]/(3*1cm))

Now we can just plug that into the Google calculator to get a result.

Voila!
3 550 016.71 miles

(put whatever units you want for an alternative answer)