earth's circumference

[réjean]

Hi all!I was at a party last night and someone asked this question;supposing that the earth is a perfectly round sphere and you add an extra 1 feet to it's circumference how much room would you get allaround the planet.My first instinct was that adding an infinitisimal number to the whole number would be in the billionth of a digit or so. She replied that it would be 2 inches all around. Which I had a hard time accepting so this morning I went to wikipedia and got some numbers and punched them in. Here is my result;In two widely-used geodetic standards, the Equator is modeled as a circle whose radius is a whole number of metres. In 1976 the IAU standardized this radius as 6,378,140 metres (20,925,656 ft), subsequently refined by the IUGG to 6,378,137 metres (20,925,646 ft) and adopted in WGS-84, though the yet more recent IAU-2000 has retained the old IAU-1976 value. In either case, the length of the Equator is by definition exactly 2π times the given standard, which to the nearest millimeter is 40,075,016.686 metres (131,479,713.54 ft) in WGS-84 and 40,075,035.535 metres (131,479,775.38 ft) in IAU-1976 and IAU-2000.[1]So taking the last figure (131,479,775.38 ft)now assuming that pi = 3.141592652x pi= 6.2831853so c ( circumference ) 1577757304.56 divided by 2x pi= 251107871.909098031 (radius)and 1577757304.56 + 12 ( inches or one foot )=1577757312.56divided by 2x pi= 251107875.580241124so 251107875.580241124 minus 251107871.909098031 equals3.671143093 ( inches )How can that be? And this could be a good topic for any holiday parties going on.

Edited December 14, 2010 by réjean

[lewmur]

Hi all!I was at a party last night and someone asked this question;supposing that the earth is a perfectly round sphere and you add an extra 1 feet to it's circumference how much room would you get allaround the planet.My first instinct was that adding an infinitisimal number to the whole number would be in the billionth of a digit or so. She replied that it would be 2 inches all around. Which I had a hard time accepting so this morning I went to wikipedia and got some numbers and punched them in. Here is my result;In two widely-used geodetic standards, the Equator is modeled as a circle whose radius is a whole number of metres. In 1976 the IAU standardized this radius as 6,378,140 metres (20,925,656 ft), subsequently refined by the IUGG to 6,378,137 metres (20,925,646 ft) and adopted in WGS-84, though the yet more recent IAU-2000 has retained the old IAU-1976 value. In either case, the length of the Equator is by definition exactly 2π times the given standard, which to the nearest millimeter is 40,075,016.686 metres (131,479,713.54 ft) in WGS-84 and 40,075,035.535 metres (131,479,775.38 ft) in IAU-1976 and IAU-2000.[1]So taking the last figure (131,479,775.38 ft)as the circumference and mutliplying it by 12 I get 1577757304.56 inches[/i]now assuming that pi = 3.141592652x pi= 6.2831853so c ( circumference ) 1577757304.56 divided by 2x pi= 251107871.909098031 (radius)and 1577757304.56 + 12 ( inches or one foot )=1577757312.56divided by 2x pi= 251107875.580241124so 251107875.580241124 minus 251107871.909098031 equals3.671143093 ( inches )How can that be? And this could be a good topic for any holiday parties going on.

I'm assuming that by "room" you are speaking of "surface area." IOW, if you increased the circumference of the earth by one foot, what would be the increase in total surface area? The formula for the surface area of a sphere is 4 x Pi x the radius squared. Plug your figures into that formula.But for a close approximation, just multiply 1 foot x the circumference in feet for the added number of square feet of surface area. 20,925,646 SqFt Edited December 13, 2010 by lewmur

[réjean]

She meant the extra radius one would get, by her gesture of the space between her thumb and her forefinger ( as in I am crushing your head which was a joke from the kids in the hall of old times if some canadian remember ).

[lewmur]

She meant the extra radius one would get, by her gesture of the space between her thumb and her forefinger ( as in I am crushing your head which was a joke from the kids in the hall of old times if some canadian remember ).

Hmmm!!! How does "room" translate to "radius"? But, assuming she did mean "radius", to solve that, the size of the earth is irrelevant. Just use the formula C=2 x Pi x radius squared where C becomes 12" and solve for the radius.edit: "room" implies at least two dimensions "radius" only one.. Edited December 14, 2010 by lewmur

[réjean]

Hmmm!!! How does "room" translate to "radius"? But, assuming she did mean "radius", to solve that, the size of the earth is irrelevant. Just use the formula C=2 x Pi x radius squared where C becomes 12" and solve for the radius.edit: "room" implies at least two dimensions "radius" only one..

Thanks for the info lewmur! I appreciate it.We were at a party and among laypeople so when she said " room " and used her finger and thumb she meant the distance between the globe ( in this instance the 'earth' ) and the " new " circumference.Now about " Just use the formula C=2 x Pi x radius squared where C becomes 12" and solve for the radius.1. where does the " squared" come from?2. why does the size of the earth do not matter is really my main question? If I could use a rope to go all around the planet in a tight squeeze ( assuming that it is a perfect sphere ) and I add an extra digit ( foot, inch, metre, milimetre, ) to it and then measure the new radius then you mean that it doesn't matter if my rope was 3 fingers long or 3 million fingers.

[lewmur]

Thanks for the info lewmur! I appreciate it.We were at a party and among laypeople so when she said " room " and used her finger and thumb she meant the distance between the globe ( in this instance the 'earth' ) and the " new " circumference.Now about " Just use the formula C=2 x Pi x radius squared where C becomes 12" and solve for the radius.1. where does the " squared" come from?2. why does the size of the earth do not matter is really my main question? If I could use a rope to go all around the planet in a tight squeeze ( assuming that it is a perfect sphere ) and I add an extra digit ( foot, inch, metre, milimetre, ) to it and then measure the new radius then you mean that it doesn't matter if my rope was 3 fingers long or 3 million fingers.

The "squared" comes from the little elf who sometimes takes over my fingers when I'm typing. Everyone knows pie 'r round.The size of the earth doesn't matter because you are looking for the "difference." The size cancels itself out. Edited December 14, 2010 by lewmur

[réjean]

The "squared" comes from the little elf who sometimes takes over my fingers when I'm typing. Everyone knows pie 'r round.The size of the earth doesn't matter because you are looking for the "difference." The size cancels itself out.

Obviously this is what I didn't get from her question. Because if I use a rope around a basketball and I make it 10 times longer and make a new circle around the same basketball the distance between the rope and the ball would be a lot larger than the ratio between the rope and the new radius.and you are implying that if I add a mere foot ( 30 cm or so ) to the rope that goes around the whole planet I would have a new gap of about 3.67 inches all around? Edited December 14, 2010 by réjean

[lewmur]

Obviously this is what I didn't get from her question. Because if I use a rope around a basketball and I make it 10 times longer and make a new circle around the same basketball the distance between the rope and the ball would be a lot larger than the ratio between the rope and the new radius.and you are implying that if I add a mere foot ( 30 cm or so ) to the rope that goes around the whole planet I would have a new gap of about 3.67 inches all around?

I'm saying that if you add 12" to the circumference of the earth or a basketball, (increase the circumference by 12") the increase in the radius would be the same. Edited December 14, 2010 by lewmur

[V.T. Eric Layton]

This is similar to the "Rope Around the World" math puzzler that has been around since the early 1700s. My niece won a prize in math class for demonstrating this. Her teacher didn't believe it at first and thought it was some sort of trick. Another teacher assured him that it was correct; not a trick at all. It's just a function of geometry. Circles are unique things in the universe.From Ten of the greatest: Maths puzzles --> Daily Mail online4. ROPE AROUND THE EARTH PUZZLE, 1702This gem from 1702 shows how simple intuition may fail us. Imagine you're given a rope that tightly encircles the equator of a basketball. How much longer would you have to make it for it to be one foot from the surface at all points?Next, imagine we have the rope around the equator of an Earth-sized sphere - making it around 25,000 miles long. How much longer would you now have to make it for it to be one foot off the ground all the way around the equator?The surprising answer is 2pi (or approximately 6.28) feet for both the basketball and the Earth. If r is the radius of the Earth, and 1 + r is the radius in feet of the enlarged circle, we can compare the rope circumference before (2pir) and after (2pi(1 + r)).

[Tushman]

How can that be? And this could be a good topic for any holiday parties going on.

I don't know about all this math mumbo jumbo. Just give me my vodka.

[réjean]

Ok! I get it. Thanks a lot people. I appreciate it. I am not going to try changing the size of the earth. Wait for me Tushman I am coming for that drink.P.S. I forgot to say that this is a great site Eric. Ièll be reading it tonight if we still have power.It is quite stormy here

Edited December 14, 2010 by réjean

[V.T. Eric Layton]

It's not stormy here, but it sure is COLD! BRRRRRR!

[Tushman]

It's not stormy here, but it sure is COLD! BRRRRRR!

Winter is upon us for sure. I wasn't ready for the cold. I thought we would 'ease' into it... but this arctic blast has got me hibernating. The older I get, the more I hate the cold. If I had money, I would throw away everything I have & move to more moderate climates like Northern CA or even Seattle. Alas, even just finding a job right now is difficult. Edited December 15, 2010 by Tushman

[V.T. Eric Layton]

Alas, even just finding a job right now is difficult.

I know all about it. I'm at 28 months unemployed.

[Tushman]

I know all about it. I'm at 28 months unemployed.

That has got to be tough... I'm sorry to hear that Eric. I wish you luck in the new year. I am praying and hoping that things will turn around quickly. I was sending out a lot of resumes this past fall and barely got 1 return phone call. All of a sudden, I got 2 interviews this week. I'm hoping the trend continues. This economy has to shape up sooner or later.

[ChipDoc]

I lost my job this past July and, though I was only unemployed for three months, that was plenty bad. Oddly enough, I got the new job through Craigslist. It's a great job too - not only am I having a great time but I'm making more as a part-timer here than I was making as a full-timer at the last place. I think it's all a matter of luck and timing.

[V.T. Eric Layton]

That has got to be tough... I'm sorry to hear that Eric. I wish you luck in the new year. I am praying and hoping that things will turn around quickly. I was sending out a lot of resumes this past fall and barely got 1 return phone call. All of a sudden, I got 2 interviews this week. I'm hoping the trend continues. This economy has to shape up sooner or later.

Yeah... I quit counting at about 600 applications. I had two call backs and one interview from those 600 or so. Pretty bad. Sucks being in the job market when you're over 45. It doesn't help that my career (electronics service) hasn't been done in this country for about a decade or so either. *sigh* Thanks for the good wishes, Tushman. I'm hoping something will break for me in the new year. It's got to. I don't have any more options.

[sunrat]

Just to clear up a bit of dodgy maths here (Eric was right):Circumference of a circle=2πr or πd: so the change in radius would be: delta C/2π Delta C is the change in circumference, so for 12 inches the radius would change by just under 2 inches.Area of a circle = π x rsquared:Surface area of a sphere = With thanks to Wikipedia here and here.

[V.T. Eric Layton]

Way to get back on topic, SR!

[sunrat]

Way to get back on topic, SR!

Someone had to give the OP a definitive answer.

[FuzzButt]

So taking the last figure (131,479,775.38 ft)now assuming that pi = 3.141592652x pi= 6.2831853so c ( circumference ) 1577757304.56 divided by 2x pi= 251107871.909098031 (radius)and 1577757304.56 + 12 ( inches or one foot )=1577757312.56divided by 2x pi= 251107875.580241124so 251107875.580241124 minus 251107871.909098031 equals3.671143093 ( inches )How can that be? And this could be a good topic for any holiday parties going on.

Math is off in you addition of 12 to 1577757304.56. It should equal 1577757316.56 not 1577757312.56

[lewmur]

Math is off in you addition of 12 to 1577757304.56. It should equal 1577757316.56 not 1577757312.56

Considering the fact that the number was totally irrelevant to the solution, what possessed you to check it out to the nth place? I opened this post thinking I might read something of significance.

[Cluttermagnet]

"It was a dark and stormy night..."Oh, wait- wrong thread. "My eyes glazed over..."Nope. Oh, here:Figure of the EarthDid you know the Earth is oblate? Yep, it's gotta chubby waistline you can duckdown behind and hide. So the Earth is actually slightly smaller than it appears inyour rear view mirror... ;)

[b2cm]

How can that be? And this could be a good topic for any holiday parties going on.

If the she had the figure and the right circumference ...

[sunrat]

i've decided to end the speculation.i started in reykjavik iceland and am headed due-east with a pedometer.we'll just see how big around this earth is!

At least if you head due east from Iceland it won't be as far to walk as if you do it at the equator!

[V.T. Eric Layton]

Don't fall off the edge, temmu. Remember, the earth is flat.

[sunrat]

Don't fall off the edge, temmu. Remember, the earth is flat.

I forgot about that. As explained by that great astronomer, Terry Pratchett:It consists of a flat disc (complete with end-of-the-world drop-off and consequent waterfall) resting on the backs of four huge elephants (Great T'phon, Tubul, Berilia, and Jerakeen) which are in turn standing on the back of an enormous turtle (Great A'Tuin) as it slowly swims through space.

[ross549]

psand -quit- imagining temmu in black swimming trunks.like now, stop.

Are you sure they are trunks? My imagination is showing a speedo.

[Urmas]

... so i'll just swim across to denmark here, ooo! the water's c-o-l-d!

The water? Better stay on the Atlantic side of the Danish straits:

FINNISH ICE REPORTNo 8723.2.2011The ice grows thicker.In the northern Bay of Bothnia there is 50–75 cm thick fast ice approximately to Kemi 2 and Oulu 3. In the central and southern Bay of Bothnia, 35–65 cm thick fast ice in the archipelago. From Kemi 2 to south of Kemi 1 there is 30–40 cm thick level ice. Farther out, in places heavily ridged, consolidated drift ice, which in the northern part is 30–60 cm thick and in the southern part 20–50 cm thick.In the Quark, 20–50 cm thick very close ridged ice north of Nordvalen. South of Nordvalen, 15–30 cm thick level ice and 20–40 cm thick, ridged very close ice. In the Vaasa archipelago, 30–65 cm thick fast ice.In the Sea of Bothnia, 30–70 cm thick fast ice in the archipelago. Farther out there is first an approximately 25–50 nautical miles wide zone of 15–30 cm thick level ice and farther west 15–40 cm thick ridged very close ice. In the southern Sea of Bothnia, in places 20–40 cm very close ice.In the Sea of Åland, 10–25 cm thick, in places rafted, level ice. Approximately along the line Simpnäsklubb–Mariehamn there is a 3 nautical mile wide brash ice barrier. South of this line there is thin level ice and new ice approximately to the lighthouse Svenska Björn.In the Archipelago Sea, 20–55 cm thick fast ice and level ice to Utö.In the Northern Baltic Proper there is new ice and thin close ice approximately to the line Häradskär–Kopparstenarna–about 55 nautical miles west of Ventspils. South of Bogskär there are areas of open water.In the Gulf of Finland, 20–60 cm thick fast ice in the archipelago. Farther out, 15–40 cm thick, in places rafted, level ice. Off the Estonian coast, 20–40 cm thick compact ice and new ice. East of Gogland there is 30–45 cm thick, ridged very close ice. In the Bay of Vyborg and the Bay of St. Petersburg, 35–75 cm thick fast ice.Icebreakers:Kontio, Otso, Sisu and Nordica assist in the Bay of Bothnia. Botnica assist in the Sea of Bothnia. Fennica, Urho and Voima assist in the Gulf of Finland.Restrictions to navigation:Minimum ice class and deadweight required of assisted vessels:Tornio, Kemi, Oulu, Raahe, Kokkola and Pietarsaari, IA of more than 4000 tons in deadweight, and per port at least 2000 tons to load or unload or both together.Vaasa, IA of more than 2000 tons.Kaskinen, Pori, Rauma and Uusikaupunki, IA and IB of more than 2000 tons.Naantali, Turku, Hanko and Koverhar, IA and IB of more than 2000 tons and IC of more than 3000 tons.Inkoo, Kantvik, Helsinki, Porvoo, Loviisa, Kotka and Hamina, IA and IB of more than 2000 tons.Vessels bound for ports with traffic restrictions in the Gulf of Bothnia shall report to ICE INFO center on VHF channel 84 when passing the Svenska Björn lighthouse.The traffic separation schemes off Hankoniemi peninsula, Off Porkkala Lighthouse and Off Kalbådagrund Lighthouse in the Gulf of Finland as well as in the Sea of Åland are temporarily out of use due to ice conditions.

[Guest LilBambi]

BRRRRRRRRRRRR!!!!!!!!This is the reason I live in Virginia! Bad enough here. Had enough of winters up north here in the USA, but that is downright nasty cold, urmas!!!!