Plank Lengths

[beso]

How many Planck lengths would it take to cover the side of a square Barn?

[jim360]

African or European barn?

[ToraToraTora]

Plank or Planck?

[jim360]

There is a unit in particle physics called a barn, which has an area 10^(-28) m^2. One Planck length meanwhile is 1.616255*10(-35) metres. So the answer is 6,187,143 Plancks :)

[douglas9401]

Does the above allow for expansion?

[LadyCG]

My calculations match Jim's precisely.

[jim360]

I'd hope not, LadyCG, because like a moron I forgot to take a square root of a barn to get the side length. Should be what I gave times about 10^14. Oops!

[Zacs-Master]

Don't forget, for a circular barn, you only need Pi to 15 decimal places.

[goodgoalie]

Consider your doctorate rescinded :)

[jim360]

And that's only for the side length -- to get to cover the entire bar you then square that.

Oh, for an edit function... or I could just be cheeky and delete my wrong answers :P

But, anyway:

A length of Plancks along the bottom of a barn = 6.187*10^20 = 619 million trillion plancks.

A whole barnside of Plancks = 3.8*10^41 square Plancks. Or ten times that if you imagine Planks of Plancks as in a 10:1 length:width ratio.

Hope that all clears that up, and remember don't hesitate to get in touch if you want me to screw up something simple again :)

[beso]

Jim is on the right track but not quite correct.

As a Barn is a unit of area, a square Barn can only mean a square with the area of a Barn. The length of a side of that square would be the square root of 10^-28 or 10^-14 metres.

The answer is 10^14 times Jim's answer.

[jim360]

Hey, be fair, I got it right after I fixed it!

[ichkeria]

Two short Plancks=thickness of Barn

And I stand by that …

[The Builder]

Max Planck's parents were barely 5 feet tall. If that helps ;o)

[jno]

what exactly do you do with a square barn if it only exists in two dimensions? I suppose you could hold barn dances on it.

[ToraToraTora]

jim surely a cross section of the nucleus of a Uranium atom would be circular?? Assuming U235 that's 1.2fm = 4.45fm². I'll let you work out how many square plancks that is!

[mibn2cweus]

42 pretty much had to fit into the answer to this question . . . somehow.

[jim360]

// jim surely a cross section of the nucleus of a Uranium atom would be circular?? //

I didn't feel like replying to this yesterday, apologies for the delay.

Just want to make a couple of points, hopefully useful in future:

1. It's probably a fair assumption that a nucleus is roughly spherical in shape, but it's also worth being careful about what assumptions you make. Atomic nuclei are very complicated, and also dynamic, objects, so they may even change shape, undulating as the particles within the nuclei move around. Although that said I don't do nuclear physics so maybe I have just spouted a whole lot of bilge.

2. But anyway, for a calculation like this, you should barely care about the difference between squares and circles. A circle with diameter D has area (pi/4)*D^2, compared with a square of side D having area D^2, and pi/4 is about 79%. Obviously a 20% difference is a lot if you are trying to get an *exact* answer, but if you're just looking for a rough-and-ready approximation, a chance to work out what the approximate size of things are, then you may as well take pi/4=1 and be done with it. Approximation is your ally!