Reddit, what is the most powerful image you have ever seen?

tokerjoker11 · 2012-05-25T11:40:22.120572-07:00

I didn't know the iPhone could develop thirty-four million Newtons of thrust.

b3n · 2 days

No, too bloody stuffy at night.

DrToonz · 2 days

I think Cracked compared it to us finding a planet that was 80% acid, and the air contain acid vapour, and the acid occasionally fell from the sky, and the inhabitants were walking bags of acid, and our battle plan was to engage in hand-to-hand combat naked.

Ryugi · 6 days

You mean it's not ghosts!?

GalacticBagel · 7 days

It's worth pointing out that in your example, the 'Windows' text is not anti-aliased, but the 'mac/lunix/phones & tablets' text is. This is almost certainly application-controlled, and not dictated by Windows.

Windows can and does anti-alias fonts, but it also tries to align letters to the pixel grid to make them a bit sharper. Mac fonts look smoother but blurrier than Windows fonts, and I don't prefer one over the other.

Shakennapoleon · 12 days

Your birth sounds like a toned-down version of Kim Jong Il's.

Diggie1225 · 16 days

I sure hope he capitalised "string".

snowglider_wants_ram · 20 days

Have your render layer deliver a "Color" pass.

Render > Layers > check the box that says Color.

LonelySteve · 1 month

I had an awesome dream last night that I was battling an evil dictator and his army of helicopters and giant zeppelins, so I guess I'm fresh out of luck.

JimmySchaps · 1 month

I see, I was considering a non-spinning ring for simplicity. I thought the main disagreement was about whether the compressional force in the ring had a vertical component (which it doesn't), and I've calculated it's magnitude for a stationary ring (which was the original question).

For a ring that's spinning, the calculation needs a trivial modification. We balance y-forces so that their sum is the centripetal acceleration a times the mass:

w - 2F_y = ma = w/g * a

WΘ/π - 2FΘ = (W/πg) * a

2F = W/π - (W/πg) * a

F = (W/2π)(1 - a/g)

So if the ring is spinning fast enough that centripetal acceleration equals acceleration due to gravity, the ring is not under compression, as you'd expect.

JimmySchaps · 1 month

I contend that premise is incorrect

It isn't a premise, it's a conclusion. The calculation that follows it is the proof.

What do you mean by 'horizontal' and 'vertical'? Remember that 'up' depends on the local direction of the gravitational force, which changes around the ring.

You are equating a horizontal force to a vertical force

No, I've equated forces acting in the y-direction, which is the direction that runs top-to-bottom in your diagram.

JimmySchaps · 1 month

Remember here that the force is only locally horizontal, which is important because 'up' is a different direction at different points around the ring.

Consider the arc section in your diagram. For clarity, I'm going to call the direction from the centre of mass of that arc to the centre of the Earth the y-direction.

I argue that forces balance in the y-direction when each of the forces in blue is locally horizontal and has magnitude W/2.

If the arc covers an angular distance of 2Θ (i.e. this arc represents 2Θ/2π of the ring) then the y-component of each of each of the forces drawn in blue is

F_y = F sin(Θ) = FΘ

where we are assuming Θ is small (i.e. we are considering a small section of the arc).

The weight w of this small section is going to be

w = W*2Θ/2π = WΘ/π

where W is the weight of the entire ring. We can consider this force to be acting entirely in the y-direction because we are looking at a small section of the ring.

So we balance y-forces:

2F_y = w

2FΘ = WΘ/π

F = W/2π

JimmySchaps · 1 month

Yeah, I did gloss over that a bit. Intuitively I was thinking of it like a lever, and then justified my answer with a somewhat rigorous energy argument. I've explained it here.

Thanks for doing the maths, though. The fact that we used two different methods to get the same answer gives me confidence in my hand-wavy reasoning. I've edited my post to make it clearer.

JimmySchaps · 1 month

I don't think you're hanging a picture analogy is accurate. I think the compressive force in the ring acts entirely horizontally, and has no locally vertical component.

If you imagine looking at, say, a 1° arc of the ring, the horizontal compressive force is pushing in from either end. Because the arc has a curve, these horizontal forces are pushing the section away from the center of the Earth.

I can't imagine how there could be a vertical component to the force. Apart from it not seeming necessary, it doesn't square with Newton's third law. Where's the reaction force?

As for the magnitude of the force, the ring has to compress by 2*pi times the distance it falls, so the force resisting compression can be 2*pi times less than the force causing it to fall (similare to a lever or wedge). Thus, The compressional force in the ring will be the weight of the ring divided by 2*pi.

A more rigorous argument would be to imagine a ring with compression perfectly balancing weight, and noticing that if the radius of the ring were to change by some infinitesimal amount, the total potential energy of the ring (gravitational plus elastic) can't change.

JimmySchaps · 1 month

Yup, orbital velocity around the Earth is about 8 km/s, so that's how fast you'd need to spin the ring.

Wouldn't help the gravitational instability, though.

JimmySchaps · 1 month

Carbon nanotubes are much stronger in tension than in compression, and they still snap at pressures one tenth as big as what we're talking about here.

JimmySchaps · 1 month

Assuming the ring was infinitely strong, yes.

Other people have dealt with the ring being gravitationally unstable. I'll deal with how strong it needs to be.

For the ring to fall to Earth, it's radius needs to decrease by 1 mile (I'll assume, to make things simple, that the Earth is a perfect sphere). Because the circumference of a circle is 2*pi* r, the circumference needs to decrease by 2pi*(1 mile).

This ring is going to be 25,000 miles around, but to drop to the ground it needs to only shrink by 6.3 miles.

The compressional force in the ring will be the weight of the ring divided by 2*pi, i.e. F=mg/(2*pi) (edit: see bottom of post for why). The pressure in the ring will be this force divided by the cross-section, i.e. P=mg/(2*pi*A)

The mass of the ring will be the density d times the volume: m = d*V = d*A*c where c is the circumference.

So the pressure in the ring will be:

P = d*A*c*g/(2*pi*A) = dcg/2pi

If the ring is made of steel (d about 8000 kg/m³ ) the pressure in the ring will be

P = 8000 kg/m³ * 25000 miles * 9.8 m/s² / 2pi = 500 GPa

This is about the highest pressure achieved in a laboratory, higher than the pressure in the Earth's core, and over 1000 times higher than the compressive strength of steel.

tl;dr It would crumple like tissue paper under it's own weight.

Edit: I kinda skipped over how I calculated what the compression force in the ring would be. Here I explain my reasoning to parallellogic, and here tomsing98 gives a thorough derivation (thanks!).

baconability · 1 month

It's pretty easy to find the area of a segment of a circle. The overlap area is just going to be union of two segments, one from each circle.

MasterJuanB · 1 month

Phil Tippett.

portero36 · 1 month

It's probably an image of the bright camera flash, caused by a reflection somewhere inside the lens assembly of the camera.

Notice how if you look at where the camera flash is relative to the centre of the picture, the 'eye' is directly opposite to it.

These sorts of images happen all the time, but someone better versed in the workings of cameras will have to explain the specifics.

Faranya · 1 month

Assuming you want to keep the moon at it's current distance, rather than bring it closer so it's in a natural geostationary orbit, you'd need to accelerate it so that it performs one orbit every 24 hours, rather than every 27 days.

Obviously, the Earth's gravity won't be nearly strong enough to hold such a super-fast moon in circular orbit, so our tether will need to have enormous tensile strength to stop the moon from flying away. But how strong?

Acceleration due to circular motion is given by a=v² /r. Currently, the acceleration of the moon is

(1.02 km/s)² / 385000 km = 0.00271 m/s²

This is equal to the acceleration due to the Earth's gravity. The force that the Earth's gravity exerts on the moon is therefore the moon's mass times this acceleration (F=ma):

F = 0.00271 m/s² * 7.3410²² kg = 1.9910²⁰ N

If we want v to increase by a factor of 27, the acceleration (and therefore the force) will increase by a factor of 27² = 729. So the tensile strength of the cable will need to be 728 times stronger than the current weight of the moon.

That's 1.45*10²³ N.

That's huge.

If we wanted to make our tether out of steel, which has a tensile strength of about 250 MN/m² , it would need a cross section of 6 billion square kilometres, which would make it about 6 times wider than the Earth.

089786 · 1 month

"After careful consideration I have decided... not to endorse your park."

"So have I."

Melner · 2 months

Relativity only requires that each frame be locally equivalent to an inertial frame.

stwange · 2 months

Stellarium, free planetarium software. All OSs. It's brilliant.

lopfynotcuber · 2 months

I think you're confusing Ian Holm with Derek Jacobi.

ghazwozza

TROPHY CASE

ghazwozza

TROPHY CASE

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