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The Flying Manhole


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I am sure most of the people on this forum know about the manhole cover launch off the top of a nuclear test at 66 km/s in 1957.

https://vm.tiktok.com/ZMePHarH7/ 

There is some debate about whether the manhole cover actually made it to space. At that velocity, it would have left the atmosphere in less than a second, so it would not have had time to melt. On the other hand, it well could have disintegrated from the forces...except for the fact that (a) it didn’t disintegrate when it was hit by the force that launched it and (b) it’s a one-time chunk of steel so if anything can survive, that would be it.

I would use Newton’s approximation but I feel like at 66 km/s it probably breaks down. 

I have seen several conflicting statements from Brownlee about it.

What do you think? Would a one-tonne steel disc 4” thick, launched from a Nevada desert at 66 km/s, be able to reach space? Or would it disintegrate?

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900 * 66 0002 / 2 = 2*1012 J ~= 0.5 kt = 500 t.

Quote
Pascal-B August 27, 1957 22:35:00.0 NTS Area U3d 17px-WMA_button2b.png37.04903°N 116.0347°W 1,229 m (4,032 ft) - 150 m (490 ft) underground shaft,
safety experiment
  300 t   [1][12][13][14][16] Shaft safety experiment, failed. Sent the shaft cap weighing several hundred pounds (2 tons) at velocity very roughly pre-calculated as 66 km/s (41 mi/s); popular claims of it reaching space are disputed, see section above.

Total yield 300 t.

Either it was not 66 km/s fast, or the plate kinetic energy was 1.7 times greater than the charge total energy.

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Sadly the guy I would ask about this died almost 20 years ago (a friend's dad, I think was test division, then X div, LANL, used to do detectors at the Nevada test site back in the day).

Wonder if he ever talked to my friend about it.

They caught the plate moving in 1 frame of high speed imaging. Camera (Rapatronic?) took a frame every millisecond.

The 66km/s was a back of the envelope calc done that assumed a vacuum, etc—an off the cuff remark by Brownlee, which he well knew was not anything like accurate.

Total bomb energy was 1.255x1012J, so if it gave all that energy to 900kg, that's 52.8 km/s. Clearly that's not a thing, but the point of the test was an explosion underground with a hole that would prevent fallout (oddly enough). Testing was actually successful in that regard. The plug could certainly have left as a substantial velocity, then the question is just how fast, would it disintegrate from heating in the atmosphere, etc.

Also, some articles some how say he said it was moving "Like a bat"

Not sure if bat out of hell is a problem on this forum, but in case it gets edited somehow:

Spoiler

 

 

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My guess is maybe half that, so maybe 33 km/s. 

I think a good analogue to think about is when a metallic meteor hits the atmosphere going that fast. A large couple-ton rock seems to me like it'd be large enough to survive, though maybe not if it broke into smaller pieces. Also, meteors usually enter Earth through the top of the atmosphere, not the bottom, so the circumstance is different there.

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The cap was on top of a 150 m deep (high, if look from below) shaft (working as a barrel) and pushed by the evaporated 2 t heavy concrete collimator above the charge,

At the first link they estimate  the physically possible terminal velocity of the gas flow ~52.8 km/s.

Also they presume total energy spend on the collimator heating as 1/3 of total yield, so ~100 t of TNT (but as far as I can see, never use it).

Imho, this sounds not enough reasonable because:

1. The collimator was evaporated by X-ray, not by push, so it should be expanding spherically.
So, it should spend at least a half of its energy on pushing against walls and down.
Of course, later this heat should pass out as hot gases, but not together with its immediate expansion flow. Like a smoke after shot.
This makes to think that the collimator push energy was not greater than ~50 t TNT .

2.
The shaft is 1.2 m wide (see the 2nd link) and 150 m deep.
So, the shaft volume is ~170 m3, ~200 times greater than a 2m3 of concrete could have.
So, the gas cloud of evaporated collimator should expand up to 200 times before reaching the cap.
This should adiabatically cool the gas cloud and decrease its pressure, so unlikely the cap could receive greater part of the original collimator energy.

3. Not all energy of the collimatorpush could be received by the cap.
Unlike a bullet which is accelerating together with the powder gas expansion, from the very beginning of the barrel, the cap was just lying on top.
So, once it had raised a little above the edge of the shaft, the gas flow should start expanding radially, so the cap would receive somewhat like its size-to-altitude ratio of the total energy..

This makes to think that the cap actually received 300 / 3 / 2 / 2 / 2 ~10 t TNT of energy, and its actual speed was ~66 * (10/500)1/2 ~= 10 km/s.
Then, it definitely could not reach the end of the atmosphere and unlikely could melt. They should check again the farmers' backyards.

https://nuclearweaponarchive.org/Usa/Tests/Plumbob.html

https://plane-encyclopedia.com/cold-war/operation-plumbbob-pascal-b-cap/

Edited by kerbiloid
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4 hours ago, sevenperforce said:

Would a one-tonne steel disc 4” thick, launched from a Nevada desert at 66 km/s, be able to reach space? Or would it disintegrate?

And if it didn't disintegrate and did reach space at something between 33–66 km/s, what next?

Wikipedia says the Pascal-B test was conducted at 10:35 PM local time, so assuming a vertical shaft, the impulse would have been applied radial-out in Earth's frame of reference, which means they picked the wrong time of day to send it falling into the sun or into a circular retrograde solar orbit to hit the Earth again.

Ignoring Earth's gravity for a moment, Pythagoras says Earth's 29 km/s orbital speed plus 66 km/s radial-out gives a diagonal 72 km/s. Attempting to dig the semimajor axis out of vis-viva, Google's built in calculator is telling me that 1/(2/(1 au)-(72km/s * 72km/s)/(1.32712440018e11 km^3 s^-2)) is -38,921,523 km. So it left the solar system?

Edited by HebaruSan
Oops, 22:35 isn't 11:35 PM
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1 hour ago, tater said:

They caught the plate moving in 1 frame of high speed imaging. Camera (Rapatronic?) took a frame every millisecond.

And every millisecond has a begin and an end of the frame sliding.
So, was it at 0.00021 or 0.00029  s ? The difference may change the calculated speed.

10 minutes ago, HebaruSan said:

So it left the solar system?

'Oumuamua is what it has splitted from another world.

Or what they had shot in response.

Upd.
And why whas it visible just once?
Because the gas flow was much faster.
A frame later it was below the gas and dust cloud.

Edited by kerbiloid
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Let’s try going from first principles. The actually mass of the cap was 900 kg, not 2 tonnes as sometimes reported. With a 300 tonne TNT equivalent yield, that’s 1.26e12 J. 100% conversion of yield energy into kinetic energy (which is of course impossible) would yield:

KE = 1/2 * m * v^2

v = sqrt(2*KE/m)

v = sqrt(2.8e10) m/s

v = 167.3 km/s

I think my math is correct. That’s far, far more than the 66 km/s quoted, so I don’t think we have any concerns there. Also, remember that the cap was welded in place, so you would have had to build up pressure behind it....not that THAT would have taken very long. 

I am more curious about whether it would survive. You might assume that the drag involved would be sufficient to obliterate literally anything, but if the nuclear blast wave was able to accelerate it to that speed even faster, then that would tend to suggest otherwise. 

Ordinary drag mechanics don’t really apply, I don’t think. It would have been tumbling, but how many times could it realistically have tumbled before crossing the majority of the atmosphere? It would have been in near-empty space in less than 1/3 of a second.

2 minutes ago, cubinator said:

What's the minimum possible speed assuming it got just out of view in one frame?

66 km/s.

Unfortunately we don’t know the height of the image in frame.

The primary mode of heating for a blunt body passing through the atmosphere at hypersonic speeds is going to be convective. But convective heating between the rapidly-formed plasma and a steel plate is going to require time. It seems that there simply wouldn’t have been enough time for that convective heating to take place.

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20 minutes ago, sevenperforce said:

It seems that there simply wouldn’t have been enough time for that convective heating to take place.

That's why I think it would have made it out of the atmosphere, if it had survived the initial blast, deformation, acceleration, and drag from the atmosphere.   If it made it through those initial moments, I don't think it would have had enough time to heat enough to be destroyed.

 

Edited by Gargamel
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1 hour ago, sevenperforce said:

The actually mass of the cap was 900 kg, not 2 tonnes as sometimes reported.

900 kg is a mass of the steel cap on top of the 150 m deep and 1.2 m wide vertical shaft above the charge.

2 t is a mass of the concrete collimator right above the charge, at the lower end of the shaft.

After the collimator had evaporated under X-ray of the charge, the former-concrete gas/plasma cloud filled the shaft as a cannon barrel and pushed the cap lying on top.

1 hour ago, sevenperforce said:

v = sqrt(2.8e10) m/s

v = 167.3 km/s

v = sqrt(1.26e12 / (2 * 900)) = sqrt(2.8e9) m/s

v = 37,4 52.9 km/s

1 hour ago, sevenperforce said:

I think my math is correct. That’s far, far more than the 66 km/s quoted,

The math lost an order of magnitude. The theoretical limit is below  the 66 km/s.

Also it magically assumes that the charge had sent all its yield as a narrow direct beam right into the cap.
When the charge burst was spherical, the evaporated collimator initial expansion was spherical, the shaft angular size is small while its volume is 200 times greater than initial volume of the evaporated collimator cloud, and that the gas cloud certainly began expanding radially once the cap elevated just a little.

1 hour ago, sevenperforce said:

Also, remember that the cap was welded in place, so you would have had to build up pressure behind it....not that THAT would have taken very long. 

The welding seal is thin. The push unlikely noticed it.

Edited by kerbiloid
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3 minutes ago, tater said:

450 because m/2, else 2*KE/900, @kerbiloid

Yes, I corrected.

Anyway it has nothing common with the real picture.

2 minutes ago, magnemoe said:

How thick was this plate, 900 kg  and around 8 ton / m^3 so 0,11 m^3 plate 

4 in.

P.S.
Has anybody at least watched these two links with detailed description?  

1 hour ago, kerbiloid said:

Ctrl-F, "pascal-b"

Edited by kerbiloid
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28 minutes ago, kerbiloid said:

4 in.

P.S.
Has anybody at least watched these two links with detailed description?  

Ctrl-F, "pascal-b"

An 4" plate should survive both the blast and the atmosphere, however I agree in that it would face insane drag who would fast slow it down, they never found it as it could easy end up 20 km away.
The weird part is that an thinner plate welded on might be deformed into an dome and might be stable 

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3 minutes ago, magnemoe said:

The weird part is that an thinner plate welded on might be deformed into an dome and might be stable 

Still enough good to put it on a pickup, carry to home, and weld something useful on top, using it as a ballast and support, lol.

Edited by kerbiloid
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7 minutes ago, tater said:

There's gotta be decent calculations on Orion concepts, right?

The lower bound is Orion with a 900kg pusher plate and no other spacecraft.

Afaik, it was a source of inspiration.

After that they were experimenting with membranes of various h/d ratios.

Also the Orion would be hit with a jet, not with a pancake. That was the idea of the thin membrane.
Thin membranes were giving jets, thick ones were giving pancakes of plasma.

The original purpose was not to hit Orion, but to send as much as possible towards a target.

Got it, you mean the pusher plate, not the membrane.
Yes, it demonstrated that a plate could keep the shape

Edited by kerbiloid
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There's about 10 tons of air above every square metre. The plate at 100mm thick (4in) has an area of ~1.25m2 or ~1.2m diameter.

It'd therefore need to move between 11t and 1.1t out of the way on its way out of the atmosphere depending on how it tumbled.

If it survived it would necessarily be going a lot slower by the time it reached space.

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6 minutes ago, RCgothic said:

There's about 10 tons of air above every square metre. The plate at 100mm thick (4in) has an area of ~1.25m2 or ~1.2m diameter.

It'd therefore need to move between 11t and 1.1t out of the way on its way out of the atmosphere depending on how it tumbled.

If it survived it would necessarily be going a lot slower by the time it reached space.

Yes, hence Newton’s approximation. Depending on the tumbling characteristics it may or may not have been able to punch through the atmosphere. It’s hard to even know where to start to guess at the combination of forces in that kind of situation.

1 hour ago, tater said:

v=sqrt[(1.255E12)/450]=52.8 km/s

It's x109, not 1010.

Good catch, I totally miscounted the zeroes. 

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https://en.m.wikipedia.org/wiki/Impact_depth

Newton's Impact depth approximation implies it'd struggle to escape the atmosphere.

It has to move through all space occupied by all the mass in its way, which must move faster than it to get out of the way.

But by conservation of momentum it can't give more velocity than it starts with to an equal or greater mass without coming to a complete halt.

 

The approximations look reasonable. Blunt body. High velocity. Non-cohesion of the impacted material.

Edited by RCgothic
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years of playing ksp has led me to the knowledge that things can burn up while leaving too, provided they are launched fast enough. the fate of the manhole cover is probably such.

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