For Questions That Don't Merit Their Own Thread

[magnemoe]

3 minutes ago, Gargamel said:

I would say this is more correct than mess, as we are talking a bar type area for recreation and socialization, not strictly for meals and such. 

I am imagining this:

more than this:

 

On an military base the above would be the officers club, the lower the mess, not sure if any ships has officer clubs. 
on civilian ships you will have a dining room and an saloon, typically observation deck. This might be combines like you find bar at one end of dining rooms in many hotels. 

It depend a lot on size, you start with an mess room who can also be used for gathering, as size grow you can add more rooms. Note that passenger ships is an bit different in that part of the revenue is to have the passengers eat and drink. 
 

[kerbiloid]

Thank you all for your responses!

I meant a module d~4 m, L~7 m with a dining/meeting room for a crew of 6 in a high-orbital combat station (a mass launch platform), unified with a standard interplanetary ship with crew of 4-8, and an ancestor for a civilian orbital base with crew of 4.

So, they eat and meet there, and use as a standalone vault to wait for a rescue ship if can't use the rescue capsule.
It has a small observation cupola on the end derived from an airlock chamber with an emergency docking port to let a rescue expedition dock, so it's also an improvised backup docking module.
So its also an auxilliary observation post and a map room.

So, I will call it "Mess Module".

Edited December 18, 2018 by kerbiloid

[ARS]

If we build a spaceship that's large enough (Like a gigantic mothership-level that measures in kilometers), could the ship being crushed by it's own "gravity"? (How to say this? basically it has enough mass to be "rounded by it's own gravity" and crush the entire ship). If that so, does a Death Star (which is already spherical) could be crushed into smaller sphere in this case?

[p1t1o]

3 hours ago, ARS said:

If we build a spaceship that's large enough (Like a gigantic mothership-level that measures in kilometers), could the ship being crushed by it's own "gravity"? (How to say this? basically it has enough mass to be "rounded by it's own gravity" and crush the entire ship). If that so, does a Death Star (which is already spherical) could be crushed into smaller sphere in this case?

Short answer yes.

Slightly longer answer: yes, but itd have to be stupendously huge & massy. A quick google says the radius at which an astronomical body falls into a spherical shape from a irregular, lumpy, potato shape by its own gravity is around 3-600km. But it is based on mass, not size, so thats solid rock, 3-600km across. A spacecraft will be very much less dense, overall, than solid rock and therefore could be significantly larger before its gravity overcame its structural rigidity. Structural strength also plays a part, an ice ball forms a sphere at a lower limit than something rocky for example.

The death star was about 150km across so could likely keep its shape. The second death star was apparently somewhere between 160 and 900km across so it probably starts to push at the limit, unless their structural materials are significantly stronger.

[Gargamel]

14 hours ago, p1t1o said:

The second death star was apparently somewhere between 160 and 900km across so it probably starts to push at the limit, unless their structural materials are significantly stronger.

But those ranges assume a certain density for gravity to do it's thing.   The Death star would have been mainly hollow, given people live inside, and hollow structures use less material to build than more solid structures.    The Death Star would probably be buoyant given it's low density, but then again, Saturn would also float.    Because of all the people space in the Death star, I would figure the diameter would have to be a few orders of magnitude bigger for gravity to start overwhelming the structural supports of the vessel. 

[K^2]

I don't know about a "few orders of magnitude", but yeah, definitely a lot larger than a chunk of rock. The fact that construction materials are likely to be selected for their compressive strength is likely only to improve that. 1000km across without collapsing into a homogenous ball is probably doable even with materials we know today. Especially, if you keep your heavier stuff close to the center, and move all of your hollow spaces, like living quarters, to the outside.

That said, SW routinely uses antigrav suspension and the ships and stations have artificial gravity. How that plays out in the construction of a megastructure I'm not really sure, nor can I be sure with such soft sci-fi world, but it sounds like they'd have a few tricks for supporting the weight of the outer shells without having to resort to brute strength of construction materials. So it's something that can be hand-waved away even if it was far beyond the structural strength of any material.

[kerbiloid]

If take a homogeneous sphere of a uncompressible fluid of total mass M and total radius R, then pressure at radius r is:

p(r) = (3 * G * M²)/(8 * pi * R⁴) * (1 - r²/R²);

So, in the center: p(0) = (3 * G * M²)/(8 * pi * R⁴)

say, M = (4/3)*pi*R³ * density.

p(0) = (3 * G * ((4/3)*pi*R³ * density)²)/(8 * pi * R⁴) = (3 * G * (4/3)² * pi² *R⁶ * density²)/(8 * pi * R⁴) =   (2/3) * pi * G * R² * density² .

Irreversible deformations begin if 
flow_stress < p = (2/3) * pi * G * R² * density² .

R * density .> sqrt((3 * flow_stress) /(2 * pi * G));

Say, flow stress of steel ~= 200 MPa.

R * density >  sqrt((3 * 2*10⁸) /(2 * pi * 6.67*10^-11)) ~= 1.2*10⁹.

R = 1.2*10⁹/ density.

 

If take a cylindric space station M = 20 t, d=4, L = 12 m, its average density is ~= 20 000 / (12 * pi * 4² / 4) ~= 132 kg/m³.
But as such big ship should be made of steel, not aluminium, 132 * 7800/2700 ~= 400 kg/m³.

A rectangular heavy tank: M = 60 t, w=4 m, h ~= 3 m, L=8 m. Density = 60 000 / (4 * 3 * 8) ~= 600 kg/m3.

A battleship or an aircraft carrier: density ~= 1030 kg/m³, lol.

So, 

A solid steel ball then R ~= 1.2*10⁹ / 7800 ~= 150 km.
A battleship-dense ball R ~= 1200 km.

So, as the death star can be neither steel-dense, nor battleship-dense, we can take its density ~sqrt(7800*1000) ~= 3000 kg/m³ and critical radius ~= 400 km.

But as you have to keep your size undercritical, Rmax ~= 300 km.

***

Its centrifugal acceleration = angular_velocity² * radius = (2 * pi)² * radius / (rotation_period²) ~= 6*10⁶ / rotation_period².

To keep it <= 1 g you have to have the rotation period > sqrt(6*10⁶ / 10) ~= 800 s.

Edited December 19, 2018 by kerbiloid

[K^2]

3 hours ago, kerbiloid said:

If take a homogeneous sphere of a uncompressible fluid

Except it's not a fluid. Stress is distributed through the structure, not concentrated at its center. Because of that, the structural limit is going to be larger by a considerable factor.

[p1t1o]

3 hours ago, kerbiloid said:

A rectangular heavy tank: M = 60 t, w=4 m, h ~= 3 m, L=8 m. Density = 60 000 / (4 * 3 * 8) ~= 600 kg/m3

A battleship or an aircraft carrier: density ~= 1030 kg/m³, lol.

Wait, these cant possibly be true, surely they should be swapped around?

[kerbiloid]

51 minutes ago, K^2 said:

Except it's not a fluid.

Except under pressure every metal first is a fluid, and this causes the irreversible plastic deformation, my guru. That's why the "flow stress" term exists.

We don't need to crash the death star here, we need to know when it starts getting deformed.

47 minutes ago, p1t1o said:

Wait, these cant possibly be true, surely they should be swapped around?

I've taken an oversized battle tank to make a rough estimation. Anyway, a battleship is what we need.

Edited December 19, 2018 by kerbiloid

[K^2]

8 hours ago, kerbiloid said:

Except under pressure every metal first is a fluid, and this causes the irreversible plastic deformation, my guru. That's why the "flow stress" term exists.

Cool. All the rooms inside your Death Star have just collapsed, because it is a fluid. The whole point of exercise is to find the size before it becomes a fluid. Size at which solid stress is still the dominant force.

The stress tensor in this problem is diagonal in polar spherical due to symmetries, but it's not degenerate. Outer layers may actually be stretched radially and compressed horizontally, so even the sign of terms isn't the same. Taking it to be a scalar pressure (fully degenerate stress) is a gross underestimate of structural strength.

In the words of Han Solo, "That's no moon. That's not even a station. That's a ball of homogeneous incompressible fluid with no structure whatsoever. These Imperials are getting really weird."

[kerbiloid]

14 minutes ago, K^2 said:

Cool. All the rooms inside your Death Star have just collapsed, because it is a fluid. The whole point of exercise is to find the size before it becomes a fluid. Size at which solid stress is still the dominant force.

The stress tensor in this problem is diagonal in polar spherical due to symmetries, but it's not degenerate. Outer layers may actually be stretched radially and compressed horizontally, so even the sign of terms isn't the same. Taking it to be a scalar pressure (fully degenerate stress) is a gross underestimate of structural strength.

In the words of Han Solo, "That's no moon. That's not even a station. That's a ball of homogeneous incompressible fluid with no structure whatsoever. These Imperials are getting really weird."

Feel free to demonstrate a complete exact solution of an estimation problem with a zillion node matrix of tensions.

Of course you know I see that not... huge solid bodies under stressed conditions act like a very viscous liquid. Such are planets, asteroids, so on.
That's why the planet definition includes the "hydrostatic equilibrium".

P.S.
Do you know that you don't need all digits of pi? First several ones is enough.

Edited December 19, 2018 by kerbiloid

[K^2]

You decide to estimate lethal dose of a poison. You give your test subject 1kg, and subject dies. You then procede to remove poison bit by bit to see when subject comes back to life and insist it will still be a fair estimate.

In technical terms, it's called histeresis, and it's not a small error.

And yeah, I'll write up a proper derivation when I get to an actual computer. The math isn't that hard. You just have to have studied material strength (сопромат). Standard course for engineers and physicists, but does go well beyond high school physics. Though, the later (and common sense) should have taught you to understand why hydrostatics is a bad estimate for when a structure collapses.

[magnemoe]

13 hours ago, p1t1o said:

Wait, these cant possibly be true, surely they should be swapped around?

An battleship need to have an density less that water to be useful or even getting out of the dry dock. WW2 battleship actually float pretty high 

Estimates 500 kg / m^3, and it makes sense, first they was huge ships they also had to be seaworthy.
Some ships: obvious example is the Monitor floated very low by design, it however was not designed to be used at open sea and was slow. it could also not many hits who penetrated the armor. 

Now in space you could make an station inside an hollowed out asteroid it would obviously be heavy as most of the mass is the armor, it would also move like an asteroid and be more like an fortified island.
 

[K^2]

1 hour ago, magnemoe said:

Now in space [...]

Of course. I think p1p1o's point is that kerboloid didn't even pause about quoting density 3% higher than that of water for something that's intended to float. The fact that average density can be much higher for a space station goes without saying. Although, if you're trying to build for extreme size, you're probably going to end up with considerations not unlike these of a warship. Make it too light, and there isn't enough armor. Make it too heavy and it collapses on itself. So using a battleship as a reference is probably the right call. Not realizing that the number has to be less than 1,000kg/m³, not so much. And looking at other numbers, "60 000 / (4 * 3 * 8) ~= 600" ??? 24*8 is closer to 200 than 100. By quite a bit. I went after core assumptions, but somebody should probably check the rest of that math. (Ok, that's actually me failing to multiply 4 * 3.)

Edited December 20, 2018 by K^2

[magnemoe]

24 minutes ago, K^2 said:

Of course. I think p1p1o's point is that kerboloid didn't even pause about quoting density 3% higher than that of water for something that's intended to float. The fact that average density can be much higher for a space station goes without saying. Although, if you're trying to build for extreme size, you're probably going to end up with considerations not unlike these of a warship. Make it too light, and there isn't enough armor. Make it too heavy and it collapses on itself. So using a battleship as a reference is probably the right call. Not realizing that the number has to be less than 1,000kg/m³, not so much. And looking at other numbers, "60 000 / (4 * 3 * 8) ~= 600" ??? 24*8 is closer to 200 than 100. By quite a bit. I went after core assumptions, but somebody should probably check the rest of that math.

Still tanks sink, exception is some light tanks who is designed for it. 

Else I agree however minor plantets and moons tend to be mostly ice and ice on earth become plastic 30-50 meter down, as gravity is lower they obliviously need to go far deeper for ice to become plastic but something like ceres would stand up even with kilometer thick armor as it would use steel or better rather than ice as structure. 

[K^2]

7 minutes ago, magnemoe said:

Still tanks sink, exception is some light tanks who is designed for it.

Well, a real tank wouldn't fill out a 4 x 3 x 8 meters block, either. That's probably why the density comes out so low. The volume that actually excludes water is going to be many times less. That's generally what happens when you start making loose assumptions without bothering to check them. 

Edited December 20, 2018 by K^2

[kerbiloid]

9 hours ago, K^2 said:

You decide to estimate lethal dose of a poison. You give your test subject 1kg, and subject dies. You then procede to remove poison bit by bit to see when subject comes back to life and insist it will still be a fair estimate.

You have strange ideas about the lethal dose estimation.
You should take some amount of mice, estimate the lethal threshold by a dichotomy, then divide the mice into groups and estimate LD50 and LD100 using different concentrations of the solutions for different groups.

Also the flow stress has nothing common with the lethal dose. It's just a physical value measured with a very simple equipment (just a mechanical press with a recorder).
You would study this on the 1st semester of engineering faculty.

9 hours ago, K^2 said:

And yeah, I'll write up a proper derivation when I get to an actual computer.

As I was sure, it was nothing more than "i-know-everything-better-than-you" sci-sounding blah-blah, zaum' and bragging.

If the engineers of XIX-XX were waiting for "an actual computer" we would still riding horses.

9 hours ago, K^2 said:

The math isn't that hard. 

Believe me, it is, and incredibly.
Mathematical physics equations are the theme of the 4th year of a university, and you arre the first person who says they are simple.
Though, for a person who is sure that he knows everything about the universe, of course no problem.

Your "proper derivation" of course will not just use mechanical properties of materials for normal conditions, I believe in you.

In a university you would study the words like "estimation" and "approximation". Nobody needs an exact solution for a thing with fuzzy given data, it's anyway useless.

9 hours ago, K^2 said:

You just have to have studied material strength (сопромат).

Uryupinsky polytech finished?
(Your understanding of metrology ensures me, btw.)
Сопромат is the 1st semester, and it has nothing common with this task beyond a rough estimation which I have made.

For a proper derivation you need something like this

Spoiler

, not just this

Spoiler

 

5 hours ago, K^2 said:

kerboloid didn't even pause about quoting density 3% higher than that of water for something that's intended to float.

Sea water density is ~1030 kg/m³, professor.

So, the upper limit of the battleship density is ~1030.

But it's obvious that I put such exact value just for fun. (Though, it's really ~1030).

5 hours ago, K^2 said:

The fact that average density can be much higher for a space station goes without saying.

You welcome, here it goes with saying.
The 300 km radius sphere should have thicker inner walls than the battleship. At least because you don'r want it to be elastic.
Near the upper limit of the size you actually have a steel ball with air bubbles of rooms.
So, 3000 looks like a not very bad compromise between the battleship and the steel ball.

5 hours ago, K^2 said:

Make it too heavy and it collapses on itself

Our Earth is made of membranes, I guess.
No. The deeper you go - the more rare the caverns. Below several kilometers there can be no caverns at all.
And you have to use the flow stress value to estimate the cavern max depth or the mountain max height. 

We here speak about a planet-sized body, no matter if its natural or artificial, not about a 2-storey market building like you're probably used to have deal with. 

4 hours ago, K^2 said:

Well, a real tank wouldn't fill out a 4 x 3 x 8 meters block, either. That's probably why the density comes out so low. The volume that actually excludes water is going to be many times less. That's generally what happens when you start making loose assumptions without bothering to check them

That's what's happening when you don't use this value in further calculations and bring it just for comparison of a value scale.

P.S.
You've forgotten you usual mantra about the "people who were teaching you physics" and "I was living on a rokkit baze", btw.

Edited December 20, 2018 by kerbiloid

[Exploro]

To fellow KSP'ers who either are mechanical engineers or students I  was wondering what would be a simple but worthwhile project recommendation you would make to mechanical engineer in potentia to conduct while on Christmas break?

Edited December 22, 2018 by Exploro

[Nightside]

If computers start counting at 0 ie 1,2,3 = 0,1,2 then how does a computer count a value of zero?

[Green Baron]

17 minutes ago, Nightside said:

If computers start counting at 0 ie 1,2,3 = 0,1,2 then how does a computer count a value of zero?

Indexing of data structures in the c family of languages starts at 0 for the first element and ends at n-1 for the last one.

-----------------

The representation of numbers depends on the data type.

Integers are quite simple: an unsigned integer zero indeed has 0 bits set. A signed one, usually in two's complement, usually is a "plus" zero, meaning all bits 0, but may be a "minus" zero (all bits 1). Let's leave it to the compiler and trust in its wisdom :-)

Floating point ... idk. We'll need somebody with deeper knowledge than mine for that one. The representation of floats is not that difficult to get, but how a zero is represented, if signed or not, idk. To state that it "may depend on compiler and architecture" is surely not incorrect :-)

Edited December 30, 2018 by Green Baron

[magnemoe]

1 hour ago, Green Baron said:

Indexing of data structures in the c family of languages starts at 0 for the first element and ends at n-1 for the last one.

-----------------

The representation of numbers depends on the data type.

Integers are quite simple: an unsigned integer zero indeed has 0 bits set. A signed one, usually in two's complement, usually is a "plus" zero, meaning all bits 0, but may be a "minus" zero (all bits 1). Let's leave it to the compiler and trust in its wisdom :-)

Floating point ... idk. We'll need somebody with deeper knowledge than mine for that one. The representation of floats is not that difficult to get, but how a zero is represented, if signed or not, idk. To state that it "may depend on compiler and architecture" is surely not incorrect :-)

Indexes start on zero. However in some languages strings start on 1, that is an s.findfirst('t'), there s="test" return 1. This is mostly legacy and many used first byte for string length. 
Floating points uses one bit for signing some for the number and others for exponential. 

One fun effect is that signed integer an tip over and become negative. This can happens if an program run for an very long time or get higher values than tested for. 
Not uncommon if you uses exploits in games. Say you earn 2 billion $ in an game there one million is very much money. 
Once you pass 2,147,483,647 you are in negative as it rolled over. 

[Gargamel]

On 12/30/2018 at 5:20 PM, Nightside said:

If computers start counting at 0 ie 1,2,3 = 0,1,2 then how does a computer count a value of zero?

I think we've gone too deep here with the answers.  

Computers do not think 0=1.   To a computer 1=1, and 0=0. 

When you have a list of objects or things, say a list of values, the first element's index in that list is 0.  If you have 15 elements in the list, the last elements index is 14, but there are still 15 items in the list.  

I believe it's a way to save memory space, back when memory was at a premium.     If we have 8 items in a list, we can number them items 0-7, so when you list them in binary it will be 000, 001, 010, 011, 100, 101, 110, 111.   But if we started at an index of 1 and go up to 8, the last one would have to be 1000, and we'd have a useless value of 0000.   So by starting at 0,  it is just more efficient to count items that way.  

So your List could be {Square, Orange, Cloud, Elephant, Circle, Hippo, Snake, Ball}, Then the item in index 0 is Square, index 4 is Circle, etc.  It doesn't matter what the value of the actual item at index X is, that is just the computer's way of numbering the items in your list.    So in your Example, your list contains {1,2,3}.  The value of the item at index 0 is 1, and index 2 is 3. 

But that's just if you list the numbers within a list/array.  The indexes do not represent what is actually in that location, the index should be thought of as an address within that list/array.  The actual value of any number is it's actual value.   A computer can count the value of 0. 

[IncongruousGoat]

5 hours ago, Gargamel said:

I believe it's a way to save memory space, back when memory was at a premium.

Nope. In C (and C++), it's because list indexing is just syntactic sugar for pointer arithmetic/dereference (for array a, a[3] is semantically equivalent to *(a + 3)), which in turn is because arrays devolve to pointers when passed to or returned from functions and can in many ways be thought of as a special case of pointers. In other languages, it's because indexing from 0 makes a lot of code a lot neater, most notably element lookup in multidimensional arrays (which are usually flattened to 1d arrays at compile time).

Edited January 4, 2019 by IncongruousGoat

[kerbiloid]

The item index is just an offset from the buffer beginning. So, it's by definition 0 for the very first item.

(Unless it's Pascal, and the language author wants to show off)

Edited January 4, 2019 by kerbiloid