Skyler4856

For Questions That Don't Merit Their Own Thread

Recommended Posts

26 minutes ago, mikegarrison said:

I've read George Dyson's book, but it was quite a while ago. One thing I don't recall is the following:

If you have a 10,000 tonne nuclear pulse spaceship, how do you steer it? What kind of RCS would it take to turn such a ship around in order to decelerate?

Use standard rocket engines like raptor for that :)
Its not like its an very nimble ship in any way. 

Share this post


Link to post
Share on other sites
6 hours ago, mikegarrison said:

I've read George Dyson's book, but it was quite a while ago. One thing I don't recall is the following:

If you have a 10,000 tonne nuclear pulse spaceship, how do you steer it? What kind of RCS would it take to turn such a ship around in order to decelerate?

I can't remember if they at least once mention or draw any attitude control or steering system at all.
Though, they definitely rotate the ship at least on Mars/Earth/Saturn aerobraking or Mars landing.

Probably, they were first of all focused on the propulsion unit, so even when they draw the rest of the ship in details (including a shower cabin), and draw the attached Martian gliders (having nothing common with Space Shuttles like in 3rd party illustrations), these parts interest them only as "How large at all should it be?".

Also, the Orion is a project of professional bombists, not rocketeers, lol

***

In Ch. XVII and XVIII of the book the author describes the late design of the propulsion unit like this.

In the engine room there are barrels with the shock absorber's anti-ablator (oil) and coolant (water or ammonia, though later he mentions random natural hydrocarbons).
Of course there are upper parts of pistons, pipes, and other steampunk stuff.

On every shot (2..4 Hz) the coolant (water or ammonia) cools the plate and gets vaporized.
The coolant vapor gets back by pipes and pushes the next propulsion charge through the barrel in the center of the plate, blowing out of the barrel next to the plate. (So, the coolant is expendable).

The anti-ablation oil gets spreaded on the working surface of the plate to prevent its erosion.

The charge booms, the tungsten plasma jet and spherical plasma cloud hit the plate.
The oil layer (and as I can get, the coolant cloud as well) soften the hit.

The plate moves back (or forward, depending on if you are the plate), the two-stage pistons cascade soften the accelerations 1002 times down to 1 g.

The next portion of the coolant in the plate cooling system gets vaporized, gets back and shoots another charge...

***

In Ch. XVIII he describes that the late conception of the Saturn trip included both aerobraking in the Saturn atmosphere with the heatproof absorber plate (and another one on return) and ISRU on Enceladus.
They knew that Enceladus density is low, so it's made of ice, ammonia, and random hydrocarbons, so they were going to mine them and prepare propellant for the return flight.

He mentions that every shot mass was 1/3 of the nuke, and 2/3 of the propellant. So, by ISRU they could make the ship much lighter, taking just nukes.
He doesn't describe in details if the "propellant" is that "coolant" or something else, but I guess yes, as what else could it be.

So, per 1 t of nukes they were anyway spending 2 t of water/ammonia.

***

Describing the Mars landing he mentions original plans with the Orion staying in orbit while a lander lands.

The later design included one Orion getting down to "several thousand feet" altitude on its nukes, then jettisonning the propulsion section at all, and landing on chemical rockets to be a base for the crew of 50 right now and 20 staying there till the next Martian trip.

So, unlikely they were going to use chemical engines regularly, say for steering, as they sound here as something additional.

***

So, I guess, if they were ever having built the Orion, they would use the coolant exhaust for its steering.
Just additional set of pipes and nozzles around the ship.

When the ship is being propelled, there is a lot of steam exhaust. Partially use it to push the charges, partially throw out from the radial nozzles.
(Another cool steampunk analogy, a ship in clouds of hot steam).

When the ship stays in orbit, unlikely a ??000 t heavy ship needs accurate maneuvers, so it doesn't need any other significant RCS system. Just wait the orbit turn with proper ship direction to start.

So, my bet is on steam nozzles.

Edited by kerbiloid

Share this post


Link to post
Share on other sites

I know that with Orion the whole idea is that more mass is actually better, but still, water is crazy valuable in space. The idea of just spraying it out as RCS propellant seems daft. You could get a much higher ISP RCS system with almost anything else.

1 hour ago, kerbiloid said:

Also, the Orion is a project of professional bombists, not rocketeers, lol

So true! If you have a hammer, everything looks like a nail.

Share this post


Link to post
Share on other sites
23 minutes ago, mikegarrison said:

The idea of just spraying it out as RCS propellant seems daft.

They anyway spray it out, just from the rear pipe.
They anyway don't need so much steam to throw the charge, so they can just redistribute the exhaust.

Also if have the RCS steam nozzles, on the aerobraking you can steer the ship just using the aerobrake heat vaporizing the same coolant in same radiator, with same nozzles.

23 minutes ago, mikegarrison said:

You could get a much higher ISP RCS system with almost anything else.

You get this for free, and you don't need active steering during the acceleration. Just make the ship don't shake.

Also in orbit (when the nuke is off) you can just use a boiler to rotate, lol.
It also can be a nuclear boiler, you anyway need it onboard.

Edited by kerbiloid

Share this post


Link to post
Share on other sites
Spoiler
5 hours ago, mikegarrison said:

1564497122-20190730.png

Water critical point = 647 K.
Thermal speed ~= sqrt(3 * 8.3 * 647 / (18*10-3)) ~= 950 m/s.

Say, exhaust speed is ~900 m/s. (Without heating, compressing or so on)

As by funny coincidence they don't mention exact values of a charge for the 4000 t Orion, let's take the 1 kt charge from page 113 for a 10-m wide one.
Mass = 311 lb ~= 140 kg.

In XVIII chapter they briefly describe the ISRU requirements, and that 1 t of charges spends 2 t of coolant.

So, let's ~280 kg of coolant per blast. Let's spend 150 kg for steering.

Blast frequency is 2 Hz, though magazines could provide 4 Hz, let's take 2 Hz.

So, 300 kg/s of steam for steering.

Thrust = 300 * 900 ~= 270 kN = 27 tf of total RCS thrust while accelerating.

Can't find right now the 10-m version mass, but as the 135 ft, i.e. 40 m Orion masses 4000 tons (btw, which "tons" ? metric or kilopound?), let's take the 10-m Orion 500 (metric) tonnes heavy.

So, T/W of RCS = 27/500 ~= 0.05.

Let's take a 20 t ship.
20 t of mass * 0.05 of T/W / 4 blocks = 0.25 tf = 250 kgf per RCS block.

TKS (mass 18) has 5x40 kgf RCS per block ~= 200 kg, for maneuvers.

So, a steam-powered RCS looks absolutely normal for Orion acceleration.

***

Just to rotate the ship in orbit you don't need such thrust, so you can wait for a proper orbit turn, then adjust the ship with a boiler to produce some low thrust steam flow.
Then the very first propulsion charge provides you with full RCS thrust.

***

Partial aerobraking (for braking, not for landing) doesn't require active maneuvering, so probably the heat from the air is enough to heat enough steam just to keep angle of attack.
If not - use the boiler.

***

The exhaust-powered steering is absolutely normal, and widely used. See R-29RM for details.

Edited by kerbiloid

Share this post


Link to post
Share on other sites

Was the Curiosity backshell+heatshield volume pressurized?

Or there was vacuum inside until aerobraking?

Edited by kerbiloid

Share this post


Link to post
Share on other sites
21 hours ago, satnet said:

LightSail 2 has successfully demonstrated solar sailing. It has raised its apogee by 2 kilometers using a solar sail for thrust, a momentum wheel for orientation, and electromagnetic torque rod for desaturation.

http://www.planetary.org/blogs/jason-davis/lightsail-2-successful-flight-by-light.html

Cool, I didn’t know about the rod.

Share this post


Link to post
Share on other sites

Does a spacecraft movement vector is affected by the direction of the nozzle or the direction of gas that's expelled from that nozzle? For example, if we outfit the spacecraft engine nozzle with target-type thrust reverser, the moment the thrust reverser deflector doors deployed, the thrust reversal (in theory) takes effect and the doors block the expelled gas in the end of the engine. In this case the gas passes through the inner surface and travels frontward to provide force opposite to the heading of the spacecraft. Assuming the spacecraft is on stable orbit, should the craft:

A. Move faster (velocity change in respect of gas expelled from nozzle)

B. Move slower (velocity change in respect of gas deflected frontward)

C. No change (velocity change of nozzle gas is counteracted by deflected gas)

Assume the reverser is able to reverse 100% of gas from the nozzle, which one is going to happen?

Share this post


Link to post
Share on other sites
11 hours ago, ARS said:

Does a spacecraft movement vector is affected by the direction of the nozzle or the direction of gas that's expelled from that nozzle? For example, if we outfit the spacecraft engine nozzle with target-type thrust reverser, the moment the thrust reverser deflector doors deployed, the thrust reversal (in theory) takes effect and the doors block the expelled gas in the end of the engine. In this case the gas passes through the inner surface and travels frontward to provide force opposite to the heading of the spacecraft. Assuming the spacecraft is on stable orbit, should the craft:

A. Move faster (velocity change in respect of gas expelled from nozzle)

B. Move slower (velocity change in respect of gas deflected frontward)

C. No change (velocity change of nozzle gas is counteracted by deflected gas)

Assume the reverser is able to reverse 100% of gas from the nozzle, which one is going to happen?

If you reversed the trust you will trust in the opposite direction so B if spacecraft is pointing with nose in direction of travel. 

You will obviously loose part of the trust because the gas will slow down a bit because of the deflection. 
This is never done as you can just rotate the craft. 

Share this post


Link to post
Share on other sites

Even in the Chronicles of Riddick they just jettison whole engines to brake, instead of reverse, lol.

Spoiler

 

 

Share this post


Link to post
Share on other sites
1 hour ago, kerbiloid said:

Even in the Chronicles of Riddick they just jettison whole engines to brake, instead of reverse, lol.

  Hide contents

 

 

Dropping engines and using them as anchors, well it was an new one. :) 

I once drove an tractor without brakes, you could brake with the engine but I was in highest gear so not so effective, dropping the harrow onto the road worked well however and I could just raise it again then slow enough.  

Share this post


Link to post
Share on other sites
20 minutes ago, magnemoe said:

as anchors

As rocketchutes, they follow the ship into the hangar.

Edited by kerbiloid

Share this post


Link to post
Share on other sites

This one may seem a bit odd out of context, so bear with me:

 

Let's say that there's a sphere 70 meters in diameter orbiting as far from the sun as saturn, with a density and thermal conductivity comparable to living flesh. How much extra thermal energy (In W or kW) does the interior of this sphere need to produce to maintain an average temperature somewhere in the range of 10-40 degrees celsius?

Share this post


Link to post
Share on other sites

Thermal conductivity doesn't play role, as this is stable equilibrium.

SolarPower:= SunLuminosIty / (4 * pi * DistanceFromSun2) * (pi * SphereRadius2) = (SunLuminosIty * SphereRadius2) / (4 * DistanceFromSun2) ;

(SolarPower + InternalPower) = 4 * pi * SphereRadius2 * StefanBoltzmannConst * SurfaceTemperature4;

InternalPower = 4 * pi * SphereRadius2 * StefanBoltzmannConst * SurfaceTemperature4 - SolarPower
=  4 * pi * SphereRadius2 * StefanBoltzmannConst * SurfaceTemperature4 - (SunLuminosIty * SphereRadius2) / (4 * DistanceFromSun2)
=  SphereRadius2 * (4 * pi * StefanBoltzmannConst * SurfaceTemperature4 - SunLuminosIty / (4 * DistanceFromSun2))
=  702 * (4 * pi * 5.67032*10-8 * (283..313)4 - 3.86*1026 / (4 * (9.54 * 149.6*109)2))
=  4900 * (7.126*10-7 * (6.414..9.598)*109 - 3.86*1026 / (4 * 2.037*1024))
=  4900 * ( (4570.6 .. 6839.5) - 47))
(so, the solar income is almost negligible)
= (2.22 ..  3.35) * 107 W = 22..34 MW;

(All values in SI units).

Check: 34*106 / (4 * pi * 702 * 5.67032*10-8)0.25 - 273.16 ~= 41°C.

So, to keep its surface temperature +10..40°C, an internal heat source 22..34 MW is required.
To keep the internal temperature distribution uniform, not layered, the heat source should be equipped with a cooling system to redistribute the heat inside the sphere, or the heat source itself should be multiple and uniformly distributed. Human body uses both ways.

Edited by kerbiloid

Share this post


Link to post
Share on other sites
On 6 August 2019 at 2:59 PM, kerbiloid said:

= (2.22 ..  3.35) * 107 W = 22..34 MW;

(All values in SI units).

Check: 34*106 / (4 * pi * 702 * 5.67032*10-8)0.25 - 273.16 ~= 41°C.

So, to keep its surface temperature +10..40°C, an internal heat source 22..34 MW is required.
To keep the internal temperature distribution uniform, not layered, the heat source should be equipped with a cooling system to redistribute the heat inside the sphere, or the heat source itself should be multiple and uniformly distributed. Human body uses both ways.

Thanks.

 

 

So, another question I have is this: Let's say I've got a rocket engine design that works by heating up liquid mercury and shooting it out the back, like an NTR but much hotter. If the mercury is heated to 100 million K, what should the expected exhaust velocity be?

Share this post


Link to post
Share on other sites
1 hour ago, ChrisSpace said:

Thanks.

No, No No....   You don't get to posit a "ball of flesh" orbiting by Saturn and not give a little context there!  :D  

1 hour ago, ChrisSpace said:

If the mercury is heated to 100 million K, what should the expected exhaust velocity be?

Cheeky answer is 0, as I don't think there's a material capable of holding gaseous mercury (or gaseous anything) at that temp.    Also, wouldn't you start to encounter some breakdowns in the fundamental forces at these energy levels?  Ie similar to the plasma/goo immediately after the Big Bang?

Share this post


Link to post
Share on other sites
1 hour ago, ChrisSpace said:

a rocket engine design that works by heating up liquid mercury and shooting it out the back, like an NTR but much hotter. If the mercury is heated to 100 million K, what should the expected exhaust velocity be?

As no solid chamber could withstand this temperature, the only way you can use is some kind of magnetic nozzle, and the magnetic field or beam heater.
So, probably you can speak just about an average thermal velocity corresponding to the given temperature.

v = sqrt(3 * R * Temperature / MolarMass) = sqrt(3 * 8.31441 J/(mol*K) * 108 K / 0.2 kg/mol ) ~= 112 km/s

No reason to use a heavy metal except in electric engines.

Share this post


Link to post
Share on other sites
1 hour ago, Gargamel said:

No, No No....   You don't get to posit a "ball of flesh" orbiting by Saturn and not give a little context there!  :D

It's part of a spaceborne creature design I'm working on as part of Spaceception's novel project. Since the creature's main body is basically a sphere, I figured this'd be a good way to determine what kind of metabolic rate it would need to have to maintain it's core body temperature.

 

And as for my second question, I'm trying to figure out how the [REDACTED] the Morrigan-class patrol destroyer from The Expanse can hold over 28000 metric tons of propellant in it's little body, as stated on the wiki. Liquid Mercury is the densest liquid I can think of, so...

1 hour ago, kerbiloid said:

As no solid chamber could withstand this temperature, the only way you can use is some kind of magnetic nozzle, and the magnetic field or beam heater.
So, probably you can speak just about an average thermal velocity corresponding to the given temperature.

v = sqrt(3 * R * Temperature / MolarMass) = sqrt(3 * 8.31441 J/(mol*K) * 108 K / 0.2 kg/mol ) ~= 112 km/s

No reason to use a heavy metal except in electric engines.

Well, that's a problem. Is there any other really-high-density propellant (over 10g/cm^3) that could do better? What about Ruthenium or Rhodium or Molybdenum?

Edited by ChrisSpace

Share this post


Link to post
Share on other sites
33 minutes ago, ChrisSpace said:

Well, that's a problem. Is there any other really-high-density propellant (over 10g/cm^3) that could do better?

Unlikely pure elements, as all of them have great molar mass.

Maybe a compound of a relatively heavy element with a lot of atoms of a light element.

H - unlikely, they are mostly gases, even PbH4.
He - No. Just no.
Li - Unlikely, no "lythide" found.
Be - Unlikely and rare, both Be and beryllides.
B - Maybe. Googling "borides density" gives: W2B5 (density 14.8), Mo2B5 (7.48), and Fe2B (7.3).
Molar masses: W = 184, Mo = 96, Fe = 56.
C - Maybe. WC (15.6)
N - Unlikely, as mostly attaches several other atoms, and density is < 4.
O - Maybe. iron oxides, density ~5.5, averag atomic mass of atoms ~33, widespread.
F - Unlikely, as fluorides solid density <= 4, they are usually gaseous.

Probably, the least evil is a mixture of iron oxides. But anyway their atoms thermal speed ~250 km/s. Or dense, or fast. But then it can eat rocks.

Share this post


Link to post
Share on other sites
50 minutes ago, kerbiloid said:

Probably, the least evil is a mixture of iron oxides. But anyway their atoms thermal speed ~250 km/s.

Is this just an estimate or did you calculate it?

Share this post


Link to post
Share on other sites
2 minutes ago, ChrisSpace said:

Is this just an estimate or did you calculate it?

average atomic mass ~= (56 * 2 + 16 * 3) / (2 + 3) ~= 32.

v = sqrt(3 * 8.3 * 108 / 32*10-3) ~= 278 km/s

But it's a rough estimation as should calculate iron and oxygen separately, then combine.

Though, I guess more or less appropriate, especially as the temperature is arbitrary.

Share this post


Link to post
Share on other sites

Imagine a mini-submarine, which has a neutral buoyancy. If it's fitted with powerful engine and "wings" (with "control surface"), could it be steered like aircraft underwater? (As long as the engines are running, since water slows it down a lot when it's stopped) If it's possible, could we "dogfight" with it underwater? (Not necessarily firing, just getting on other's rear)

Share this post


Link to post
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.