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6 hours ago, kerbiloid said:

This one.

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3132BC1E00000578-0-image-a-44_1455461335

 

I say it depend, bicycles pretty much require roads while horses work in most terrain. The downside of horses is that they are harder to transport especially on ships and require care and upkeep. 
Finally while bicycles has been used most famously by the Japanese taking Singapore horses has been used way more in the time since bicycles become common. 

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30 minutes ago, magnemoe said:

The downside of horses is that they are harder to transport especially on ships and require care and upkeep. 

These ones don't know such words, that's why they are very useful. Also they are greater on mountains.

 

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From the moon surface.

Delta-V ~= 2 km/s
ISP ~= 3.5 km/s
Mass ratio (together with tanks and engines) = exp(2/3.5) * 1.1 ~2.
Fuel mass ~= ISS mass ~= 420 t.
Fuel density ~ 1 t/m3.
Tanks volume ~420 m3.
ISS truss size ~= 73 x 109 m.
ISS vertical  projection area = 73 * 109 ~= 8000 m2.
Average fuel storage height ~= 420 / 8000 = 0.05 m = 5 cm.

So, if put a truss platform under it, fix the extended parts, and properly distribute tank+engine assemblies below, why not.

***

From LEO - why not at all?

Edited by kerbiloid
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2 hours ago, Flying dutchman said:

Would it be possible to boost the iss into orbit of the moon?

Troublesome. High-acceleration thruster would tear it apart. Low-acceleration thruster will lead to months in the van Allen belts, frying its electronics.

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1 hour ago, Flying dutchman said:

How about somewhere in between?

Somewhere in between is generally a nuclear rocket. Cramming the station behind a shadow shield will be problematic.

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3 hours ago, Flying dutchman said:

Would it be possible to boost the iss into orbit of the moon?

How much are you willing to spend?

There's nothing in physics that says you can't, and if you have a large enough budget, you can get it done.

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16 minutes ago, DDE said:

Somewhere in between is generally a nuclear rocket. Cramming the station behind a shadow shield will be problematic.

But is does not have to be a nuclear rocket. I think the iss would be able to survive one rl10 engine with a little reinforcement

 

Alternatively use the onboard engine and add extra fuel tanks.

 

Or extend the orbit to just above the first belt and use ion thrusters the rest of the way

Edited by Flying dutchman
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Regarding nuclear pulse propulsion, the idea of it, by my understanding, is to convert the nuclear potential energy of the propulsion units into kinetic energy for the ship. So, if you know the amount of kinetic energy successfully imparted to the ship by a propulsion unit and mass of ship you can find the delta-v imparted to the ship by that propulsion unit as per the equation v=sqrt(2KE/m), right?

If the rocket equation is not applicable to NPP ships then my only thought to find the ship's delta-v budget is to add up the delta-v provided by each propulsion unit, accounting for the changing mass of the ship as each propulsion unit is expelled. So, the equation would then be:

aZfVKhu.png

Where:

E_K is the kinetic energy

m_wet is the ship's wet mass

n_units is the total number of propulsion units

and 

m_units is the mass of each propulsion unit

That's what makes intuitive sense to me, anyway. 

Edited by HaplessBystander
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18 hours ago, HaplessBystander said:

Regarding nuclear pulse propulsion, the idea of it, by my understanding, is to convert the nuclear potential energy of the propulsion units into kinetic energy for the ship. So, if you know the amount of kinetic energy successfully imparted to the ship by a propulsion unit and mass of ship you can find the delta-v imparted to the ship by that propulsion unit as per the equation v=sqrt(2KE/m), right?

The rocket equation depends on the conservation of momentum.  In a NPP, this breaks down into two parts: the pusher plate conserving the momentum of the blast it is hit with, and the pusher plate and rest of the rocket conserving momentum as the pusher plate is pushed back toward the rocket and then pushed back to the initial position.  I'm not really sure if it still holds (it probably should, as the pusher plate forms an elastic body.  But I'm not sure I can justify the same assumptions as a rocket).

 

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On 3/18/2020 at 6:03 PM, HaplessBystander said:

Regarding nuclear pulse propulsion, the idea of it, by my understanding, is to convert the nuclear potential energy of the propulsion units into kinetic energy for the ship. So, if you know the amount of kinetic energy successfully imparted to the ship by a propulsion unit and mass of ship you can find the delta-v imparted to the ship by that propulsion unit as per the equation v=sqrt(2KE/m), right?

If the rocket equation is not applicable to NPP ships then my only thought to find the ship's delta-v budget is to add up the delta-v provided by each propulsion unit, accounting for the changing mass of the ship as each propulsion unit is expelled. So, the equation would then be:

Doesn't work this way. Kinetic energy is frame-dependent. Adding 1m/s to something stationary takes a lot less energy than to something that's already going 100m/s. This is why you balance momentum, not energy, when deriving rocket formula.

And honestly, rocket formula is still a very good estimate for NPP. You can play off the fact that sum 1/n is very closely approximated by ln(n) to within a correction term. So as long as you can estimate the ISP of your pulses, you can simply treat this as a continuous burn rather than series of pushes. And for estimating ISP you could go with energy conservation approach, but you'd be using energy to find the velocity of the propellant. Of course, you'd have to know how much energy you'd straight up lose in NPP, and I suspect, it's a rather large fraction.

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8 hours ago, kerbiloid said:

No, not the snorkeling equipment. The wooden log that's usually carried on the back of the tank that traced back as far as the British Mark series'

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4 hours ago, ARS said:

The wooden log that's usually carried on the back of the tank that traced back as far as the British Mark series

From here:
"Mark IVs were also the first tanks fitted with unditching beams by field workshops. A large wooden beam, reinforced with sheet metal, was stored across the top of the tank on a set of parallel rails. If the tank became stuck, the beam was attached to the tracks (often under fire) and then dragged beneath the vehicle, providing grip."

Spoiler

In use:
MarkIVTankWithUnditchingBeam1917.jpg

"If a tank got stuck in the mud and the tracks were finding it hard to gain purchase (a common hazard in the artillery-churned no-mans-land of the Western Front), the unditching beam would be chained to both tracks. As the tracks travel under the tank, the beam would dig into the mud, providing a much better purchase, and then the tank would pull itself over the beam that has now dug into the ground. Once the tank had completely passed over the beam, it could be moved to the front end of the tank again, and the process repeated until the tank reached more solid ground."
From here.

And for extra awesomeness, in use:

Spoiler

 

 

Edited by razark
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9 hours ago, razark said:

<snikt>

Incidentally, this is about the only method for self-recovering a tank that "bellies out", ie: gets raised up by an obstruction (treestump, antitank wedge etc) under its belly so a/the track/s cannot contact the ground or gain purchase. Usually these days it has been replaced with a length of cable, seen here strung along the side:

Challenger_2_Tank_Firing_a_Shell_MOD_451

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On 3/21/2020 at 4:10 AM, K^2 said:

Doesn't work this way. Kinetic energy is frame-dependent. Adding 1m/s to something stationary takes a lot less energy than to something that's already going 100m/s. 

What I'm taking away from this is I've failed to account for the relative velocities between the ship and the propulsion unit and the faster the propulsion unit is ejected the more kinetic energy it will have to impart to the ship to produce the same delta-v. Will the aforementioned additive approach still work if the equation is altered to account for this? If so, how do I alter the equation to make it work?

If that wasn't the intended takeaway, I'm lost. 

Edited by HaplessBystander
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1 hour ago, HaplessBystander said:

What I'm taking away from this is I've failed to account for the relative velocities between the ship and the propulsion unit and the faster the propulsion unit is ejected the more kinetic energy it will have to impart to the ship to produce the same delta-v. Will the aforementioned additive approach still work if the equation is altered to account for this? If so, how do I alter the equation to make it work?

Velocity of the tungsten jet = 30..60, up to 150 km/s.

Velocity f the charge ~10-1 km/s

Doesn't matter at all.

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On MAZ-based transporter-erector launcher vehicles, what's the purpose of the that secondary cabin separated from the driver's cabin? Does it serve any functional purpose or just for passengers? What about it's civillian version?

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1 hour ago, ARS said:

On MAZ-based transporter-erector launcher vehicles, what's the purpose of the that secondary cabin separated from the driver's cabin? Does it serve any functional purpose or just for passengers? What about it's civillian version?

LIke this?
TOX01401b_1024x1024.jpg?v=1571271529

Pretty sure its the rocket control room at the center. 
 

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