Recent content by spirochete

The concept is sometimes called "Worldhouse". It is the rationale I use for the Size 1 worlds in the Spinward Marches with breathable atmosphere - canopied "terrarium" worlds.

By taking on less fuel and cargo. The lighter load will net more acceleration for the same thrust level, allowing you to reach orbit.

He was also a submarine commander in U-571. He failed his survival roll in that one as well.

Attempt to capture & contain it. LBB 3 shows antimatter as a TL 17 energy source, but in reality M=E/C², therefore antimatter takes far more energy to create than it would yield in a reactor. It just isn't practical. However, a harvest-able antimatter resource like this comet would be an...

I have an Excel worksheet that is an expanded Tech Level table that would give you an idea of the capabilities at different levels. It's a semi-comprehensive catalog of invention and innovation by category and tech level up to TL 23. The problem is, I have no idea of where to upload it to.

The "turbine" terminology is misleading. Real-world MHD generators aren't really "turbines" with rotating blades like a gas turbine. They're more like a dynamo with a gas armature.

Correct. Complete BS since they're very low thrust. Drives of this sort are suitable only for long duration burns with very low acceleration. Pitifully underpowered for orbit ascent from the tiniest moonlets. Thus far, any propulsion system with high efficiency has low thrust. The...

Anytime :) Just remember, the units are "earths", where earth = 1 for any of the variables. I don't remember. I *do* remember the MMWR is generally proportional to the radius of a planet and inversely proportional to its mass, e.g., a planet with twice the mass-radius ratio as Earth retains...

The units in the formulas are "earths". For example: r = sqrt(m/g) , so if m = 0.25 earth masses and g = 0.54 earth gravities, then r = sqrt(0.25/0.54) = 0.68 earth radii

No. Increasing a planet's size subjects it to greater tidal braking.

The appropriate radius is 1.795 earth radii, which gives a diameter around 22,866 km. Mass is a tad under 10 earth-masses, gravity 3.1 G. This is the upper limit for a solid silicate-nickle-iron exoplanet. The minimum molecular weight retained is less than 2, meaning it retains hydrogen in its...

Ditto. It' basic Newtonian mechanics. F = ma, therefore a = F/m. This way, a type A free trader with 1G nominal acceleration can liftoff from a 1.2 G planet by reducing its load (payload AND fuel). Less burden = more acceleration.

What if the payload is entirely internal? If acceleration is determined by displacement (non-newtonian), then I can load the cargo bay with 700 metric tons of lead and still get 3G acceleration. The ship occupies the same volume regardless of its internal cargo.

It's just the raw cross-section; it should be easy to approximate by placing a frontal silhouette over a deckplan grid and counting the squares. The streamlining and skin friction are incorporated into the drag coefficient.

Top speed is reached when the aerodynamic drag force equals thrust. Fd = pv²CdA/2 Where - Frontal area (cross-section) A - Drag coefficient Cd - Velocity v - Atmospheric mass density p See http://en.wikipedia.org/wiki/Drag_equation