Re: [Spaceships] Designers' Championship I: ground to LEO
 

Note: I know nothing of actual spaceflight.

While it's neither in RAW nor ground-rules, it seems reasonable to somehow subtract (some factor of) planetary gravity at least when comparing launch-times. But 2G (3-1) is just four times higher than 0,5G (1,5-1) which should make a 1,5G thrust ship viable, shouldn't take more than 6x4=24 minutes, still not that high.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

While I can't say anything about if your right or not, I can say it's neither in RAW nor in the ground-rules.

In spaceships there is no scramjets. You may be perfectly right, but in the contest of the competition it wouldn't be fair to switch jet-fuel for 4K a pop with hydrogen for 2K.

This brings me to my conclusion. The ground-rules need updating. More precise rules for calculating fuel-consumption with/without wings, ram-jets and variable thrust. If there is a ~20-25% increase in fuel-efficiency if you increase thrust from 1,5G to 3G, as your numbers indicate, we need some way of actually calculating that.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

It seems reasonable, but is a flawed reading of the principles. Delta-V is the change in velocity produced by an engine. There is a minimum dV needed to reach a given orbit, or escape a planet altogether. This is the integral or sum of the change in velocity from ground to orbit. The rate at which this change in velocity is achieved is irrelevant, the dV is the same. The dV of an engine is an integral of the change in velocity produced by the thrust, that is the sum of thrust over all the burn time. This is why GURPS Spaceships doesn't talk about the effects of increased acceleration on fuel efficiency in planetary lift-off. As long as you have enough thrust to overcome gravity, its the total Delta-V that counts, and has to be greater than the dV needed to reach orbit.

For comparison, note that a craft with wings or contragrav and only 0.1G acceleration would be able to achieve orbit, given enough dV and patience.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

This is fundamentally misunderstanding the mathematics. Ignoring the effects of wings, the total dV needed to reach orbit is a function of gravity. The design of your craft has no effect at all on the dV needed. If your passengers are patient, 1.1G and plenty of dV is far better than 3G and barely enough dV. Because of the way velocity changes aggregate, the rate of change of velocity does not affect the gravitational effects that have to be overcome. For the same dV, more thrust means that you lose the same amount of gravity (gravitational potential energy) faster. It doesn't mean less gravity (gravitational potential energy).

Re: [Spaceships] Designers' Championship I: ground to LEO
 

Could someone take a look at this draft, for a TL 10 antimatter thermal lighter? All costs are in thousands:

Front
1 Armor (AML/SL) $60 dDR 3
2 Passenger (2) $10
3 Passenger (2) $10
4 Passenger (2) $10
5 Passenger (2) $10
6 Passenger (2) $10
0 Control Room $60

Central
1 AC Hydrogen Fuel $10
2 AC Hydrogen Fuel $10
3 AC Hydrogen Fuel $10
4 Armor (AML/SL) $60 dDR 3
5 Passenger (2) $10
6 Passenger (2) $10
0 Passenger (2) $10

Rear
1 ATR Ram-Rocket $750
2 ATR Ram-Rocket $750
3 ATR Ram-Rocket $750
4 ATR Ram-Rocket $750
5 ATR Ram-Rocket $750
6 ATR Ram-Rocket $750

Winged $150

Thrust 1.2 G SM 5
Air Speed 2739 mph dST/dHP 20
0.76 mps Hnd 4/0
Orbit DV 4.6 mps SR 5/4
Fuel DV 3.84 mps HT 12
Fuel Cost $16.67 $/mps TL 10
Fuel to Orbit $63.99 Loaded Wgt 30 tons
Time to Orbit 10.35 Length 45 ft
Cost $4,940

VC Per Trip Fixed Annual
Op Ins $0.74 Ins $49.4
Service $0.45 Crew $537.6
Fuel $63.99 Dep+ROC $444.6
Total $65.18 Total $1,031.6

Flights/Year 2400
Passengers 12 /trip
28800 /year
Total Costs $157,468
Cost/Passenger $5.47

Re: [Spaceships] Designers' Championship I: ground to LEO
 

<deleted>

I'm sorry, that was uncalled-for.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

Notice that if you go straight up the flight path angle is zero, the sine of which is also zero. Then your losses are zero, irrespective of your acceleration. In other words, this is energy used to accelerate your craft horizontally, not energy used to "fight" gravity. The gravitational potential energy of an object in orbit is the same regardless of how it got there. All other things being equal, all you need to give your orbiter is that much energy. Real life is undoubtedly never quite that simple.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

If you go straight up, your flight path angle is 90 (it's measured from horizontal), and your losses are maximized.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

Well, if you really want to mess with things, I'm sure we can integrate over the path of the rocket, factoring in losses due to gravity at a given height, losses due to atmospheric drag dependent on speed, latitude, and altitude, and variations depending on the latitude and direction of launch. It would be a good exercise in whether or not I can still calculate an integral that complex, but, we can probably just forget about atmospheric drag if pretty much every design uses a ram-rocket or jet component, and estimating losses due to gravity for LEO insertion is pretty simple too.

I've had another thought too - I'm not sure if it's occurred to anybody to take passenger comfort into account - most people don't appreciate being accelerated at rates approaching 3G for an extended period of time. For example, you don't want to have to resuscitate Grandma on her way to visit her son who's working in Luna-3 Colony. It might be that a slower, though lossier lighter is preferable to a faster one.

Re: [Spaceships] Designers' Championship I: ground to LEO
 

I hadn't spotted that. I was taking my figures from VE2/Vx1, but that also has fission reactors requiring a 2 year refuel cycle. So I guess my original figures stand.