Cost of an ATV

[Timerover51]

Originally Posted by LBB 3
An air/raft can reach orbit in several hours (number of hours equal to planetary size digit in the UPP

In theory, it would be possible. It take about 4 kilowatts per pound to put something into orbit around the Earth. Assuming 8,000 to 10,000 pounds for an air/raft, that would be 8 to 10 Megawatts over 8 hours, so about a Megawatt an hour. However, your energy conversion from Fusion to Electricity would have to be really good to generate that much power in what has to be a fairly small space, like a cubic meter or so. Plus, your fusion plant would have to be extremely small as well.

I am not exactly a fan of very small fusion plants.

 

[Cryton]

An air/raft can reach orbit in several hours (number of hours equal to planetary size digit in the UPP

Yep, that's the Black letter. It always gave me visions of the intro scene from Heavy Metal the Movie, you know, the one of the Space suited driver doing re-entry in his 'vette.

Good times

 

[kilemall]

Current pricing of a new Cessna 172- $300K.

https://en.wikipedia.org/wiki/Cessna_172

Beechcraft Bonanza, $700K.

https://en.wikipedia.org/wiki/Beechcraft_Bonanza

For the low tech version of the grav speeder, very light jets typified by the Honda jet, $1-3 million-

https://en.wikipedia.org/wiki/List_of_very_light_jets

A suborbital 600 kilocredit gravjeep looks in the ballpark to me.

 

[AnotherDilbert]

In theory, it would be possible. It take about 4 kilowatts per pound to put something into orbit around the Earth. Assuming 8,000 to 10,000 pounds for an air/raft, that would be 8 to 10 Megawatts over 8 hours, so about a Megawatt an hour.

Um, it was a few decades since I did orbital mechanics, but an orbit takes a specified energy. MW is a measurement of effect, not energy.

With anti-grav we can probably ignore the velocity required to sustain orbit, but to rendezvous with something in orbit we have to match its velocity.

 

[aramis]

Um, it was a few decades since I did orbital mechanics, but an orbit takes a specified energy. MW is a measurement of effect, not energy.

With anti-grav we can probably ignore the velocity required to sustain orbit, but to rendezvous with something in orbit we have to match its velocity.

4.25 km/s delta V Surface to LEO.

Plus whatever change in velocity, roughly doubled. MW = kg*(m/s)^2

 

[AnotherDilbert]

Um, no?

Energy J = kg m² / s²
Effect W = kg m² / s³ = J / s

Let me give it a try:
To achieve LEO seems to take a delta-v of about 8 - 10 km/s (I think including the velocity needed)
Energy = mv² / 2 ≈ 9000² / 2 ≈ 40 MJ to lift 1 kg into orbit. So to lift a 4000 kg air/raft it takes 4000 kg × 40 MJ/kg = 160 GJ.

1 h is 3600 s, 10 h is 36000 s. So it would take 160 000 000 000 / 36 000 ≈ 4,4 MW for 10 h to achieve 160 GJ.

4,4 MW is about 6000 hp.

 

[aramis]

LEO requires 4.24 km/s total delta V. Known pblished quantity. If your equation shows otherwise, then you've got it wrong.

E_J=MV^2; m=kg, E=J=1WS, M in kg, V in m/s. Known quantity.
V of 4240 m/s gives KW=17977600× M_kg.
E_{GJ = Mtons×Vkm² so you need roughly 18 GW×Second per ton of mass. (As 1J=1W × 1 sec) VERY straightforward algebra. Very basic HS physics.}

 

[AnotherDilbert]

I think we are in the same ballpark...

LEO requires 4.24 km/s total delta V. Known pblished quantity. If your equation shows otherwise, then you've got it wrong.

I got my number from https://en.wikipedia.org/wiki/Low_Earth_orbit, so I cannot guarantee it. I think it includes the orbital velocity whereas your number did not include it? In such case, we basically agree.

E_J=MV^2; m=kg, E=J=1WS, M in kg, V in m/s. Known quantity.

Kinetic energy = mv²/2?

V of 4240 m/s gives KW=17977600× M_kg.

E = mv²/2 = m × 17 977 600 / 2 = m × 8 988 800 J or about 9 MJ/kg? Definitely not kW.

E_GJ = M_tons×V_km²
so you need roughly 18 GW×Second per ton of mass. (As 1J=1W × 1 sec)

18 GWs/ton = 18 GJ/ton = 18 MJ/kg (excluding orbital velocity?)
I came to 40 MJ/kg including orbital velocity, we are not far away from each other.

VERY straightforward algebra.
Very basic HS physics.

Given that our vehicle is almost constant mass I hope we can use Newton instead of Tsiolkovsky.

I think delta-v is defined using Tsiolkovsky and is not a velocity even if it has the dimension m/s?

I suspect we have both simplified the problem a bit...

 

[AnotherDilbert]

This article seems about right for a theoretical minimum, ignoring petty details like the atmosphere:
http://www.wired.com/2015/09/whats-special-low-earth-orbit/

In a graph it gives a value of 30 - 60 MJ/kg depending on orbital height. LEO would be around 35 MJ/kg.

 

[AnotherDilbert]

If we use 35 MJ/kg a fully loaded air/raft at 8500 kg (4000 kg + 4000 kg cargo + 4 people) would take 35 MJ/kg × 8500 kg = 297 500 MJ ≈ 300 GJ to reach orbit. To do this in 8 hours (as Earth is a size 8 planet) it would take 300 000 000 000 J / ( 8 × 3600 ) s ≈ 10 416 667 W ≈ 10 MW or about 14 000 hp.

Using Striker we can see that a 10 MW power plant would be 5 m³ and mass 20 000 kg at TL 9, impossible to fit in a 4 000 kg air/raft.

So, an air/raft cannot mount a power plant with power enough to push it to orbit, grav vehicles cannot work this way. Since they do work in Traveller, they must work in a different manner.



The "reasonable" assumption would be that anti-grav vehicles actually create negative gravity, allowing them to simply fall away from the planet.