3 ms·

A closed tube could maybe have negligible air friction, if you can keep it moving.

snogglethorpe 14y ago
An evacuated tube could have negligible air-friction—but an evacuated tube is very different than what's being discussed as it meets none of the apparent criteria (low-cost, low-maintenance, easily constructed, precision-insensitive). Also, Musk explicitly said it's not that ("But Elon has said there’s no vaccuum tubes involved").
- hotpockets 14y ago
  Nah I am saying move the air at the same speed as the trains.
  - snogglethorpe 14y ago
    You still get friction then, just between the air and the tube walls. The bad effects of air-friction would seem to still be present, and maybe much greater, because you'd have to move a huge mass of air and the friction would be over a much greater surface area (isolating the moving air to within a small portion of the tube does not look simple to me).
    - hotpockets 14y ago
      Also, this article mentions that it is a tube of some sort: http://www.businessweek.com/articles/2012-09-13/elon-musk-the-21st-century-industrialist#p5 http://www.businessweek.com/articles/2012-09-13/elon-musk-th...
    - hotpockets 14y ago
      I tried a calculation. It appears there is an optimal gas velocity in between 0 and V_train. It depends on a lot of factors that I didn't spend much time trying to come up with realistic values. If I weight things heavily in my favor, you can save over 50% power requirements by moving the air at around ~120m/s. If I weight things in your favor there is a pretty paltry power savings at around ~10m/s. If you care, here was my method: f=0.006; %wall friction factor inside tube, also skin friction outside a train rho=1.2; %air density [kg/m^3] R=1.5; %tube radius [m] Vg=0:200; %velocity of tube gas [m/s] Vt=250; %velocity of train [m/s] L=500e3; %length of tube [m] Cd=0.8; %drag coefficient, form drag trainfill=0.1; %percent of tube filled by train airfill=1-trainfill; %percent of tube filled by air ncars=1000; %number of cars in tube Pg=pi/4fVg.^3LairfillR; %power required to move gas through tube Pt_fm=ncars0.5rhopiR^2Cd(Vt-Vg).^3; %power required to overcome form drag on each car Pt_sk=pi/4f(Vt-Vg).^3LtrainfillR; %power required to overcome skin friction drag Pg_annulus=0; %since the mass of air is so small between the train and tube wall, assume 0. Pt=Pg+Pt_fm+Pt_sk; %total power requirements
- thedrbrian 14y ago
  Has he ruled out running the trains in a lighter gas like hydrogen or helium?
  - hotpockets 14y ago
    Interesting, but then you have to worry about keeping oxygen in the train, and not having leaks?

An evacuated tube could have negligible air-friction—but an evacuated tube is very different than what's being discussed as it meets none of the apparent criteria (low-cost, low-maintenance, easily constructed, precision-insensitive). Also, Musk explicitly said it's not that ("But Elon has said there’s no vaccuum tubes involved").

hotpockets 14y ago
Nah I am saying move the air at the same speed as the trains.
- snogglethorpe 14y ago
  You still get friction then, just between the air and the tube walls. The bad effects of air-friction would seem to still be present, and maybe much greater, because you'd have to move a huge mass of air and the friction would be over a much greater surface area (isolating the moving air to within a small portion of the tube does not look simple to me).
  - hotpockets 14y ago
    Also, this article mentions that it is a tube of some sort: http://www.businessweek.com/articles/2012-09-13/elon-musk-the-21st-century-industrialist#p5 http://www.businessweek.com/articles/2012-09-13/elon-musk-th...
  - hotpockets 14y ago
    I tried a calculation. It appears there is an optimal gas velocity in between 0 and V_train. It depends on a lot of factors that I didn't spend much time trying to come up with realistic values. If I weight things heavily in my favor, you can save over 50% power requirements by moving the air at around ~120m/s. If I weight things in your favor there is a pretty paltry power savings at around ~10m/s. If you care, here was my method: f=0.006; %wall friction factor inside tube, also skin friction outside a train rho=1.2; %air density [kg/m^3] R=1.5; %tube radius [m] Vg=0:200; %velocity of tube gas [m/s] Vt=250; %velocity of train [m/s] L=500e3; %length of tube [m] Cd=0.8; %drag coefficient, form drag trainfill=0.1; %percent of tube filled by train airfill=1-trainfill; %percent of tube filled by air ncars=1000; %number of cars in tube Pg=pi/4fVg.^3LairfillR; %power required to move gas through tube Pt_fm=ncars0.5rhopiR^2Cd(Vt-Vg).^3; %power required to overcome form drag on each car Pt_sk=pi/4f(Vt-Vg).^3LtrainfillR; %power required to overcome skin friction drag Pg_annulus=0; %since the mass of air is so small between the train and tube wall, assume 0. Pt=Pg+Pt_fm+Pt_sk; %total power requirements
thedrbrian 14y ago
Has he ruled out running the trains in a lighter gas like hydrogen or helium?
- hotpockets 14y ago
  Interesting, but then you have to worry about keeping oxygen in the train, and not having leaks?