I think we can tolerate about 0.5g of deceleration, which shifts the direction of gravity by 30 degrees. So, about 5ms⁻².  To get to interesting speeds such as 500ms⁻¹ (about mach 1.5), it would take 100 seconds, and 25,000m (25km, or about 15 miles).  Assuming maximum acceleration and deceleration, it would take 3 minutes and 20 seconds to travel over 50km.  At first approximation, separating stations by 50km looks doable.<p>But that&#x27;s not the whole story.  If we have intermediate stations <i>at all</i>, we need to ensure all cars have the same speed when they get to share the same main tube.  We need acceleration tubes the same way we need acceleration lanes in regular roads. (How to plug those tubes to the main one is left as an exercise to whoever builds this.) Anyway, those tubes need to be 25km long.<p>But that doesn&#x27;t mean they have to be separated by that much.  You could have a station every 5km if you&#x27;re willing to run 5 acceleration lanes in parallel (and pay for them).<p>This gets even easier once you get close to the destination: even in the main tube, cars need to decelerate, starting at 25km from their destination.  10km away for instance, the speed is already down to a little over 300ms⁻¹.  To join the lane at <i>that</i> point, you need only 10km to accelerate.  Likewise, if you need to join 5km before the destination, you need to accelerate for 5km as well.<p>In other words, you could very well have stations as close as 5, 10, or 20km from the urban centre.  You won&#x27;t go very fast, but remember that at those speeds and accelerations, it&#x27;s going to take less than 5 minutes anyway.<p>---<p>If trains didn&#x27;t have to stop at <i>every</i> station, and used acceleration and deceleration rail-roads where appropriate, their average speed would be much higher.  First, they can&#x27;t for 2 reasons: rail-roads are expensive and take land (so does an Hyperloop tube).  Second,  passengers need to go in and out.  This means many stops for the same train.  Hyperloop&#x27;s uses individual, so it doesn&#x27;t have that problem.