How much time and distance would a maglev train take to accelerate to the top speed of 700 km/h?

Question

So I have decided that in my story on Mars, I'll have magnetic levitation trains. Now I'm calculating travel times and, considering passenger comfort, the question is:

How quickly could a passenger train accelerate while not causing significant discomfort? And how much time and distance would it take to achieve the top speed of 700 km/h?

The total travel distance is 1'000 km, in a straight line.

Answer 1

About 27 kilometers over 4 minutes 45 seconds

Let's first answer the passenger comfort question.

Humans generally feel acceleration, not necessarily velocity. Wikipedia provides that the Shanghai Transrapid accelerates at 300 km/h (186 mph) in 2 minutes and 15 seconds, or approximately 0.617 m / s2. Incidentally, the Shinkansen N700 Series, a high-speed conventional train, accelerates at 0.720 m/s2. Between the two, we can presume the 0.70-0.75 m/s2 range as reasonable for human passengers.

The Newtonian equations of motion offer much in terms of helping you calculate the time and distance of a moving object. They are:

\begin{aligned}v&=at+v_{0}&[1]\\r&=r_{0}+v_{0}t+{\tfrac {1}{2}}{a}t^{2}&[2]\\r&=r_{0}+{\tfrac {1}{2}}\left(v+v_{0}\right)t&[3]\\v^{2}&=v_{0}^{2}+2a\left(r-r_{0}\right)&[4]\\r&=r_{0}+vt-{\tfrac {1}{2}}{a}t^{2}&[5]\\\end{aligned}

• v is final velocity
• v0 is initial velocity
• a is acceleration
• t is time
• r is final distance
• r0 is initial distance

For the purposes of this response, I will use this calculator. Using the numbers provided:

• 0.6 m/s/s provides a displacement of approximately 31,500 meters and 324 seconds
• 0.7 m/s/s provides a displacement of approximately 27,000 meters and 278 seconds
• 0.8 m/s/s provides a displacement of approximately 23,600 meters and 243 seconds

By this calculation, about 27 kilometers over circa 4 minutes 45 seconds will be sufficient to comfortably accelerate to full speed.

EDIT: as some have mentioned, the Hong Kong MTR Rotem EMU claims an acceleration of 1 m/s/s. This calculation would yield 19 km over 194 secs (3 min 14 seconds). Ultimately given the nature of this question it is up to you which estimate to use specifically.

Answer 2

Maximum comfortable train acceleration doesn't depend on the top speed. So, just look at the data of existing trains for a reference.

For example the Shinkansen

Another feature of the N700 is that it accelerates more quickly than the older 700 series Shinkansen trains, with a maximum acceleration rate of 2.6 km/h/s (0.72 $m/s^2$).

With that in mind, calculating how long does it take to reach a certain speed and to complete a given distance is trivial high school math on the known formulas

$v=a\cdot t$

$s=1/2 a \cdot t^2$

Answer 3

All the other answers use totally valid math (I guess).

But the passengers in your story are martians. They are used to a 3.72076 m/s2 gravity.

How this affect their tolerance to acceleration is ... whatever suits you until proven wrong.

Also earth newcommers may enjoy an increased gravity

Answer 4

Comfortable acceleration for humans on Mars

Gravity on Mars is about 38% of gravity on Earth.

If the inhabitants of Mars are human colonists, they probably want to be subjected regularly to an acceleration closer to that of Earth, in order to maintain a healthy human body and avoid health issues such as loss of bone; atrophy of antigravity muscles; decreased blood volume; loss of blood circulation, leading to intolerance to standing up. Reference: long-term effect of low gravity on the human body.

• Gravity on Earth has magnitude: $g_E = 9.81 \,m/s^2$.
• Gravity on Mars has magnitude: $g_M = 3.72 \,m/s^2$.
• If your train accelerates with horizontal acceleration $a$, then the total force felt by the passenger will have magnitude: $$\sqrt{{g_M}^2 + a^2}$$

So now we can solve the equation $\sqrt{{g_M}^2 + a^2} = g_E$ in order to determine a very comfortable $a$.

This equation corresponds to the following diagram, in which the passenger in the train on mars feels a total acceleration with same magnitude as the acceleration of Earth gravity, although the acceleration will be oblique and closer to horizontal than vertical:

Solve for $a$: \begin{align*} \sqrt{{g_M}^2 + a^2} & = g_E \\ a & = \sqrt{{g_E}^2 - {g_M}^2} \\ a & = \sqrt{{9.81}^2 - {3.72}^2} & m/s^2 \\ a & = 9.08 & m/s^2 \\ \end{align*}

Time and distance to reach top speed of 700 km/h

From this comfortable value of $a$, you can deduce the time $t_0$ and distance $x(t_0)$ necessary to go from speed $v(0) = 0$ to speed $v(t_0) = 700\,km/h = 194.4 \,m/s$.

This is as simple as : \begin{align*} v(t) & = a t \\ x(t) & = \frac 1 2 a t^2 \\ \end{align*} Thus the time is: \begin{align*} t_0 & = \frac{v(t_0)}{a} \\ t_0 & = \frac{194.4}{9.08} & s \\ t_0 & = 21.4 & s \\ \end{align*} And the distance is: \begin{align*} x(t_0) & = \frac{1}{2} \, a \, {t_0}^2 \\ x(t_0) & = \frac{1}{2} \times 9.08 \times {21.4}^2 & m \\ x(t_0) & = 2083 & m \\ \end{align*}

The train will accelerate for about 21s, over about 2.1km.

Jerk

In practice you probably want the acceleration to change smoothly from 0 to max acceleration down to 0 again. A brutal change in acceleration is called a jerk and as the name suggests, it is quite unpleasant. A jerk of up to 1g/s (9.8 m/s³) shouldn't be harmful, but unpleasant. If this train is part of a military operation, or if the passengers are well-seated and the luggage is well-secured, you could theoretically reach top acceleration in one second. But if this is a passenger train where passengers expect a modicum of comfort, you should probably spread the jerk over a longer period. I would probably assume 30 seconds and 3 km instead of 21 seconds and 2.1 km.

Voluntary acceleration

I started my answer by stating that humans on Mars probably wanted to be subjected regularly to high acceleration. Some humans are probably willing to pay for that, even when they don't have to travel. However, with the acceleration of 9 m/s mentioned above, you only get 21s of acceleration before you reach the maximum speed of 700km/s.

To subject your humans to high acceleration for longer periods of time, I can think of two solutions:

• a circular track, in which case you can have a constant acceleration, oriented in the radius of the circle;
• alternate between high acceleration and high deceleration, in which case you need to rotate the seats of the passengers inside the vehicle, so they are always facing in a comfortable direction.

Answer 5

There's a Tom Scott video https://youtu.be/4ZX9T0kWb4Y about a maglev train in japan.

He doesn't give the acceleration, but the train accelerates from 195km/h at 39.47km from the destination to 503km/h at 31.4km. Thats a speed difference of 308km/h in 8.07km.

If i calculated this correctly that's an acceleration of about 1.03m/s^2.

Tom Scott describes this as significant, but not unpleasant, so that seems a good guideline.

Of course, with some disregard for passenger comfort, or appropriate acceleration couches, you could certainly up this significantly, keeping in mind that martians will be accustomed to only a third of our gravity.

Answer 6

Affect of reduced Martian gravity

Mars has a surface acceleration of about 3.7 m/s/s - or about 38% of that on earth.
The acceleration figures used below and in most or all other answers are based on terran experience. The tolerable figures in Mars much lower gravity may perhaps be lower or even much lower.
Native born martians would never experience 1g in "walking about". If musculature and bone development were adapted to their local environment I'd expect martians to be tall, of thinner bone structure, and perhaps weaker due to typical object "weighing" far less.

Earthies staying for any period, or permanently, might be expected to also be less tolerant of higher g forces. For a good answer on this you need input from biological experts. Scaling figures given here and in other answers down in g by about 2.5 and up in time by 2.5 may be in order.

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Time to speed can be determined form the formula

• T = V / A
  (similarly) V = A x T

km/h = m/s x 3.6
So

• Seconds = km/h /(m/s x 3.6)

So 700 km/h at 1 m/s/s = 700/(1 x 3.6)= 200 seconds.

Distance = 0.5 x acceleration x time_squared.
Here metres = 0.5 x 1 m/s/s x (200 seconds squared)
Distance to 700 km/h = 20 kilometres.

T is time in seconds
V is velocity in m/s
A is acceleration in m/s/s (metre per second per second)

km/h = m/s x 3.6
Terran acceleration due to gravity is about 10 m/s/s (closer to 9.8, but 10 is a useful figure)

If the above acceleration is scaled down to 0.4 m/s/s then time to speed is about 500 seconds, and distance to speed is about 125 kilometres ! Much further due to distance rising with time-squared.

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This page from the "Tramway & Urban transport site says:

Modern trams have rapid acceleration – 1.3 m/s2 is often used – so to reach 50km/h on a level surface will take 11 seconds. To calculate the tractive effort required to attain this, we multiply the acceleration rate by the vehicle’s mass in tonnes. In our example, this gives 78kN, to which we add the rolling resistance to give 81kN – far greater than the 3kN required at a constant speed.

The Hong Kong MTR trains are the fastest accelerating ones I've met at a claimed 1 m/s/s maximum in normal use. Singapore are similar. I'd say from experience that the deceleration is higher - standing passengers need to brace against deceleration and maybe each other. This is well accepted as "part of the ride" and not seen as at all remarkable or uncomfortable.

Image form here Wikipedia - Mass Rapid Transit (Singapore)

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I have travelled in the Shanghai Maglev / Transrapid on 'quite a few' occasions. It has a posted max speed of 430 km/h - well above the 300 km/h suggested elsewhere here.

Acceleration is "pleasant and noticeable".
Higher acceleration would not be unpleasant and would add to the experience, but would make walking in the aisles, which you can do at any time without restriction, more hazardous. Given the figures suggested by others for actual accelerations I feel that 1 m/s/s would be very comfortable.

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In a system with short distances as well as longer ones, increased interstation acceleration would make a significant difference to trip times.
If passengers are seated accelerations well above 1 m/s/s would not be uncomfortable. Braking while facing in the forward direction would be more noticeable than acceleration which presses you into your seat.

Motor vehicle accelerations of several 10ths of a g are very much "part of the ride". If harnessed a 0.5g forward facing deceleration would be significant but bearable.

My late lamented MR2 sportscar long ago exceeded these levels in 2 and sometimes 3 dimensions. These levels may have been above "acceptable comfort levels" but for a belted in occupant were never painful.

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Desired speeds and resultant power:

Circumference of Terra and Mars are close to 40,000 and 20,000 km respectively. Once there are enough population centres (say Kim Stanley Robinson Mars trilogy scenarios) then settlements will still be very widespread.
Distances of 1000 km would be liable to be common between clusters of population and the ability to support end to end trips of 5000-10,000 km highly desirable - if not always practical.
It's likely that IF rail could be fast and energy economic that rail could be the predominant transport mode. "Air"craft may be viable with eg thermal transfer engines but lack of atmospheric oxidant will make that hard. SO an eg 1000 km/h train could be immensely attractive.

The greatly less dense Martian atmosphere helps decrease power requirements due to atmospheric drag - the predominant energy sink at high speed.
Martian atmospheric density is about 0.02 kg/m^3 (as opposed to 1.2 at earth sea level so power at 1000 km/h (280 m/s) will be about
Power = Cd x 280^3 x 0.5 = 220 kW per square metre of frontal area at Cd of 1.
Assume Say Cd=0.3 and A of 10 m^2 giving power of 660 kW - say 1000 kW actual. This is for a "not too long " train - friction and body drag add with length. See

AERODYNAMICS OF HIGH-SPEED TRAINS - their use of Cd is unfamiliar to me. I'll try to resolve. My figure is useful as used.

How a 1000 km/h train would be propelled is very much "a problem for the student". The low density atmosphere would make aerodynamic propulsion difficult. Wheeled propulsion at 1000 km/h would be 'interesting'. MAGLEV could be attractive but the energy and rail construction issues are also challenging. (10,000 km of MAGLEV track is very far from trivial).

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ENERGY SOURCES:

Kim Stanley Robinson's Mars trilogy Red / Green / Blue Mars provides some good ideas re Mars development.
This story's scenarios will differ, but things can be learned from KSR's tales. He had massive AI controlled machinery available. He operated near the edge of it can be done it will be done. They made synthetic diamond to build space elevators when needed. They ... !!!

Time period for this story is unknown.

Mars has substantial Uranium and Thorium in some areas.
Mars had/(has?) natural reactors that bred U233 and went critical and ...!!!
Importation of U/Th is feasible.
Construction of reactors on planet is highly feasible longer term.
KSR's people used Rickover reactors - a hat tip to US Navy Admiral Rickover. – See ref at end to "Rickover reactor" for lots of interesting real world stuff.

Mars has massive volcanic activity (but no tectonic plates). Potential energy is large - but not much good for transport.

Solar insolation is far lower than earth's - about 100-200 Wh/m^2 peak compared to Terra's 1 kW/m^2 typical. Daily kWh/m^2 is typically 1.5-2.5 which is better than I'd have expected.

An electric train is OK over 100s of km. Over 1000's of km with little parallel infrastructure is harder. May still be viable. Very high voltage allows long distance feeds at low loss.

Long term, nuclear seems attractive - with maybe thermal propulsion despite the thin atmosphere.

Elon and others propose Methane synthesis for rocket fuel.

KSR's people drill very very very deep "Moholes" to tap thermal energy.

Martian wind is high velocity but low energy, mostly.

KSR has vast water reserves in aquifers just waiting to green the planet. One can hope :-). Copious water does seem to exist. But not that much. Probably. –

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EVIDENCE FOR A LARGE, NATURAL, PALEO-NUCLEAR REACTOR ON MARS.

Wikipedia - Ore resources on Mars

Solar Radiation on Mars NASA 1989

Hyman Rickover his reactors may yet be seen on Mars :-)

Rickover Reactor - Google search