Extremely simple would be to use a linear acceleration with a cap:

thrust = distance*constant

if thrust>thrustmax
thrust = thrustmax

This might work, but it depends on your other controls. Does no thrust slow down the ship? (friction) If not, you''ll need negative thrust to get to that position.

Have the ship start breaking in when within a certain radius of the goal maybe? And once it''s slow enough, just turn the thrusters off. And maybe zero the velocity(if it''s slow enough the user won''t notice, and the ship won''t drift away).

Of course the radius is depending on the velocity. Something like a linear of quadratic function of velocity. Fiddle around.

Well, that shouldn''t be too hard. Just thinking aloud here:

position(t+dt) = position(t) + dt*velocity(t);
velocity(t+dt) = velocity(t) + dt*acceleration(t);

Since your thrus is the acceleration is constant, and you want to arrive with velocity essentially zero you should start braking:

velocity(t_end) = 0;
0 = velocity(t_begin) + dt*thrust;
dt = -velocity(t_begin)/thrust;

Now you know how much before you get to the target you should reverse the thrust: dt

But how far away is that?

position(t_end) = position(t_begin) + dt*velocity(t_begin);

position(t_end) is known, zo is velocity(t_begin) and dt, so you can calculate position(t_begin). That''s when you should reverse the thrust. Please note that since I''ve only calculated this formula once here (hence dt could be in the range of seconds) this will not be accurate. It''s only a linear function, and does not take into account that the velocity will become lower after you start braking. You should be able to do that with an integration.

Good luck!

with constant acceleration you have (x0...initial position, v0...initial velocity)

x = x0 + v0*t + a/2 * t^2

if you''re braking

x = x0 + v0*t - a/2 * t^2

you want to come to a stop at x1 by starting to brake at x0:

x1 = x0 + v0*t - a/2 * t^2

and the velocity should be zero there:

v1 = v0 - a * t

t = (v0 - v1)/a

put that into the above equation for x1
and you''ll find the position where you have to start braking

if you want to accelerate first and the to break the first thing that comes to my mind isget to equations for the position in accelerating and in decelerating mode and look where they are intersecting.


your current position is x2, so by acceleration you get

x = x2 + v0*t + a/2 * t^2

when decelerating towards x1 you want v = 0 at x1 (at time t_end)

v = a * (t_end - t)

by integration you get

x = a * t_end * t - a/2 * t^2 + c

with c the integration constant; x should be x1 for t = t_end

x1 = a * t_end^2 - a/2 * t_end^2 + c = a/2 * t_end^2 + c

c = x1 - a/2 * t_end^2

so for the position at a specific time

x = a * t_end * t - a/2 * t^2 + x1 - a/2 * t_end^2

now you set the two x positions equal

x2 + v0*t + a/2 * t^2 = a * t_end * t - a/2 * t^2 + x1 - a/2 * t_end^2

t^2 * a + t * (v0 - a*t_end) + x2 - x1 + a/2 * t_end^2 = 0

we solve for t

t = (a*t_end - v0 ± sqrt(a^2*t_end^2 - 2*a*t_end*v0 + v0^2 - 4 * a * (x2 - x1) - 2 * a^2 * t_end^2))/(2*a)

now this t obviously depends on t_end which we do not yet now. let''s say we want only a single solution so the expression under the square root should be zero

t_end^2*(-a^2) - t_end * 2*a*v0 + v0^2 - 4*a*(x2 - x1) = 0

t_end^2 + t_end * 2*v0/a - v0^2/a^2 + 4*(x2 - x1)/a = 0

let''s solve for t_end

t_end = -v0/a ± sqrt(v0^2/a^2 + v0^2/a^2 + 4*(x1-x2)/a)

that should be the time we need to get there...

the time when we have to switch from accelerating to decelerating is

t1 = (a*t_end - v0)/(2*a)

and the position

x1 = x2 + v0*t1 + a/2 * t1^2

let''s check the velocities at x1:

from the acceleration

v1 = v0 + a * t1 = v0 + a * (a*t_end - v0)/(2*a)
= v0/2 + a*t_end/2

from deceleration we have

v1 = a * (t_end - t1) = a*t_end - a*(a*t_end - v0)/(2*a)

= a*t_end/2 + v0/2

so we''ve got the same velocities at time t1 - this will work!
(there would be different velocities if we wouldn''T have chosen the t_end in a way that there''s only one solution for the switching point - it''s best imagine this in a graph with x vertical and t horizontal with the acceleration curve (a parabola with a minimum at x2) from x2 < x0 and the deceleration curve (a parabola with maximum at x0). the parabolas are fixed in size and the second one can be shifted along the t axis (this means varying the t_end value). they''ve to match exactly in one point, if you shift it too far to the right (positive t) there''s no intersection and therefore no real solution for the equation, if you shift it too far to the left there are 2 solutions and you''d have to change velocity instantanously)