Calculate point of rotation start for spacecraft with linear acceleration

Question

I hope someone can help me with my problem.

I would like to move a spaceship from A to B. I want the spaceship to accelerate to the middle of the path, then rotate 180° and then decelerate.

Now the point where the spaceship finishes the acceleration can't be exactly the middle of the way, because the rotating also needs time (in my case a fixed value). The point must therefore be shortly before the middle.

How can I calculate this?

My attempt so far:

Calculate v for s/2. Subtract a * (time for rotate/2) from v. Now calculate with v the point s.

The result does not look wrong at first sight, but the spaceship does not reach exactly the target point during the deceleration, but always stops shortly before it.

I hope you can understand the description of my problem.

I am very happy about ideas. Thank you.

Answer 1

The total distance $s_0$ (which you know is 1000 m according to your plot) is: $$ s_0=s_{acc} + s_{rotate} + s_{dec} $$ with $s_{acc}$ beig the distance from A to the blue point and $s_{dec}$ the distance from red point to B and $s_{rotate}$ the distance needed for rotating the spacecraft. Because of the symmetry we know that $s_{acc}=s_{dec}$, meaning: $$ s_0= 2 \cdot s_{acc} + s_{rotate} $$

Furthermore, we know that $$s_{acc}= \frac{1}{2}\cdot a \cdot t_{acc}^2$$ and that the end speed (between blue and red point) is $$v_{end} = a \cdot t_{acc}$$ With $v_{end}$ the distance between blue and red point is: $$ s_{rotate} = v_{end} \cdot t_{rotate}=a \cdot t_{acc} \cdot t_{rotate} $$ Now we get one equation $$ s_0 = a \cdot t_{acc}^2 + a \cdot t_{acc} \cdot t_{rotate} $$ with only one unknown ($t_{acc}$). Solve this for $t_{acc}$ and use it to calculate $v_{end}$ as well as $s_{acc}$ and $s_{rotate}$.

Answer 2

Suppose the total distance is D, the rotation time is T, and the acceleration is a, and the acceleration time is t.

$D = \frac{1}{2}at^{2} + atT + \frac{1}{2}at^{2}$

This is because the deceleration phase is a mirror image of the acceleration phase, which reaches velocity $at$ to move a distance $atT$ during rotation.

Simplifying,

$D/a = t^{2}+tT$.

This is a quadratic you can solve for $t$, the acceleration time.

$-T/2 \pm \sqrt{(T/2)^{2}+D/a}$

Let's take the positive root, $t = \sqrt{(T/2)^{2}+D/a}-T/2$.

The associated distance is $\frac{1}{2}at^{2}$