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## Powered landing, practical considerations (1-dim)

There was some excellent commentary in the last post, and that has partly been one inspiration in writing this post.

In the last part, the absolute minimum efforts by the launch vehicle making a vertical landing were determined. On top of these, there are lots of slightly more practical ones. This is still a one-dimensional vertical analysis.

## Effects of a Time Delay

Let us see what a delay of t_delay before starting the main deceleration would do to a landing vehicle. The craft would travel $t_{delay} v_{ter}=y_m$ extra distance before starting deceleration. That would mean that the actual reaching of zero speed would be y_m behind the original deceleration target. If we assume the original target was on ground level, then the new target would be underground. The difference between the targets is $y_m=0.5 a t_m^2$. (We used the trick of counting backwards in time from the underground target.) We can solve the imaginary underground time duration, $t_m=sqrt(2 y_m/a)$. And the speed at ground level thus is $v_m = t_m a = sqrt(2 y_m a) = sqrt (2 a v_{ter} t_{delay})$

### Example Vehicle

Our example vehicle, with v_ter=100 m/s and a=20 m/s^2 would stop in 5 seconds and 250 meters normally. If it traveled without any margins and would miss the engine ignition by one second (t_d=1 s) and 100 meters (y_m=100 m), it would impact the ground at over 60 m/s (130 mph, 220 km/h), or roughly two thirds of its original terminal velocity of 100 m/s. That would destroy it completely.

If we used a gentler and more lossy 10 m/s^2 deceleration (and started the engines higher accordingly, but still had the unfortunate one second delay), then the impact speed would be about 40 m/s. Still far too much.

Even a realistic 0.1 s of delay would cause a 20 to 14 m/s impact, depending on planned acceleration. Complete crash, remotely possibly survivable.

The results might look unintuitive. That’s because in a constant deceleration, the most time is spent on the end, low speed distances, while most distance is covered with high speeds, early on. So if the deceleration is cut short even by a short percentage in distance, the time is cut short by a much bigger percentage, so the vehicle does not have enogugh time to brake much. Running out of a little distance you run out of a lot of time. This in total makes the vertical landing a very time sensitive problem!

It is also the reason why in the classical driving school example, even if you lock-brake and the car overshoots the pedestrian crossing by just a few meters, it was still going quite fast when you crossed it. The car’s speed doesn’t change much per meter when you start braking, but in the end half it changes a lot because it’s going slower and there are many more seconds per meter and thus the force has more time to effect. The drag is not dependent on velocity so in a sense it functions like a rocket engine. Accelerating with the car’s engine is the opposite as it generates constant power, not constant acceleration.

So, if we anyway design the vehicle to survive some vertical touchdown speed v_touch , then we can calculate the margin for the fast deceleration directly to pad in seconds. $t_d = v_{touch} / sqrt (2 a v_{ter})$. With our example vehicle it could be 4 m/s, equal to a drop from 0.8 meters height. Then the acceptable time delay would be 0.063 s or 63 milliseconds. That’s probably too small of a margin.

## More Realistic Approaches

So, in reality, one can not have a direct high deceleration to pad. One could have multiple other approaches and ways to think about it. These are the most obvious to me:

1. Put the the deceleration aim target (hover) some distance above the ground. Layne Cook told that was an approach the DC-X used.
2. Stop the high deceleration at some determined velocity and switch to a low deceleration. For example at 10 m/s.
3. Use some more elaborate forecast of the vehicle position and throttle accordingly to always have a plan to reach very slow speed at touchdown. I think Armadillo is using something like this approach.

These are handled below.

### (more…)

Deceleration from terminal velocity to hover usually requires about 5-10% of total landed mass as propellants. The amount is directly proportional to the vehicle’s terminal velocity. Also, the higher acceleration, the better. The powered landing penalty fraction is $N_{penalty}=\frac{2g}{3a}$, so landing at 2 gee acceleration (3 gee felt) gives a penalty of 0.3. The required delta vee is $\Delta v = v_{ter} (1+N_{penalty} )$ where v_ter is terminal velocity. And the required impulse $I=m \Delta v$ and propellant mass $m_{prop}=I/v_{ex}$.