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## Powered landing, analytical solution

There is a long standing dispute whether winged landing or vertical rocket powered landing is more efficient and overall better for reusable rockets on earth. 🙂
Powered landing can be looked at in a broad overview. The post documents what I doodled in my notebook some day in the summer.

## Summary

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 $m_{prop}=I/v_{ex}$.

## Derivation

Initially, when the craft is descending through the lower atmosphere at terminal velocity, it is at a steady state where gravity and drag balance. A subsonic approximation with a drag number b is used to get: $F_{drag} = m g = b_{drag} v_{ter}^2$ (Eq. 1)

When the rocket engine is ignited, the vehicle starts decelerating: $m a(t) = F_{rocket}(t) + b_{drag} v(t)^2 - m g$ (Eq. 2)

Note that a is real acceleration. The felt acceleration would be one g more since this happens at no significant horizontal velocity or distance from earth.

Note also that since powered landing uses so little propellants, we can consider the mass of the vehicle, m, to be constant.

We can get the drag number b from the terminal velocity and thus get rid of it: $m a (t) = F_{rocket}(t) + \frac{m g}{v_{ter}^2} v(t)^2 - m g$ (Eq. 3)

Now, if we assume the thrust of the rocket to be constant, and use a(t) = v(t)’, we get a not nice equation (arc tans and all) because of the second order of v(t) in the drag term.
But! If we assume a constant deceleration, ie, a(t)=a, by throttling the rocket accordingly, the problem becomes a bit different: $F_{rocket}(t) = m ( a + g - g \frac{v(t)^2}{v_{ter}^2} )$ (Eq. 4)

Now, since we have a constant acceleration: $v(t) = v_{ter}-at$ (Eq. 5)

We combine 4 and 5 to $F_{rocket}(t) = m (a + g - g ( \frac{v_{ter}-at}{v_{ter}} )^2 )$ (Eq. 6) $F_{rocket}(t) = m (a + \frac{2ag}{v_{ter}}t - \frac{a^2g}{v_{ter}^2}t^2)$ (Eq. 7)

If we integrate this with regards to t from 0 to t=v_ter/a, we get the total impulse I to slow down the vehicle to a standstill (or hover, really). $I=\int_{0}^{v_{ter}/a}F_{rocket}(t)dt$ (Eq. 8 )

This yields $I = m v_{ter} (1 + \frac{2g}{3a})$ (Eq. 9)
also known as the formula of Powered Landing Impulse from now on.

Simple, isn’t it?!

The first term, 1, tells about the terminal velocity momentum that always has to be nulled, and the second term 2g/3a is gravity loss with aerodynamic gain taken into account.

What we can see from this is that the landing impulse is directly proportional to the terminal velocity, which was not so obvious beforehand.

We can still calculate further parameters of the landing penalties:

The delta vee can be calculated (again we use the constant mass approximation): $\Delta v=I/m =v_{ter} (1 + \frac{2g}{3a})$ (Eq. 10)
And the propellant mass as well: $I=m_{prop} v_{ex}$ so $m_{prop}=I / v_{ex}$ (Eq. 11)

One last simplification is the Powered Landing Penalty Fraction which is the fraction of delta vee going over the ideal delta vee with infinite acceleration. This is simply: $N_{penalty}=\Delta v - v_{ter} = \frac{2g}{3a})$ (Eq. 12)

## Example

For a 1000 kg vehicle with a terminal velocity of 100 m/s, and real landing acceleration of 2 gees (about 20 m/s^2), the penalty is $\frac{2g}{3a} = \frac{2g}{3 \cdot 2g} = 1/3 \approx 0.3$ powered langing impulse is thus $I = m v_{ter} (1 + N_{penalty}) = 1000 kg \cdot 100 m/s (1+1/3) \approx 1.3E5 Ns$

and $\Delta v = 100 m/s (1+N_{penalty}) \approx 130 m/s$

With an ISP of 250 s, the exhaust velocity would be roughly 2500 m/s. Thus the amount of propellant needed would be: $m_{prop}=I / v_{ex} = \frac{1.3E5 Ns}{2.5E3 m/s} = 50 kg$

The amount of propellant is surprisingly low, 5% of the total mass. Thus we can get away with not using the rocket equation or changing mass. The vehicle feels 3 gees here (because of the one additional earth gravity gee besides the two acceleration gees), so it might be a little hard on a crew.

If the deceleration would have been done at only one gee, then the penalty would have been 0.7, delta vee would have been 170 m/s and the impulse would have been 1.7E5 Ns. The propellant mass would have been 70 kg, or about 7% of the total landed mass.

## Graphs

Included are a few general graphs where you can overview the situation with a quick glance.  ## Further Notes

It can be noted that these formulas are probably not very accurate (the drag portion, the total mass portion just to mention a few), but nevertheless take into account most of the important variables in the powered landing problem, and an elaborate finite step analysis can be avoided when doing preliminary design. (I have done such in Simulink). The low terminal velocity and reasonable acceleration are of paramount importance to minimize mass penalties from a powered landing. Also, the craft will require some propellants for hover, as the braking has to be done to a certain height from the surface and a final low velocity landing must be done from there. Height measurements are not sufficiently accurate and the time responses of engine ignition and throttle are too slow so a direct landing to a pad is probably not realistic. This all will be handled in the second part of the powered landing analysis. 🙂

A similar simple result can probably not be reached for launch, because it is most often thrust limited (full throttle), the mass changes significantly (higher fuel fraction) and drag fights against the vehicle as well (drag helps in landing) and the acceleration is to ever higher speeds. Thus it seems unlikely that there is a simple analytical equation taking into account drag and gravity losses for launch, even though there is one for landing.

### 14 Responses

1. on Wednesday 2007.10.03 at 17:10 | Reply Jonathan Goff

Gravityloss,
Cool post. I think I may have a new addition to my blogroll… It would be interesting to see the results of a more detailed model. We’ve done some low-level BOTE analyses of powered landing, and it’s always interesting to see other approaches to modeling the problem.

~Jon

2. on Thursday 2007.10.04 at 09:34 | Reply gravityloss

Thanks!
I’ll dig up my old simulink model some day and make a comparison.
The drag bit is one thing though that’s probably hard to model, or I at least only have somewhat sketchy knowledge of it. What I think is the first cut, is to assume that the problem is completely subsonic, ie below transonic which is mach 0.6-0.8. That means roughly below 200-240 m/s… There are also some other tradeoffs to mention.

But all that is material for the next post on the subject. 🙂

3. on Thursday 2007.10.04 at 11:20 | Reply Andy Clark

Just found the blog from Hobbyspace:RLV etc.

Nice piece of work. I will peruse it carefully.
Looking forward to more.

Do you have a mathematical editor here or do I have to import the math stuff?

4. on Thursday 2007.10.04 at 11:31 | Reply gravityloss

Thanks.
I’m not sure what you mean, I used the inbuilt latex plugin of wordpress. That changes latex equations into pictures. When writing a blog post, you write starting with a dollar sign and keyword latex and ending in a dollar sign and it compiles the latex code in between and makes a small picture of it.
It’s pretty easy although the parser doesn’t give any useful error messages but just puts a pic with a coloured “can’t compile” placeholder if you make an error in the code.

So if you want to use the equations in any code of yours, you have to copy them by hand, since they are pictures. But they are quite simple so there’s no harm. 🙂

5. on Thursday 2007.10.04 at 17:37 | Reply Doug Jones

You may be able to eliminate the hover if you add a finagle factor to the acceleration of about 1.3:

a = 1.3*2S/V if I recall correctly.

This excess acceleration ramps down linearly during descent and brings the thrust to hover at touchdown. You can add an offset altitude and velocity target to allow for measurement uncertainties.

As for the Latex text, right-click & look at the properties of the generated images- the alt text has the original code, you can just highlight, copy, & paste.

6. on Friday 2007.10.05 at 00:11 | Reply Layne Cook

I stumbled on this blog a little by accident. I hope you don’t mind me adding a few cents worth. I was deeply involved in a sometimes-contentious VL vs. HL debate at McDonnell Douglas during the X-33 proposal back in the mid nineties. It got a little nasty at times, and when we picked a VL for the proposal, Boeing pulled out and went home. We won that battle, but lost the war (to Lockheed).

I like your powered landing impulse and penalty formulae. They offer a quick “absolute minimum” propellant estimate. As you noted, it will be larger after real-world effects are factored in. Another thing you have to budget for is error in the lateral channel. If the lateral guidance and navigation errors can’t be nulled by engine startup time, then that usually means you’re going to have to stay in the air a little longer. For the DC-X, we targeted a point about one body length above the landing target at the end of the high-thrust deceleration, then dropped the last forty or so feet at a constant 3 or 4 feet/sec.

The landing propellant requirement can be an emotional thing too, especially if somebody’s riding in the vehicle! We found our engine startup altitude constantly being raised by the need to allow for contingencies such as engine startup failure (and having to start a backup) or allowing time and space for a crew to punch out if they wanted.

But, in my opinion, VL is still the best way to go for many RLV missions. It always comes down to the complete mission requirements. But its good to see you guys working on it.

7. on Saturday 2007.10.06 at 18:29 | Reply gravityloss

Doug, yeah, cool, that alternate text thing works, didn’t know that. I don’t know what the S in your equation is, but yeah it makes sense to do it that way, by reducing the throttle smoothly to enable just hover in the end. (In my solution the throttle is increased to maintain deceleration, since the drag helps less and less at lower velocities.)
I’ll write about that more later and an alternative.

And Layne, big thanks, it’s so damn cool to have someone from the DC-X project and X-33 competition commenting on the first real post of my blog!
The lateral channel I had not thought about so much, but it makes a lot of sense of course.
One body length.. 🙂 That’s quite bold. But maybe your guidance was so good (F-15 laser gyro, right?)
All good info about the startup altitude raising too, great stuff, I will take this in the input for my next post on the subject.

8. […] 7th, 2007 by gravityloss There was some excellent commentary in the last post, and that has partly been one inspiration in writing this […]

9. […] very first post in the whole blog derived the analytical solution for delta vee in a vertical powered landing that […]

10. on Friday 2009.04.10 at 03:35 | Reply Matt Jessick

>>Layne Cook
I stumbled on this blog a little by accident. I hope you don’t mind me adding a few cents worth. I was deeply involved in a sometimes-contentious VL vs. HL debate at McDonnell Douglas during the X-33 proposal back in the mid nineties. It got a little nasty at times, and when we picked a VL for the proposal, Boeing pulled out and went home. We won that battle, but lost the war (to Lockheed).
————–

I worked on the proposal from the Boeing side, working actuator req. I recall sitting beside the controls guys

11. on Friday 2009.04.10 at 03:42 | Reply Matt Jessick

(sorry, typing too fast with emacs fingers)
… at Boeing who were considering an optional aero controlled way to flip the vehicle around rapidly and catching it so that the back end could be pointed down for landing. Another group somewhere was concerned with analyzing whether it was reasonable for the engine to remain firing during this sort of gyration. Sadly, I don’t recall the conclusion for that concern…

12. on Thursday 2010.09.16 at 06:14 | Reply Igor Rozenberg

G’day gravityloss,

Thank you for an elegant solution – I used it to estimate solid propellants for new Russian manned spacecraft (TPK NP).
Would you be so kind to reference a NASA report you used for analytical solution?
Regards from DownUnder.

13. on Thursday 2010.09.16 at 23:54 | Reply gravityloss

NASA report? I think it goes further back than that, maybe Leibnitz or Newton. 😉 It’s simple enough that it was derived from first principles.

14. on Thursday 2010.09.16 at 23:56 | Reply gravityloss

Marti Sarigul-Klijn also had derived something like this earlier, I ran into it in one of his papers later after I had written this post.