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## Drag: Loss in Ascent, Gain in Descent, and What It Means for Scalability

Jon Goff had an analysis (no equations or numbers though!) of an air-launched reusable SSTO idea, furthermore he promised to look at other reusable orbital launch vehicle ideas in the future. I thought I’d drop a few lines in regards to almost all architectures.

The summary is that big rockets have little drag losses going up but coming down they come too fast and need some tricks.

## Drag in Ascent

The total negative impulse of drag during ascent is to a big order well approximated to be proportional to the frontal area of a rocket.

$I_{drag} = \int C_d A \rho_{air}(t)v(t)^2dt \propto A$

Drag loss as delta vee is a more ideal, but even more useful variable. It is simply the integral of drag force per mass.

$\Delta v_{drag} =\int \frac{C_d A \rho_{air}(t)v(t)^2}{m(t)} dt$

Of course, mass at any point is directly proportional to takeoff mass.

$m(t) \propto m_{glow}$

And thus

$\Delta v_{drag} \propto \frac{A}{m_{glow}} = \frac{A}{A \cdot l \cdot \rho_{rocket}} = \frac{1}{l \cdot \rho_{rocket}}$

So the profound result is that drag loss as delta vee is directly proportional to frontal area per liftoff mass, or in the case of a constant shape rocket, inversely proportional to rocket length l and rocket density rho.

It is NOT proportional to frontal area alone. This means that if you have a rocket that is, say, 50 m tall at takeoff, and you make an enlarged version and double it’s diameter while not changing the density, the drag loss delta vee stays the same. Drag loss grows in Joules or Newton seconds of course, but the rocket’s energy and total impulse grow as much at the same time, because the mass grows – both frontal area and thus mass are quadrupled.

What this of course means is that shorter rockets will always suffer from worse drag losses. Increasing density and decreasing the drag coefficient might help a little but probably not that much. Of course, in real life the acceleration profile and trajectory mean a lot too. Here is some data gleaned from this thread at nasaspaceflight.com and from “Space Propulsion Analysis and Design” by Ronald Humble.

• Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s
• Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s
• Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s
• Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s
• Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!)
• Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s

The Saturn V being by far the tallest has a much lower drag loss than the rest. Though it also had the slowest acceleration in the low atmosphere.

## Drag in Descent

The very first post in the whole blog derived the analytical solution for delta vee in a vertical powered landing that takes into account the drag force. The total landing delta vee is:

$\Delta v_{land} = v_{ter}(1+\frac{2g}{3a})$

This means the system is sensitive to terminal velocity. And terminal velocity is determined by terminal mass, drag coefficient and frontal area:

$m_{ter} g = C_d A \rho_{air} v_{ter}^2$

and

$v_{ter} = \sqrt{(m_{ter} g) / (C_d A \rho_{air} ) }$

and

$v_{ter} = \sqrt{ (l \rho_{ter} g) / (C_d \rho_{air} ) }$

Meaning that the terminal velocity is inversely proportional to the square root of the length of the rocket l and the density of the almost empty rocket, rho_ter. This means that if density is held constant (a reasonable first assumption), as well as shape (similar C_d), the terminal velocity is doubled when the rocket size (length) is quadrupled. And this means more delta vee needed for landing. It doesn’t matter if the rocket is lengthened and widened or only lengthened. If the constant density is held, the increased length (or depth) just means more mass per frontal area.

So, rocket size presents a scalability problem for the powered landing. Bigger rockets have bigger terminal velocities and need more landing deceleration propellants.

## Scalability

What does this mean? Well, a rocket should be tall and dense for ascent to minimize drag losses. And it should be short and fluffy to minimize powered landing requirements. Slenderness or stubbiness is not necessary, but absolute tallness is what counts here, if we assume a constant cross section.

What can be done to help? Well, obviously, the rocket can have small frontal area in ascent but deploy a speed brake when descending to lower terminal velocity. From looking at TGV rockets’ materials, they already have thought of this. But it seems to me they are the first to actually do it. It’s weird that for example Armadillo or Masten hasn’t talked about speed brakes before. They could be very simple structures. The first idea that pops to mind self-evidently is a system that has brake plates on the surface of the rocket that are hinged at the top. During ascent the airflow pushes the brake to stay closed, but during tail-first descent the airflow raises the brake plate up until it hits a builtin stop at 90 degrees. Then it’s perpendicular to the airflow and exerts a maximum drag force. Quadrupling frontal area leads to halving of terminal velocity and thus halving of landing propellant needs.

Let us assume a 1000 kg empty vehicle and braking deceleration of 20 m/s^2 or 2 gees. Penalty is 0.3 and thus total delta vee 1.3 v_ter. Furthermore we can assume an ISP of 255 s or v_ex 2500 m/s.

Entering at 300 m/s (transonic at sea level), the needed braking propellant mass is about 1000 kg * 390 m/s / 2500 m/s = 156 kg. If we use a drag device to quadruple the area (double frontal diameter) the terminal velocity drops to half and propellant needed to 78 kg. Thus we can estimate that the drag device can be mass-useful if it weighs less than 78 kg. On one hand, it introduces additional failure modes, but on the other, it can be deployed pretty early and reduces timing and sensor criticality a lot by decreasing the terminal velocity.

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