Author Topic: Falcon 9-R gravity drag into Tsiokolvsky equation?  (Read 15307 times)

Offline pagheca

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Hi,

this is my first post here, so please accept my apologies if there is something not correct...

I was doing some delta-v back-of-the-envelope evaluations for the Falcon 9-R rocket 1st stage. In various sites I found that the delta-v due to gravity drag is evaluated as ~1200 m/s for rockets of this class.

1. However, the 1st stage separation seems to happen at 100 Km of altitude that is about 1/2 orbit altitude. Therefore, I think only half of that value should be taken in account. Is that correct?

On the other side, I was thinking that, as most of the trajectory is probably vertical till that altitude, there should be a simple way to evaluate the gravity drag by including it into the Tsiokolvsky equation.

As well known the equation can be obtained by integrating the differential equation:
[I assume anyone trying to answer this is familiar with the symbols used here]

SUM(all forces acting on the rocket) = m dV / dt + v_exh * dm / dt           [eq. 1]

As SUM(...) = 0 in vacuum:

dV = - v_exh * dm / m

Integrating this, we obtain the simplified rocket equation:

ln (m1/m0) = - DV / v_exh.

or

m1 = m0 * exp( - DV / v_exh)

Now, if the rocket is ascending vertically and we do not consider the atmospheric drag, the only force to be added on the left of [eq. 1] is

F = mg

(g is ~ constant for the first hundreds Km)
Now, after integrating, one obtains:

m1 = m0 * exp ( (- g * Dt - DV) / v_exh ).

note that if Dt is small (aka very high thrust) or g = 0, the result is the usual Tsiokolvsky eq. Dt is also h / <v>, where <v> is the average speed.

2. Now, for a 3 mins ascent before stage separation, for example, 3 * 60 sec * 9.8 m/s^2 ~ 1750 m/s. This looks too much. Am I wrong? Including the gravity turn should add to this result.

3. I also understand that I should add the potential energy aka sqrt(2 * g * h) to this. Correct?
« Last Edit: 10/23/2013 12:49 am by pagheca »

Offline QuantumG

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Re: How to include gravity drag into Tsiokolvsky equation?
« Reply #1 on: 10/22/2013 10:41 pm »
Welcome to the forum!

The question of how much delta-v each stage imparts to achieve orbit does not have a fixed answer. It depends on how dissimilar the stages are and there's other trade-offs.

For Falcon-9, I think the delta-v for each stage is probably similar.. around 4500 m/s.

For Falcon-9R, I imagine it's still similar but it has a higher delta-v requirement.. around 4800 m/s?

We have some fantastic trajectory simulation people on the forum, they can probably give you a better estimate.
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Offline deltaV

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Re: How to include gravity drag into Tsiokolvsky equation?
« Reply #2 on: 10/22/2013 11:14 pm »
1. Conservation of energy is obnoxious to apply in this case. Just solve the differential equation as you did. Do not randomly add sqrt(2 g h).

2. After the gravity turn gravity is acting perpendicular to the desired motion rather than opposite it. This makes it a lot easier to fight; just aim the thrust a little below horizontal. Because of the behavior of sin and cos near zero, and the many gees of thrust available at this point of the first stage trajectory, you can get an adequate amount of gravity fighting (the sin) with only a minor reduction in the forward thrust (the cos).

3. You won't get a precise figure for the gravity drag without doing a simulation of the launch. If you don't want to do a simulation you can instead use the "Schillings" tool to estimate payload of a launch vehicle to an arbitrary orbit: http://www.silverbirdastronautics.com/LVperform.html . It uses historical data to estimate the sum of the various losses (drag, steering, gravity) for a given launch vehicle.

Offline pagheca

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Re: How to include gravity drag into Tsiokolvsky equation?
« Reply #3 on: 10/22/2013 11:49 pm »
Thanks a lot for your quick comments deltaV and QuantumG!

I forgot to state clearly that I'm interested to the 1st stage only delta-v, not to the overall earth-to-orbit transfer...

Now, a few further comments:

1) I think that for the Falcon 9-R the Delta-v is smaller rather than higher. This because while the 9 version is staging at Mach 10, the 9-R separates at Mach 6, that is a big difference in speed. I guess larger than the additional weight of the landing gear etc. that someone evaluate at 2.5 - 4.5 metric tons (I guess closer to the first).

2) as SpaceX is planning to do RTLS mission soon or later, I assume that the gravity turn can't happen too early otherwise the 1st stage would requires a big additional delta-v just to return back to base. By going straight as much as he can and turn later, the manoeuvre should save some fuel.

3) I was aware of the Shilling tool. The problem is that it doesn't allow me as far as I understand, to make the calculation for the 1st stage only. Moreover, I want to understand why the value calculated is so high.

4) I found that someone computed the gravity drag due to the change in potential energy as sqrt(g*h), forgetting the 1.4 factor. Any idea why? Just rounding? http://bit.ly/17eHaMw for an example...

5) I couldn't find the altitude for the staging. Some website mention 100 Km, but it is not clear if this refers to the Mach 10 or the Mach 6 separation. Any timeline I found for the recent Falcon 9 v1.1 launch doesn't mention the altitude, only the time T+.

Any additional help or comment welcome.

Offline QuantumG

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Re: How to include gravity drag into Tsiokolvsky equation?
« Reply #4 on: 10/22/2013 11:54 pm »
There's some numbers in this that you might like to use:


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Offline Chris Bergin

Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #5 on: 10/22/2013 11:57 pm »
Welcome to the site's forum. Not sure why so many people are reporting this, it's clearly about F9-R.
« Last Edit: 10/23/2013 12:16 am by Chris Bergin »
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Offline pagheca

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #6 on: 10/23/2013 12:45 am »
The movie confirms staging on the 1st flight of v1.1 happened at Mach 6 at an altitude of ~100 Km.... Thx!
 

Offline pagheca

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #7 on: 10/23/2013 01:41 am »
Sorry - one more:

According to this comment these are some typical drags:

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

Taken straight from "Space Propulsion Analysis and Design". The author of the comment also states that "gravity losses are based upon the amount of time thrusting directly against gravity rather than orthogonal to it, drag is based on angle of attack". Does it means that if I escape a celestial body following a spiral I can virtually reduce to zero this component? So, if I could use sort of "mass driver" propulsion on an atmosphere-free body like the Moon I can spare the gravity loss at once? Or I should assume that, whatever I do, this affect only the istantaneous delta-v, not the total amount that remain the same and is proportional to g*Dt (see above)?

Thanks again to anyone can help me to better understand this topic!
« Last Edit: 10/23/2013 01:43 am by pagheca »

Offline Jcc

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #8 on: 10/23/2013 01:57 am »
Pegasus reduces gravity loss by air launch and winglets that provide some lift in the air. This method will scale out to probably the maximum with Stratolaunch. [ edit: maybe Skylon? ]

I think bigger rockets avoid  a lot of friction losses by exiting the atmosphere as soon as possible, and if they didn't, friction increases as the square of v, it would get very large quickly, causing excessive heating, etc.
« Last Edit: 10/23/2013 02:01 am by Jcc »

Offline LouScheffer

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #9 on: 10/23/2013 02:17 am »

 Does it means that if I escape a celestial body following a spiral I can virtually reduce to zero this component?

You can't eliminate it, but you can much reduce it.   Apollo did exactly this on launching from the moon, switching to a nearly horizontal trajectory almost immediately after takeoff.  The astronauts commented that it seemed quite weird, tilted way over and picking up speed without much increase in altitude.

Say your thrust to weight at takeoff is 1.4.  Then you start by thrusting at a 45 degree angle.  Your vertical acceleration is 1.41/sqrt(2)-g, so vertically you are hovering, as the horizontal component of acceleration (also thrust/sqrt(2) for this case) builds up speed.  As you burn off fuel and the T/W becomes larger, instead of increasing altitude you tilt over more, keeping just above the surface at all times.  Eventually you reach orbital speed at 0 altitude.   Gravity losses are not 0, but are much better than going straight up at first, since the majority of the thrust at all times is going to increasing your orbital velocity.  Also your mgh energy is 0 since you never gain altitude.
« Last Edit: 10/23/2013 11:16 am by LouScheffer »

Offline kkattula

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #10 on: 10/23/2013 09:58 am »
Note that for large vehicles the ratio of surface area to mass is much smaller, so the drag forces are small compared to the thrust of the engines. This is part of the reason for the small Saturn V drag losses. The other reason being the low T/W at liftoff which delayed higher speeds to higher altitudes at the cost of higher gravity losses.

This also allows larger vehicles to be 'fatter'. Of course SpaceX went to an extended tank for F9 1.1 first stage, rather than a wider tank, because it's extremely expensive to change to manufacturing a wider diameter, and relatively inexpensive to just make longer barrel sections.

However, consider how much time & effort goes into making sure launch vehicles survive Max Q. I have often wondered whether it wouldn't be worth designing a first stage that 'eats' a few hundred m/s of extra gravity losses, in order to stay sub-sonic until at higher altitude (say 15 to 20 km). Especially for a reusable first stage.


Offline Jcc

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #11 on: 10/23/2013 10:43 am »
In case I wasn't clear, the air launch vehicles turn a liability (atmosphere) into an asset, which allows them to spend more time in the atmosphere without penalty. The reason why a medium or heavy lift vehicle can't do that is because the carrier aircraft would need to be impossibly large.

True a bigger rocket has proportionally less surface area per volume, but increase in friction as the square of velocity dominates at high velocity. I am pretty sure the launch profiles for the vehicles lifted above are close to optimal despite the fact that friction losses seem small compared to gravity. Total losses are minimized.

Offline LouScheffer

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #12 on: 10/23/2013 11:27 am »

So, if I could use sort of "mass driver" propulsion on an atmosphere-free body like the Moon I can spare the gravity loss at once? Or I should assume that, whatever I do, this affect only the istantaneous delta-v, not the total amount that remain the same and is proportional to g*Dt (see above)?

Yes, a mass driver on the moon would suffer no gravity losses.  You calculate the desired speed for an transfer orbit with the surface at perigee and the final desired orbit at apogee, subtract the the rotational speed of the body, and that's exactly what your mass driver needs to produce.  In particular, for a zero-altitude, surface-skimming orbit on a perfectly spherical, air-less, non-rotating body, the delta-V you need is exactly the orbital speed.

Offline pagheca

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #13 on: 10/23/2013 01:07 pm »
Quote
Yes, a mass driver on the moon would suffer no gravity losses.

I'm not completely convinced...

The point is that at each orbital velocity corresponds an orbital radius. It is completely clear to me that if one reaches the ground orbital speed he can orbit the moon at zero altitude (forget the mountains of course). The problem is: what happens if you increase again the speed to gain some altitude and transfer the vehicle to an higher orbit? Should you include only the sqrt(2*g*h) factor or also a (small, almost negligible) gravity drag factor depending on the altitude change? How to compute it for a general burst?

When I say that I'm not completely conviced, I'm not saying you are wrong. My suspect is that the problem is the vertical component of the velocity during the ascent that you have to include anyway to change orbit.

When you are vertically ascending you ARE on an orbit (everyone realise soon or later that even while someone jump from the ground he is - for a fraction of a second - orbiting the Earth in free fall and you around the apogee of the orbit :) ). The problem is that that orbit is intersecting the Earth. So, you have to someway rotate and add some tangential velocity component. THAT rotation can't be done in a perfectly efficient way: some energy will be thrown away. Is THAT energy the one that must be added to the delta-v.

So, because at some point, even with a mass driver on the Moon, you will need to change orbit, that slightly small vertical component will be thrown away. But the smaller it is, the more time you will need to gain altitude, and so there will be no gain at the end of the day.

Am I correct? I would like to see this mathematically but, unlike for the mgh factor or the delta-v one, I'm unable to write a general equation in vacuum able to account for that.

And thanks again for any comment past and future! I'm not a rocket scientist as you understood. Just someone trying to figure out a second order math behind the simplified Tsiolkovsky equation...

« Last Edit: 10/23/2013 01:15 pm by pagheca »

Offline go4mars

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #14 on: 10/23/2013 01:39 pm »
Might have meant to type drag losses.
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Offline pagheca

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #15 on: 10/23/2013 02:15 pm »
I just found a former discussion (2009) following this very same issue.

This paper has just most of what I was looking for.

However, it is not obvious that the equations there will also work in the case of reusable first stage. They may require some adjustments.
« Last Edit: 10/23/2013 03:06 pm by pagheca »

Offline cambrianera

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #16 on: 10/24/2013 12:15 pm »
2. Now, for a 3 mins ascent before stage separation, for example, 3 * 60 sec * 9.8 m/s^2 ~ 1750 m/s. This looks too much. Am I wrong? Including the gravity turn should add to this result.

It is too much because the formula:
m1 = m0 * exp ( (- g * Dt - DV) / v_exh ).
is valid only for a pure vertical path.
Leaving vertical path and building up transverse velocity makes gravity losses smaller.
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Offline pagheca

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Re: Falcon 9-R gravity drag into Tsiokolvsky equation?
« Reply #17 on: 10/24/2013 01:04 pm »
Thanks. Actually I came to the same conclusion. This is a sort of "upper limit" for the gravity loss. Any deviation from verticality tends to reduce the gravity loss, that seem to be proportional to the product between the Dv and the gravity vectors.

If you or any other reader can kind address any of the questions I disseminated in this discussion, your answers would be very welcome.

pagheca

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