Author Topic: Solar thermal beats all for space flight  (Read 31358 times)

Offline Archibald

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Re: Solar thermal beats all for space flight
« Reply #20 on: 06/23/2011 06:18 pm »
As I see it, STP might be a competitor to SEP if, someday, there are COTS-like missions to a L1 or L2 Gateway.
I can see a private company selecting STP for such mission...
Han shot first and Gwynne Shotwell !

Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #21 on: 06/23/2011 06:53 pm »
I wouldn't call it a "loss" at all.  It's a diminished gain from the Oberth effect.  If you put 3km/s worth of impulse, you still get at least 3km/s worth of delta-v from it.  It's just that you don't get the big Oberth effect boost you would have gotten if you had applied all of the impulse near perigee.

I'd call it a loss.  If in LEO I do a 3-km/s instantaneous burn, then I get 3 km/s of delta-V, just like the rocket equation says.  If I stretch the burn out over a significant amount of time, I'm going to need a greater mass ratio to get to the same final state.

Offline Andrew_W

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Re: Solar thermal beats all for space flight
« Reply #22 on: 06/23/2011 07:41 pm »
I don't think failure to utilize the Oberth effect can be called a loss, if you use it you get a gain in delta V above what you get from the rocket equation, if you don't use it you still get all the delta v of the equation.

IssacKuo was right in his first comment that using a combination, a chemical rocket at perigee to capitalize on the Oberth effect, and then using the STP to accelerate further on a voyage to Mars could give you 15km/s rather than the mere 12 I was getting with just STP.

I don't think the idea of storing energy in the form of heat for STP would be practical, unless the heat sink had other essential functions, the usable energy stored would be small compared to that in rocket propellants.

Proponent, I wouldn't know what to do with your program, but thanks for the offer.
« Last Edit: 06/23/2011 08:07 pm by Andrew_W »
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
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Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #23 on: 06/24/2011 04:52 am »
I don't think failure to utilize the Oberth effect can be called a loss, if you use it you get a gain in delta V above what you get from the rocket equation, if you don't use it you still get all the delta v of the equation.

Actually, now that I think about it, it's obvious that there is a perfectly standard definition of gravity loss.  It's simply the integral over the powered portion of flight with respect to time of the local acceleration of gravity times the sine of the angle between the flight path and the local verticalhorizontal.  This loss is an amount which needs to be subtracted (along with drag and steering losses) from the ideal delta-V computed according to the rocket equation to give the actual delta-V.  By this standard definition, a long escape burn at low-thrust deep in the Earth's gravity leads to a large loss.  Now it's true that some of that gravity loss is really lost, because by raising the altitude of the escaping spacecraft, the long burn has reduce the velocity needed to escape.  But it's still less efficient than an impulsive burn.
« Last Edit: 06/24/2011 06:47 am by Proponent »

Offline Andrew_W

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Re: Solar thermal beats all for space flight
« Reply #24 on: 06/24/2011 05:06 am »
When you're in orbit there is no local vertical, all of the delta v gained goes into increasing the orbital velocity and energy of the spacecraft, it's not wasted in the fighting of gravity in the way a launch vehicle suffers gravitational loss before achieving orbital velocity.
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
Wilbur Wright

Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #25 on: 06/24/2011 05:12 am »
There's no magical change in flight dynamics as a vehicle achieves orbital velocity:  there is still a local vertical, defined by the direction of the local gravitational field, and doing work against gravity is still a loss.  The magnitude of the loss in orbit is generally smaller than near the ground, because the trajectory is further from the local vertical, but it's still there.

Offline Andrew_W

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Re: Solar thermal beats all for space flight
« Reply #26 on: 06/24/2011 05:17 am »
So as long as a vehicle in orbit applies thrust perpendicular to the local gravitational field there is no gravity loss.
« Last Edit: 06/24/2011 05:18 am by Andrew_W »
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
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Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #27 on: 06/24/2011 05:24 am »
Yes, that's true, but then there will be steering losses, which arise anytime the thrust vector doesn't coincide with the flight path.

Offline kkattula

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Re: Solar thermal beats all for space flight
« Reply #28 on: 06/24/2011 05:47 am »
So as long as a vehicle in orbit applies thrust perpendicular to the local gravitational field there is no gravity loss.

AiUI, what's referred to here as gravity loss is due to the burn not being instantaneous. i.e. most of the burn is done progressively further away from the ideal point. The lower the thrust, the worse the effect.


Page 19 has a handy graph of ignition T/W vs TLI  gravity losses :

http://esamultimedia.esa.int/docs/exploration/ReferenceArchitecture/Final%20ReviewJan09/04_Human_moon_mission_version9esa120109.pdf


Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #29 on: 06/24/2011 06:03 am »
AiUI, what's referred to here as gravity loss is due to the burn not being instantaneous. i.e. most of the burn is done progressively further away from the ideal point. The lower the thrust, the worse the effect.

Yes, although it is possible to have a long burn with no gravity loss by always thrusting horizontally.

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Page 19 has a handy graph of ignition T/W vs TLI  gravity losses :

http://esamultimedia.esa.int/docs/exploration/ReferenceArchitecture/Final%20ReviewJan09/04_Human_moon_mission_version9esa120109.pdf

Thanks!

Page 20 is interesting too.  The last three lines show losses for a particular vehicle for three different steering laws.  I believe tangential direction refers to always thrusting horizontally, eliminating gravity losses increasing steering losses.  In the gravity, thrust is always aligned with the flight path, so there are no steering losses but there are gravity losses.  Finally, there's the best of the three, a fixed inertial direction: choose an orientation with respect to the stars and hold it.  As usual, a compromise is better than optimization with respect to any single variable.
« Last Edit: 06/24/2011 06:11 am by Proponent »

Offline IsaacKuo

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Re: Solar thermal beats all for space flight
« Reply #30 on: 06/24/2011 06:20 am »
Actually, now that I think about it, it's obvious that there is a perfectly standard definition of gravity loss.  It's simply the integral over time of the local acceleration of gravity times the sine of the angle between the flight path and the local vertical.  This loss is an amount which needs to be subtracted (along with drag and steering losses) from the ideal delta-V computed according to the rocket equation to give the actual delta-V.

This is wrong.  By the standard definition of gravity drag, there is a large gravity drag loss from nearly hovering regardless of the flight path.  For example, consider a rocket ship which is landing on the Moon.  It requires a minimum of 2km/s to kill off its orbital speed, but then it takes some time nearly hovering as it carefully approaches the landing site.

It might spend a few seconds to impulsively kill off the 2km/s, and then carefully descend vertically after that.  According to your definition, the gravity loss is zero.  The flight path is vertical, so the sine of the angle between the flight path and local vertical is zero.  Regardless of how long it takes to land, according to your definition gravity loss is zero.

But in reality, gravity drag losses increase the longer the lander takes to descend.  If it lands almost immediately, the total delta-v required was 2km/s--almost no gravity drag loss.  If it slowly descends for ten minutes, the total delta-v required increases to 3km/s--meaning a gravity drag loss of 1km/s.

Now, maybe you mean the cosine of the angle from vertical instead of the sine.  That might make a little more sense because the definition you gave gives a constantly increasing gravity loss for an object in circular orbit (which makes no sense).

In that case, you're still wrong.  Simply replace the slow vertical descent with a slow horizontal flight to line up with the desired landing site, followed by a quick vertical descent down to the landing site.  Again, your definition fails.

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By this standard definition, a long escape burn at low-thrust deep in the Earth's gravity leads to a large loss.  Now it's true that some of that gravity loss is really lost, because by raising the altitude of the escaping spacecraft, the long burn has reduce the velocity needed to escape.  But it's still less efficient than an impulsive burn.

Actually, a long escape to parabolic at low thrust can be EXACTLY the same delta-v as a fast escape burn to parabolic, as long as the burns take place at perigee.  From a 7.5km/s circular orbit to parabolic escape requires 3.1km/s at perigee regardless of whether it is done all at once with a high acceleration thruster or done over several months of perigee thrusts using a low acceleration thruster.

Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #31 on: 06/24/2011 06:35 am »
Good point.  A couple of errors in my definition.  It should be the integral with respect to time over the powered portion of the trajectory of the local gravity times the sine of the angle between the flight path and the local horizontal.  LMK what's wrong with that. :)

Actually, a long escape to parabolic at low thrust can be EXACTLY the same delta-v as a fast escape burn to parabolic, as long as the burns take place at perigee.

Yeah, I think we've been over that before, hence the previous discussion of passes through van Allen Belts, boil-off and mitigants.
« Last Edit: 06/24/2011 06:36 am by Proponent »

Offline IsaacKuo

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Re: Solar thermal beats all for space flight
« Reply #32 on: 06/24/2011 06:41 am »
IssacKuo was right in his first comment that using a combination, a chemical rocket at perigee to capitalize on the Oberth effect, and then using the STP to accelerate further on a voyage to Mars could give you 15km/s rather than the mere 12 I was getting with just STP.

This is exactly opposite of what I was proposing.  You use solar thermal first, to get up to a highly elliptical orbit.  Then you use the chemical rocket thruster at perigee to maximize the Oberth effect boost on the escape trajectory.  It only takes 450m/s delta-v to get from a highly elliptical orbit to a Hohmann MTO.  You can get a v_inf of 8.8km/s at a cost of only 3.1km/s from the chemical rocket thruster.

So, you use solar thermal to provide 3.2km/s to get from LEO to the highly elliptical orbit and then use a chemical rocket to boost at perigee by 3.1km/s.

If you try to use solar thermal all the way, then you kiss the Oberth effect goodbye after the first 3.2km/s.  To get the same v_inf of 8.8km/s, you need to provide 3.2km/s + 8.8km/s = 12.0km/s delta-v from the solar thermal thruster.

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I don't think the idea of storing energy in the form of heat for STP would be practical, unless the heat sink had other essential functions, the usable energy stored would be small compared to that in rocket propellants.

If you don't use heat storage, then you can't store energy from the entire orbit for use at perigee.  This means that you either need a far bigger and heavier solar concentrator (MUCH heavier since it needs to be stiff during accelerations), or you need to use a continuous spiraling path to make up for the much lower acceleration.  A spiraling escape path requires twice as much delta-v to reach escape, and the resulting circular orbit has a high perigee which rules out the use of an Oberth effect boost.  Thus, the delta-v to get to 8.8km/s v_inf from LEO is over 15km/s.  That's not good!

Offline IsaacKuo

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Re: Solar thermal beats all for space flight
« Reply #33 on: 06/24/2011 06:55 am »
Good point.  A couple of errors in my definition.  It should be the integral with respect to time over the powered portion of the trajectory of the local gravity times the sine of the angle between the flight path and the local horizontal.  LMK what's wrong with that. :)

The counterexample I gave of a lunar lander approaching the landing site in different ways still stands.  Consider a lunar lander that spends some time flying horizontally to line up over the desired landing spot.  According to the above definition, it suffers zero gravity loss, because the sin of the angle is zero.  In reality, the lander racks up increasing gravity drag losses the longer it takes to land.

Also, consider the classic gravity drag example of a rocket booster getting up to orbital speed.  In the simplified case, the rocket booster starts off on top of a mountain on an otherwise spherical airless world.  Thus, the rocket simply flies horizontally until it builds up orbital speed.  Again, according to your definition it suffers zero gravity loss.  But in reality, it suffers more gravity loss the longer it takes to reach orbital speed.  Suppose gravity is 10m/s/s and the thruster provides enough thrust for 14.14m/s/s.  For the first second of flight, the rocket thruster is pointed at a 45 degree angle in order to maintain altitude.  Its horizontal acceleration is only 10m/s/s.

So, for each second of flight, the thruster provides 14.14m/s worth of impulse but the ship only accelerates by 10m/s.  This is 4.14m/s worth of gravity drag loss per second, for the first seconds of flight.  As the rocket ship builds up to speed, it essentially starts to experience lift.  This allows the thruster to be tilted more horizontally, increasing horizontal thrust while reducing vertical thrust.  This reduces gravity drag losses until eventually the rocket ship reaches orbital speed.  At orbital speed, the thruster no longer needs to be tilted downward at all in order to maintain altitude.  All of the thrust goes into horizontal acceleration, meaning zero gravity losses from this point onward.

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Actually, a long escape to parabolic at low thrust can be EXACTLY the same delta-v as a fast escape burn to parabolic, as long as the burns take place at perigee.

Yeah, I think we've been over that before, hence the previous discussion of passes through van Allen Belts, boil-off and mitigants.

Your supposed definition of gravity loss needs to somehow be successful in this situation.
« Last Edit: 06/24/2011 07:06 am by IsaacKuo »

Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #34 on: 06/24/2011 07:01 am »
The counterexample I gave of a lunar lander approaching the landing site in different ways still stands.  Consider a lunar lander that spends some time flying horizontally to line up over the desired landing spot.  According to the above definition, it suffers zero gravity loss, because the sin of the angle is zero.  In reality, the lander racks up increasing gravity drag losses the longer it takes to land.

Also, consider the classic gravity drag example of a rocket booster getting up to orbital speed.  In the simplified case, the rocket booster starts off on top of a mountain on an otherwise spherical airless world.  Thus, the rocket simply flies horizontally until it builds up orbital speed.  Again, according to your definition it suffers zero gravity loss.  But in reality, it suffers more gravity loss the longer it takes to reach orbital speed.  Suppose gravity is 10m/s/s and the thruster provides enough thrust for 14.14m/s/s.  For the first second of flight, the rocket thruster is pointed at a 45 degree angle in order to maintain altitude.  Its horizontal acceleration is only 10m/s/s.

So, for each second of flight, the thruster provides 14.14m/s worth of impulse but the ship only accelerates by 10m/s.  This is 4.14m/s worth of gravity drag loss per second, for the first seconds of flight.  As the rocket ship builds up to speed, it essentially starts to experience lift.  This allows the thruster to be tilted more horizontally, increasing horizontal thrust while reducing vertical thrust.  This reduces gravity drag losses until eventually the rocket ship reaches orbital speed.  At orbital speed, the thruster no longer needs to be tilted downward at all in order to maintain altitude.  All of the thrust goes into horizontal acceleration, meaning zero gravity losses from this point onward.

Those are steering losses (a component of thrust normal to the flight path) rather than gravity losses.

Offline IsaacKuo

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Re: Solar thermal beats all for space flight
« Reply #35 on: 06/24/2011 07:06 am »
Those are steering losses (a component of thrust normal to the flight path) rather than gravity losses.

Whether you like it or not, it's the actual definition of gravity drag that everyone else other than you uses.  The example of a rocket booster which must tilt its thrust downward to maintain altitude is the classic textbook example of gravity drag.

Offline Andrew_W

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Re: Solar thermal beats all for space flight
« Reply #36 on: 06/24/2011 07:12 am »
"This is exactly opposite of what I was proposing."

You were suggesting a chemical rocket at perigee, how is me saying:"a chemical rocket at perigee to capitalize on the Oberth effect"  the opposite of what you were suggesting?

If the STP has a delta v of 12km/sec, how do you propose using that delta v before perigee? How do you plan to take full advantage of the higher Isp of STP if you are going to restrict its use to only the ~3km/s delta v required to move into a highly elliptical orbit?

"If you don't use heat storage..."

That's not a refutation of my point which was that you can't physically store enough heat energy for it to be worth a damn, maybe you're planning to run your car to town on a pile of hot bricks in the boot, if so you won't be getting very far.
« Last Edit: 06/24/2011 07:13 am by Andrew_W »
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
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Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #37 on: 06/24/2011 07:14 am »
Those are steering losses (a component of thrust normal to the flight path) rather than gravity losses.

Whether you like it or not, it's the actual definition of gravity drag that everyone else other than you uses.  The example of a rocket booster which must tilt its thrust downward to maintain altitude is the classic textbook example of gravity drag.

If a booster directs a component of thrust downward, there is a gravity loss (which is negative, so it becomes a gain).  If the booster is rising, there will also be a steering loss.  Show us the textbooks' definitions of gravity and steering losses.

Offline Proponent

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Re: Solar thermal beats all for space flight
« Reply #38 on: 06/24/2011 07:21 am »
"If you don't use heat storage..."

That's not a refutation of my point which was that you can't physically store enough heat energy for it to be worth a damn, maybe you're planning to run your car to town on a pile of hot bricks in the boot, if so you won't be getting very far.

Fun fact:  As it happens, an attempt was made to use hot bricks to power the steam-propelled trains of London Underground between stations, when the system opened in the 1860s.  The idea was to heat the bricks up in the locomotive's firebox when the train was in a station, where ventilation was provided, and then damper down the fire make steam from the bricks in the unventilated tunnels.  Apparently it didn't work very well, and the burning coal between stations made for some rather unpleasant rides until electric trains were introduced about 20 years later.

Offline Andrew_W

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Re: Solar thermal beats all for space flight
« Reply #39 on: 06/24/2011 07:28 am »
If you accelerate you ship to escape velocity plus 1km/s in a circular low Earth orbit you'll escape Earth at far more than 1km/s.

Does that still count as the Oberth Effect?
I confess that in 1901 I said to my brother Orville that man would not fly for fifty years.
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