Author Topic: Basic Rocket Science Q & A  (Read 272241 times)

Offline clongton

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Re: Basic Rocket Science Q & A
« Reply #620 on: 06/24/2011 10:10 AM »
I have a case of insomnia (my wife goes in for surgery tomorrow, and I am 2400 miles away), so I felt like tackling this.

Let us know how things go.
Chuck - DIRECT co-founder
I started my career on the Saturn-V F-1A engine

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #621 on: 06/24/2011 03:57 PM »
Best wished, no matter how "small" the surgery is.

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #622 on: 06/24/2011 08:02 PM »
Best wished, no matter how "small" the surgery is.


Let us know how things go.

Everything was a success, no issues. Thank you both for your concern, it means a lot.

As for my late night rocket science, I hope I adequately answered where Jensen went astray. High temperature is good, but so is low molecular weight, and it's a trade between the two.

Incidentally, nuclear rockets are governed by the same equation. The nice thing with nuclear is not so much that you can have higher temperatures. Your materials limits mean that they are about the same chamber temp as a chemical rocket.

Where nuclear pulls ahead is your molecular weight and temperature are decoupled. You can use pure hydrogen, which is how you get 1000s of Isp. That's for solid core, of course. If you're talking about more exotic nukes, you can bump up temp too (I'm a fan of the nuclear lightbulb concept, myself).
Don't flippantly discount the old rules of this industry. Behind each one lies a painful lesson learned from broken, twisted hardware. Learn those lessons, and respect the knowledge gained from them. Only then, see if you can write new rules that will meet those challenges.

Offline clongton

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Re: Basic Rocket Science Q & A
« Reply #623 on: 06/24/2011 08:13 PM »
Best wished, no matter how "small" the surgery is.


Let us know how things go.

Everything was a success, no issues. Thank you both for your concern, it means a lot.

As for my late night rocket science, I hope I adequately answered where Jensen went astray. High temperature is good, but so is low molecular weight, and it's a trade between the two.

Incidentally, nuclear rockets are governed by the same equation. The nice thing with nuclear is not so much that you can have higher temperatures. Your materials limits mean that they are about the same chamber temp as a chemical rocket.

Where nuclear pulls ahead is your molecular weight and temperature are decoupled. You can use pure hydrogen, which is how you get 1000s of Isp. That's for solid core, of course. If you're talking about more exotic nukes, you can bump up temp too (I'm a fan of the nuclear lightbulb concept, myself).

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.
« Last Edit: 06/24/2011 08:16 PM by clongton »
Chuck - DIRECT co-founder
I started my career on the Saturn-V F-1A engine

Offline strangequark

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Re: Basic Rocket Science Q & A
« Reply #624 on: 06/24/2011 08:31 PM »

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Indeed, I've also wondered about that design using synthetic diamond. If you're not familiar with the CVD process for creating diamond, it's interesting. You make a low pressure, low temp carbon gas, and allow it to deposit on a preexisting diamond. The upshot is the technology has the potential to create relatively large, relatively arbitrary shapes out of diamond. Aside from temp resistance (as long as it's not in an oxidizer...), diamond has phenomenal thermal conductivity.

Don't flippantly discount the old rules of this industry. Behind each one lies a painful lesson learned from broken, twisted hardware. Learn those lessons, and respect the knowledge gained from them. Only then, see if you can write new rules that will meet those challenges.

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #625 on: 06/24/2011 08:43 PM »

Glad to hear that.

I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Indeed, I've also wondered about that design using synthetic diamond. If you're not familiar with the CVD process for creating diamond, it's interesting. You make a low pressure, low temp carbon gas, and allow it to deposit on a preexisting diamond. The upshot is the technology has the potential to create relatively large, relatively arbitrary shapes out of diamond. Aside from temp resistance (as long as it's not in an oxidizer...), diamond has phenomenal thermal conductivity.



What would it's transparency be to hard ultraviolet?

Offline IsaacKuo

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Re: Basic Rocket Science Q & A
« Reply #626 on: 06/24/2011 08:51 PM »
I too am a complete fan of the nuclear lightbulb. That's why I said "a process to create pure silica" over in the thread about what do we want to see developed. That material is completely transparent to the thermal energy of a nuclear reaction so whatever thermal energy is left over from dealing with the pure hydrogen just goes out thru the glass engine into space. The isp would climb into the tens or even hundreds of thousands of seconds, and the thrust would climb right along with it.

Even if the bulb walls are completely transparent, they will still absorb incredible amounts of heat from direct contact with the reactor fuel.  If you had some magical method of preventing direct contact with the reactor fuel, then the bulb walls would be superfluous.  (Unfortunately, magnetic fields aren't strong enough, which is the reason why transparent bulb walls are considered necessary in the first place.)

The bulb walls need to be kept actively cooled to below melting point, which means a high rate of heat transfer if the reactor gas is hot.  If you try to use your propellant to do this cooling, then you need a very high mass rate.  Your performance ends up limited to solid core NTP Isp--which makes the whole thing pretty pointless.

So instead, you need a closed loop cooling system with large radiators to keep the bulb walls from melting while also maintaining a propellant consumption rate low enough to provide high Isp.  This radiator system limits your potential power/weight ratio.  As such, you're never going to have the same sort of high thrust potential as solid core NTP.  And since you're power limited, the potential thrust is at best inversely proportional to the Isp.

All in all, the nuclear lightbulb concept is technologically challenging with numerous potential pitfalls.  It represents a high technology risk compared to nuclear electric, and the potential performance benefit over nuclear electric is questionable.  While it might have an order of magnitude better power/weight ratio than nuclear electric, it might not.  And if not, then it's just a massive R&D investment wasted on something which is inferior to nuclear electric (which is far less expensive to develop, can use a variety of dense storable propellants, provides plentiful electrical power for other mission hardware, and can have much higher Isp).

My bet is that the inherent technology risk of nuclear lightbulb is high enough that we never find out how hard or easy it would have been for us to develop.

Offline Antares

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Re: Basic Rocket Science Q & A
« Reply #627 on: 06/26/2011 03:44 AM »
Nuclear lightbulb /= basic rocket science
If I like something on NSF, it's probably because I know it to be accurate.  Every once in a while, it's just something I agree with.  Facts generally receive the former.

Offline baldusi

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Re: Basic Rocket Science Q & A
« Reply #628 on: 07/07/2011 09:20 PM »
I guess is a silly question, but with tanks and such structures, is the isogrid first machined and then formed, or first formed and then machined?
A related question it, is formed and then welded, or welded and then formed?
All this was from seeing the Dragon's interior. It would seem that they formed the parts, machined them, and then friction welded. Because you can see the rims of each part. I didn't knew you could do friction welding in a curved surface like the Falcon's, though.
I guess that from a structural side the ideal process would be to weld->form->machine. But it's exactly opposite of how is easier to machine.

Offline Antares

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Re: Basic Rocket Science Q & A
« Reply #629 on: 07/08/2011 12:07 AM »
Someone will correct me if I'm wrong.  Machined in a flat sheet then formed to a curved tank.  Usually formed then welded, but not always - depends on other factors.
« Last Edit: 07/08/2011 12:10 AM by Antares »
If I like something on NSF, it's probably because I know it to be accurate.  Every once in a while, it's just something I agree with.  Facts generally receive the former.

Offline neutrino78x

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Re: Basic Rocket Science Q & A
« Reply #630 on: 07/08/2011 03:03 AM »
Hey guys (men and women)... if you were in a stanford torus of the right dimensions to provide an acceleration equal to that of Earth gravity (9.81 m/s), but you were in an evacuated compartment, that is to say, one in which is there is no matter whatsoever except you and a spacesuit, would you still feel the acceleration pushing you against the outside edge of the torus (the "floor")? Or does there have to be air in the compartment to transfer the force? I think the answer is that you would just float, but I wanted to check what others think, especially if they have a college degree!

btw we assume that you start out in the middle of the compartment, like maybe you were hooked up to a tether or something, and then they evacuated all the air.

--Brian

Offline Sparky

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Re: Basic Rocket Science Q & A
« Reply #631 on: 07/08/2011 03:41 AM »
Assuming you weren't touching the "floor", and there were no obstacles within your path to bump into, you should be able to float, but only if the entire torus were evacuated (or at least a single corridor along its length). If you were floating inside a compartment, you would be able to experience Microgravity only until the wall bumped into you.

I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

Offline lcs

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Re: Basic Rocket Science Q & A
« Reply #632 on: 07/08/2011 04:08 AM »
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

Offline STS-134

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Re: Basic Rocket Science Q & A
« Reply #633 on: 07/08/2011 04:16 AM »
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".

Offline lcs

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Re: Basic Rocket Science Q & A
« Reply #634 on: 07/08/2011 04:21 AM »
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".

In any case, the answer to neutrino's question is that even in a vacuum, if you were suspended in the middle of the torus and suddenly released you would fall to the floor, because you would be moving in a straight line at the velocity of the torus, while the torus floor rushed up to meet you as it curved away from your straight line trajectory. 

Offline STS-134

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Re: Basic Rocket Science Q & A
« Reply #635 on: 07/08/2011 04:42 AM »
I imagine that even with atmosphere, if a person were to jump at the right speed, in the right direction, they could get very impressive hang time until just the air resistance slowed them enough to fall to the floor.

In other words, orbital velocity inside the torus is just the v=omega x R.  For the Stanford torus v=pi*1.6 km /60 sec = 94 meters/sec.  I don't think jumping into orbit will be an Olympic event. 

94 m/s = about 210.7 mph.  At that speed, air resistance will VERY QUICKLY cause you to begin rotating again with the torus, and get pushed to the "floor".

In any case, the answer to neutrino's question is that even in a vacuum, if you were suspended in the middle of the torus and suddenly released you would fall to the floor, because you would be moving in a straight line at the velocity of the torus, while the torus floor rushed up to meet you as it curved away from your straight line trajectory. 

Well, it depends on the initial condition (velocity).  If your velocity is such that you are initially rotating along with the torus, you will NOT feel any acceleration (because you won't be accelerating, you'll just be moving along a straight line in the non rotating reference frame of the torus hub).  You will be weightless, but as time goes on you'll see the "floor" accelerate toward you at something less than 1.0g, until you eventually hit the floor.  The rate at which you acclerate toward the floor depends on your height above the "floor" when you are released from the tether.  The higher up you are (the closer you are to the "hub"), the slower the "floor" will "accelerate" toward you.  This is basically equivalent to being held above the floor on Earth, and released in an evacuated chamber.  You will accelerate toward the floor at 1.0g, and you will experience weightlessness until you hit the floor.

If, on the other hand, you are inside an evacuated torus, AND stationary with respect to the hub at the center of the torus, you'll just float.

Offline MP99

Re: Basic Rocket Science Q & A
« Reply #636 on: 07/15/2011 08:05 AM »
...until you drift into the floor that's moving past at 210MPH.

I suspect the initial contact wouldn't be too painful if the closure rate is very low, but it would set you spinning, and add a small component of lateral velocity.

This will ensure another contact, increasing lateral velocity and spin rate.

Each time the process repeats the increasing lateral velocity increases the effective g forces of the "fall" towards the floor. By the time you're travelling at 105MPH, you've having 1/2g impacts with the floor at a relative velocity of 105MPH while tumbling violently. Suspect that's not survivable.

cheers, Martin

Offline Warren Platts

Re: Basic Rocket Science Q & A
« Reply #637 on: 07/25/2011 05:03 AM »
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci

Offline IsaacKuo

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Re: Basic Rocket Science Q & A
« Reply #638 on: 07/25/2011 02:39 PM »
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

The maximum bonus factor is infinite, in the case of an infinitesmal delta-v at perigee compared to delta-v at apogee (or v_inf, in the case of a parabolic orbit).  It's not a matter of a clean "bonus factor", because the multiplier goes down with higher delta-v.

Quote
Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

No.  Even if you start from a parabolic escape trajectory and perform an 11km/s burn at an Earth grazing perigee, the resulting v_inf is still only sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+11)^2-11.1^2) = 19.11km/s.  This is a "bonus factor" of only 19.11/11 = 1.73.

The "bonus factor" is higher for smaller delta-v.  For example, a delta-v of .45km/s gives a v_inf of 3.19km/s.  This is a "bonus factor" of 3.19/.45 = 7.09.

A way to calculate the effect geometrically is to use a right triangle.  The base is escape velocity; the hypotenuse is escape velocity plus delta-v.  The other side of the triangle is velocity at infinity.  When delta-v is small compared to escape velocity, the "bonus factor" is high.  When delta-v is large compared to escape velocity, the "bonus factor" is low.

Quote
Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?

I think so, if you're willing to accept very long trip times (on the order of many years).  However, it would be better to stage at a highly elliptical Mars orbit.  This gives you all of the benefit without any of the wait time.  Similarly, staging at a highly elliptical Earth orbit gives you less of a wait time than EML1 or EML2.

Offline Warren Platts

Re: Basic Rocket Science Q & A
« Reply #639 on: 07/25/2011 05:47 PM »
Q1: Roughly, what is the maximum bonus factor one can expect from the Oberth effect when injecting into Low Mars orbit (or Sun-Mars Lagrange point) fully propulsively with high thrust chemical rockets?

The maximum bonus factor is infinite, in the case of an infinitesmal delta-v at perigee compared to delta-v at apogee (or v_inf, in the case of a parabolic orbit).  It's not a matter of a clean "bonus factor", because the multiplier goes down with higher delta-v.

Quote
Q2: Given a chemical, single-stage MTV with 11 km/sec nominal delta v, if staged from EML1 or EML2, could one get an effective 44 km/sec total delta v using the Oberth effect when slingshotting around Earth and then braking into Mars orbit?

No.  Even if you start from a parabolic escape trajectory and perform an 11km/s burn at an Earth grazing perigee, the resulting v_inf is still only sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+11)^2-11.1^2) = 19.11km/s.  This is a "bonus factor" of only 19.11/11 = 1.73.

The "bonus factor" is higher for smaller delta-v.  For example, a delta-v of .45km/s gives a v_inf of 3.19km/s.  This is a "bonus factor" of 3.19/.45 = 7.09.

A way to calculate the effect geometrically is to use a right triangle.  The base is escape velocity; the hypotenuse is escape velocity plus delta-v.  The other side of the triangle is velocity at infinity.  When delta-v is small compared to escape velocity, the "bonus factor" is high.  When delta-v is large compared to escape velocity, the "bonus factor" is low.

Quote
Q3: When departing Mars, can one get a significant Oberth bonus if the ERV was staged at Mars's Sun-Mars L-point?

I think so, if you're willing to accept very long trip times (on the order of many years).  However, it would be better to stage at a highly elliptical Mars orbit.  This gives you all of the benefit without any of the wait time.  Similarly, staging at a highly elliptical Earth orbit gives you less of a wait time than EML1 or EML2.

Thanks for the thoughtful remarks Isaac. The way you express the math is very clear.

Of course the delta v would be split up into two separate burns, but using Oberth's equation, I can see even that won't make that much of a difference:

sqrt((v_esc+delta_v)^2 -v_esc^2) = sqrt((11.1+5.5)^2-11.1^2) = 12.3km/s

Bonus factor: 12.3 / 5.5 = 2.2

The maximum warp for a single-stage craft with a usable payload isn't going to be much more than 11 km/sec nominally. With an Oberth bonus, we can't expect much more than to double that to maybe 20-22 km/sec. Still, that's not bad for a chemical rocket.

As for elliptical orbits versus LaGrange points: I don't see why you think the "wait time" is going to be a big deal--after all, EML1 or 2 is only a week or so from Earth. Since Mars gravity is so weak compared to the Sun, the Sun-Mars L1 point is going to be fairly close to Mars. Also, Lagrange points are ideal locations for propellant depots. (Not to mention that EML1/2 is close to Lunar propellant sources ;) ).
"When once you have tasted flight, you will forever walk the earth with your eyes turned skyward, for there you have been, and there you will always long to return."--Leonardo Da Vinci