Profile overview for adeldor.
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This user has mostly submitted to the following subverses (showing top 5):

13 submissions to Spaceflight

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3 submissions to DIY

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This user has so far shared a total of 34 links, started a total of 4 discussions and submitted a total of 59 comments.

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Submissions: This user has upvoted 55 and downvoted 0 submissions.

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3 highest rated comments:

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 0 points 7 points (+7|-0) ago

The lowest viable orbit is around 530,000 ft @ 18,000 mph. Against that, the boost from being launched at 35,000 ft @ 500 mph is small. Even the advantage afforded to the nozzle design by the thinner air is minor. So carrier height and speed aren't the reasons for choosing air launch.

The big advantages with air launch are:

  • flexibility of orbital inclination without having to overfly populated areas,

  • bypassing the current burden of paperwork required for a vertical launch,

  • ability to fly around bad weather, thus improving schedule reliability.

ETA: tying the booms together might actually cause problems. Being free to flex individually is important to structural stability. I have no doubt Scaled Composites investigated the design very well beforehand. It has a good working model already - White Knight 2.

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 0 points 7 points (+7|-0) ago

I have some background in this subject. I'll attempt to illustrate the minor advantage offered by air launch through the use of a simplified example. Apologies in advance for the math.

This illustrative example is simplified by eliminating staging and bumping Isp to high levels (don't want too much math in a mere Voat post :-) ). Note: without staging, this simplification works significantly in favor of air launch.

If you start with Tsiolkovsky's infamous Rocket Equation

ΔV = Ispg0ln(mi/mf)

and rework it to determine final mass at burnout, you end up with

mf = mi/e(ΔV/(Ispg0))

where

ΔV is total change in velocity

Isp is specific impulse of the rocket (assume 400 seconds, simplified choice for this example)

g0 is gravitational force (9.81m/s2)

mi is initial mass of rocket including propellant and payload (25,000 kg)

mf is final mass of depleted rocket including payload

If the ΔV required for orbit from a ground launch is 9,500 m/s (about right if one includes gravity and atmospheric losses), it would result in a final mass to orbit of

25,000/e(9,500/(400x9.81)) = 2,221 kg.

Now subtract the 500 mph (about 220 m/s), and subtract another 80 m/s for lower gravity and atmospheric losses (rounded number for ease), the final mass to orbit from the air launch becomes

25,000/e(9,200/(400x9.81)) = 2,397 kg.

That's a difference of 176 kg, or an increase of about 8%.

Even with the large advantage being given to air launch in this example (no staging), the difference in mass lofted to orbit is minor. Of course, there are many real-world complications that would fuzz this example, but the first order approximation shown here holds, regardless.

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 0 points 3 points (+3|-0) ago

Well that's a terse response! :-)

Might you show where I went wrong?

3 lowest rated comments:

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 1 points -1 points (+0|-1) ago

I'm afraid it doesn't work like that. Regardless of method, the same kinetic energy must be imparted to the vehicle in order to reach the necessary altitude and orbital velocity. Given the numbers above, the air launch provides ~7% of altitude, and ~3% of orbital velocity. Note that the orbital velocity component needs about an order of magnitude more energy, yet the air launch assists there even less.

It might seem counterintuitive, but It's a direct consequence of the physics described by Tsiolkovsky's equation (given above). The equation is fundamental to reaching orbit, and is the foundation of all launch systems, regardless of topology. If you wish to learn more about how it works, a quick search found this introductory public course presented by MIT. If that doesn't help, Google turned up quite a few other resources.

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 0 points 0 points (+0|-0) ago

Good eye! Isp is defined as ve/g0 - where Ve is the rocket motor's exhaust velocity, and g0 is 9.81m/s2. So when plugging Isp into Tsiolkovsky's equation instead of Ve, it's necessary to use that definition of g0 as used in the calculation of Isp, regardless of the rocket's eventual altitude.

So, were one calculating all this on Mars, g0 still would be 9.81m/s2, despite Mars' gravitational acceleration on the surface being 3.71m/s2!

Isp is SPECIFIC impulse. The unit is given in seconds. It's not really a duration measurement, but much more a measure of rocket motor efficiency.

The world's largest airplane is a launch platform. Private space goes where NASA wouldn't. submitted by Dfens to technology

adeldor 0 points 0 points (+0|-0) ago

Of course! :-)