So, who wants to win 1 million Euro?

[raven]

What makes aerospikes particularly interesting is that the surrounding atmosphere acts as the nozzle wall, allowing them to operate with greater efficiency at a wider range of altitudes,which is great for, say, a Single Stage to Orbit vehicle.

[Heiwa]

OK Heiwa, you want to do a simple energy difference calculation? Fine. Forget burning the fuel. Just imagine that the spacecraft dumps all that fuel overboard in a non-propulsive way. Its mass decreases, it's velocity remains unchanged. Its kinetic energy therefore has decreased. Where did that energy go? How did the kinetic energy of the spacecraft change? The answer to that might help you with the answer to your original issue.

Where did that energy go? It was dumped! What are you trying to say? This discussion is getting sillier and sillier. Like the post about space navigation by sextant and compass and charts at high g (like in a WWII bomber) while swinging into Moon orbit or that weak structures like tin boxes can slow down from 11 200 m/s to 100 m/s (re-entry) by friction/turbulence without burning up. Sorry, you have to do much better to earn topic!

[Inanimate Carbon Rod]

Where did that energy go? It was dumped!

Jason clearly referenced kinetic energy. So answer his question and say how the kinetic energy changed.

Hint 1: Fuel != kinetic energy.
Hint 2: Since you're clearly pretending to be an engineer and clearly have little grasp of physics I'll help you a little more and tell you that the symbol "!=" means "does not equal"

[Heiwa]

Where did that energy go? It was dumped!

Jason clearly referenced kinetic energy. So answer his question and say how the kinetic energy changed.

Hint 1: Fuel != kinetic energy.
Hint 2: Since you're clearly pretending to be an engineer and clearly have little grasp of physics I'll help you a little more and tell you that the symbol "!=" means "does not equal"

?? Kinetic energy (J) per mass unit (kg) is just a function of velocity v (m/s) squared (v²) and when v is unchanged the kinetic energy (per mass unit) is unchanged. What are you trying to say? Instead of asking stupid question try to explain what you want to say.
Can you, e.g. explain re-entry. You are aboard the famous International Space Station, ISS, that according NASA is orbiting Earth every 90 minutes at 400 000 m altitude (almost vacuum) at 7 200 m/s velocity and you want to go down to Earth. It means you have to go down 400 000 m and slow down from 7 200 m/s to 0 m/s speed. How to do it?

Do you jump into a little capsule with a little rocket engine to slow you down? Yes, apparently you do that and the result is that you arrive at 120 000 m altitude but that the velocity then has increased to 9 000 m/s as some potential energy of the capsule has become kinetic energy = greater velocity. It is like diving from the 10 m board. It gets faster the closer you get to the water.

At 120 000 m altitude there is a thin atmosphere with nitrogene and oxygene atoms that you collide with and ... MAGIC ... suddenly you slow down to 100 m/s (at say 5 000 m altitude) and deploy a parachute and land. In a desert in Kazakstan. Where nobody lives. In the middle of nowhere!

Imagine that - you slow down from 9 000 m/s to 100 m/s just by colliding with atoms. But why don't you slow down to 0 m/s by colliding with atoms? Let me ask a stupid question or two? Why do you need a parachute at the end? What is wrong with colliding with atoms to the end?

When you dive from a 10 m board you do not need a parachute.

[grmcdorman]

I just had a thought that should underscore the absurdity of Heiwa's "analysis”.

Let's take an imaginary spacecraft that has two identical engines, arranged directly opposite each other, and which do not impart any turning moment, i.e. they can't change the craft's spin or attitude.

Ivan is the operating engineer, and he decides to fire both engines for the same, identical duration at the same time.

Let's say the craft has a mass of 1,000 kg and a velocity of 1,000 m/s before the engines fire.

That means its KE is 1/2 × 1000 × 1000 × 1000, or 5×10^5 J (if I've got my units right).

Let's also say the burn consumes 500 kg.

So afterwards, the craft is still travelling at the same velocity - same speed, same direction - but it now has half the KE (1/2 × 500 × 1000 × 1000 = 2.5 × 10^5 J).

Where did the KE go, Heiwa?

[Halcyon Dayz, FCD]

An engineer who has never heard of terminal velocity or even doesn't understand why it exists?

A bright 8-year-old could figure that one out.

[LunarOrbit]

Can you, e.g. explain re-entry. You are aboard the famous International Space Station, ISS, that according NASA is orbiting Earth every 90 minutes at 400 000 m altitude (almost vacuum) at 7 200 m/s velocity and you want to go down to Earth. It means you have to go down 400 000 m and slow down from 7 200 m/s to 0 m/s speed. How to do it?

My layman's explanation: you fire your engine in a retrograde direction until your periapsis is far enough inside the atmosphere to use atmospheric drag to slow your speed to terminal velocity, and then you use parachutes to finish the job.

It is like diving from the 10 m board. It gets faster the closer you get to the water.

You don't constantly accelerate, eventually you reach terminal velocity (the point where the downward force of gravity and the air resistance balance out).

At 120 000 m altitude there is a thin atmosphere with nitrogene and oxygene atoms that you collide with and ... MAGIC ... suddenly you slow down to 100 m/s (at say 5 000 m altitude) and deploy a parachute and land. In a desert in Kazakstan. Where nobody lives. In the middle of nowhere!

Would you rather they dropped the Soyuz capsule in the middle of a major city?

Imagine that - you slow down from 9 000 m/s to 100 m/s just by colliding with atoms. But why don't you slow down to 0 m/s by colliding with atoms?

You would reach 0 m/s if you collided with the atoms that make up the ground.

Let me ask a stupid question or two?

I haven't stopped you from doing so before, so why start now?

Why do you need a parachute at the end? What is wrong with colliding with atoms to the end?

Because air resistance isn't enough (on Earth) to slow you down to a safe (survivable) speed. But we'll add "terminal velocity" to the many things you don't understand (but still consider yourself an expert in).

When you dive from a 10 m board you do not need a parachute.

The key points here are the fact that you're only diving from 10m and can't reach a high speed, and the fact that you're diving into water and not into the ground.

[Noldi400]

Of course, you are light years off topic about the Heiwa €1M Challenges...

Nonsense.  You have claimed to be a qualified and skilled engineer.  You have offered a substantial prize for anyone who can refute your findings, which you characterize as having come from a rigorous engineering background.  Your personal qualifications and expertise are therefore very much part of the question, and they will be investigated by any means possible.

^{-                      ---- SNIP FOR SPACE ----}

After all that, you really can't figure out why you provoke such a strong reaction among people with legitimate qualifications and expertise?

Well spoken.

"Das ist nicht nur nicht richtig, es ist nicht einmal falsch!"   - Wolfgang Pauli

Edit: Oh, and Heiwa? Lunar Orbit Insertion (swinging into moon orbit")  hardly qualifies as a High-G maneuver;  Delta-V of 889.1 m/s over 362 seconds would be an average acceleration of 2.456 m/s² - about a quarter of a g.

I'm no engineer, but I can certainly do simple arithmetic.

[cjameshuff]

?? Kinetic energy (J) per mass unit (kg) is just a function of velocity v (m/s) squared (v²) and when v is unchanged the kinetic energy (per mass unit) is unchanged. What are you trying to say? Instead of asking stupid question try to explain what you want to say.

It's a function of velocity and mass...something that is not unchanged in the described situation.

Imagine that - you slow down from 9 000 m/s to 100 m/s just by colliding with atoms. But why don't you slow down to 0 m/s by colliding with atoms? Let me ask a stupid question or two? Why do you need a parachute at the end? What is wrong with colliding with atoms to the end?

So, it looks like we can add terminal velocity to the list of physical phenomena that Heiwa/Anders Björkman is unfamiliar with.

[sts60]

Where did that energy go? It was dumped! What are you trying to say?

Yes, it was dumped - that is exactly the point.  According to your layman's energy-balance technique, since V_f = V_i in this scenario, the kinetic energy of the vehicle did not change - while even you acknowledge it did. 

His scenario was intended to demonstrate to you how your method doesn't work.  And it doesn't work; real engineers have been telling you this for quite a while now.  Now that it has been presented to you in the simplest possible form, do you finally get it?

This discussion is getting sillier and sillier. Like the post about space navigation by sextant and compass and charts

The Apollo navigation solutions were all computed beforehand with allowances for variances in actual mass, performance, event times, and so on.  They were updated constantly on the ground by some of the most powerful computers available.  In any case, I already provided you detailed references on the methods and means of Apollo navigation.  Feel free to provide an informed criticism, if you have one; all you have at this point is an appeal to ignorance, which is rejected as unfounded and uninformed.

at high g

The LOI burn duration was 357.5 seconds long with a velocity change of 2917.5 ft/sec, for an average acceleration of 8.2 ft/sec², or approximately 1/4 G.   "High g"?  Really?  How do you survive living in a 1 G environment?  Or would you simply care to admit your mistake?

(like in a WWII bomber)

Wrong again.  Or do you really believe bomber navigators took Sun sights while their ships were maneuvering to avoid FW-190s? 

while swinging into Moon orbit

Wrong again.  The spacecraft computer managed the burn, including attitude control.

or that weak structures like tin boxes

Wrong again.  The CM featured an outer stainless-steel honeycomb structure around the inner pressure vessel, an aluminum honeycomb structure - a very common construction technique in air- and spacecraft.  Your characterization is nothing more than an appeal to ridicule, but once again serves only to make you look ridiculous because - once again - you have no idea what you are talking about.

can slow down from 11 200 m/s to 100 m/s (re-entry) by friction/turbulence without burning up.

Wrong in many ways: first, not by turbulence, as has been explained to you already.  Second, by use of an ablative heat shield - standard engineering practice, as has been explained to you already.  Third, the absolute maximum G load at any point during any Apollo reentry was 7.19 during Apollo 16 reentry - less than that commonly encountered during aerobatics or air combat.

Sorry, you have to do much better to earn topic!

You've already been proven wrong on almost everything you've said in this thread.  All you accomplished in this post was to demonstrate further ignorance. 

Now, I ask you again to address the three main issues with your "challenge":

1. You are offering money you don't have, for a challenge you have defined poorly and has no proper adjudication.

2. Your primary calculation is completely wrong because you don't understand energy balances.

3. You have no idea what you are talking about, and no apparent interest or ability in relieving your own ignorance.

In light of these issues, I ask again - do you have any intention of actually learning anything at all, or are you just trolling?  Because right now, the bit-bucket yawns wide.

[ka9q]

Yes.  The faster you can throw mass, the better you are.

Yes, but putting maximum energy into your exhaust isn't what you're really trying to do. You want to put maximum momentum into your exhaust so that in pushing on it, by Newton's Third Law, you'll impart an equal and opposite momentum to your spacecraft. That's how rockets work.

Momentum is simply proportional to velocity:

p = mv

So if you double the velocity of your exhaust, you double the momentum you get out of each kilogram of propellant, and you'll only have to carry half as much. Cool.

But there's a problem. Pushing on any moving object takes energy, but that energy increases as the square of the velocity:

Ke = 1/2 m v²

So doubling your velocity doubles your momentum rate, but quadruples the power required. And the energy available to eject each kilogram of propellant is limited by the energy stored in that kilogram.

Solving for velocity we get

v = sqrt(2 Ke/m).

If we have an ideal rocket that somehow converts all of the stored energy in its propellants into kinetic energy of the exhaust, then Ke/m is the energy stored in each unit of propellant. We can't throw it any faster than v because, as the saying goes, "Cap'n, we haven't the power"!

This is rocketry's "dismal equation". If you want to minimize the propellant weight you have to carry, you have to eject it at a high velocity, so you want the most energetic ones you can find (i.e., with the greatest energy per unit mass). Chemicals like this tend to be unstable or toxic or hard to handle or expensive -- or all four.

But suppose the energy to eject the propellant came from something other than the propellant itself? That's what's neat about nuclear thermal or electric propulsion. They separate the two combined roles of a chemical propellant: reaction mass to push on, and stored energy to push on it with.The reaction mass can be chosen to be cheap and/or easy to store and handle and/or nontoxic and/or for other purposes, such as good radiation shielding. You can take a lot of energy from a nuclear reactor or big solar panels and put it all into a tiny amount of propellant, ejecting it really really fast, and you'll get by with much less propellant. That's how ion engines, especially, get their extremely high performance. But because so much energy is required to eject each unit of propellant, the mass flow rates tend to be very low, and so is the thrust. So you have to be patient.

One non-chemical engine with a lot of promise is the nuclear thermal rocket, because reactors can produce a lot of power in a very small volume and produce thrusts comparable to chemical rockets. One was actually developed and tested in the early 1960s but was then cancelled. None have ever been flown, but it is probably an enabling technology for interplanetary human space flight.

[sts60]

At 120 000 m altitude there is a thin atmosphere with nitrogene and oxygene atoms that you collide with and ... MAGIC

No, it's called "aerodynamic drag", known to laymen like you - let's drop the pretense that you are any kind of engineer - as "friction".

... suddenly

Not "suddenly".  And the literature has reams and reams of research into atmospheric braking; in fact, there are file cabinets full of it - decades of work - in my office, in this case pertaining to the atmospheric reentry of isotope heat sources.

you slow down to 100 m/s (at say 5 000 m altitude) and deploy a parachute and land. In a desert in Kazakstan. Where nobody lives. In the middle of nowhere!

Or you glide to a stop at the Shuttle Landing Facility on Merritt Island after making a precision approach - in fact, I was at the first night landing at KSC.  Where you land depends on which kind of blunt lifting body technology you choose to employ - the tradeoffs of which are the kinds of things engineers do.

Imagine that - you slow down from 9 000 m/s to 100 m/s just by colliding with atoms. But why don't you slow down to 0 m/s by colliding with atoms?

Terminal velocity, as any reasonably-educated high-school student should know, happens when aerodynamic drag balances gravitational acceleration.   Again, let's drop the silly pretense that you are an engineer when you display such ignorance.

Let me ask a stupid question or two? Why do you need a parachute at the end? What is wrong with colliding with atoms to the end?

For the same reason a skydiver needs a parachute.  And, if you choose a glider-type approach, you don't need a parachute at all - the Shuttle only added a drag chute on rollout as an extra safety measure, and had landed just fine without one.  Again, you have no idea what you are talking about.

When you dive from a 10 m board you do not need a parachute.

When you dive from a 1000 m platform you do.  Let alone from 10⁵ m.

After making yet another staggeringly ignorant post, Heiwa, and having your head handed to you yet again, one has to ask - don't you ever get embarrassed by being so routinely and spectacularly wrong?  Me, I would be embarrassed, and would regroup and try to understand what I was talking about rather than keep making a fool of myself, but you just keep right on talking.

Don't you have any desire to actually learn anything, rather than to stubbornly repeat your errors and keep demonstrating your ignorance?

Or are you just saying stupid things to get attention?

[sts60]

One non-chemical engine with a lot of promise is the nuclear thermal rocket, because reactors can produce a lot of power in a very small volume and produce thrusts comparable to chemical rockets. One was actually developed and tested in the early 1960s...

Well, as it turned out it was merrily using the reactor core as additional reaction mass.  But that's another story.

[Not Myself]

Imagine that - you slow down from 9 000 m/s to 100 m/s just by colliding with atoms. But why don't you slow down to 0 m/s by colliding with atoms?

You do.  That's why when the engines on an aeroplane go out, it just hangs there, and they have to send someone up to rescue the occupants.

[Peter B]

Heiwa, as I am not an engineer I imagine you are not interested in reading what I have to say, but I think it is still worth saying: even I can see that there is nothing inherently implausible in the ideas of: (a) using the atmosphere to slow down, (b) having a spacecraft's trajectory altered by gravity, (c) firing a rocket engine to change a spacecraft's trajectory, or (d) being able to calculate the effects of these accurately enough to send a spacecraft to the Moon and successfully retrieve it.

As mentioned, I'm not an engineer, but I'm familiar enough with the concept of separating the direction the spacecraft is travelling from the direction it's pointing. One of the games I play with my sons when they're riding in a shopping trolley is to spin the trolley as I walk down the aisle of the supermarket. Depending on the trolley, in the best case I can make the trolley complete a 360 degree spin and have the handle return to my hands without me having to change the speed or direction I'm walking.

Accordingly, the idea that a spacecraft could be made to face backwards to use its engine to slow down into orbit or to head for land just seems completely unremarkable. I seriously don't understand how someone claiming to be an engineer could fail to Get It.

Incidentally, Heiwa, your objections are all very interesting, but you then have to explain how NASA has possession of ~380 kilograms of material from the Moon. But perhaps that's better suited to another thread.