Rosetta's Philae probe makes historic landing on comet

Quote from: markjo on December 09, 2014, 06:51:55 AM

So you can't think of any technical or engineering reasons why the Orion mission couldn't be carried out?  Good to know.

There is no scientific descriptions off any kind - incl. peer reviewed thesises and articles - how anything dropping from the sky at 100 000 m altitude with an initial velocity of say 7 500 m/s can slow down by friction in the atmosphere and land safely on the ground. Meteorites will simply burn up. Capsules of any kind will rotate around themselves, break up and disintegrate ... and burn up. The suggestion that a heat shield will assist is just a fairy tale. But if you can show me a heat shield that has passed the test, please do so. Try the Apollo 11 PICA heat shield patented 1965. Why didn't it burn up at re-entry?

It never re-entered anything!

Dropping a capsule with parachutes from a plane at low altitude is just plane stupid. What does it prove?

Quote from: Heiwa on December 09, 2014, 07:26:07 AM

Capsules of any kind will rotate around themselves, break up and disintegrate ... and burn up.

Are you saying that a combination of reaction control thrusters and aerodynamic design aren't sufficient to control a capsule's attitude?

Second of all, you do realize that when a capsule is de-orbited or when the shuttle is de-orbited, they do multiple retrograde burns to slow down right? Or else.. you know.. they would still be in orbit.

Quote from: Jet Fission on December 09, 2014, 11:54:55 AM

Second of all, you do realize that when a capsule is de-orbited or when the shuttle is de-orbited, they do multiple retrograde burns to slow down right? Or else.. you know.. they would still be in orbit.

Of course I know this. And neither a capsule nor a Shuttle has sufficient fuel aboard to do retrograde burns to slow down at re-entry. They are going much too fast. You cannot stop any object with speed >7 000 m/s using a rocket engine/fuel aboard and friction against atmosphere just burns you up.

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ! And the same amount of energy is required to slow it down to 0 m/s. Imagine the amount of energy required to stop an 8 500 kg Orion capsule.

I am amazed at all participants believing in space travel above and that re-entry is possible.

By the way, I can't find where your website is listed on your profile.  Could you please give me a link to it, because i am really interested to know why you think that a space shuttle is aerodynamically unstable.  It is really back heavy because of the engines, which is why the wings have to be so far back to be at the center of mass, which makes the shuttle look like it can't fly, but there are " class="bbc_link" target="_blank" rel="noopener noreferrer">videos of space shuttles landing and spectators are invited to these events.

Quote from: mikeman7918 on December 09, 2014, 06:37:31 PM

By the way, I can't find where your website is listed on your profile.  Could you please give me a link to it, because i am really interested to know why you think that a space shuttle is aerodynamically unstable.  It is really back heavy because of the engines, which is why the wings have to be so far back to be at the center of mass, which makes the shuttle look like it can't fly, but there are " class="bbc_link" target="_blank" rel="noopener noreferrer">videos of space shuttles landing and spectators are invited to these events.

Thanks for asking! Start reading chapters 1.12+ and 2.3+ at http://heiwaco.com/moontravel.htm and you'll find out my thoughts about capsules and shuttles returning from space trying to land on Earth ... and why it is not possible.

Thanks for asking! Start reading chapters 1.12+ and 2.3+ at http://heiwaco.com/moontravel.htm and you'll find out my thoughts about capsules and shuttles returning from space trying to land on Earth ... and why it is not possible.

You did mention that you had no idea what a plasma trail is, so I will explain that first.  Plasma is a state of matter that you get by heating up a gas, it is any matter that is so hot that the atomic nuclei can't hold on to any electrons and it also glows, and it is basically the glowing part of a fire.  Objects reentering the atmosphere leave a trail of plasma because of all of the heat involved that glows and can be seen from the ground, it's the reason that meteors (shooting stars) appear to glow as they streak across the sky. 

One thing that you demonstrated a lack of understanding on is how the space shuttle's windows do not melt during reentry, and the reason for that is that the nose of the shuttle takes a lot of the beating because of the angle that it reenters at and also the windows are at an angle and so they don't really hit a lot of the incoming air and the air they do hit hardly get's deflected and pressurized, and the rapid pressurization is where all of the heat comes from.  You also never mentioned why you thought that the picture from the ISS of the shuttle reentering was faked other then you thought that it looked like it and it conflicts with your theory.

Quote from: mikeman7918 on December 09, 2014, 07:52:38 PM

You did mention that you had no idea what a plasma trail is, so I will explain that first.  Plasma is a state of matter that you get by heating up a gas, it is any matter that is so hot that the atomic nuclei can't hold on to any electrons and it also glows, and it is basically the glowing part of a fire.  Objects reentering the atmosphere leave a trail of plasma because of all of the heat involved that glows and can be seen from the ground, it's the reason that meteors (shooting stars) appear to glow as they streak across the sky. 

One thing that you demonstrated a lack of understanding on is how the space shuttle's windows do not melt during reentry, and the reason for that is that the nose of the shuttle takes a lot of the beating because of the angle that it reenters at and also the windows are at an angle and so they don't really hit a lot of the incoming air and the air they do hit hardly get's deflected and pressurized, and the rapid pressurization is where all of the heat comes from.  You also never mentioned why you thought that the picture from the ISS of the shuttle reentering was faked other then you thought that it looked like it and it conflicts with your theory.

Plasma - according Wikipedia, LOL, it is loosely described as an electrically neutral medium of unbound positive and negative particles (i.e. the overall charge of a plasma is roughly zero) and I doubt it leaves a trail after a Shuttle. Plasma trails are just SF nonsense - sometimes/often I am quite ironic and sardonic about the NASA lingo.

If you believe the Shuttle cockpit windows can resist hitting atmosphere at 7 000 m/s velocity producing a plasma trail, be my guest. Solid water hitting the flat bow of a ship at 10 m/s velocity makes a lot of noise, I know from experience. The Shuttle structure would not resist 300-400 m/s velocity in air and break apart at a little higher speed. It is 78 000 kg of junk! Probably less but junk all the way.

I have read the specified articles on your website and I have noticed some issues with your reasoning.

On the first article, there was a distinct lack of math and instead there were a lot of "I have the feeling", which makes most of your conclusions there based off of an assumption rather then any actual evidence.  You also failed to state your sources for the information that needed it the most like the "NASA can't explain this" statements.  Math is an amazing thing because you can use it to put logic on paper, so please use that instead of your feelings.

Quote from: mikeman7918 on December 09, 2014, 07:52:38 PM

You did mention that you had no idea what a plasma trail is, so I will explain that first.  Plasma is a state of matter that you get by heating up a gas, it is any matter that is so hot that the atomic nuclei can't hold on to any electrons and it also glows, and it is basically the glowing part of a fire.  Objects reentering the atmosphere leave a trail of plasma because of all of the heat involved that glows and can be seen from the ground, it's the reason that meteors (shooting stars) appear to glow as they streak across the sky. 

One thing that you demonstrated a lack of understanding on is how the space shuttle's windows do not melt during reentry, and the reason for that is that the nose of the shuttle takes a lot of the beating because of the angle that it reenters at and also the windows are at an angle and so they don't really hit a lot of the incoming air and the air they do hit hardly get's deflected and pressurized, and the rapid pressurization is where all of the heat comes from.  You also never mentioned why you thought that the picture from the ISS of the shuttle reentering was faked other then you thought that it looked like it and it conflicts with your theory.

Plasma - according Wikipedia, LOL, it is loosely described as an electrically neutral medium of unbound positive and negative particles (i.e. the overall charge of a plasma is roughly zero) and I doubt it leaves a trail after a Shuttle. Plasma trails are just SF nonsense - sometimes/often I am quite ironic and sardonic about the NASA lingo.

If you believe the Shuttle cockpit windows can resist hitting atmosphere at 7 000 m/s velocity producing a plasma trail, be my guest. Solid water hitting the flat bow of a ship at 10 m/s velocity makes a lot of noise, I know from experience.

The Shuttle structure would not resist 300-400 m/s velocity in air and break apart at a little higher speed.

It is 78 000 kg of junk! Probably less but junk all the way.

It is 78 000 kg of junk! Probably less but junk all the way.

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ!

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ!

Can you do the math for a starting mass of 1 kg at 0m/s  and a final mass of 0.5 kg at 7000 m/s considering a constant rate of mass loss ?

And by the way, what would be the final speed and energy of a rocket weighting 1000 kg , including  500 kg of fuel, burning it with an ejection speed of 4500 m/s ? (commmon value for oxygen/hydrogen rocket engines)

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ!

Can you do the math for a starting mass of 1 kg at 0m/s  and a final mass of 0.5 kg at 7000 m/s considering a constant rate of mass loss ?

And by the way, what would be the final speed and energy of a rocket weighting 1000 kg , including  500 kg of fuel, burning it with an ejection speed of 4500 m/s ? (commmon value for oxygen/hydrogen rocket engines)

Quote from: Antonio on December 10, 2014, 12:47:55 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ!

Can you do the math for a starting mass of 1 kg at 0m/s  and a final mass of 0.5 kg at 7000 m/s considering a constant rate of mass loss ?

And by the way, what would be the final speed and energy of a rocket weighting 1000 kg , including  500 kg of fuel, burning it with an ejection speed of 4500 m/s ? (commmon value for oxygen/hydrogen rocket engines)

That doesn't matter so much as the shuttle didn't reduce it's speed to 0 m/s for rentry.

Assuming a fully fuel laden shuttle massing around 585,000 kg and carrying 103,000 kg of LH₂ in a low earth orbit at 7,800 m/s. H₂ has a combustion energy value of 141,000 KJ/kg. The shuttle needs to reduce it's orbital velocity from 7,800 m/s to around 4,700 m/s for re-entry.

So:
Energy required for de-orbit = 1/2 x 585,000 x (7,800-4700)² = 2.81 TJ

Energy available = 103,000 x 141,000 = 14.5 TJ

That's enough energy for 5 de-orbit burns. Even if you reduce the fuel to 25% of capacity it would still amount to over 3 TJ of energy, still enough to deorbit without even correcting the mass of the shuttle.

Quote from: Antonio on December 10, 2014, 12:47:55 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

Do you know how much energy is required to accelerate a mass 1 kg to 7 000 m/s speed? Right 24.5 MJ!

Can you do the math for a starting mass of 1 kg at 0m/s  and a final mass of 0.5 kg at 7000 m/s considering a constant rate of mass loss ?

And by the way, what would be the final speed and energy of a rocket weighting 1000 kg , including  500 kg of fuel, burning it with an ejection speed of 4500 m/s ? (commmon value for oxygen/hydrogen rocket engines)

That doesn't matter so much as the shuttle didn't reduce it's speed to 0 m/s for rentry.

Assuming a fully fuel laden shuttle massing around 585,000 kg and carrying 103,000 kg of LH₂ in a low earth orbit at 7,800 m/s. H₂ has a combustion energy value of 141,000 KJ/kg. The shuttle needs to reduce it's orbital velocity from 7,800 m/s to around 4,700 m/s for re-entry.

So:
Energy required for de-orbit = 1/2 x 585,000 x (7,800-4700)² = 2.81 TJ

Energy available = 103,000 x 141,000 = 14.5 TJ

That's enough energy for 5 de-orbit burns. Even if you reduce the fuel to 25% of capacity it would still amount to over 3 TJ of energy, still enough to deorbit without even correcting the mass of the shuttle.

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

And by the way, what would be the final speed and energy of a rocket weighting 1000 kg , including  500 kg of fuel, burning it with an ejection speed of 4500 m/s ? (commmon value for oxygen/hydrogen rocket engines)

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

Quote from: Heiwa on December 12, 2014, 02:39:24 AM

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

lol

So you haven't done the maths?

Quote from: JimmyTheCrab on December 12, 2014, 04:23:26 AM

Quote from: Heiwa on December 12, 2014, 02:39:24 AM

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

lol

So you haven't done the maths?

No, I have done the maths in my head as usual - one result 5362, one 5256. About the same. Please use your brain in your head when calculating. How else can you do maths?

Quote from: JimmyTheCrab on December 12, 2014, 04:23:26 AM

Quote from: Heiwa on December 12, 2014, 02:39:24 AM

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

lol

So you haven't done the maths?

No, I have done the maths in my head as usual - one result 5362, one 5256. About the same. Please use your brain in your head when calculating. How else can you do maths?

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

Quote from: JimmyTheCrab on December 12, 2014, 04:23:26 AM

Quote from: Heiwa on December 12, 2014, 02:39:24 AM

Quote from: Antonio on December 12, 2014, 02:18:21 AM

Quote from: Heiwa on December 12, 2014, 01:51:32 AM

Quote from: Antonio on December 12, 2014, 12:47:14 AM

Quote from: Heiwa on December 09, 2014, 06:16:43 PM

If the initial speed of mass 1000 kg was 7000 m/s, I suggest the final speed of the remaining 500 kg is 5362 m/s (after dropping off 500 kg fuel as hot gases), i.e. a 23.4% speed reduction. It is not easy to brake or slow down in space.

Care to give the maths for this result please?

Pls do it yourself! Tip - deduct the kinetic energy of fuel, which is lost into space while braking, from the original kinetic energy of the 1000 kg mass to obtain the kinetic energy of the mass 500 kg with no fuel left aboard.

I'm finding very different results, that's why I'm kindly requesting your math, as I suspect some mistakes from it

There is only one result - 5362 m/s (or maybe 5256 m/s?). If you get other results (sic) you are mistaken. Try again!

lol

So you haven't done the maths?

No, I have done the maths in my head as usual - one result 5362, one 5256. About the same. Please use your brain in your head when calculating. How else can you do maths?

Pencil, paper and calculator Heiwa. Thats the only way to do it so you can then go back over it and verify you didnt make any errors.

No wonder you come up with such wildly inaccurate numbers.

Quote from: Lemmiwinks on December 12, 2014, 10:18:23 AM

Pencil, paper and calculator Heiwa. Thats the only way to do it so you can then go back over it and verify you didnt make any errors.

No wonder you come up with such wildly inaccurate numbers.

Paper and pen? Calculator?? Inaccurate numbers??? What's wrong with a fresh brain and using it for simple calculations? Anyway, plenty of people above think they know the answer and are therefore invited to present the calculations to reach it. It is simple astrophysics:
A. Calculate the kinetic energy of the total mass 1000 kg spacecraft at 7000 m/s velocity relative Earth in space (mv²/2).
B. Calculate the kinetic energy of the fuel mass 500 kg after it has been ejected from the spacecraft as hot gas at velocity 4500 m/s relative the spacecraft polluting the space at 2500 m/s velocity relative Earth and having slowed down the space craft to unknown speed.
A-B is then the kinetic energy of the 500 kg spacecraft after braking (ejection of 500 kg fuel as hot gas), speed of which is easy to find (sqr((2(A-B)/500))).