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Disposal of radioactive materials

Posted by: slycker - Thu Dec 02, 2004 10:33 pm
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Disposal of radioactive materials 
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Post    Posted on: Thu Dec 23, 2004 3:58 pm
Ekkehard Augustin on December 16 wrote:
To return to "tangential velocity" and the second component: In the case of an ellipse the gravity vector isn't vertical to the tangential velocity - as I explained earlier in this thread. But the parallelogram still is there and the two components can be separated. All is more difficult to calculate because the coordinate system has no orthogonality except at aphelion and perihelion and the gonality of the coordinate system is changing permanently because the angle between the two vectors is changing permanently.

I believe you are handling the second component incorrectly, but since you didn’t post any math about how the second component would be calculated it would be hard for me to point out any error in it. However, if your calculation of both the tangential and second component results in a spiral path then there is an error in it someplace.

(EDIT) I may be wrong in my statement above. I see from looking back over the previous posts that you have not yet actually calculated an orbit into the Sun. I will be very interested to see the result of your calculation. I would also like to see your work. That means you should show each step like I have done with my math examples.


Last edited by campbelp2002 on Thu Dec 23, 2004 11:03 pm, edited 3 times in total.



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Post    Posted on: Thu Dec 23, 2004 4:11 pm
Ekkehard Augustin wrote:
Mathematics can convince me because I do understand mathematics very good - so that's really no problem and has never been.
….
Please point directly to the error or mistake I made seen from your point of view - that would be helpful for this discussion.

Your error is having the idea that a spiral orbit is possible. Conservation of energy says it is not possible.
Your error is attempting to use complicated vector math to prove your point. Complicated math is often used incorrectly to prove false statements. Here is an example:
http://mcraefamily.com/MathHelp/JokeProofFactoring.htm
Errors in such math can be found, but in our case it would be easier if you just understood the physics of orbits as related to conservation of energy better. Or if you understood the funnel example better.

(EDIT) If the result of your calculation is a correct orbit I will agree you understand the math of orbits. But you can't possibly understand the physics of orbits if you still believe an object can spiral into the Sun.


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Post    Posted on: Fri Dec 24, 2004 9:59 am
Hello, Peter,

you posted a chain of answers now which increases the time required to answer to it all and I have to establish sprreadsheets etc. to calculate my mathematics for millions of seconds.

So I do answer to your first anser only: The thrust, push or pulse will point to the sun directly. Your question after km and meters is puzzling me a little bit - 7 km/s is identical to 7,000 meters per second. I don't calculate miles if that were the real question.

Regarding your new topic you posted it doesn't have to do with the sun but with a gravitational center in principle and there really is agravitational center that a vehicle had been spiralling down to. The ferries to the moon had velocity relative to the lunar surface after releasing from Apollo and while going down to the surface. Short before going down entirely the velocity relative to surface had been slowed down.



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Post    Posted on: Fri Dec 24, 2004 2:03 pm
OK, I thought it was toward the Sun.
Now the Earth's orbit around the sun is at about 29.5 km/s and your 7 km/s is 90 degrees away from the Earth's motion so it is simple to get the resulting speed in a coordinate system centered on the Sun. It is:

SQRT(29.8^2 + 7^2) = 30.6 km/s

So your speed in a Sun centered coordinate system is 30.3 km/s.
Do you Agree?

The spacecraft is still near Earth so it is about 150,000,000 km from the Sun.
Do you agree?

Now we can can calculate the orbital energy about the Sun from the orbital energy equation.

E = (V^2 / 2) - (U / R)

E = potential + kinetic energy
V = speed (not velocity because direction is not included)
R = distance from the Sun
U = the standard gravitational parameter = Gm = 132,712,440,000 for the Sun (according to http://en.wikipedia.org/wiki/Standard_g ... _parameter)

This equation is a standard orbital mechanics equation I got from:
http://en.wikipedia.org/wiki/Specific_orbital_energy
Do you accept the validity of this equation?

Entering the numbers above gives:
E = (30.6^2 / 2) – (132,712,440,000/150,000,000)
E = 468.18 – 884.7496
E = -416.5696

Notice that the energy is negative. This is always the case for orbits below escape velocity.

Now when you calculate your orbit I want you to tell me the speed and distance from the sun at some point near the Sun so I can plug the values into this equation. If your speed and distance do not give the same E then your orbit has violated conservation of energy.


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Post    Posted on: Sun Dec 26, 2004 4:23 pm
Hello, Peter,

I explained the way I calculate the second component and you did the calculation and posted the results. Your result has been correct - but you never said that my mathematics were incorrect and you never posted the step of my mathematical way that you consider to be incorrect and you never posted a mathematical proof of the error.

That's one point to list here. There are others too: 1. My issue that a spiral course is possible is not the beginning is not the beginning of my considerations etc. but the result of theoretical conclusions without calculating numbers. I used simply vectors - and I did it like this: The movement along an orbit can be explained by two vectors - the one pointing to the sun and the other tangential to the orbit just at the object orbiting. Both vectors ae in equilibrium. Now this equilibrium can be destroyed in favor for the vector pointing to the sun. As long as no new equilibrium is reached the object will approach to the sun but it will not go straight to the sun because the second vector that has been tangential to the orbit still is there. But sun will pull the object to itself by the second vector pointing to the sun. This vector is increasing while the object approaches sun - the vector is increasing by square or by cube (one of my books says by square). If this vector becomes larger than the tangential vector after a sufficiently short period no equilibrium seems to be possible. I did get spirals earlier but only used grahics that time instead of numbers. The question isn't if spirals are possible but wether they are possible in the case of vehicles launched from earth - i.e. under the real concrete conditions Earth and the solar system are providing to vehicles.

To repeat it - it's not an idea but a result which has to be checked. I don't believe it - I have seen a spiral. Now I'm checking it.

You say "If the result of your calculation is a correct orbit I will agree you understand the math of orbits" - that's no valid criterion. It's dogma and ideology - the result itself cannot be the criterion of right or wrong - the manner and the way of calculation and the logicy are the only valid criteria.

Please leave away te funnel - the funnel isn't a system in whih gravity etc. are working like in the solar system - a funnel is an illustratin only an analogon like acceleration is an analogon to gravity seen from Theory of Relativity.

Your energy calculation seems to be wrong. I didn't recalculate it because I'm working on my own calculations but you didn't include all energies involved and yu didn't include the interchanges of energy that are going on while launching or accelerating a vehicle.

You seem to be trying to argue based on constraints that don't exist in my mathematics - cosntraints not of physical nature but constraints concerning your assumptions about what I'm calculating.

Using my mathematics I have ready three steps - representing one second each. I have to write some text around each step.

Orbital energy is just the energy of an object in orbit - leaving an orbit means changes of energy. And everyone knows that objects can leave their orbits. They cannot do that of their own - but I never stated that they will do that. I allways stated an external impulse.

One of the most recent examples of a vehicle spiraling is Smart1 - it spiraled away from Earth and it spiraled down to a low lunar orbit. The drive didn't work constantly but gave only a pulse time by time..



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Post    Posted on: Mon Dec 27, 2004 8:16 am
Step One:

A vehicle being on Earth is orbiting sun by the velocity of Earth which means that the two components of velocity are identical to the two components of Earth's velocity.

Bradley C. Edwards in a special study to be read at NIAC's website proposed a space elevator going up to an altitude of 100,000 km to be able to launch vehicles deep intothe interplanetary space from that altitude. Scientists recently achieved significant progresses in the production of nanocarbontubes and said that mass production of these tubes is closer than ten years.

So I assume that the vehicle carrying the radioactive materials will be borught up to 100,000 while being docked to the space elevator. Then it will be orbiting Earth but it will keep orbiting Sun by the velocity of Earth and its two velocity components still will be identical to Earth's components.

The diameter of the 100,000 km high orbit then is Pi * (2 * 100,000 km + 12,756 km) = 668392,69 km (rounded). The outer edge of the elevator where the vehicle will be waiting to be launched will orbit Earth by 7.74 km/s (rounded).

Now if the vehicle is released in the correct moment to be launched towards the sun directly it will get a vertical velocity vector of 7.74 km/s in the first second - an acceleration of 7.74 km/s^2.

This only one of several alternatives - interesting too are Magbeam and the pulsed fusion drive and combinations. But the elevator can be used only at earth and only at launch. And it has the advantage of launching a vehicle without consumption of propellant. For this reason I consdier the elavator first. The vehicle gets additional impulse and energy while Earth because of the Energy Conservation Theorem and the Impulse Conservation Theorem loses a little bit Energy and a little bit impulse.

To get the velocity etc. of the vehicle after this launch I recalculated the values for Earth. I used a year of 365.256 days equal to 31558118.4 seconds which means the Earth to go an angel of 0.0000114075242204554° per second, two angles of 89.9999942962378000000° at the begiining and the end of the distance the Earth is going per second. The cosinus of this angle multiplied by Earth's orbital velocity of 29,8 km/s as hypothenusis is resulting in 0.0000029665729715044 km/s vertical and the sinus of this angle multiplied by 29.8 km/s as hypothenusis is resulting in 29.7999999999998000000 km/s tangential.

The vertical 0,0000029665729715044 km/s have to be added to the acceleration got by the elevator. I used the non-rounded acceleration and got 7.7360294319335000000 km/s which again rounds up to 7.74 km/s vertical - tangetial has not been modified and so is constant.

So after this first second the vehicle is going to 75.45° (rounded) by a resulting velocity of 30.79 km/s (rounded). The resulting velocity hasn't been increased very much in comparison to Earth's orbital velocity - but the angle is more than 14° closer than the angel of Earth and that really is much. It means that the vehicle is approaching to the sun now which Earth is not doing.

This is the result after the first second. I will continue with the next second later - perhaps I will consider special results requiring a yet to me unknown number of seconds first and then return to second-by-second-calculation.



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EDIT: Peter, supposedly I know the reason why you assumed earlier in this post that I had to use the cosinus of twice the angle between the direction - the vertical velocity I got by trigonometrics is half the acceleration of Earth towards the sun. I'm thinking about it.


Last edited by Ekkehard Augustin on Wed Dec 29, 2004 1:25 pm, edited 1 time in total.



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Post    Posted on: Tue Dec 28, 2004 2:08 pm
Step Two:

The angle of more than 75° is much closer to the direction of sun than the angle between this direction and the direction Earth is moving to but compared to the maximum elongation of Venus which is 46,16° (rounded) it still may be have to be reduced once more at least.

Acceleration by the space elevator isn't possible anymore - so something else has to be made use of. I prefer one of the two alternatives I listed.

To save propellant I prefer Magbeam which can be used near Earth only. By Magbeam velocities of 11,6 km/s (rounded) can be achieved. The vehicle would get a vertical component of 19,29 km/s (rounded) vertical then. The tangential velocity again will be kept constant.

Result: The vehicle would be going to 57,08° (rounded) and by 35,50 km/s velocity now. Again the angle has been become much closer - by more than 18° - and the resulting velocity is a littlebit higher than that of Venus now. The angle has to be reduced more.

But there is a problem I have with Magbeam - I don't know the number of seconds required to achieve those 11,6 km/s which I simply added here to the vertical velocity got after the first second.

So I will use the third alternative - the pulsed fusion drive the propellent for the vehicle does carry itself. This drive has been part of the britisc Daedalus project wqhich had a predecessor called Orion in the fifties and sixties of the last century in the US. This Orion would have been able to accelerate by 1 g - 0,00981 km/s^2 which would add up to 35,316 km/s velocity after an hour if the initial velocity were 0 km/s. My informations concerning Daedalus are sounding as if much higher acceleration would be possible - the interstellar Daedalus could achieve 0,13 c within four or five years whreas Orion is said to be interplanetary.

I will not use the simple adding-up or multiplication but calculate the acceleraton of 1 g second by second plus the gravitational acceleration and EDIT the result at the end of this post when it is ready.

And please list what's logically-mathematically wrong with my mathematics and calculations if you detect an error.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)

EDIT: I decided to let the vehicle go without further acceleration for the first time. Additional acceleration will occur when required. If thevehicle would go straight by the angle of more than 75° its course would be tangential to an orbit having a stance of a little bit less than 146 million km after going a little bit less than 15° angle at the side of sun. The time required would be a little bit less than 1.3 million seconds.

But the distance of 146 million km to sun is achieved much earlier (less than half the 1,3 million seconds) because of the gravitational acceleration.

The distance reached after the 1.3 millions seconds really is between 135,045,309 km and 135,045,292 km. The vertical velocity then is a little bit less than 16 km/s, the resulting velocity along course is a little bit than 34 km/s and the gravitational acceleration is 0.00000728271292680563 km/s^2 to 0.0000072827146523346 km/s^2.

The angle to the original line going to the sun at launch of the vehicle is a little bit less than 62° now.

There may be an error involved in my issues - I have to check that now.

To be checked:

1. All the vertical velocity vectors seem to be calculated as if they were parallel to the initial Sun-Earth-Line. If the check shows that this is really so then there are three alternatives - a) recalculation; b) assumption that the direction of the total vertical vector is permanently readjusted by control engines (seems to be problematic); c) recalculation of the non-gravitational portion of the vectors only because the gravitational portion seems to be calculated correctly.


Last edited by Ekkehard Augustin on Tue Jan 04, 2005 2:13 pm, edited 3 times in total.



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Post    Posted on: Fri Dec 31, 2004 4:27 pm
Hi Ekkehard,

I just got back from vacation and it will take some time for me to read all your posts.


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Post    Posted on: Fri Dec 31, 2004 5:39 pm
No problem - I have to repeat my calculations of the second step because the excel file became unusable beacuse it had become to large while it was calculating the secind 65536 seconds.

So there is no second result yet. But I remeber that after the first 65536 seconds the permanetly working pulsed fusion drive had accelerated the vehicle up to more than 650 km/s and the angle between the direction the vehicle was going to and the driection to sun in the first step was 2° only. The vehicle had gone less than 10 million km then - but I don't remeber the number now.

I have been calculating one special alternative a litte bit - the engine isn't required to work permanently.

You see - it's taking time and I had very much to do for the company during the last days.



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Post    Posted on: Sat Jan 01, 2005 5:54 am
Wow.
I have not studied your posts in detail yet but it sounds like you are proposing using several powerful methods to get very high velocities to throw the waste into the Sun. Such power could easily throw it into some other star I think, which is how this long discussion started.


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Post    Posted on: Sat Jan 01, 2005 1:48 pm
Ekkehard Augustin wrote:
I used a year of 365.256 days equal to 31558118.4 seconds which means the Earth to go an angel of 0.0000114075242204554° per second, two angles of 89.9999942962378000000° at the begiining and the end of the distance the Earth is going per second. The cosinus of this angle multiplied by Earth's orbital velocity of 29,8 km/s as hypothenusis is resulting in 0.0000029665729715044 km/s vertical and the sinus of this angle multiplied by 29.8 km/s as hypothenusis is resulting in 29.7999999999998000000 km/s tangential.

Have you tried using 8766.144 hours for the year and 107280.0 km/h as the Earth's orbital velocity? Using one hour instead of one second increments would allow you to complete your calculations 3600 times faster.


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Post    Posted on: Sat Jan 01, 2005 5:08 pm
Throwing the waste into another star this way still will require more time and more propellant than throwing it into the sun.

And the goal is to destroy the waste as quick as possible and by hying as much control of it as possible. This wouldn't be the case if it would be moved to another star the environment of is unknown yet and may require corrections of the course - the other star may have a yet undetetcted planet hosting intelligent life for example.

If I would use hours instead of seconds this wou.d mean to use much larger accelerations, significantly less exactness and less opportunities for fine adjustments.

The alternative I was considering when my excel file became unusable is allowing for less velocity, less consumption of propellant and better calculations. May be I can use a formular by which each of us can calculate each second himself - it doesn't make sense to list tenthousands or millions of seconds here.

But I have to have some looks into it before considering it to be correct and posting it above in the post about the second step by EDIT.



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Post    Posted on: Sat Jan 01, 2005 10:20 pm
Ekkehard Augustin wrote:
Throwing the waste into another star this way still will require more time and more propellant than throwing it into the sun.

No, it would take LESS propellant. That was my whole point in this thread. It takes LESS energy to escape the solar system than to force a object down into the Sun.

However I agree it would take more time. But you said in an earlier post that you didn't care how long it took.


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Post    Posted on: Sun Jan 02, 2005 4:32 pm
There is a misunderstanding - the use of a pulsed fusion drive will require more propellant for throwing it into another star. The maximum possible velocity of a pulsed fusion drive according to Ulrich Walter is 0,12c to 0,14c. The vehicle would need a few decades than to Barnard's Star for example.

Because this never will be required by the pulsed fusion drive to thrwo the waste into the sun the amount of propellant required is much less.

I really don't care how long it will take - I'm doing my calculations only to see one course at least by which the package at the end will fall into the sun without the vehicle going into the sun too.

And I think that we must have as much control as possible over the package and the vehicle - which seems to be possible within our solar system only.



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Post    Posted on: Sun Jan 02, 2005 4:56 pm
There is a misunderstanding and I am not sure how it can be resolved.

You keep bringing up objections that have nothing to do with what I am saying, so I assume you do not understand what I am saying.

Sending waste into the Sun is a bad idea because it takes so much propulsion. I have no argument with any of the OTHER reasons why you want to send waste into the Sun. I only object to the idea that it does not require much propulsion.

I will say it AGAIN. It takes more energy to send an object into the Sun than it does to escape the solar system. The method of propulsion does not matter. Very high speed is not needed. The Pioneer and Voyager space craft have already escaped the solar system. They did not use exotic propulsion or travel very fast, but they have escaped. The same space craft would have required more propulsion or a bigger gravity assist from Jupiter to hit the Sun.


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