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Can you move in space?


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5 hours ago, MPMin said:

Thus F = 0.02/10

In addition to questions by other members above:

Is the rig lifted into space using conventional methods and then accelerated by force that is provide by the rig?

I think you are mixing things. Force is not related to time the way you use it. If I push with 1N of force for 1 miniute or for 1 hour does not matter, it is still 1 Newton. Maybe you are looking for some other quantity?

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9 hours ago, Mordred said:

Secondly 0.02 N won't move a 250 kg craft

In the context that’s been set here, are you sure? 

3 hours ago, Ghideon said:

In addition to questions by other members above:

 Is the rig lifted into space using conventional methods and then accelerated by force that is provide by the rig?

I think you are mixing things. Force is not related to time the way you use it. If I push with 1N of force for 1 miniute or for 1 hour does not matter, it is still 1 Newton. Maybe you are looking for some other quantity?

Yes the rig is lifted into space by conventional methods and then accelerated by the force generated by the rig. 

Perhaps I am missing something or not understanding the basic concept of physics, I thought that a = F/m, where a constant force is applied to a mass it causes it to accelerate, the longer you accelerate a given mass the faster it goes over time thus time does matter. If this is not correct please explain what it should be. 

 

6 hours ago, Strange said:

I think you should read back. The force on the craft is not the same as the force on the wires.

As the wires are part of the craft, how is this so? 

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Positive.

Look up the definition of the unit Newton. "the SI unit of force. It is equal to the force that would give a mass of one kilogram an acceleration of one meter per second per second, and is equivalent to 100,000 dynes."

 

I seem to recall an earlier post comparing wind blowing on a rock lol. Even in space you would require the correct amount of force to move your mass. 

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2 minutes ago, Mordred said:

Look up the definition of the unit Newton. "the SI unit of force. It is equal to the force that would give a mass of one kilogram an acceleration of one meter per second per second, and is equivalent to 100,000 dynes."

Doesn’t your reference mean that a=F/m? 

1ms/s = 1N / 1kg? 

8x10^-6ms/s = 0.002N/250kg? 

Also, if you apply this force over time it will continue to gain momentum.

 

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Perhaps more

27 minutes ago, MPMin said:

Doesn’t your reference mean that a=F/m? 

1ms/s = 1N / 1kg? 

8x10^-6ms/s = 0.002N/250kg? 

Also, if you apply this force over time it will continue to gain momentum.

 

And you wish to counter a combination of the solar wind and Earth's gravity with this ?

Please note Strange,'d comment

Edited by Mordred
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36 minutes ago, Mordred said:

Positive.

Look up the definition of the unit Newton. "the SI unit of force. It is equal to the force that would give a mass of one kilogram an acceleration of one meter per second per second, and is equivalent to 100,000 dynes."

 

I seem to recall an earlier post comparing wind blowing on a rock lol. Even in space you would require the correct amount of force to move your mass. 

Sorry Mordred. That simply is not true. Unless by "correct amount" you mean anything greater than none at all. There is no minimum requirement.

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There is when other forces are present. Ie gravity solar winds etc.You may as well have the astronaut blow through a tube it would probably generate more force.

Edited by Mordred
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4 minutes ago, Mordred said:

You may as well have the astronaut blow through a tube it would probably generate more force.

Definitely.

Just leaving the lights on and blocking all the windows but one would generate more force.

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On 7/17/2019 at 3:04 AM, Strange said:

(As action and reaction are equal and opposite, I suppose you can think of it the other way round. That would be like saying that a rocket goes because the burning fuel pushes on the front of the reaction chamber but not on the back where the exhaust is(*). Which is not wrong, but is just ... weird. So if you want to think that the craft moves because the EMP pushes on the wire, I guess you can. That doesn't stop it being hopelessly inefficient.)

As you said yourself, it’s action and reaction, it’s just a point of view - are you now able to explain how the force on the wires OR the reaction force created by the wires is not the same as the force on the craft? Or do we need to spend another ten pages of my statements being deliberately taken out of context to try to invalidate what I’ve been saying all along! 

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Ok, an attempt to use some very simplified calculations to show that the "invention" will, even if allowed to operate beyond theoretical maximum, generate negative thrust. Assumptions: the invention "magnetic drive" is solar powered requiring a solar panel. Hence radiation pressure* is needed to be accounted for. The drive operates in areas in the solar system where a solar sail such as the one on IKAROS**, is usable.

The radiation pressure experienced by a totally absorbing body is [math]P= \frac{I}{c} [/math] where I is the incident irradiance (usually in W/m2) and c is the speed of light in vacuum.

If the body is reflecting all the incoming radiation back, it will create a recoil; the pressure is doubled. In reality an ideal flat area reflects 90%***

Let’s say a space vehicle with mass m1 has a solar sail with area A. The sail is reflecting 90% of incoming radiation. The acceleration due to the pressure on the sail is [math]a= \frac{1.9IA}{m_{1}c } [/math].

Let’s now say another version of the space vehicle is created. It has same properties but is also equipped with the magnetic drive proposed in this thread. The drive adds mass [math]m_{drive}[/math] and has a solar panel with area A instead of the sail. Panel is 100% effective absorbing all incoming radiation. The acceleration due to the pressure on the panel is [math]a= \frac{IA}{(m_{1}+m_{drive})c } [/math]. 

We assume a few things about the drive to simplify calculations. The solar panel converts 100% of incoming radiation to electricity that is running the drive. The drive is, with zero loss, converting the electricity to EMPs sent out in 360 degrees around a cable. Part of that radiated energy is caught by some means by a second cable, exactly how is unknown and unimportant. We define the amount of EMP radiation caught as [math]\gamma [/math]. Since EMPs are radiated 360 degrees we can see that basically ¼ goes up, ¼ down, ¼ back, ¼ forward****. It is not possible but let’s assume the second cable is able to catch all radiation going in forward direction. That means has the theoretical maximum of [math]\gamma=1/4[/math]. In other words, at least 75% of the energy is wasted, escaping in symmetrical amounts and not affecting the vehicle’s movements.

 

Total acceleration due to the pressure on the panel and the magnetic drive is then   [math]a_{drive} = \frac{IA+IA \gamma }{(m_{1}+m_{drive})c } [/math]

 

Now if   [math]\frac{a_{drive}}{a}>1[/math] then the magnetic drive is adding value. Otherwise not.  

 [math]\frac{a_{drive}}{a}= \frac{\frac{IA+\frac{IA}{4} }{(m_{1}+m_{drive})c }}{\frac{1.9IA}{m_{1}c }}=\frac{\frac{1\frac{1}{4}}{(m_{1}+m_{drive}) }}{\frac{1.9}{m_{1}}}=\frac{1\frac{1}{4}m_{1}}{1.9(m_{1}+m_{drive}) } \approx 0.66\frac{m_{1}}{m_{1}+m_{drive} } [/math] 

So the drive, driven by a solar panel, will slow down the vehicle, not add thrust, compared to a “simple” solar sail of same size as the solar panel.

I had no intentions turning this post in to a textbook, so lots of things covered in earlier posts are left out. Corrections and questions welcome!

*)https://en.wikipedia.org/wiki/Radiation_pressure

**)https://en.wikipedia.org/wiki/IKAROS

***)https://en.wikipedia.org/wiki/Solar_sail. I have not had time to track down better resources. Basic physics seems ok but some parts of article are old

****) simplification to avoid trigonometry


 

Edited by Ghideon
format. Missing word. Disclaimer added. "of same size." added. More clarifications.
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6 minutes ago, Ghideon said:

So the drive will slow down the vehicle, not add thrust, compared to a “simple” solar sail.

That’s all very impressive but how have you concluded this without a figure for the reflective surface area of the craft? 

 

Edited by MPMin
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4 minutes ago, MPMin said:

That’s all very impressive but how have you concluded this without a figure for the reflective surface area of the craft? 

The area of the sail is the same as the area of the solar panel, hence cancelled in the calculations. (edit: added "of same size" to previous post to clarify, good point)

Edited by Ghideon
grammar, misread question
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1 minute ago, Ghideon said:

Yes

 The area of the sail is the same as the area of the solar panel, hence cancelled in the calculations. 

Ok, I get what you are trying say, but to help me understand it better, could you please tell me what the reflective surface area of the craft (or sail, same effect) would have to be to accelerate a 250kg mass at 8x10^-6ms/s with solar radiation? 

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1 minute ago, MPMin said:

Ok, I get what you are trying say, but to help me understand it better, could you please tell me what the reflective surface area of the craft (or sail, same effect) would have to be to accelerate a 250kg mass at 8x10^-6ms/s with solar radiation? 

What value did you get when you tried? Or, where did you get stuck?

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2 minutes ago, Ghideon said:

What value did you get when you tried? Or, where did you get stuck?

I didn’t try. Are you able to provide a numerical answer to my question? 

14 minutes ago, Mordred said:

You are aware the solar wind is of the order 10^-9 Newtons/ m^2 at 1au right ?

This is why I didn’t try 

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3 minutes ago, MPMin said:

I didn’t try. Are you able to provide a numerical answer to my question? 

Something like this;

[math]9,08 \mu N/ m^{2} * a=250kg*8* 10^{-6}  \Rightarrow a\approx 220m^{2}[/math]

That is the theoretical maximum if sail is operating at 100% efficiency near earth orbit where the irradiance (solar constant) value is 1361 W/m2. Note that in reality you account for angle of the sail and direction of travel etc. 

https://en.wikipedia.org/wiki/Solar_sail

 

Edited by Ghideon
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Here is a far more practical design to give you some better ideas and some useful factors to consider. Including formulas.

M2P2 solar magneto sphere spacecraft. 

https://www.google.com/url?sa=t&source=web&rct=j&url=https://pdfs.semanticscholar.org/14f7/2f939b203de184adc83d6473541ce629741a.pdf&ved=2ahUKEwiDipKm7b_jAhV1MH0KHUV9BvoQFjAAegQIAxAB&usg=AOvVaw2X_EaixugX6lDoOPTRZYsP

35 kg of harvested propellant is minor. Some of the speeds attainable with this design is significant

Far better details on this link

https://www.google.com/url?sa=t&source=web&rct=j&url=http://earthweb.ess.washington.edu/space/M2P2/m2p2.PDF&ved=2ahUKEwj118Sh77_jAhUKip4KHS5dCakQFjABegQIBxAC&usg=AOvVaw39j3Uq8yNC1ZieSw31VHUg

Considering that (according to this paper ) it's viable to reach speeds up to 50 to 75 km/s it's a far better alternative than chemical thrusters. (Assuming the papers accuracy)

Even if this design doesn't agree with the OP some of the data within the article will be handy example being the data with regards to the solar winds which maintain significant speeds up to 80 AU

3 hours ago, J.C.MacSwell said:

That is equivalent to claiming no force is required. If unbalanced forces are already present, it's already accelerating.

You still require significant acceleration to combat solar winds and gravity. Think of paddling against a strong river current in a canoe... As a simple analogy. Not to mention the sheer size of our solar system. You don't want to be slower than Voyager 1 but ideally faster.

Edited by Mordred
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