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peron
2010-Jul-14, 05:12 PM
If I have a 10km sail at L5, how much protons from the sun will hit it's surface?
I know their is about 82 ions per cubic inch, how much of those are protons I have no idea.

Some help, even better if someone can show me how to do the calculation (I suck at math.)

Thanks.

tusenfem
2010-Jul-14, 05:36 PM
The solar wind ion density is a few per cubic centimeter (at the moment 4)
The solar wind speed near Earth is about 350 km/s
So, now you can calculate the particle flux through a square cm.
The proton component of the solar wind is roughly 95%.

So, I guess that would solve something.....

peron
2010-Jul-15, 12:15 AM
So, what does the speed of the ions have to do with anything?

Andrew D
2010-Jul-15, 12:25 AM
I suck at math.

Math is about effort, not ability! No one was born a mathematician, don't give up on yourself!

peron
2010-Jul-15, 02:13 AM
Well, I'm trying to calculate a new idea I have. I was thinking capturing the protons in the solar wind to make hydrogen, now I'm trying figure out how much protons I would need to capture to make a reasonable amount of hydrogen.

tusenfem
2010-Jul-15, 06:45 AM
So, what does the speed of the ions have to do with anything?

Well the speed will give you the flux of protons through a square centimeter, i.e. how many protons per second will "impact" one square centimeter of your solar sail, so it is rather important in your quest.

tusenfem
2010-Jul-15, 06:46 AM
Well, I'm trying to calculate a new idea I have. I was thinking capturing the protons in the solar wind to make hydrogen, now I'm trying figure out how much protons I would need to capture to make a reasonable amount of hydrogen.

Of course then you also need to capture or generate electrons to neutralize the hydrogen ions.

Dale Botha
2010-Jul-15, 09:25 AM
Math is about effort, not ability! No one was born a mathematician, don't give up on yourself!

I absolutely agree! Sometimes those who suck at math are missing some vital earlier fundamental which makes it all seem like giberish!

peron...If you want to do the math you need to know (or assume) a few things first:

Energy = mass x velocity (per particle)

mass of the particles: assume the mass of a single proton
velocity = 350 000 meters/second

I'm not sure what "flux" is but then you would need to know how many protons per second.

4 per second per cubic centimeter which at 350 000 000 cm/s is 350 000 000 x 4 = 14 000 000 000 per second per square centimeter.

1.4 x 1010 X mass of proton X speed = momentum transfered per second...

now I'm a little lost as to how to convert this type of interaction in to f=ma type of situation...what is the missing equation with time=t as a variable?

anybody?

Ken G
2010-Jul-15, 12:54 PM
Well, I'm trying to calculate a new idea I have. I was thinking capturing the protons in the solar wind to make hydrogen, now I'm trying figure out how much protons I would need to capture to make a reasonable amount of hydrogen.Then a fact that might interest you is that the energy carried by the luminosity of the Sun is about the same as the rest energy carried in its wind. Since you can only get about 1% of the rest energy of hydrogen through fusion, that means even if you are going to fuse that hydrogen, you can still only get 1% of the energy that already strikes Earth in the form of sunlight. If you are planning to extract the chemical energy from the hydrogen by combining it with oxygen, that's about a million times less energy than fusion. So the "hydrogen-powered car" kind of chemical energy you could extract from the solar wind that strikes Earth is about a hundred million times less than the sunlight that strikes Earth. So that's a pity, but there you have it.

peron
2010-Jul-15, 07:06 PM
I absolutely agree! Sometimes those who suck at math are missing some vital earlier fundamental which makes it all seem like giberish!

peron...If you want to do the math you need to know (or assume) a few things first:

Energy = mass x velocity (per particle)

mass of the particles: assume the mass of a single proton
velocity = 350 000 meters/second

I'm not sure what "flux" is but then you would need to know how many protons per second.

4 per second per cubic centimeter which at 350 000 000 cm/s is 350 000 000 x 4 = 14 000 000 000 per second per square centimeter.

1.4 x 1010 X mass of proton X speed = momentum transfered per second...

now I'm a little lost as to how to convert this type of interaction in to f=ma type of situation...what is the missing equation with time=t as a variable?

anybody?

Flux is the amount of particles passing a area at a given distance, governed by the inverse square law.

Dale Botha
2010-Jul-15, 11:26 PM
Ok, cool! Do you know to converted this particles per second type of equation to the more familiar f=ma equation to work out the actual force generated and thereby work out the acceleration achieved?

Ken G
2010-Jul-16, 02:15 AM
There's really no need to apply f=ma here, the wind by this point is more or less just streaming at constant speed. If you want the particle flux, it's easier just to measure it, and we have satellites that do just that.

mugaliens
2010-Jul-16, 04:28 AM
If you are planning to extract the chemical energy from the hydrogen by combining it with oxygen, that's about a million times less energy than fusion. So the "hydrogen-powered car" kind of chemical energy you could extract from the solar wind that strikes Earth is about a hundred million times less than the sunlight that strikes Earth. So that's a pity, but there you have it.

Meanwhile, allowing for the solar-->electricity-->hydrogen conversion losses (total, approximately 90%), available land masses (1%), you're still looking at a solar-->electricity-->hydrogen converation capability that's 100,000 times greater than trying to capture hydrogen from the solar wind.

Ken G
2010-Jul-16, 05:03 AM
Yes, it's never going to be feasible. And if we find a way to fuse hydrogen, we'll have plenty of energy to extract it from regular stuff (like water).

peron
2010-Jul-16, 06:05 PM
Ok, cool! Do you know to converted this particles per second type of equation to the more familiar f=ma equation to work out the actual force generated and thereby work out the acceleration achieved?

No I do not.


There's really no need to apply f=ma here, the wind by this point is more or less just streaming at constant speed. If you want the particle flux, it's easier just to measure it, and we have satellites that do just that.

Oh, is their a link to this information?

Thanks.

Ken G
2010-Jul-17, 12:53 AM
Oh, is their a link to this information?Start anywhere (Google?), what I said rests on nothing but the most basic possible information about the solar wind. A long way from the Sun, there's nothing left to either accelerate or slow the flow, so it is traveling radially at pretty constant speed. There's no way to calculate what the particle flux should be, we don't understand well enough what processes put heat into that pressure-driven flow. So we just measure it, and you've already been given average values. What's more, you can skip over the particle flux and go right to the rest energy flux-- which is similar to the energy flux in sunlight, as I mentioned. So the idea is not feasible, because it's much easier to extract energy from sunlight than from the rest energy of solar wind particles. But it's a cute thought, because you are right that the big problem with "hydrogen (chemical) energy" is the absence of a decent source of free hydrogen.

peron
2010-Jul-17, 02:19 AM
Start anywhere (Google?), what I said rests on nothing but the most basic possible information about the solar wind. A long way from the Sun, there's nothing left to either accelerate or slow the flow, so it is traveling radially at pretty constant speed. There's no way to calculate what the particle flux should be, we don't understand well enough what processes put heat into that pressure-driven flow. So we just measure it, and you've already been given average values. What's more, you can skip over the particle flux and go right to the rest energy flux-- which is similar to the energy flux in sunlight, as I mentioned. So the idea is not feasible, because it's much easier to extract energy from sunlight than from the rest energy of solar wind particles. But it's a cute thought, because you are right that the big problem with "hydrogen (chemical) energy" is the absence of a decent source of free hydrogen.

Yeah, in ten cubic miles, there is about -20 sextillion kg of protons by weight. At least thats what my calculation shows.
I thought I had a ground breaking idea. :) Well, I guess you can't win all of the time.

Thanks.