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Question about energy at one wavelength:

Question about energy at one wavelength:

Science

twhitehead

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Originally posted by sonhouse
Krist, who uses such outdated units anyway?
http://mentalfloss.com/article/55895/countries-havent-adopted-metric-system

Next we need to agree on . vs ,

Were you able to view any of my music links I posts over there?
Some of them. To be honest, I am not a big music fan so I didn't watch them all the way through.

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Originally posted by twhitehead
http://mentalfloss.com/article/55895/countries-havent-adopted-metric-system

Next we need to agree on . vs ,

[b]Were you able to view any of my music links I posts over there?

Some of them. To be honest, I am not a big music fan so I didn't watch them all the way through.[/b]
Ok, back to science then๐Ÿ™‚

Hey, I am totally with you on Metric units. Americans are stubborn, set in their ways and just want to keep adding 9 and 3/32 to 5 and 14/128ths forever.

Just like the prohibition against marijuana.

Look how well THAT turned out.....

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Originally posted by twhitehead
Americans ......
I managed to confuse this Greek chap who runs a bar by giving an area as about a metre by a foot.

twhitehead

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Originally posted by sonhouse
Hey, I am totally with you on Metric units. Americans are stubborn, set in their ways and just want to keep adding 9 and 3/32 to 5 and 14/128ths forever.
And nothing is stopping them. The secret is to put metric in the education system so the the next generation grows up with it. Most people who deal with the units in their jobs have to learn both systems anyway, and everyone else can continue with the old units till the next generation replaces them.

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Originally posted by twhitehead
And nothing is stopping them. The secret is to put metric in the education system so the the next generation grows up with it. Most people who deal with the units in their jobs have to learn both systems anyway, and everyone else can continue with the old units till the next generation replaces them.
The thing is that which units I use tends to be a function of what I'm looking at. For science I'll use natural units (c = h = k = 1), for an engineering problem (if I were to look at one) I'd use SI, for computation I'd use abstract units (as is standard), and for every day stuff like cooking I'll use pounds and ounces.

I have the Wikipedia page for light year open as I wanted the conversion to Astronomical Units and noticed something. Entertainingly it's defined in terms of the Julian year (365.25 days) as opposed to the Gregorian year (365.2425 days), this system was adopted in 1984 long after anyone continued to use the Julian calendar...

twhitehead

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Originally posted by DeepThought
The thing is that which units I use tends to be a function of what I'm looking at. For science I'll use natural units (c = h = k = 1), for an engineering problem (if I were to look at one) I'd use SI, for computation I'd use abstract units (as is standard), and for every day stuff like cooking I'll use pounds and ounces.
Well I grew up with and use the metric system in daily life. Even for cooking, the food is marked in grams and litres. My parents were comfortable with both. Most kids wouldn't even know what you were talking about if it wasn't metric.

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Originally posted by twhitehead
Please refrain from using different units all the time lets stick to a particular unit, in this case W/m^2
My figure is based on DeepThoughts origional post where he says that Sirius:
At 1AU the radiance is 1,368 W/m^2
I then use the inverse square law and the distance between the sun and Sirius.

I think I have the distance from Sirius to the sun in ...[text shortened]... e 800 Kilowatts/M^2.

One step at a time. If my first figure is wrong, the rest will be too.[/b]
Just noticed this - the posts were coming thick and fast earlier - I'll check your spreadsheet.

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Originally posted by twhitehead
Please refrain from using different units all the time lets stick to a particular unit, in this case W/m^2
My figure is based on DeepThoughts origional post where he says that Sirius:
At 1AU the radiance is 1,368 W/m^2
I then use the inverse square law and the distance between the sun and Sirius.

I think I have the distance from Sirius to the sun in ...[text shortened]... e 800 Kilowatts/M^2.

One step at a time. If my first figure is wrong, the rest will be too.[/b]
Right I've found three problems. The first is in box B14, you have a typo and are using tanh rather than tan - this makes no difference numerically because of the small value.

In boxes B29 and B30 you are dividing where you should be multiplying - although the ratio comes out right anyway (as you're doing the wrong operation with the same number so the error cancels).

twhitehead

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Originally posted by DeepThought
Right I've found three problems. The first is in box B14, you have a typo and are using tanh rather than tan - this makes no difference numerically because of the small value.
Not a typo. I am working out an angle.

In boxes B29 and B30 you are dividing where you should be multiplying - although the ratio comes out right anyway (as you're doing the wrong operation with the same number so the error cancels).
OK, so we now agree, a ratio of about 17,000

Next I will work out the slightly further out point where there should be approximately double the radiation.

And then I will think about which figures would be fun to put in a program.

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OK, it was a typo, should have been atan not tanh ๐Ÿ™‚

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Originally posted by twhitehead
Not a typo. I am working out an angle.

[b]In boxes B29 and B30 you are dividing where you should be multiplying - although the ratio comes out right anyway (as you're doing the wrong operation with the same number so the error cancels).

OK, so we now agree, a ratio of about 17,000

Next I will work out the slightly further out point where there ...[text shortened]... le the radiation.

And then I will think about which figures would be fun to put in a program.[/b]
We've already got that, it's the 117 km figure (for a 1 m radius collector disk). The reason for this is that the top of our current annulus will be the centre of the annulus for the point where the amount of radiation is doubled relative to the first focus.

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Originally posted by DeepThought
We've already got that, it's the 117 km figure (for a 1 m radius collector disk). The reason for this is that the top of our current annulus will be the centre of the annulus for the point where the amount of radiation is doubled relative to the first focus.
What is the ration 17,000?

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Originally posted by sonhouse
What is the ration 17,000?
Light from Sirius hitting the collector disk divided by light from the Sun hitting the collector disk at the point of first focus.

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Originally posted by DeepThought
Light from Sirius hitting the collector disk divided by light from the Sun hitting the collector disk at the point of first focus.
So you are saying 0.005 watts / meter squared from sol vs 81 watts at first focus?

if Sirius puts out 25X sol, that would amount to about 33,000 watts per meter squared at 1 AU. Sirius is about 529,000 AU from Sol. That is a factor down about 3.5E-12 and that times 33,000 is about 511 watts per square meter at first focus, where there is about 4.3E9 meters available around the disc of the sun to focus at 542 AU. 33,000 (1 AU) times 3.5E12 = 1.1E-7 Watts/M^2 at sol times 4.3E9 and I get about 500 watts concentrated at 542 AU.
It also seems to be at the other end, the power goes to 15,000 watts/M^2 suspiciously close to half the number at 1 AU (33,000 watts/m^2)
You want to go over these figures?

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Originally posted by sonhouse
So you are saying 0.005 watts / meter squared from sol vs 81 watts at first focus?

if Sirius puts out 25X sol, that would amount to about 33,000 watts per meter squared at 1 AU. Sirius is about 529,000 AU from Sol. That is a factor down about 3.5E-12 and that times 33,000 is about 511 watts per square meter at first focus, where there is about 4.3E9 mete ...[text shortened]... iciously close to half the number at 1 AU (33,000 watts/m^2)
You want to go over these figures?
Yes, the power ratio at the collector - focused radiance from Sirius divided by unfocused direct radiance from the Sun.

You seem to be out by a factor of two for the power from Sirius at the point of first focus - the annulus is 1/2 a metre wide for the first focus which I'm guessing is what the problem there is (for a focus point more than 117 km further out on a 1m radius collector that would go up to 1 m wide, see earlier in the thread).

I don't understand what you mean by "at the other end" - are you talking about the equivalent effect of light from the Sun being focused by Sirius?

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