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My Quark images with a soft processing approach WIP though


Guest Ewan

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Just done a few figures.

 

Assuming I have your scope and imager details correct...
Technosky 152 F5.9 with 135 ERF giving an effective f28.7
Sony IMX174 cmos with 11.3 x 7.1mm 2.3 megapixel sensor 

 

Your field of view is about 0.173 degrees x 0.105 degrees which gives a section of the sun about 282000 miles x 171000 miles.

 

At 1936x1216 pixels you are working at an image scale of about 0.3 arcseconds per pixel which equates to a best resolution of about 140 miles per pixel on the sun!

 

So to be able to examine the surface of the sun at 93 million miles away and see detail down to just 140 miles across is incredible. :2thumbsup:

 

Now, looking back at those fantastic images, we can see them in a slightly different light knowing roughly what scale we are viewing these structures and features at. :)

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On Saturday, April 22, 2017 at 19:47, Tweedledee said:

Just done a few figures.

 

Assuming I have your scope and imager details correct...
Technosky 152 F5.9 with 135 ERF giving an effective f28.7
Sony IMX174 cmos with 11.3 x 7.1mm 2.3 megapixel sensor 

 

Your field of view is about 0.173 degrees x 0.105 degrees which gives a section of the sun about 282000 miles x 171000 miles.

 

At 1936x1216 pixels you are working at an image scale of about 0.3 arcseconds per pixel which equates to a best resolution of about 140 miles per pixel on the sun!

 

So to be able to examine the surface of the sun at 93 million miles away and see detail down to just 140 miles across is incredible. :2thumbsup:

 

Now, looking back at those fantastic images, we can see them in a slightly different light knowing roughly what scale we are viewing these structures and features at. :)

Don't spoke you could show me the maths for doing that could you ?

Didn't realise the scale was that close, really makes you think.

 

I have put my imaging details into Genika but it seemed to have averted back to the included custom setup which doesn't match mine, I did recently update a few bits so maybe that caused it.

Atb

Ewan

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3 hours ago, Ewan said:

Don't spoke you could show me the maths for doing that could you ?

Didn't realise the scale was that close, really makes you think.

 

I have put my imaging details into Genika but it seemed to have averted back to the included custom setup which doesn't match mine, I did recently update a few bits so maybe that caused it.

Atb

Ewan

OK, you asked for it. :)

 

Expect a few small rounding errors to creep in with these rough figures.

 

There are various ways you can do these calculations that will give the same answer.

 

I'm trying to explain this like Professor Cox does, but minus the stick, the sand and the exotic location. :D

 

Get the focal length of the system...
135mm x 28.7 = 3875mm

 

Note that...
1 radian is 360 degrees / 2pi = 57.3 degrees

 

Get the image scale of the system at prime focus by taking the focal length and dividing it by a radian (I'm sure that like me, you're now thinking "aren't radians just brilliant" :D )...
3875mm / 57.3 degrees = 67.6mm per degree

 

Find the image size spanned by the sensor which is 11.3mm wide...
11.3mm / 67.6mm per degree = 0.17 degrees

 

Find out how many miles are spanned by a degree of sky at the distance of the sun.
The sun is about 865000 miles in diameter and its apparent diameter is 0.53 degrees in the sky...
865000miles / 0.53 degrees = 1632075 miles per degree

 

The sensor is spanning just 0.17 degrees of the sky as found above. So we can now find out how many miles the sensor is spanning on the sun...
0.17 degrees x 1632075 miles per degree = 280000 miles

 

To find the miles per pixel resolution, divide the number of miles shown across the sensor by the number of pixels across the sensor...
280000 miles / 1936 pixels = 144 miles

 

Hope you can follow it.

 

You can adapt the above techniques to working out all sorts of parameters for focal systems, eyepieces and camera chips. I'll leave you to take the superb pictures 'cos I don't have a clue. :rolleyes:

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19 hours ago, Tweedledee said:

OK, you asked for it. :)

 

Expect a few small rounding errors to creep in with these rough figures.

 

There are various ways you can do these calculations that will give the same answer.

 

I'm trying to explain this like Professor Cox does, but minus the stick, the sand and the exotic location. :D

 

Get the focal length of the system...
135mm x 28.7 = 3875mm

 

Note that...
1 radian is 360 degrees / 2pi = 57.3 degrees

 

Get the image scale of the system at prime focus by taking the focal length and dividing it by a radian (I'm sure that like me, you're now thinking "aren't radians just brilliant" :D )...
3875mm / 57.3 degrees = 67.6mm per degree

 

Find the image size spanned by the sensor which is 11.3mm wide...
11.3mm / 67.6mm per degree = 0.17 degrees

 

Find out how many miles are spanned by a degree of sky at the distance of the sun.
The sun is about 865000 miles in diameter and its apparent diameter is 0.53 degrees in the sky...
865000miles / 0.53 degrees = 1632075 miles per degree

 

The sensor is spanning just 0.17 degrees of the sky as found above. So we can now find out how many miles the sensor is spanning on the sun...
0.17 degrees x 1632075 miles per degree = 280000 miles

 

To find the miles per pixel resolution, divide the number of miles shown across the sensor by the number of pixels across the sensor...
280000 miles / 1936 pixels = 144 miles

 

Hope you can follow it.

 

You can adapt the above techniques to working out all sorts of parameters for focal systems, eyepieces and camera chips. I'll leave you to take the superb pictures 'cos I don't have a clue. :rolleyes:

Blimey !, I see why now I have not looked at this before but I shall give it a go on my next captures. Thank you for taking the time out for doing that ?

 

Atb

Ewan 

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  • 1 month later...

Just love all the above pictures, never seen our star as up close and personal before.

 

Thanks for sharing and thanks to Pete for the maths lesson.

 

:thumbsup:

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