I have a Lunt solar 60 and two 290 mm camera. i cannot image the whole dic though. i bought a focal reducer but it wont focus, wha should I get if its possible at all? Thanks

# imaging whole disc

### #1

Posted 18 May 2024 - 06:53 PM

### #2

Posted 18 May 2024 - 08:31 PM

Hi,

Take your scope's focal length and divide by 110. This gives you your disc image diameter in mm. The camera sensor needs to be a little larger than the disc image in mm in its shortest dimension.

Very best,

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### #3

Posted 18 May 2024 - 08:36 PM

Hi,

Take your scope's focal length and divide by 110. This gives you your disc image diameter in mm. The camera sensor needs to be a little larger than the disc image in mm in its shortest dimension.

Very best,

This can be seen as a simple question, but where does the 110 come from?

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### #4

Posted 18 May 2024 - 09:30 PM

The 533 chip with a 2X barlow works nicely with the 60mm Lunt for a full disk. I have the Player One Saturn-M SQR and really like it with my 60mm Lunt.

- Skywatchr likes this

### #5

Posted 19 May 2024 - 02:11 AM

This can be seen as a simple question, but where does the 110 come from?

I had the same question.

- Look at the sky 101 likes this

### #6

Posted 19 May 2024 - 08:10 AM

This can be seen as a simple question, but where does the 110 come from?

I had the same question.

Solar disc angular size is 0.517 degrees or 31.02 arc-minutes or 1861.2 arc-seconds.

Angle in arc-seconds = radians * 206,264.81

1 radian = 57.2958 degrees

1 degree = 0.0174533 radians

I'll use degrees as its just more common.

s = (a*f)/(57.2958)

s = size in mm

a = angular size of subject

f = focal length of objective

s = (0.517 degrees)(f)/(57.2958 degrees)

Notice if we only look at angular size in degrees here, we have a relationship that is much easier to spot and they're like units. And all that is left is the size and focal length without values. So a relationship is there, a ratio. Let's rearrange.

(s)(57.2958 degrees) = (0.517 degrees)(f)

57.2958 degrees / 0.517 degrees = **110.823**

(s)(110.823) = f

Or, rewritten: **f / 110.823 = s**

So, a focal length of 1000mm / 110.823 = 9.023mm disc diameter as an example, based on angular sizes and the focal length of this objective.

That's the disc diameter, proms extend beyond the diameter of the disc, so always assume you need slightly more space around the disc image.

+++++++++++++++++++++++++++++++++++++++

Let's verify the above math with a simple ratio from a circle and triangle shape with some presumable known values (diameter of the Sun and distance of Sun to Earth).

If the solar diameter is approximately known to be about 1.39 million kilometers and the since from Earth to the Sun is known to be approximately 151.37 million kilometers, we can use this as a ratio.

151.37 million km / 1.39 million km = **108.89**

That's *very* close to the 110.823 that I calculated above. So, just another way to get a crude means to establish an approximate disc size based on focal length from these values using ratios.

+++++++++++++++++++++++++++++++++++++++

Why do we need this information? Two reasons with our gear:

1 - We need it so we know what size sensor will fit the disc image without needing a mosaic. The sensor needs to be a little larger than the disc image (keep in mind room for prominences, don't cut it close). I would just add 1mm as a minimum to the size difference needs. Ie, if the disc image is 3.8mm diameter then I would try to use a sensor around 5mm as the shortest dimension or larger for the disc image.

2 - This determines the blocking filter size need for a full disc to pass without vignetting. Focal length determines disc image size. Well, we easily calculate disc image size above. The blocking filter diameter needs to be larger than the disc image size to pass it without vignetting. Same size needs as the camera sensor. Ideally a little larger than the disc image to pass everything, like proms, without vignetting. So if your disc image is 3.8mm, then a 5~6mm blocking filter, or larger, will comfortably pass this disc image without vignetting. A larger diameter blocking filter does nothing special. It's a minimum sized thing as needed. It does start to matter in a few instances where the disc image is larger, or where binoviewers or any other gear is used that requires more inward focus (reducers come to mind). It starts to matter because when you use a lot of inward focus, the blocking filter size is pushed back away from the disc image at focus and instead its deeper into the wider unfocused light cone and might start to vignette. A larger blocking filter is useful with larger camera sensors too, to avoid vignetting, especially if a reducer is involved. This starts to become noticeable with larger sensors, like 1.1" class stuff that are commonly larger than small blocking filters are.

Lastly, regarding blocking filter size, if you look through an h-alpha system and it seems like you're getting "black outs" and looking through a straw or tunnel unless you're perfectly aligned with the disc image, this is due to the blocking filter being very small. This is common with 5mm blocking filters, or similar. When using a larger, much larger blocking filter, this phenomena is less or eliminated. So it can be useful to get a larger blocking filter if you want it to be very comfortable to view from many positions at the eyepiece without experiencing black outs at the eyepiece. So while a minimum can work, its a different experience. Getting a 10~15mm blocking filter goes a long way for comfort of viewing. If one does imaging only, getting a smaller one based on disc image size is fine.

Very best,

**Edited by MalVeauX, 19 May 2024 - 12:38 PM.**

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### #7

Posted 19 May 2024 - 08:16 AM

I have a Lunt solar 60 and two 290 mm camera. i cannot image the whole dic though. i bought a focal reducer but it wont focus, wha should I get if its possible at all? Thanks

What reducer and where is it placed?

I've used a cheap 0.5x reducer with my asi290mm.

- skydivephil likes this

### #8

Posted 19 May 2024 - 04:07 PM

What reducer and where is it placed?

I've used a cheap 0.5x reducer with my asi290mm.

Yeah thats what i used but it just wont focus.

### #9

Posted 19 May 2024 - 04:09 PM

Hi,

Take your scope's focal length and divide by 110. This gives you your disc image diameter in mm. The camera sensor needs to be a little larger than the disc image in mm in its shortest dimension.

Very best,

Thanks

### #10

Posted 19 May 2024 - 04:12 PM

Yeah thats what i used but it just wont focus.

Focal reducers will require more *inward* focus, so you may find it limited due to not having enough inward focus with the diagonal eating up a lot of the pathway. A way to reduce this is to unthread the eyecup from the diagonal and directly thread the camera to it (they're both T2 threads) to gain that extra distance.

Very best,

### #11

Posted 19 May 2024 - 08:02 PM

Angle in arc-seconds = radians * 206,264.81

where are you getting the 206,264.81 from?

I was able to follow the rest of your math (although it did hurt my poor old fuddled brain).

So, my Lunt 60 has a focal length of 500mm.

500/110.823 = 4.51mm (apparent diameter of the Sun). I bought the Lunt with the B1200 blocking filter (12mm). So, the blocking filter shouldn't either limit the ability to get the full solar disc, or have any real vignetting issues. Correct?

I plan(ned) to use the Lunt 60 for both imaging and HA visual, so that larger blocking filter size is most welcome for me (I have tunnel vision issues with TeleVue Naglers - the 12mm type IV was especially horrible for me).

Many thanks as usual Marty for taking the time to explain things to us newbies.

For me, it's hard cos I have to unpack everything (camera, scope, mount), assemble (not easy even with a smaller/lighter HEQ5pro when you have lower back issues) and then disassemble and pack everything back up.

- MalVeauX likes this

### #12

Posted 19 May 2024 - 09:16 PM

where are you getting the 206,264.81 from?

I was able to follow the rest of your math (although it did hurt my poor old fuddled brain).

So, my Lunt 60 has a focal length of 500mm.

500/110.823 = 4.51mm (apparent diameter of the Sun). I bought the Lunt with the B1200 blocking filter (12mm). So, the blocking filter shouldn't either limit the ability to get the full solar disc, or have any real vignetting issues. Correct?

I plan(ned) to use the Lunt 60 for both imaging and HA visual, so that larger blocking filter size is most welcome for me (I have tunnel vision issues with TeleVue Naglers - the 12mm type IV was especially horrible for me).

Many thanks as usual Marty for taking the time to explain things to us newbies.

For me, it's hard cos I have to unpack everything (camera, scope, mount), assemble (not easy even with a smaller/lighter HEQ5pro when you have lower back issues) and then disassemble and pack everything back up.

Hi,

206,264 and some endless change (because pi is endless) is just a relationship ratio for converting with arc-seconds between radians and degrees. It's all based on circles because half of a sine wave is 180 degrees of a circle. A circle (360 degrees) has 2*pi radians. So if we divide 180 by pi, we get 57.2957. That's per radian. So 1 radian = 57.2957 degrees. So if a circle has 2*pi radians, that's 2*pi= 6.283 radians. Well, if 57.2957 degrees per radian, then 6.283 * 57.2957 = 360 degrees. There's our circle again. Degrees and radians are really just describing an arc, such as arc-minute.

Well, if an arc minute has 60 arc-seconds in it, then its 60*60 = 3600 arc-seconds, per degree. We just solved that there's 57.2957 degrees per radian. So to get arc-seconds per radian, we do 3600*57.2957 = **206,264**

Re: your 12mm blocking filter diameter will not vignette nor limit your 4.51mm disc image size from the focal length of your objective. Correct.

Very best,

**Edited by MalVeauX, 19 May 2024 - 10:12 PM.**

- dpastern and Look at the sky 101 like this

### #13

Posted 20 May 2024 - 12:34 AM

Hi,

206,264 and some endless change (because pi is endless) is just a relationship ratio for converting with arc-seconds between radians and degrees. It's all based on circles because half of a sine wave is 180 degrees of a circle. A circle (360 degrees) has 2*pi radians. So if we divide 180 by pi, we get 57.2957. That's per radian. So 1 radian = 57.2957 degrees. So if a circle has 2*pi radians, that's 2*pi= 6.283 radians. Well, if 57.2957 degrees per radian, then 6.283 * 57.2957 = 360 degrees. There's our circle again. Degrees and radians are really just describing an arc, such as arc-minute.

Well, if an arc minute has 60 arc-seconds in it, then its 60*60 = 3600 arc-seconds, per degree. We just solved that there's 57.2957 degrees per radian. So to get arc-seconds per radian, we do 3600*57.2957 =

206,264

Re: your 12mm blocking filter diameter will not vignette nor limit your 4.51mm disc image size from the focal length of your objective. Correct.

Very best,

Ah, now that makes sense to this old man. It has been many years since I went to school and I can barely remember any of the maths. I don't even know if we learned this in high school or not (despite being in the advanced maths stream).

Many thanks again.

I am off to try my hand at a bit more solar imaging with the R10, but this time stacking both the 1.4x and 2x TC for 1344mm focal length (I don't expect it to go very well).

- MalVeauX likes this