How does camera rotation on a Newt relate to, say, Stellarium?

[BrendanC]

Hi all,

Not really sure how to express this, but I'll try.

I'd like to be able to get my camera roughly in the right orientation before a shoot, so that if, for example, I need the frame to be rotated at 75 degrees to frame an object nicely according to Stellarium, I can get the camera somewhere near that before I start. Then I can just take a shot, plate solve, rotate the camera, iterate until we're all set.

I have a nice 'flat' position for my camera in which it's aligned along the scope ie so that the top/bottom longer edges of the camera are aligned with the scope lengthways (a Newtonian - which makes this more complicated, in my mind, on an equatorial mount). In fact, you can see that position in my avatar, I just realised. I've been trying to get shots, and then plate solve them, to find out what this flat position actually equals in terms of degrees rotation according to the equatorial grid. Then it should be a simple calculation to say 'this is your flat position on the scope, which equals X degrees in the sky, so to rotate it to Y degrees, just rotate it Z degrees on the scope'.

However, I've not had enough nights to do this recently, and I'm lazy, so I thought what the hey, perhaps someone can tell me whether this is possible on the SG forum! Before I waste time, effort and tears on trying to work out something that seems to be easy but turns out to be hard/impossible.

I'm assuming it would be a consistent rotation conversion no matter where the camera is pointing? Or, does it depend on where in the sky I'm pointing and/or where on the planet I am?

If so, does anyone happen actually to know what the rotation conversion might be? As in, has anyone done this before?

Am I making any sense here?

Thanks, Brendan

Edited February 21, 2022 by BrendanC

[Varavall]

46 minutes ago, BrendanC said:

Hi all,

Not really sure how to express this, but I'll try.

I'd like to be able to get my camera roughly in the right orientation before a shoot, so that if, for example, I need the frame to be rotated at 75 degrees to frame an object nicely according to Stellarium, I can get the camera somewhere near that before I start.

I have a nice 'flat' position for my camera in which it's aligned along the scope ie so that the top/bottom longer edges of the camera are aligned with the scope lengthways (a Newtonian - which makes this more complicated, in my mind). I've been trying to get shots, and then plate solve them, to find out what this flat position actually equates to in terms of degrees rotation according to the equatorial grid. Then it should be a simple calculation to say 'this is your flat position on the scope, which equals X degrees in the sky, so to rotate it to Y degrees, just rotate it Z degrees on the scope'.

However, I've not had enough nights to do this recently, and thought what the hey, perhaps someone can tell me whether this is possible on the SG forum! Before I waste time, effort and tears on trying to work out something that seems to be easy but turns out to be hard/impossible.

I'm assuming it would be a consistent rotation conversion no matter where the camera is pointing? Or, does it depend on where in the sky I'm pointing and/or where on the planet I am?

If so, does anyone happen actually to know what the rotation conversion might be? As in, has anyone done this before?

Am I making any sense here?

Thanks, Brendan

Hi

I don't know about Stellarium, but with NINA you can set it so it will plate solve and tell you how many degrees to turn the camera for framing. Look here https://nighttime-imaging.eu/docs/master/site/tabs/equipment/rotator/ Maybe that'll help!

[BrendanC]

Sure, I can do the same (kind of) in APT - shoot, plate solve, it tells me the degrees I'm out by, rotate, iterate.

What I'm trying to do is work out how much I should rotate the camera before I go outside! I really like using Stellarium to plan the framing and timing, so, let's say I want to do M42, and I find that, according to Stellarium, the best framing for that is to rotate the frame by 90 degrees. Now, given that my camera is usually in its nice 'neutral' position so that it's aligned with the scope, is it possible to convert the Stellarium rotation into camera rotation? That is, if Stellarium says 90 degrees, by how much from the neutral position do I rotate my camera?

Or, to put it another way, how should I rotate my camera on my scope so that it's at zero degrees when I take a shot? That might be an easier way to express this actually.

My instinct tells me it should be a simple case of taking shots in the neutral position, at several different spots in the sky, and if they all agree then I'm right in that there is a constant conversion factor and it should be fairly easy.

But as with so many things in this game, I'm also prepared to believe that it isn't easy! I'd also rather not waste time having to take shots and figure this out myself. So, I just wondered if someone could, you know, just tell me!

Edited February 21, 2022 by BrendanC

[iapa]

50 minutes ago, BrendanC said:

Sure, I can do the same (kind of) in APT - shoot, plate solve, it tells me the degrees I'm out by, rotate, iterate.

What I'm trying to do is work out how much I should rotate the camera before I go outside! I really like using Stellarium to plan the framing and timing, so, let's say I want to do M42, and I find that, according to Stellarium, the best framing for that is to rotate the frame by 90 degrees. Now, given that my camera is usually in its nice 'neutral' position so that it's aligned with the scope, is it possible to convert the Stellarium rotation into camera rotation? That is, if Stellarium says 90 degrees, by how much from the neutral position do I rotate my camera?

…….

But as with so many things in this game, I'm also prepared to believe that it isn't easy! I'd also rather not waste time having to take shots and figure this out myself. So, I just wondered if someone could, you know, just tell me!

Short answer: it depends. 

very short answer: don’t 

longer: possibly you could 

Assuming that

1. the camera sensor long edge is parallel to the OTA, AND

2. the light path to the sensor through the focuser is parallel to the ground

turn on the sensor view in Stellarium, and you can change the rotation on there to get an idea how much to rotate the camera.

BUT, if condition 2 does not apply, and it won’t if you have balanced on RA, DEC and with  OTA pointing vertically. (I followed Dion’s tutorial on youtube…. Either the Quattro 10 review, or the Quattro 10 tweaking  tutorial.) So you need to take that that rotation into account too.
 

need wee guddle with 3D geometry to be sure, but my first thought is that the rotation you will want is the difference between the rotation from Stellarium and the rotation of the OTA.
 

I'm in the plate solving camp for this one.

[Varavall]

1 hour ago, BrendanC said:

Sure, I can do the same (kind of) in APT - shoot, plate solve, it tells me the degrees I'm out by, rotate, iterate.

What I'm trying to do is work out how much I should rotate the camera before I go outside! I really like using Stellarium to plan the framing and timing, so, let's say I want to do M42, and I find that, according to Stellarium, the best framing for that is to rotate the frame by 90 degrees. Now, given that my camera is usually in its nice 'neutral' position so that it's aligned with the scope, is it possible to convert the Stellarium rotation into camera rotation? That is, if Stellarium says 90 degrees, by how much from the neutral position do I rotate my camera?

Or, to put it another way, how should I rotate my camera on my scope so that it's at zero degrees when I take a shot? That might be an easier way to express this actually.

My instinct tells me it should be a simple case of taking shots in the neutral position, at several different spots in the sky, and if they all agree then I'm right in that there is a constant conversion factor and it should be fairly easy.

But as with so many things in this game, I'm also prepared to believe that it isn't easy! I'd also rather not waste time having to take shots and figure this out myself. So, I just wondered if someone could, you know, just tell me!

If the framing doesn't fit the target, I tend to do a mosaic and crop the image rather than upset the balance of the rig. But if there's a few hundred quid lying around, a Pegasus rotator will solve the problem every time 😉 Well, on thinking about it, rotating the filter wheel along with the camera must alter the balance even a little, so perhaps problem created 🤔

Edited February 21, 2022 by Varavall
Added comment

[BrendanC]

Thanks @iapa, I think I'm in agreement. The geometry is fairly mind-bending, and I think you're right that really it's down just to plate solving and getting it right in the field.

The real reason I ask is that I'm building a funky spreadsheet that helps with planning, and it just occurred to me that this might be another nice feature of it, ie so that I could type in the rotation according to Stellarium, and it would just tell me how the camera should look - just to get it close-ish to the right rotation, and make the final tweaks with plate solving a bit quicker. But I'm thinking the theory won't fit the practice.

I'm still going to take some shots with the camera at neutral, panning across the sky, solving and then noting the rotation, and it might proving interesting, who knows?

[iapa]

The math is pretty straight forward, well….. I thought so a few decades ago.

Get focus sorted, and polar aligned, then plate solving takes a couple of seconds.

So, why go to that effort? Other than the "interest". 

if you have an old image with no meta data you could use astrometry.net to upload image and get coordinates of the image centre, field of view and rotation.

given I’ve had less the 49 hrs in the past 4 months i’d rather spend my time taking actual images at night than using the time for that type of data gathering.

Personal choice 

 

[BrendanC]

Not sure I'm explaining myself very well here.

What I'm trying to do is in fact increase the time I spend imaging, as you say, by getting my camera closer to the proper orientation before I even go outside with the scope. That way, I spend less time shooting, plate solving, rotating, repeating. It's not just for theoretical interest.

I wouldn't benefit from plate solving old images, because I wouldn't know how the camera was oriented when I took them.

I'll just do as I said, align the camera with the scope, take some pics in various parts of the sky, plate solve them, see what the rotation figures say.

Edited February 23, 2022 by BrendanC

[iapa]

11 minutes ago, BrendanC said:

Not sure I'm explaining myself very well here.

What I'm trying to do is in fact increase the time I spend imaging, as you say, by getting my camera closer to the proper orientation before I even go outside with the scope. That way, I spend less time shooting, plate solving, rotating, repeating. It's not just for theoretical interest.

I'm not interested in plate solving old images, because I wouldn't know how the camera was oriented. 

I'll just do as I said, align the camera with the scope, take some pics in various parts of the sky, plate solve them, see what the rotation figures say.

Sorry, I understand that you want some method to determine the rotation you need to apple to the camera for specific targets - I was just pointing to an option available for any existing images so you can integrate that to your spreadsheet.

I use either an ASIAIR to manage everything (I have enough ZWO kit to warrant that), or i use Sequence Generator Pro on a stick PC for non-ZWO kit. Both have a cost though. But everything is integrated.

[BrendanC]

I want a method to determine the rotation of my camera, for a given DSO, before I even start plate solving, so that I can get closer to the actual rotation needed, and spend less time actually outside plate solving.

I knew it was going to be difficult to explain this!

While planning, if I frame an object in Stellarium, I don't know for sure what, for example, 35 degrees rotation means I need to rotate my camera by - or, even, if this is consistent given the position of the object in the sky at a given time.