Hi Im doing a spreadsheet for the forum. I need to know what is the max exposure time you can use with the em1/em5 to avoid star trails with a fixed tripod mounted camera. I need to know the shutter speed seconds. Focal length of lens and object or declination any user has used which just avoids trails. The spreadsheet will output limiting magnitudes for stars and nebulae as well as max exp time at any declination to avoid trails. User inputs iso. Focal length and lens aperture as well as declination. Because of weather I cannot calibrate myself at the moment. Any help appreciated. Cheers Stuart
Use this formula to determine the longest exposure time in seconds for any given focal length. 600 / crop factor / marked focal length. Example: EM-5 camera using mZuiko 12mm lens. 600 / 2 / 12 = 25 seconds
Hi. Thanks for your reply. I think you will find this will over estimate the exp time for the small photo sensor size in the em1/em5. Its more like 150/focal length but I wanted to calibrate it with a users image ideally. Cheers Stuart
I think Sammyboy is right, although I've heard 500/fl. Did some shots last night with the Samyang 14mm f2.8 on a Metabones Speedbooster wideopen. Used 20secs because I have never done this before. Single shot. PM1, ISO 800. I guess I need to learn about stacking software next
Curious, where did this formula come from? At 15s I notice just hints of elongation in the stars with the Panny 20mm. I notice none at 15s with the 17mm.
The factor of 600 is for larger photosites. But for olympus em1 the photosites are smaller so the formula is too optimistic. Stuart
I learned it in college publics class in the 60's long before digital, where crop factor was the film format size.
Question: how long can you expose so that the image moves no more than one pixel? If you know this number, you can scale it to fit your image size and your smear tolerance. For example, if a planet image is 100 pixels wide and you decide you can tolerate 10% elongation, then you can allow it to move 10 pixels during exposure. From http://en.wikipedia.org/wiki/Angle_of_view, the angle of view α for a lens of focal length f and sensor dimension d is α = 2atan(d/2/f). Atan means the arctangent trig function. Let's use the horizontal direction for simplicity. Then for µ4/3, d = 17.3mm. So for a 50mm lens, α ≅ 19.6°, where ≅ means approximately equal. For an OM-D EM-5, there are 4608 pixels in the horizontal direction. Therefore, each pixel spans 19.6°/4608 ≅ 0.0043°. This assumes a 100% crop. If you scale the image to 1024 pixels horizontally, then each pixel spans 19.6°/1024 ≅ 0.019°. In 24 hours a stellar object spins 360° about the camera if it is perpendicular to the direction of the North Star. The closer the object is to the North Star or the opposite direction to the south, the less it will move. Let's use 360° as a worst case. 24 hours is 24×60×60 = 86400 seconds. So the object moves 0.0043° in 0.0043/360×86400 ≅ 1.03 seconds. Therefore with a 50mm lens you can expose for about 1 second without smearing more than 1 pixel, assuming a 100% crop. This seems like an easy number to remember. The arctangent isn't quite linear but it's close enough. So if you use a 200mm lens as I recently did, you could expose for 1/4 second for a 1-pixel smear. I happened to do just that, but it was pure luck. Brian
I did the same calcs for the 20mm (effective) and got ~3.5 secs. I'm guessing from an IQ point of view, 20secs is fine if you are showing a scaled down version, but looking again at 100% at the image I posted above I can see why I thought 20secs were fine even at 100% from a quick glance. 'The Southern Cross' is off a little way to the right along with the 'Two Pointers', which means that the celestrial south pole is well inside the image. So some stars are moving much faster than others in the image. But then what is considered good enough? ~3.5s is for one star to move one entire photosite. What if some stars are almost near a photosite boundary and will cross it even at smaller than these calculated exposure times?
I think the number works out to 2.5 seconds for a 20mm lens since 50/20 = 2.5. That's for a one pixel change for a star roughly overhead. I think you can assume that the camera system takes care any digitization artifacts. The antialias filter should yield a smooth transition among pixels. I used one pixel as an example just because it is easy to imagine. If you have a sharp lens, you should be able to see a half-pixel location shift of a pinpoint object. The dot will just get divided among two or three pixels. With my lenses, tripod, and technique, it will get smeared a lot more than that! Brian
Angle of view of the 20mm effective is about 63deg horizontally according to my calcs. 63deg/19.6deg = 3.3ish. Unless I've fluffed something somewhere
You should have seen all the fluffs I made the first time I posted my formula! I had to delete the entire post and start over. I just plugged in 20mm to the formula and got 2.4 deg. I'm not sure what our difference is, but it doesn't matter that much. I would use the formula only as a first approximation to set your initial shutter speed. Once you take a photo you'll probably want to expose longer or shorter based on how it turned out. Brian
Hi. Ive got a formula which allows for diffraction. Seeing and drift. Can I post an excel spreadsheet on this forum? As stars arnt recorded as infinitely small points their image is spread out normally over several pixels. The formula ive used produces min exposure times for large prints. Obviously for the web images will be smaller so exp times can be increased a bit. Stuart
Heh - yeah found my 'fluff' and now I'm getting same values as you... yay (opens beer) That sounds very interesting mate. Does your spreadsheet also take into account the circle of confusion and enlargement issues? Presumably some guidance is there when you say for large prints. I don't mind Excel spreadsheets, but maybe its better to make a Google Sheet and share it publically - ie https://docs.google.com/spreadsheet/ Edit: found a bit on the web that the old 600/focal length rule is based on the image quality of a typical 8" x 10" enlargement.
Hi. Id prefer to use excel as I have created it already. Yes airey disc allowed for. The enlargement would cover big prints. Users can adapt to their own requirements. Wondering if I can post excel on this site? As I wanted to share it. Stuart
Hi, Ive put the spreadsheet on my website. Click on the link and goto bottom of page under photo gallery to download. Any limiting magnitude data from users would be helpful. Big prints I mean over 24 inches. Notes at bottom of spreadsheet. http://myweb.tiscali.co.uk/pryer/ Feedback welcome. Cheers stuart