pedro said:
jrista said:
pedro said:
@jrista: Did I get that wrong with rule of 600? I thought the calculation 600:lens length refers to its LONG end? So the 16-35 won't give you more than about 16 sec of exposure. Therefore I like the 5D3 which allows me to crank up the ISOs significantly compared to my trust rusty 30D. Cheers, Pedro
It simply referrs to the focal length you are using. Doesn't matter if the lens is prime or zoom...a zoom is nothing more than a lens that lets you change the selected focal length without swapping lenses. If I use the 16-35 @ 16mm, then the rule of 600 would logically apply to 16mm, not 35mm.
@jrista: oh, didn't know that then. great! thanks for the explanation. then I already have my (at least) 16-24 once I purchase the lens. and that's plenty compared to a phantom lens that might surely be sold at twice the price of the 16-35 should it ever reach the shelves...glad to learn this...that gives me at least 35 sec at the wide end then...wow. 8) So taking a picture at let's say ISO 3200 or even 6400 means capturing way more light than at ISO 800 on a 30D...!!!
Actually, to get really technical, the rule of 600 really is kind of a bad way to figure out how long you can expose a night sky with digital sensors. You'll find a lot of anecdotes, such as use the "FF effective focal length on APS-C" and whatnot. None of it really applies, since what actually matters is pixel density, which is independent of form factor.
Here is a better way to figure out what you can really handle (for "native resolution" presentation, anyway...you could transform for web sized with a scale factor): Determine the arc degrees per pixel for a given focal length and pixel size.
If we assume a 12mm lens on a FF sensor, that gives us a horizontal FoV of approximately 111°. If we assume we are shooting with a 5D III, that means we have a pixel row of 5760 pixels, or ~52 pixels per degree. The reciprocal of that gives us the number of degrees per pixel, which in this case is 0.0192°/px. To calculate how many seconds it will take a star to traverse one pixel, divide that number by the arc degrees per second a star moves across the sky thanks to the rotation of the earth (360° / (24h * 60m/h * 60s/m) = 360° / 86400s = 0.0042°/s). At 12mm, a 5D III will experience a
time on pixels period of 0.0192°/px / 0.0042°/s, or ~4.6s/px.
So, assuming a
time on pixels period of 4.6s/px on the 5D III at 12mm, we can figure out how long, in total, we might be able to expose for by determining the number of pixels we are willing to let a star traverse before we assume trailing will be visible. For maximum quality, or the ability to see absolutely zero startrailing at "native resolution", and even offer enough sharpness to enlarge a night sky photo, you wouldn't want a star to affect much more than 4 pixels in a 2x2 block...a 2 pixel pitch. That would mean at 12mm, you could only expose for 9.2 seconds. Thanks to atmospheric effects which causes stars to affect more than the mere 2x2 pixel grid their point light source might affect directly anyway, we can generally assume a larger star trailing pitch. In my experience, 6-9 pixels is easily good enough for native resolution output (and much more than that for web size). Assuming the factor of 9 pixels is valid, that gives us an exposure time of about 42 seconds for a 12mm lens on the 5D III. On the 7D, the same 12mm lens is the FoV equivalent to a 20mm lens on FF, which ultimately gets us to ~3.8s/px, or an exposure time of about 34 seconds for a 12mm lens on the 7D.
If we want to figure out how long to expose for web-sized images, you could make the assumption that the scale factor between native size and the final web size, times the base 9 pixel pitch, divided by two is sufficient to minimize trailing at any size:
Code:
scaledPitch = (dimNative/dimWeb * 9) / 2
I generally like to scale my images to around 900 pixels long size for the web. The ratio between native size and web size is 5760/900, or 6.5x, which when halved leads to a pixel pitch of ~29 pixels. At 12mm on the 5D III, that gives us a maximum possible exposure time of ~ 2min22sec, which still seems too long. To compensate, I assume the scaled pitch is diagonal, so computing for the horizontal or vertical pitch I get (sqrt(diagPitch^2 / 2) = horizPitch; sqrt(29^2/2) = ~20 pixels. That leaves us with a maximum exposure time of ~90 seconds for a 900 pixel web sized image (~5x7" size on the average screen).
